CSIR-UGC NET/JRF LIFE SCIENCES. Solved Papers

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1 CSIR-UGC NET/JRF LIFE SCIENCES Solved Papers

2 CSIR-UGC NET/JRF Life Sciences Solved Papers Content Contribution : Amar Ujala Education Books Authors Panel Dr. Kumar Pushkar Amar Ujala Publications Ltd. Published by Amar Ujala Publications Ltd. and printed at C-21, Sector 59, Noida (U.P.) Latest Edition ***** Price : ` 250/- ISBN : Due care and diligence has been taken while publishing this book. However, the publisher does not hold any responsibility for any mistake that may have inadvertently crept in. The publisher does not accept responsibility for any loss arising out of the use of this book. All rights reserved. Neither this publication nor any part of it may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publisher. All disputes are subject to the exclusive jurisdiction of competent courts and forums in Noida only.

3 A/UG-15 Amar Ujala Education INTRODUCTION The subject of Life Sciences is one of the optional papers for the students appearing in the CSIR UGC-NET and Lectureship in LIFE SCIENCES examination.this book CSIR UGC-NET LIFE SCIENCES MODEL SOLVED PAPER attempts to satisfy the needs of students appearing for CSIR UGC-NET LIFE SCIENCES Examinations. It is really with a great sense of pleasure and satisfaction that we are presenting this book for the aspirants of CSIR UGC-NET LIFE SCIENCES. To get maximum advantage of this book in preparation for your examination of CSIR UGC-NET and Lectureship in LIFE SCIENCES, which takes place twice in a year. Now a days, only objective type (multiple choice ) questions are being asked in the CSIR UGC-NET LIFE SCIENCES Examination instead of descriptive type or essay type of questions. In such multiple choice type of questions four probable answers (options) are given in the question paper, out of which a student has to choose the correct answer (Option). We hope that this book will prove to be highly useful for the students appearing in the CSIR UGC NET and Lectureship in LIFE SCIENCES examination. Like our numerous other books on various educational and general topics, if this book is also able to guide the aspirants to achieve their desire goal, we shall have the satisfaction that our work and labour have been well rewarded. About CSIR-UGC/NET for JRF & Lecturership - Life Sciences Examination Objective of the test This national level test is conducted to determine the eligibility of Indian nationals for the award of Junior Research Fellowships (JRF)-NET and eligibility for appointment of Lecturers (NET) in certain subject areas falling under the Faculty of Science. This test is conducted twice every year by CSIR (Council of Scientific and Industrial Research) in June and December. The notification of the June examination usually appears in February-March, while that of the December examination would appear in August-September. Eligibility Conditions M.Sc. or equivalent degree under the subjects Chemical Sciences/Atmospheric Sciences/Life Sciences/Mathematical Sciences/Physical Sciences with minimum 55% marks for General and OBC candidates; 50% for SC/ST candidates, Physically and Visually Handicapped candidates. For JRF (NET), minimum age can be 19 years and maximum age can be 28 years Upper age limit is relatable by 5 years for candidates belonging to SC/ST/OBC categories, physically/visually handicapped and female applicants. For LS (NET), there is no upper age limit, but minimum age should be 19 years as on cut off date. SCHEME OF EXAMINATION Single Paper MCQ (Multiple Choice Question) based test. The pattern for the Single Paper MCQ test shall be as given below:- The MCQ test paper of each subject shall carry a maximum of 200 marks. The exam shall be for duration of three hours. The question paper shall be divided in three parts. Part A shall be common to all subjects. This part shall be a test containing a maximum of 20 questions of General Science and Research Aptitude test. The candidates shall be required to answer any 15 questions of two marks each. The total marks allocated to this section shall be 30 out of 200. iii

4 Amar Ujala Education iv A/UG-15 Part B shall contain subject-related conventional MCQs. The total marks allocated to this section shall be 70 out of 200. The maximum number of questions to be attempted shall be in the range of Part C shall contain higher value questions that may test the candidate s knowledge of scientific concepts and/or application of the scientific concepts. The questions shall be of analytical nature where a candidate is expected to apply the scientific knowledge to arrive at the solution to the given scientific problem. The total marks allocated to this section shall be 100 out of 200. A negative marking for wrong answers, wherever required, shall 25% SYLLABUS FOR LIFE SCIENCES PART A (GENERAL PAPER) It is notified for information of all students that syllabus of Part A of Joint CSIR-UGC Test for Junior Research Fellowship and Eligibility for Lectureship has been revised. The existing syllabus and revised syllabus is as under : EXISTING SYLLABUS General Science, Quantitative Reasoning and Analysis and Research Aptitude. General Aptitude with emphasis On logical reasoning graphical analysis analytical and numerical ability quantitative comparison series formation puzzles etc. REVISED SYLLABUS The number of questions available and to be attempted in this Part A remains the same i.e. there will be 20 questions and the candidates shall be required to answer any 15 questions. Each question will be of two marks. PART B & PART C 1. MOLECULES AND THEIR INTERACTION RELAVENT TO BIOLOGY A. Structure of atoms, molecules and chemical bonds. B. Composition, structure and function of biomolecules (carbohydrates, lipids, proteins, nucleic acids and vitamins). C. Stablizing interactions (Van der Waals, electrostatic, hydrogen bonding, hydrophobic interaction, etc.). D. Principles of biophysical chemistry (ph, buffer, reaction kinetics, thermodynamics, colligative properties). E. Bioenergetics, glycolysis, oxidative phosphorylation, coupled reaction, group transfer, biological energy transducers. F. Principles of catalysis, enzymes and enzyme kinetics, enzyme regulation, mechanism of enzyme catalysis, isozymes. G. Conformation of proteins (Ramachandran plot, secondary, tertiary and quaternary structure; domains; motif and folds). H. Conformation of nucleic acids (A-, B-, Z-,DNA), t-rna, micro-rna). I. Stability of protein and nucleic acid structures. J. Metabolism of carbohydrates, lipids, amino acids, nucleotides and vitamins. 2. CELLULAR ORGANIZATION A. Membrane structure and function : Structure of model membrane, lipid bilayer and membrane protein diffusion, osmosis, ion channels, active transport, ion pumps, mechanism of sorting and regulation of intracellular transport, electrical properties of membranes. B. Structural organization and function of intracellular organelles : Cell wall, nucleus, mitochondria, Golgi bodies, lysosomes, endoplasmic reticulum, peroxisomes, plastids, vacuoles, chloroplast, structure & function of cytoskeleton and its role in motility. C. Organization of genes and chromosomes : Operon, interrupted genes, gene families, structure of chromatin and chromosomes, unique and repetitive DNA, heterochromatin, euchromatin, transposons. D. Cell division and cell cycle : Mitosis and meiosis, their regulation, steps in cell cycle, and control of cell cycle. E. Microbial Physiology : Growth, yield and characteristics, strategies of cell division, stress response.

5 A/UG-15 v Amar Ujala Education 3. FUNDAMENTAL PROCESSES A. DNA replication, repair and recombination : Unit of replication, enzymes involved, replication origin and replication fork, fidelity of replication, extrachromosomal replicons, DNA damage and repair mechanisms. B. RNA synthesis and processing : Transcription factors and machinery, formation of initiation complex, transcription activators and repressors, RNA polymerases, capping, elongation and termination, RNA processing, RNA editing, splicing, polyadenylation, structure and function of different types of RNA, RNA transport. C. Protein synthesis and processing : Ribosome, formation of initiation complex, initiation factors and their regulation, elongation and elongation factors, termination, genetic code, aminoacylation of trna, trna-identity, aminoacyl trna synthetase, translational proof-reading, translational inhibitors, post- translational modification of proteins. D. Control of gene expression at transcription and translation level : Regulation of phages, viruses, prokaryotic and eukaryotic gene expression, role of chromatin in regulating gene expression and gene silencing. 4. CELL COMMUNICATION AND CELL SIGNALING A. Host parasite interaction : Recognition and entry processes of different pathogens like bacteria, viruses into animal and plant host cells, alteration of host cell behavior by pathogens, virus-induced cell transformation, pathogeninduced diseases in animals and plants, cell-cell fusion in both normal and abnormal cells. B. Cell signaling : Hormones and their receptors, cell surface receptor, signaling through G-protein coupled receptors, signal transduction pathways, second messengers, regulation of signaling pathways, bacterial and plant two-component signaling systems, bacterial chemotaxis and quorum sensing. C. Cellular communication : Regulation of hematopoiesis, general principles of cell communication, cell adhesion and roles of different adhesion molecules, gap junctions, extracellular matrix, integrins, neurotransmission and its regulation. D. Cancer : Genetic rearrangements in progenitor cells, oncogenes, tumor suppressor genes, cancer and the cell cycle, virus-induced cancer, metastasis, interaction of cancer cells with normal cells, apoptosis, therapeutic interventions of uncontrolled cell growth. E. Innate and adaptive immune system : Cells and molecules involved in innate and adaptive immunity, antigens, antigenicity and immunogenicity. B and T cell epitopes, structure and function of antibody molecules, generation of antibody diversity, monoclonal antibodies, antibody engineering, antigen-antibody interactions, MHC molecules, antigen processing and presentation, activation and differentiation of B and T cells, B and T cell receptors, humoral and cell-mediated immune responses, primary and secondary immune modulation, the complement system, Toll-like receptors, cell-mediated effector functions, inflammation, hypersensitivity and autoimmunity, immune response during bacterial (tuberculosis), parasitic (malaria) and viral (HIV) infections, congenital and acquired immunodeficiencies, vaccines. 5. DEVELOPMENTAL BIOLOGY A. Basic concepts of development : Potency, commitment, specification, induction, competence, determination and differentiation; morphogenetic gradients; cell fate and cell lineages; stem cells; genomic equivalence and the cytoplasmic determinants; imprinting; mutants and transgenics in analysis of development. B. Gametogenesis, fertilization and early development : Production of gametes, cell surface molecules in sperm-egg recognition in animals; embryo sac development and double fertilization in plants; zygote formation, cleavage, blastula formation, embryonic fields, gastrulation and formation of germ layers in animals; embryogenesis, establishment of symmetry in plants; seed formation and germination. C. Morphogenesis and organogenesis in animals : Cell aggregation and differentiation in Dictyostelium; axes and pattern formation in Drosophila, amphibia and chick; organogenesis vulva formation in Caenorhabditis elegans; eye lens induction, limb development and regeneration in vertebrates; differentiation of neurons, post embryonic development-larval formation, metamorphosis; environmental regulation of normal development; sex determination. D. Morphogenesis and organogenesis in plants : Organization of shoot and root apical meristem; shoot and root development; leaf development and phyllotaxy; transition to flowering, floral meristems and floral development inarabidopsis and Antirrhinum. E. Programmed cell death, aging and senescence.

6 Amar Ujala Education 6. SYSTEM PHYSIOLOGY - PLANT A/UG-15 A. Photosynthesis : Light harvesting complexes; mechanisms of electron transport; photoprotective mechanisms; CO 2 fixation-c 3,C 4 and CAM pathways. B. Respiration and photorespiration : Citric acid cycle; plant mitochondrial electron transport and ATP synthesis; alternate oxidase; photorespiratory pathway. C. Nitrogen metabolism : Nitrate and ammonium assimilation; amino acid biosynthesis. D. Plant hormones: Biosynthesis, storage, breakdown and transport; physiological effects and mechanisms of action. E. Sensory photobiology : Structure, function and mechanisms of action of phytochromes, cryptochromes and phototropins; stomatal movement; photoperiodism and biological clocks. F. Solute transport and photoassimilate translocation : Uptake, transport and translocation of water, ions, solutes and macromolecules from soil, through cells, across membranes, through xylem and phloem; transpiration; mechanisms of loading and unloading of photoassimilates. G. Secondary metabolites : Biosynthesis of terpenes, phenols and nitrogenous compounds and their roles. H. Stress physiology : Responses of plants to biotic (pathogen and insects) and abiotic (water, temperature and salt) stresses; mechanisms of resistance to biotic stress and tolerance to abiotic stress 7. SYSTEM PHYSIOLOGY - ANIMAL A. Blood and circulation : Blood corpuscles, haemopoiesis and formed elements, plasma function, blood volume, blood volume regulation, blood groups, haemoglobin, immunity, haemostasis. B. Cardiovascular System : Comparative anatomy of heart structure, myogenic heart, specialized tissue, ECG its principle and significance, cardiac cycle, heart as a pump, blood pressure, neural and chemical regulation of all above. C. Respiratory system : Comparison of respiration in different species, anatomical considerations, transport of gases, exchange of gases, waste elimination, neural and chemical regulation of respiration. D. Nervous system : Neurons, action potential, gross neuroanatomy of the brain and spinal cord, central and peripheral nervous system, neural control of muscle tone and posture. E. Sense organs : Vision, hearing and tactile response. F. Excretory system: Comparative physiology of excretion, kidney, urine formation, urine concentration, waste elimination, micturition, regulation of water balance, blood volume, blood pressure, electrolyte balance, acid-base balance. G. Thermoregulation : Comfort zone, body temperature physical, chemical, neural regulation, acclimatization. H. Stress and adaptation I. Digestive system : Digestion, absorption, energy balance, BMR. J. Endocrinology and reproduction : Endocrine glands, basic mechanism of hormone action, hormones and diseases; reproductive processes, neuroendocrine regulation. 8. INHERITANCE BIOLOGY A. Mendelian principles : Dominance, segregation, independent assortment, deviation from Mendelian inheritance. B. Concept of gene : Allele, multiple alleles, pseudoallele, complementation tests. C. Extensions of Mendelian principles : Codominance, incomplete dominance, gene interactions, pleiotropy, genomic imprinting, penetrance and expressivity, phenocopy, linkage and crossing over, sex linkage, sex limited and sex influenced characters. D. Gene mapping methods : Linkage maps, tetrad analysis, mapping with molecular markers, mapping by using somatic cell hybrids, development of mapping population in plants. E. Extra chromosomal inheritance : Inheritance of mitochondrial and chloroplast genes, maternal inheritance. F. Microbial genetics : Methods of genetic transfers transformation, conjugation, transduction and sex-duction, mapping genes by interrupted mating, fine structure analysis of genes. G. Human genetics : Pedigree analysis, lod score for linkage testing, karyotypes, genetic disorders. H. Quantitative genetics : Polygenic inheritance, heritability and its measurements, QTL mapping. I. Mutation : Types, causes and detection, mutant types lethal, conditional, biochemical, loss of function, gain of function, germinal verses somatic mutants, insertional mutagenesis. vi

7 A/UG-15 vii Amar Ujala Education J. Structural and numerical alterations of chromosomes : Deletion, duplication, inversion, translocation, ploidy and their genetic implications. K. Recombination : Homologous and non-homologous recombination, including transposition, site-specific recombination. 9. DIVERSITY OF LIFE FORMS A. Principles and methods of taxonomy : Concepts of species and hierarchical taxa, biological nomenclature, classical and quantititative methods of taxonomy of plants, animals and microorganisms. B. Levels of structural organization : Unicellular, colonial and multicellular forms; levels of organization of tissues, organs and systems; comparative anatomy. C. Outline classification of plants, animals and microorganisms : Important criteria used for classification in each taxon; classification of plants, animals and microorganisms; evolutionary relationships among taxa. D. Natural history of Indian subcontinent : Major habitat types of the subcontinent, geographic origins and migrations of species; common Indian mammals, birds; seasonality and phenology of the subcontinent. E. Organisms of health and agricultural importance : Common parasites and pathogens of humans, domestic animals and crops. 10. ECOLOGICAL PRINCIPLES A. The Environment : Physical environment; biotic environment; biotic and abiotic interactions. B. Habitat and niche : Concept of habitat and niche; niche width and overlap; fundamental and realized niche; resource partitioning; character displacement. C. Population ecology : Characteristics of a population; population growth curves; population regulation; life history strategies (r and K selection); concept of metapopulation demes and dispersal, interdemic extinctions, age structured populations. D. Species interactions : Types of interactions, interspecific competition, herbivory, carnivory, pollination, symbiosis. E. Community ecology : Nature of communities; community structure and attributes; levels of species diversity and its measurement; edges and ecotones. F. Ecological succession : Types; mechanisms; changes involved in succession; concept of climax. G. Ecosystem : Structure and function; energy flow and mineral cycling (CNP); primary production and decomposition; structure and function of some Indian ecosystems: terrestrial (forest, grassland) and aquatic (fresh water, marine, eustarine). H. Biogeography : Major terrestrial biomes; theory of island biogeography; biogeographical zones of India. I. Applied ecology : Environmental pollution; global environmental change; biodiversity-status, monitoring and documentation; major drivers of biodiversity change; biodiversity management approaches. J. Conservation biology : Principles of conservation, major approaches to management, Indian case studies on conservation/management strategy (Project Tiger, Biosphere reserves). 11. EVOLUTION AND BEHAVIOUR A. Emergence of evolutionary thoughts : Lamarck; Darwin concepts of variation, adaptation, struggle, fitness and natural selection; Mendelism; spontaneity of mutations; the evolutionary synthesis. B. Origin of cells and unicellular evolution : Origin of basic biological molecules; abiotic synthesis of organic monomers and polymers; concept of Oparin and Haldane; experiment of Miller (1953); the first cell; evolution of prokaryotes; origin of eukaryotic cells; evolution of unicellular eukaryotes; anaerobic metabolism, photosynthesis and aerobic metabolism. C. Paleontology and evolutionary history : The evolutionary time scale; eras, periods and epoch; major events in the evolutionary time scale; origins of unicellular and multicellular organisms; major groups of plants and animals; stages in primate evolution including Homo. D. Molecular Evolution : Concepts of neutral evolution, molecular divergence and molecular clocks; molecular tools in phylogeny, classification and identification; protein and nucleotide sequence analysis; origin of new genes and proteins; gene duplication and divergence. E. The Mechanisms : Population genetics populations, gene pool, gene frequency; Hardy-Weinberg law; concepts and rate of change in gene frequency through natural selection, migration and random genetic drift; adaptive radiation and modifications; isolating mechanisms; speciation; allopatricity and sympatricity; convergent evolution; sexual selection; co-evolution.

8 Amar Ujala Education viii A/UG-15 F. Brain, Behavior and Evolution : Approaches and methods in study of behavior; proximate and ultimate causation; altruism and evolution-group selection, kin selection, reciprocal altruism; neural basis of learning, memory, cognition, sleep and arousal; biological clocks; development of behavior; social communication; social dominance; use of space and territoriality; mating systems, parental investment and reproductive success; parental care; aggressive behavior; habitat selection and optimality in foraging; migration, orientation and navigation; domestication and behavioral changes. 12. APPLIED BIOLOGY A. Microbial fermentation and production of small and macro molecules. B. Application of immunological principles (vaccines, diagnostics). tissue and cell culture methods for plants and animals. C. Transgenic animals and plants, molecular approaches to diagnosis and strain identification. D. Genomics and its application to health and agriculture, including gene therapy. E. Bioresource and uses of biodiversity. F. Breeding in plants and animals, including marker assisted selection. G. Bioremediation and phytoremediation. H. Biosensors. 13. METHODS IN BIOLOGY A. Molecular biology and recombinant DNA methods : Isolation and purification of RNA, DNA (genomic and plasmid) and proteins, different separation methods; analysis of RNA, DNA and proteins by one and two dimensional gel electrophoresis, isoelectric focusing gels; molecular cloning of DNA or RNA fragments in bacterial and eukaryotic systems; expression of recombinant proteins using bacterial, animal and plant vectors; isolation of specific nucleic acid sequences; generation of genomic and cdna libraries in plasmid, phage, cosmid, BAC and YAC vectors; in vitro mutagenesis and deletion techniques, gene knock out in bacterial and eukaryotic organisms; protein sequencing methods, detection of post-translation modification of proteins; DNA sequencing methods, strategies for genome sequencing; methods for analysis of gene expression at RNA and protein level, large scale expression analysis, such as micro array based techniques; isolation, separation and analysis of carbohydrate and lipid molecules; RFLP, RAPD and AFLP techniques B. Histochemical and immunotechniques : Antibody generation, detection of molecules using ELISA, RIA, western blot, immunoprecipitation, floweytometry and immunofluorescence microscopy, detection of molecules in living cells, in situlocalization by techniques such as FISH and GISH. C. Biophysical methods : Analysis of biomolecules using UV/visible, fluorescence, circular dichroism, NMR and ESR spectroscopy, structure determination using X-ray diffraction and NMR; analysis using light scattering, different types of mass spectrometry and surface plasma resonance methods. D. Statistical Methods : Measures of central tendency and dispersal; probability distributions (Binomial, Poisson and normal); sampling distribution; difference between parametric and non-parametric statistics; confidence interval; errors; levels of significance; regression and correlation; t-test; analysis of variance; X2 test; basic introduction to Muetrovariate statistics, etc. E. Radiolabeling techniques : Properties of different types of radioisotopes normally used in biology, their detection and measurement; incorporation of radioisotopes in biological tissues and cells, molecular imaging of radioactive material, safety guidelines. F. Microscopic techniques : Visulization of cells and subcellular components by light microscopy, resolving powers of different microscopes, microscopy of living cells, scanning and transmission microscopes, different fixation and staining techniques for EM, freeze-etch and freeze-fracture methods for EM, image processing methods in microscopy. G. Electrophysiological methods: Single neuron recording, patch-clamp recording, ECG, Brain activity recording, lesion and stimulation of brain, pharmacological testing, PET, MRI, fmri, CAT. H. Methods in field biology : Methods of estimating population density of animals and plants, ranging patterns through direct, indirect and remote observations, sampling methods in the study of behavior, habitat characterization-ground and remote sensing methods. I. Computational methods: Nucleic acid and protein sequence databases; data mining methods for sequence analysis, web-based tools for sequence searches, motif analysis and presentation.

9 A/UG-15 Amar Ujala Education TREND ANALYSIS LIFE SCIENCES (Trend Analysis) No. of Questions Dec June 2016 Dec June 2015 Cellular Organization Fundamental Processes Cell Communication and Cell Signaling Developmental Biology System Physiologyplant Inheritance Biology DiversityofLifeForms Applied Biology Molecules and their Interaction ix

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11 CONTENTS CSIRUGC/NET LIFE SCIENCES Model Solved Paper CSIR UGC / NET LIFE SCIENCE Solved Paper, June-2017 P1/1-21 CSIR UGC / NET LIFE SCIENCE Solved Paper, December-2016 P1/1-30 CSIR UGC / NET LIFE SCIENCE Solved Paper, Juner-2016 P2/1-26 CSIR UGC / NET LIFE SCIENCE Solved Paper, December-2015 P3/1-25 CSIR UGC / NET LIFE SCIENCE Solved Paper, June-2015 P4/1-25 CSIR UGC / NET LIFE SCIENCE Solved Paper, December-2014 P5/1-24 CSIR UGC / NET LIFE SCIENCE Solved Paper, June-2014 P6/1-22 CSIR UGC / NET LIFE SCIENCE Solved Paper, December-2013 P7/1-20 CSIR UGC / NET LIFE SCIENCE Solved Paper, June-2013 P8/1-23 CSIR UGC / NET LIFE SCIENCE Solved Paper, December-2012 P9/1-23 CSIR UGC / NET LIFE SCIENCE Solved Paper, June-2012 P10/1-22 CSIR UGC / NET LIFE SCIENCE Solved Paper,December-2011 P11/1-23 CSIR UGC / NET LIFE SCIENCE Solved Paper, June-2011 P12/1-19 xi

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13 A/UG-15 Amar Ujala Education CSIR UGC (NET/JRF) LIFE SCIENCES Solved Paper, June An ant starts at the origin and moves along the y-axis and covers a distance l. This is its first stage in its journey. Every subsequent stage requies the ant to turn right and move a distance which is half of its previous stage. What would be its coordinates at the end of its 5th stage? 31 13l (A), 8 16 (B) 13l 3l, l 3l 3l 13l (C), (D), Exp.Initially the coordinates are (0,0). It moves l units upwards towards y axis and then move right half of it. It is the 1 second stage. Now the coordinates are,1. Then it 2 moves right and covers half of distance. Now the 1 31 coordinates are,. The it again moves right and 2 4 covers half the distance the coordinates are now 31 31, 8 4 and gain it moves right and covers half distance. The final coordinates are, So option (A) is the right option. 2. In a group of siblings there are seven sisters, and each sister has one brother. How many siblings are there in total? (A) 15 (B) 14 (C) 8 (D)7 Exp. There are seven sisters and each sister has 1 brother. That means total no. of siblings are 8. So option (C) is the right option. 3. What is the average value of y for the range of x shown in the following plot? PART A P1/1 1 y x (A) 0 (B) 1 (C) 1.5 (D) 2 Exp.The range is lying between 1 and 2. So average of 1 and 2 is 1.5. Therefore, option (C) is the right option. 4. A bread contains 40% (by volume) edible matter and the remaining space is filled with air. If the density of edible matter is 2 g/cc, what will be the bulk density of the bread (in g/cc)? (A) 0.4 (B) 0.8 (C) 1.2 (D) 1.6 Exp.Let say mass of the bread is 100 grams So weight of edible matter is 40 grams. Now density = mass/volume So volume = mass/density= 100 = 50 cc. 2 Now bulk density of bread = weight of edible matter/ volume = = 0.8 So right option is option (B) 5. A board has 8 rows and 8 columns, A move is defined as two steps along a column follwed by one step along a row or vice-versa. What is the minimum number of moves needed to go from one corner to the diagonally opposite corner? (A) 5 (B) 6 (C) 7 (D)9

14 Amar Ujala Education Exp. If we draw the figure and see that there are 5 moves in which it goes from one corner to other corner diagonally. So right answer is option (A). 6. A job interview is taking place with 21 male and 17 female candidates. Candidates are called randomly. What is the minimumnumber of candidates to be called to ensure that at least two males or two females have been interviewwd? (A) 17 (B) 2 (C) 3 (D)21 Exp. Correct option is (D) because at least 2 males or 2 females have called for interview therefore it could be 10 males also or 15 females also or even could 21 males and the only option that goes in accordance with this is option (D). 7. Cumulative frequency % (Research Scholars) Number of papers Published The graph shows cumulative frequency % of research scholars and the number of papers published by them. Which of the following statements is true? (A) Majority of the scholars published more than 4 papers. (B) 60% of the scholars published at least 2 papers. (C) 80% of the scholars published at least 6 papers. (D)30% of scholars have not published any paper. Exp. It has been shown in the graph that from 40% -100% of research scholars has published more than 2 papers. So 60% of the scholars has published at least 2 papers. 8. A tells only lies on Monday, Tuesday and Wednesday and speaks only the truth for the rest of the week. B tells only lies on Thursday, Friday and Saturday and speaks only the truth for the rest of the week. If today both of them state that they have lied yesterday, what day is it today? (A) Monday (B) Thursday (C) Sunday (D) Tuesday Exp. Clearly it is Thursday because on Thursday A speaks truth that he lies 1 day before that is on Wednesday and it is given in the question. Also, B lies on Thursday that P1/2 A/UG-15 it lies on 1 day before on Wednesday because B speaks truth on Wednesday. So, the right option is (B) 9. A fair die was thrown three times and the outcome was repeatedly six. If the die is thrown again what is the probability of getting six? (A) 1 6 (B) (C) (D) Exp. In this question, we just need to find out the probability of 6 in a die and that is 1 6. Therefore right option is (A). 10. Which is the odd one out based on a divisibility test? (A) 474 (B) 572 (C) 682 (D)154 Exp. In this all the numbers except 474 are divisible by 11 so 474 is the odd one out. 11. My birthday is in January. What would be a sufficient number of questions with 'Yes/No' answers that will enable one to find my birth date? (A) 6 (B) 3 (C) 5 (D)2 Exp. Consider all answers to be yes; First question - Is your birthday in the first 15 days? Second question - Is your birthday in the first 7 days? Third question - Is your birthday in the first 4 days? Fourth question - Is your birthday in the first 2 days? Fifth question - Is your birthday on the first day? In case any answer is no, the alternate half of the question will be considered and the next question will be framed accordingly. First question - Is your birthday in the first 15 days? - No Second question - Is your birthday in the first 8 days? (16th to 24th) - No Third question - Is your birthday in the first 4 days? (25th - 28th) -No Fourth question - Is your birthday in the first 2 days? (29th to 30th) -No Fifth question - Is your birthday on the 31st? -Yes It could be 4 questions as well, but none of the options match

15 A/UG A square is drawn with one of its sides as the hypotenuse of a right angled triangle as shown in the figure. What is the area of the shaded circle? 3 cm 4 cm 25 (A) 1 cm 2 (B) 25 2 cm 25 2 (C) 3 cm 25 2 (D) 4 cm Exp. Here diameter of the circle is calculated using Pythagoras theorem in the triangle containing the sides 3 cm and 4 cm. From pythagoras theorem the diameter of circle is coming as 5 cm. Therefore the radius of the circle is 5 cm.now applying 2 the area of circle as pi r square where r is 5 2 cm. So the area comes out to be 25 pi cm square What should be added to the product of the two numbers and to make it a perfect square? (A) 9 (B) 13 (C) 19 (D)27 Exp. Here if we multiply the numbers we get the units digit as 7 and (from the given option) = 16 a perfect square. Therefore 9 is the correct answer. 14. In ABC, AB = AC and BAC = 90º; EF AB and DF AC. The total area of the shaded region is D A E B F C (A) (C) 2 AF 2 2 BC 2 (B) AF 2 (D) BC 2 2 Amar Ujala Education Exp.Here BDF is 90º because DF is parallel to AC and FEC = 90 because EF is parallel to AB. Therefore BD = DF Because B = DFB= 45º and EF = EC Because EFC = C =45º. Now area BDF = 1 2 *BD DF= 1 2 BD2 and area CEF = 1 2 FE *EC = 1 2 FE2. Now using Pythagoras theorem, BD 2 = 2 BF 2 and EF 2 = 2 CF 2. By adding BD 2 + EF 2 = 2 BF CF 2 2 BC and from there we get the answer as 2. So option (C) is the right option 15. Consider a circle of radius r. Fit the largest possible square inside it and the largest possible circle inside the square. What is the radius of the innermost circle? (A) r / 2 (B) r 2 2 (C) r / 2 (D) r / 2 Exp. Here the diagonal of the square will be 2r. From Pythagoras theorem, we get 4r 2 = 2a 2. From this a 2 2 multiplied by r. 2 r r Now radius of the inner circle is that will be. 2 2 Now again applying pythagoras theorem, we get radious of inner circle = r r that is coming out to be 2 r. That is the answer In how many ways can you place N cons on a board with N rows and N columns such that every row and every column contains exactly one coin? (A) N (B) N (N-1) (N-2) (C) N 2 (D)N N P1/3

16 Amar Ujala Education Exp.In this question it is like arranging N alphabets of a word. So the answer is N factorial Option (B) is the correct option. 17. Two identical wheels B and C move on the periphery of circle A. Both start at the same point on A and return to it, B moving inside A and C outside it. Which is the correct statement? A B C (A) B wears out times C (B) C wears out times B (C) B and C wear out about equally (D)C wear out two times B Exp. It is clear from the figure both B and C wear out equally. 18. Which of the following is the odd one out? (A) Isosceles triangle (B) Square (C) Reglar hexagon (D) Rectangle 21. The Hardy-Weinberg principle comes from considering what happens when Mendelian genes act in a population. The model predicts that there will be no change in allele frequencies when (A) Migration into the population norms at a steady rate (B) The population suffers a bottleneck (C) A rare new mutation is associated with a sharp increase in fitness (D) No evolutionary process is at work Exp. Hardy-Weinberg law states that both alleles and gesnotypic frequencies in a population remains constant from generation to generation until some specific disturbing influences are introduced like non random mating, mutation, selection, random genetic drift, gene flow and meiotic drive. 22. Which one of the following is responsible for initiation of maternal behaviour in the first time pregnant rats after parturition? (A) Higher prolactin levels in blood (B) Srimulation of sensory receptors during delivery (C) Changes of uterine volume (D) Presence of males rats PART-B P1/4 A/UG-15 Exp.In this question, we can say isosceles triangle does not contain any diagonals while others contain diagonals. So (A) is the right option. 19. find the missing word : A, AB,, ABBABAAB (A) AABB (B) ABAB (C) ABBA (D) BAAB Exp.In this, if we look at first and second A and AB in first only the first half part of AB is given. So in 3rd and 4th if we compare, only the first half part of 4th will be in g the 3rd. So right option is (C) 20. A 100 m long train crosses a 200 m long and 20 m wide bridge in 20 seconds. What is the speed of the train in km/hr? (A) 45 (B) 36 (C) 54 (D)57.6 Exp. Total length is length of the train plus length of the bridge. So, total length is = 300 m and time taken is 20 sec. Therefore, speed is = 15 m/sec = 54 km/hour. Exp. Prolactin is inducer for lactation and metalnal behavior. 23. To replace animal use in testing hepatic toxicity of a drug on trial, which one of the following would be used in vitro to be closet to the in vivo scenario? (A) Liver cells (B) Hepatic cell lines (C) Liver slices (D) Co-culture of liver parenchymal cells and Kupfer cells 24. Which of the following does not represent a strategy for phytoremediation? (A) Phytodegradation (B) Phytomining (C) Continuous removal through hyper accumulators (D) Chelate-mediated extraction of pollutants Exp. Phytomining is use of plants to extract metal compound of high economic value. 25. The word "fermentation" is used in biochemistry and microbial technology to denote different phenomena. If the former is called C and latter is called T, which of the following statements is true?

17 A/UG-15 (A) All C is T but all T is not C (B) All T is C but all C is not T (C) T is always a product of genetic engineering while C is not (D) C is always an anaerobic process, while T can be aerobic or anaerobic 26. Which of the following statements is NOT true during the infection of plant cells with Agrpbacterium? (A) The protein products of virulence genes vira and virg perceive acetosyringone (B) The virb protein forms a connection between Agrobacterium and the plant cell and facilitates T- DNA transfer into the plant (C) The T-DNA is excised and bound to VirD2 protein (D) The T-DNA, after becoming coated with VirF binds to phosphorylated VIP1, which, allows the complex to enter the plant's nucleus Exp. Agrobacterium is a genus o Gram-negative bacteria established by H.J.Conn that uses horizontal gene transfer to cause tumors in plants. 27. among exixting technologies, which of the following vector systems would you prefer to use for generating a library for 140 kb eukaryotic genomic DNA fragments, while giving due consideration to size as well as stability of the insert? (A) Phage (B) Cosmid (C) Bacterial artificial chromosome (BAC) (D) Yeast artificial chromosome (YAC) Exp. COSMID-37 to 52kb BAC kb YAC kb Lamda-below 100 kb 28. If r denotes the correlation coefficient and m denotes the slope of regression line, interchanging X and Y axes would (A) Change m but not r (B) Change r but not m (C) Change both r or m (D) Not change r or m 29. The use of biotinylated secondary antibody in ELISA (A) Increases the sensitivity of the assay but compromises the specificity (B) Increases the sensitivity of the assay without compromising the specificity (C) Does not alter either sensitivity or specificity (D) Decreases both sensitivity and specificity Exp. Biotin labeled Antibodies (Abs) increase sensitivity. They reduce cross reactivity and steric hindrance. It P1/5 Amar Ujala Education results in higher sensitivity and better specificity. 30. Which is the best method for checking mycoplasma contamination in a mammalian cell line? (A) Southern hybridization (B) ELISA (C) PCR (D) Western hybridization 31. Two pure lines of corn have mean cob length of 9 and 3 inches, respectively. The polygenes involved in this trait exhibit additive gene action. Crossing these two lines is excepted to produce a progeny population with mean cob length (in inches) of (A) 12.0 (B) 7.5 (C) 6.0 (D) Which of the following organism is widely used as a biocontrol agent in organic farming? (A) Rhizobium tropicii (B) Trichoderma virdis (C) Fusarium oxysporum (D) Nostac muscorum Exp. Trichoderma viridis, is used as a biocontrol agent in organic farming. 33. A paraphyletic group (A) Contains unrelated organisms (B) Includes the most recent common ancestor but not all of its descendents (C) Includes all the representatives of a clade but not the most recent common ancestor (D) Contains all the representatives of a clade and the most recent common ancestor Exp. A paraphyletic group contains the most recent common ancestor but not all of its descendants, e.g.-reptilia contains the last common ancestor of descendants of that ancestor but not birds and mammals. 34. Which of the following is NOT an adaptive modification in a xerophytic plant? (A) Strongly developed sclerenchyma (B) Sunken stomata (C) Sparse stomata (D) Presence of lucnar tissues Exp. Xerophytic plants are those which grown in arid condition,where evaporation is more than precipitation and hence plant has sunken stomata. 35. If milk is left open, lactose is fermented first to produce acid. This is followed by proteolytic bacterial activity which increases the ph. Ultimately milk fats degraded to produce rancidity. This is an example of (A) Ecological succession (B) Microbial antagonism

18 Amar Ujala Education (C) Interference competition (D) Microevolution Exp. Ecological succession is the phenomenon or process by which an ecological community undergoes more or less orderly and predictable changes following disturbance or initial colonization of new habitat. 36. Symbiotic biological nitrogen fixation takes place with the association between a plant and a nitrogen fixing prokaryote as shown in the following table: List of plants Nitrogen fixing 1. Soybean (i) Frankia 2. Casuarina (ii) Bradyrhizobium 3. Gunnera (iii) Anabaena 4. Azolla (iv) Nostac The correct combination is: (A) 1 - (i), 2 - (ii), 3 - (iii), 4 - (iv) (B) 1 - (ii), 2 - (i), 3 - (iv), 4 - (iii) (C) 1 - (iii), 2 - (ii), 3 - (i), 4 - (iv) (D) 1 - (iv), 2 - (iii), 3 - (ii), 4 - (i) 37. Secondary sewage treatment involves (A) Physical removal of solids from polluted water by filteration and sedimentation (B) Removal of chemical remains by precipitation (C) Removal of dissolved organic compounds by activated sludge ot trickling filter (D) Removal of microbial pathogens by chlorination or ozonization 38. Based on per molecule, which of the following gas has the most powerful greenhouse effect? (A) CO 2 (B) CH 4 (C) N 2 O (D) CFCs Exp. Carbon dioxide equivalent is highest for CFCs. (amongst given option. It is 140~11,700) 39. Sexual selection results in variation in the reproductive success of males, often due to female choice with particular phenotypes. This type of sexual selection occurs because (A) Males cannot compete with other males (B) Cost of breeding is higher for females as compared to males (C) Inappropriate mating results in a similar reduction in fitness of females and males (D) Males are limiting resource for females Exp. Cost of breeding is higher for females as compared to males. 40. Among the following events in the history of life 1. prokaryotic cell 2. eukaryotic cell P1/6 3. natural selection 4. organic molecules 5. self-replicating molecules Which is the correct chronological order? (A) 4, 5, 3, 1, 2 (B) 4, 5, 1, 2, 3 (C) 5, 4, 1, 3, 2 (D) 4, 5, 1, 3, 2 A/UG Which one of the following non covalent interactions between two non bonded atoms A and B is most sensitive to the distance between them? (A) A and B are permanent dipoles and are in volved in hydrogen bonding. (B) A and B are fully ionized and are involved in salt bridge formation. (C) A and B are uncharged and repel each other. (D) A and B are uncharged and attract each other. Exp. Non covalent interactions between two non bonded atoms A and B is most sensitive to the distance between them because A and B are uncharged and repel each other. 42. Which statement best describes the pka of amino groups in proteins? (A) pka of a amino group is higher than the pka of Î amino group. (B) pka of a amino group is lower than the pka of Î amino group. (C) pka of a amino groip is same as the pka of Î amino group. (D) pka of a amino group is higher than the pka of guanidine side chain of ariginine. Exp. Lysine. an essential amino acid, has a positively charged e-amino group (a primary amine). Lysine is basically alanine with a propylamine substituent on the b-carbon. The e-amino group has a significantly higher pk a (about 10.5 in polypeptides) than does the a-amino group. The amino group is highly reactive and often participates in a reactions at the active centers of enzymes, Proteins only have one a amino group, but numerous e amin so groups. However, the higher pk a renders the lysyl side chains effectively less necleophilic. Specific environmental effects in enzymes active centers can lower the pk a of the lysyl side chain such that it becomes reactive. Note that theside chain has three methylene groups. So that even though the terminal amino group will be charged under physiological conditions, the side chain does have significant hydrophobic character. Lysines are often found buried with only the e amino group exposed to solvent pk R = 10.5

19 A/UG-15 + NH3 CH2 CH2 CH2 CH2 +H N CH C O 3 pk 2 = 9.0 O pk 1 = What is the effect of 2, 4 dinitrohphenol on mitochondria? (A) Blocks ATP syntheses without inhibiting electron transport by dissipating the proton gradient. (B) Blocks electron transport and ATP synthesis by in hibiting ATP ADP exchange across the inner mi tochondrial membrane. (C) Blocks electron transport and proton pumping at complexes I, II and III. (D) Interacts directly with ATP synthase and inhibits its activity. Exp. 2, 4-dinitrophenol (DNP) uncouples the mitochondria by shuttling H+ ions across the inner membrane, by passing ATP synthase. DNP is a mobile ionophore, this means that it is a lipid soluble molecule that is capable of transporting ions across a biological membrane. The way it does this is by shielding the charge of the ion within its hydrophobic exterior, thereby facilitating the transport of H+ ions across a biological membrane. The mode of action provides less resistence to the H+ ion than as if it were to move through ATP synthase and thus it becomes the preferred route. Because of this, less ATP is produced inversely proportional to the concentration of DNP. Subsequently the potential energy that was stored in the concentration gradient is released as heat energy. If the concentration of DNP becomes too great, the cell will be unable to produce ATP and eventually die. 44. A protein has 30% alanine. If all the alanines are replaced by glycines. (A) Helical content will increase. (B) b sheel content will increase. (C) There will be no change in conformation. (D) The alanine substituted protein will be less struc tured than the parent protein. Exp. In internal helical positions, alanine is regarded as the most stabilizing residue, whereas gylcine. after proline, is the more destrabilizing. P1/7 Amar Ujala Education 45. The gel to liquid crystalline transition temperature (Tm)of phospholipids is dependent on the fatty acid composition. Considering this, Tm of (A) All the phospholipids will be indentical. (B) DPPC will be lowest and DOPC will be highest. (C) POPC and DOPC will be indentical and lower than DMPC or DPPC. (D) DOPC will be lowest and DPPC will be highest. 46. You have created a fusion between the trp operon, which encodes the enzymes for tryprophan biosynthesis, under the regulatory control of the lac operatir. Under which of the following condetions will tryptophan synthase be induced in the strain that carries the chimeric operator fused operons? (A) Only when both lactose and gluse are absent. (B) Only when both lactose and glucose are present. (C) Only when lactose is absent and glucose is present. (D) Only when lactose is present and glucose is absent. Exp. Lac repressor protein always binds to the operator there by blocking transcription. In the presence of glucose and lactose, the bacterium will utilize glucose as the primary Carbon source - a transcription off Lac Operon off. In the presence of lactose alone, Lactose binds to the lac repressor-transcription on. 47. Which of the following pairs of subcellular compartments is likely to have same ph and electrolyte composition? (A) Cytosol and lysosomes. (B) Cytosol and mitochondrial inter membrane space. (C) Cytosol and endosome. (D) Mitochondrial matrix and inter membrane space. 48. Regarding microtubule assembly and disassembly during cell division, which will be the most appropriate answer? (A) Once formed, kinetochore microtubules depolymerize at the plus ends throughout mitosis. (B) Once formed, kinetochore microtubules depolymerize at the plus ends throughout mitosis. (C) Kinetochore microtubules polymerize at their plus ends up to anaphase, at which point they begin to depolymerize. (D) Kinetochore microtubules polymerize at their mi nus ends up to cytokinesis, at which point they depolymerize. 49. Origin of replication usually contains : (A) GC rich sequences (B) Both AT and GC rich sequences

20 Amar Ujala Education (C) No particular stretch of sequences (D) AT rich sequences Exp. Repeated sequences are commonly present in the sites for DNA replication initiation in bacterial, archaeal, and eukaryotic replicons. Those motifs are usually the binding places for replication initiation proteins or replication regulatory factors. In prokaryotic replication origins, the most abundant repeated sequences are DnaA boxes which are the binding sites for chromosomal replication initiation protein DnaA, iterons which bind plasmid or phage DNA replication initiators, defined motifs for site-specific DNA methylation, and 13- nucleotide-long motifs of a not too well-characterized function, which are present within a specific region of replication origin containing higher than average content of adenine and thymine residues. In this review, we specify methods allowing identification of a replication origin, basing on the localization of an AT-rich region and the arrangement of the origin's structural elements. We describe the regularity of the position and structure of the AT-rich regions in bacterial chromosomes and plasmids. The importance of 13-nucleotide-long repreats present at the AT-rich region, as well as other motifs overlapping them, was pointed out to be essential for DNA replication initiation including origin opening, helicase loading and replication complex assembly. We also summarize the role of AT-rich region repeated sequences for DNA replication regulation. 50. S csubunit of E. coli RNA polymerase does not : (A) Initiate transcription and fall off diring elongation (B) Increase affinity of the core enzyme to the promoter (C) Binds to DNA, independent of the core enzyme (D) Ensures specificity of transcription by inter acting with the core enzyme Exp. Every molecule of RNA polymerase holoenzyme contains exactly one sigma factor subunit, which in the model bacterium Escherichia coli is one of those listed below. The number of sigma factors varies between bacterial species. E. coli has seven sigma factors. RNA polymerase holoenzyme complex consists of core RNA polymerase and a sigma factor executes transcription of a DNA template strand. Once initiation of RNA transcription is complete, the sigma factor can leave the complex. It long has been thought that the s factor obligatorily leaves the core enzyme and initiate transcription allowing the free s to link to another core enzyme and initiate transcrption at another site. Thus, the s cycles from one core to another. However, s does not obligatorily leave the core. Instead, the s changes its P1/8 A/UG-15 binding with the core during initiation and elongation. Therefore, the s cycles between a strongly bound state during initiation and a weakly bound state during elongation. 51. The cap binding protein (eif4e), which is involved in the global regulation of translation, is highly regurlated in eukaryotic cells. In an experiment, a researcher transfected mammalian cells with (eif4e) gene for its overxpression. Due to this, the cells will undergo : (A) Apoptosis (B) Neoplastic transformation (C) No change (D) Differentiation 52. Bacteriophage T4 infects E. coil and injects its DNA inside the cell. The transcription of viral genes occurs in three stages : immediate early, early and late. All the promotrs on viral genome are available, but the control takes place at the level of : (A) Promotes strength (B) Modification of host RNA polymerase (C) Synthesis of new RNA merases (D) Turn over rate of RNA synthesis Exp. T4 early promoter strength probed in vivo with unribosylated and ADP-ribosylated Excherichia coli RNA polymerase : a mutation analysis. 53. Gram negative bacteria, Klebsiella pneumoniae, upon infecting humans, results in severe septic shock after a few hours of infection.which of the following is not true for this type of infection? (A) Cell wall endotoxins cause overproduction of cytokinnes. (B) Septic shock can be treated by anti TNFa antibodies. (C) Recombinant bacterial proteins can be used for the treatment of septic shock (D) Recombinant TNFa receptor antagonist can be used for the treatment of septic shock 54. Which of the following is NOT associated with insulin action? (A) Increased glucose transport (B) Increased glycogen formation (C) Enhanced lipolysis in adipose tissue (D) Decreased rate of gluconeogenesis Exp. Insulin inhibited glycerol release by adipose tissue in all groups of rats with fructose in the medium. in fed rats a barely significant insulin inhibition occured also in the presence of glucose, and in fasted-refed rats it was indepedent of whether a hexose was present in the medium or not.

21 A/UG-15 The following effects of insulin on tissue of fasted-refed rats incubated in the absence of hexose were observed: Inhibition of glycerol release with a dose effect relationship between 10 and 250 m units insulin per milliliter; partial inhibition of glycogen breakdown during incubation; diminished lactic and pyruvic acid ratio. These same effects were also exerted by the antilipolytic drug 5-methylpyarazole 3-carboxylic acid, which does not stimulate glucose uptake. 55. When adenoma is converted to metastatic adenocarcinoma, which of the following compination of proteins is almost certainly to be degraded? (A) Type IV collagen and laminin (B) Fibranectin and b 2 integrin (C) Metalloprotease and serine protease (D) Elastin and selectin 56. Which of the following is considered to be a combined B and T cell deficiency? (A) Ataxia telanglectasia (B) Swiss type agammaglobulinemia (C) Wiskott Aldrich syndrome (D) Bruton's agammaglobulinemia Exp. Combined B-cell and T-cell immuno deficiencies, or SCID, is a group of medical disorders that are the result of genetic defects in both cellular and humoral immunity. The defects in humoral and cellular immunity have an early clinical presentation and, if untreated, result in a fatal outcome in the first few years of life. This article focuses only on SCID disorders and outlines recent advances in therapeutics options for patients. The profound degree of immune compromise in SCID leads to infections with bacterial, viral, and fungal pathogens that cause significant morabidity and eventually mortality in patients. Swiss type agammaglobulenemia is a type of combined immuno-deficiency syndromes. 57. The part of the embryo from which the ectoderm, mesoderm and endoderm are formed in chick is known as (A) Primitive streak (B) Hypoblast (C) Epiblast (D) Cytotrophoblast Exp. The epiblast is capable of forming all three germ layers (ectoderm, mesoderm and endoderm) during gastrulation. Epiblast cells migrate to the primitive streak and invaginate into a space between the epiblast and the hypoblast. It is a cross section through the cranial region of the primitive streak at 15 days, illustrating the invagination of epiblast cells. The invaginating epiblast cells displace the hypoblast to create the definitive endoderm. Once the definitive endoderm is established, migrating epiblast cells also form the intraem-bryonic mesoderm. The remaining epiblast P1/9 Amar Ujala Education cells, which do not migrate through the primitive streak. remain in the epiblast to form the ectoderm. Hence, the epiblast gives rise to all three layers in the embryo. 58. Which protein secreted by the amphibian orgainizer indced neural tissue formation by inhibiting Bone Morphogenetic Protein? (A) b catenin (B) Noggin (C) Armadillo (D) Cubitus interruptus 59. The homologue of b catenin in Drosophila is (A) Fushi tatazu (B) Engrailed (C) Armadillo (D) Cubitus interruptus Exp. The armadillo repeat family of proteins is defined by the existence in length (42 amino acids) and spacing (typically end-to0end with no intervening sequences), though not necessarily highly conserved in sequence. Rather, it is the structure of the arm motif that is conserved; repeats of this domain together form a positively charged groove that mediates protein-protein interactions. This is so-called 'arm motif' was originally identified in the protein product of the Drosophila segment polarity gene known as armadillo. It was subsequently determined that the vertebrate homologues of armadillo, b-catenin and plakoglobin and the Src tyrosine kinase substrate p120 catenin, also contain multiple arm repeats. Together, these four proteins initially defined the armadillo repeat family of proteins. Although the function of these four proteins all coincidentally relate to cell-cell adhesion, it is now clear that members of the armadillo repeat family have widely varied roles in cellular physiology. 60. Which of the floral whorls is affected in apetala 3/ pistillata (ap3/pt) mutants? (A) Sepals and petals (B) Petals and stamens (C) Stamens and carpels (D) Sepals and stamens 61. Spindle fibres of a mitotic cells are made up of (A) Action (B) Myosin (C) Collagen (D) Tublin Exp. Spindle fibres of a mitotic cells are made up of tubulin, a filament like protein. The filaments are arranged in such a way as to form a tubule between them. 62. Which of the following occurs in mitochondria? (A) Glycolysis (B) Glycogenolysis (C) Histolysis of tissue (D) Tricarboxylic acid cycle

22 Amar Ujala Education Exp.Mitochondria contains all necessary enzymes for Kreb s cycle in which hydrogen molecules are removed from fuel molecule (pyruvic acid) and transferred to carries molecules like NAD for further processing-in ETS. 63. Cell wall of diatoms is made up of (A) Silica (B) Carbonate (C) Calcium carbonate (D) All of these Exp. Cell wall of diatoms is made up of two halves, one half covering the other, resembling a soap as box, expitheca. Cellulosic cell wall impia-gnated with silica to form transparent silicious shell known as frustule. 64. The study related to the structure and function of cell is known as (A) Cytology (B) Histology (C) Anatomy (D) Palynology Exp. The study of structure and function of cell with their constituents is known as cytology or cell biology. 65. Pullorum disease of poultry is caused by (A) Hemophilus (B) Mycobacterium (C) Salmonella (D) Clostridium Exp. Pullorum disease of poultry is caused by Salmonella pullorum. The common symptoms are loss of appetite, diarrhoea and dysentry. 66. In albinism the absence of which pigment makes the skin and hair very light coloured (A) Carotene (B) Melanine (C) Haemoglobin (D) Chlorophyll Exp. In albinism, the absence of melanin pigment makes skin and hair very light coloured. Individuals suffering with A/UG-15 this disease are incapable of converting dihydroxy - phenyl alanine, into melanin. 67. The technique of growing plants, in water culture is called (A) Lkebana (B) Hydroponics (C) Tissue culture (D) Cell culture Exp. Hydroponic means growth of plants; in water culture rather than soil, providing all the nutrients. 68. DDT is (A) An antibiotic (B) Not a pollutant (C) A non-degradable pollutant (D) A biodegradable pollutant Exp. Such pollutants generally not degraded or degrade at a very slow pace by natural biological process. These not only accumulate but are often get biologically magnified. 69. Which of the following is produced by yeast? (A) Enzyme (B) Vitamin (C) Hormone (D) Riboflavin Exp. Yeast produces zymase enzyme. It is intracellular enzyme which ferments invert sugar into alcohol and CO IR-8 is a variety of (A) Cajanus (B) Groundnut (C) Rice (D) Maize Exp. oryza sativa (Rice) belongs to family Poaceae. IR-8 is a variety of rice cultivated in Andhra Pradesh. 71. Cyclic AMP (A) Is found only in eukaryotes (B) Stimulates sodium channels (C) Interacts with G proteins directly (D) Activates an intracellular protein kinase Exp. Cyclic-AMP derived from ATP by adenyl ate cyclase, camp removes inhibition of protein kinase by inhibitory sub unit, active protein kinase can phosphorylate another intra-cellular enzyme thereby either activating or repressing it depending upon enzyme. 72. The high solubility of amino acids in water is due to (A) Presence of side chain (B) Dipolar ion structure PART - C P1/10 (C) Unipolarity (D) Hydrophilic nature of the amino group Exp. Amino acids are soluble in water except cysteine and tyrosine. Solubility of amino acid in water is due to nonpolar structure. With the exception of proline all are insoluble in alcohol. 73. A mechanism that can cause a gene to move from one linkage group to another is (A) Translocation (B) Inversion (C) Crossing over (D) Duplication Exp. Chromosome aberration in which a segment of chromosome is shifted to a new location in the same or completely different chromosome.

23 A/UG Increase number of chromosomes occur in (A) Turner s syndrome (B) Fragile-X syndrome (C) Klinefelter s syndrome (D) retinoblastoma Exp.Klinefelter s syndrome. In human beings a chromosomal abnormality where an individual has two-x-chromosomes and only one Y-chromo-sornes (X x Y). 75. Which one of the following is a cofactor and not a coenzyme? (A) Biotin (B) Tetrahydrofolic acid (C) Copper (D) Methylcobalamin Exp. A substance which acts in conjunction with an enzyme or attaches itself to a enzyme in order to activate it. e.g. Cu. 76. Which of the common bases (A, C, G, T) of DNA has no oxygen in its structure? (A) Thymine (B) Adenine (C) Cytosine (D) Guanine Exp. N NH2 6 N N N H Adenine (6-amino purine), there is no oxygen in its structure. 77. Example of Analogous organs is (A) Wings of bird and insect (B) Forelimbs of horse and man (C) Teeth of elephant and man (D) None of the above Exp. As we know that analogous organs are different in structures but similar in functions. So, the wings of bird and insect are the best example of analogous organ. 78. Which statement is incorrect? (A) Plant viruses contain RNA (B) Animal viruses contain DNA (C) T 4 contains double strand DNA (D) TMV contains double stranded RNA P1/11 Amar Ujala Education Exp. Single stranded RNA is present in TMV (Tobacco Mosaic Virus) in the form of spirals, surrounded by capsomeres. 79. If dorsal nerve of spinal cord is broken down then its effect is (A) No impulse is transmitted (B) Impulse is transmitted but slowly (C) Impulse is transmitted fast (D) No effect on impulse Exp. Chloreplast, peroxisomes and mitochondria are involved in Photorespiration. In this process, loss of the fixed CO 2 at high temperature and O 2 concentration takes place. 80. Serum is (A) Blood minus fibrinogen (B) Lymph minus corpuscles (C) Lymph (D) Blood minus corpuscles and fibrinogen Exp. Serum is the blood plasma that separates from clot and is free from cell or fibrin. 81. Prosthetic group of an enzyme is (A) A non-protein part lightly bound to the enzyme (B) Organic compounds or inorganic ions loosely bound to the enzyme (C) The three dimensional structure of the protein forming the active site (D) Any inhibitor that forms a complex with the enzyme rendering it reactive Exp. It is a organic compound essential to function of an enzyme in the organism. It differs from co-enzyme in the fact that they are more firmly attached to enzyme protein. It is a non-protein part which bound to the enzyme. 82. Hormone that promotes growth of lateral buds, has negative effect on apical dominance is (A) Cytokinin (B) Gibberellin (C) Auxin (D) Both (B) and (C) Exp. Cytokinin promotes cell divisions and growth of lateral. buds in presence of auxins. Auxins and cytokinin pact opposite to one another in relation to apical dominance. 83. What are bile salts? (A) Charged phospholipids (B) Amphipathic cholesterol analogs with detergent properties (C) Esterified cholesterols (D) Hydrolyzed forms of triacylglycerols

24 Amar Ujala Education Exp. Fluid secreted by liver in vertebrates i.e. bile, containing 97% water, 0.7%. bile salts, 0.2% bile pigments, 0.06% cholesterol, 0.7% inorganic salts and 0.1 % fats along with little alkaline phosophatases. 84. Renal portal system is (A) Present in all vertebrates (B) Present in all chordates (C) Absent in mammals (D) Present in all mammals Exp. Renal portal system is well developed in fishes and amphibians, reduced in reptiles and birds and is absent in mammals. 85. Amphimixis is (A) Fusion of sperm with egg (B) Fusion of pronucleus of sperm with of egg (C) No fusion (D) Fusion of diploid cells Exp. Amphimixis is formulation of new individual through normal process of sexual reproduction. It is the fusion of sperm nycleus with egg nucleus. 86. Term New Systematic was given by (A) Julian Huxley (C) Linnaeus (B) Bateson (D) Darwin Exp. In 1940, the term new systematics was given by Julian Huxley. 87. Diadelphous (9 + 1) stamens occur in (A) Gramineae (B) Cucurbitaceae (C) Papilionaceae (D) Malvaceae Exp. Stamens with filaments fused and forming two groups of fused stamens. 88. Hormone controlling contraction of uterus during parturation is (A) Luteinising hormone (B) Estrogen (C) Progesterone (D) Oxytocin Exp. Oxytocin hormone from posterior pituitary lobe acting mainly on myoepithelial cells in the ducts of breasts. It causes contraction of uterus and help in birth. 89. The following is true about gluconeogenesis (A) It occurs in liver (B) It occurs in adipose tissue (C) It is inhibited by glucagon (D) It is stimulated by insulin P1/12 A/UG-15 Exp.Glyconeogenesis is the conversion of fats and proteins into carbohydrate with extreme demand of glucose. It occurs in liver. 90. The cause of the resting membrane potential is (A) Fick s law (B) Potassium ion concentration inside the cell (C) Nernst s equation (D) Donnan effect Exp. Nernst s equation: Equation relating the e.m.f. a cell and concentration or the activities of reactants and products of the cell reaction. 91. The heart rate shows variation during respiratory rhythm in most human subjects. Which one of the following statements describing the changes of heart rate during respiratory phases is true? (A) The heart rate is accelerated during expiration, but no change occurs during inspiration (B) The heart rate is accelerated during inspiration and decelerated during expiration (C) The heart rate is accelerated during expiration and decelerated during inspiration (D) The heart rate is accelerated during inspiration and no change occurs during expiration Exp. Respiratory sinus arrhythmia (RSA) is a naturally occurring variation in heart rate that occurs during breathing cycle. Heart rate increase during inspiration and decreases during expiration. Heart rate is normally controlled by centers in the medulla oblongata. One of these centers, the nucleus ambiguus, increases parasympathetic nervous system input to the heart via the vagus nerve. The vagus nerve decreases heart rate by decreasing the rate of SA node firing. Upon expiration the cells in the nucleus ambiguus are activated and heart rate is slowed down. In contrast, inspiration triggers inhibitory signals to the nucleus ambiguus and consequently the vagus nerve remains unstimulated. 92. At 17 years, a 7 feet tall human was diagnosed with gigantism caused by pituitary tumor. The condition was surgically corrected by removal of the person s pituitary gland. Doctors advised hormonal therapy. The possible hormonal therapies that would be required for survival are : (1) Thyroid hormone (2) Glucocorticoids (3) Glucagon (4) Growth hormone (5) Insulin Which one of the following combination can be used? (A) (1) and (2) only (B) (2) and (4) only (C) (1), (2) and (4) (D) (1), (3) and (5)

25 A/UG-15 Exp. He would require only Thyroid hormone and Glucocorticoids for survival. If thyroid hormones will be absent then basic metabolic rate of the body will decrease and if glucocorticoids will be absent then that individual will not be able to survive in stressful conditions. 93. Autotetrapoloids arise by the doubling of 2n complement to 4n. There are three different pairing possibilities at meiosis in tetraploids as given below : (1) Two bivalents (2) one quadrivalent (3) One univalent + one trivalent Which of the above pairings can lead to production of diploid gamete? (A) only (1) (B) (2) and (3) (C) (1) and (3) (D) (1) and (2) Exp. If 2 bivalents or one quadrivalent is present in a cell only then diploid gametes can be produced due to distribution of bivalents between the two cells. Whereas, unequal distribution of cromosomes will be observed in case of univalents and trivalents and trivalents, therefore, diploid gametes cannot be produced. 94. The following is the amino acid sequence of a part of a protein encoded by gene X.... Phe Leu Val Pro Ser Tyr Cys... A mutant for gene X is isolated following treatment with a mutagen. the amino acid sequence of the same region encoded by the mutant gene is as follows :... Phe Leu Phe Arg Arg Ile... Which of the following mutagenes is most likely to have been used? (A) 5-bromouracil (B) 2-amino-purine (C) Ethyl methanesulfonate (D) Acridine orange Exp. After leuicine the amino acids sequence is changed and that reflect an indel which is caused by acridine orange. 95. In Neurospora, The mutant stp exhibits erratic stopand-start growth. When a female of stp strain is crossed with a normal strain acting as a male, all progeny individuals showed stp mutant phenotype. However, the reciprocal cross resulted in all normal progeny individuals. These results can be explained on the basis of : (1) Maternal inheritance (2) sex limited inheritance Amar Ujala Education (3) sex influenced inheritance (4) stp mutation may be located in mitochondrial DNA The most appropriate statement or combination of the above statements for explaining the experimental results is : (A) (1) and (3) (B) (3) only (C) (1) and (4) (D) (2) and (4) Exp. When mother is stp, all offsprings are stp. When mother is normal, all offspring are normal. this shows that the character is getting transfered from mother to the offspring directly. This can happen when the stp mutation is present in extranuclear DNA i.e. mitochondrial DNA. Statements A and D, both are correct. 96. A co-transduction experiment was performed to decipher the linear order of 4 genes : a, b, c and d. Three sets of experiments were done where transductants were selected for a (Set-1) or b (Set-2) or c (Set-3) and screened for co-transduction of the other markers : Set-1 Selected for Co-transduction Frequency a b 31 a c 3 a d 89 Set-2 Selected for Co-transduction Frequency b a 22 b c 78 b d 68 Set-3 Selected for Co-transduction Frequency c a 0 c b 69 c d 43 Based on the frequencies shown above, identify the most likely order in the genome : (A) a b c d (B) b c d a (C) c d a b (D) a d b c Exp. Co-transduction frequency between a and c is least. So a and c should be farthest. 97. A hypothetical biochemical pathway for the formation of eye color in insect is given below : Substrate X Substrate Y (Colorless) (Colorless) x mutant b P1/13

26 Amar Ujala Education Colorless Intermediate mutant a x Brown pigment Colorless Intermediate organge pigment Orange-Brown Eye color (wild type) Two autosomal recessive mutants a and b are identified which block the pathway as shown above. Considering that the mutants are not linked, what will be the phenotype of the F 2 progeny if crosses were made between parents of the genotype aabb AAbb and the F1 progeny are intercrossed? (A) 9 orange-brown : 3 orange : 3 brown : 1 colorless (B) 9 orange-brown : 7 colorless (C) 1 orange : 2 colorless (D) 15 orange-brown : 1 colorless Exp. Parental cross : aabbx AAbb; F1 : AaBb; F1 self cross : AaBb AaBb, since genes are unlinked, they will get assorted independently F2 : A_B_(Organge brown) 9 : A_bb(Brown) 3 : aab_(orange) 3 : aabb (white) 1. Answer. 98. An analysis of four microsatellite markers was carried out in a family showing a genetic disorder. The results are summarized below : F1 F F1 F M1 M2 M3 M4 Based on the above, which of the markers shows linkages to the disorder? (A) M1 (B) M2 (C) M3 (D) M4 Exp. P1, 1, 2, 5, 6-all affected individual share M2 as their common marker. this means M2 is linked with the genetic disorder. 99. In group I are given 4 orders of class insecta. Match each one with a common name (Group II) and its P1/14 diagnostic characters (Group III) : Dermaptera (A) Ephemeroptera (B) Odonata (C) Group III (i) Group I Plecoptera (D) Group II Ant (E) Mayfly (F) Grasshopper (G) Damselfly (H) Stonefly (I) Earwig (J) A/UG-15 Elongate, membranous wings with net like venation, abdomen long and slender, compound eyes occupy most of head, hemimetabolous metamorphosis (ii) Elongate chewing mouthparts, thread-like antennae, abdomen with unsegmented forcepslike cerci, hemimetabolous metamorphosis (iii) Forewing long, narrow and leathery, hindwing broad and membranous, chewing mouthparts, hemimetabolous metamor-phosis (iv) Elongate abdomen with two or three tail filaments, two pairs of membranous wings with many veins, forewings traingular, short, bristlelike antennae, hemimetabolous metamorphosis (v) Adults with reduced mouthparts, elongate antennae, long cerci, nymphs aquatic with gills, hemimetabolous metamorphosis (vi) Wings membranous with few veins, well developed ovipositiors, sometimes modified into a sting, mouth parts modified for biting and lapping, holometabolous metamorphosis (A) A-J-(ii) (B) B-I-(vi) (C) C-H-(v) (D) D-F-(i) Exp. Damselfly belong to Odonata order of insecta nymphs are aquatic with gills Which of the following is a correct match of the animal with its attribute? Animal Attribute (1) Rotifer (i) Nauplius larval stage (2) Sea anemone (ii) Radial symmetry (3) Barnacle (iii) Pacudocoelomate body cavity (4) Sea urchin (iv) Water vascular system (A) (1) (iii), (2) (ii), (3) (i), (4) (iv)

27 A/UG-15 (B) (1) (ii), (2) (iii), (3) (i), (4) (iv) (C) (1) (iii), (2) (iv), (3) (i), (4) (ii) (D) (1) (iv), (2) (iii), (3) (ii), (4) (i) Exp. Rotifer-are pseudocoelomate animals. Sea animonebelong to cnideria phylum they have radial symmetry. Sea urchin-belong to Echinodemata phylum. they possess water vascular system Which of the following characteristics make Amborella the most basal living angiosperm? (A) Carpels fused by tissue connection and absence of vessel elements (B) Absence of carpels and presence of vessel element (C) Carpels free and presence of vessel elements (D) Presence of carpels and absence of vessel elements Exp. The carpels are sealed by a secretion of stick fluid, rather than developmentally fused as in most angiosperms and vessels are absent Which of the following statements is not correct? (A) Stomata are present in mosses and hornworts but absent in liverworts (B) Only the lycophytes have microphylls and almost all other vascular plants have megaphylls (C) Monocot pollen grains have three openings whereas eudioot pollen grains have one opening (D) Monocots have fibrous root system whereas eudicots have taproot Exp. Monocot pollen have one opening whereas dicot have three Following is a table showing selected characte-ristics of important fungal groups : Fungal group Characteristic a No regularly occuring septa in thallus b Perforated septa c Forms arbuscular mycorhizae on plant roots d Have zoospores with flagella In the above table, the fungal groups A, B, C and D, are respectively : (A) Chytridiomycetes, Ascomycetes, Glomeromycetes, Zygomycetes (B) Zygomycetes, Ascomycetes, Glomero-mycetes, Chytridiomycetes, (C) Ascomycetes, Zygomycetes, Glomero-mycetes, Chytridiomycetes (D) Chytridiomycetes, Zygomycetes, Ascomy-cetes, Glomeromycetes P1/15 Amar Ujala Education 104. Which of the following is a correct statement? (A) Euglenids have a spiral or crystalline rod inside flagella (B) Pheophytes have a spiral or crystalline rod inside flagella (C) Euglenids have a hairy and smooth flagella (D) Euglenids and pheophytes both have a spiral of crystalline rod inside flagella Exp. Euglenids are the one having spiral or crystalline rod inside flagellas. And pheophytes are otherwise called as brown algae and they are having hairy and smooth flagella kindly find it from the reference where it is tabulated In life history evolution there is generally a trade-off between the size and number of offspring produced. Some conditions are listed below : (1) Scarcity of food during the early stages of life (2) Provision of parental care (3) High mortality during early stages of life (4) Predator s prefernece for large sized prey What are the above two conditions that would favour production of a small number of large-sized offspring? (A) (2) and (3) (B) (2) and (4) (C) (1) and (2) (D) (1) and (3) Exp. Small sized offsprings are produced in large No. and they dont get any parental care usually. As the population of offspring is large that will lead to the scarcity of the food The possible relationship between level of disturbance and species diversity in a biological community are that species diversity : (i) is unaffected by disturbance (ii) is highest at intermediate levels of disturbance (iii) decreases exponentially with increasing levels of disturbance (iv) starts decreasing only at higher levels of disturbance (1) High Species diversity Low Low Disturbance High

28 Amar Ujala Education (2) High Species diversity Low Low (3) High Species diversity Low Low (4) High Species diversity Low Low Disturbance Disturbance Disturbance High High High Match each graph with its corresponding state-ments above : (A) (1) (iv), (2) (iii), (3) (ii), (4) (iv) (B) (1) (iii), (2) (iv), (3) (ii), (4) (i) (C) (1) (i), (2) (ii), (3) (iii), (4) (iv) (D) (1) (iii), (2) (i), (3) (ii), (4) (iv) Exp. As it is clear from the figure that in case second is showing the independent pattern and option 4th is the only optoin The diagram represents competition between species 1 and species 2 according to Lotka-Volterra model of competition : Sp 2 K 2 K 1/ K 2/ Sp 1 Given the conditions in the diagram, the predicted outcome of competition is : K 1 A/UG-15 (A) Unstable coexistence between species 1 and 2 because K 1 > K 2 / and K 2 > K 1 / (B) Unstable coexistence between species 1 and 2 because K 1 < K 2 / and K 2 < K 1 / (C) Stable coexistence between species 1 and 2 because K 1 > K 2 / and K 2 > K 1 / (D) Stable coexistence between species 1 and 2 because K 1 < K 2 / and K 2 < K 1 / Exp. The isoclines are intersecting at the stable equilibrium point Following four types of species were observed in a community : (1) Species A has a large effect on community because of its abundance. (2) Species B has a large role in community out of proportion to its abundance. (3) Status of species C provides information on the overall health of an ecosystem. (4) Significant conservation resources are allocated to species D which is single, large and instantly recognizable According to above description, species A, B C and D are called respectively : (A) Dominant, Keystone, Indicator and Flagship (B) Keystone, Flagship, Dominant and Indicator (C) Keystone, Dominant, Indicator and Flagship (D) Flagship, Dominant, Keystone and Indicator Exp. Keystone species effects the community largely irrespective to its abundance as per this B is keystone species. Thus option 1 is correct Complete the following hypothetical life table of a species to calculate the net reproductive rate R 0 : Age Number Number Age Age l x m x Class (X) Alive of dying Specific Specific (n x ) (d x ) survivor Fertility -ship The calculated R 0 will be : (A) 0.75 (B) 1.00 (C) 0.65 (D) 0.15 Exp. dx is age specific death is 200 and 300 (missing one), missing nx are 500 and 200, Now lx of each class are 1, 0.8, 0.5, 0.3, 0.2. P1/16

29 A/UG-15 Then, lxbx for each class are (1 0 = 0) + (0.8 0 = 0) + ( = 0.25) + (0.3 1 =.3) + (0.2 1 = 2) =.75 (R0) Which of the following is the correct decreasing order for the rate of decomposition of litter constituents? (A) Hemicellulose, cellulose, lignin, phenol (B) Cellulocse, hemicellulose, phenol, lignin (C) Hemicellulose, cellulose, phenol, lignin (D) Lignin, phenol, hemicellulose, cellulose 111. Which of the following is not a benefit for the female adopting polyandry? (A) Greater probability of getting all her eggs fertilized (B) Ability to receive more resources from the males (C) Ability to produce more offspring than normal (D) Improved chances of genetic compatbility with her own DNA 112. Assume that individual A wants to do an altruistic act to individual B and that the benefit and cost of doing this act are, in fitness units, 40 and 12, respectively. According to Hamilton s Rule, A should perform the altruistic act only if B is his : (A) nephew (B) niece (C) grandson or granddaughter (D) daughter or son Exp. Polyandry increases the probability of getting all the eggs fertilised, it also increases genetic compatibility and female receives more rewards in this. But it does not result in production of more offspring than normal rather increases the survival rate of those offsprings due to the parental care received Individual A performs to another individual a behavioural act which has a fitness consequences. Match the behavioural acts (a) to (e) with the correct fitness consequence ((i) to (iv)). Behavioral act Cooperation (a) Adaptive altruism (b) Spite (c) Deceit and manipulation (d) Reciprocity (e) Fitness consequence to A Gains direct fitness but after delay (i) Loses inclusive fitness (ii) Gain indirect fitness (iii) Gains direct fitness but immediately (iv) (A) (a) (iv), (b) (iii), (c) (ii), (d) (i), (e) (i) (B) (a) (i), (b) (ii), (c) (ii), (d) (iii), (e) (iv) P1/17 Amar Ujala Education (C) (a) (i), (b) (iii), (c) (ii), (d) (ii), (e) (iv) (D) (a) (ii), (b) (ii), (c) (iii), (d) (i), (e) (iv) Exp. In adaptive Alturism there is the indirect gain of fitness (inclusive fitness) and in Spite, deciet and manipulation No gain of inclusive fitness is there, these are the selfish behaviours In a random sample of 400 individuals from a population with alleles of a trait in Hardy-Weinberg equilibrium 36 individuals are homozygous for allele a. How many individuals in the sample are expected to carry at lest one allele A? (A) 36 (B) 168 (C) 364 (D) 196 Exp. # of individuals with At lest 1 A = # of AA + # of Aa; = 364 will be the number of people with at least one A Two kinds of natural selection (A and B) acting on a trait are shown in the figure below. In each, the top graph shows the trait frequency before and the bottom graph frequency after the action of natural selection : F R E Q U E N C Y A TRAIT F R E Q U E N C Y B TRAIT The kind of natural selection in A and B are : (A) A Directional, B Disruptive (B) A Netural, B Disruptive (C) A Stabilizing, B Disruptive (D) A Disruptive, B Stabilizing Exp. In stabilising selection intermediate forms are selected by nature while in disrupted both the extreme forms of the population are selected Which of the following statements is not correct regarding effect of genetic drift? (A) It alters allele freuency substantially only in small population (B) It can cause allele frequencies to change at random (C) It can lead to a loss of genetic variation within population (D) It can cause harmful alleles to become eliminated Exp. All other statements are correct moreover genetic drift is a random process. It can not change the genetic structure in any specific direction.

30 Amar Ujala Education 117. During the production of alcohol by fermentation using budding yeast, oxygen supply is kept limited. Why? (A) Budding yeasts are obligate anaerobes and cannot tolerate oxygen (B) Budding yeats lose mitochondria in the absence of oxygen (C) Budding yeast are facultative anaerobes (D) Alcohol is oxidized further in the presence of oxygen Exp. In presence of oxygen, pyruvate is completely oxidized into carbon dioxide and water. However htere is no oxidation of Alcohol. Majority of yeast species, prefer fermention or respiration. i.e. they are facultative anaerobes and they will produce alcohol in absence of oxgyen A student was asked to design a knockout cassette for specifically deleting the p53 gene from the prostate gland of mice. Which one of the following pairs of cassettes will ensure deletion of the gene? TSP Cre p53 (A) Cre p53 p53 (B) TSP Cre p53 (C) TSP Cre p53 (D) TSP = Tissue specific parameter Cre = Cre recombiness = kap sites Exp. Cre should be under TSP and the gene p53 should be surrounded by direct repeats of lox as direct repeats causes deletion Following are certain statement regarding somatic hybridization, a technique used for plant improvement : (1) Protoplasts of only sexually compatible plant species can be fused (2) Hybrids are produced with variable and asymmetric amounts of genetic material of parental species (3) Protoplast fusion permits transfer of gene block or chromosomes (4) Genes to be transferred need to be identified and isolated Which one of the following combinations of the above statements is correct? (A) (1) and (3) (B) (2) and (3) (C) (1) and (4) (D) (2) and (4) Exp. Protoplast of sexually incompatible can also be fused (POMATO) and genes are not be isolated and identified for transfer. A/UG Small molecular weight ocmpounds affect the activity of luciferase differently. the oil/water solubility of various compounds is one property important for its effect on luciferase. the straight line in the graph was obtained by plotting the activity data with 50 different compounds. the luciferase activity, of two new derivatives of Benzene (A and B) are shown below : Luciferase activity B A Oil/water partitioning coefficient Which of the following statement is correct? (A) A is phosphate and B is amine (B) A is methyl and B is amine (C) A methyl and B is propyl (D) A is amine and B is phosphate Exp. A partition coefficient is the ratio of concentrations of a compound in the two phases of a mixture of two immiscible liquid at equilibrium. Higher values indicate that compound is more hydrophobic. According to graph derivative A is hydrophobic and derivative B is hydrophilic. So a correct combination of the functional groups is option 2 (methyl is hydrophobic and amine is hydrophilic) A researcher was repeating a FACS experiment but somehow got confused with the labeling of the tubes. There are four tubes, one control, C (with no fluorescent label), one standard 1, S1 (with FITC label), one standard 2, S2 (with PE label) and the last one test, T (which should be FITC positive). Given below is the result of the FACS experiment : (1) PE (2) PE FITC FITC P1/18

31 A/UG-15 (3) PE (4) PE FITC FITC What should be the correct labeling? (A) a, S2; b, S1; c, T; d, C (B) a, S1; b, T; c. C; d, S2 (C) a, S2; b, S1/T; c, C; d, S1/T (D) a, S1/T, b, S2; c, S1/T; d, C Exp. Control-zero fluorescence figure d that has minimum fluorescence, S1-has only FITC label so could either be figure a or c, S2 has onlype label sois figure b, test also is FITC positive so can either be figure a or c. (corresponding labels will show more fluorescnece with respect to their region towards X-axis for FITC and towards Y-axis for PE) Nichrome coated stainless steel electrodes were implanted in a rat brain for chronically recording the electrical activity of deep brain structures. During a study of 3 months the intensity of electrical signals gradually decreased. The following statements may explain the cause of this observation : (1) The depositions of metallic iron from the electrode tips caused degeneration of some neurons (2) The gradual accumulation of microglia at the electrode tips increased the resistnace of electrodes (3) The neurons at the electrode tips were hyperpolarized gradually (4) The threshold for firing action potential in the neurons at the electorde tips was increased due to prolonged presence of electordes Which one of the following is correct? (A) (1) only (B) (1) and (2) (C) (3) only (D) (3) and (4) Exp. Under evaluation Amar Ujala Education 123. In an attempt to detect protein expression profile in a cell, Western blot technique is employed. Expression of two new proteins is to be followed by probing with respective high affinity antibodies (raised in rabbit). Unfortunately, the two proteins were found to comigrate in SDS-PAGE profile. Under this situations, using one dimensional SDS-PAGE and by Western blot, which one of the following is the best way to demonstrate the presence of both the proteins? (A) Develop Western blots with their antibodies in the same gel (B) Prior to doing SDS-PAGE/Western blot, one protein could be removed by immuno precipitating in the cell extracts (C) Silencing the expression of one protein at a time by sirna and performing Western blotting (D) Subjecting the technique of stripping/reprobing of the gel after transferring to nitrocellulose membrane while doing Western blotting Exp. Researchers have used been using stripping or reprobing techniques to screen a single blot with multiple antibodies. Protocols have been specifically formulated to dissociate and remove all antibodies from the membrane immobilized proteins without destroying the antigenic binding sites or removing the protein A chromatin immuno precipitation (ChIP) assay was performed to determine specific transcription factor binding sites on the promoter of a gene. Pull down was done using either IgG or antibodies against c- myc. A DNA containing c-myc binding regions was used as a control for PCR amplification (input). Which one of the following PCR representations of DNA is correct? (A) (B) (C) (D) Input IgG Anti-C-myc 125. the number of seeds in the fruit of a plant species, H0 : µ = 30. A random sample of 9 fruits gives the mean number of seeds as 24 with a standard deviation of 6.12 (a) What are the confidence limits for the sample mean? (b) Would you reject or accept the null hypothesis at 95% confidence level? P1/19

32 Amar Ujala Education (A) (a) 18 and 30, (b) reject the hypothesis (B) (a) 20 and 28, (b) reject the hypothesis (C) (a) 20 and 28, (b) reject the hypothesis (D) (a) 18 and 30, (b) reject the hypothesis Exp. The sigma M = sigma/(n)^-1/2. Sigma = 6.12 and N = 9. Sigma M would come out to be Since the Confidence interval = 95 percent, hence range of C.I. can be calculated as 1.96*2.04 = 4. Therefore actual CL = 24-4 to = 20 to 28. And since mu = 30 does not fall in the CI and hence will reject the null hypothesis Lungs are enclosed in (A) Pericardium (C) Perichondrium (B) Peritonium (D) Pleural membrane Exp. Lungs are enclosed in plueral membrane which consists of two layers; the visceral and parietal. The parietal pleura lines the interior of the chest wall and upper surface of the diaphragm. The visceral pleura attached to the lung itself, completely covering its surfaces except at the hilium The enzyme restriction endonuclease (A) Cuts RNA strand (B) Cuts double strand of DNA (C) Joins the strands of DNA (D) Cuts single strand of DNA Exp. The enzyme restriction endonuclease helps in unwinding of double strand of DNA by cutting at specific site during DNA replication Which of the following is a test cross? (A) WWww (B) WW-ww (C) WW+ww (D) WWxww Exp. A test cross is a cross between heterozygous F 1 hybrid and double recessive homozygous e.g. Ww ww to show whether FI is homozygous or heterozygous Chromosome complement with (2n l) is called (A) Monosomy (B) Nulisomy (C) Trisomy (D) Tetrasomy Exp. Chromosome complement with (2n 1) is called monosomy i.e. lack one complete chromosome. These could be easily produced in polyploids. A polyploid has several chromosomes of same type, and therefore, this loss can be easily tolerated Nucleotide constituent of RNA are (A) Adenine, guanine, cytosine, uracil (B) Thymine, cytosine, guanine, uracil P1/20 (C) Adenine, guanine, cytosine, thymine (D) Cytosine, adenine, aric acid, guanine A/UG-15 Exp. Nucleotide constituents of RNA are adenine, guanine, cytosine, uracil. Mainly RNA, found in all organisms would perform different, important functions during protein synthesis Known sequence of DNA that is used to find complimentary DNA strand is (A) Vector (B) Plasmid (C) DNA probe (D) Recombinant DNA Exp. Known sequences of DNA that is used to find complementary DNA strand is DNA probe. These probes are faster and sensitive, so that many conventional diagnostic tests for viruses and bacteria involving culturing of the organism, are being fast replaced by antibody and DNA probe assays Cranial nerves numbering iv, v, vi are (A) Trochlear, trigeminal abducens (B) Trochlear, trigeminal facial (C) Auditory, facial, trochlear (D) Auditory, trochlear, facial Exp. Cranial nerve numbering iv, v, vi are trochlear, trigeminal and abducens. The nerves coming out from the brain constitute the cranial nerves. Each cranial nerve originates from a distinct separate root Gonadotropins are secreted from (A) Gonads (B) Post pituitary (C) Anterior pituitary (D) Hypothalamus Exp. Gonadotropins are secreted frorn anterior pituitary. Gonadotropins-a group of hormones which regulate certain activities of the gonads i.e. FSH and LH Insulin is secreted by (A) Spleen (B) b-cells of pancreas (C) a-cells of pancreas (D) Mucosa oesophagus Exp. Insulin is secreted by B-cells of pancreas. A rise in blood sugar stimulated B-cells to secrete the hormone. The hormone enhances the uptake of glucose by tissues like muscle and adipose tissue, probably by enhancing the transport of the sugar across the cell membrane Book lungs are respiratory organs of (A) Mollusca (B) Mammals (C) Arachnida (D) Earthworm

33 A/UG-15 Exp. Book lungs are respiratory organs of Arachnida, (The scorpion). The respiratory system consists of four pairs of cuticular book lungs or pulmonary sacs. One pair of them lies inside each mesomeric segment from 3 to Micturition reflex is related to (A) Urination (B) Ovulation (C) Copulation (D) Earthworm Exp. Micturition reflex is reflected to urination. Expelling of urine from urinary bladder is called micturition DNA is associated with highly basic- proteins called (A) Histones (B) Non-histones (C) Albumins (D) All of these Exp. Histones are basic proteins which contain basic amino acids i.e. arginine and lysine. DNA is complexed with histone protein in the form of deoxyribonucleo proteins or chromatin material During DNA replication, the term leading strand is applied to the one which strand (A) 5' 3' direction continuously (B) 5' 5' direction continuously (C) 5' 3' direction discontinuously (D) 3' 5' direction discontinuously Exp. In DNA molecule, out of the two strands, one strand replicates continuously in 5' 3' direction and other grows discontinuously and in pieces in 3' 5' direction. The former is known as leading strand and later as lagging strand and the pieces as okazaki fragments A distinct mechanism that usually involves a short segment of DNA with remarkable capacity to move from on location in a chromosome to another. this is called (A) DNA replication (B) DNA transposon (C) DNA hybridization (D) DNA recombination Exp. DNA transposons are small DNA fragments carry a copy of certain host genes which can move or jump from one location to another on he chromosome Chromosomal number are reduced to half in (A) Meiosis (B) Mitosis (C) Binary fission (D) Parthenogenesis Exp. Chromosomal number are reduced to half in meiosis. This reduced chromosome number, whether reduced at P1/21 Amar Ujala Education the time of formation of the sex cells or at any other time during life cycle is called haploid condition Linkage was discovered by (A) Muller (B) De vries (C) Morgan (D) Mendel 142. The characters of birds without exception is (A) Omnivorous (B) Flying wings (C) Beak without teeth (D) Lay eggs with calcareous shell Exp. The character of birds without exception is beak without teeth. Birds constitute a well defined group of vertebrate animals. As a class they form a more homogeneous group than any other class of vertebrates Physiological evidence through blood test. reveals that turtles are more closely related to (A) Mammals than to crocodiles. lizards and mammals (B) Snakes than to crocodiles. lizards and mammals (C) Crocodiles than to lizards. snakes and mammals (D) None of the above Exp. Physiological evidence through blood test, reveals that turtles are more closely related to crocodiles than to lizards, snakes and mammals. Turtle is usually applied to aquatic or semi- aquatic forms and resembles to crocodile in possession of (A) palate (B) immovable quadrate (C) proteas etc Fresh water bony fishes maintain water balance by (A) Excreting a hypotonic urine (B) Excreting salt across their gills (C) Excreting waste in the form of uric acid (D) None of the above Exp. Fresh water bony fishes maintain water balance by exerting hypotonic urine i.e. containing fewer nonpenetrating particles on one side of semipermeable membrane than on the other side thus exerting less osmotic pull If you are given an insect spider, a peripatus and crab based on which character you can identify an arachnida from other? (A) Sense organ s (B) Number of wings (C) One pair of legs (D) Four pairs of legs Exp. An arachnida is characterized by four pairs of legs. Cephalothorax with 2 chelicerae and 2 pedipalps.

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35 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, December-2016 PART-A 1. Two cockroaches of the same species have the same thickness but different lengths and widths. Their ability to survive in oxygen deficient environments will be compromised if (A) their thickness increases, and the rest of the size remains the same. (B) their thickness remains unchanged, but their length increases. (C) their thickness remains unchanged, but their width decreases. (D) their thickness decreases, but the rest of the size remains unchanged. Exp. As the main respiratory openings called spiracles are present along the lateral side of the body of the cockroach,so if it increases length wise then the probability of surviving in the oxygen deficient condition will be more with the respiratory openings. 2. The bar chart shows number of seats won by four political parties in a state legislative assembly seats party Which of the following pie-charts correctly depicts this information? (A) (C) (B) (D) Exp. Total no seats=140 Seats won by 1st party=50/140*100=35.7% of the total seats seats won by 2nd and 4th party are the same which will be=35/140*100=25%of total seats seats won by 3rd party=20/140*100=14.2% of total seats If we arrange this % in pie chart the best explanatory ans is B. 3. The random errors associated with the measurement of and Q are 10% and 2%, respectively. What is the percentage random error in P/Q? (A) 12.0 (B) 9.8 (C) 8.0 (D) In how many distinguishable ways can the letters of the word CHANCE be arranged? (A) 120 (B) 720 (C) 360 (D) 240 Exp. Total no letters=6 No of C = 2 Total no of ways of arrangement= 6! divided by 2!= Find out the missing pattern ? (A) 7 (B) (C) 7 (D) Exp. If we move square in clockwise direction, then according to question answer will be 1. In first square, 2+5=7, in second square it will be 7*2 = 14 (answer 1), in thrid, it is 18/6=3 and in fourth square, 9-7=2. 6. Seeds when soaked in water gain about 20% by weight and 10% by volume. By what factor does the density increase? (A) 1.20 (B) 1.10 (C) 1.11 (D) 1.09 Exp. Density=m/V d2= (12/11)*d1 because after soaking there is 20%increase in weight and 10% increase in volume P1/1

36 Amar Ujala Education 7. Retarding frictional force, on a moving ball, is proportional to its velocity,. Two identical balls roll down identical slopes (A & B) from different heights. Compare the retarding forces and the velocities of the balls at the bases of the slopes. A B (A) f A > f B ; V A > V B (B) f A > f B ; V B > V A (C) f B > f A ; V B > V A (D) f B > f A ; V A > V B Exp. When sphere rolls down, potential energy is converted into kinetic energy, mgh = 1/2mv 2, here velocity will be proportional to height of ball, hence Va>Vb. Also retardational force f=kv 2, where f is proportional to v, hence, Fa>Fb - option 1 8. A river is 4.1 km wide. A bridge built across it has 1/7 of its length on one bank and 1/8 of its length on the other bank. What is the total length of the bridge? (A) 5.1km (B) 4.9km (C) 5.6km (D) 5.4km 9. OA, OB, and OC are radii of the quarter circle shown in the figure. AB is also equal to the radius. A A/UG A hall with a high roof is supported by an array of identical columns such that, to a person lying on the floor and looking at the ceiling, the columns appear parallel to each other. Which of the following designs conforms to this? (A) (B) (C) (D) ROOF FLOOR ROOF FLOOR ROOF FLOOR ROOF FLOOR 12. Which of the following graphs correctly shows the speed and the corresponding distance covered by an object moving along a straight line? O B? C What is angle OCB? (A) 60º (B) 75º (C) 55º (D) 65º Exp. Given Triangle OAB is an equilateral triangle. So Angle AOB = 60 degree.since AOC= 90 degrees, then angle BOC = 30 degree. Given OB and OC are radii that means it has same length. So triangle OBC is an Isosceles triangle. Therefore angle OCB= = 150/2 = Intravenous (IV) fluid has to be administered to a child of 12 kg with dehydration, at a dose of 20 mg of fluid per kg of body weight, in 1 hour. What should be the drip rate (in drops/min) of IV fluid? (1mg = 20 drops) (A) 7 (B) 80 (C) 120 (D) 4 Exp. For 1Kg= 20mg, For 12Kg= 240 mg. 240 mg= 4800 drops/ hour, 4800/60 = 80 drops/min P1/2 (A) (B) (C) speed, distance speed, distance speed, distance 0,0 0,0 0,0 speed time distance time time

37 A/UG-15 (D) speed, distance 0,0 time Exp. As the distance covered with time increases, this will increase the slope of the speed -time graph as well but once distance graph becomes parallel to time axis speed time graph will show a negative slope. 13. A normal TV screen has a width to height ratio of 4:3, while a high definition TV screen has a ratio of 16:9. What is the approximate ratio of their diagonals, if the heights of the two types of screens are the same? (A) 5:9 (B) 5:18 (C) 5:15 (D) 5:6 Exp. TV screen is like rectangle and diagonal of any rectangle can be calculated by the formula Diagonal = ew 2 +h 2 Diagonal of 1st TV screen=5 diagonal of other screen is Comparing numerical values, which of the following is different from the rest? (A) The ratio of the circumference of a circle to its diameter. (B) The sum of the three angles of a plane triangle expressed in radians. (C) 22/7. (D) The net volume of a hemisphere of unit radius, and a cone of unit radius and unit height. Exp. 1. circumference/diameter of a circle= 2 r/2r= 2. sum of three angles of any triangle is invariably equal to the straight angle, also expressed as 180, radians 3. 22/7= 4. volume of hemisphere/vol of cone= 2/3 r^3/ r^3=2/3 not equal to pie. 15. If a person travels x% faster than normal, he reaches y minutes earlier than normal. What is his normal time of travel? 100 (A) 1 y minutes x x (B) 1 y minutes 100 y (C) 1 x minutes 100 (D) x minutes y P1/3 Amar Ujala Education Exp. Normally considering, distance = speed* time Modified case, (speed + x*speed/100) = distance/(time-y) or, speed (1+x/100) = distance/(time-y) Substituting for distance, speed (1+x/100) = speed*time/ (time-y) on cancelling speed on RHS and LHS, 1+x/100 = time/time-y time = time-y +x(time-y)/ time = 100time -100y +x time -xy x time = 100y + xy time = (100/x +1)ymins This is the original time needed 16. A and B walk up an escalator one step at a time, while the escalator itself moves up at a constant speed. A walks twice as fast as B. A reaches the top in 40 steps and B in 30 steps. How many steps of the escalator can be seen when it is not moving? (A) 30 (B) 40 (C) 50 (D) 60 Exp. With speed of constant escalotor the ratio of speed of A and B = 2:1. A reaches top in 40 steps. Therefore the number of Steps B walked = 40 x ½ = 20. Thus B has to take = 20 steps to reach to the top. But B takes = = 10 steps only. Thus remaining is the speed of the escalator I.e = 10 steps. Speed of B: Speed of escalator = 10:10 = 1:1 Therefore Speed of A : Speed of B : Speed of escalator 2:1:1 So speed of B = speed of escalator Which means when escalator is not moving then B has to climb double the steps it moves when escalator is moving 30 2=60steps 17. Twoironspheresofradii12cmand1cmaremelted and fused. Two new spheres are made without any loss of iron. Their possible radii could be (A) 9 and 4 cm (B) 9 and 10 cm (C) 8 and 5 cm (D) 2 and 11 cm Exp. Total volume of two sphere = 4/3 (R^3+r^3), where R and r are the radii respectively. First case - volume = 4/3 (12^3+1^3) = 4/3 (1728+1) = 4/ 3 (1729), to avoid loss of iron, total volume must be equal to sum of the volume of the two new spheres Here as 4/3 is constant, R^3+r^3 must be = 1729 (radii of new spheres) out of given option, 9^3+10^3 = = 1729, hence spheres must be having 9 and 10 cm radii each.

38 Amar Ujala Education 18. A man buys alcohol at Rs. 75/cL, adds water, and sells it at Rs.75/cL making a profit of 50%. What is the ratio of alcohol to water? (A) 2:1 (B) 1:2 (C) 3:2 (D) 2:3 Exp. CP = 75/cL rs Profit = 50 % ; SP = = Quantity sold = 112.5/75 = 1.5; Therefore ratio of alcohol: water = 2:1 19. The sum of digits of a two-digit number is 9. If the fraction formed by taking 9 less than the number as numerator and 9 more than the number as denominator is 3/4, what is the number? (A) 36 (B) 63 (C) 45 (D) 54 Exp. Addition of all the given four option will give 9. Like 3+6=9, 6+3=9, 4+5=9 and 5+4=9. But according to A/UG-15 question, we need to select one option and making a fraction of 3/4 by that option in such a way that taking 9 less than the number as numerator and 9 more than the number as denominator. So option two, 63 is correct, because for numerator 63-9= 54 and for denominator 63+9 = 72, so the fraction formed by this will be 54/72 =3/ The distance between X and Y is 1000 km. A person flies from X at 8 AM local time and reaches Y at 10 AM local time. He flies back after a halt of 4 hours at Y and reaches X at 4 PM local time on the same day. What is his average speed for the duration he is in the air? (A) 500 km/hour (B) 250 km/hour (C) 750 km/hour (D) cannot be calculated with the given information Exp. Given Distance between X and Y = 1000 km.letatravels from X at 8 am and reaches Y at 10 am = 2 hrs After halting 4 hrs, he started from Y at 2 pm and reaches Xat4pm=2hrs Average speed = 500 kms PART-B 21. RNA editing, a post-transcriptional process, is achieved with the help of guide RNA (g-rna). Which one of the following statements about the process is NOT true? (A) g-rna dependent RNA editing happens in the kinetoplast DNA (B) g-rna is involved in chemical modification of t- RNA (C) This process involves insertion or deletion of uridines (D) Sequences edited once may be re-edited using a second g-rna Exp. The guide RNA, containing correct nucleotides as complementary sequences, acts as a catalyst and restore transcripts to be functional molecules.the guide RNA and protein complexes, which perform editing, are called Editosomes. The composition of Editosomes, a 20s complex, are an endonuclease, an RNA ligase and one Uridyl transferase. In all cases the editing of pre-ed RNA starts from its 3 end and progressively moves towards its 5 end. For this the guide RNA associates with mrna using anti parallel base pairing mechanism.g-rna dependent RNA editing happens in the kinetoplst DNA,some trna of mitochonria. 22. Telomerase, a RNA-protein complex which completes the replication of telomeres during DNA synthesis, is a specialised (A) RNA dependent DNA polymerase (B) DNA dependent DNA polymerase (C) DNA dependent RNA polymerase (D) RNA dependent RNA polymerase Exp. The synthesis of telomeres is mainly achieved by the cellular reverse transcriptase telomerase, an RNAdependent DNA polymerase that adds telomeric DNA to telomeres. 23. Consider a short double-stranded linear DNA molecule of 10 complete turns with 10.5 bp/turn. The ends of the DNA molecule are sealed together to make a relaxed circle. This relaxed circle will have a linking number of (A) 105 (B) 20.5 (C) 10.0 (D) 10.5 Exp. Linking number= Number of base pairs/ bp per turns =(10.5 * 10)/10.5=10 P1/4

39 A/UG In the above signalling cascade, which one of the following molecules is denoted by B? (A) STAT 5 (B) SMAD 6 (C) GSK3 (D) SMAD 4 Exp. GSK3b is a cytoplasmic protein, STAT5 and SMAD 6 themselves get phosphorylated. They can be A. But B denots the co-smad which is required for transport of phosphorylated SMAD-6 into the nucleus. So the B is SMAD The secondary antibodies routinely used for the detection of primary antibodies in western blotting experiment are (A) anti-allotypic (B) anti-idiotypic (C) anti-isotypic (D) anti-paratypic Exp. A secondary antibody is an antibody directed against an Immunoglobulin (antibody) molecule. Under optimal conditions, a secondary antibody specifically binds the species and class (isotype) of the primary antibody in an immunoassay (indirect detection). In some cases, it is possible to make an antibody that is directed against the antigen binding site of another antibody (i.e., the antigen binding site is the epitope). This type of antibody is called an anti-idiotype antibody.* In this case, the antigen binding site of the anti-idiotype antibody can be similar in structure to the original antigen (they both recognise the same antibody). Amar Ujala Education 26. Which of the following events will NOT usually lead to transformation of a normal cell into a cancer cell? (A) Gain of function of oncogenes (B) Loss of function of tumor suppressors (C) Gain of function of genes involved in nucleotide excision repair (D) Loss of function of pro-apoptosis related genes Exp. DNA repair enzymes are like tumor suppressor genes in which loss of function mutation result in cancer. 27. Which one of the following is a food borne toxin? (A) Tetanus toxin (B) Botulinum toxin (C) Cholera toxin (D) Diptheria toxin Exp. Clostridium botulinum can exist as a heat-resistant spore, and can grow and produce a neurotoxin in under processed, home-canned foods. 28. Excess oxygen consumed after a vigorous exercise is (A) to pump out lactic acid from muscle. (B) to increase the concentration of lactic acid in muscle. (C) to reduce dissolved carbon dioxide in blood. (D) to make ATP for gluconeogenesis. Exp. Option 4- After exercise has stopped, extra oxygen is required to metabolize lactic acid; to replenish ATP, phosphocreatine, and glycogen; and to pay back any oxygen that has been borrowed from hemoglobin, myoglobin (an iron-containing substance similar to hemoglobin that is found in muscle fibres), air in the lungs, and body fluid. 29. Which one of the following describes the primary function of flippases? (A) Help in increasing lipid-protein interaction in the outer leaflet of the bilayer (B) Move certain phospholipids from one leaflet of the membrane to another (C) Localize more negatively charged membrane proteins in the lipid bilayer (D) Cause uncoupling of v-snares and t-snares after fusion of incoming vesicle with target membrane Exp. The typically distinct phospholipid composition of the two leaflets of a membrane bilayer is generated and maintained by bi-directional transport (flip-flop) of lipids between the leaflets. Specific membrane proteins, termed lipid flippases, play an essential role in this transport process. Energy-independent flippases allow common phospholipids to equilibrate rapidly between the two monolayers and also play a role in the biosynthesis of a variety of glycoconjugates such as glycosphingolipids, P1/5

40 Amar Ujala Education N-glycoproteins, and glycosylphosphatidylinositol (GPI)-anchored proteins. ATP-dependent flippases, including members of a conserved subfamily of P-type ATPases and ATP-binding cassette transporters, mediate the net transfer of specific phospholipids to one leaflet of a membrane and are involved in the creation and maintenance of transbilayer lipid asymmetry of membranes such as the plasma membrane of eukaryotes. Energy-dependent flippases also play a role in the biosynthesis of glycoconjugates such as bacterial lipopolysaccharide. 30. Mitotic cyclin-cdk activity peaks in M phase. This is because (A) Mitotic cyclin is synthesised only in M phase. (B) Threshold level of mitotic cyclin accumulates only in late G2. (C) Cyclin subunit is activated by phosphorylation only in M phase. (D) The kinase subunit is activated by dephosphorylation only in M phase. Exp. Mitotic cyclin-cdk complexes are synthesized during the S phase and G2, but their activities are held in check by phosphorylation at inhibitory sites until DNA synthesis is completed. Once activated by dephosphorylation of the inhibitory sites, mitotic cyclin-cdk complexes phosphorylate multiple proteins that promote chromosome condensation, retraction of the nuclear envelope, assembly of the mitotic spindle apparatus, and alignment of condensed chromosomes at the metaphase plate. 31. Choose the most appropriate ph at which the net charge is zero for the molecule from the data shown below: COOH H + 3 N CH pka CH2CH 2COOH COO HN CH 2 pka CH CH COO 2 2 COO H + 3 N CH CH2CH 2COOH pka COO H + 3 N CH CH CH COO 2 2 (A) 2.02 (B) 2.91 (C) 5.98 (D) 6.87 Exp. From the diagram, form 2 has net charge zero, so, pka = (pk1 + pk2)/2 = 2.91 at which most of the molecules will be neutral P1/6 A/UG Choose the correct statement about peptides in the Ramachandran plot. (A) Peptides that are unstructured will have all the backbone dihedral angles in the disallowed regions. (B) It is not possible to conclude whether a peptide adopts entirely helix or entirely beta sheet conformation. (C) The occurrence of beta turn conformation in a peptide can be deduced. (D) The sequence of a peptide can be deduced. Exp. The ß-turn was originally identified, in model building studies. He proposed three distinct conformations based on phi,psi values (designated I,II and III) along with their related turns (mirror images) which have the phi, psi signs reversed (I,II and III ), each of which could form a hydrogen bond between the main chain C=O(i) and the N-H(i+3). 33. Equilibrium constant Keq of a reaction is a ratio of product to substrate concentrations. The relation between Keq andfreeenergychangeinareaction G is as follows G= RTln K eq Reaction A and Reaction B have Keq values of 10 and 100, respectively. Which of the following statements is correct with respect to G? (A) G of A = G of B (B) G of A > G of B (C) G of B > G of A (D) G of A G of B Exp. If G is negative, then K>1, which means that the reaction will be spontaneous in the forward direction when all species are present in standard concentrations. If G>0 then the reverse reaction is spontaneous from standard conditions. Here Putting value of Keq ofawe get delta G = -2,303RT, while putting value of Keq of B we get deta G = - 2x2.303RT. Hence delta G of Ais higher than B 34. The gel to liquid crystalline phase transition temperature in phosphatidyl choline (PC) lipids composed of dioleoyl (DO), dipalmitoyl (DP), disteroyl (DS) and palmitoyl oleoyl (PO) fatty acids in increasing order will be (A) DOPC > DPPC > POPC > DSPC (B) DSPC > DPPC > POPC > DOPC (C) DPPC > DSPC > DOPC > POPC (D) POPC > DPPC > DOPC > DSPC

41 A/UG-15 Exp. Transition temperatures DPPC - ~ C and DSPC - ~ C. POPC bilayers undergo a gel-to-liquid crystalline phase transition around 2 C. (DOPC) has a transition temperature of 16.5 ºC. so increasing order is DSPC>DPPC>POPC>DOPC 35. Which of the following is NOT an example of transmembrane transport between different subcellular compartments? (A) Transport from cytoplasm into the lumen of the endoplasmic reticulum (B) Transport from endoplasmic reticulum to the Golgi complex (C) Transport from stroma into thylakoid space (D) Transport from mitochondrial intermem-brane space into the mitochondrial matrix Exp. Tansport from ER to Golgi complex is a type of vesicular transport, not transmembrane transport. Rest all are examples of transmembrane transport. 36. Which of the following are NOT transcribedbyrna polymerase II? (A) mirna and some snrna (B) mirna and snorna (C) mrna and snorna (D) trna and 5S rrna Exp. trna and 5S rrna are synthesized by RNA polymerase III 37. Apical ectodermal ridge induction is essential for tetrapod limb development. Which one of the following is NOT essential for the formation of a functional limb? (A) Tbx genes and Wnt (B) Androsterone (C) Apoptotic genes (D) Fibroblast growth factor Exp. Tetrapode limb development is a complex process of oragnogenesis which initiate with the formation of a concentration gradient of Retinoic acid which lead to the expression of different set of hox genes from anterior to posterior somites.tbx genes are essential for differentiation of the hind limbs and fore limbs and FGF plays a very important role in the development of AER during limb development. 38. Which one of the following statements is WRONG? (A) The megasporocyte develops within the megasporangium of the ovule (B) Megasporocyte undergoes meiosis to produce four haploid megaspores (C) All the four megaspores undergo several mitotic divisions to form female gametophyte in most angiosperms (D) Female gametophyte is haploid P1/7 Amar Ujala Education Exp. Diploid mother cell divides by meiosis and cytoplasmic division to produce four haploid megaspores; only one of these will survive, the other three disintegrateremaining megaspore undergoes mitosis three times without cytoplasmic division! to produce a single cell with eight nuclei.all megaspores do not form female gametophyte 39. Certain proteins or mrnas that are regionally localized within the unfertilized egg and regulate development are called (A) gene regulators. (B) morphometric determinants. (C) cytoplasmic determinants. (D) mosaic forming factors. Exp. In drosophila,unfertilized egg contains certain maternal mrnas which are located in different regions of egg and play an important role in the formation of anteriorposterior and dorsal-ventral symmatry in embyro.these maternal mrnas are called cytoplasmic determinants. 40. Cell to cell communication is important in development of an organism. The ability of cells to respond to a specific inductive signal is called (A) Regional specificity of induction (B) Competence (C) Juxtracrine signalling (D) Instructive interaction Exp. Competence ability of cells to respond to inductive signal 41. Which one of the following compounds is NOT apart of alkaloid class of secondary metabolites? (A) Lignin (B) Indole (C) Tropane (D) Pyrrolidine Exp. Indole, Tropane and pyrrolidine..all are nitrogen contaiuniung compounds used as a precursors for alkaloid whereas lignin is phenolic compound and main component of wood.lignified walls are much less palatable to insects and other animals than are nonwoody plants and are much less easily digested by fungal enzymes than plant parts that contain only cells with primary cellulose walls. 42. Which one of the following best describes the symplast pathway of water flow from the epidermis to endodermis in a plant root? (A) Water moves through cell walls and extracellular spaces without crossing any membrane (B) Water travels across the root cortex via the plasmodesmata (C) Water crosses the plasma membrane of each cell in its path twice, once on entering and once on exiting (D) Transport across the tonoplast

42 Amar Ujala Education Exp. Water moves through the root layers from the epidermis upto xylem via either the symplastic or apoplatic pathway. In the symplastic pathway, water moves through the cytoplasm of each cell via plamodesmata of the cortical layer. 43. The herbicide, dichlorophenyldimethylurea, is an inhibitor of (A) shikimate pathway for biosynthesis of aromatic amino acids. (B) electron transport from P680 to P700. (C) branched chain amino acid pathway. (D) electron transport from P700 to ferredoxin. Exp. herbicides, DCMU among them, act by blocking electron flow at the quinone acceptors of photosystem II, by competing for the binding site of plastoquinone that is normally occupied by QB 44. Vasopressin secretion does NOT increase with (A) exercise (B) an increase in extracellular fluid volume (C) standing (D) vomiting Exp. Vasopressin or ADH helps in water conservation in the body and thus is secreted under conditions like sweating, diarrohea, vomitting etc. Vassopressin secretion is also affected by body posture with a progressive increase in secretion while standing. Moreover, a decrease in extracellular volume and not an increase will cause ADH secretion. 45. Which one of the following does NOT occur due to stimulation of baroreceptors? (A) Bradycardia (B) Hypotension (C) Venodilation (D) Vasoconstriction Exp. Baroreceptors or pressoreceptors, which are located in the wall of each internal carotid artery at the carotid sinus, and in the wall of the aortic arch. After the baroreceptor signals have entered the medulla, secondary signals eventually inhibit the vasoconstrictor centre of the medulla and excite the vagal centre. The net effects are vasodilatation of the veins and arterioles throughout the peripheral circulatory system and decreased heart rate and strength of heart contraction. 46. Serum has essentially the same composition as plasma EXCEPT that it lacks (A) Albumin (B) Stuart-Prower factor (C) Antihemophilic factor (D) Hageman factor Exp. Experiments were performed on common clotting by mixing serum or plasma from stuart and prower showing it did not correct the abnormility mutually and thus it proved that 2 commonly lack this factor. Stuart & Prower is also called clotting factor X. P1/8 A/UG Which type of cells located in gastric glands is responsible for the release of histamine? (A) Mucous neck cells (B) Enterochromaffin-like cells (C) Chief cells (D) Parietal cells Exp. Histamine in the stomach occurs in endocrine cells (socalled enterochromaffin-like (ECL) cells. Local release of histamine is thought to be necessary for the stimulation of parietal cells. they are located basally in the oxyntic gland area, in the chief-cell-rich region 48. Which one of the following plant derived signalling molecule induces hyphal branching of arbuscular mycorrhizal fungi, a pheno-menon that is observed at the initial stages of colonization by these fungi? (A) Salicylic acid (B) Abscisic acid (C) Strigolactones (D) Systemin Exp. Strigolactones are signaling compounds made by plants. They have two main functions: first, as endogenous hormones to control plant development, and second as components of root exudates to promote symbiotic interactions between plants and soil microbes. however Abscisic acid, systemin and salicylic acid are predominantly involved in plant -pathogen inteaction. 49. If non-disjunction occurs in meiosis I, which of the following scenario is most likely to occur? (A) Two gametes will be n+1 and two will be n-1 (B) One gamete will be n+1, two will be n and one will be n-1 (C) Two gametes will be normal and two will be n-1 (D) Two gametes will be normal and two will be n+1 Exp. Nondisjunction means that a chromosome pair failed to separate during the meiotic division. This will create one daughter cell with an extra chromosome and another daughter cell with one too few chromosomes. If the nondisjunction occurs during the first meiotic division (meiosis I), all the gametes derived will be abnormal. Thus half the gametes will have an extra copy of chromosome and other half of the gametes derived from this primary gametocyte will lack both copies of chromosome. 50. Which of the following is true for cells harbouring F plasmid? (A) Their F plasmid is non-functional. (B) They exhibit increased rates of transfer of all chromosomal genes. (C) They are merodiploids. (D) They fail to survive as the chromosomal origin of replication is inactivated.

43 A/UG-15 Exp. A merodiploid is an essentially haploid organism that carries a second copy of a part of its genome. The term was originally used to describe both unstable partial diploidy, such as that which occurs briefly in recipients after mating with an Hfr strain, and the stable state, exemplified by F-prime strains. 51. Maternal inheritance of coiling of shell in snail (Limnaea peregra) is well established. The dextral coiling depends on dominant allele D and sinistral coiling depends upon recessive allele d. AfemaleF1 progeny of dextral (Dd) type is crossed with a male sinistral snail. What will be the ratio of heterozygous: homozygous individuals in its F2 progeny? (A) 3:1 (B) 1:1 (C) 1:3 (D) 1:2:1 Exp. The inheritance pattern for certain nuclear genes in which the genotype of the mother directly determines the phenotype of her offspring is termed as Maternal effect. genotypes of the father and offspring do not affect the phenotype of the offspring. 1 DD: 2 Dd: 1 dd -all are dextral the ratio between heterozygous and homozygous will be 1: Which of the following mutagens is most likely to result in a single amino acid change in a gene product? (A) Acridine orange (B) X-rays (C) Ethylmethane sulphonate (EMS) (D) Ethidium bromide Exp. Induction of point mutations by ethylmethane sulfonate (EMS), a commonly used mutagen. (a) EMS alkylates guanine at the oxygen on position 6 of the purine ring, forming O6- ethylguanine (Et-G), which base-pairs with thymine. that results in a protein in which one amino acid is substituted for another. other mutragens given like X-rays causes thymine dimerization whereas acridine orange and ethidium bromide are inetrcalators that get bindorporated into the replicating strands. thus EMS can substitution altered base resulting tin new amino acid. 53. Which one of the following statements supports the concept of trade-off in the evolution of life history traits? (A) Level of parental care and clutch size are positively correlated (B) Animals maturing early tend to live longer (C) An increase in seed size is usually associated with a decrease in seed number (D) Allocation of higher energy for reproduce- tion leads to higher population growth P1/9 Amar Ujala Education Exp. A trade-off in evolution means one trait cannot increase without a decrease in another (or vice versa). Such a situation can be caused by a number of physical and biological mechanisms. Clutch size is large for animals that give less parental care thus ensure maximum survival by high number of young ones at single reproduction cycle. The size and number of eggs that, that an organism produce is common example of trade off. Depending on the organism, this trade-off can be caused by a limitation in the amount of energy available, the amount of time available to produce eggs or the amount of space available to hold eggs. 54. A plot of dn/dt as a function of population density yields a (A) rectangular hyperbola. (B) negative exponential curve. (C) positive rectilinear curve. (D) bell-shaped curve. Exp. Plot of dn/dt or change in the population size with time, first increases with increase in the population size, reaches maximum value at half of the carrying capacity after that decreases as the population size approaches carrying capacity. 55. For a species having logistic growth, if K = 20,000 and r = 0.15, the maximum sustainable yield will be (A) 450 (B) 1500 (C) 3000 (D) Which of the following is a correct ranking of ecosystems based on the root: shoot ratio of plants? (A) Tropical wet forest > Tropical dry forest > Temperate grassland > Tropical grassland (B) Temperate grassland > Tropical grassland > Tropical wet forest > Tropical dry forest (C) Tropical dry forest > Tropical wet forest > Tropical grassland > Temperate grassland (D) Temperate grassland > Tropical grassland > Tropical dry forest > Tropical wet forest Exp. The root: shoot ratio describes the proportion of plant biomass allocated to roots. Roots are sometimes a more than 50% contributor of net primary productivity (grassland). root:shoot ratio will be hiest incase of temperate grassland ( ) follwed by tropical grassland (1.89) follwed by temperate dry forest (boreal ) as 0.39 and least will be with tropical wet forest(amazon) Which of the following periods is known as Age of Fishes? (A) Devonian (B) Jurassic (C) Cambrian (D) Carboniferous

44 Amar Ujala Education Exp. Age of Fishes. In Devonian time, from about 415 to 355 million years ago, fishes of many different types swam and hunted in the seas. Lobe-finned fishes ancestors to the amphibians and the early sharks made their appearance by this time. 58. Which of the following is NOT an assumption of the Hardy-Weinberg model? (A) Population mates at random with respect to the locus in question (B) Selection is not acting on the locus in question (C) One allele is dominant and the other is recessive at this locus (D) The population is effectively infinite in size Exp. Hardy-Weinberg equation is a mathematical equation that can be used to calculate the genetic variation of a population at equilibrium. it states If an infinitely large, random mating population is free from outside evolutionaryforces (i.e. mutation, migration and natural selection), then the gene frequencies will not change over time and the frequencies in the next generation will be p2 for the AA genotype, 2pq for the Aa genotype and q2 for the aa genotype. 59. Which of the following geological periods is characterized by the first appearance of mammals? (A) Tertiary (B) Cretaceous (C) Permian (D) Triassic Exp. By the middle of the Eocene (45 MYA) all the major groups of mammals alive today had come into existence, though not necessarily as we know them now. Primates for instance have been around since the beginning of the Paleocene, 65 MYA, but the distant bipedal ancestors of man only occurred for the first time 5 MYA. The same applies to all the other major groups. The Jurassic and Cretaceous (150 million years) was the age of the Dinosaurs, a long reign as a dominant animal group. The Tertiary, however, has been and still is the age of mammals. During the 2nd half of the Eocene (12 million years) the Oligocene (23 million years) and the Miocene (18 million years) the mammals have been dominant. Though they are still the dominant group of animals on the planet it is worth noting that over the last 10 million years, 6 of the 24 major mammal groups to come out of the Eocene have died out. This is 25% of the existing mammals then and is important because for the previous 20 million years before that no major groups died out at all. 60. An alga having chlorophyll a, floridean starch as storage product and lacking flagellate cells belongs to the class (A) Phaeophyceae. (B) Chlorophyceae. (C) Rhodophyceae. (D) Xanthophyceae Exp. Rhodophyceae will have all the properties mentioned P1/10 A/UG Which of the following is NOT true for monocots? (A) Sieve tube members with companion cells (B) Vasculature atactostelic (C) Tricolpate pollen (D) Vascular cambium absent Exp. Monocots have monosucate pollens and tricolpate pollens are a characteristic of eudicots, a type of dicots 62. Individuals occupying a particular habitat and adapted to it phenotypically but not genotypically are known as (A) Ecophenes. (B) Ecotypes. (C) Ecospecies. (D) Coenospecies. Exp. Ecotype is a distinct type of race of plant or animal forms that are particularly adapted phenotypically and to the environmental conditions. They carry their distinct genetic differences 63. Different leads are used to record ECG of humans. Which one of the following is NOT unipolar leads? (A) Augmented limb leads (B) V1 and V2 leads (C) Standard limb leads (D) VR and VL leads Exp. Standard limb lead is not unipolar lead, rest all are unipolar 64. The presence and distribution of specific mrnas within a cell can be detected by (A) Northern blot analysis (B) RNase protection assay (C) in situ hybridization (D) real-time PCR Exp. In Situ Hybridization (ISH) is a technique that allows for precise localization of a specific segment of nucleic acid within a histologic section using a complementary DNA/ RNA molecule. RNase protection assay is a technique used for transcription start-site localisation. Real-time PCR and Northern blotting techniques are used to detect the presence of mrna molecules within a tissue, however do not provide exact cellular localisation 65. In which of the following mating systems there is likely to be NO conflict of interest over reproductive success between the sexes? (A) Polyandry (B) Monogamy (C) Promiscuity (D) Polygamy Exp. Promiscuity is the practice of having casual sex frequently with different partners or being indiscriminate in the choice of sexual partners. Polyandry and polygamy is a marital arrangement in which a woman has several husbands and man has many females respectively. In contrast monogamy supports one man-one women relation.

45 A/UG Which one of the following analytical techni-ques does NOT involve an optical measurement? (A) ELISA (B) Microarray (C) Flow cytometry (D) Differential Scanning Calorimetry Exp. Of all the options given, only Differential scanning calorimetry does not involve optical measurement. Differential scanning calorimetry or DSC is a thermoanalytical technique in which the difference in the amount of heat required to increase the temperature of a sample and reference is measured as a function of temperature. On the other hand, ELISA is a technique used to detect and quantify peptides, proteins, antibodies etc. in samples and measurement is usually based on changes in optical density of the solution. Flow cytometry is a technology that is used to analyse the physical and chemical characteristics of particles in a fluid as it passes through at least one laser. Cell components are fluorescently labelled and then excited by the laser to emit light at varying wavelengths and fluorescence is measured. Microarray is a technique used to study gene expression in cells/tissues using fluorescent-tagged complementary probes. 67. Which genes have been introduced in Bollgard II cotton to get resistance against cotton bollworm, tobacco budworm and pink bollworm? (A) cry1ab + cry1ac (B) cry1ac + cry2ab (C) cry1ab + cry2ab (D) cry9c + cry2ab Exp. Bollgard II contains two Bacillus thuringiensis genes Cry1Ac and Cry2Ab. Bollgard cotton varieties were introduced in 1996 and contain only one gene (Cry 1Ac) for fruit feeding, leaf feeding caterpillar control. Research trials evaluating the Bollgard transgenic Bt gene technology have determined these varieties to be highly effective against tobacco budworms. Bollgard II contains two Amar Ujala Education genes (Cry 1Ac and Cry 2Ab) and is are more effective against tobacco budworm and bollworm compared to Bollgard. 68. An optical measurement of a protein is taken both before and after digestion of the protein by a protease. In which of the following spectroscopic measurements the signal change, i.e., before vs after protease treatment, could be the maximum? (A) Absorbance at 280 nm(b) Circular dichroism (C) Absorbance at 340 nm(d) Fluorescence value Exp. Proteins in solution absorb ultraviolet light with absorbance maxima at 280 and 200 nm.amino acids with aromatic rings are the primary reason for the absorbance peak at 280 nm. Peptide bonds are primarily responsible for the peak at 200 nm. Secondary, tertiary, and quaternary structure all affect absorbance, therefore factors such as ph, ionic strength, etc. can alter the absorbance spectrum. 69. The tetanus vaccine given to humans in the case of a deep cut is a (A) DNA vaccine (B) recombinant vector vaccine (C) subunit vaccine (D) toxoid vaccine Exp. Tetanus toxin is attenuated bacterial toxin thus called the toxoid vaccine. this vaccine works by exposing to a small dose of the bacteria (or a protein from the bacteria), which causes the body to develop immunity to the disease. This vaccine will not treat an active infection that has already developed in the body. 70. The electrospray ionization spectrum of a mixture of two peptides show peaks with m/z values 301, 401, 501, and 601. The molecular weights of the peptides are (A) 1200 and 1250 (B) 1200 and 1500 (C) 1350 and 1500 (D) 1250 and From the following statements, A. For a reaction to occur spontaneously the free energy change must be negative B. The interaction between two nitrogen molecules in the gaseous state is predominantly electrostatic C. By knowing bond energies, it is possible to deduce whether the bond is covalent bond or hydrogen bond D. Hydrophobic interactions are not important in a folded globular protein pick the combination with ALL WRONG statements. (A) AandB (B) Band C (C) CandD (D) BandD PART-C P1/11 Exp. Hydrogen bonds are much weaker than covalent bonds. They have energies of 1 3 kcal mol-1 (4 13 kj mol-1) compared with approximately 100 kcal mol-1 (418 kj mol- 1) for a carbon-hydrogen covalent bond. The hydrophobic effect is considered to be the major driving force for the folding of globular proteins 72. A researcher investigated a set of conditions for a protein with an isoelectric point of 6.5 and also binds to calcium. This protein was subjected to four independent treatments: (i) ph 6.4, (ii) 10% glycerol, (iii) 10 mm CaCl2, (iv) 40% ammonium sulphate. This was

46 Amar Ujala Education followed by centrifugation and estimation of the protein in the supernatant. The results are depicted in the graph below: %protein in supernatant a b c d Which of the following treatments best represents the results shown in the graph? (A) a = ammonium sulphate, b = glycerol, c = ph 6.4, d = CaCl2 (B) a = CaCl2, b = glycerol, c = ammonium sulphate, d = ph 6.4 (C)a=pH6.4,b=CaCl2, c = ammonium sulphate, d = glycerol (D) a = CaCl2, b = ph 6.4, c = glycerol, d = ammonium sulphate Exp. At ph=pi the net charge on protein will be zero thus all the protein will be in precipitate and the % protein in supernatent will be less. Similarily Ammonium sulphate will precipitate the protein by salting out method. But in glycerol and CaCl 2 (minute concentration) protein will remain in supernatent only. 73. In the biosynthesis of purine: 6 C 7 N 1N C5 C8 2C C4 N N 3 9 (A) All N atoms, C4 and C5 are from aspartic acid (B) Nl is from Aspartic acid; N3 and N9 are from Glutamine side-chain; N7, C4 and C5 are from Glycine (C) Nl is from Aspartic acid; N3 from Glutamine sidechain; N9 from N attached to C of Glutamine; N7, C4andC5arefromGlycine (D) Nl is from Glutamine; N3 from Glutamine sidechain; N9 from N attached to C of Glutamine; N7, C4andC5arefromGlycine 74. From the following statements, A. Hydrogen, Deuterium and Tritium differ in the number of protons B. Hydrogen, Deuterium and Tritium differ in the number of neutrons C. Both Deuterium and Tritium are radioactive and decay to Hydrogen and Deuterium, respectively A/UG-15 D. Tritium is radioactive and decays to Helium E. Carbon-14 decays to Nitrogen-14 F. Carbon-14 decays to Carbon-13 pick the combination with ALL correct statements. (A) A,BandF (B) B,DandE (C) A,CandD (D) C,EandF Exp. Hydrogen Deuterium and Tritium are isotopes that are having same number of proton but differ in the number of neutrons, Tritium radioactive isotope contains one proton and two neutrons in its nucleus and decaying into helium-3 through âe decay. Carbon-14 goes through radioactive beta decay: 75. The following are four statements on peptide/ protein conformation: A. Glycine has the largest area of conforma- tionally allowed space in the Ramachandran plot of and B. A 20-residue peptide that is acetylated at the N- terminus and amidated at the C- terminus has = º(+5), = 30º(+5) for all the residues. It can be concluded that conform- ation of the peptide is helix-turn-strand C. The allowed values of, for amino acids in a protein are not valid for short peptides D. A peptide Acetyl-A 1 -A 2 -A 3 -A 4 -CONH 2 (A 1 -A 4 are amino acids) adopts a well defined -turn. The dihedral angles of A 2 and A 3 determine the type of -turn Choose the combination of correct statements. (A) AandB (B) Band C (C) AandD (D) CandD Exp. Statement A & D is correct as Glycine can take up all conformations as it is the most flexible amino acid available. Regarding statement D, The C-terminus made as an amide, and having the N-terminus blocked by acetylation blocks the end which can be beneficial because the peptide is then more likely to behave like, or be recognized, as if it were a part of the whole protein from which the sequence was chosen. A type II -turn is a four-residue turn characterized by the following three criteria (for properties of -turns : (1) torsion angles 2= 60, 2 = +120, 3 = +90, 3 = 0 ; (2) the residue at position 3 is almost always a glycine; and (3) a C-O H-N hydrogen bond is formed between the main-chain oxygen atom of the residue at position one and the main-chain nitrogen atom of the residue at position four. 76. A researcher was investigating the substrate specificity of two different enzymes, X and Y, on the same substrate. Both the enzymes were subjected to treatment with either heat or an inhibitor which inhibits the enzyme activity. Following are the results obtained where a=inhibitor treatment, b=heat treatment and c=control. P1/12

47 A/UG Activity Protein X c a b a b 0 0 [S] [S] Which of the following statements is correct? (A) Only protein X is specific for the substrate, S (B) Only protein Y is specific for the substrate, S (C) Both X and Y are specific for the substrate, S (D) Both X and Y are non-specific for the substrate, S Exp. Adding inhibitor or heat destroys specificity towards the substrate and affects the activity of both the enzymes 77. A is an inhibitor of chloroplast function. The production of O 2 and the synthesis of ATP are measured in illuminated chloroplasts before and after addition of A as shown below A added O Production 2 A added 10 Protein Y Time (min) Time (min) Which statement is correct? (A) A inhibits the reduction of NADP+ (B) A inhibits the proton gradient and the reduction of NADP+ (C) A inhibits the proton gradient but not the reduction of NADP+ (D) A inhibits neither the proton gradient nor the reduction of NADP+ Exp. According to the graphical represntation the inhibitor A is inhibiting the action of ATP synthesis and not the production of O2 and it is obvious if there is an inhibitory effect on ATP synthesis the proton gradient will also get inhibited as because there is presence of ATPase proton pumps that generates the protons or H+ and it does not inhibit the action of NADP During cell cycle progression from G 1 to S, cyclin D- CDK4 phosphorylates Rb and reduces its affinity for E2F. E2F dissociates from Rb and activates S-phase gene expression. Overexpression of protein A arrests G 1 phase progression. Which of the following statements is TRUE? (A) A inhibits Rb-E2F interaction ATP Synthesis c P1/13 Amar Ujala Education (B) A inhibits CDK4 activity (C) A phosphorylates E2F (D) A degrades Rb Exp. Inhibition of CDK4 activity by Protein A will lead to the arrest of cell cycle in G1 phase only as active CDK4 is required for the progression. Active CDK4 causes the phosphorylaton of Rb, make E2f free so that E2f can bind with the promoter of s pahse cyclins and lead their synthesis 79. Cells in S-phase of the cell cycle were fused to cells in the following stages of cell cycle: (a) G 1 phase, (b) G 2 phase, (c) M phase. These cells were then grown in medium containing tritiated thymidine. Maximal amount of freshly labelled DNA is likely to be obtained in S- phase cells fused with (A) G 1 phase cells (B) G 2 phase cells (C) M phase cells (D) Both G 1 and G 2 phase cells Exp. Fusion of S-phase cells with G1-phase cells led to the rapid induction of DNA synthesis in the G1 nuclei which lead to introduction of labelled thymidine in nucleus of cells. 80. Addition of the antibiotic cephalexin to growing E. coli cells lead to filamentation of the cells, followed by lysis. Cephalexin is an inhibitor of (A) protein synthesis (B) DNA synthesis (C) peptidoglycan synthesis (D) RNA polymerase Exp. Cephalexin, like the penicillins, is a beta-lactam antibiotic. By binding to specific penicillin-binding proteins (PBPs) located inside the bacterial cell wall, it inhibits the third and last stage of bacterial cell wall synthesis. Cell lysis is then mediated by bacterial cell wall autolytic enzymes such as autolysins; it is possible that cephalexin interferes with an autolysin inhibitor. 81. Fluorescently tagged protein was used to study protein secretion in yeast. Fluorescence was observed in: (a) the Golgi (b) the secretory vesicles (c) the rough ER. Which of the following describes best the sequence in which these events occur? (A) (a) (b) (c) (B) (b) (c) (a) (C) (c) (a) (b) (D) (c) (b) (a) Exp. Protein is synthesized on ribosomes of Rough ER then the vesicles containing the protein carry it from the rough endoplasmic reticulum and move to the nearby Golgi apparatus and then it moves through the entire Golgi apparatus, secretion vesicles containing the protein bud off.

48 Amar Ujala Education 82. In order to ensure that only fully processed mature mrnas are allowed to be exported to cytosol, premrnas associated with snrnps are retained in the nucleus. To demonstrate this, an experiment was performed where a gene coding a pre-mrna with a single intron was mutated either at the 5 or 3 splice sites or both the splice sites. Given below are a few possible outcomes: A. Pre-mRNA having mutation at both the splice sites will be retained in the nucleus because of the presence of bound snrnps. B. Pre-mRNA having mutation at both the splice sites will be exported to cytosol because of the absence of bound snrnps. C. Pre-mRNA mutated at either 3 or 5 splice sites will be retained in the nucleus because of the presence of bound snrnps. D. Pre-mRNA mutated at either 3 or 5 splice sites will be exported to cytosol because of the absence of bound snrnps. Choose the correct combination of the possible outcomes: (A) Band C (B) AandD (C) BandD (D) AandC Exp. a gene encoding a pre-mrna with a single intron that is efficiently spliced out was mutated to introduce deviations from the consensus splice-site sequences. Mutation of either the 52 or 32 invariant splice site at the ends of the intron resulted in pre-mrnas that were bound by snrnps to form spliceosomes; however, RNA splicing was blocked and the pre-mrna was retained in the nucleus. In contrast, mutation of both the 52 and 32 splice sites in the same pre-mrna resulted in efficient export of the unspliced pre-mrna. In this case, the premrnas were not efficiently bound by snrnps. 83. Telomerase, a protein-rna complex, has a special reverse transcriptase activity that completes replication of telomeres during DNA synthesis. Although it has many properties similar to DNA polymerase, some of them are also different. Which one of the following properties of telomerase is different from that of DNA polymerase? (A) Telomerase requires a template to direct the addition of nucleotides (B) Telomerase can only extend a 3 -OH end of DNA (C) Telomerase does not carry out lagging strand synthesis (D) Telomerase acts in a processive manner Exp. Telomerase like DNA polymerase has processivity, require 3 OH end and requires a template to direct addition of nucleotides. A/UG In eukaryotes, a specific cyclin dependent kinase (CDK) activity is required for the activation of loaded helicases to initiate replication. On the contrary, this CDK activity inhibits the loading of helicases onto the origin of replication. Considering the fact that during each cycle, there is only one opportunity for helicases to be loaded onto origins and only one opportunity for these loaded helicases to be activated, which one of the following graphs best depicts this CDK activity in G1 and S phases of the cell cycle? (A) (B) (C) CDK activity CDK activity G 1 S (D) CDK activity CDK activity G 1 S G 1 S Exp. As CDK activity inhibiting the loading of helicase it shoule be inactivated at G1 phase (So that helicase can be loaded to start S phase) and as it is required for activation of loaded Helicase, it will become more active to proceed S phase. 85. Polysome profiling of cells treated with three hypothetical translation inhibitors is shown in the plots below. These three inhibitors are (i) CHP leaky inhibitor of translation (ii) LTM arrests ribosome at the initiation codon (iii) PTM inhibits ribosome scanning (a) OD 260 (b) OD 260 (c) OD S 60S 80S Sucrose density fraction 40S 60S 80S Sucrose density fraction 40S 60S Sucrose density fraction G 1 Polysome S P1/14

49 A/UG-15 Match the polysome profile to the inhibitor (A) (i) a; (ii) b; (iii) c (B) (i) b; (ii) c; (iii) a (C) (i) c; (ii) b; (iii) a (D) (i) a; (ii) c; (iii) b 86. In mammals, CG rich sequences are usually methylated at C, which is a way for marking genes for silencing. Although the promoters of housekeeping genes are often associated with CpG islands yet they are expressed in mammals. Which one of the following best explains it? (A) Methylation of cytosine does not prevent the binding of RNA Pol II with the promoter, so housekeeping genes are expressed (B) During housekeeping gene expression, the enzyme methyltransferase is temporarily silenced by mirna, thus shutting down global methylation (C) Unlike within the coding region of a gene, CG rich sequences present in the promoters of active genes are usually not methylated (D) As soon as the cytosine is methylated in the promoter region, the enzymes of DNA repair pathways remove the methyl group, thereby ensuring gene expression Exp. Many human genes have regions abundant in CpG (called islands ) around their promoters, and in particular virtually all housekeeping genes. When genes are active, these CpG islands are unmethylated - hence in most tissues housekeeping genes have unmethylated CpG islands, something else is keeping them unmethylated. 87. In an experiment, red blood cells were subjected to lysis and any unbroken cells were removed by centrifugation at 600g. The supernatant was taken and centrifuged at 100,000g. The pellet was extracted with 5M NaCl and again centrifuged at 100,000g. Which of the following proteins would be present in the supernatant? (A) Band 3 (B) Glycophorin (C) G protein-coupled receptor (D) Spectrin Exp. All given proteins are integral proteins of RBCs except Spectrin which is a peripharal proteins. Transmembrane proteins will be removed with the debrices while spectrin will remain in the supernatent. 88. In order to study the intracellular trafficking of protein A, it was tagged with GFP (A-GFP). Fluorescence microscopy showed that A-GFP co-localizes with LAMP1. In the presence of bafilomycin A, an inhibitor of H+-ATPase, A-GFP does not co-localize with LAMP1. Amar Ujala Education Instead, it co-localizes with LC3 puncta. Which one of the following statements is TRUE? (A) A-GFP targets to the ER in the absence of bafilomycin A. (B) Autophagy is required for trafficking of A-GFP to lysosomes. (C) Bafilomycin A facilitates targeting of A- GFP to the ER. (D) Bafilomycin A facilitates targeting of A- GFP to the mitochondria Exp. Befilomycin A1 is a known inhibitor of the late phase of autophagy. Bafilomycin A1 prevents maturation of autophagic vacuoles by inhibiting fusion between autophagosomes and lysosomes. Bafilomycin A1 acts by inhibiting vacuolar H+ ATPase (V-ATPase). 89. In animals, four separate families of cell-cell adhesion proteins are listed in Column A and their functional characteristics are given in Column B. A B a. Integrin (i) Lectins that mediate a variety of transient, cell-cell adhesion interact-ions in the blood stream b. Cadherin (ii) Contains extracellular Iglike domains and are mainly involved in the fine tuning of cell-cell adhesive interactions during development and regeneration. c. Ig-super-family (iii) Mediates Ca 2+ -dependent strong homophilic cell-cell adhesion. d. Selectin (iv) Transmembrane cell adhesion proteins that act as extracellular matrix receptors Which one of the following is the correct combination? (A) a (i), b (ii), c (iii), d (iv) (B) a (ii), b (iii), c (iv), d (i) (C) a (iii), b (iv), c (i), d (ii) (D) a (iv), b (iii), c (ii), d (i) Exp. Selectins: Cell surface lectins which mediate the binding of leukocytes to endothelial cells, Integrins are proteins that function mechanically, by attaching the cell cytoskeleton to the extracellular matrix (ECM). 90. A student treated cancer cells with an anti-cancer drug and performed western blot analysis. Which one of the following blots is the best representation of untreated control (C) and treated (T) samples? P1/15

50 Amar Ujala Education Cleaved Parp Cyclin D1 p53 Sonic hedgehog Loading control C T (A) (B) (C) (D) Exp. Increased level of cleaved PARP, p53 as a result of anticancerous drug treatment may lead to the induction of apoptosis in any cancer cells. At the same time over expression of Cyclin D1 and Sonic hedgehog (Shh) signaling pathway is the sign of cancer tissue. 91. In Trypanosomes, a 35 base leader sequence is joined with several different transcripts making functional mrnas. The leader sequence is joined with the other RNAs by (A) a specific RNA ligase (B) the process of trans-splicing (C) a nucleophilic attack caused by a free guanine nucleotide (D) a nucleophilic attack caused by a 2 OH of an internal A present in the leader sequence Exp. Tryanosomes do show a different type of pre-mrna splicing called Transplicing. Trans-splicing involves cutting and joining of two different transcripts into a functional RNAs; it is like cut and paste together. 92. Following are the list of some of the pathogens (column A) and the unique mechanisms they employ for evading immune response (column B). A B a. Trypanosoma (i) Capable of employing unusual genetic processes by which they generate extensive variations in their variant surface glycoproteins (VSG) b. Plasmodium (ii) Capable of continua-lly falciparum undergoing maturatio-nal changes in transfor-mation to different forms which allow the organism to change its surface molecules c. Haemophilus (iii) Capable of evading influenzae immune response by frequent antigenic changes in its hemagglutinin and neuraminidase glyco-proteins C T C T C T A/UG-15 Which of the following is the correct match between the organisms and their respective mechanism to evade immune response? (A) a (i), b (ii), c (iii) (B) a (ii), b (iii), c (i) (C) a (iii), b (i), c (ii) (D) a (i), b (iii), c (ii) Exp. T. brucei - In order to protect itself from host defenses, the parasite decorates itself with a dense, homogeneous coat (~10^7 molecules) of glycoprotein known as the variant surface glycoprotein (VSG). P. falciparum - Plasmodium, this is accomplished via the dual purpose P. falciparum erythrocyte membrane protein 1 (PfEMP1). PfEMP1 is encoded by the diverse family of genes known as the var family of genes (approximately 60 genes in all). The diversity of the gene family is further increased via a number of different mechanisms including exchange of genetic information at telomeric loci, as well as meiotic recombination. The PfEMP1 protein serves to sequester infected erythrocytes from splienic destruction via adhesion to the endothelium. Moreover, the parasite is able to evade host defense mechanisms by changing which var allele is used to code the PfEMP1 protein. Like T. brucei, each parasite expresses multiple copies of one identical protein. However, unlike T. brucei, the mechanism by which var switching occurs in P. falciparum is thought to be purely transcriptional H. influenzae - similar antigenic changes in glycoproteins 93. Two steroid hormone receptors X and Y both contain a ligand binding domain and a DNA binding domain. Using recombinant DNA technology, a modified hybrid receptor H is prepared such that it contains the ligand binding domain of X and DNA binding domain of Y. Three sets of cells over-expressing receptors X, Y and H were then treated separately either with hormone X or with hormone Y. Assuming that there is no crossreactivity, which one of the following graphs best represent the receptor-ligand binding in each case? Cells over expressing receptor (A) X Y H %receptor ligand binding Hormone X Hormone Y P1/16

51 A/UG-15 (B) (C) (D) %receptor ligand binding %receptor ligand binding %receptor ligand binding Hormone X Hormone Y Hormone X Hormone Y Hormone X Hormone Y Exp. It is due to specificity and selectivity of receptor-ligand binding. Hybrid receptor H has ligand binding domain of steroid receptor X, therefore it can bind only to hormone X. So in case of excess hormone X receptor X and Receptor H will show more % receptor ligand binding while in excess of hormone Y only receptor Y will show recptor ligand binding. 94. A protein X is kept in an inactive state in cytosol as complexed with protein Y. Under certain stress stimuli, Y gets phosphorylated resulting in its proteasomal degradation. X becomes free, translocates to nucleus and results in the transcription of a gene which causes cell death by apoptosis. Stress stimuli were given to following four different cases. Case A : Protein Y has a mutation such that phosphorylation leading to proteasomal degradation does not occur. Case B : Cells are transfected with a gene which encodes for a protein L that inhibits the translocation of protein Y to the nucleus. Case C : Cells are transfected only with empty vector used to transfect the gene for protein L. Case D : Cells are treated with Z-VAD-FMK, a broad spectrum caspase inhibitor. Which one of the following graphs best describes the apoptotic state of the cells in the above cases? Y-axis represents % apoptotic cells. (A) (B) (C) (D) A B C D A B C D A B C D Amar Ujala Education A B C D Exp. There is cytoplasmic complex of protein X and Y. protein X is effector, Protein Y is regulator,. In case A: when protein Y is always remain associated with X that means there will be less apoptosis. Case B: Due to presence of Protein L, protein X (ques says protein Y) fails to reach nucleus that time also apotosis will be less. Case C: there is no change in normal pathway so apoptosis will be there and in case D: as the caspase inhibitor are present apoptosis can not proceed. 95. Consider the following events which occur during fertilization of sea urchin eggs. A. Resact/Speract are peptides released from the egg jelly and help in sperm attraction. B. Bindin, an acrosomal protein, interacts in a speciesspecific manner, with eggs. C. A respiratory burst occurs during cross- linking of the fertilization envelope, where a calciumdependent increase in oxygen levels is observed. D. IP3, which is formed at the site of sperm entry, sequesters calcium leading to cortical granule exocytosis. P1/17

52 Amar Ujala Education Which of the above statement(s) is NOT true? (A) OnlyC (B) AandC (C) OnlyD (D) BandD Exp. The small peptide RESACT acts as a chemoattractant for sea urchin sperm. In the sea urchin, contact with egg jelly initiates the acrosome reaction, which is a calciummediated process. The acrosomal vesicle fuses with the plasma membrane, releasing enzymes from the tip of the sperm that aid digestion of egg jelly. At the same time, bindin is deposited on the surface of the acrosomereacted sperm. One of the hallmarks of fertilization is that it is species-specific: sperm from one species cannot or usually do not fertilize eggs of another. Sea urchin eggs synthesize H 2 O 2 in a respiratory burst at fertilization, as an extracellular oxidant to crosslink their protective surface envelopes. The slow block to polyspermy in sea urchin egg is mediated by PIP2 second messenger system. 96. Following statements were given regarding the decisions taken during the development of mammalian embryos: A. The pluripotency of the inner cell mass (ICM) is maintained by a core of three transcription factors, Oct 4, Sox 2 and Nanog. B. Prior to blastocyst formation each blastomere expresses both Cdx 2 and the Oct 4 transcrip- tion factors and appears to be capable of becoming either ICM or trophoblast. C. Both ICM and trophoblast cells synthesize transcription factor Cdx 2. D. Oct 4 activates Cdx 2 expression enabling some cells to become trophoblast and other cells to become ICM. Which of the above statements are true? (A) AandB (B) AandC (C) BandD (D) Band C Exp. Nanog, Oct4, and Sox2 are the core regulators of ESC pluripotency. Oct 4 expression is abundant and uniform in all cells of the embryo throughout the morula stage. However, as the outer cells of the embryo differentiate into the trophectoderm (TE), Oct 4 expression becomes downregulated and restricted to cells of the ICM in the blastocyst. 97. Apoptosis during early development is essential for proper formation of different structures. In C. elegans, apoptosis is accentuated by ced-3 and ced-4 genes, which in turn are negatively regulated by ced-9 and eventually EGL-1. When compared to mammals, functionally similar homologues have been identified. Accordingly, which one of the following statements is NOT correct? (A) CED-4 resembles Apaf -1 A/UG-15 (B) CED-9 resembles Bcl-XL (C) CED-3 resembles caspase-3 (D) CED-4 resembles caspase-9 Exp. In C elegans ced-9 blocks the apoptotic protein similarily as Bcl-XI in mammals, so CED-9 is homolog of Bcl-XI, while CED -3 is act as the effector caspase like mammalan caspase-3. CED-4 is the initiator protein functions to activate effector apoptotic protein CED-3 like in mammals Apaf-1 functions to activate effector caspase 9. So all given option are correct except option 4 as CED-4 is homolog of mammalian Apaf Which one of the following statements regarding B cell receptor (BCR) and T cell receptor (TCR) is NOT true? (A) TCR is membrane bound and does not appear as soluble form as does the BCR (B) Unlike BCR, most of the TCR are not specific for antigen alone but for antigen combined with MHC (C) In order to activate signal transduction, BCR associates itself with Ig- /Ig- whereas TCR associates with CD3 (D) The antigen binding interactions of BCR is much weaker than TCR Exp. BCR binds to Ag with high affinity than TCR. BCR comprised of memrane bound ig which binds to surrounding Ag with high affinity. High binding affinity of BCR is attributed to affinity maturation, isotypic switching which enhances the binding affinity of lg in BCR with surrounding Ag. The other three statements are correct. 99. In case of amphibians, the dorsal lip cells and their derivatives are called as Spemann - Mangold organizer. Following statements related to the organizer were made: A. It induces the host s ventral tissues to change their fates to form a neural tube and dorsal mesodermal tissues. B. It cannot organize the host and donor tissues into a secondary embryo. C. It does not have the ability to self- differentiate into dorsal mesoderm. D. It has the ability to initiate the movements of gastrulation. E. Both -catenin and Chordin are produced by the organizer. Exp. Spemann referred to the dorsal lip cells and their derivatives (notochord, prechordal mesoderm) as the organizer because they induced the host s ventral tissues to change their fates to form a neural tube and dorsal mesodermal tissue (such as somites), and they organized host and donor tissues into a secondary embryo with clear anterior-posterior and dorsal-ventral axes and have the ability to induce movements of gastrulation. P1/18

53 A/UG Driesch performed the famous pressure plate experiment involving intricate recombination with an 8-celled sea urchin embryo. This procedure reshuffled the nuclei that normally would have been in the region destined to form endoderm into the presumptive ectoderm region. If segregation of nuclear determinants had occurred, the resulting embryo should have been disordered. However, Driesch obtained normal larvae from these embryos. The possible interpretations regarding the 8-celled sea urchin embryo are: A. The prospective potency of an isolated blastomere is greater than its actual prospective fate. B. The prospective potency and the prospective fate of the blastomere were identical. C. Sea-urchin embryo is a harmoniously equipotential system because all of its potentially independent parts interacted together to form a single embryo. D. Regulative development occurs where location of a cell in the embryo determines its fate. Which of the above interpretation(s) is/are true? (A) OnlyA (B) OnlyD (C) OnlyAandB (D) A,CandD Exp. Driesch had demonstrated that the prospective potency of an isolated blastomere (those cell types it was possible for it to form) is greater than its prospective fate (those cell types it would normally give rise to over the unaltered course of its development). Second, Driesch concluded that the sea urchin embryo is a harmonious equipotential system because all of its potentially independent parts functioned together to form a single organism. Third, he concluded that the fate of a nucleus depended solely on its location in the embryo Read the following statements related to plant-pathogen interaction A. Systemic acquired resistance is observed following infection by compatible pathogen B. Induced systemic resistance is activated following infection by compatible pathogen C. A bacterial infection can induce effector triggered immunity (ETI) leading to hypersensitive response locally D. NPR1 monomers that are released in cytosol due to salicylic acid accumulation is rapidly translocated to nucleus Which combination of above statements is correct? (A) A,BandC (B) A,CandD (C) A,BandD (D) B,CandD Exp. SAR is a systemic resistance response that occurs following an earlier localized exposure to a pathogen while ISR is an activated resistance process that is activated by biological or abiotic factors and is dependent on the physical or chemical barrier of the host plant, and its action is characterized by no direct killing or inhibition of the pathogen instead through induction of plant disease resistance to disease prevention and control purposes. P1/19 Amar Ujala Education 102. Given below are statements describing various features of solute transport and photoassimilate translocation in plants. A. Apoplastic phloem loading of sucrose happens between cells with no plasmo- desmatal connections B. Growing vegetative sinks (e.g., young leaves and roots) usually undergo symplastic phloem unloading C. Movement of water between the phloem and xylem occurs only at the source and sink regions D. Symplastic loading of sugars into the phloem occurs in the absence of plasmodesmatal connections Select the option that gives a combination of correct statements: (A) OnlyAandC (B) OnlyBandC (C) OnlyBandD (D) OnlyAandB Exp. Plants can use either the symplastic or apoplastic pathway for sugar loading into phloem. Cells that lack cytoplasmic connections i.e. plamodesmata opt for the apoplastic pathway. Movement of water between xylem and phloem occurs usually only at the source where water moves into sieve elements through osmosis building the internal pressure which aids in movement of sugar through the phloem from source to sink (Pressure-flow machanism). Young leaves which act as sinks usually undergo the symplastic mode of phloem unloading Given below are names of phytohormones in column I and their associated features/effects/ functions in column II. I II A Auxin (i) Delayed leaf senescence B Gibberellins (ii) Epinastic bending of leaves C Cytokinin (iii) Polar transport D Ethylene (iv) Removal of seed dormancy Select the correct set of combinations from the options given below: (A) A (iii), B (ii), C (iv), D (i) (B) A (iv), B (iii), C (i), D (ii) (C) A (iii), B (iv), C (i), D (ii) (D) A (i), B (iv), C (iii), D (ii) Exp. Auxin is the only plant growth hormone that is transported polarly (unidirectional), i.e. IAA is synthesized primarily in the apical bud and is transported polarly to root. Gibberelins remove seed dormancy and promote seed germination, it stimulates the production of various hydolases, notably alpha-amylase by aleurone layers of germination cereal grains. Cytokinin delay leaf senescence. Ethylene mimics high concentration of auxin by inhibiting stem growth and causing epinasty (a downward curvature of leaves).

54 Amar Ujala Education 104. If in a blood transfusion, type A donor blood is given to a recipient having type B blood, the red blood cells (RBCs) of donor blood would agglutinate but the recipient s RBCs would be least affected. These observations can be explained in the following statements: A. Agglutinins in recipient s plasma caused agglutination by binding with type A agglutinogens B. The agglutinins of donor blood was diluted in recipient s plasma resulting in low agglutination C. Low titre of anti-a agglutinins is the cause of low agglutination of recipient s RBCs D. High agglutination of donor RBCs is the outcome of high titre of anti-b agglutinins Which of the above statement(s) is/are INCORRECT? (A) OnlyA (B) AandB (C) OnlyB (D) CandD Exp. When someone has type A blood (with type A antigens), their plasma has type B antibodies and vice versa. Therefore, a reaction occurs when the antigens on the red blood cells of te donor blood react with the antibodies in the recipient s plasma. When a transfusion reaction does occur, an antibody attaches to antigens on several red blood cells. This causes the red blood cells to clump together and plug up blood vessels The arterial pressure usually rises and falls 4 to 6 mm Hg in a wave like manner causing respiratory waves. The probable mechanism of these waves has been proposed in the following statements: A.Themorenegativeintrathoracicpressureduring inspiration reduces the quantity of blood returning to the left side of the heart causing decreased cardiac output. B. The changes of intrathoracic pressure during respiration can excite vascular and atrial stretch receptors which affect heart and blood vessels. C. The activity of medullary respiratory centres can influence the vasomotor centre. D. The respiratory waves are outcome of the oscillation of the central nervous system ischemic pressure control mechanism. Which of the above statement(s) is/are NOT appropriate? (A) OnlyA (B) AandB (C) Band C (D) OnlyD 106. The uptake of nitrous oxide (N 2 O) and carbon monoxide (CO) in the blood of lung alveolar capillary relative to their partial pressure and the transit time of red blood cell in capillary is shown in the figure below: P1/20 Alveolar Partial Pressure Partlal pressure In blood Start of capillary End of capillary Time in capillary(sec) A/UG-15 NO 2 CO The reasons for difference in the pattern of alveolar gas exchange of N 2 O and CO have been proposed in the following statements: A. N 2 O does not chemically combine with proteins in blood but equilibrate rapidly between alveolar gas and blood B. CO has high solubility in blood C. CO has high solubility in the alveolar capillary membrane D. The dispersion of N 2 O between alveolar gas and blood is considered as diffusion limited. Which of the above statement(s) is/are INCORRECT? (A) Only A (B) A and B (C) Only C (D) C and D Exp. Nitrous oxide is a perfusion-limited gas while carbon monoxide is a diffusion-limited gas. In Perfusion-limited Gas Exchange, the rate at which gas is transported away from functioning alveoli and into tissues is principally limited by the rate of blood flow through the pulmonary capillaries and thus across the alveolar membrane. Nitrous oxide displays a rapid diffusion rate acros the alveolar membrane and is not bound by any proteins within blood. Consequently, the gas partial presure of nitrous oxide within blood rises rapidly following the transport of only minimal amounts of the gas. As a result, the partial pressure of nitrous oxide equilibrates rapidly as blood traverses the pulmonary capillaries, thus eliminating the partial pressure gradient across the alveolar membrane long before blood reaches the end of the capillaries. While in diffusion-limited gas exchange, the rate at which gas is transported away from functioning alveoli and into tissues is principally limited by the diffusion rate of the gas across the alveolar membrane. Because hemoglobin has very high CO-binding capacity, any CO diffusing into the blood is rapidly bound by hemoglobin and thus cannot contribute to the blood partial pressure of CO. Consequently, by the time blood traverses across the

55 A/UG-15 pulmonary capillaries, the blood partial pressure of CO is still very low compared to that in the alveolar space, meaning a large partial pressure gradient remains even in the end-capillary blood. Because the rate of CO diffusion across the membrane is the principal factor limiting further blood transport of carbon monoxide away from the lungs, pulmonary exchange of carbon monoxide is referred to as Diffusion-limited Individual and overlapping expression of homoeotic genes in adjacent whorls of a flower determine the pattern of floral organ development. In an Arabidopsis mutant, floral organs are distributed as follows: Whorl 1 (outer most) carpel Whorl 2 stamens Whorl 3 stamens Whorl 4 (inner most) carpel Loss of function mutation in which one of the following genes would have caused the above pattern of floral organ development? 1. APETALA 2 2. APETALA 3 3. PISTILLATA 4. AGAMOUS Exp. Three classes of mutations were identified in Arabidopsis. This is class A mutation. Flower with these mutations have (unfused) carpels instead of sepals in whorl 1 and stamens instead of petals in whorl 2. The pattern of organ from (outside to inside) is carpel, stamen, stamen, carpel. The genes containing these mutations were named APETALA1 (AP1) andapetala2 (AP2) In photosynthetic electron transport, electrons travel through carriers organized in the Z-scheme. The following are indicated as directions of electron flow: (A) P680 PQ A PQ B Cytb 6 f Pheo PC P700 (B) P700 A 0 A 1 FeS X FeS A FeS B Fd (C) P680 Pheo PQA PQB Cytb6f PC P700 (D) P700 A1 A0 FeSB FeSA FeSX Fd Which one of the following combinations is correct? (A) AandB (B) Band C (C) CandD (D) AandD Exp. On excitation-by the absorption of a photon-p680* rapidly transfers an electron to a nearby pheophytin a. Pheophytin a is a chlorophyll a molecule. The electron is then transferred to a tightly bound plastoquinone at the QA site of the D2 subunit. The electron is then transferred to an exchangeable plastoquinone located at the QB site of the D1 subunit. The arrival of a second electron to the QB site with the uptake of two protons from the stroma produces plastoquinol, PQH2. Plastoquinone (PQ) carries the electrons from PSII to the cytochrome bf complex. Amar Ujala Education This complex functions as plastoquinol-plastocyanin oxido-reductase. The final stage of the light reactions is catalyzed by PSI. Upon excitation-either by direct absorption of a photon or exciton transfer-p700* transfers an electron through a chlorophyll (A0) and a bound phllyoquinone (A1) to a set of 4Fe-4S clusters. From these clusters the electron is transferred to ferredoxin (Fd). The flow of electron occur from water to photosystem-ii, from photosystem-ii to photosystem-i and PS-I to NADP+ and arrangement is described as Z- scheme Phytochrome-mediated control of photomorphogenesis is linked to many other gene functions. The following statements are made on the mechanism of phytochrome action: a. Phytochrome function requires COP1, an E3 ubiquitin ligase that brings about protein degradation b. COP1 is slowly exported from the nucleus to the cytoplasm in the presence of light c. HY5 is targeted by COP1 for degradation in the presence of light d. HY5 is a transcription factor involved in photomorphogenetic response Which one of the following combinations is correct? (A) A,BandC (B) B,CandD (C) A,BandD (D) A,CandD Exp. ABD are correct statements photoreceptors ubiquitination by COP1 E3 desensitizes siganlling pathway. These mutants look light grown in the dark and therefore are thought to function as negative regulators of light signal transduction pathways (i.e. the signal transduction pathways are active in the absence of light). In the dark, COP1 is present in the nucleus, but in the light it is only found in the cytoplasm. COP1 targets HY5 for degredation by CNS thus only C is wrong statement The C 4 carbon cycle is a CO 2 concentrating mechanism evolved to reduce photorespiration. The following are stated as important features of the C 4 pathway: A. The leaves of C 4 plants have Kranz anatomy that distinguishes mesophyll and bundle sheath cells. B. In the peripheral mesophyll cells, atmospheric CO 2 is fixed by phosphoenol pyruvate carboxy- lase yielding a four-carbon acid. C. In the inner layer of mesophyll, NAD-malic enzyme decarboxylates four-carbon acid and releases CO 2. D. CO 2 is again re-fixed though Calvin cycle in the bundle sheath cells. Which one of the following combinations is correct? (A) B,CandD (B) A,BandC (C) A,BandD (D) A,CandD P1/21

56 Amar Ujala Education Exp. The leaves of C4 plants demonstrate an unusal anatomy involving two different types of choloroplast containing cells-mesophyll cells and bundle sheath cells. This feature is called Kranz anatomy. The mesophyll cells surrounds the bundle sheath cells which in turn surrounds the vascular tissue. Initial CO2 fixation occur in outer mesophyll cells using enzyme PEPcarboxylase by the production of 4 carbon compound i.e. oxaloacetic acid. The OAA then makes malate (4 carbons). Malate enters bundle sheath cells and releases the CO2. This released CO2isusedinCalvincyclethatisfixedbyRubiscoinC3 cycle External pressure given on a mixed nerve causes loss of touch sensation while pain sensation remains relatively intact. On the other hand, application of local anaesthetics on the same nerve, induces loss of pain sensation keeping touch sensation least affected. These observations can be explained by the following statements: A. External pressure causes loss of conduction of impulses in small diameter sensory nerve fibres B. Local anaesthetics depress the conduction of impulses in large diameter sensory nerve fibres C. Touch-induced impulses are carried by fibre type A D. Fibre type C is responsible for pain sensation Which of the above statement(s) is/are INCORRECT? 1. A and B 2. C and D 3. Only C 4. Only D Exp. Mixed nerves contain both afferent and efferent axons, and thus conduct both incoming sensory information and outgoing muscle commands in the same bundle. The sensations transmitted by type A delta fibeers include pressure, touch, cold and cold-induced pain sensations, is transmitted along type C fibers The probable effects of lesion of left optic tract on the vision of a human subject are given below. Identify the correct statement. (A) Blindness in the left eye but the visual field of right remains intact. (B) Blindness in the right half of the visual fields of both the eyes. (C) Blindness in the left half of the visual field of left eye and blindness in the right half of the visual field of right eye. (D) Blindness in the left half of the visual field of both the eyes. Exp. Lesions in the optic tract correspond to visual field loss on the left or right half of the vertical midline, also known as homonymous hemianopsia. A stroke, or injury to the right side of the brain would cause loss of the left visual field of each eye, and an injury to the left side of the brain would cause loss of the right visual field of each eye. A/UG Inversions are considered as cross-over suppressors because (A) Homozygous inversions are lethal and thus they do not appear in next generation (B) Inversion heterozygotes, i.e., one copy having normal chromosome and its homologue having inversion, does not allow crossing over to occur as they cannot pair at all (C) Due to inversion present, four chromosomes take part in the pairing and crossing over events and make the structure difficult for separation and gamete formation (D) The pairing and crossing overs do occur in inversion heterozygotes but the gametes having cross over products are lethal Mutant strain Mutagen treatment EMS Proflavin A + B + C Exp. Inversions are a special type of mutation in which a piece of chromosomal DNA is flipped 180 degrees. For an inversion to occur, two breaks occur in a chromosome, the region between the breaks gets inverted, and the ends of the region get rejoined to the rest of the chromosome. In paracentric inversions, the inverted region contains a centromere. Paracentric inversions are called crossover suppressors. *an inverted region will prevent pairing for mechanical reasons, *If crossing over occur, the products of crossing over are never recovered because the resulting chromosomes are nonfunctional. *an acentric (no centromere) and dicentric (two centromere) chromatid are obtained acentric is lost Three met E. coli mutant strains were isolated. To study the nature of mutation these mutant strains were treated with mutagens EMS or proflavins and scored for revertants. The results obtained are summarized below: (+ stands for revertants of the original mutants and stands for no revertants obtained) Based on the above and the typical mutagenic effects of EMS and proflavin, what was the nature of the original mutation in each strain? (A) A- Transversion B- Insertion or deletion of a single base C- Deletion of multiple bases (B) A- Transition B- Transversion C- Insertion or deletion of a single base (C) A- Insertion or deletion of a single base B- Transition C- Deletion of multiple bases (D) A- Transition B- Insertion or deletion of multiple bases C- Transversion P1/22

57 A/UG-15 Exp. During DNA replication, DNA polymerases that catalyze the process frequently place thymine, instead of cytosine, opposite 06-ethylguanine. Following subsequent rounds of replication, the original G:C base pair can become ana : T pair (a transition mutation). Hence B has to be transition. Proflavin causes mutations by inserting itself between DNA bases, typically resulting in insertion or deletion of a single base pair. So revertant of mutant A might have deleted the base pair which might have been inserted in the first mutation. In case multiple bases were deleted, mutation could not be reverted by either EMS or proflavine The following pedigree shows the inheritance pattern of a trait. From the following select the possible mode of inheritance and the probability that the daughter in generation III will show the trait. (A) X-linked recessive, probability is 1/2 (B) X-linked recessive, probability is 1/4 (C) Autosomal recessive, probability is 1/2 (D) Autosomal recessive, probability is 1/3 Exp. The trait cannot be X-linked as it father to son transmission is not possible so must be an autosomal trait. The normal parent must be heterozygous; Aa (other wise the trait would not be seen in the II generation) I aa X Aa Aa X aa II aa Aa X aa Aa Aa and aa.. thus the probabilty is 1/ A pair of alleles govern seed size in a crop plant. B allele responsible for bold seed is dominant over b allele controlling small seed. An experiment was carried out to test if an identified dominant DNA marker (5kb band) is linked to alleles controlling seed size. A plant heterozygous for the marker and the alleles was crossed to a small seeded plant lacking the 5kb band. 100 progeny obtained from the cross were analysed for the presence and absence of the DNA marker. The results are tabulated below: Phenotype Plant with Plant with bold seed small seed Present Absent Present Absent No. of pro-geny showing presence or absence of DNA marker Amar Ujala Education Based on the above observations which one of the following conclusions is correct? (A) The DNA marker assorts independently of the phenotype (B) The 5kb band is linked to the allele B (C) The 5kb band is linked to the allele b (D) The DNA marker assorts independently with bold seed but is linked to the small seed trait 117. The following diagram represents steroidogenic pathway in the Zona Glomerulosa of the adrenal cortex: Cholesterol A Pregnenolone ser B ser 11-Deoxycorticosterone Mitochondria C Mitochondria 18 (OH) Corticosterone Mitochondria Aldosterone What do A, B and C represent, respectively? (A) ser, Progesterone, 11(OH) cortisol (B) Mitochondria, Progesterone, Corticosterone (C) Mitochondria, 3 -pregnenolone, 11(OH) cortisol (D) ser, Progesterone, Corticosterone Exp. Cortical mitochondria contain mixed oxidases which convert cholesterol to pregnenolone. Further pregnenolone converted to progesterone. 11-Deoxy corticosterone converted to corticosterone The following scheme represents deletions (1-4) in the rii locus of phage T 4 from a common reference point: Deletion 1 Deletion 2 Deletion 3 Deletion 4 (The bars represent the extent of deletion in each case). Four point mutations (a to d) are tested against four deletions for their ability (+) orinability(-) togive wild type (rii+) recombinants. The results are summarized below: P1/23

58 Amar Ujala Education a b c d Based on the above the predicted order of the point mutations is (A) b-d-a-c (B) d-b-a-c (C) d-b-c-a (D) c-d-a-b Exp. Deletion 4(longest deletion) result in the ability of point mutation c to produce wild type phenotype(c+) deletion 3-result in c+ and a+ deletion 2-result in c+, a+ and b+ deletion 1- c+, a+, b+ and d+ so the correct order will be d-b-a-c (if the point mutation is very close to the end of the deletion, but not within sequences deleted in the recipient, then the frequency of (+) recombinants will be low) 119. Following are the plots representing biological rhythms at different time points depicted as: SR = Sunrise; N = Noon; SS = Sunset; MN = Midnight A B C D SR N SS MN SR N SS MN SR SR N SS MN SR N SS MN SR SR N SS MN SR N SS MN SR A/UG-15 SR N SS MN SR N SS MN SR Which of the plot(s) represent the ultradian biological rhythm(s)? (A) Plot B (B) Plots A and C (C) Plots C and D (D) Plot D Exp. These occur more than one in a 24 hour cycle and at night time. We shall consider the stages of sleep. As you should be aware, a typical night s sleep takes you from stage1to4thenbackto2andfinallyintorem.this whole cycle then repeats itself three or four more times during the night, each cycle lasting about 90 minutes. There are a number of similar cycles during the daytime too. Sometimes these are referred to as diurnal. Examples include eating (approximately every four hours), smoking and drinking caffeine (in those addicted), and urination A population of non-poisonous butterflies have the same colour pattern as some highly poisonous butterflies. Assume that the population of nonpoisonous butterflies is higher than the population of poisonous butterflies. Given this, what will be the impact of this mimicry on the fitness of the population of the poisonous butterflies in the presence of the predator? (A) It will lower the fitness, that is, fitness of the mimic is negatively frequency-dependent (B) It will increase the fitness, that is, fitness of the mimic is positively frequency-dependent (C) It will not affect the fitness, that is, fitness of the mimic is frequency independent (D) It will increase the fitness, that is, fitness of the mimic is negatively frequency-dependent Exp. It is the case of Batesian mimicry, when one species called mimic (non-poisonous butterfly) resembles a second species called the model (poisonous butterfly), that is protected by aposematic coloration. In this mimicry, the mimic benefits but the model may actually find its survival threatened, specially if the harmless mimic becomes too common. The model, on the other hand, is disadvantaged, alongn with the dupe. At higher frequency there is also a stronger selective advantage for the predator to distinguish mimic from model. For this reason, mimics are usually less numerous than models, an instance of frequency dependent selection. However, some mimetic populations have evolved multiple forms (polymorphism), enabling them to mimic several different models. This affords them greater protection, a concept in evolutionary biology known as frequency dependent selection 121. Given below is a graphical representation of changes in morphological features over a period of geological time scale, where population A accumulates heritable morphological changes and give rise to a distinct species B. Population B splits into a distinct species B2 TIME B 2 Lineage 2 A Lineage 1 MORPHOLOGY Which of the above lineages represent the pattern of speciation by cladogenesis? (A) Lineage 1 (B) Both the lineages 1 and 2 (C) Lineage 2 (D) Neither of the lineages 1 and 2 B P1/24

59 A/UG-15 Exp. Both the lineages shows speciation by cladogenesis as Cladogenesis occurs because reproductive-isolating mechanisms prevents two sub-populations from interbreeding. This reproductive barrier could be the result of 1) the isolation of one portion of the population by some physical barrier; 2) a sub-population becoming established ina new ecological niche not previously occupied by the species; or 3) a polymorphism becoming established in a population before it moves to a new ecological niche Red hair is a recessive trait in humans. In a randomly mating population in Hardy-Weinberg equilibrium, approximately 9% of individuals are red-haired. What is the frequency of heterozygotes? (A) 81% (B) 49% (C) 42% (D) 18% Exp. To estimate the frequency of alleles in a population, we can use the Hardy- Weinberg equation. For a population in genetic equilibrium: p+q = 1.0 (The sum of the frequencies of both alleles is 100%). (P+q) 2 = 1 so p 2 + 2pq + q 2 =1 p 2 = frequency of AA (homozygous dominant) 2pq = frequency of Aa (heterozygous) q 2 = frequency of aa (homozygous recessive) here we have q2 = 9/100 = 0.09 q = 0.3 p = 1; thus 0.7 frequency of heterozygotes = 2pq = = 0.42 OR 42% 123. Interrupted mating experiments were performed using three different Hfr strains (1-3). The three strains have different combinations of selectable markers. The time of entry for markers for each strain is shown in the table below : Strain Time of entry Hfr #1 met thr strr phe pro (5 ) (17 ) (25 ) (30 ) (45 ) Hfr #2 strr pur pro his met (15 ) (28 ) (35 ) (45 ) (55 ) Hfr #3 pro his met strr phe (2 ) (12 ) (22 ) (42 ) (47 ) Using the above data, predict the correct seque-nce of markers on the E. coli chromosome. (A) met-thr-strr-phe-pro-purr-his (B) purr-pro-his-met-thr-strr-phe (C) strr-purr-his-met-phe-pro-strr (D) his-met-phe-thr-pro-strr-purr Exp. By the given time intervals between markers of all strains, wecancalculatethedistancebetweenthesemarkers which will be like this... met-thr-12, thr-str-8, str-phe-5, phe-pur-8, pur-pro-5, pro-his-10 nd his-met-10 so by Amar Ujala Education arranging the markers with resppect to their distance we will get the order of markers. Becaue these are in circular DNA form so there transfer depend on matting time. Order will be... met-thr-str-phe-pur-pro-his or in either way... pur-pro-his-met-thr-str-phe 124. Peripatus is an interesting living animal having unjointed legs, nephridia, haemocoel, trachea, dorsal tubular heart, claws, jaws, continuous muscle layers in body wall. This is considered as a connecting link between (A) Nematoda and Annelida: continuous muscle layers in body wall, unjointed legs and nephridia being nematode characters while haemocoel, trachea and dorsal tubular heart being annelid characters (B) Annelida and Arthropoda: unjointed legs and nephridia being annelid characters while claws, jaws, haemocoel, trachea and dorsal tubular heart being arthropod characters (C) Arthropoda and Mollusca: unjointed legs and nephridia being mollusca characters while claws, jaws, trachea and dorsal tubular heart being arthropod characters (D) Nematoda and Arthropoda: continuous muscle layers, unjointed legs and nephridia being nematode characters while claws, jaws, trachea and dorsal tubular heart being arthropod characters. Exp. Peripatus is the living connecting link between arthropods and annelids. Its arthropods characters are claws, jaws, haemocoel, tracheae and dorsal tubular heart. The annelidian characters are continuous muscle layers in the body wall, unjoined legs and nephridia The following schematic diagram represents secondary growth in the angiosperms. Apical Meristem Cortex Pith Rays Procambium Secondary Pholem Secondary Xylem Based on the above scheme, which of the following options represents the correct identity of cambia labelled as A, B, C and D. (A) A Inter-fascicular, B Fascicular, C Vascular, D Cork (B) A Fascicular, B Inter-fascicular, C Vascular, D Cork (C) A Cork, B Inter-fascicular, C Fascicular, D Vascular (D) A Cork, B Fascicular, C Inter- fascicular, D Vascular Exp. The vascular cambium produces secondary xylem and secondary phloem, and the cork cambium (phellogen) produces cork cells A B C D P1/25

60 Amar Ujala Education 126. The table below lists the major fungal groups and their characteristics: Fungal Groups Characteristics A Ascomycota (i) Hyphae aseptate, coenocytic; asexual reproduction by sporangiophores B Chytrids (ii) Hyphae aseptate, coenocytic; asexual reproduction by zoospores C Glomeromycetes (iii) Hyphae aseptate, coenocytic; no sexual spores D Zygomycetes (iv) Hyphae septate or unicellular; asexual reproduction by conidia Which one of the following options represents the appropriate match between the fungal group and their characteristics? (A) A (ii), B (iii), C (i), D (iv) (B) A (iv), B (ii), C (iii), D (i) (C) A (i), B (iv), C (iii), D (ii) (D) A (ii), B (iv), C (iii), D (i) Exp. In Ascomycota, asexual reproduction occur by conidia, In Chytrids, asexual reproduction occur by zoospores, In Glomeromycetes, sexual spores are not formed and in Zygomycetes, asexual reproduction occur by Formation of sporangiophores As a biologist, you want to classify three taxa, A, B and C. You have the information on three traits, p, q and r. The trait that is ancestral is counted 0 and the trait that is derived is counted as 1. The distribution of traits found in three taxa is given below p q r A B C Based on the above table, the following cladograms were drawn: A p q B r C A B r p q p C q r A p (a) (b) (c) B p q Based on trait distribution and the principle of parsimony, select the correct option. (A) Both a and b cladograms are possible C r A/UG-15 (B) Only b cladogram is possible (C) Only c cladogram is possible (D) Only a cladogram is possible 128. Given below are some pathogens and diseases of humans, animals and plants. A Bordetella pertussis (i) Lyme disease of humans B Tilletia indica (ii) Grain rot in rice C Borrelia burgdorferi D Anaplasma marginale (iii) Karnal bunt of wheat (iv) Whooping cough in humans E Burkholderia glumae (v) Hemolytic anemia in cattle Which one of the following is the correct match between the pathogen and disease caused? (A) A (iv), B (iii), C (i), D (v), E (ii) (B) A (iv), B (v), C (i), D (ii), E (iii) (C) A (iii), B (iv), C (v), D (i), E (ii) (D) A (ii), B (v), C (i), D (iii), E (iv) Exp. Borrelia burgdorferi is causative agent of lyme disease, they are diderm bacteria, Bordetella pertussis is causitive agent of whooping cough, Karnal bunt is fungal disease of wheat cause by smut fungus tilletia indica, anaplasma marginale cause hemolytic anaemia in cattle Given below are statements pertaining to organisms belonging to three domains of life. Identify the INCORRECT statement. (A) Unlike Bacteria and Eukarya, some Archaeal membrane lipids contain long chain hydrocarbons connected to glycerol molecules by ether linkage. (B) Peptidoglycans are absent in the cell wall of Archaea (C) Proteobacteria include many species of bacteriochlorophyll-containing, sulphur using photoautotrophs. (D) Mycoplasma, a group of low GC content, gram positive bacteria that lack cell wall, belong to the same family as the gram positive Mycobacteriaceae Exp. Archaea are prokaryotic cells which are typically characterized by membranes that are branched hydrocarbon chains attached to glycerol by ether linkages. The presence of this ether containing linkages in Archaea adds to their ability of withstanding extreme temperature and highly acidic conditions. Extreme halophiles - i.e. organisms which thrive in highly salty environment, and hyperthermophiles - i.e. the organisms which thrive in extremely hot environment, are best examples of Archaea. Bacteria and the Eukarya, the Archaea have membranes composed of branched hydrocarbon chains (many also containing rings within the hydrocarbon chains) attached to glycerol by ether linkages. P1/26

61 A/UG-15 The cell walls of Archaea contain no peptidoglycan. Archaea are not sensitive to some antibiotics that affect the Bacteria, but are sensitive to some antibiotics that affect the Eukarya. Mycoplasma and mycobacterium do share some similarity but are different. Mycoplasma refers to a genus of bacteria that lacks a cell wall. Mycobacteria are acid-fast Gram positive bacteria that have a very thick, protective, waxy cell wall. They belong to family Mycoplasmataceae and Mycobacteriaceae respectively 130. You observed that two species of barnacles, species 1 and species 2, occupy upper and lower strata of intertidal rocks, respectively. Only when species 2 was removed by you from the lower strata, species 1 could occupy both the upper and lower strata. From the choices given below, what would be your inference from these observations? (A) Upper strata of the intertidal rock is the realized niche of species 1 (B) Upper strata of the intertidal rock is the fundamental niche of species 1 (C) Species 1 and species 2 exhibit mutualism (D) Species 1 can compete out species 2 Exp. The entire niche that species is capable of using based on its physiological tolerance limits and resource need is called the fundamental niche. The actual niche the species occupies is its realized niche. A fundamental niche is the resources an organism or population is theoretically capable of using under ideal circumstances. However biological constraints such as competition restrict organisms to their realized niche. So in presence of species 2, species 1 is only able to occupy upper strata of intertidal rock In a natural system, a species producing large numbers of offsprings, with little or no parental care, generally exhibits which one of the following kind of survivorship curves? (A) (B) (C) % SURVIVORS % SURVIVORS TIME TIME (D) % SURVIVORS % SURVIVORS TIME TIME P1/27 Amar Ujala Education Exp. This is the example of type-iii survivorship curve, also called early-loss which produces a concave curve. These curves are characteristic of organisms that produce large numbers of offsprings. Here, death rate is very high among the young because offspring receive little or no parental care. Example - Most insects Match the correct local names of temperate grasslands with their geographical range. Geographical range Local name of the grassland (i) Asia (a) Pampas (ii) North America (b) Prairies (iii) South America (c) Steppes (iv) South Africa (d) Veldt (A) (i) c, (ii) b, (iii) d, (iv) a (B) (i) c, (ii) b, (iii) a, (iv) d (C) (i) d, (ii) b, (iii) a, (iv) c (D) (i) b, (ii) c, (iii) a, (iv) d Exp. These all are temperatre grassland. Steppes belong to Asia, Prairies belong to North America, Pampas belong to South America and Veldt belong to South Africa Following is a hypothetical life table for a species. Age Number Number Age Age class alive dying specific specific (x) (nx) (dx) survivorship fertility (l x ) (m x ) Which one of the following is the correct net reproductive rate (Ro)? (A) 0.0 (B) 0.3 (C) 0.7 (D) Which one of the following statements is true for the trends of Dissolved Oxygen (DO) and Biological Oxygen Demand (BOD) in a water stream receiving pollutants from a point source? (A) In septic zone, both DO and BOD levels remain stationary (B) In recovery zone, both DO and BOD levels increase rapidly (C) In decomposition zone, DO level drops rapidly, whereas BOD level remains more or less stable (D) In septic zone, DO level decreases and BOD level increases whereas in recovery zone DO increases and BOD decreases

62 Amar Ujala Education Exp. In Recovery zone the stream tries to recover its former appearance. Most of the organic matter has been settled as sludge, B.O.D. falls and the D.O. content rises above 40%, microscopic aquatic life reappear. Mineralization is active and products such as nitrates, sulphates, and carbonates are formed, whereas septic zone is marked by heavy pollution characterized by absence of dissolved oxygen, and BOD increases in this zone Following are the graphical representations of various hypotheses proposed for explaining the possible relationships between species richness (X-axis) and community services (Y-axis). (a) (c) Which of the following options is the correct match between the graphical representations and the hypotheses? (A) (a) Redundancy, (b) Keystone, (c) Rivet, (d) Idiosyncratic (B) (a) Idiosyncratic, (b) Rivet, (c) Keystone, (d) Redundancy (C) (a) Rivet, (b) Redundancy, (c) Idiosyncratic, (d) Keystone (D) (a) Rivet, (b) Keystone, (c) Redundancy, (d) Idiosyncratic Exp. Certain hypothetical relationships between diversity and ecological processes are there : Rivet : all species contribute to the integrity of an ecosystem in a small but significant way such that a progressive loss of species steadily damages ecosystem function; shown by graph 1. Redundant : the contribution of additional species is redundant above a critical level. Idiosyncratic hypothesis : ecosystem function changes unpredictably as species richness changes. Keystone hypothesis says a single species has remarkable effect on ecosystem Agrobacterium Ti plasmid vectors are used to generate transgenic plants. The following are examples of vir gene-encoded proteins that are important for the transfer of T-DNA into plants: (b) (d) A/UG-15 (A) Vir E, a single-stranded DNA binding protein (B) Vir D2 that generates T-strands (C) Vir A that senses plant phenolic compounds (D) Vir F which directs T-complex proteins for destruction in proteasomes Which one of the following combinations of proteins functions inside the plant cells? (A) OnlyAandC (B) A,BandC (C) OnlyBandC (D) A,BandD Exp. The transfer and integration of T-DNA into the genome of plant is mediated by genes in the virulence region which are grouped into operons vir ABCDEFG, Vir A, VirG genes are expressed constitutively. Transfer of T- DNA is initiated by the product of the vir D1 and vird2 genes. Vir D1 and Vir D2 act as endonuclease. Protein vir D2 cuts phosphodiester bond and remain covalently attached to 5 end of the nicked DNA. The nicked DNA is then displaced 5 to 3 form the plasmmid, producing single strand T-DNA. T-DNa is coated with ssdna binding protein Vir E which protect T-DNA against nuclease and target the DNA to plant cell nucleus. Vir F mediate the ubiquitination and subsequentant proteasomal degradation of target protin after its implication in T- DNA translocation. This explanation clearly depicts that Vir E,Vir D2 & Vir F work intracellularly, hence this justifies the answer A researcher is investigating structural changes in a protein by following tryptophan fluorescence and by circular dichroism (CD). Fluorescence and CD spectra of a pure protein were obtained in the absence of any treatment (a), in the presence of 0.5 M Urea (b), upon adding acrylamide, a quencher of tryptophan (c) and upon heating (d). The data are shown below: Fluorescence a b c d Wavelength Ellipticity Wavelength Which one of the following statements is correct? (A) CD is more sensitive to structural changes than fluorescence (B) Fluorescence is more sensitive to structural changes than CD (C) Both methods are equally responsive to structural changes (D) Acrylamide alters the secondary structure of the protein d c b a P1/28

63 A/UG-15 Exp. CD spectroscopy is more sensitive than fluorescence spectroscopy. In CD spectroscopy when analysis is carried out in far UV region (180 nm to 280) it tells about the secondary structure such as alpha helix and beta sheets conformations of proteins. Additionally CD analysis also tells whether beta sheets structure is in parallel or anti-parallel orientation. Near UV CD analysis reveals the information about tertiary structures present in protein. However in tryptophan quenching is just based upon change in the emission of trypthophan fluorescence. It is true that it tells about changes in the structure of protein in presence of treatment but this method just depends upon trypthophan emission not really tells about secondary or tertiary structure as that of observed in CD spectroscopy. Tyrpthophan quenching is not suitable for the proteins which are lacking in trpthophan Polynucleotide kinase (PNK) is frequently used for radiolabeling DNA or RNA by phosphorylating 5 - end of non-phosphorylated polynucleotide chains. Which of the following statement about PNK is NOT true? (A) PNK catalyzes the transfer of ±-phosphate from ATP to 5 -end of polynucleotide chains (DNA or RNA). (B) PNK has 3 -phosphatase activity (C) PNK is inhibited by small amount of ammonium ions (D) PNK is a T4 bacteriophage-encoded enzyme Exp. In the forward reaction, PNK transfers the gamma phosphate from ATP to the 5 end of a polynucleotide (DNA or RNA). It is a product of the T4 bacteriohage, PNK also has 3 phosphatase activity. PNK is inhibited by small amounts of ammonium ions. So incorrect statement is option A gene encoding for protein X was cloned in an expression vector under the T7 RNA polymerase promoter and lac operator. Cells were induced by the addition of 1 mm IPTG at 37ºC for 6 h. Cells were lysed and fractionated into insoluble bodies and cell-free supernatant by centrifugation. Protein X is present in the insoluble bodies. Which one of the following strategies would you use to express protein X in the soluble fraction (cell-free supernatant)? (A) Increase the duration of induction with 1mM IPTG (B) Grow cells at lower temperature after induction with 1mMIPTG (C) Increase the concentration of IPTG (D) Grow cells at higher temperature after induction with 1 mm IPTG Exp. An experiment was conducted with E. coli where lowering the incubation temperature for the expression of fusion gene under 25 C prevented the formation of inclusion Amar Ujala Education body, optimally at 25 C mm of IPTG was sufficient for the soluble expression of fusion gene at 25 C. Details can be found from reference link Engineering of metabolic pathways in plants can be achieved by introduction and over expression of appropriate candidate gene(s) using transgenic technology. The figure given below represents a biochemical pathway in plants where a precursor molecule A is converted into products T and X through a series of enzymatic reactions. Enzymes 1-5 are involved in this pathway. Scientists attempted to increase the level of X by introducing an additional copy of the gene for enzyme 5 under transcriptional control of a strong constitutive promoter. However, the developed transgenic plants did not display a proportionate increase in the level of X. '1' '2' '3' A C D E '4' '5' T X The following statements were proposed for explaining the above results: (A) Enzyme 4 has greater affinity for D than enzyme '3 (B) Feedback inhibition of enzyme 5 by compound X (C) Substrate limitation for enzyme 5 Which of the above statements could represent probable reasons for NOT obtaining a proportionate increase in the amount of X in the transgenic plants? (A) OnlyC (B) Only A and B (C) OnlyA (D) A,BandC Exp. All statements seem probable reasons 141. A single copy homozygous transgenic plant containing the transgene A for fungal resistance was subsequently re-transformed with another gene B for conferring resistance to salt-stress. The selection marker genes used for both the transformation experiments were different. Transgenic plants obtained following the retransformation experiment were screened for salt-stress resistance and single copy events were identified by Southern hybridization. These single copy events were self-pollinated. In the event of the two T-DNAs (containing the A and B transgenes) getting integrated in unlinked locations in all the transgenic plants, the phenotypic ratios among the T1 progeny would be: (A) 3 (Fungal resistant + Salt-stress resistant): 1 (Fungal resistant) (B) 1 (Fungal resistant): 2 (Fungal resistant + Salt-stress resistant): 1 (Salt-stress resistant) (C) 3 (Salt-stress resistant): 1 (Fungal resistant) (D) 1 (Fungal resistant): 1 (Salt-stress resistant): 1 (Fungal resistant + Salt stress resistant) P1/29

64 Amar Ujala Education 142. You are inserting a gene of 2kb length into a vector of 3kb to make a GST fusion protein. The gene is being inserted at the EcoRI site and the insert has a HindIII site 500bp downstream of the first codon. You are screening for the clone with the correct orientation by restriction digestion of the plasmid using HindIII plus BamHI (H+B) and HindIII plus PstI (H+P). The map of the relevant region of the vector is shown below: BamHI EcoRI Pst I Promoter GST Terminator Given below is the pattern following restriction digestion of plasmid isolated from four independent clones (A,B,CorD) A B C D H+B H+B H+B H+B H+B H+B H+B H+B Which of the plasmids shown above represents the clone in the correct orientation? (A) A (B) B (C) C (D) D 143. The frequency of M-N blood types in a population of 6129 individuals is as follows: Blood type Genotype Number of individuals M L M L M 1787 MN L M L N 3039 N L N L N 1303 The frequency of LN allele in this population is (A) (B) (C) (D) Exp. Frequency of the L N allele will be the total number L N L N homozygotes added to half of the number of L M L N heterozygotes all divided by the total number of alleles sampled; /( )= P1/30 A/UG Mayfair genes (hypothetical) consist of a super family of transcription factors. They are found in 4 clusters in mammals; in 2 clusters in insects; and in a single cluster in an ancestor to insects. These data are consistent with all of the following explanations EXCEPT: (A) Two successive genome duplication events occurred between ancestral organism and vertebrates (B) The first duplication may have taken place before divergence of vertebrates (C) Exon shuffling exclusively produced such cluster (D) Whole genome duplications could lead to such observations Exp. This question can be explained through the following example- The developmental control genes containing an Antennapedia-type homeobox are clustered in insects and vertebrates. Subsequent duplication events generated a cluster of at least five homeobox genes in the last common ancestor of insects and vertebrates. At the time of the insect/vertebrate divergence two of the resulting genes had already undergone a further duplication, while the Abdominal-B precursor remained singular. Later in the evolution of vertebrates and insects, gene duplications in all three of the original clases occurred Fluorescence recovery after photobleaching (FRAP) is a method to estimate the diffusion of molecules in a membrane. Fluorescently labelled molecules such as i. a receptor tagged with green fluorescent protein (GFP) ii. a receptor labelled with GFP which interacts with cytoskeleton iii. a labelled lipid iv. a labelled protein that binds to the membrane surface are photobleached and the recovery profiles (a-d) were obtained to estimate their diffusion coefficients. The following data were obtained: b a c d Which one of the combinations is correct? (A) a = i; b = ii (B) b = iii; a = iv (C) c = iii; d = iv (D) d = ii; b = i Exp. As the receptor is attached with cytoskeleton, it should take more recovery time.

65 A/UG-15 Amar Ujala Education CSIR JRF (NET) LIFE SCIENCES Solved Paper, June Of the following, which is the odd one out? (A) Cone (B) Torus (C) Sphere (D) Ellipsoid Exp. The option (A) is correct as other three are 3D representation of a circle while cone is not. 2. "My friend Raju has more than 1000 books said Ram. "Oh no, he has less than 1000 books", said Shyam. "Well, Raju certainly has at least one book". said Geeta. If only one of these statements is true, how many books does Raju have? (A) 1 (B) 1000 (C) 999 (D) 1001 Exp. If Raju has 1 book. Statements of Geeta and Shyam are correct. If Raju has 1000 books, statement of Geeta is correct. If Raju has 999 books, statements of Geeta and Shyam are correct. If Raju has 1001 books, Statements of Geeta and Ram are correct. 3. It takes 5 days for a steamboat to travel from A to B along a river. It takes 7 days to return from B to A. How many days will it take for a raft to drift from A to B (all speeds stay constany)? (A) 13 (B) 35 (C) 6 (D) 12 Exp. Let d be the distance between A and B, b the speed of the boat in still water and c the speed of current. Then d/ (b + c) = 5 days...(1) d/ (b c) = 7 days...(2) First add (1) and (2) d [(b c) + (b + c)] / [(b + c) + (b c)] = 12 days 2db/ [(b + c) (b c)] = 12 days db/ [(b + c) (b c)] = 6 days...(3) Now, substract (1) from (2) d[(b+c) (b c)]/(b c) (b c)=2days 2dc / [(b + c) (b c)] = 2 days dc / [(b + c) (b c)] = 1 day...(4) Divide (3) by (4) b/c=6b=6c Substitute in (1) d/7c = 5 days d/c = 35 dsays. PART-A P2/1 4. An infinite number of identical circular discs each of radius 1 are rightly packed such that the centres of the 2 discs are at integer values of coordinates x any y. The ratio of the area of the uncovered patches to the total area is (A) 1 - /4 (B) /4 (C) 1- (D) Exp. Area = pie r 2 =pie/4 Ratio = Volume of liquid/volume of cavity = pie/34 pie = 7/8.50 for gas (1 pie/4) when we calculate the ratio of Liquid w.r.t. cavity. 5. N is a four digit number. If the leftmost digit is removed. the resulting three digit number is 1/9 th of N. How many such N are possible? (A) 10 (B) 9 (C) 8 (D) 7 Exp. Let a be the left most digit and l, m, n be other digits, So N = almn lmn = 9 lmn. ( 3 digit number is 1/9 of 4 digit number or 4 digit number is 9 times 3 digit number) 1000 a = 8 lmn, lmn = 125 a possible values of a = 1, lmn = 125 N = 1125 a = 2, lmn = 250, N = (250 a = 2250) a = 3, lmn = 375 Similarly, N = 4500, 5625, 6750, 8750 for a = 4, 5, 6, 7. If a = 8 lmn = 1000 which is not a 3 digit and hence not possible. So 7 such N are possible option (D) is correct answer. 6.

66 Amar Ujala Education Which of the following inferences can be drawn from the above graph? (A) The total number of students qualifying in Physics in 2015 and 2014 is the same (B) The number of students qualifying in Biology in 2015 is less than that in 2013 (C) The number of Chemistry students qualifying in 2015 must be more than the number of students who qualified in Biology in 2014 (D) The number of students qualifying in Physics in 2015 is equal to the number of students in Biology that qualified in 2014 Exp. According to the graph given, number of chemistry students is more in 2015 qualifying the test as per in the year 2014 biology test. 7. What is the minimum number of moves required to transform figure 1 to figure 2? A move is defined as removing a coin and placing it such that it touches two other coins in its new position. (A) 1 (B) 2 (C) 3 (D) 5 Exp. Only 2 moves are required. As coin can be removed from its place and it is not essential that the coin should be touching two more coins during the move, a coin can be picked up and moved. Consider the desire circle with the blank centre at the right coin of second row. Four coins are alrealy in position i.e. top row coin, left coin of second row, middle coin of third row, right most coin of third row. Remaining two coins can be picked up one by one and placed to complete the circle. 8. Which of the following best approximates sin (0.5 0 )? (A) 0.5 (B) 0.5X 90 (C) 0.5X 180 (D) 0.5X 360 Exp. Sin 0.5 = which is equal to 0.5 p/ What comes next in the sequence? (A) (B) (C) (D) P2/2 A/UG-15 Exp. Alphabets A, B, C and D where the upper half is represented, so the option (C) will be the correct answer of the upper half is E. 10. How many times starting at 1:00 pm would the minute and hour hands of a clock make an angle of 40 0 with each other in the next 6 hours? (A) 6 (B) 7 (C) 11 (D) 12 Exp. Each hour it makes 40 degree twice and in six hours it will make 2 6 = 12 fourty degree angle but at 1'O clock it will make only one 40 degree angle at 1 : 12, So it will be 12 1 = Which of the following statements is logically incorrect? (A) I always speak the truth (B) I Occasionally lie (C) I occasionally speak the truth (D) I always lie Exp. Logically, I alway lie is incorrect. there are two possibilities that either the sentence is true or it is false. If the sentence is true then I always lie is not a lie it s true. If it is false then that means I always lie is true that means the person cannot always lie. It is an inconsistency, a paradox. 12. AB and CD are two chords of a circle subtending 60 0 and respectively at the same point on the circumference of the circle. Then AB : CD is (A) 3 :1 (B) 2 :1 (C) 1 : 1 (D) 3 : 2 Exp. Length of chord = 2 X radius X sin q/ 2, length of AB = 2 r sin 30 = r, Length of CD = 2r sin 60 = 2r 5q. rt 3/2 = r Xsqrt3, HenceAB:CD=1:sqrt Find the next figure 'D" (A) (C) Exp. It is nothing but the rotation pattern of the given figure. Ist and 2nd related with each other by 180 degree switching horizontally. In the same way 3rd and 4th will be related with each other by 180 degree switching vertically. (B) (D)

67 A/UG A student appearing for an exam is declared to have failed the exam if his/her score is less than half the median score. This implies (A) 1/4 of the student appearing for the exam always fail. (B) If a student scores less than 1/4 of the maximum score, he/she always fails. (C) If a student scores more than 1/2 of the maximum score. he/she always passes. (D) it is possible that no one fails. Exp. According to question, half of median i.e. 1/2 1/2 = 1/ 4. So ill student secures less than 1/4 of maximum score, he/she will fail. 15. A solid contains a spherical cavity. The cavity is filled with a liquid and includes a spherical bubble of gas. The radii of cavity and gas bubble are 2 mm and 1 mm, respectively. What proportion of the cavity is filled with liquid? 1 (A) (B) Amar Ujala Education 17. Brothers santa and Chris walk to school from their house. The former takes 40 minutes while the latter, 30 minutes. One day Santa started 5 minutes earlier than Chris. In how many minutes would Chris overtake Santa? (A) 5 (B) 15 (C) 20 (D) 25 Exp. Santa started his journey 5 minutes early, so according to the question, the difference in later and former is 10 minutes. therefore = 15 minutes. 18. The diagram shows a block of marble having the shape of a triangular prism. What s the maximum number of slabs of 10 X 10 X 5cm 3 size that can be cut parallel to the face on which the block is resting? (C) 5 8 (D) Exp. Volume of spherical cavity = 4.3 pr 3 =4.3p mm 3. Volume of spherical bubble = mm 3. Proportion of cavity filled with fluid = cavity filled with fluid/total volume of spherical cavity = (Vol. of spherical cavity vol. spherical bubble) / total volume = 4/3 (8 1)/4/3 (8) = 7/8 16. The relationship among the numbers in each corner square is the same as that in the other corner squares. Find the missing number. (A) 10 (B) 8 (C) 6 (D) 12 Exp. After adding all the numbers in each corner, the sum is 44, in the last corner if 6 is put at the missing place, the total sum will be 44 in that corner as well. Hence option (C) is correct. 7 8 P2/3 (A) 50 (B) 100 (C) 125 (D) 250 Exp. Vlume of marble = 1/ = Now divide by one volume i.e = 500. So, 62500/500 = Fill in the blank: F2,..., D8, C16, B32, A 64. (A) C4 (B) E4 (C) C2 (D) G16 Exp. F2, E 4, D 8, C 16, B The setof numbers (5, 6,7, m, 6,7, 8, n) has an arithmetic mean of 6 and mode (most frequently occurring number) of 7. Then m x n = (A) 18 (B) 35 (C) 28 (D) 14 Exp. Mean = sum of number/number of items =(5+6+7+m n)/8. =6 39+m+n=48 m + n = 9. Hence either of the following combination of m and n is possible = (1, 8), (2, 7), (3, 6), (4, 5). Since mode is 7, this means either m or n will be 7 as it has to be the most frequently occurring numbers. So 2, 7 is the combination which maches our question and 2 7 = 14. Hence option (D) is the correct answer.

68 Amar Ujala Education PART-B A/UG Entry of enveloped viruses into its host cells is mediated by: (A) Only endocytosis (B) Both endocytosis and phagocytosis (C) Both endocytosis and membrane fusion (D) Only pinocytosis Exp. Both mechanisms are used by enveloped virus to make entry into host cells. 22. Penicillin acts as a suicide substrate. Which one of the following steps of catalysis does a suicide inhibitor affect? (A) K 1 (B) K 2 (C) K 3 (D) K 4 Exp. Suicide inhibition in irreversible reaction that occurs when an enzyme binds a substrate analogue and forms an irreversible comoplex with it through a covalent bond druing the normal catalysis reaction and does not allow the enzyme to form products. Mechanistically, penicillin forms a penicilloyl-enzyme complex with a serine residue found in glycopeptide trans-peptidase (an enzyme involved in the formation of peptidoglycan in bacteria) forming an ester, which is stable indefinitely, thereby rendering the enzyme inactive and inhibiting its ability too form products, Hence, here, penicillin inhibits reaction having equilbrium constant. k Histone deacytalase (HDAC) catalyses the removal of acetyl group from N-terminal of histones. Which amino acid of histone is involved in this process? (A) Lysine (B) Arginine (C) Asparagine (D) Histidine. Exp. HDAC are a class of enzymes that remove acetyl groups (O = C-CH3) from an e-n-acetyl lysine amino acid on a histone, allowing the histones to wrap the DNA more tightly. 24. The solubility of gases in water depends on their interaction with water molecules. Four gases i.e. carbon dioxide, oxygen sulphur dioxide and ammonia are dissolved in water. In terms of their solubility which of the following statements is correct? (A) Ammonia > Oxygen > Sulphur dioxide > Carbon dioxide. (B) Oxygen > Cabon dioxide > Sulphur dioxide > Ammonia P2/4 (C) Sulphur dioxide > Oxygen > Ammonia > Carbon dioxide (D) Ammonia > Sulphur dioxide > Carbon dioxide > Oxygen Exp. Solubility of gases in liquids is based on Henrys law. Solubility of Ammonia = 52.9 Solubility of sulphur dioxide = 11.28, Solubility of Carbon dioxide = 0.169, Solubility of oxygen = Error-free repair of double strand breaks in DNA is accomplished by (A) non-homologous end-joining. (B) base excision repair (C) homologous recombination. (D) mismatch repair. Exp. Mismatch repair system and excision repair are mainly involved in the repair of damage to single strand of DNA., while homologous recombination and NHEJ are mainly involved in the repair of double strand breaks, Out of these 2, NHEJ is known to be an error-prone process. So, we are left with homologous recombination. 26. Lateral diffusion of proteins in membrane can be followed and diffusion rate calculated by (A) Atomic force microscopy (B) Scanning electron microscopy (C) Transmission electron microscopy (D) FRAP Exp. Leteral diffusion can be tracked by a process called fluorescence recovery after photo bleaching (FRAP). This process is abailable because the use of fluorescence labelling allows the tracking of the molecules. The cell surface will be labelled first with a chromophore, then analysed under a fluorescence microscope on one section (illuminated area). On this specific site, the fluoresced molecules are destroyed by bleaching them (use of laser) and watching if they leave or enter the illuminated area., If the molecules are mobile, it has two different states, bleached or unbleached. If the molecule is leaving the illuminated area, this means that the molecule is bleached. If the molecule is entering the illuminated area, this means that the molecules is unbleached. The unbleached molecules help to increase the fluorescence intensity. Lateral diffusion can also be measured by a complementary strategy know as fluorescence loss in photo-bleaching (FLIP). In this technique, a small area is continuously bleached and the fluorescent proteins are bleached as they diffuse into it. Eventually, the number

69 A/UG-15 of fluorescent proteins will decrease and will result in all bleached proteins. From both FRAP and FLIP, we can calculate the diffusion coefficient from the bleached proteins. Although these two techniques seem promising, it has some drawbacks. One problem is that individual movement of each protein cannot be observed because there are too many bleached/fluoresecent proteins. For example, one cannot tell whether each individual protein is immobile or if it's only restricted to a small area, perhaps by cytoskeleton impediment. In order to circumvent this problem, a technique known as Single-particle tracking can be used. In this technique, an individual protein is labelled by antibodies and are colored by fluorescent dye or small gold specks. The movement of these proteins are then recorded by video microscopy. Using this technique allows one to observe the diffusion pathway of a protein periodically. 27. E. coli is being grown in a medium containing both glucose and lactose. On depletion of glucose, expression of B- galactoside will (A) remain unchanged (B) increase (C) decrease (D) initially decrease and then increase Exp. When glucose is present, there is no need for the expression of the enzymes required for the metabolism of lactose. But once glucose if depleted and only lactose is present in the medium, this lactose is taken up by the cell, immediately converted to allo-lactose which acts as the inducer of Lac Operon, Beta-galatosidase being the enzyme of lacoperon, hence is induced i.e., its expression increases. 28. RNA interference is mediated by both si RNA and mirna, Which one of the following statement about them is NOT true? (A) Both sirna and mirna are processed by DICER. (B) Both sirna and mirna usually guide silencing of the same genetic loci from which they originate. (C) mirna is a natural molecule while sirna is either natural or a synthetic one. (D) mirna, but not sirna is processed by Drosha. Exp. sirna usually regulate the gene loci from which they are expressed, while mirna regulate genes other than which express them. 29. Which one of the following statements about receptorenzyme is FALSE? (A) A receptor-enzyme has an extracellular ligand binding domain. a transmembrane domain and an intracellular catalytic (enzyme) domain. (B) Many types of receptor enzymes are found in animals. (C) The signal transduction pathways of receptor enzyme involve phosphory lation cascades. P2/5 Amar Ujala Education (D) Receptor-enzymes interact directly with intracellular G-proteins. Exp. A receptor enzyme has a dual activity that means, it can function both as an enzyme as well as a receptor. For example receptor tyrosine kinase. is a receptor enzyme, it has an extracellular ligand binding domain, a transmembrane domain and an intracellular catalytic domain. Signal transduction pathway through this receptor enzyme involves phosphorylation cascades., but they donot interact directly with G proteins, for eg,. ras is a G protein. It interacts indirectly with Receptor tyrosine kinase with the help of GRB 2,GRB 2 is a protein which interact directly (throuth its SH 2 domain) with phosphotyrosine residues, present in the cytosolic domain of the receptor (which got phosphorylated through the receptor's intrinsic kinase activity). Now, the GRB2 activates sos which acts as a guanine nucleotide exchange factor and helps in exchaning GDP for GTP in the ras protein (which is a G protein). So, in this way the receptor enzyme is interacting indirectly with the G protein (ras). 30. Which of the following is NOT true for cholesterol metabolism? (A) HMG-CoA reductase is the key regulator of cholesterol biosynthesis. (B) Biosynthesis takes place in the cytoplasm. (C) Reduction reactions use NADH as cofactor. (D) Cholesterol is transported by LDL in plasma. Exp. Reduction reactions mainly use NADPH as cofactor. 31. Which one of the following best defines an oncogene? (A) An oncogene never codes for a cell cycle protein, which promotes cell proliferation. (B) Oncogenes are always involved in inherited forms of cancer. (C) An oncogene codes for a protein that prevents a cell from unergoing apoptosis. (D) An oncogene is a dominantly expressed mutated gene that renders a cell advantageous towards survival. Exp. Oncogene is basically mutated form of proto-oncogene protooncogenes are the genes which code for proteins required for cell proliferation and provide cell survival signals for cell. 32. Which of the following bacteria has subcellular localization in lysosomes? (A) Salmonella typhi (B) Streptococcus pneumonie (C) Vibrio cholerae (D) Nycobacterium tyberculosis

70 Amar Ujala Education Exp. Mycobacteria (including M. tuberculosis and Mycobacterium leprae) grow inside phagocytic vacuoles even after extensive fusion with lysosomes. Mycobacteria have a waxy, hydrophobic cell wall containing mycolic acids and other lipids, and are not easily attacked by lysosomal enzymes. 33. Following are some of the characteristics of MHC class I and class II molecules except one which is applicable only for MHC class I. Identify the appropriate statement. (A) The are experssed constitutively on all nucleated cells. (B) They are glycosylated polypeptides with domain structure. (C) They are involved in presentation of antigen framents to cells. (D) They are experssed on surface membrane of B cells. Exp. MHC 1 glycproteins are present on almost every cell. Whereas MHC II are only present on specialised antigen presenting cells. 34. The- COOH group of cellular amino acids can form which of the following bonds inside the cell? (A) Ether and ester bonds. (B) Ester and amide bonds. (C) Amide and ether bonds. (D) Amide and carboxylic anhydride bonds. Exp. Peptide bond is amide bond and all amino acids form peptide bond. Ether bond does not involve carboxylic group, and there is no instance for the presence of anhydride bond in proteins. On the other hand, it has been found that methyl esterification involves the conversion of carboxylic acids to methyl esters and this methyl esterification has been associated with the membrane localization of proteins. 35. Predominant interaction between phospholi-pids that stabilize a biological membrane include (A) hydrogen bonds and covalent interactions. (B) van der Waal and ionic interactions. (C) hydrophobic interactions and hydrogen bonding. (D) covalent and hydrophobic interactions. Exp. The hydrophobic tail interacts by hydrophobic interactions, the polar head group by hydrogen bonds and the amide in the alpha helices of the transmembrane helices by hydrogen bond. 36. Labelling of membrane spanning domain of any integral membrane protein in a given plasma membrane vesicle (without disrupting its structure) is successfully carried out by : (A) immunochemical methods. (B) metabolic labelling with radioisotopes. (C) hydrophobic photoaffinity labelling. (D) limited proteolysis followed by metabolic labelling. P2/6 A/UG-15 Exp. Hydrophobic photo affinity labelling is an useful technique for identification of transmembrane segments of proteins and protein subunits associated with membranes. 37. Bones of vertebrates are drived from embryonic (A) ectoerm (B) epiderm (C) mesoderm (D) endoderm Exp. Bones are derived from mesoderm. 38. During development if a cell has committed to a particular fate,itissaidtobe (A) pluripotent (B) totipotent (C) determined (D) differentiated Exp. Commitment for a particular fate is known as determination totipotency and pluripotency is associated with undifferentiated cells. 39. Sperm cell behaviour during double fertilezation in Arabidopsis can be stated as follows. ldentify the INCORRECT statement : (A) Pollen tube bursts and discharges sperm cells. (B) Sperm cells produce pollen tubes and enter into female gametophyte. (C) The receptive antipodal cells break down when pollen tube enters the female gametophyte. (D) One sperm nucleus fuses with the egg cell and the other fuses with the central cells. Exp. It is the pollen grain that forms pollen tube and not the sperm cells. 40. The initial dorsal-ventral axis in amphibian embryos is determined by (A) The point of sperm entry. (B) gravity. (C) the point of contract with the uterus. (D) genetic differences in the cells. Exp. Entry fo seprm determined the rotation of cortical cytoplasm and cortical rotation determine the formation of dorsal ventral symmetry. 41. Which one of the follwing photoreceptors plays a role in day length perception and circadian rhythms? (A) Zeitupe family (B) Cryptochromes (C) Phototropins (D) UV resistance locus 8 Exp. Circadian clock, the endogenous oscillator must be entrained to the daily light a cycles of the external environment. Cryptochromes and phytochromes, are known to be the photoreceptors associated with this whereas phototropin is involved in phototrophism.

71 A/UG Which one of the following plant hormones use the twocomponent histidine kinase receptor system for signal transduction? (A) Auxin (B) Gibberellin (C) Cytokinin (D) Abscisic acid Exp. Cytokinin, a plant hormone, uses a multistep phosphorelay system to transduce its signal from membrane-bound receptors to the nucleus to mediate transcription of cytokinin response genes. 43. Rhizobial genes that participate in legume nodule formation are called nodulation (nod) gens. The nod/ D- encoded protein (A) is an acetyl transferase that adds a fatty acyl chain to the Nod factor (B) binds to the nod box and induces trans-cription of all nod genes. (C) catalyzes the linkage of N-acetyl glucosa-mine residues. (D) influences the host specificity of Rhizo-bium. Exp. Rhizobial genes that participate in nodule formation are called nodulation (nod genes). The nod genes are classified as common nod genes or host-specific nod genes. The common genes are presnet in all rhizobial strains and are named as nod A nod B and nod C., nod D is constitutive genes that induces transcription of other nod genes. 44. The cell bodies of sympathetic preganglionic neurons are located in : (A) Intermediolateral cell columan of spinal cord (B) Posterior cell columen of spinal cord (C) Celiac ganglion (D) Paravertebral ganglion. Exp. The symphtetic division (thoracolumbar outflow) consists of cell dodies in the lateral horn of the spinal cord (intermediolateral cell columns) from T 1 to L The transport of fructose into the enterocytes is mediated by : (A) Sodium-dependent glucose transporter 1 (SGLT 1) (B) glucose tranporter 5 (GLUT5) (C) SGLT 2 (D) GLUT 4 Exp. The passage of both the glucose and the galactose, chemically very similar, is mediated by a transporter calles SGLT 1, acronym of Sodium-dependent Glucose co Transporter 1, which carries out the co-transport of one molecule of glucose/galactose together with two sodium ions (then energy will be spent for removing sodium from the cell and so keeping the gradient along which other sodium may enter again). Fructose, chemically different from glucose and galactose, enters P2/7 Amar Ujala Education into the cells by facilitated diffusion, also called passive transport (no energy expenditure) by transporter GLUT 5 (acronym of Glucose Transporter type 5). 46. Insulin increases facilitated diffusion of glucose in muscle cells by : (A) Phosphorylation of glucose transporters. (B) Translocation of glucose transporter containing endosomes into the cell membrane. (C) inhibition of the synthesis of mrna for glucose transporters. (D) dephosphorylation of glucose transporters. Exp. Vesicles containing glucose transporters are mobilized to the plasms membrane by insulin stimulation, thereby effecting glucose transport into the cell. 47. The di-and tripetides are transported in the enterocytes by peptide transporter I that requires : (A) Na (B) Ca ++ (C) H + (D) CI Exp. The human peptide transporter 1, PEPT1 is primarily responsible for the absorption of dietary di and tripeptides from the small intestinal lumen, PEPT1 is a sodiumindependent symporter that catalyzes electrogenic uphill peptide transport (i.e. sequence-independent di- and tripetides) by coupling of substrate translocation to the movement of H + with the transmembrane electrochemical proton gradient providing the driving force. 48. Which one of the following is the correct oder of electron transport during light reaction in the thylakoid membrane of chloroplast? (A) P680 Cytochrome b f PC PQ 6 (B) P680 PC Cytochrome b f PQ 6 (C) P680 PQ PC Cytochrome b f 6 (D) P680 PQ Cytochrome b f PC 6 Exp. PSII-680, transfers electron to pheophytin, which then transfers electrons to the acceptors Q A and Q B, which are plastoquinones (PQ). The cytochrome b6 f complex transfers electrons to plastocyanin (PC), a soluble protein, which in turn reduces P700 + (oxidized P700). 49. Which one of the following statements is INCORRECT? (A) Quantitative inheritance results in a range of measurable phenotypes for a polygenic trait. (B) Polygenic traits often demonstrate continuous variation. (C) Certain alleles of quantitiative trait loci (QTL) have an additive effect on the character/trait. (D) Alleles governing quantitiative traits do not segregate and assort independently. Exp. Alleles governing quantitative traits assort independently. But since multiples genes present in different loci effect a single phenotype, that's why they show continuous variation. Statement D is incorrect.

72 Amar Ujala Education 50. What is the genotype of a male Drosophila fly that has yellow body colour and red eyes. Brown (y) is dominant over yelow (y) and red (w) is dominant over white (w) Both are carried on X chromosome. (A) X w+y Y (B)X wy Y (C) X wy+ y (D) X wy+ X wy+ Y Exp. In Drosophila, the normal male has XY as the pair of sex chromosome. Since the male Drosophila in question has yellow body color (coded by y allele) and red eye color (coded by w+ allele), so the correct representation of the above genotype will be X^ Option A is correct. w+y. 51. Which one of the following statements is Incorrect? (A) Loss of genetic variation occurs within a small population due to genetic drift. (B) The number of deleterious alleles present in the gene pool of a population is called the genetic load. (C) Genetic erosion is a reduction in levels of homozygosits. (D) Inbreeding depression results from increased homozygosity for deleterious alleles. Exp. Genetic erosion refers to the process in which a plant or animal species faces a gradual or drastic diminishing or complete loss of its unique gene pool. In genetic erosion, there is a loss of genetic diversity. Heterozygosity determines genetic diversity in the gene poll. So. Genetic erosion will be reduction in heterozygosity and not homozygosity. So Statement C is incorrect. 52. A mouse carring two allelesof insulin-like growth factor II (IgF2) is normal in zize; whereas a mouse that carries two mutant alleles lacking the growith factor is dwarf. The size of a heterozygous mouse carrying on the parental origin of the wild type allele Such pattern of inheritance is known as (A) Sex-linked inheritance (B) Genomic imprinting (C) Gene-environment interaction (D) Cytoplasm inheritance Exp. People inherit two copies of their genes one from their mother and one from their father. Usually both copies of each gene are active, or turned on, in cells. In some cases, however, only one of the two copies is normally turned on. Which copy is active depends on the parent of origin : some genes are normally active only when they are inherited from a preson's father; others are active only when inherited from a person's mother. This phenomenon is known as genomic imprinting. Since, the question states that the phenotype of the heterozygote depends on the parent origin of the wild type allele, so it is genomic imprinting. Option B is correct. 53. In which ecosystem is the autotroph-fixed energy likely to reach the primary carnivore level in the shortest time? (A) Temperate decioduous forest P2/8 A/UG-15 (B) Grassland (C) Ocean (D) Tropical rain forest Exp. In terrestrial system high biomass is found at lower trophic level and less biomass at higher trophic level, this leads to less energy transfers at successive stages, Whereas in ocean system has less primary producers than of higher trophic levels., phytoplankton's are more edible than terrestrial counterparts. Thus in ocean energy will be transferred in shortest time. 54. The utilization or consumption efficiency of herbivores is highest in (A) Plankton communities of ocean waters. (B) mature temperate forests. (C) managed grasslands. (D) managed rangelands. Exp. The highest consumption efficiency is found on plankton communities of ocean water that is found to be around 60-99% in comparison to forest, grassland and range lands. That shows it as , and respectively. 55. During which of the following major mass extinction events, over 95% of the marine species disappeared from the planet Earth? (A) Ordvician (B) Devonian (C) Permian (D) Triassic Exp. Permian-Triassic extinction event (End permian) : Earth's largest extinction killed 57% of all families, 83% of all genera and 90% to 96% of all species (53% of marine families, 84% of marine genera about 96% of all marine species and an estimated 70% of land species, including insects). 56. Fossils of the same species of fresh water reptiles have been found in South America and Africa. Based on the current understanding, which of the following is the best possible explanation for this patern. (A) The same species originated and evolved independently in these two places. (B) Species migrated from Africa to establish new populations in South America. (C) Species migrated from South America to establish new populations in Africa. (D) South America and Africa were joined at some point in Earth's history. Exp. Remains of Mesosaurus, a freshwater crocodile-like reptile that lived during the early Permian (between 286 and 258 million years ago), are found solely in Southern Africa and Eastern South America. It would have been physiologically impossible for Mesosaurus to swim between the continents. This suggests that South America and Africa were joined during the Early Permian.

73 A/UG Which of the following global hotspots of biodiversity has the highest number of endemic plants and vertebrates? (A) Sundaland (B) Tropical Andes (C) Brazil's Atlantic Forest (D) Mesoamerican forests Exp. To qualify as a biodiversity hotspot a region must meet two strict criteria : it must contain at least 0.5% or 1,500 species of vascular plants as endemics, and it has to have lost at least 70% of its primary vegetation. The correct option for this is 2-tropical andes. 58. Which of the following is NOT a prediction arising out of Wilson-Mac Arthur's Theory of lsland Biogeography? (A) The number of species on an island should increase with its size/area. (B) The number of species should decrease with increasing distance of the island from the source pool. (C) The turnover of species should be common and frequent. (D) Species richness on an island should be related to its average distance to the neighbouring islands. Exp. Distance between neighbouring islands was not a factor that was considered to be affecting species richness in island biogeography theory. 59. For a population growing exponentially with a growth rater r, its population doubling time is (A) (N O X2)/r (B) In 2/r (C) In 2 (D) In r X2 Exp. When exponentially growing population doyubles its final value is twice the initial value, that can be used in formula as " 2Ao = Ao (E) lrt. On divinding both side by Ao we get 2 = e^rt. After rearranging for T, we get answer as option B. 60. Match the following larval forms with the phyla that they occur in Larva Phylum (a) Amphiblastula (i) Mollusca (b) Nauplius (ii) Echinodermata (c) Glochidium (iii) porifera (d) Bipinnaria (iv) Arthropoda (v) Annelida (A) a-iii, b-iv, c-i, d-ii (B) a-iv, b-iii, c-i, d-v (C) a-ii, b-v, c-iv, d-i (D) a- v, b-i, c-ii, d-iii Exp. Mollusca larval form glochidium... echinod- ermata larval form bipinnaria..only option with both of these. P2/9 Amar Ujala Education 61. Which of the following National parks has the highest density of tigers among protected areas in the world? (A) Jim Corbett (B) Kaziranga (C) Keoladeo Ghana (D) Manas Exp. The density of tigers at Kaziranga is tigers per 100 sq. km. The highest in any known tigher habitat. Previously, this status was held by the Corbett Tiger Reserve in northern India which had 19.6 tigers per 100 sq. km. 62. Which of the following statements is NOT true regarding the closer affinity of Archaea to Eukarya than to Bacteria? (A) Both Archaea and Eukarya lack peptideglycan in their cell walls. (B) The initiator amino acid for protein synthesis is methionine in both Archaea and Eukarya. (C) Histones associated with DNA are absent in both Archaea and Eukarya. (D) In both Archaea and Eukarya the RNA polymerase is of several kinds. Exp. Stability and DNA-binding properties have been established for archaeal histones from mesophiles, thermophiles and hyperther-mophiles. Most archaeal histones are simply histone folds that are stabilized by dimer formation. 63. Which of the following is wild relative of wheat? (A) Triticum moncoccum (B) Triticum compactum (C) Triticum vulgare (D) Triticum boeoticum Exp. Wild Einkorn, Triticum boeotictim Einkorn, Triticum monococcum-a primitive kind the cultivation of which goes back to prehistoric times. 64. TILLING is a reverse genetics approach used in functional genomics. Which one of the following is used for TILLING? (A) T-DNA tagging by Agrobacterium mediated transformation. (B) Transposon tagging using Ac/DS elements. (C) Mutagenesis with ethylmethane sulponate. (D) Protoplast transformation by electro-poration. Exp. TILLING (Targeting induced Local Lesions in Genomes) is a method in molecular biology that allows directed indentification of mutations in a specific gene. The method combines a standard and efficient technique of mutagenesis using a chemical mutagen such as Ethyl methane Sulphonate (EMS) with a sensitive DNA screening-technique that identifies single base mutations (also called point mutations) in a target gene.

74 Amar Ujala Education 65. Which of the following is NOT an attribute of a species that makes it vulnerable to extinction? (A) Specialized diet (B) Low dispersal ability (C) Low trophic status (D) Variable population density Exp. Low trophic status will ensure large population of individuals that will not allow its extinction. Thus the option C is not correct. Enikom or one-tyrained wheat is a primitive kind the cultivation of which goes back to prehistroric times. Both winter and spring forms ocur, Spikes are awned, slender, narrow, flattened, and fragile, Spikelets contain only a single fertile flower and thus produce only one seed. Seeds are pale red, slender, flattened, almost without crease, and remain in the spikelets after threshing. Einkorn is little grown at present and not at all in the United States. 66. Which one of the following can be analysed using Surface Plasmon Resonance method? (A) Radiolabelled DNA probes. (B) Protein structure. (C) Optical density of a solution (D) Label-free bimolecular interaction. Exp. The binding of biomolecules results in the change of the refractive index on the sensor surface, which is measured as a change in resonance angle or resonance wavelength. 67. Which one of the following will be observed when auxin to cytokinin ratio is increased in the culture medium during organogenesis from tobacco pith callus? (A) Adventitious roots will form. (B) Adventitious shoot will form (C) There will be no root formation. (D) There will be no shoot formation. Exp. Cytokinin Ratio Regulates Morphogenesis in Cultured Tissues high auxin : cytokinin ratio in the tumor cells causes the proliferation of roots instead of undifferentiated callus tissue. When either gene for auxin biosynthesis is muated it stimulates the proliferation of shoots. 68. A and B are two enantiomeric helical peptides. Their chirality can be determined by recording their. (A) Circular dichroism spectrum. (B) UV spectrum. (C) Fluorescence spectrum. (D) Edman sequencing. Exp. Out of these techniques only Circular dichroism spectroscopy is used to determine chirality of molecules. Chiral molecules have the property to rotate polarised P2/10 A/UG-15 light either towards right or towards left. Circular dichroism measures the difference in the absorption of left-handed circularly polarised light (L - CPL) and right handed circularly polarised light (R-CPL) and occurs when a molecule contains one or more chiral chromophores (Light-absorbing groups). While, UV light absorption does not depend upon the presence of chiral molecules, rather it depends on the presence of certain functional groups in the molecule, which can absorb in UV range. Similarly fluorescence spectrum has nothing to do with chiral molecules, it totally depends upon the presence of certain groups in the molecule which can produce fluorescence. Also, Edman sequencing involves labelling of N-terminal of a protein with phenyl isothiocyanate, irrespective of the fact that amino acid at the N-terminal is chiral or not. 69. Which one of the following statements is correct for amplified fragment length polymorphism (AFLP)? (A) PCR using a combination of random and genespecific primers. (B) PCR amplification followed by digestion with restriction enzymes. (C) Digestion of DNA with restriction enzymes followed by one PCR step. (D) Digestion of DNA with restriction enzymes followed by two PCR steps. Exp. AFLP involves total restriction digestion of genomic DNA, after which selective amplification of restriction fragments take place by PCR. This amplification by PCR involves 2 stpes : Pre-amplification and final selective amplification. 70. The use of kruskat Wallis test is most appropriate in which of these cases? (A) There are more than two groups and each group in normally distributed. (B) There are more than two groups and the distribution in each group in not normal. (C) There are two groups and each group is normally distributed. (D) There are two groups and the distribution in each group in not normal. Exp. The Kruskal-Wallis H test (sometimes also called the one-way ANOVA on ranks ) is a rank-based nonparametric (Distribution free) test that can be used to determine if there are statistically significant differences between two or more groups of an independent variable on a continuous or ordinal dependent variable. So it wont matter whether distribution is normal or not.

75 A/UG-15 PART-C Amar Ujala Education 71. The trunover number and specific activity of an enzyme (molecular weight 40,000 D) in a reaction (V max = 4μmol of substrate reacted/min, enzyme amount = 2μg) are. (A) 80, 000/min, 2 x 10 3 mol substrate/min (B) 80,000/ min, 2 x 10 3 mol substrate/second (C) 40, 000/min, 1 x 10 3 mol substrate/min (D) 40,000/min, 2 x 10 3 mol substrate/min Exp. Given that Vmax = 4umol/min, Molar mass of enzyme = 2ug Molecular weight of enzyme = To Calculate; 1) Turn over number (kcat) and 2) Specific activity of enzyme 1. Turn over number : The formula for turn over num ber is kcat = Vmax/Et.where Et is the total enzyme concentration and Et = molar mass of the enzyme/ mol wt of the enzyme Et = 2/40000 = umol. Substituting the values of Vmax and Et in the formula, we get Kcat = (4umol/min) umol = 80000/ min. 2. Specific activity of the enzyme: The formula for spe cific activity of enzyme is : Enzyme units/total en zyme (mg), Where enzyme units (EU) = umol of the substrate converted introproduct per minute and it is given by Vmax. So EU = 4 umol/min Total enzyme (in mg) = Substituting the values of EU and total enzyme in the formula, we get Sp. Activity = (4)/ u mol/min. 72. In a mitochondrial respiration experiment, a researcher observed the following profile of oxygen consumption upon addition of following compounds at times I, II and III. (a) ADP + pi (b) Dinitrophenol, an uncoupler (c) Cyanide (d) Succinate P2/11 Which of the following describes the profile appropriately? (A) I- b; II- d; III-e (B) I-a; II-d; III-c (C) I-a; II-e; III-c (D) I-a; II-c; III-b Exp. Repeated additions of ADP increase O 2 consumption until either ADP or O 2 is depleted if an excess of ADP is added, followed by oligomycin, then O 2 consumption ceases. If uncoupler is now added the rate of repiration (O 2 consumption) increases dramatically as the proton gradient collapses as quickly as it forms. 73. The standard free energy change (^Go) per mole for the reaction A B At 30 0 C in an open system is-1000 cal/mole. What is the approximate free energy change (G) when the concentration of A and B are 100 micromolar and 100 millimolar, respectively? (A) 3160 (B) 316 (C) (D) 3160 Exp. Given Standard delta G = 1000 cal/mol converting it into joules will give 4184 J/mole, After putting all the values in the formula Free energy change = standard delta G + RTln K will give 13246j/mole now converting it into ca/mol will give = 3.16 cal/mol as standard answer are mentioned in kj/mol or kcal/mol hence converting that to kcal will give = 3160 kcal/mol. 74. Indicate the names of the following molecules (A) A = isocitrate, B=a-ketoglutarate, C = oxaloactate, D=citrate (B) A = citrate, B=isocitrate, C= -ketoglutarate, D=oxaloacetate (C) A = isocitrate, B=citrate. C= -ketoglutarate, D=oxaloacetate (D) A = citrate, B=isocitrate, C = oxaloactate, D=a-ketoglutarate Exp. A-citrate B-isocitrate C-alpha-ketoglutarate and D- oxaloactate.

76 Amar Ujala Education 75. A researcher has developed a program to evaluate the stability of a protein by substituting each amino acid at a time by the other 19 amino acids. For a protein, a researcher has observed the following changes in stability upon substitution of amino acids in loops, helices, sheets, protein core and on the protein surface. Substitutions in. (a) loops are more tolerant (b) sheets are more tolerant (c) Core is less tolerant (d) helices are less tolerant (e) surface is more tolerant Which of the above statements are correct? (A) (a) and (c) (B) (c) and (d) (C) (b) and (e) (D) (a) and (b) Exp. As per the graph given in diagram the stability is increasing with respect to number of amino acids for loops and surface. 76. Indicate which one of the following statements about nucleic acids and protein structures is correct. (A) Hydrogen bonding between the bases in the major and minor grooves of DNA is absent. (B) Both uracil and thymine have a methyl group but at different positions. (C) The backbone dihedral angles of a-helices and B- Sheets are very similar. Only the hydrogen bonding pattern is different. (D) A -turn is formed by four amino acids. The type of -turn is determined by the dihedral angles of the second and third amino acid. Exp. The first three options are incorrect, because hydrogen bonding is always there amongst the bases present in major and minor groove of DNA, thyminre has methyl group at 5th position but uracil has no methyl group in its structure, The backbone dihedral angles of alpha helix and beta pleated sheets is not same rather quite different, Hence, we are left with option D only. 77. It is well established that "Band 3" protein of red blood cell membrane is solely responsible for Cl - transport across membrane. A lysine group in the Cl - binding site of "Band 3" is crucial for this event. Keeping this in mind what is the most appropriate way to load and retain a small anionic fluorescent probe (x) inside the red blood cells (RBCs) suspended in phosphate buffered saline (PBS) ph 7.4 P2/12 A/UG-15 (A) Incubate the RBCs with x in phosphate buffered saline (PBS, ph 7.4) at 37 C for 30 min. (B) Incubate the RBCs with x in PBS at 4 0 C for 30 min. (C) Incubate the RBCs with x in Hepes sulfate buffer (ph 7.4) at 37 0 C for 30 min (D) Incubate the RBCs with x in hepes sulfate buffer (ph 7.4) at 37 0 C for 30min followed by treatmeny with a NH 2 group modifying agent (covalent modification) 78. Influenza virus (IV), a well known enveloped animal virus, enters its host cells through membrane fusion process catalyzed by haemagluttinin (HA) protein inside endosomes at 37 0 C. HA is localized in the lipid bilayer membrane of the IV as an integral membrane protein and is responsible for binding and fusion of IV membrane with the endosomal membrane of host cells. Upon binding, IV is internalized into host cells through receptor mediated endocytosis followed by fusion of the IV membrane with endosome membrane catalyzed by HA. In a situation if we wish to fuse IV membrane with its host cells (deficient in endocytosis) at the plasma membrane, mention the correct condition out of the following. (A) Pre-treat IV in ph 5.0 followed by its binding and fusion with host cells at ph 7.4 and 37 0 C (B) Allow the IV to bind and fuse with host cells at ph 7.4 and 37 0 C (C) IV and host cells are allowed to bind and fuse at ph 5.0 and 37 0 C (D) IV is subjected to incubation at 60 0 C for 30 minutes and allowed to bind and fuse with host cells at ph 5.0 and 37 0 C 79. Ministellites are used as marker for identi-fying individuals via DNA fingerprinting as the alleles may differ in the number of repeats. From the Southern blot shown below identify the progeny (A, B, C and D) for the given parents (M = mother, F = Father) (A) A,B,CandD (B) A,BandD (C) A and D only (D) B, C and D Exp. Except C, All the other progeny A, B and D receives alleles form both M and F.

77 A/UG Both sphingomyelin and phosphoglycerides are phospholipids. Which one of the following statements is NOT correct? (A) While one has a fatty acid tail attached via an ester bond, in another, the fatty acid tail is attached via an amide bond. (B) The hydrophilicity of both is dependent on the phosphate group and othe head groups attached to the phosphate group. (C) Only one of them may contain a carbon carbon double bond (C = C). (D) Both may have choline as head group. Exp. The hydrophobicity of sphingomyelins and phosphoglycerides does not depend upon the presence of phosphate group and type of head group attached to the phosphate group (head groups are polar in nature), but rather it depends upon the type of fatty acids attached to them. 81. Glycophorin of red blood cell (RBC) membrane spans the membrane only once and the N-terminal is projected extracellularly and the C-terminal is exposed to the cytosolic side. With the help of antibodies (labelled with fluorophors) against N-terminal and C-terminal peptides, orientation of glycophorin across membrane can be verified. Which one of the following statements is correct? (A) Intact RBC can be labelled with C-terminal antibody. (B) Permeabilized RBC can be labelled with C-terminal antibodies as well as N-terminal antiobodies. (C) Intact RBC canot be labelled with N-terminal antibodies. (D) Inside out ghost of RBC can be labelled with N- terminal antibodies. Exp. Intact RBC can be labelled with N terminal Antibodies an N terminal of Glycophorin facing outside Inside out ghost of RBC reverse the arrangement of glycophorin i.e. N terminal will become cytoplasm facing. So the best answer is B. 82. As topoisomerases ply an important role during replication, a large number of anticancer drugs have been developed that inhibit the activity of these enzymes. Which of the following statements is NOT true about topoismerases as a potential anticancer drug target? (A) As cancer cells are rapidly growing cells, they usually contain higher level of topoisomerases. (B) The transient DNA breaks created by topoisomerases are usually converted to permanent breaks in the genome in the presence of topoisomerase targeted drugs. Amar Ujala Education (C) As cancer cells often have impaired DNA repair pathways, they are moe susceptible towards toposiomerase targeted drugs. (D) The drugs which specifially traget topoisomerases. usually do not affect normal fast growing cells. Exp. Rest three options are correct for cancer cells. 83. Some errors occur during DNA replication that are not corrected by proof reading activity of DNA polymerase. These are corrected by specialized repair pathways. Defect in the activities of some of the following enzymes impair this process. (a) DNA polymerase III and DNA ligase (b) AP endonuclease and DNA glycosidase (c) Mut S and Mut L (d) RecA and RecF Defect in which of the above enzymes impair the process? (A) (a), (b) and (c) (B) (d) and (b) (C) (a) and (d) (D) (a) and (c) Exp. DNA Polymerase III and DNA Ligase are the most important enzymes, are required at the later stages of almost all repair system present in a. Similarly, AP endonuclease and DNA glycosylase are the important enzymes of excision repair system and MutS and MutL are the primary of mismatch repair system. So, if any of them is non-functional, the accuracy of the repair system would be compromised. All these repair systems. So, if any of them is non-functional, the accuracy of the repair system would be compromised. All these repair systems i.e., excision repair, mismatch repair, etc are needed to repair the damage, when the damage is not too extensive, for eg., due to lack of proof reading activity. Rec A is not included here as an option, because RecA is mainly an enzyme of SOS Repair system and SOS system is activated when DNA damage is extenstive and occurs due to a number of factors (which include lack of proof reading activity along with many other reasons.) 84. E. coli was grown in three different experimental conditions. In one, it was grown in medium containing glucose as carbon source; in the second in medium containing both glucose and galactose; and in third was infected with phage. Match the curves shown below to the treatment P2/13

78 Amar Ujala Education (A) a is grown in glucose; b is grown in glucose and galactose; c is infected with phage (B) a is grown in glucose and galactose; b in glucose; c is infected with phage (C) a is infected with phage; b is grown in glucose and galactose; c in glucose (D) a is infected with phage; b is grown in glucose; c in glucose and galactose. Exp. In presence of glucose, normal growth curve will be followed having all phases. In presence of glucose and galactose the bacteria will follow diauxic type growth pattern where firstly it will use glucose as a C source then it will switch towards galactose after a little half in the growth. 85. Each aminoacyl-trna synthetase is precisely able to match an amino acid with the trna containing the correct corresponding anticodon. Most organisms have 20 different trna synthetases, however some bacteria lack the synthetase for charging the trna for glutamine (trna Gin ) with its cognate amino acid. How do these bacteria manage to incoporate glutamine in their proteins? Choose the correct answer. (A) Glutamine is not present in the newly synthesized bacterial protein. Post translational modification converts glutamate to glutamine at the required sites. (B) In these bacteria, the aminoacy I trna synthetase pecific for trna glutamate (trna glu )alsocharges trna gin with glutamine. (C) In these bacteria, the aminoacyl trna synthetase specific for trna giu also charges trna gln with glutamate. A second enzyme then converts the glutamate of the charged trna gln to glutamine. (D) In these bacteria. the aminoacy I trna synthetase charges trna glu with either glutamate or glutamine according to their requirement during protein synthesis. Exp. It's not that bacteria do not contain glutamine, they do contain glutamine, but they do not contain amino acyl trna synthetase which charges the trna with glutamine. So, during translation, when trna reads the codon for glutamine on mrna, trna synthetase specific or glutamate charges trna with glutamate. After that trna Aminotransferase converts that glutamate to glutamine, by transferring amide group to glutamate., Same is the case for asparagine. 86. One of the cellular events that TOR, a kinase, positively regulates is the rate of rrna synthesis. TOR regulates the association of a transcription factor to a PolI subunit. When TOR is inhibited by the drug rapamycin, the transcription factor dissociates from Pol I.A yeast strain P2/14 A/UG-15 is engineered, which expresses a fusion of the transcription factor and the Pol I subunit. The level of rrna synthesis is monitored in these cells using pulse labelling following rapamycin addition for the times indicated below. the transcript profile of rrna observed for the wild type cells is given below : t(min): Identify the pattern expected in the engineered starin. (A) (B) (C) (D) Exp. Its only TOR that is senstiive to rapamycin, and once it is inhibited, the association between pol I and transcription factor is lost, But, if we create a fusion protein consisting of poll and transcription factor, the process of rrna synthesis will not be affected by rapamycin, because here TOR plays no role and rapamycin inhibits only TOR, it has no effect on Pol 1 or transcription factor. 87. Transposons can be primarily categorized into two types, DNA transposons and retrotrans-posons. Given below is some information regarding the above. (a) Eukaryotic DNA transposons excise themselves from on place in the genome and integrate into anothers site. (b) Retrotransposons are RNA sequences that are first reverse transcribed into cdna and then integrate into the genome. (c) Retrotransposons move by a copy and paste mechanism through an RNA intermediate. (d) As DNA transposons move via a cut and paste mechanism, there can never be an increase in the copy numeber of a transposon. Which of the Statements (s) is/are true? (A) a and c (B) b and d (C) b only (D) d only Exp. Retrotransposons move by copy paste mechanism by RNA intermediate and DNA transposons move through cut and paste mechanism.

79 A/UG An eukaryotic cell undergoing mrna synthesis and processing was incubated with 32 p labelled ATP, with the lable at the -position. Where do you think the radioactive isotope will appear in the mature mrna? (A) 32 P will not appear in the mature mrna under any circumstances because andy phosphates are released during trans-cription. (B) Phosphate groups of the phoshodiester backbone of the mrna will be uniformly labelled as only a phosphates are released during transcription. (C) 32 P will appear at the 5' end of the mrna if only it has "A" as the first nucleotide. (D) No 32 P will appear in the mature mrna because the 5'-terminal phosphate of an "A" residue will be further removed during the capping process. Exp. It is always the alpha phosphate of nucleotide that gets incorporated in the growing chain of mrna during the process of transcription and beta and gamma phosphate are always lost. Here, beta phosphate is labelled, since beta phosphate is always lost, radioactivity will not be incorporated into mrna. 89. Dose-dependence of retinoic acid treatment supports the notion that a gradient of retinoic acid can act as a morphogen along the proximo-distal axis in a developing limb. Following are certain facts related to the above notion. (a) Treatment with high level of retinoic acid causes a proximal blastema to be respecified as a distal blastema and only distal structures are regenerated. (b) Treatment with high level of retinoic acid causes a distal blastma to be respecified as a proximal blastema and regeneration of a full limb may be initiated. (c) Treatment with retinoic acid affects only distal blastemas and causes them to form only proximal structures. (d) Treatment with high level of retinoic acid causes any blastema to form only distal structures. Which one of the following is correct? (A) (b) and (d) (B) Only (c) (C) (a) and (c) (D) Only (b) Exp. During amphibian limb regeneration, retinoic acid reset the distal cell position to more proximal values resulting in proximalization of regeneration with additional humurus, radius and unla. 90. According to the ABC model of floral development in Arabidopsis as shown below, P2/15 Amar Ujala Education Several genes/transcription factors e.g. API, AP2, AP3, AG etc., are involved. Which one of the following statements is correct? (A) Apetala 2 (AP2) transcripts expressed during sepal and petal development. (B) Agamous AG is considered as class A gene. (C) AP1 expressed during carpel development. (D) AP3 expressed during sepal development. Exp. Class A genes (AP2) alone specify sepal development. Class A genes and class B genes (AP3 and Pl) together specify petals, Class B and class C (AG) genes are necessary for stamen formation, class C genes alone specify carpel formation. 91. A virus infects a particular cell type, integrates its genome into a site that contains a protooncogene, transforms the cell and increases the level of a protein 'X', which increases cellular proliferation. A compound 'P' is known to increase the level of tumor suppressor proteins in that cell type whereas a compound 'Q' helps in stmulating a protein 'Z' that can ind to 'X' rendring it inactive, Which one of the following graphs correctly represents the mode of action of 'P' and 'Q' (A) (B) (C) Exp. P is a tumor supprssor protein which will decrese the proliferation. In the same way Q is going to increase conc of Z which will bind with X (oncoprotein) and blocks its activity. So in presence of either P or Q % cell proliferation will be less but in absence of these two, X will be highly active and proferation will increase. And in presence of both the proliferation will be minimum. So best explanatory graph here is C. 92. In an experiment peritoneal macrophages were isolated from strain A of guinea pig. These cells were then incubated with an antigen. After the antigen pulsed macrophages processed the antigen andd presented it on their surface, these were mixed with T cells from (i) strain A or (ii) strain B (a different strain of guinea pig) or (iii) F1 progeny of strain AXB. T cell proliferation was measured in response to antigen pulsed macrophages. T cells of which strain of guinea pig will be activated? (A) Strain A only (B) Strain B only (C) Strain A and F1 progeny (D) Strain B and F1 progeny Exp. For both strain A and F1 progeny.

80 Amar Ujala Education 93. Cadherins mediate Ca2+ -dependent cell-cell adhesion and play an important role in embryonic development by changing the adhesive properties of cell. aggregation of nerve cells to form an epithelium is correlated with the appearance of N-cadherins on cell surface and vice versa. N-CAM (neural cell adhesion molecules) belongs to Ig-SF (immunoglobulin super family) and involved in fine tuning of adhesive interactions. In order to see the effect of mutations of N-cadherin and N-CAM, two sets of mice were senerated. Set A-mice with mutation in N-cadherin and set B-mice with mutation in N-CAM. Which of the following results is most likely to occur? (A) Mice of both set A and Set B will die in early development. (B) Mice of set A will die in early development but mice of set B will develop normally and show mild abnormalities in the development of nervous system. (C) Mice of Set A will show mild abnormalities in the development of nervous system whereas mice of set B will die early in development. (D) Mice of both set A and Set B develop normally as other cell adhesion molecules will compensate for the mutations. Exp. Set A mice deficient in N-Cadherin will die early in devlopment and Set B Mie deficient in N-CAM will show few abnormalities as other cell adhesion molecules will serve the same function. 94. If you remove a set of cells from an early embryo, you observe that the adult organism lacks the structure that would have been produced from those cells. Therefore, the organism seems to have undergone. (A) Autonomous specification. (B) Conditional specification. (C) Morphogenic specification. (D) Syncytial specification. Exp. If a particular blastomere is removed from an embryo early in its development, that isolated blastomere will produce the same cells that it would have made if it were still part of the embryo. Moreover, the embryo from which that cell is taken will lack those cells (and only those cells) that would have been produced by the missing blastomere. This type of specification. 95. Which one of the following combinations is the correct pairing of ligands with their receptors? (i) FGF (a) Patched (ii) Hedgehog (b) Frizzled (iii) Wnt (c) Receptor tyrosine kinase (A) i-c, ii-a, iii-b (B) i-a, ii-c, iii-b (C) i-b, ii-c, iii-a (D) i-c, ii-b, iii-a Exp. Want ligand-frizzled receptor, hedgehog-patched and or FGF-PTKs. P2/16 A/UG Match the two columns following asexual reproduction of plants and apomixes: (a) Agamospermy (i) No seed Formation (b) Clonal propagation (ii) Seed formation (c) Embryo sae formed (iii) Diplospory from nucellus or integument of the ovule (d) Gametophyte develops (iv) Apospory witout fertilization from unreduced megaspore (A) a-(i), b-(ii), c-(iii), d-(iv) (B) a-(ii), b-(iii), c-(iv), d-(i) (C) a-(ii), b-(i), c-(iii); d-(iv) (D) a-(ii), b-(i), c-(iv),d-(iii) Exp. In Diplospory gametophytes develops without fertilization from unreduced megaspore. Only option D is showing the correct match (D-iii). 97. Cancer is often belived to arise from stem cells rather than fully differentiated cells. Following are certain views related to the above statement. Which one of the following is NOT correct? (A) Stem Cells do not divide and therefore require fewer changes to become a cancer cell. (B) Cencer stem cells can self-renew as well as generate the non-stem cell populations of the tumor. (C) Teratocarcinomas prove tumors arise from stem cells without further mutations. (D) Stemness genes can often function as oncogenes. Exp. Cancer stem cells (CSCs) are cancer cells (found within toumors or haematological cancers) that possess characteristics associated with normal stem cells, specifically the ability to give rise to all cell types found in a particular cancer sample. CSCs are therefore tumorigenic (tumor-forming), perhaps in contrast to other non-tumorigenic cancer cells. CSCs may generate tumors through the stem cell processes of self-renewal and differentiation into multiple cell types. Such cells are hypothesized to persist in tumors as a distinct population and cause relapse and metastasis by giving rise to new tumors. 98. Given are certain facts which define determination' of a developing embryo. a. Cells have made a commitment to a diffecrentiation program. b. A phase where specifice biochemical actions. occur in embryonic cells. c. The sell cannot respond to differentiation signals. d. A phase where inductive signals trigger cell differentation. Which of the above statements best define determination? (A) b and d (B) a and c (C) Onlya (D) Only b

81 A/UG-15 Exp. The determination of different cell types (cell fates) involves progressive restrictions in their developmental potentials, When a cell chooses a particular fate. it is said to be determined, although it still "looks" just like its undetrmined neighbours. Determination implies a stable change-the fate of determined cells does not change. 99. What would happen as a result of a transplantation experiment in a chick embryo where the leg mesenchyme is placed directly beneath the wing apical ectodermal ridge (AER)? (A) Distal hindlimb structures develop at the end of the limb. (B) A complete hindlimb will form in the region where the forelimb should be. (C) The forelimb would form normally. (D) Neither a forelimb nor a hindlimb would form since the cells are already determined. Exp. If an extra AER is grafted onto an existing limb bud, supernumerarystructures areformed, usually toward the distal end of the limb Immunaoglobulins have therapeutic applications in cancer treatment, infection clearance and targeted drug delivery. For this reason, immunoglobulins are briefly cleaved by the enzyme pepsin. Following are some of the statement regarding the brief digestion of immunoglobulin by pepsin. (i) F (ab) 2 fragment is generated which retains the antigen binding activity. (ii) F (ab) fragment having antigen binding activity and the crystallisable Fc fragment are generated. (iii) The fragment generated on incubation with a proper antigen forms a visible precipitate. (iv) The fragment generated is incapable of forming a visible precipitate on incubation with a proper antigen. Which of the above statements are correct? (A) (i) and (ii) (B) (i) and (iii) (C) (i) and (iv) (D) (ii) and (iii) Exp. The correct option is D (i and iii) as pepsin digestion produces one F (ab)2 fragment and Fc fragment. The Fab is intact with its antigen binding ability whereas the Fc fragments are digested in so small fragment that are not discoverable A diabetic patient has a high blood glucose level due to reduced entry of glucose into various peripheral tissues in addition to other causes. There is no problem of glucose absorption. however. in the small intestine of these Amar Ujala Education patients. The following statements are put forward to explain this observation: (a) Glucose is transported into the cells of musceles by glucose transporters (GLUTs) which are influenced by insulin receptor activation. (b) Glucose transport into the enterocytes is mediatedby sodium-dependent glucose transporters (SGLTs) (c) which are not dependent on insulin. Glucose molecules are transported in the small intestine by facilitated diffusion. (d) The secondary active transport of glucose occurs in muscles. Which one of the above statement (s) is/are Incorrect? (A) Only (a) (B) (a) and (b) (C) Only (c) (D) (c) and (d) Exp. Here option C and D are false, because transport of glucose across small intestine is not by facilitated diffusion, but rather its active transport mediated by certain transporters. The transporter that carries glucose and galactose into the enterocyte is the sodium-dependent hexose transporter, known more formally as SGLUT-1. Also transport of glucose into muscles is not secondary active transport which needs energy other than ATP, but rather it is an active process mediated by carrier proteins, which use energy from ATP Following are certain statements that describe plantpathogen interactions : (a) Hemibiotorphic pathogens are characterized by initially keeping host cells alive followed by extensive tissue damage during the later part of the infection. (b) Effectors are molecules present in host plants that act against the pathogen attack. (c) Plants possess pattern recognition receptors (PRRs) that perceive microbe-associated molecular patterns (MAMPs) present in specifice class of microorganisms but are absent in the hosts. (d) Phyloalexin production is a common mechanism of resistance to pathogenic microbes in a wide range of plants. Which one of the following combinations is correct? (A) (a), (b) and (c) (B) (a), (c) and (d) (C) (b), (c) and (d) (D) (a), (b) and (d) Exp. Effectors are pathogen molecules (not plant molecules) which makes statement b absolutely false. According to given option B is the only one not containing statement b. P2/17

82 Amar Ujala Education 103. Several transport steps are involved in the movement of photosynthate from the chloroplates. Following are certain statements regarding the transport of photosynthate : (a) Pentose phosphate formed by photosynthesis during the day is transported from the chloroplast to the cytosol, where it is converted to sucrose. (b) Carbon stored as starch exits the chloroplast at night primarily in the form of maltose and is converted to sucrose in cytosol. (c) During short distance transport. sucrose moves from producing cells in the mesophyll to cells in the vicinity of the sieve elements in the smallest veins of the leaf. (d) In the process of pholoem loading, sugars are transported into phloem parenchyma cells. Which one of following combinations of above statements is correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (a) and (d) Exp. Pentose phosphate formed during day moves from chloroplasts to cytosol and is converted to sucrose. Starch in the form of maltose exits the chloroplast at right and is converted to sucrose in the cytosol Which one of the following options correctly relates the source gland/organ with its respective hormone as well as function? Source Hormone Function. gland (A) Thyroid Thyroxine Regulates blood calcium level (B) Anterior Oxytocin Contraction of pituitary uterine muscles (C) Posterior Vasopressin Resorption of Pituitary water in distal tubules of nephron. (D) Corpus Estrogen Supports luteum Pregnancy Exp. Vasopressin, also known as antidiuretic hormone (ADH), is a neurohypophysial hormone found in most mammals. Vasopressin regulates the body's retention of water by acting to increase water reabsorption in the kidney's collecting ducts, the tubules which receive the very dilute urine produced by the functional unit of the kidney, the nephrons. It is derived from a preprohormone precursor that is synthesized in the hypothalamus and stored in vesicles at the posterior pituitary. Function of the thyroid gland is to take iodine Oxytocin produced in posteriorpitutary the corpus luteum produces progesterone. A/UG The membrane potential in a giant squid axon recorded interacellularly at the resting condition ( 70 mv) was reversed at the peajk of action potential ( + 35 mv) after stimulation of the nerve fibre with a threshold electrical stimulus. This overshoot of the membrane potential has been explained in the following proposed statements: (a) TherapidincreaseinNa + -conductance during early phase of action potential causes membrane potntial to move toward the equilibrium potential of Na ( + 45 mv). (b) The Na + -conductance quickly decreases toward resting level after peak in the early phase and Na + - ions are not able to attain its equilibrium potential within this short time. (c) The conductance of K + at the carly phase of action potential is increased and that leads to the reversal of membrane potential. (d) The increase of K + - conductance due to stimulation of nerve occurs before the changes of Na + conductance is initiated and thus causes overshoot at the peak of action potential. Which one of the following is correct? (A) a only (B) a and b (C) c only (D) c and d 106. The following statements are made to describe auxin signal transduction pathway, from receptor binding to the physiological response : (a) Auxin response factors (ARFs) are nuclear proteins that bind to auxin response elements (Aux REs) to activate of repress gene transcription. (b) AUX/IAA proteins are secondary regulators of auxininduced gene expression. Binding of AUX/IAA proteins to the ARF protein blocks its transcription regulation. (c) A uxin binding to TIRI/AFB promotes ubiquitin mediated degradation and removel of AUX/IAA proteins. (d) Auxin biding to auxin respone factos (ARFs) causes their destruction by the 26S proteasome pathway. Which one of the foloowing combinations of above statements is correct? (A) a, b and c (B) a, c and d (C) b, c and d (D) a, b and d Exp. D is incorrect because neither Auxin binds ARF nor ARF gets degraded by protea -some. P2/18

83 A/UG A person showed the symptoms of diarrhea, gas and pain whenever milk was consumed. The doctor advised the person to take curd instead of milk and subsequently the symptoms mostly disappeared due to this change of dairy product. The following statements are proposed to explain this observation: a. The person has deficiency in the intestinal sucrasemaltase b. The person has deficiency in the inestinal lactase c. The person has deficiency in the intestinal lactase d. The bacteria in curd contain lactase Which one of the following is true? (A) a only (B) a and b (C) c only (D) c and d Exp. Pereson having lactose intolerance and curd bacteria also contain lactase Ribulose bisphosphate carboxylase (Rubisco) catalyzes both carboxylation and oxygenation of ribulose-1, 5- bisphosphate. The latter reaction initiates a physiological process known as 'photorespiration'. The following are certain statements on photorespiration: a. The active sites on Rubisco for carbo-xylation and oxygenation are different. b. One of the steps in photorespiration is conversion of glycine to serine. c. 50% of carbon lost in chlorplast due to oxygenation is recovered through photo-respiration. d. The pathway of photorespiration involves chloroplast, peroxisome and mitochondria. Which one of the following combinations of above statements in correct? (A) a and c (B) a and d (C) b and d (D) c and d Exp. Following mechanism are involved in photorespiration; 1. Conversion of 2 phosphoglycolate to glycine. 2. Decarboxylatin of two molecules of glycine to serine, carbon dioxide and ammonia 3. Serine is then converted into 3 phosphoglycerate, which re-enters calvin cycle. This pathway for pho torespiration involves chloroplast, peroxisomes and mitochondria A majority of humans with normal colour vision was found to be more sensitive to red light in Rayleigh match where the subject mixed variable amount of red and green light to match monochromatic orange. Which one of the following statements in NOT true to explain the observation? (A) There are variations in the sensitivity of long-wave cone pigments. (B) The short-wave cone opsin in red-sensitive subjects is different from other. P2/19 Amar Ujala Education (C) The absorption curve of long wave cone pigment peaks at 556 nm in red sensitive subjects while it peaks at 552 nm in others. (D) The long-wave cone opsin in redsensitive subjecfts is different in primary structure from that of others. Exp. Cones of human retina are responsible for vision in bright light and for color vision. Colour vision is due to the absorption of light by three classes of cone photoreceptors which are. The short wave sensitive (or blue), middle wave sensitive (green), and long wave sensitive in (red). In the question, it is asked to find out the possible reason for red light sensitivity in majority of humans with normal color vision and the photoreceptors for red light vision are long wave sensitive and not short wave sensitivity (which is given in option B). So, option B is not true to explain the observation Constitutive photomorphogensesis (COPI) protein, an E3 ubiquitin ligase, regulates the turnover of proteins required for photo-morphogenic development. Following are certain independet statements related to the function of COPI protein : a. In light, COPI along with SPAI adds ubiquitin tags to a subset of nuclear proteins. b. The proteins ubiquinated by COPI and SPAI are targeted for degradation by the 26S proteasome. c. In dark COPI is slowly exported to the cytosol from nucleus. d. The absence of COPI in the nucleus permits the accumulation of transcriptional activators necessary for photomorphogenic development. Which one of the following combinations is correct? (A) a and c (B) a and d (C) b and c (D) b and d Exp. COP1 is a protein that inhibits photo morphogenesis. In the dark, COP1 is present in the nucleus, but in the light is is only found in the cytoplasm. COP1 contains a ringginger, which is a feature of many E3 ubiquitin ligases, which are involved in targeting proteins for 26S proteosome-mediated degradation. COP1 interacts with HY5, and in the dark HY5 protein levels show a decline that is dependent on COP1 and CNS. Thus it is hypothesized that in the dark, COP1 targets HY5 for degradationbycns.asmentioned,cry1and2are constitutively nuclear and also interact with COP1. It is hypothesized that CRY binding inhibits COP1 activity. Another factor that functions in phytochrome signal transduction, SPA1, also interacts with COP1. Thus all photo morphogenetic signal transduction pathways appear to converge on COP1. Inhibition of COP1 may then allow accumulation of HY5 and the response to light.

84 Amar Ujala Education 111. Action potentials were recorded intracellularly from different parts of mammalian heart and these are shown below. Which one of these has been recorded from sinoatrial node? (A) (C) (B) (D) 112. Light reactions of photosynthesis are carried out by four major protein complexes: Photosystem I (PSI), photosystem II (PSII), the cytochrome b6f complex and ATP synthase. the following are certain statements on PSI: a. PSI reaction centre and PSII reaction centre are uniformly distributed in the granal lamellage and stromal lamellae. b. The electron donor for the P700 of PSI is plastocyanin and electron acceptor of P700* is a chlorophyll known as A 0. c. The core antenna and P700 are bound to two key proteins PsaA and PsaB. d. Cyclic electron flow occurs from the reducing side of PSI via plastohydro-quinone and B 6 f complex. This supports ATP synthesis bnut does not reduce NADP + Which one of the following combinations of the above statements is correct? (A) a,bandc (B) a,candd (C) a,b andd (D) b,c andd Exp. The correct option is D (B C D). Photosystem II is located predominantly in the stacked regions of the thylakoid membrane; photosystem I and ATP synthase are found in the unstacked regions protruding into the stroma. Other options are incorrect Fruit colour of wild Solanum nigrum is controlled by two alleles of a gene (A and a). The frequency of A, P = 0.8 and a, Q = 0.2. In a neighbouring field a tetraploid genotype of S. nigrum was found. After critical examination five distinct genotypes were found; which are AAAA, AAAa, AAaa, Aaaa and aaaa. Following Hardy Weinberg principle and assuming the same allele A/UG-15 fequency as that of diploid poulation, the numbers of phenotypes calculated within a population of 1000 plants are close to one of the following: AAAA: AAAa: AAaa: Aaaa: aaaa (A) 409:409: 154: 26:2 (B) 420:420: 140: 18:2 (C) 409: 409: 144: 36:2 (D) 409: 420: 144: 25: 2 Exp. In case of tetaploid species, the Hardy weinberg law states that the total genotype frequency for such population will be (p + q)^ 4 = 1. The frequencey of individual genotypes can be calculated by expanding (p + q)^ 4, with is equal to p^4+4p^3q +6p^2+4pq^3+q^4. p = 0.8 and q = 0.2 (given). Total population 1000 (given). # of AAAA genotype is this population = p^4 * 1000 = 0.8 * 0.8 * o.8 * 1000 = 409. # of AAAa = 4p^3q ^1000 = 4 * 0.8 * 0.8 * 0.8 * 0.2 * 1000 = 409. # of AAaa = 6p^2q^2 * 1000 =6 * 0.8 * 0.8 * 0.2 * 0.2 * 1000 = 154. # of Aaaa = 4pq^3 * 1000 = 4 * 0.8 * 0.8 * 0.2 * 0.2 * 0.2 * 1000 = 26. # of aaaa =q^4 *1000 = 0.2 * 0.2 * 0.2 * 0.2 *1000 = 2. So, AAAA: AAAa: AAaa: Aaaa: aaaa = 409 : 409 : 154 : 26 : 2. Answer will be option A Two interacting genes (independently assorting) were involoved in the same pathway. Absence of either genes function leads to absence of the end product of the pathway. A dihybrid cross involving the two genes is carried out. What faction of the F 2 progeny will show the presence of the end product? (A) 1/4 (B) 3/4 (C) 9/16 (D) 15/16 Exp. The correct option is C. It is case of complementary gene interation where at least single copy of dominant allele of both gene is required to give the product. Whereas absence of any of the gene will give same phenotype only. The ratio obtained is 9:7 where 9 out of 16 shows the phenotype Five bacterial markers were followed for a cotransduction experiment. The following table documents the observations of this experiment. '+' denotes cotransduction and '-' dentoes tack thereof: 'ND' stands for not determined. Pick the correct order in which the genes are arranged on the bacterial chromosome. (A) Str-gal-leu-arg-met (B) leu-met-arg-str-gal (C) leu-str-met-gal-arg (D) arg-gal-str-leu-met P2/20

85 A/UG-15 Exp. Closer the genes are on the chromosome, higher is their cotransduction frequency. As the distance between the genes increases, their respective cotransduction frequency decreases. In the data displayed in the table, gal cotransduces with arg and str, but arg doesn't cotransduce with str. These two observations can be true only if gal is present in between arg and str. So, their sequence will be arg- gal-str,. Now, leu cotransduces with str and met, so leu should be present between arg and met. Hence, the sequence of the 5 genes will be arg-galdtr-leu-met. And this sequence satisfies our last row of table as well, i.e., doesn't contranduce with arg, but str cotra sduces with leu (as they are present next to eaxch other). So, answer will be option D A three point test cross was carried out in Drosophila melanogaster involving theree adjecent genes. X, Y and Z, arranged in the same order. The distance between X to Y is 32.5 map unit (mu) and that betwen X to Y is 20.5 map. The coefficient of coincidence= What is the percentage of double recombinants in the progeny obtained from the testcross? (A) ~6% (B) ~8% (C) ~12% (D) ~16% Exp. Coefficient of coincidence = observed double crossovers (?)/expected double crossovers (Edco) expected double crossovers (Edco) = [recombination frequency in region 1 (map units / 100] [recombination frequency in region 2 (map units / 100] i.e., Edco = 32.5/ /100 = Coefficient of coincidence = (given). So observed double crossovers = Coefficient of coincidence expected double crossovers = = Or 5.92% or ~ 6% Poplar is a dioecious plant. A wild plant with 3 genes AABBCC was crossed with a triple recessive mutant aabbcc. The F1 male hybrid (AaBbCc) was then back crossed with the triple mutant and the phenotypes recorded are as follows: AaBbCc 300 aabbcc 100 aabbcc 16 AabbCc 14 AaBbcc 65 aabbcc 75 aabbcc 310 Aabbcc 120 The distance in map unit (mu) between A to B and B to C is (A) 25 and 17 mu, respectively (B) 33 and 14 mu, respectively (C) 25 and 14 mu. respectively (D) 33 and 17 mu, respectively P2/21 Amar Ujala Education Exp. Ab + ab = ( ) recombinants /total offsprings = 250 recom/1000 total = 0.25 A-B = 25 mu Bc + bc = ( ) recombinants/ Total offsprings = 170 recom/1000 total = 0.17 B-C = 17 mu A male mouse cell line has a large translocation from X chromosome into chromosome 1. When a GFP containing transgene is inserted in this chromosome 1 with translocation, it is often silenced. Howver when inserted in the other homologue of chromosome I that does not contain the translocation, it is almost alway expressed. Which of the following phenomenon best describes this effect? (A) Genome imprinting (B) Gene balance (C) Sex-specific expression (D) Dosage compensation. Exp. Dosage Compensation is a term that describes the processes by which organisms equalize the expression of genes between members of different biological sexes. Because sex chromosomes contain different numbers of genes, different species of organisms have developed different mechanisms to cope with this inequality. Replicating the actual gene is impossible, thus organisms instead equalize the expression from each gene Match the following human diseases with their causal organisms (a) Sleeping Sickness (i) Trypanosoma cruzi (b) Chagas disease (ii) Trypanosoma brucei (c) Elephantiasis (iii) Borrelia burgdorfei (d) Lyme disease (iv) Wuchereria bancrofti (A) a-(ii); b-(iv); c-(iii); d-(i) (B) a-(i); b-(ii); c-(iv); d-(iii) (C) a-(ii); b-(i); c-(iv); d-(iii) (D) a-(ii); b-(iv); c-(i); d-(iii) Exp. Trypanosoma cruzi-chagas disease Trypanosoma brucei- Sleeping Sickness wuchereria bancrofti-elephantiasis Borrelia burgdorferi-lymes Disease With reference to the phylogenetic tree presented above, which of the following statements is true?

86 Amar Ujala Education (A) Amphibians, reptiles, birds and mammals share common ancestor. (B) Birds are more closel related to reptiles than to mammals. (C) Cartilagenous fishes are the ancestore of amphibians. (D) Lampreys and mammals are not related. Exp. All the other options are wrong except A. B is wrong because birds are in fact closer to mammals than to reptiles. C, is wrong because cartilaginous fishes are not the ancestors of amphibians, they share a common ancestor with the ancestors of amphibians (the bony fishes) D. Lampreys and mammals are related as they are part of the same phylogenetic tree. Thus, the option D is wrong too Match major events in the history of life with Earth's geological period. Event Geological Period (a) First reptiles (i) Quarternary (b) First mammals (ii) Tertiary (c) First humans (iii) Cretaceous (d) First amphibians (iv) Triassic (v) Carboniferous (vi) Devonian (A) a-(v); b-(i); c-(ii); d-(v) (B) a-(v); b-(iv); c-(i); d-(vi) (C) a-(vi); b-(iv); c-(ii); d-(vi) (D) a-(iii); b-(i); c-(vi); d-(v) Exp. First Reptiles- Carboniferous First Mammals Triassic First Humans Quaternary First Amphibians- Devonian The approximate P:B (Net primary Production: Biomass) ratios in four different ecosystmes (A, B, C, D) are A-0-29; B-0.042, C-16.48; D-82 The four ecosystms are (A) A-Ocean: B-Lake; C-Grassland; D-Tropical forest. (B) A-Grassland; B-Tropical forest; C-Ocean; D-lake (C) A-tropical forest; B-Ocean; C-Grassland; D-lake (D) A-Grassland; B-Ocean; C-Lake; D-Tropical forest. Exp. The correct option will be B. The ratio for tropical rain forest will be lowest because most of the plant biomass is accumulated in non-photosynthetic tissue. And for ocean the net productivity is highest. For grassland it higher as they do not accumulate much biomass as a result of herbivory If gypsy moth egg density is 160 at time t and 200 at t + 1, what will be its value at time t +3, assuming that egg density continues to increase at constant rate? (A) 250 (B) 280 (C) 312 (D) 390 Exp. To find the net growth rate (R), we determine the ratio of N (t + 1): N (t) = 200 ; 160 = 1.25 now to find out egg densityatn(t+3)=n(t)*r^3 = 160* (1.25)^ 3 = Hence egg density at N(t + 3) = 312. A/UG Based on the table given below, which of the following option represents the correct match? Category Planty Species (a) Critically (i) Chromolaena endangered odorata (b) Vulnerable (ii) Dipterocarpus grandiflorus (c) Extinct (iii) Euphorbia mayuranthanii (d) Invasive (iv) Sarace asoka (A) a-(i); b-(iv); c-(iii); d-(ii) (B) a-(ii); b-(iii); c-(iv); d-(i) (C) a-(i); b-(iv); c-(ii); d-(iii) (D) a-(ii); b-(iv); c-(iii); d-(i) Exp. Chromolaena odorata is an invasive species and Saraca asoca is a species with vulnerable status. D is the only option that satisfies both the conditions For the following invertabrate structures/ organs, identify their major function and the animal group in which they are found : Nematocyst (A), Protonephridia (B), Malpighian Tubules (C) Radula (D) (A) A-Porifera, Skeletal Support; B-Mollusca, excretion C-Inscta, respiration; D-Anthozoa, prey capture (B) A-Anthozoa, prey capture; (B-planaria, excretion C- Mollusca, excretion; D-Insecta, food processing (C) A-planaria, excretion; B-Mollusca, respiration; C- Insecta, respiration; D-porifera, prey capture (D) A-Anthozoa, prey capture; B-Planaria, excretion; C- insecta, excretion; D-Mollusca, food processing Exp. The Malpighian tubule system is a type of excretory and osmoregulatory system found in some insects, myriapods, arachnids, and tardigrades. The radula (plural radulae or radulas) is an anatomical structure that is used by mollusocs for feeding. sometimes compared to a tongue both these conditions are satisfied only in option D Following is a cladogram showing phylogenetic relationships among a group of plants: In the above representation, A, B, C, and D respectively represent (A) xylem and pholem embryo flower. seed. (B) embryo, xylem and phloem, seed, flowres. (C) embryo, xylem and phloem, flower. seed. (D) xylem and phloem, flower embryo, seed. P2/22

87 A/UG-15 Exp. Plants are calssified into two main groups: the bryophytes (nonvascular plants) and the tracheophytes (vascular plants). Both groups have multicellular embryos. The bryophytes consist of the liverworts, hornworts, and mosses All other plants, including the ferns, gymnosperms, and angiosperms, are calssified as tracheophytes. These possess specialized vascular tissues phloem and xylem to transport sugars, water and minerals throughout their bodies. The oldest true gymnosperms, which produce seeds rather than spores, first appeared about 35 milionyears ago. The evolution of seeds, with their hard, resilient coats, was almost certainly a key factor in the success of the group. Pines are gymnosperms A particular behavioural variant affects fitness of an organism. The relationship between the frequency of the variant in the population and fitness are plotted below. In which of these cases is the behavioural variant most likely to reach a frequency of I? (A) (C) (B) (D) (A) Only b (B) Only C (C) b and d (D) a and d Exp. In option A fitness is not affected by frequency of the behavioural variant. In option C (both b and d) again as b is present this option cannot be right. In option D (both a & d) here in 'a' fitness decreases with increase in frequency of the behavioural variant. So behavioural variant can never reach the frequency of 1. The only option that seems most likely is option B Match the following associations involved in dinitrogen fixation with their representative genera Associations Genera a. Heterotrophic (i) Azotobacter Nodulate b. Heterotrophic (ii) Frankia Non-nodulate c. Phototrophic (iii) Nostoc associative d. Phototrophic (iv) Rhodospirillum free-living Amar Ujala Education (A) A-(ii); B-(i); C-(iv); D-(iii) (B) A-(iii); B-(i); C-(iii); D-(iv) (C) A-(i); B-(ii); C-(iii); D-(iv) (D) A-(ii); B-(i); C-(iii); D-(iv) Exp. Refer Table No 9.1 Organism and association involved in di nitrogen fixation The coefficient of relatedness between individuals A and B, A and D, and between D and C is (A) 0.5, 0.25, respectively. (B) 0.5, 0.5, 0.25 respectively. (C) 0.5, 0.25, 0.75 respectively (D) 0.125, 0.5, 0.5 respectively. Exp. The correct option is A, And B are father daughter.. so related to 50%, A and D are grandfather daughter so related with = 0.25., finally D and C have a aunt and neice related ness so related with = The birth rates (b) and death rates (d) of two species land 2 in relation to population density (N) are shown in the graph. Which of the following in NOT true about the density dependent effects on birth rates and death rates? (A) Birth rates are density-dependent in species I and density independent in species 2. (B) Death rates are density-dependent in both the species. (C) Density-dependent effect on birth rate is stronger in species I than in species 2. (D) The density dependent effects on death rates are similar in both the species. P2/23

88 Amar Ujala Education Exp. There are many kinds of densify-dependent processes, which can be differentiated according to effect, cause, response, meachanism and severity, The negative effects of density dependence are usually better known than the positive ones. Negative density-dependent effects occur if a vital rate decreases as density increases, positive effects occur if both increase. None of this is seen in birth rates of species 1. Thus we can say that birth rate of species one is density independent and death rates for species 1 and 2 are dependent on density. And birth rate for species 2 is dependent on density too. But the density dependent effects are not similar for both species which is visible from the figure With reference to the graph given below identify the optimal territory size. A/UG-15 (B) Species A shows random distribution, species B shows uniform distribution and species C shows clumped distribution. (C) Species A and B show clumped distri-bution, whereas species C shows uniform distribution. (D) Species A shows clumped distribution species B shows random distribution, and species C shows uniform distribution. Exp. Clumped: If mean < variance observed distribution is tending toward clumped Random; If mean = variance, observed distribution matches the distribution expected under the Poisson probability density function (PDF) Uniform : If mean > variance observed distribution is tending toward uniform In a lake subjected to progressive eutro-phication, temporal changes in the magnitude of selected parmaters (A, B, C, D) are shown in the graph. (A) A (B) B (C) C (D) D Exp. An optimal territory size exists where the net benefit, the difference between benefit gained and cost of defence, is greatest Consider an autosomal locus with two alleles A 1 and A 2 at frequencies of 0.6 and 0.4 respectively. Each generation. A 1 mutates to A 2 at a rate of μ=1x10-5 while A 2 mutatestoa1atarateof=2x10-5. Assume that the population is infinitely large and no other evolutionary force is acting. The equilibrium frequency of allele A 1 is (A) 1.0. (B) 0.5. (C) (D) Exp. Rate for forward mutation = 1 10^ 5. Rate of reverse mutation=2 10^ 5. Equilibrium frequency of A 1 allele = rate of reverse mutation/rate of (forward + reverse mutation). Equilibrium frequency of A 1 = 2 10^-5/ [ 2 10^-5) + (1 10^-5)] = 2/3 = The following table shows the mean and variance of population densities of species A, B and C. Statistic Spec- Spec- Spec. ies A ies B ciesc Mean X Variances Based on the above, which ofthe following statements is correct? (A) Species A and B show uniform distribution, whereas species C shows clumped distribution. P2/24 The parmeters A, B, C, D are : (A) A-Green algal biomass, B-Cyanobacterial biomass, C-Dissolved Oxygen concen-tration, D-Biological Oxygen Demand (B) A-Biological Oxygen Demand, B-Cyanobacterial biomass, C-Dissoloved Oxygen concentration, D- Gren algal biomass (C) A-Biological Oxygen demand, B-Green algal biomass, C-Cyanobacterial biomass, D-Dissolved Oxygen concentration (D) A-Cyanobacterial biomass, B-Biological Oxygen Demand, C-Green algal biomass, D-Dissolved Oxygen concentration. Exp. Algal blooms are one of the more insidious consequences of eutrophication. In addition to being unsightly and smelly, masses of blue-green algae can literally choke the life out of a lake or pond by depriving it of much needed oxygen. At first glance this may seem like something of a paradox- since blue-green algae undergoes photosynthesis, it should produce more oxygen than it consumes. However, after large concentrations of algae have built up, aerobic processes such as respiration and the decomposition of dead algal cells becomes increasingly significant. Under extreme conditions a eutrophic lake of pond may be left entirely devoid of fish. The overall process will increase the BOD of the lake and DO will be decreased.

89 A/UG For two species A and B in competions, the carrying capacities and competittion coefficients are K A =150 K B =200 =1.0 =1.3 According to the lotka-volterra model of interspecific competition, the outcome of competition will be (A) Species A wins. (B) Species B wins. (C) Both species reach a stable equilibrium. (D) Both species reach an unstable equilibrium Performance of biosensor is evaluated by their response to the presence of an analyte. The physiological relevant concentration of analyte is between 10μM and 50μm. Which among the following biosensor responsesis best? (A) (C) (B) (D) Exp. At low concentration of analyse, the response is less but as soon as the concentration reaches min threshold value the response also increases It is hypothesized that the mean (μo) dry weight of a female in a Drosophila population is 4.5 mg In a sample of 16 female with Y=4.8 mg and s=0.8 mg, what dry weight values would lead to rejection of the null hypothesis at p=0.05 level? (take to 0.05 =2.1). (A) Values lower than 4.0 and values higher than 5.6 (B) Values lower than 3.20 and values higher than 6.40 (C) Values lower than 4.38 and values higher than 5.22 (D) Values lower than 3.22 and values higher than From statements on protein structure and interactions detailed below. indicate the correct statment. (A) The concentration of a tryptophan containing protein can be determined by monitoring the fluoresence spectrum of the protein. (B) A peptide with equal number of Glu and lys amino acids can show multiple charged species in its electrospray ionization mass spectrum. (C) The circular dichroism spectrum of a protein shows predominantly helical conformation. Analysis of its two dimensional NMR spectrum shows predominantly -Structure. P2/25 Amar Ujala Education (D) Binding constant can be determined by two interacting molecules by the technique of surface plasma resonance only if there is strong hydrophobic interactions betwen them. Exp. Statemetnt B, C and D are not correct because electron spray ionization mainly involves generation of positively and negatively charge ions by bombarding neutral molecules with electrons. But if the sample is something which is having equal number of positively and negatively charge amino acids, then there are chances that when bombarded with electrons, positively charged residues absorb electrons and become neutral and negatively charged residues donate electrons and become neutral and neutral molecules are of no use in electron spray ionization. Also circular dichroism spectroscopy is mainly involved in the determination of chirality of macromolecules and Two-dimensional NMR gives information about the total structure of the protein, not predominantly about beta structure. Surface Plasmon resonance is a technique, that mainly gives information about binding interactions between different molecules for eg.., receptor-ligand interactions and the presence of hydrophobic interaction is not a pre-requisite for two interacting molecules to be determined by surface Plasmon resonance. Though statement 1, in itself is a correct statement, i.e., the concentration of a protein can be determined by monitoring the fluorescence due to presence of tryptophan, but this statement does not correlate with the question where it is asked that which technique helps to determine structure of protein, fluorescence plays a role in quantification of protein but not the structure. So, here, none of the options correlate with the question, but out of all the options, option A seemtobemoreappropriate In an effort to produce gene knockout mice, a gene targeted homologous recombination was tried with the exogenous DNA containing neor gene (confer G-418 resistance) and tk Hsv gene (confers sensitivity to the cytotoxic nucleotied analog ganciclover) if the neor gene was inserted within the target gene in the exogenous DNA and considering that bot homologous and non&homologous recombination (random integration) is taking place, which one of the following statements is NOT correct about the possible outcome of the experiment? (A) Cells with non-homologous insertion will be sensitive to ganciclovir. (B) Non-recombinant cells will be sensitive towards G- 418 and resistant to ganciclovir. (C) Homologous recombination will ensure that cells will be resistant to both ganciclovir and G-418 (D) Homologous recombinants will grow in G-418 containing media but will be senstive towards ganciclovir.

90 Amar Ujala Education Exp. To produce a knock out mice the neo^r andtk^hsv are introduced in a way that target gene is interrupted by the neo^r. the chance of getting gene targeted insertion is less (homologous) as compared to random insertion (nonhomologous), neomycin will kill all non-recombinant cell, gancyclovir will kill non homologous with tk^hsv, whereas only homologous recombinants will survive the selection process Radioimmuno assay (RIA) can be employed for the detection of insulin in blood plasma. For this, 125 I- labelled insulin is mixed and allowed to bind with known concentration of antiinsulin antibody. A known volume of patients blood plasma is then added to the conjugate and allowed to compete with the antigen binding sites of antibody. The bound antigen is then separated from unbound ones and the radiocativity of free antigen is then measured by gamma counter. Following are some of the statements made aobut this assay. (i) The ratio of radioactive count for unbound antigen to the bound one is more at the end of reaction. (ii) The ratio of radioactive count for unbound antigen to the bound one is less at the end of reaction. (iii) For a diabetic patient, the radioactive count for free antigen is less than that for a normal individual. (iv) For a diabetic patient, the radioactive count for free antigen is more than that for a normal individual. Which of the abnove statements are true? (A) (i) and (iii) (B) (i) and (iv) (C) (ii) and (iii) (D) (ii) and (iv) Exp. Diabetic patients have low or nil amount of insulin in their blood so when the patient serum will be added in the reaction, very less bound labelled insulin will be displaced. So the concentration of free/unbound insulin will be less in comparison to the bound one A researcher wants to obtain complete chemical information; i.e.; head groups and fatty acids of phospholipids from liver tissues. Phospholipids have fatty acids of different lengths and unsaturation and also the head groups are of different chemistries. Which of the following combination of techniques would provide complete chemical description of phos-pholipids? (A) Only thin layer chromatography (TLC) (B) TLC and gas chromatography (C) Paper and thin layer chromatography (D) Only paper chromatography Exp. After separating the different phospholipids by thin layer chromatography. The composition of fatty acids can be determined by subjecting them to Gas chromatography, where fatty acids are derivatized to convert them into volatile molecules before carrying out their separation In a typical gene cloning experiment, by mistakes a researcher intoduced the DNA of interest within ampicilin A/UG-15 resistant gene instead of lac z gene. The competent cells were allowed to take up the plasmid and thn plated in he media containing ampicilin, X-gal and IPTG and subjected to blue whit screening. Considering all plasmids were recombinant which one of the following statement correctly describes the outcome of the experiment? (A) The bacteria which took up the plasmids would grow and give blue colonies. (B) The bacteria which took up the plasmids would not grow. (C) The bacteria which took up the plasmids would form white colonies. (D) All of the bacteria would grow and give white colonies. Exp. Cells which took up the plasmid will be ampicillin sensitive One hundred independent populations of Drosophila are established with 10 individuls in each population, of which one individual is in each population, of which one individual is of Aa genotype and the other nine are of AA genotype. If random genetic drift is the only mechanism acting onn these populations, then after a large number of generations the expected number of populations fixed for the 'A" allele is (A) 75 (B) 50 (C) 25 (D) The sequence of the peptide KGLITRTGLKR can be unequivocally determined by. (A) Only Edman degradation. (B) Amino acid analysis and MALDI MS/MS mass spectrometry. (C) MALDI MS/MS mass spectrometry. (D) MALDI mass spectrometry after treamment of the peptide with trypsin. Exp. If the identity of the protein is desired, usually the method of in-gel digestion is applied, where the protein spot of interest is excised, and digested proteolytically. The peptide masses resulting from the digestion can be determined by mass spectrometry using peptide mass fingerprinting Molecular polymorphic markers are already known with respect to tobacco mosaic virus (TMV) resistance in tobacco. Among these, which marker system you will select that will be simple, economic and less time consuming: (A) RAPD (B) RFLP (C) AFLP (D) EST-SSR P2/26

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92 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, December-2015 PART-A 1. A shopkeeper purchases a product for `100 and sells it making a profit 10%. The customer resells it to the same shopkeeper incurring a loss of 10%. In these dealings the shopkeeper makes (A) no profit,no loss (B) ` 11 (C) ` 1 (D) ` 20 Exp. SP = 10% = 110, Reselling price = % 110 = =99 Hence shopkeeper gets = 11/- extra 2. A vessel is partially filled with water. More water is added to it at a rate directly proportional to time ie.., dv t dt. Which of the following graphs depicts correctly the variation of total volume V of water with time t? (A) (C) v 0 t v 0 t (B) (D) Exp. Since, vessel is already filled with some volme of water (hence the graph will not start with Origin, thus neglecting option 1 & 4) & its under filling process by a constant rate (Hence option 2 is also neglected) 3. The triangle formed by the linesy y = x, y =1 xand x = 0 in a two dimensional plane is (x and y axes have the same scale ) (A) isosceles and right-angled v 0 v 0 t t (B) isosceles but not right-angled (C) right-angled but not isosceles (D) neither isosceles nor right-angled Exp. since, x =0, hence its a straight line at Y axis, (hence its a right angled triangle) Line 1, y = x, Line 2, y =1 x, Equating line1 & 2 we get, x =1 x, => x =1/2 =0.5, putting this value in line 2 equation, we get y =0.5, Hence it is isoscales with 2 sides of same dimension. 4. Statement A : The following statement is true. Statement B : The preceding statement is false. (A) Statement A and B are always true (B) Statement A and B can be true if there is at least one statement between A and B (C) Statements A and B can be true if there are at least two statements between A and B (D) Statements A and B can never be true, indipendently 5. The number of squares in the above figure is (A) 30 (B) 29 (C) 25 (D) 20 Exp. There are 20 small sqares & 10 big squares made by 4 small squares each P3/1

93 Amar Ujala Education 6. A person walks downhill at 10 km/h, uphill at 6 km/h and on the plane at 7.5 km/h. If the person takes 3 hours to go from a place A to another place B, and I hour on the way back, the distance between A and B is (A) 15 km (B) 23.5 km (C) 16 km (D) Given data is insufficient to calculate the distance Exp. Data is insufficient because by any means the person can't walk all these three distances given in the answers in one hour. 7. A bird leaves its nest and flies away. Its distance x from the nest is plotted as a function of time t. Which of the following plots cannot be right? (A) x nest t (B) x nest x x (C) (D) nest t nest t Exp. time can never be decreased. 8. A car is moving at 60 km/h. The instantan-eous velocity of the upper most points of its wheels is (A) 60 km/h forward (B) 120 km/h forward (C) 60 km/h backwar (D) 120 km/h backward Exp. At the top of the wheel velocity will be equal to 2v means 120 km/hr. At the bottom of the wheel velocity will be -v at the top it will be v so add these vector quantities it will be equal to 2v. 9. A living cell has a protoplasm which is water based and demarcated by a lipid bilayer membrane. If a cell is pierced up to 1 th of its diameter with a very sharp 5 needle, after taking the needle out (A) no effect will be observed (B) protoplasm will leak out from the hole made by the needle for a few minutes until the cell heals the wound (C) protoplasm will keep on leaking out till the cell is dead (D) the cell will burst like a balloon Exp. No effect will be observed, as soon as needle will come out the cell lipid bilayer will be formed again. t A/UG If D + I + M = 1501 C+I+V+I+L = 157 L+I+V+I+D = 557 C + I + I + C = 207 What is V + I + M? (A) Cannot be found (B) 1009 (C) 1006 (D) Density of a rice grain is 1.5 g/cc and bulk density of ricc heap is 0.80 g/cc. If a l litre container is completely filled with rice, what will be the approximate volume of pore space in the container? (A) 350 cc (B) 465 cc (C) 550 cc Exp. porosity (D) 665 cc = 1 bulk density/density = cc 12. Four circles of unit radius each are drawn such that each one touches two others and their centres lie on the vertices of a square. The area of the region enclosed between the circles is (A) 1 (B) 2 (C) 3 (D) 4 Exp. Area of square encolsed by each circle = (1/4) r 2 (since 1/4th area of circle is overlapping with sqaure), Since there are 4 squares, hence area of square under all 4 circles =4 (1/4) r 2 = r 2 Since r = 1 unit, hence area under 4 circles = Total area of square = (side) 2 =2 2 = 4, hence left out area =4 13. A turtle starts swimming from a point A located on the circumference of a circular pond. After swimming for 4 meters in a straight line it hits point B on the circumference and swims for 3 meters in a straight line and arrives at point D diametrically opposite to point A. How far is point D from A? (A) 3 m (B) 4 m (C) 7 m (D) 5 m Exp. It will make triangle of height 3 meter, base 4 meter and hypotenuse 5 m. P3/2

94 A/UG A film projector and microscope give equal magnification. But a film projector is not used to see living cells because (A) a living cell cannot be placed in a film projector (B) the viewer s eye is close to a microscope whereas it is far away from the projector s screen (C) a microscope produces a virtual image whereas a projector produces a real image (D) a microscope produces has greater resolving power than a projector Exp. A projector can have the same magnification but in any case resolution or the ability to differentiate between closly placed objects will always be higher in case of microscope. So thats why all other options are wrong, only fourth one is correct. 15. In each of the following groups of words is a hidden number, based on which you should arrange them in descending order.pick the correct answer (E) Papers I Xeroxed (F) Wi -Fi veteran (G) Yourself ourselves (H) Breaks even (A) H,F,G,H (B) E,G,F,H (C) H,F,G,E (D) H,E,F,G Exp. E. Six F. Five G. Four H. Seven Descending order gives H, E, F, G 16. Five congruent rectangles are drawn inside a big rectangle of perimeter 165 as shown. What is the perimeter of one of the five rectangles? (A) 37 (B) 75 (C) 15 (D) 165 Exp. Let l represent the length of one of the smaller rectangles, and let w represent the width of one of the smaller rectangles, with w < l. From the large rectangle, we see that the top has length 3w, the right has length l + w, the bottom has length 2l, and the left has length l + 2. Since the perimeter of the large rectangle is 165, we know that, 165 = 3w + l + w + 2l + l + w or 165 = 5w + 4l. From the top and bottom of the large rectangle, we know that 3w = 2l, or l=1.5 w. Plugging that into the first equation, we get 165 = 5w + 4 (1.5) w, 165 = 11w, w =15, L = 1.5 w = 22.5, P = 2*15 + 2*22.5, and the answer is B 17. At one instant, the hour hand and the minute hand of a clock are one over the other in between the markings for 5 and 6 on the dial. At this instant, the tip of the minute hand Amar Ujala Education (A) is closer to the marking for 6 (B) is equidistant from the markings for 5 and 6 (C) is closer to marking for 5 (D) is equidistant from the markings for 11 and 12 Exp. When the time and minute hand will be on the top of each other between 5 and 6, the time will be in between 5:25 to 5:27. So in that case minute hand tip will be closer to A cubical cardboard box made of 1cm thich card board has outer side of 29 cm. A tight-fitting cubical box of the same thickness is placed inside it, then another one inside it and so on. How many cubical boxes will be there in the entire set? (A) 29 (B) 28 (C) 15 (D) 14 Exp. Dimension of box = cm^3, Since, the thickness of the paper is 1 cm, hence the next box which will be fitted in will have to be of dimension cm^3 & similarly we conclude that there will be 14 boxes which can be fitted inside. 19. There are two buckets A and B. Initially A has 2 litres of water and B is empty. At every hour l litre of water is transferred from A to B following by returning 1 2 litre back to A from B helf an hour later. The earliest A will get empty is in (A) 5 h (B) 4 h (C) 3 h (D) 2 h Exp. So after one hour bucket b will be having 1 liter and A will be having 1 liter. After one and half hour A = 1.5 liter, B = 1/2 liter. After 2 and half hour, A = 0.5 liter, B = 1.5 liter. After 3 hours A = 1 liter, B = 1 liter. After 4 hours A = 0, B = 2 liter. 20. Secondary colours are made by a mixture of three primary colours, Read, Green and Blue, in different proportions; each of the primary colours comes in 8 possible levels. Grey corresponds to equla proportions of Red, Green and Blue. How many shades of grey exist in this scheme? (A) 8 3 (B) 8 (C) 3 8 (D) 8 3 Exp. Here colours are constant, levels are variable. So probability will be equal to 3 8. P3/3

95 Amar Ujala Education PART-B A/UG Enzymes accelerate a reaction by which one of the following strategies? (A) Decreasing energy required to form the transition state (B) Increasing kinetic energy of the substrate (C) Increasing the free energy difference between substrate and the product (D) Increasing the turn over number of enzymes Exp. Enzymes Accelerate Reactions by Facilitating the Formation of the Transition State 22. The genome of a bactrium is composed of a single DNA molecule which is 10 9 bp long. How many moles of genomic DNA is present in the bacterium? [Consider Avogadro No = ] 1 1 (A) (B) (C) (D) Exp. 1 molecule = moles Avagadro number =1/ The ionic strength of a 0.2 M Na 2 HPO 4 solution will be (A) 0.2 M (B) 0.4 M (C) 0.6 M (D) 0.8 M Exp. Ionic strength = number of ions produced after ionization, so in this case it will be doubled = 2*0.2 M =0.4M 24. Glycophrin having one highly hydrophobic domain is able ot span a phospholipid bilayer membrane only (A) once (B) twice (C) thrice (D) four times Exp. Glycophorin has a segment of 19 hydrophobic residues (residues 75 to 93) forms an helix that traverses the membrane bilayer once. 25. A cell line deficient in salvage pathway for nucleotide biosynthesis was fed with medium containing 15 N labelled amino acids. Purines were then extracted. Treatment with which one of the following amino acids is likely to produce 15 N labelled purines? (A) Aspartic acid (B) Glycine (C) Glutamine (D) Aspartamine Exp. Since cell line is deficient in Salvage pathway so it will follow de novo pathway for purine synthesis.in de novo pathway the amine group and nitrogen is given by all three amino acids. 26. Given below are events in the cell cycle. (a) Phosphorylation of lamin A, B, C P3/4 (b) Phosphorylation of Rb (Retinoblastoma protein) (c) Polyubiquitination of securin (d) Association of inner nuclear membrane proteins and nuclear proe complex proteins with chromosomes. Which one of the following reflects the correct sequence of events in the mammalian cell cycle? (A) a b c d (B) b c d a (C) c a b d (D) b a c d Exp. Phosphorylation of Rb protein is the 1st step during G1- S phase transition. Then Mitotic CDKs Promote Nuclear Envelope Breakdown by phosphorylation of laminins. Polyubiquitination of securin cause the separation of sister chromatids and finally the nuclear membrane and pore complex reappear. 27. It takes 40 minutes for a typical E. coli cell to completely replicate its chromosome. Simultaneous to the ongoing replication, 20 minutes of a fresh round of replication is competed before the cell divides. What would be the generation time of E. coli growing at 37ºC in complesx medium? (A) 20 munutes (B) 40 minutes (C) 60 minutes (D) 30 minutes Exp. The natural environment of E.coli cells is the lower intestine of a warm-blooded animal. Its optimal growth temperature is 37 0C. The doubling time or generation time for most E.coli strains in a rich medium at 37 0C is between 20 to 40 minutes. As its given that the fresh round of replication took 20 minutes it is understandable that the division process will take more time than 28. An antibiotic that resemble the 3 end of a charged trna molecule is (A) Streptomycin (B) Sporsomycin (C) Puromycin (D) Tetracycline Exp. Puromycin is a well-known inhibitor of protein synthesis in vivo as well as in cell-free systems.its structure, analogous to aminoacyl-trna, leads to the premature release of unfinished polypeptide chains as polypeptidyl-puromycin derivatives. 29. Coupling of the reaction centers of oxidative phosphorylation is achieved by which one of the following? (A) Making a complex of all four reaction centers (B) Locating all four complexes in the inner membrane. (C) Uviqunones and cytochrome C. (D) Pumping of protons. Exp. Coupling of reaction centers of ETC complexes occurs by combining Ubiquinones and Cytochrome C.

96 A/UG Which one of the following chemicals is a DNA intercalator? (A) 5-Bromouracil (B) Ethyl methane sulfonate (C) Acridine orange (D) U V Exp. The ability of a "vital" dye, acridine orange (AO), to intercalate into the DNA of living cells was investigated by quantitative intensified fluorescence microscopy and digital imaging under various conditions of dye concentration, excitation light intensity, and ionic concentration 31. In eukaryoric replication, helicase loading occurs at all replicators during (A) G 0 phase (B) G 1 phase (C) Sphase (D) G 2 phase Exp. Loading of the replicative DNA helicase at origins of replication is of central importance in DNA replication. In eukaryotic cells, helicase loading is tightly restricted to the G1 phase of the cell cycle. 32. Which of the following is NOT a second messenger? (A) Cyclic GMP (B) Diacyglycerol (C) Inositol triphosphate (D) Phosphatidyl inositol Exp. Phosphatidyl inositol is not a second messenger. IP3 and DAG these two second messengers are from phosphatidyl inositol 4,5 bisphosphate 33. Cytotoxic T cells express (A) C D8 market and are class II MHC restricted (B) CD 4 marker and are class I MHC restricted (C) CD 4 marker and are class II MHC restricted (D) CD 8 marker and are class I MHC restricted Exp. CD8 T cells recognize antigen that is expressed on the surface of class I MHC molecules, and function mainly as cytotoxic T cells. 34. Which of the following is NOT a cell adhesion protein? (A) Cadherin (B) Selectin (C) Immunoglobulin (lg) superfamily (D) Laminin Exp. Cadherin,selectin,immunoglobulin super family are all cell adhesion poteins.but laminin is an extracellular matrix protein.basal lamina the very good example of extracellular matrix is having the protein laminin 35. The mutaion in an oncogene falls under which of the following classes? (A) Loss of function mutation (B) Frame shift mutation (C) Gain of function mutation (D) Dominant negative mutation P3/5 Amar Ujala Education 36. -Amanitin is a fungal toxin which inhibits eucaryoricd RNA polymerases. The three eucaryotic RNA polymerases show differential sensitivity order (higher to lower ) is correct in respect of sensitivity towards a- amanitin? (A) RNA POL III > RNA POL > RNA POL I (B) RNA POL II > RNA POL III > RNA POL I (C) RNA POL I > RNA POL I > RNA POL II (D) RNAPOLII>RNAPOLI>RNAPOLIII Exp. Alpha amanitin phalloidin poisonous mushroom. It will inhibit RNA polymerases. RNA polymerase 1 is insensitive to alpha amanitin. RNA Pol2 is highly sensitive.while RNA pol3 is sensitive in high concentration 37. For which one of the following physiological studies 12 CO 2 and 13 CO 2 are used? (A) Estimate the rate of photosynthesis (B) Determine rate of photorespiration (C) The ratio of C 4 and CAM pathways of CO 2 fixation (D) The ration of C 3 and C 4 pathways of CO 2 fixation Exp. Stable isotopes of carbon is used to distinguish between C3,C4 and CAM plants. These plants differs in their ability to assimilate different carbon isotopes. 38. Floral organ development is controlled by overlapping expression of A class, B class and C class genes in different whorls. In an Arabidopsis mutant, the flowers had sepals, sepals, carpels and carpels in the four whorls. Mutation in which one of the following is the cause for the mutant phenorype? (A) A class gene alone (B) B class B class genes (C) A and B class genes (D) C class gene alone Exp. Class A will give carpels instead of sepals and stamens instead of petals,class A and B will give petals, class B will give sepals instead of petals and carpels instead of stamen, Band C will give stamens,class C will have petals instead of stamen and C alone produces Carpels. B is mutated then sepals sepals carpel carpel will be the result. 39. Migration of individual cells from the surface into the embryo s interior is termed as (A) ingression (B) involution (C) invagination (D) delamination Exp. Ingression. The migration of individual cells from the surface layer into the interior of the embryo.

97 Amar Ujala Education 40. Gibberellic acid (GA) controls seed germi-nation by directing breakdown of the stored starch. In which one of the following tissues of the barley seed, a-amylase gene is induced in response to GA? (A) Endosperm (B) Coleoptile (C) Aleurone layer (D) Embryo Exp. Aleurone layers in grains of long-day plants such as Barley, secretes more a-amylase and had a higher responsiveness to GA3 which can be measured by a- amylase secretion. Storage of the grains increased both the percentage of germination and the responsiveness of the aleurone to GA Phenylalanine, a precursor of most of the phenolics in higher plants is a product of which one of the following pathways? (A) Shikimic acid pathway (B) Malonic acid pathway (C) Mevalonic acid pathway (D) Methylerythritol pathway Exp. Phenyl alanine a precursor for phenolics and Alkaloids originated from Shikimic acid 42. A lveolar cells of the lung arise from which one of the following layer (s)? (A) Mesoderm (B) Endoderm (C) Ectoderm (D) Both ectoderm and endoderm Exp. Alveoli arise from endoderm only. 43. In chick, development of wing feather, thigh feather and claws depends on epithelial specificity conferred by induction from mesenchymal components from different sources of the dermins. This may be attributed to? (A) Autocrine interaction (B) Regional specificity of induction (C) Receptor activation by hormones (D) Inactivation of genetic interactions Exp. Chichkdevelopment of wings, feathers and claws is decided by regional specificity of induction. In this case mesenchyma is going to play instructive role for the differentiation of epithelial layer. 44. Which one of the following is NOT involved with the pacemaker poktential of heart? (A) h -channel (B) Transient calcium channel (C) Long-lasting calcium channel (D) f -channel Exp. h-channel is not a pacemaker potential of heart. P3/6 A/UG Which one of the following neurotrans-mitters is secreted by the pre-ganglionic neurons of sympathetic nervous system? (A) Epinephrine (B) Acetylcholine (C) Dopamine (D) Norepinephrine Exp. Preganglionic neurons of sympathetic nervous system secrete Acetylcholine as their neurotransmitter 46. The photosynthetic assimilation of atmospheric CO 2 by leaves yield sucrose and starch as end products of two gluco-neogenic pathways that are physically separated. Which one ojf the following combination of cell organelles are involved in such physical separation of the process? (A) Sucrose in cytosol and starch in mitochondria (B) Sucrose in chloroplasts and starch in cytosol (C) Sucrose in mitochondria and starch in cytosol (D) Sucrose in cytosol and starch in chloroplasts Exp. Most of the triose phosphate synthesised in chloroplasts is converted to either sucrose or starch. Starch accumulates in chloroplasts, but sucrose is synthesised in the surrounding cytosol. 47. You are asked to identify the stage of estrus cycle in vaginal smear of a mouse containing large number of leukocytes and very few nucleated epithelial cells. Which one of the following will be the correct stage of estrous cycle? (A) Early estrus, late proestrus (B) Late estrus, early metestrus (C) Late metestrus, early diestrus (D) Late diestrus, early proestrus 48. A diabetic patient developed metabolic acidosis resulting in deep and rapid breathing which is called (A) Kussmaul breathing (B) Cheyne-Stokes respiratory pattern (C) Apneustic breathing (D) Periodic breathing Exp. Kussmaul breathing is a deep and labored breathing pattern often associated with severe metabolic acidosis, particularly diabetic ketoacidosis (DKA) but also kidney failure. It is a form of hyperventilation, which is any breathing pattern that reduces carbon dioxide in the blood due to increased rate or depth of respiration. 49. Mutation in gene 'X' leads to lethality in a haploid organism. Which one of the following is vest suited to analyse the fune 'X'? (A) Pleiotropic mutants (B) Temperature-sensitive mutants (C) Recessiv mutants (D) Mutants with low penetrance

98 A/UG-15 Exp. Pleotropic mutation is a mutation that has effects on several different characters. So the best way to analyse the function of gene X will be by creating pleotropic mutants. Since the question doesn't mention anything about the differential expression of the gene X under different temperature, so we can't say conclusively that temperature sensitive mutants will help in analysing the gene's function. As the mutation is lethal in haploids, so it will be lethal in recessive mutants as well, therefore we can't see its function in recessive mutants. Same is the case with mutants with low penetrance. So option (A) is most appropriate. 50. The following pedigree chart shows inheritance of a given trait The trait can be called (A) autosomal dominant (B) autosomal recessive (C) X-lindked dominant (D) sex limited Exp. Since the children of affected parents of generationn I are both normal, that means trait cannot be a recessive trait (because children of homozygous recessive parents are also homozygous recessive). So the trait here is a dominant one and both the parents of gen I are heterozygous for it. Father affected with X-linked domiannt trait will always have affected daughters. Since that is not the case here, so it is a Autosomal dominant trait. Answer is option (A). 51. Following is a hypothetical biochemical pathway responsible for pigmentation of leaves. The pathway is controlled by two independently assorting genes 'A' and 'B' encoding enzymes as shown below. Mutant alleles 'a' and 'b' code for non-functional proteins. A X Y Z (White substrate) (Yellow intermediate) B (Green product) What is the expected progeny after selfing a plant with the genotype AaBb? (A) Green (9) : White (4) : Yellow (3) (B) Green (9) : Yellow (4) : White (3) (C) Green (9) : Yellow (6) : White (1) (D) Green (9) : White (7) P3/7 Amar Ujala Education Exp. Cross : AaBb AaBb; in next generations following general genotypes will be seen- (i) A _ B _ : phenotype Green as A and B both present. Proportion 9/16 (ii) A _ bb : Phenotype yellow as A is present alone, so only yellow intermediate is formed. Proportion 3/16. (iii) aab _ : Phenotype white. As A itself is absent, so the yellow intermediate itself can't be formed. Proportion 3/16. iv) aabb: since Both A and B is absent, again phenotype will be white. Proportion 1/16. So the final proportion will be Green (9) : White (4) : Yellow (3). 52. In a heterozygous individual for a given gene, if a crossing over has occurred between the gene locus and the centromere of the chromosome, the segregation of the two alleles of the given gene will occur during meiosis at (A) either anaphase I or anaphase II (B) anaphase I only (C) anaphase II only (D) both anaphase I and II Exp. When crossing over occurs between gene locus and centromere in a heterozygous individual, then the segregation of the 2 alleles will occur in second meiotic division i.e. anaphase II, when the sister chromatids separate. 53. Which of the following statements about evolution is NOT true? (A) Evolution is the product of natural selection. (B) Evolution is goal-oriented. (C) Prokaryotes evol faster that eukaryotes. (D) Evolution need not always lead to a better phenotype Exp. Evolution is a random process that takes place continuously but very slowly over the years of time. not every evolutionary changes is beneficial. we can not moniter the process as it takes million of years to take place and we can see only the resultant phenotype only if it survives. For example, modern days birds have supposed to be evolved from reptiles and acheoptyrx is the link organism we all know but no one has seen the intermediate form other than this.thus evolution is continuous process of selection. 54. Identify the correct match between the animal (flatworm, earthworm, roundworm)and its boby cavity type (acoelomate, coelomate, pseudoco-elomate (A) Roundworm-pseudocoelomate;Earthworm - acoelomate; Flatworm -acoelomate (B) Roundworm-acoelomate; Earthworm -coelomate; Flatworm-acoelomate (C) Roundworm-pseudocoelomate; Earthworm - coelomate; Flatworm-acoelomate

99 Amar Ujala Education (D) Roundworm-coelomate; Earthworm- pseudocoelomate; Flatworma-coelomate Exp. Animals possessing body cavity/coelom- Annelida (Earthworm), the body cavity not lined with mesodermpseudocoelom, Aschelminthes/Nematodes (Roundworm), animals without body cavity- Acoelom- Playthelminthes (Flatworm) 55. According to 2014 IUCN Red List, which of the following vertebrate classes has the largest percentage of threatened species? (A) Mammals (B) Birds (C) Reptiles (D) Amphibians Exp. The IUCN Red List of Threatened Species has identified amphibians as being the most threatened vertebrate group assessed so far, with around 41% at risk of extinction. 56. Which gas does ONT contribute to global warming through its grecnhouse effect? (A) Nitrous oxide (B) Methane (C) Carbon dioxide (D) Nitric oxide Exp. The primary greenhouse gases in Earth's atmosphere are water vapor,carbon dioxide, methane, nitrous oxide, and ozone 57. Most members of bryophyte phylulm Antho-cerophyta are characterized by (A) gametophyte with single chloroplast per cell and multicellular rhizoids sporohyte without stomata (B) gametophyte with single chloroplast per cell and unicellular rhizoids sporophyte with stomata (C) gametophyte with multiple chloroplasts per cell and unicellular rhizoids sporophyte without stomata (D) gametophyte with single chloroplast per cell and multicellular rhizoids sporophyte with stomata Exp. Anthocerophyta are group of non-vascular plants charaterised by presence of greem gametophyte with single large chloroplast and sporophyte with stomata. Rhizoids are presenet which are usually unicellular. 58. Which one of the following gymnosperm phyla produces motile sperms, bears ovulate and microsporangiate cones on separate plants and has fleshy, coated seeds? (A) Coniferophyta (B) Cycadophyta (C) Ginkgophyta (D) Cnetophyta Exp. Ovulate & microsporangiate cones on separate plants; fleshy-coated seeds 59. A red coloured tubular flower without any odour is most lidely to be pollinated by (A) beetles (B) bees (C) butterflies (D) birds Exp. birds do not have a good sense of smell, so bird-pollinated flowers usually have little odor. P3/8 60. In the following equations (a) dn/dt = rn (b) Nt = N 0 e rt A/UG-15 (c) dn/dt = rn K N K (d) dn/dt = rn N/K exponential population growth is described by (A) (a) and (b) (B) (a) only (C) (c) only (D) (b) and (d) Exp. Exponential word problems almost always work off the growth / decay formula, A = Pert, where "A" is the ending amount of whatever you're dealing with (money, bacteria growing in a petri dish, radioactive decay of an element highlighting your X-ray), "P" is the beginning amount of that same "whatever", "r" is the growth. dn/dt = rn is the logarithmic form of formula. 61. Which isotope below is best suited for metabolic labelling of glyceraldehyde-3-phospho-dehy-drogenase? (A) 14 C (B) 125 I (C) 32 P (D) 131 I Exp. radioisotpes are used in non-clinical studies to assess drug absorption, distribution, metabolism, and excretion (ADME). Some isotopes that are currently inuse are H3,C14,P32, and I131but the C14 is the choice of most of the ADME studies. 62. Which one of the following conditions is NOT likely to favour male monogamy? (A) When the male has to guard his mate against mating by another male (B) When the mate wants to spend more time for foraging (C) When the male has to assist the mate in brood and nestling care (D) When the female guards her mate against seeking other females to mate Exp. Animals are monogamous because it was the only way they could guard their mates and thus their breeding rights. Males stayed with one female to ensure their young were not killed by another male. It s one thing to sire a litter of pups and another to ensure that they actually survive to reproduce to carry on their genetic lineage. 63. Neomycin phosphortransferase gene, fre-quently used as a selection marker during plant trans formation, inactivates which one of the following antibiotics? (A) Hygromycin (B) Ampicillin (C) Streptomycin (D) Kanamycin Exp. The neomycin phosphotransferase gene (nptii) conferring resistance to kanamycin and neomycin is one of the antibiotic resistance genes commonly present in GM plants.

100 A/UG Which among the following is the simplest method to estimate the concentration of glycerol in an aqueous solution of glycerol? (A) UV absorption spectroscopy (B) Gas chromatography (C) ph measurement (D) Viscosity measurement Exp. As the conc. of glycerol will be increased in aquous solution viscosity will also be inccreased. By making a standard plot it can be easily determined. 65. The origin and diversification of Angio-sperms was during which geological period? (A) Permian (B) Triassic (C) Jurassic (D) Cretaceous 66. Application of gene therapy in clinical trials did NOT succeed due to (A) poor integration of a genc in the host genome (B) lack of expression of integrated gene in cells (C) degradation of gene inside the cell (D) activation of oncogenes consequent to integration of the gene Exp. The most common reason forfailure of clinical trials of gene therapy trials is due to inefficient gene transfer that also creates issued for regulation of gene expression and every succeeding steps. 67. Which one of the following would contribute to intrinsic fluorescence to a protein? (A) aromatic amino acids (B) disulfide bonds (C) charged amino acids (D) branced chain amino acids Exp. Aromatic amino acids gives intrinsic fluoroscence in ultraviolet region (280 nm). Intrinsic fluoroscence is due to presence of benzene or aromaticc rings in these amino acids. Amar Ujala Education 68. For identification of three proteins moving together (as a single band) upon loading in a single lane of a SDS- PAGE gel, the best method is (A) one step Western blot (B) NMR spectroscopy (C) Western blot followed by stripping and reprobing (D) ESR spectroscopy Exp. Reprobing a Western blot save times and conserves sample while allowing optimization to be peformed as needed. Reprobing also allows the same blot to be probed for different target proteins. 69. A gene expressing a 50 kda protein from an eukaryote was cloned in an E. coli plasmid under the lac promoter and operator. Upon addition of IPTG, the 50 kda protein was not detected. Which one of the following explains the above observation? (A) The cloned sequence lacked the Kozak sequence (B) E. coli does not make proteins larger than 40 kda (C) Differences in codon preference (D) 50 kda protein contains a nuclear localization signal Exp. The classic cloning vectors are plasmids, phages, and cosmids, which are limited to the size insert they can accommodate, taking up to 10, 20, and 45 kb, respectively. 70. Which one of the following techniqes will you use to identify more than 1000 differentially expressed genes in normal and tumor tissues in one single experiment? (A) RAPD (B) Genome sequencing (C) ChIP assay (D) Transcriptome analysis Exp. transcriptome databases aid in the identification of genes that are differentially expressed in distinct cell populations. PART-C 71. Which one of the follwing statements is correct? (A) In all L-amino acids, only the C carbon atom is chiral (B) Deoxyribose is optacally inactive (C) The specific rotation of sucrose will be The sum of the specific rotations of D-glucose and D-fructose (D) phosphatidyl cholines isolated from biological membranes is optically active Exp. Phosphatidylcholine is optically active because middle carbon of glycerol backbone is attached to four different functional groups. 72. Membrane proteins are synthesized on endoplamic reticulum and membranes. One hypothesis for membrane protein sorting is hydrophobicity matching i.e.,the proteins with a shorter transmembane portion would partition into thinner membrane. you are given the following three observation (a) It was found that tranmembrane protions of proteins in Golgi Membrames shorter than those in plasma membranes (b) presence of cholesterol increases the thickness of (c) the bilayer the phospholipid composition of Golgi and plasma membranes are same (A) Protins in plasma membrane have longer transmembranes protion that proteins in Glogi membrane P3/9

101 Amar Ujala Education (B) proteins in Golgi membranes have longer transmembrane protion than protein in plasma membranes (C) proteins of both Golgi and plasmamembrans have same length of transmembrane protion (D) Cholesterol is more in Golgi membrane than in plasma membrane Exp. As already been mentioned in the observations that golgi has shorter proteins in its membrane. Cholesterol increases the thickness of membrane. Phospholipids content is same in both the cases, so only first option is correct. 73. Four single amino acid mutants(atod)ofaproteinin the epitope-region of a monoclonal atibody X were made and expressed in E. coli. the lysates from the four E. coli culture expressing these forur proteins were run or an SDS-PAGE gel and subsequently transferred to nitrocellulose membrane and Western blotted using a monoclonal antibody X raised against the wild type protein. The results are presented in the figure blow a b c d wt Monoclonal Load control The four single mutation, upon sequencing, whre found to be Valine (V) to Alanine (A); Glycine (G) to Proline (P); Alanine (A) to Aspartic acid (D) and isoleucine (I) to leucine (L). Which one of the following statements is correct? (A) b is due to V A and c is due to G P (B) b is due to G P and c is due to V A (C) b is due to I L and a is due to A D (D) c is due to V A and a is due to I L Exp. Single point mutations are run on gel and stained by western blotting, so similar type of mutations will give higher intensity band. Completely different type of mutations will not give any band on western blot. So only first seems to be correct. 74. The exact backbone dihedral angles in a folded protein can be obtained by (A) deconvolution of its circular dichoism spectra obtained at different ph and temperature (B) estimating the numbe of protons that exchange with deuterium on treating the protei with D 2 O (C) forming fiberes of the protein and analyzing the fibre diffraction pattern (D) analysis of the crystal structure of the protein obtaned by X-ray diffraction at high resolutions Exp. Exact dihedral angles can be determined by High resolution X-ray crystal structure only. NMR can also provide us dihedral or 3D structural information, but resolution will be less as compared to X-ray crystal structures. CD and fibers of proteins can t help in the measurement of dihedral angles. P3/10 A/UG the following are the statements about pyruvate kinase (PK). (a) ATP is an allosteric inhibitor of PK (b) Fructose 1,6 biphosphate is an activator of PK (c) ADP is an allosteric inhitor of PK (d) Alanine is an allosteric modulator of pk Whichs of the above statements (s) are true? (A) A,B,C (B) A,B,D (C) B,C,D (D) only A Exp. ATP is a negative allosteric inhibitor. fructose 1,6- bisphosphate, an intermediate in glycolysis, enhance enzymatic activity. Alanine, a negative allosteric modulator 76. A practical class was going on where the students were demonstrating ATP syn-thesis in vitro using active mitochondria, Some students added one of the following to their tubes (a) Dinitrophenol (DNP), an uncouper (b) Mild acidification of the medium (c) Glutilferone, that permeabilizes both the membranes (d) An outer membrane permeable H + quencher compound, Elilak In which one of the above, ATP synthesis will be detected? (A) A (B) B (C) C (D) D Exp. DNP is uncoupler so ATP will not be formed in this case. Glutilferon, if both the membranes will be permeabilized there will be no inner-membrane no ATP formation. If H+ ions will be quenched than also they will not be transported to mitochondrial matrix and no ATP formation will occur. So left with only option B, mild acidification of the medium. 77. A culture medium contains two carbon sources, one is prefrred carbon source (glucose) and the second is a non- preferred source (lactose). Which one below is correct regarding the nature of growth curve of E.coli cultured in this medium? (A) Growth curve will be same as when grown in presence of only glucose. (B) Growth curve will be same as when grown in presence of only lactose. (C) A lag phase will be oserved between the two exponential phases. (D) Two lag phases will be observed between the two exponential phases. Exp. When the culture is given two carbons, diauxic growth curve is obtained. The first growth phase occurs when cells use a preferred carbon source, and the second growth phase begins when the less-preferred carbon source is used after the preferred carbon source has been depleted. Diauxic growth curve is characterized by the appearance of two exponential growth phases separated by a lag phase called diauxic lag.

102 A/UG Which one of the following statements correctly applies to proteins which are translated on the rough endoplasmic reticulum? (A) Cytoplasmic proteins which are targeted to the nucleus in response to hormone stimuli (B) Proteins targeted to lyusosomes, plasma membrane and cell exterior (C) Proteins which are targeted to the nucleus through endoplasmic reticulum lumen as the lumen is in direct conne-ction with the inter membrane space of the nucleus (D) All proteins which get targeted to peroxisomes Exp. proteins for plasma membrane, lysosome, and cell exterior are synthesised by RER.after that it is transported in vesicles 79. Lipid ragts are rich in both sphingolipids and cholesterol. Cholesterol plays a central role in raft formation since lipid rafts apparently do not form in its absence.why do you think cholesterol is essential for the formation of lipid rafts? (A) Cholesterol decreases the mobility of sphingolipids in the lipid bilayer (B) Large head groups of sphingolipids repel each other in presence of cholesterol (C) Cholesterol interacts with fatty acid tails in the membrane (D) The planar cholesterol molecules are postulated to fill the voids that form underneath the large head groups of the sphingolipids Exp. Cholestrol in lipid rafts interacts with sphingolipids and decrease the mobility of sphingolipids. Interaction is not important, decrease in mobility is important for lipid rafts formation. 80. Following is the domain organization of three proteins that are targeted to the mitochondria. (a) Mature protein (b) Matrix protease cleavage site Matrix protease cleavage site Hydrophobic region (c) Interacts with outer membrane pore Based on the domain organization in the above figure and assuming the left box to ber having the mitochondrial sorting singnal, predict the most likely sub-compartment of the mitochondria in which the protein will be found. (A) A in matrix; B in inner membrane; C in inter-membrane space P3/11 Amar Ujala Education (B) A in inner membrane; B in inter-membrane space; C in outer membrane (C) A and B are in matrix; C in outer membrane space (D) A in matrix; B and C are in inter- membrane space Exp. As A protein has matrix protease site, simply it means it has signal sequence to get transported to Matrix only. B has signal sequence for matrix protease as well as inner membrane,sofirstitwillreachmatrixthanmatrix protease will cleave this signal sequence, than it will be imported to inner membrane where inner-membrane protease will cleave inner membrane signal sequence. C has signal sequence for the interaction with outer membrane transporter so C will be imported to intermembrane space. 81. You have labelled DNA in a bacterium by growing cells in medium containing either 14 N nitrogen or the heavier isotope 15 N. Furthermore, you have isolated pure DNA from these organims, and subjected it to CsCl density gradient centrifugation leading to their separation of light ( 14 N) and heavy ( 15 N) forms of DNA to different locations in the centrifuge tube. In the next experiment, bactera were grown first in medium containig 15 N, so that all the DNA made by cells will be in heavy form. Then these cellswere transferred to medium containing only14n and allowed the cells to divide for one generation. DNAs were extracted and centrifuged as above in the CsCl gradient. A hybrid DNA band was observed at a position located between and equidistant from the 15N and 14N DNA bands. Based on the above observation, which one of the following conclusions is correct? (A) Replication of DNA is conservative (B) Replication of DNA is semi- conservative (C) Replication of DNA is dispersive (D) Replication by rolling circle mode Exp. one round of replication,hybrid DNA will be generated.one strand is having N15 and other is having N14.The band will be inbetween N15 and N14 densities. 82. The frequency of cells in a population that are undergoing mitosis (the mitotic index) is a convenient way to estimate the lengthe of the cell cycle. In order tomeasure the cell cycle in the liver slices are prepared and stained to easily identify cells undergoing mitosis. It was obserbved that only 3 out of 25,000 cells are found to be undergoing mitosis. Assuming that M phase lasts 30 minutes, calculate the approximate length of the cell cycle in the liver of an adult mouse? (A) 46 hours (B) 50 hours (C) 42 hours (D) 21 hours Exp. The sum of all cells in phase as prophase, metaphase, anaphase and telophase, respectively; N total number of cells. Mitototic index (MI) is 3/ = 1.2 % From the cell cycle, 1.2% is mitotic and the rest will obviously be interphase. So, 1.2% is 30 minutes, so 100% (length of total cell cyle) is 2500 minutes (42hours)

103 Amar Ujala Education 83. Although ribonucleoside triphosphates (rntps) are present at approximate 10-fold higher concentration than deoxyribo-nucleoside triphosphates (dntps) in the cell, but they are incorporated into DNA ar a rate that is more than 1000-fold lower than dntps. This is because (A) DNA Polymerase cannot discriminate between dntps and rntps. But as soon as rntps are incroporated in the DNA chain, they are hydrolyzed due to the presence of 2-OH group (B) DNA polymerase efficiently discrimi-nates between rntpsanddntps,butassoonasrntpsareincised by the proof reading activity of DNA polymerase (C) DNA polymerase efficiently discrimi-nates between rntps, because its nucleotide binding pocket cannot accommodate a 2-OH on the incoming nucleoride (D) DNA polymerase cannot discriminate between rntps and dntps. Since the rate of transcription in cell is 10 6 htimes faster than replication, it cannot compete with RNA polymerase for rntps. Exp. DNA Polymerase efficiently discriminates between rntps dntps, because its nucleotide binding pocket cannot accommodate a 2'-OH on the incoming nucleotide. 84. The mismatch repair activity of E.coli reparis misincorporated bases which is not removed by the proofreading activty of DNA polymerase. However, while doing so, it has to decide which strand of the DNA is newly synthesized and which one is parental. Mismatch repair system does it by which one of the following ways? (A) It recognzes nearby GATC sequence. (B) It recognixes any nearby palindromic sequence. (C) It recognises a specific repetitive sequence. (D) It recognises the hemi-methylated GATC sequence nearby. Exp. Methylation is the process by which ecoli does this repair. 85. Enlisted belw are differnt types of RNAs produced in the cell (Column A) and their functions (Column B), but not in the same order. Column A Column B (a) Sn RNAs (i) turn off gene expression by directing degradation of selective mrnas. (b) si RNAs (ii) regulate geme expressopm by blocking translation of selective mrnas. (c) mi RNAs (iii) function in a variety of processes including splicking of pre-mrna (d) Sno RNAs (iv) used to process and chemically modify rrnas. P3/12 A/UG-15 (A) a-(iv), b-(ii), c-(i), d-(iii) (B) a-(iii), b-(i), c-(ii), d-(iv) (C) a-(iv), b-(i), c-(ii), d-(iii) (D) a-(iii), b-(ii), c-(i), d-(iv) Exp. mirna blocks translation by making double stranded region.while si RNA degrades mrna itself. snrna help in the processing or splicing of RNA. Sno RNA chemically modifie the rrna. 86. In prokaryotes, the initiatior t-rna is first charged with a methioine, floolwed by the addition of a formyl group to the methinine by the enzyme Met-tRNA transformylase. Given below are several statements in this context (a) All prokaryotic proteins have formyl methionine at their amino-terminal end. (b) Deformylase removes the formyl group from the amino terminal methionine. (c) All prokaryotic proteins have methio-nine at their amino terminal end. (d) Aminopeptidases often remove amino terminal methionine. (e) Aminopeptidases remove amino terminal formyl methionine. Which fo the above statement (s) are most likely to be true? (A) A only (B) B and C (C) E only (D) B and D Exp. Deformylation is the only process if N terminal amino acid is methionine.but in prokaryotes half of proteins is undergone aminopeptidase action.it will remove the N terminal methionine and next amino acid will become the first one. 87. A hypothetical operon involved in the synthesis of an amino acid X is ON (transcribing) in the presence of low levels of X The symbols a, b and c (in the table below) represents a structural gene for the synthesis of X (X - synthase), the operator region and gene encoding the repressor- but not necessarily in that order. From the following data, in which superscripts denote wild type or defective genotype, identify which are the genes for X-synthase, operator region and the repressor Strain Genotype X-synthase activity in the presence of Low Level High level of X of X 1. a b + c + Detected Detected 2. a + b + c Detected Detected 3. a + b c Not detected Not detected 4. a + b + c + /a b c Detected Not detected 5. a + b + c /a b c + Detected Not detected 6. a b + c + /a + b c Detected Detected

104 A/UG-15 The respective genes for X synthase, the operator region and repressor are (A) a,b,c (B) c,a,b (C) b,c,a (D) b,a,c Exp. Consider a well-known and understood operon which has some similarities to this hypothetical operon. trp operon where trp(x) will bind to the represson when in high number and repress the activity of the enzymes involved in synthesis of trp.here in case 1 in absence of wild type operator, repressor cannot bind in low or high presence of X, so the X-synthase activity will be detected at lower and higher level of X. In case 2 wild type operator is present, repressor cannot bind because wild type form of repressor is not expressing, so the X- synthase activity will be detected at lower and higher level of X. In case 3 wild type operator is present, repressor cannot bind because wild type form of repressor is not expressing, neither is wild type synthase, so the X-synthase activity will not be detected at lower and higher level of X. In case 4 it is a merozygote and operator, repressor, synthase wild type forms are present, repressor can bind, so the X-synthase activity will be detected at lower but not at i her level of X because at higher levels of X, X will interact with repressor and bind to the oppressor. In case 5 it is a merozygote and operator and synthase wild type forms are present on one chromosome and repressor wild type is present on the other chromosome, repressor can bind as it s a protein it can interact with other chromosome s operator region, so the X-synthase activity will be detected at lower but not at higher level of X because at higher levels of X, X will interact with repressor and bind to the oppressor. In case 6 it is a merozygote and repressor and synthase wild type forms are present on one chromosome and operator wild type is present on the other chromosome, repressor can bind as it s a protein it can interact with other chromosome s operator region but the X synthase gene and operator region are present on two separate chromosomes and as operator is a genetic element it cannot affect the activity of a gene present on another chromosome, so the X-synthase activity will be dete cted at ower and at higher level of X. 88. A protein has 4 equally spaced trypsin sensitive sites which results in peptide fragments A 1,A 2,A 3,A 4,andA 5 upon digestion with trypsin. The peptides A 2 and A 5 represents N-terminal and C-trminal fragments respectively. Now you are asked to synthesise this protein. At time t = 0 you added all the 20 amino acids labelled with 14 C and initiated the synthesized. At time t = 4, full length protein is synthesized. If you stop the P3/13 Amar Ujala Education synthesis of the protein in time t = 1 and digest the protein with trypsin, which peptide will have maximum 14 C label than others? (A) A 3 (B) A 1 (C) A 4 (D) A 2 Exp. Initial synthesis will lead to formation of N terminal region so when it will be digested it will be producing peptide A2 fragment 89. Which one of the following statements about the nuclear receptor superfamily is NOT true? (A) The receptors are always cytosolic, where they remain associated with heat-shock proteins and have variable ligand binding domains in the N-terminal region (B) The receptors have characteristic repeat of the C 4 zine-finger motif (C) The receptors are either homodimeric or heterodimeric, and in the absence of their hormone ligand, the hetero-demeric receptors repress transcription, whe bound to their response elements (D) The receptors have a unique N-terminal region of variable length and may contain a nuclear localization signal between the DNA-and ligand-binding domains Exp. Not all the Nuclear receptors are cytosolic, type I nuclear receptors are in the cytosol, Type II receptors, are retained in the nucleus, Type III receptors function similarly to type I receptors except that the organization of the HRE differs and type IV receptors instead bind as monomers to half-site HREs. 90. Physical attachment between cells is very important in imparting strength in tissues. Various physical cell junctions in vertebrate epithelial tissues are classified according to their primary functions. Enlisted below in column A is the major function of a particular junction and column B enlists cell junctions, but not in the same order (a) Seals gap between 1. Desmo-somes epithe lial cells (b) Connects actin 2. Hemidesmo-somes filarment bundle in one cell with that in the next cell (c) Connects inter- 3. Tight junction mediate fila-ments in one cell to those in the next cell (d) Anchors interme- 4. Adherens junction diate filaments in a cell to extracellular matrix

105 Amar Ujala Education Choose the correct combination. (A) a (1),B (2),C (3),D (4) (B) A (2),B (3),C (4),D (1) (C) A (3),B (4),C (1),D (2) (D) A (4),B (1),C (2),D (3) Exp. Tight junction seals the gap,adherens junction connects actinfilament,desmosomes connect intermediate filaments within two cell,hemidesmosomes connect extracellular matrix 91. G-protein coupled receptors (GPCR) consist of three protein subunits a, b and c. In unstimulated state, a subunit is GDP bound and GPCR is inactive. When GPCR gets acts like guanine nucleotide exchange (GEF) factor and induces a-subunit to release its bound GDP allowing GTP to bind in its place. In order to regulate G- protein activity by regulating GDP/GTP concentration, a subunit acts as (A) GTPase (B) cgmp-specific phosphodiesterase (C) GDP kinase (D) camp-specific phosphodiesterase Exp. Galpha subunits have an intrinsic GTPase activity.thats why it is converted into GDP form.otherwise it will be in active form only. 92. Cellular level of tumour suppressor protein p53 is maintained by the ubiquitin ligase protein, Mdm2. Over expression of Mdm2 was found to convert a normal cell into cancer cells by destabilizing p53 Another protein p19 ARF inhibitsthe activity of Mdm2 thus stabilizing p53 Loss of p19 ARF function also converts normal cells into cancer cells. Based on the above information, which one of the following statements is correct? (A) Both MDM2 and p19 ARF are oncogenes. (B) Both MDM2 andp19 ARF are tumour suppressor genes. (C) MDM2 is an oncogene but p19 ARF is a tumor suppressor gene. (D) p19 ARF is an oncogene but MDM2 is a tumor suppressor gene. Exp. Gain of function mutation will results in oncogene.loss of function mutation will leads to cancer only in the case of tumor supressor gene.so mdm2 Iis oncogene and p16 is tumor suppressor gene. 93. The relation between cellular immune response generated against hepatitis C virus is the critical determinant of the outcome of infection. Given below are the representative figures of cellular immune response in column I and various outcome of infection in column II. (a) (b) (c) Column-I Response Response Response O x Time Time Time O Viral load x T Cell response A/UG-15 Column-II 1. Acute 2. Resolution 3. Chronic Choose the best possible combination (A) a (2),b (3),c (1) (B) a (1),b (3),c (2) (C) a (3),b (2),c (1) (D) a (1),b (2),c (3) Exp. Acute infection is the early stage of infection, viral load will increase and than as T-cell response will increase viral load will be dropped down. In resolution time of infection virus will be removed completly from the body,as it can be observed in C plot only. Chronic case is related to weakened immune system and persistantcy of viral in the body, it can be observed in plot B. 94. There are various subclasses of antibosies found in body fluids and body secretion. Many different functions may be attributed to these subclasses. Given below in column I is major functions of different subclasses and column II consists of the name of the suclass. Column I Column II (a) Binds to macrophages by F c (i) IgA (b) Binds to mast celss and (ii) IgD basophils (c) First B cell receptor (iii) IgD (d) No major specific function (iv) IgD known other than antigen binding (e) Protector of mucous membrane (v) IgD Select the correct combination (A) a (i), b (ii), c (iii), d (iv), e (v) (B) a (ii), b (iii), c (iv), d (v), d (i) (C) a (iii), b (iv), c (v), d (i), e (ii) (D) a (iv), b (iii), c (v), d (ii), d (i) P3/14

106 A/UG-15 Exp. First B cell receptors are Ig M and Ig A are secretory Igs present in secretions like mucous,ig E are the major class of whichareboundtomastcellsandbasophills (hypersensitivity) Ig G are the major class of antibody which act as receptors on macrophages and Ig D known for no specific function till now. 95. Instructive and permissive interactions are two major modes of inductive interaction during development. The following compares some properties of cell lines and cord blood stem ells, Cell lines, which are stred in liquid nitrogen, can be retrived for experiments, where they behave as per their original self. Cord blood can also be retrieved from liquid nitrgrn for procuring stem cells. Unlike cell lines, the stm cells can be additionally induced to undergo differentiation into desired lineages, which are very different from their original self. The behaviour of cell lines add stem cells is analogous to which of the interactions? (A) Both cell lines and stem cells show instructive interaction (B) Cell lines show instructive interaction whereas stem cells show instructive interaction (C) Cell lines show permissive interaction whereas stem cells show instructive interaction (D) Both types of cells show permissive instruction Exp. Cell lines show permissive interaction whereas stem cells show instructive interaction. 96. Following are certain statements regarding morphogen gradients and cell specification. (a) Morphogens are always transcription factors (b) Morphogens can be paracrine factors that are produced in one group of cells and travel to another population of cells (c) When the concentration of a morpho- gen drops below a certain threshold, cells stop differentiating and never get determined to another fate (d) Morphogen gradients are involved in conditional specification Which combination of the above state- (A) (a) and (b) (B) (b) and (d) (C) (c) and (d) (D) (a) and (c) Exp. A morphogen is a signaling molecule that acts directly on cells to produce specific cellular responses depending on its local concentration. Paracrine signaling is a form of cell-cell communication in which a cell produces a signal to induce changes in nearby cells, altering the behavior or differentiation of those cells. Conditional Specification : In contrast to the autonomous specification, this type of specification is a cell-extrinsic process that relies on cues and interactions between cells or from concentration-gradients of morphogens. Amar Ujala Education 97. Successful fertilization in sea urchin demands specific interaction between proteins and receptors of sperms and eggs. In view of the above, which one of the following combinations is correct? (A) Bindin in acrosomes and bindin receptors on egg vitelline membrane (B) Bindin in egg membrane and bindin receptors in acrosomes (C) Resact on egg jelly and bindin on sperm membrane (D) Proteasomes on egg membranes and complex sugars on sperm membranes Exp. The acrosomal protein mediating this recognition is called bindin.species-specific bindin receptors on the egg, vitelline envelope, or plasma membrane 98. Following statements are made in relation the five widely recognized stages of Arobidopsis embryogenesis (a) The fusinon of haploid egg and sperm takes place in Globular stage (b) Rapid cell division in two regions on either side of (c) thje future shoot apex forms Heart stage The cell elongation throughout the embryo axis and further development result in Torpedo stage (d) The embryo loses water and becomes metabolically inactive in the Zygotic stage Which combination of the above state-ments is correct? (A) A and B (B) B and C (C) C and D (D) D and A Exp. Globular stage: The first zygotic division gives rise to the apical and the basal daughter cell. The apical embryo undergoes many rounds of division to form a globular structure called protoderm. Heart stage: The rapid cell divisions on two side of the future shoot apex (tips) form a heart-shaped structure.torpedo stage: is resulted from cell elongations throughout the embryo axis and the further development of dicotyledons. 99. The following are statements regarding the development and maintenance of anterior and posterior compartments in each segment of Drosophila (a) Expression of wingless and engrailed is activated by pair-rule genes (b) Continued expression of wingless and engrailled is maintained by interaction between the cells expressing Engrailed and Wingless proteins (c) Hedgehog is expressed in wingless expressing cells and forms short range gradient (d) Hedgehog is a transecription factor (e) Engrailed is a secretory factor and binds with the patched receptor of the wingless expressing cells P3/15

107 Amar Ujala Education Which one of the following combinations of above statements is correct? (A) Cand E (B) C,DandE (C) DandE (D) AandB Exp. Engrailed is activated by even skipped and fuzhitarazu.both are pair rule genes.wingless also activated by pair rule gene.but inhibited by even skipped and fuzhitarazu.expression of these maintained by cell signalling 100. In C. elegans, an anchor cell and few ghpodermal cells take part in the formation of vulve. The experiments performed to understand the role of these cells in vulva formation and the results obtained are as follows - If tha anchor cell is killed by laser beam, hypodermal cells do not participate in vulva formation and no vulva develops. - If six hypodermal cells closely located with anchor cell (called vulval procur-sor cells ) are killed, no vulva develops. - If the three central vulval precursors are destrayed, the three outer cells, which normally form hypodermis, take the fate of vulval cells instead. Following are certain statements regarding vulva formation: (a) Anchor cell acts as an inducer (b) Six hypodermal cells with the potentiality to form vulva, form an equivalence group (c) Three, out of six, hypodermal cells participate in in vulva formation (d) The central cell functions as the 1º cell and the two cells on both side act as the 2º cells (e) The 1º cell secretes a short range juxtacrine signal Which combinations of the above state- ments have been derived from the above exprimental results? (A) A,BandC (B) A,BandD (C) DandE (D) B,DandE Exp. Based on the given experimental observation, one can derive that anchor cells acts as an inducer, six hypodermal cells with potentiality to form vulva, and three hypodermal cells form vulva Following are certain statements regard-ing terpene class of secondary metabolites in plans (a) Isopentenyl diphosphate and its isomer combine to form larger terpenes. (b) Diterpenes are 20 carbon compounds. (c) All terpenes are dirived from the union of 4-carbon elements. (d) pyrethroids are monoterpene esters. Which one of the following combination of above statements is correct? (A) (a), (b) and (c) (B) (a), (b) and (d) (C) (b), (c) and (d) (D) (a), (c) and (d) A/UG The nodulation (nod) genes are classified as common nod genes or host specific nod genes. Some statements related to such classification are given below (a) noda is a common nod gene and nodc is a host specifc gene. (b) nodb is a common nod gene and nodp is a host (c) specifc gene. nodq is a common nod gene and noda is a host specifc gene. (d) nodh is a common nod gene and nodq is a host specifc gene. Choose the correct answer from the above statements: (A) (a) and (b) (B) (c) and (d) (C) (a) only (D) (b) only Exp. Common nod genes are present among all species of rhizobium while host specific are specific for some species Following are certin statements regarding CO 2 assimilation in higher plants (a) The action of aldolase enzyme during Calvin-Benson cycle produce fructose 1, 6-bisphosphate. (b) The conversion of glycine to serine takes place in mitochondria during C2 Oxidative phorosynthetic (c) carbon cycle. During C4 carbon cycle, NAD-malic enzyme releases the CO2 from the 4-carbon acid, pyruvate. (d) Malic acid during crassulacean acid metabolism (CAM) is stored in mitochondria during dark and released back to cytosol during day. Which one of the following combinations of above statements is correct? (A) (a), (b) and (c) (B) (a), (c) and (d) (C) (b), (c) and (d) (D) (a), (b) and (d) Exp. D option is wrong as malic acid is stored is vacuoles not in mitochondria and then released into cytosol. option A B and C are correct with respect to C4,C2 pathways Many factors related to the role of abscisic acid (ABA) in contributing to drought, cold and salt resistance in plants are listed below (a) The transcription factors DREN1 and DREB2 bind to the cis-acting elements of the promoter of ABAresponsive genes in an ABA- dependent manner. (b) ABA induces many genes such as LEA and RD29. (c) ABA-responsive genes contain six-nucleotide ABRE elements in the promoter. (d) Nine-nucleotide dehydration- responsive elements (DRE) are present in ABA-responsiv genes. Which one of the following combinations of the above statements is correct with respect to ABA? (A) (a), (b) and (c) (C) (b), (c), and (d) (B) (a), (c) and (d) (D) (a) only P3/16

108 A/UG-15 Exp. Two independent families of DREB proteins, DREB1 and DREB2, have been reported to function as transacting factors in two separate signal transaction pathways under low temperature and dehydration. The ABAresponsive element (ABRE) contains the palindromic motif CACGTC with the G-box ACGT core element and A 9-bp conserved sequence, TACGACAT, termed dehydration responsive element (DRE), is essential for regulation of dehydration-responsive gene expression. The DRE has been demonstrated to function as a cisacting element involved in the induction of rd29a expression by low temperature stress. There are many genes which are induces by ABA, LEA and RD 29 are one of them Examples of many factors that regulate plant height in response to gibberellic acid (GA) are listed below (a) Binding of a GA-bound repressor to the promoter of DELLA domain-containing GRAS protein gene and blocking its expression. (b) Binding of the GA-reccptor complx to GRAS. (c) Directing GRAS for ubiquitination and degradation by the 26S protea- some. (d) Micro RNA directed down regulation of the GRAS protein expression. Which one of the following combinations is correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (b) (D) (a) and (d) Exp. GA action results in the degradation of a group of transcriptional regulators known as DELLA proteins that form a subgroup of the GRAS family of proteins. DELLA proteins are named for a conserved domain within the N terminus that is unique to this subgroup and is necessary for GA-induced degradation. Binding of GA to its soluble, nuclear receptor, GID1, causes a conformational change in the protein that promotes its association with the N-terminal domain of the DELLA protein, enabling, in turn, interaction with an SCF ubiquitin ligase, such that the DELLA is ubiquitinated, and thus targeted for degradation via the 26S proteasome Ethylene is an important plant hormon that regulates several aspects of plant growth and development. Some state-ments are given below in relation to ethylene signalling pathways (a) Unbound ethylene receptors work as positive regulators of the response pathway. (b) There are more than two ethylene receptorrs known (c) to date. The crboxy-terminal half of the ethylene receptor, ETRI (Ethylene- response 1), contains a domain homologous to histidine kinase cataytic domain. (d) EIN2 (Ethylene-insensitive 2) encodes a transmembrane protin. The ein 2 mutation promotes ethylene responses in both seedlings and adult Arabi-dopsis plants. Amar Ujala Education Which combination of the above state-ments is correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (b) (D) (d) and (a) Exp. B and C are correct.there are more than two receptors are there for ethylene and ETR1 contains domain homologous to histidine kinase activity D statement is incorrect as EIN2 mutation leads to loss of ethylene responsiveness throughout plant development After hemorrhage, a subject develops hypo- volemia and hypotension. Following are some of the statements regarding homeostatic measure taken by the body after hemorrhage (a) Increased release of vasopression (b) Increased water retention and reduced plasma osmolality (c) Increased rate of afferent discharge from lowpressure receptors of vascular system (d) Decreased rate of afferent discharge from high pressure receptors of vascular system Which one of the following in NOT correct in this condition? (A) Only (a) (B) (a) and (b) (C) Only(c) (D) (b)and(d) Exp. Vasopressin increases peripheral vascular resistance (vasoconstriction) and thus increases arterial blood pressure. This effect appears small in healthy individuals; however it becomes an important compensatory mechanism for restoring blood pressure in hypovolemic shock such as that which occurs during hemorrhage. Age Vasopressin level will be increased and is secreted from the posterior pituitary gland in response to reductions in plasma volume, and in response to increases in the plasma osmolality A visitor to a region of hot climate is more distressed by the than the regular resident. by heat than the regular resident. Within A few weeeks, the visitor is more comfortable with the heat and capacity for work is increased. Following are some of the explanations given by a researcher regarding acclimatization to heat. (a) Sweating begins at a lower body temperature (b) Blood flow through skin is high for any body temperature (c) There is rise in resting body temprature (d) Vasoconstriction start at a lower body temperature Which one of the following is NOT true? (A) Only (a) (B) (a) and (b) (C) Only (c) (D) (c) and (b) Exp. All the other statements are correct. Only Satement D is wrong. Since hydrostatic pressures of GC and TC are different so the osmotic pressure of them will also be different. P3/17

109 Amar Ujala Education 109. The difference in circulation between glomerular capillaries (GC) and true capillaries (TC) are described by a researeher in the following statements (a) The hydrostatic pressure in GC is higher than that in TC (b) The endothelial cells are fenestrated in GC but not in (c) TC Both filtration and fluid movement into capillary takes place in TC but only filtration occurs in GC. (d) The plasma colloid osmotic pressures in both the ends of GC or TC are similar. Which one of the following is NOT corract (A) Only (a) (B) (a) and (b) (C) (b) and (c) (D) Only (d) 110. When rods of retina kept in dark, were exposed to light, photoreansduction occurred. Following are some explations given by a researcher regarding phototransduction (a) Activation of transducin (b) Inhibition of cgmp phosphodiesterase (c) Closure of Na + channels (d) Hyperpolarization of rads Which one did NOT occur in photo- transduction (A) only (a) (B) only (b) (C) (a) and (c) (D) (c) and (d) Exp. All the other statements are correct. Only Satement B is wrong. as during phototrans-duction cgmp phosphodiesterase is activated, not inhibited The afferent nerve fibres of a stretch reflex were electrically stimulated and thecontraction of themuscle innervated innervated by efferent fibres was recorded. The time points of the nerve stimulation and response of the muscle. Which one of the following time durations will be probabel value for the observed synaptic delay? (A) msec (B) 0.5 msec (C) 0.5 sec (D) 5.0 msec Exp. Values of 0.5 ms are typical in this kind of experiment. The delay represents the time it takes for Ca2+ ions to enter the presynaptic terminal and bring about the subsequent release of neurotransmitter, followed by diffusion across thecleftand activation of receptor sites on the postsynaptic membrane A convenient and reasonably reliable indicator of the time of ovulatio is usually a rise in the bassibly because progesterone is thermogrnic. Of the four situations given below, which one is ideal for ensuring pregnancy agter intercourse? Body temperature Progesterone Follicle-stimulating hormone (A) (B) (C) (D) Relative levels Relative levels Relative levels Relative levels Time Time Time Time A/UG-15 Exp. Follicle stimulating hormone is produced by the pituitary gland. In women, this hormone stimulates the growth of ovarian follicles in the ovary before the release of an egg from one follicle at ovulation. In option A, the progesterone and body temperature is very high, at the same time a FSH level is low, so it might be the ideal for pregnancy. In option B, the progesterone level goes up but the temperature is low. Its wrong. In option c, the FSH level is increased, So its wrong. In option D, when progesterone level is high body temperature is very low, so its wrong In an organism, expression of gene X is induced in the presence of a hormone. Genetic analysis showed that the hormonal signal is transduced through two proteins Let 1 and Let 2. The expression of gene X was Let2. The lines over-expressing (OE) the active Lit proteins or combination of both. Results of expression of gene X in presence of the hormone is summarized below Lines Expression of Gene X WT + + Ket1OE ++++ Ket2OE ++++ Ket1 KO Let2 KO Let1 OE Let2 KO Let1KO ++++ Let2 OE P3/18

110 A/UG-15 Based on the above, which one of the following pathways best fits the observatio made? (A) Let1 Hormone Activates both Let2 Both combinationally activate Gene X expressed (B) Hormone activates Let 1 activates Let2 activates gene X (C) HOrmone activates Let2 activates Let1 activates gene X (D) Hormone let2isa repressor activates Let1 of gene'x' Let2 inactivates fene X 114. Two siblings who ingerit 50% of the genome from the mother and 50% from the father show lot of phenotypic differences. Which one of the parents will maximally of the parents will maximally contribute to this difference? (A) Mutaion (B) Recombination (C) Independent assortment (D) Environment Exp. The event that occurs during the gameto-genesis of the parents that creates differences in the phenotypes of the siblings is Recombination. answer is option Consider the following hypothetical pathway X substance A phenotype A H substance B phenotype H allele convert X substance to H substance h allele cannot convert X to H substance and leads to phenotype O A allele converts H substance leading to A phenotype a allele cannot convert H substance B allele converts H substance leading to B phenotype b allele cannot convert H substance An individual with A phenotype when crossed with that of B phenotype has a progeny with O phenotype. Which one of the following orosses can lead to the above observation? B Amar Ujala Education (A) Aahh BbHH (B) AsHh BBHh (C) AaHh BBHH (D) AAHH BbHh 116. Three somatic hybrid cell lines, designated as X, Y and Z, have been scred for presence or absence of chromosomes 1 through 8, as well as for their product the hypothetical gene product A, B, C and D as shown in the following table : Hybrld Human chromosoes Gene products cell present expressed lines A B C D X Y Z Which of the following option has most appropriately assigned chramosomes for each of the given genes? (A) Geme A on chromosome 5, Gene B on chromosome 3, Gene C on chromo-some 8 and Gene D on chromosome 1 (B) Gene A on chromosome 5 and Gene B on chromosome 3 only (C) Gene D on chromosome 8, Gene C on chromosome 1, Gene B on chromo-some 5 and Gene A on chromosome 4 (D) Gene A on chromosome 5, Gene B on chromosome 3 and Gene D on chromo-some 1m Exp. Since gene product A is present in hybrid Y and Z but not in X and same is the pattern shown by chromsome 5, so A is present on chromosome 5. In the same way, gene product B and chromsome 3 are present in X and Z but not in Y, so B is on chromosome 3. Gene product D is present in hybrids X, Y and Z, same as chromosome 1, so D is on chromosome 1. So, A-5, B- 3 and D-1. Now, gene product C is absent in all the cell lines same as chromsome 8 but that doesn't give its conclusive position as gene for C may be present on chromsome 8 or any other chromsome (the one which has not been analyzed). So we are not sure about the position of gene C from the given data. So answer is option D The following diagram shows meiotic pairing in an inversion heterozygote and a point where single crossing over has A A B C C B G O O G D F D F E E P3/19

111 Amar Ujala Education occurred The resulting gametes produced may have (a) The chromosome having normal gene sequence (b) the chromosome having inverted gene sequence (c) a dicentric chromosome with duplication and deletion (d) an acentric chromosome having duplication and deletion (e) the chromosome having duplication and deletion Which of the following combination will be most appropriate for the diagram shown : (A) A,B,CandD (B) A,BandE (C) B,C,DandE (D) A,C,DandE Exp. Diagram given in the question is of pericentric inversion as it includes centromere. When crossing over happens through the pericentric inverted segment in an inversion heterozygote, the gametes formed may be viable (the chromatids that do not participate in crossing over) and non viable (because of deletion of some genes and duplication of some genes). One of the viable gamete will have the correct gene sequence and the other viable gamete will have inverted sequence. So the statements A,BandEarecorrect.Answerisoption Of the following which one of the indivi-duals will NOT necessarily carry the allele responsible for the mentione trait? (A) A woman in a family where an uatosomal dominant trit is segregating and her mother and son are affected. (B) A daughter of a man who is affected by an X-linked dominant trait dominant trait (C) Afatherofachildwhoisaffectedwithanautosomal recessive trait (D) A father of a boy affected with X-linked recessive trait Exp. Since X-linked traits never transfers from father to son, so the father of a boy affected with X-linked recessive trait need not be caryying the responsible allele. So the answer is option Which of the following is the correct match of the algal group with ist food reserve? Algal Gropu Carbohydrate Reserve (a) Bacillariophceae (i) Oil (b) Xanthophyceae (ii) Floriden starch (c) Phaeophyceae (iii) Laminarin (d) Rhaeophyceaecear (iv) Chrysolaminarin (v) Starch (A) a-(iv), b-(i), c-(iii), d-(ii) (B) a-(ii), b-(i), c-(iii), d-(iv) (C) a-(iv), b-(i), c-(ii), d-(v) (D) a-(i), b-(v), c-(iii), d-(ii) P3/20 A/UG-15 Exp. Bracillariophyceae has food reserves in form of chrysolaminirin. Xanthophyceae has food reserves in form of oil. Phaeophyceae has food reserves in form of laminirin. Rhodophyceae has food reserves in form of Floridian starch A research conducts a standard test to identify enteric bacteria (A, B, C) on the basis of their biochemical properties. The result is given in the following table Bacteria Test A B C Indole + Methy Red + + +/ Voges-Proskaure + Based on the above, the identified bacteria B and C are most probably (A) Enterobacter, Salmonella, Escherichia. (B) Escherichia, Salmonella, Enterobacter. (C) Salmonella, enterobacter, Escherichia. (D) Eschrichial, Enterobacter, Salmonella. Exp. This test is important for coliform group of bacteria. The indole test screens for the ability of an organism to degrade the amino acid tryptophan and produce indole.the indole test is used as a classic test to distinguish Indole positive E coli/from enterobactor and klebsiella. Voges Proskauer or VP is a test used to detect acetoin in a bacterial broth culture where VP positive organisms include Enterobacter, Klebsiella, Serratia Following is a cladogram of the major taxonomic groups of the angiosperms A B C D E Monocots Eudicots Groups A-E represent respectively : (A) Astrobaileyales, Nymphaedales, Amborellales, Chlornthacean, Magnoliids (B) Amborellales, Astrobaiteyales, Nymphae-dales, Magnoliids, Chloranthaceae (C) Amborellales, Nymphedales, Astrobaileya-les, Magnoliids (D) Amborellales, Nymphaedales, Chlorantha-ceae, Magnoliids, Astrobaileyales Exp. This is the cladogram of the basal angiosperm taxa within the context of the flowering plants.

112 A/UG The following are some imprortant features which are commonly associated with animal development (a) Position of anus development with respect to blastopore (b) Method of cell division (c) Mechanism of coelom formation (d) Cleavage pattern during egg development Based on the above, which one of the following combinations differentiate the development of deuterostomes from that of protostomes? (A) (a), (b) and (c) (B) (b), (c) and (d) (C) (a), (c) and (d) (D) (a) and (b) Exp. The majority of coelomate invertebrates develop as protostomes ("first mouth") in which the oral end of the animal develops from the first developmental opening, the blastopore. In the deuterostomes ("second mouth": cf. Deuteronomy, "second book of the law"), including Echinodermata and the ancestors of the Chordata, the oral end of the animal develops from a second opening on the dorsal surface of the animal; the blastopore becomes the anus.the Deuterostomes also show indeterminate development, in which each of the cells of the eight-cell embryo if separated remain capable of developing as complete organisms, in contrast to determinate development, in which the development fate of each cell in the adult organism has already been determined The phylogenetic tree of amniote vertebrates is given in the following diagram Ancestral amniote Reptiles The groups labelled A, B, C, D are Mammals (A) A-Snakes, B-Turtles, C-Birds, D-Mammals (B) A-Snakes, B-Turtes, C-Mammals, D-Birds (C) A-Turtles, B-Birds, C-Snakes, D- Mammals (D) A-Birds, B-Turtles, C-Snakes, D-Mammals Exp. Mammals are an outgroup for this phylogenetic tree. and crocodiles are more closer to birds than they are to snakes. and snakes birds and crocodiles are closely related to each other than they are to turtles. Option 3 satisfies all three conditions. A B C D Amar Ujala Education 124. The following are matches made between adult animals and their larval forms (a) Copepods-Nauplius (b) Sea cucumber-zoea (c) Cea urchin-echinapleuteus (d) Crabs-Auricularia (e) Star fish-bipinnaria (f) Brittle star-ophiopleuteus Which one of the combinations below reflects INCORRECT matches? (A) (a) (c), (e) (B) (b) and (d) (C) (b) only (D) (f) only Exp. Eggs of Copepods hatch into nauplius larvae, The first stage of larval development of sea cucumber is known as an auricularia, Echinoderm feeding larvae include dipleurula type (holothuroid auricularia (sea cucumbers); asteroid bipinnaria (starfish)) and pluteus type (ophiuroid ophiopluteus (brittle stars); echinoid echinopluteus (sea urchin)) larvae. Zoea larvae are members of the crab group Which of the following statements about the birth rates (b 1,b 2 ) and death rates (d 1,d 2 ) of species 1and 2 indicated in the figure is Not true? Birth rates/death rates d Population density (A) Birth rates of species 1 anr density-independent. (B) Death rates of both species are dinsity-dependent (C) Birth rates of species 2 are density- dependent (D) Density-dependent effects on death rates are similar for both the species Exp. There are many kinds of density-dependent processes, which can be differentiated according to effect, cause, response, mechanism and severity. The negative effects of density dependence are usually better known than the positive ones. Negative density-dependent effects occur if a vital rate decreases as density increases, positive effects occur if both increase. None of this is seen in birth rates of species 1. Thus we can say that birth rate of species one is density independent and death rates for species 1 and 2 are dependent on density. And birth rate for species 2 is dependent on density too. But the density dependent effects are not similar for both species which is visible from the figure. 1 d 2 b 2 b 1 P3/21

113 Amar Ujala Education 126. Important chemical reactions involved in nutrient cycling in ecosystems are given below (a) NO 2 NO A/UG Which of the following X - Y relation-ships does Not follow the pattern shown in the graph? (b) N 2 NH 3 (c) NH 4 NO 2 (d) NO 3 N 2 The organisms associated with these chemical reactons are (A) a - Nirrosomonas b- Pseudomonas c - Notoc d - Nitrobacter (B) a - Pseudomonas b- Nitrobacter c - Nostc d - Nitrosomonas (C) a - Nitrobacter b -Nostoc c - Nitrosomonas d - Pseudomonas (D) a - Nostoc b - Nitrosomonas c - Nitroacter - d - Pseudomonas Exp. Step a (Nitrobacter) & c (Nitrosomonas) show Nitrification, Step b. N2 fixation (Nostoc) step d. denitrification (Pseudomonas) A population is growing logistically with a growth rate (r) of 0.15/week, in an environment with a carrying capacity of 400. What is the maximum growth rate (No. of individuals/week) that this population can achieve? (A) 15 (B) 30 (C) 22.5m (D) 60 Exp. In the question the following information has given: K = 400, r = 0.15, Maximum population growth at K/2 Therefore, the maximum population size =K/2 = 400/2 = 200 dn/dt = rn[1-n/k] - this is the logistic growth equation dn/dt = (0.15)(200) [1 - (200)/400)] dn/dt = 15 individuals/week 128. In a field experiment, autorophs are provided a 14Clabelled carbon compound for photosynthesis. Radioactivity Radioactivity (14C) levels were then monitored at regular intervals in all the trophic levels. In which ecosystem is the radioactivity likely to be detected fastest at the primary carnivore level? (A) Open ocean (B) Desert (C) Deciduous forest (D) Grassland Exp. The ecosystem with the highest Net Primary productivity is the fastest one to get the radioactivity. Therefore, the order will be Decidous forest, then grassland, then desert, and finally open ocean. P3/22 Y X (A) Number of prey killed (Y) in relation to prey density (X) (B) Photosynthetic rate (Y) in relation to light intensity (X) (C) Species richness (Y) in relation to area (X) (D) Tree spechness (Y) in relation to actual evapotranspiration 130. Following table shows the number of individuals of five tree species in a community Tree species Np. of Individuals A 50 B 20 C 20 D 05 E 05 Based on the above, the Simpson' s diversity (ds) Index of the community will be (A) (B) (C) (D) Exp. The tables for question 131 and 129 have been mistakenly exchanged. There are more similarities in A and B as in they have say species 1 and species 2 present in both the communities and also, there is absence of species 3 and 5. So they are similar in 4 aspects. Presence of 1 and 2 and absence of 3 and 5. For communities B and C they have 2 common aspects presence of species 1 and absence of species 4. While communities C and A are only similar in 1 aspect presence of species Following table shows the presence (+) or abse nce ( ) of five species in three communities ( A, B, C) Community Species A B C

114 A/UG-15 Based on the above, Which of the following is the correct order of similarity between two pairs of communities? (A) A and B > B and C > A and C (B) A and B > A and C > B and C (C) B and C > A and B > A and C (D) A and C > A and B > B and C Exp. Calculating simpson s index through the formula of Simpson s diversity index = S n(n - 1)/N(N-1) where n is the total number of organisms of a particular species. N is the total number of organisms of all species, the answer that we get will be but here in the questions only option that is there gives the value for Simpson Index Approximation which in this case is In several populations, cach of size N=20, it genetic drift results in a change in the relative frequencies of alleles, (a) What is the rate of increase per generation in the proportion of populations in which the allele is lost or fixed? (b) What is the rate of decrease per gencration in cach allele frequency class between 0 and 1? The correct answer for A and B is (A) a 0.25, b (B) a 0.025, b (C) a , b (D) a , b 0.25 Exp. Effect of genetic drift during evolution: Calculate by formulae : Proportion of populations in which allele is fixed or lost increases at a rate of 1/4N per generation = 1/4 20 = Each class of allele frequencies between 0 and 1 decreases at a rate of 1/2N per generation =1/2 20= Individual A can derive 'fitness' cosat of 50 units in doing so. Following Hamilton' s Rule, A should help B ONL if B is has (A) brother or sister (B) first cousin only (C) cousin or uncle (D) nephew or niece Exp. C<r *B This means than in our case where C is 50 and B is 160 the coefficient of relatedness r should be 50% and not lower. R is 0.5 for brother and sister. But it will be lesser for first cousin (12.5 %), for cousin or uncle (25 %) and for nephew or niece (25%). In all the other cases the outcome will be that the cost of the act will be greater than the benefit. And hence only in the case where the receiver is brother or sister of the actor, that the actor will perform the altruistic act The ''Red Queen Hypothesis'' is related to (A) the mating order in the harem of a polygamous male (B) the elimination by deleterious mutations by sexual reproduction P3/23 Amar Ujala Education (C) mate selection process by a female in a lik (D) the evolutionary arms race between the host and the parasite Exp. The original idea is that coevolution could lead to situations for which the probability of extinction is relatively constant over millions of years (Van Valen 1973). The gist of the idea is that, in tightly coevolved interactions, evolutionary change by one species (e.g., a prey or host) could lead to extinction of other species (e.g. a predator or parasite), and that the probability of such changes might be reasonably independent of species age In a population of effective population size Ne, with rate of neutral mutation the frequency of heterozygotes per nucleotide site at equilibrium between mutation and genetic drift calculated as 2Ne o 4Ne o (A) 2N 3 (B) 4N 1 e 0 Ne o 4Ne o (C) 4Ne 0 1 (D) (4Ne 0 1) Exp. Mutation balances the loss of heterozygosity that results from drift. Prediction of heterozygosity (H) in a small population In the absence of mutation, the probability of being autozygous is : Ft = (1/2N) + (1 (1/2N)Ft- 1 (where t is the present generation) But either of the two autozygous alles could mutate, so the probability that BOTH will NOT mutate is (1 u 0 ) 2 Hence, Ft = (1-u)2 [(1/2N) + {1 - (1/ 2N)}Ft-1] At equilibrium, Ft = Ft 1; this reduces to Feq = 1/(1 + 4Neu 0 ) Because H = 1 Feq, then H = (4Neu 0 )/(1+4Neu 0 ) 136. As cancer progresses, several genome rearrangements including translocation, deletion, duplications etc. If these rearrangements are to be identified, which of the following teehniques would be most suitable? (A) RAPD (B) Microarray (C) Multi-colouer FISH (D) Flow cytometry Exp. Microaaray is bacically perform with mrna pool. Flowcytometery is used for the separation of cells on the basis of their size, complexity etc A student noted the following points regarding Agrobacterium tumefaciens (a) A tumefaciens is a gram - negative soil bacterium (b) Opine cataboism genes are present in T-DNA region of T-plasmid e 0

115 Amar Ujala Education (c) Opines are synthesized by condensation of amion acids ketoacids or amion acids and sugars (d) A callus culture of crown gall tissue casued by A tumefaciens in plants can be multiplied without adding phyto-hormones Which one of the combinations of above statements is correct? (A) (a) (b) and (c) (B) (a), (b) and (d) (C) (b), (c) and (d) (D) (a), (c) and (d) Exp. Opines catabolism genes are not present in T-DNA region of Ti plasmid A mixture of two proteins was subjected to following three chromatographic columns: a) Cation exchange, b) Size exclusion (Sep-hadex 100) and (c) Reverse phase. Following elution profiles were obtained Cation exchange Size exclusion A B B A NaCl Absorption Time Absorption Absorption Time Reverse Phase B A Acetonitrile Time Which of the following statements is correct? (A) A is larger and more hydrophodio that B (B) B is more anionic and more hydro- phodic that A (C) A is more hydrophobic and smaller that B (D) A is more cationic and smaller that B Exp. Based on the information below, one can understand. Molecule A is smaller since it elute second in the size exclusion chromatography so option 1 is not correct. Molecule B is more cationic since it elute second in the cation exchange chromatography so Option 2 is not correct.based on reverse phase chromatography A is more hydrophobic and smaller than B (Size exclusion) So option 3 is correct. Molecule A is less cationic since it elute first in the cation exchange chromatography so Option 4 is not correct Glucose in the blood is detected by four different methods (a, b, c and d) the sensitivity and range of detection of glucose by these for methods is shown below. Clinically relevant concentration of glucose in blood is between mg\dL Signal a b c Glucose mg/dt d A/UG-15 Which of the following method is most appropriate? (A) a (B) b (C) c (D) d Exp. In method A the detection starts before 80 and remains at constant value between In method C the detection starts showing significant differences among concentrations after 120 and remains at constant value from , so its cant be selected. In method D the detection starts after 200 so other values before 200 can t be measured. B is the only option where all the concentrations can be observed on the graph The following statements are related to plant tissue culture (a) Friable callus provides the inocuium to form cellsuspension cultures. (b) The process known as habituation to the property of callus loosing the requirement of auxin and \or (c) cytokinin during long term culture Cellulase and pectinase enzymes are usually used for generating protoplast cultures. (d) During somatic embryo development, torpedo stsge embryo is formed before heart stage embryo. Which one of the following combinations of above staetements is correct? (A) (a), (b) and (c) (B) (a), (b) and (d) (C) (a), (c) and (d) (D) (a), (c) and (d) Exp. A is right as the friable callus can be used as an inoculum to produce cell suspension cultures. B is correct as habituation is a condition where plant can grow in absence of auxin and cytokinin. C is correct as cellulose and pectinase are polysachride degrading enzymes that are used to create protoplasts. D is wrong since during somatic embryo development heart stage embryo is formed first followed by torpedo shaped embryo 141. Which one of the following statements is correct? (A) Electrospray ionixation mass spectrum of a compound can be obtained only if it has a net positive charge at ph 7.4 P3/24

116 A/UG-15 (B) Helical content of a truptophan containig peptide can be obtained by examinig the fluorescence spectrum of tryptophan (C) The occurrence of beta sheet in a protein can be inferred from its circular dichroism spectrum (D) The chemical shift spread for a compo-und is more in its 1 H NMR spectrum as compared to its 13 CNMR spectrum Exp. circular dichroism spectroscopy is particularly good for determining whether a protein is folded, and if so characterizing its secondary structure, tertiary structure, and the structural family to which it belongs 142. A researcher is studying the subcellular localization of a particular protein X in an animal cell. The researcher performs successive centrifugation at increasing rotor speed. The researcher start spinning the cellular homogenate at 600g for 10 min, collects the pellet, spins the supernatant at 10,000g for 20 min, collects the pellet, spins the supernatant at 100,000g for 1 hour, collects both the pellet and the final supernatant. On subjecting various pellets and the final supematant to Western blotting with anti-protein-x antibody, the protin X is observed to be maximally expressed in pellet aftar centrifugation at 10,000g. Based on the above observation, what will be the most likely localization of protein X. (A) Nucleus (B) Ribosomes (C) Mitochondria (D) Microsomes Exp. For a typical cell homogenate, a 10 min. spin at low speed ( g) yields a pellet consisting of unbroken tissue, whole cells, cell nuclei, and large debris. The low speed pellet is traditionally called the nuclear pellet. A 10 min. spin at a moderately fast speed, yielding forces of 10,000 to 20,000 g brings down mitochondria along with lysosomes and peroxisomes. Therefore the second pellet in the traditional cell fractionation scheme is called the mitochondrial pellet Fluorescence recovery after photobleach-ing in live cells is used to determine (A) co-localization of proteins (B) distance between two organelles (C) diffusion of proteins (D) nucleic acid compactness Amar Ujala Education Exp. techniques have been developed to investigate the 3- dimensional diffusion and binding of molecules inside the cell; they are also referred to as FRAP You have transiently expressed a new protein (for which no antibody is availale) in a cell line to estavlist structure function relationship. Which one of the folowing strategies is the nost straight forward way to examine the expression profile of this new protein? (A) By metabolic labelling using 35S labelled amino acids (B) Making a GFP fusion protein with this new protein (C) Immunoprecipitating this protein with the help of another protein for which antibody is available (D) Running SDS-PAGE and identify the protein Exp. Metabolic labeling mostly used to track the movement and track the protein. Immun-opprecipitation and sdspage is not possible since antibody is not available. In cell and molecular biology, the GFP gene is frequently used as a reporter of expression.moreover it is used as a proof-of-concept that a gene can be expressed throughout a given organism During an experiment, a student found increased activity of a protein, for which there were three possible explanations, iz, increased expression of the protein ncreased phosphorylation, or increaseo interaction with other effector proteins. After conducting several experiments, the student concluded that increased activity was due to increased phosphorylation which one of the following experiments will NOT support/ provide the correct explanation drawn by the student? (A) Westerm blot analysis (B) Analysis of transcription rate (C) Mass spectroscopy (D) Phospho amino acid analysis Exp. Transcription is the first step of gene expression, in which a particular segment of DNA is copied into RNA (mrna) by the enzyme RNA polymerase. Protein phosphorylation is a posttranslational modification of proteins in which an amino acid residue isphosphorylated by a protein kinase by the addition of a covalently bound phosphate group. So the analysis of transcription rate experiment will not help the to conclude that the increased activity is due to increased phosphorylation P3/25

117 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, June In each of the following groups of words is a hidden number, based on which you should arrange them in ascending order, Pick the right answer : (a) Tinsel event (b) Man in England (c) Good height (d) Last encounter (A) (a), (b), (c), (d) (B) (c), (b), (d), (a) (C) (a), (c), (d), (b) (D) (c), (d), (b), (a) Exp. Hidden number are A = Eleven, B : Nine, C : Eight, D : Ten. So, according to arrangement in ascending order C, B, D, A. 2. Let m and n be two positive integers such that m+n+mn= 118 : Then the value of m+nis : (A) not uniquely determined (B) 18 (C) 20 (D) 22 Exp. If m + n + mn = 118 then mn = 118 (m + n)when22is substituted for m + n value of mn will be can be written as 16 6 that is value of m and n, hence is 22. At the same time (16 6) will be AB is the diameter of a circle. The chord CD is perpendicular to AB intersecting it at P, If CP = 2 and PB =1, the radius of the circle is C A D (A) 1 (B) 2.5 (C) 2 (D) 5 Exp. Mark a point O (in the centre) on AB. Join OC and OD. Hence OC, OD, OB will be radius of circle = X Let OP = Y. Since OB is XOP can be writtne as X-1. Apply pythogarus theorem to triangle OCP.X^2 = 4 + (X 1)^ 2. By solving the equation X will be 2.5 which is the radius of the circle. P B PART-A P4/1 4. What is angle x in the schematic diagram given, below? 130 x (A) 60 (B) 50 (C) 40 (D) 30 Exp. Let the third angle in the triangle containing the X be Y. Second angle would be = 50. From the other triangle, 180 Y + 50 = 130 Y = 100, if Y = 100, X = From a group of 40 players, a cricket team of 11 players is chosen. Then, one of the eleven is chosen as the captain of the team. The total number of ways this can be done is : m [ n below means the number of ways n objects can be chosen from m objects] (A) 11 (B) (C) (D) 10 Exp. We can select 11 players from 40 by 40 C through 11 ways. We can choose 1 captain out of 11 in 11 C 1 way. Total number of ways 11 C 1 40 C 11 = C 11 C 11 C 1 = 11!/ 10! 1! = Starting front a point A you fly one mile south, then one mile East, then one mile North which brings you back to point A, Point, A is NOT the North pole. Which of the following MUST be true? (A) You are in the Northern Hemisphere (B) You are in the Eastern Hemisphere (C) You are in the Western Hemisphere (D) You are in the Southern Hemisphere

118 Amar Ujala Education Exp. If we consider the southern hemisphere, there is a ring may the south pole that has a circumfernece of one mile. So what if we were standing at any point one mile north of this ring. If we walked one mile south, we would be on the ring. Then one mile east would bring us back to some point on the ring (since its circumference is one mile). One mile north from that point would bring us back to the point where we started from. 7. A 3 m long car goes past a 4 m long truck at rest on the road. The speed of the car is 7 m/s. The time taken to go past is (A) 4/7 s (B) 1 s (C) 7/4 s (D) 10/7s Exp. Total distance = = 7, speed = 7, time = Jar W contains 40 white marbles and jar B contains 40 black marbles. Ten black marbles from B are transferred to W and mixed thoroughly. Now, ten randomly selected marbles from W are put back in Jar B to make 40 marbles in each jar. The number of black marbles in W : (A) would be equal to the number of white marbles in B (B) would be more than the number of white marbles in B (C) would be less than the number of white marbles in B (D) cannot be determined from the information given Exp. If the batch of 10 marbles we moved from B to A contained K black ones and 10-k white ones, then A, at the end has K black marble and B has 10 (10 K) k white marbles. Thus the two numbers are the same. 9. Consider 3 parallel strips of 10 m width running around the Earth, parallel to the equator; A 1 at the Equator, A 2 at the Tropic of Cancer and A 3 at the Arctic Circle. The order of the areas of the strips is : (A) A 1 <A 2 <A 3 (B) A 1 =A 2 >A 1 (C) A 1 >A 2 =A 3 (D) A 1 >A 2 >A 3 Exp. the area would be maximum at equator as the radius is maximum. 10. If an S ef I gh M then ns? (A) T (B) A (C) L (D) K Exp. an when reversed becomed Na that is sodium hence S. similarly ef = Fe i.e., iron = 1, gh = Hg = mercury and ns = tin len T. 11. How many non-negative integers less than 10,000 are there such that the sum of the digits of the number is divisible by three? (A) 1112 (B) 2213 (C) 2223 (D) 3334 Exp /3 = P4/2 A/UG Of three persons A, B and C, one always lies while the others always speak the truth. C asked A, Do you always speak the truth, yes or no? He said something that C could not hear. So, C asked B, what did A say? B replied, A said No. So, who is the liar? (A) A (B) 8 (C) C (D) cannot be determined Exp. Either B can lie or A. 13. A B Two ants, initially at diametrically opposite points A and B on a circular ring of radius R, start crawling towards each other, The speed of the one at A is half of that of the one at B. The point at which they meet is at a straight line distance of 3R (A) R from A (B) 2 from A 3R (C) R from B (D) 2 from B Exp. The distance between A and B = pir = 3.14 R. As the speed is in ratio of 2 : 1 and hence the distance covered would be in ratio 2 : 1, 1.2, R from A and 2R from B. 14. Medats Won Total Year 54 Gold Based on the graph, which of the following statements is NOT true? (A) Number of gold medals increased whenever total number of medals increased (B) Percentage increase in gold medals in 2010 over 2006 is more than the corresponding increase In total medals (C) Every time non-gold medals together account for more than 50% of the total medals

119 A/UG-15 (D) Percentage increase in gold medals in 2010 over 2006 is more than the corresponding increase in 2002 over 1998 Exp. Percentage increase in 2010 = 20/ percentage increase in 2002 = 15/ A pyramid shaped toy is made by tightly placing cubic blocks of cm 3. The base of the toy is a square 4 4cm 2. The width of each step is 0.5 cm. How many blocks are required to make the toy? (A) 30 (B) 34 (C) 36 (D) 40 Exp. The base consists of 4 4, the next layer would be 3 3 (as 0.5 has to be step), next would be 2 2, 1 1 Total= = Information in DNA is in the form of sequence of 4 bases namely A, T, G and C, The proportion of G is the same as that of C, and that of A is the same as that of. Which of the following strands of DNA will potentially have maximum diversity (i.e., maximum information content per base)? (A) length 1000 bases with 10% G (B) length 2000 bases with 10% A (C) length 2000 bases with 40% T (D) length 1000 bases with 25% C Exp. The maximum diversity is possible when all the bases occur in the same proportion. 17. Two plane mirrors facing each other are kept at 60 to each other, A point is located on the angle bisector. The number of images of the point is (A) 6 (B) 3 (C) 5 (D) Infinite Exp. n = (360/the ta) 1, the ta = 60, n =5. Amar Ujala Education 18. I bought a shirt at 10% discount and sold it to a friend at a loss of 10%, If the friend paid me Rs, for the shirt, what was the undiscounted price of the shirt? (A) Rs. 900 (B) Rs. 800 (C) Rs (D) Rs Exp. Let the price be x, after discount it would be 0.9x A loss of 90% incurred, so final s.p. = x =0.81x 0.81x = 729 or x = A single celled spherical organism contains 70% water by volume. If it loses 10% of its water content, how much would its surface area change by approximately? (A) 3% (B) 5% (C) 6% (D) 7% Exp. The answer is most likely option 2. Surface are changes by 5% volume of sphere = 4/3 pi r^3. Since 7% is lost (10% of 70%), new volume 0.93 earlier volume. So the radii are also in the proportion r^3 = 0.93 R^3 (consideering r to be earlier radius and R new radius after loss of water) r = R. Surface area = 4 pi r^2. So it will change by a factor of (0.976)^2 = SO the change is approximately 5%. 20. Suppose (1) x =4 (2) Then x 4=x (as both sides are zero) (3) Therefore (x 4)=(x 4)(x +4) Cancelling (x 4)frombothsides (4) 1 = (x +4) (5) Then x = 3 Which is the wrong step? (A) 1to2 (B) 2to3 (C) 3to4 (D) 4to5 Exp. X-4 can not be equal to X^2 4^2, it should be (x 4)^2. PART-B 21. A 1% (w/v) solution of a sugar polymer is digested by an enzyme (20 μg, MW = 200,000). The rate of monomer sugar (MW = 400) liberated was determined to have a maximal initial velocity of 10 mg formed/min. The turnover number (min 1 ) will be (A) (B) (C) (D) Exp. Keat is the turnover number calculated as kcat = Vmax/ Et where, Vmax = maximum velocity, Et = total enzyme concentration. In the question we should convert the concentration terms to molarity. Vmax = /400 = ,Et= / = Kcat = /10 10 = /min. 22. In an alpha helical pypeptide, the back bone hydrogen bonds are between (A) NH of n and CO of n + 4 amino acids (B) CO of n and NH of n + 3 amino acids (C) CO of n and NH of n + 4 amino acids (D) NH of n and CO of n + 3 amino acids Exp. The backbone H-bonds are between Co of n and NH of n + 4 amino acids in an alpha helical polypeptide. P4/3

120 Amar Ujala Education 23. Following are three single stranded DNA sequences that form secondary structures. (a) ATTGAGCGATCAAT (b) ATTGAGCGATATCAAT (c) AGGGAGCGATCCCT Based on their stability, which one is correct? (A) (a)=(b)=(c) (B) (c)>(a)>(b) (C) (b) > (c) = (a) (C) (b) > (c) > (a) Exp. In this case the DNA sequence has the highest amount of GC content among the three and hence would have maximum stability 24. Which one of the following enzymes is NOT part of pyruvate dehydrogenase enzyme complex in glycolysis pathway? (A) Pyruvate dehydrogenase (B) Dihydrolipoyl transferase (C) Dihydrolipoyl dehydrogenase (D) Dihydrolipoyl oxidase Exp. Dihydrolipoyl oxidase is not a part of pyruvate dehydrogenase enzyme complex in glycolysis pathway. 25. What phenotype would you predict for a mutant mouse lacking one of the genes required for site-specific recombination in lymphocytes? (A) Decrease in T cell counts (B) Immunodeficient (C) Increase in T cell counts (D) Increase in B cell counts Exp. Autation in genes required for site specific recombination results in reduced or absence of mature T and B cells. Immature cells are still present resulting in immune deficient condition. 26. The key determinant of the plane of cytokinesis in mammalian cells is the position of (A) chromosomes (B) central spindle (C) centrioles (D) per-prophase band 27. Beating of cilia is regulated by (A) actin (B) myosin (C) cofilin (D) nexin Exp. Beating of cilia is carried through nexindyenin microtubule activity. All the other options are associated with microfilaments. 28. Cystic fibrosis transmembrane conductance regulator (CFTR) is known to control the transport of which ion? (A) Ca +2 (B) Mg +2 (C) HCO 3 (D) CI Exp. The transport of chloride ions helps control of the movement of water in tissues, it is necessary for the production of thin and freely moving mucus. P4/4 A/UG In bacteria, heat-shock response is primarily controlled by (A) Sigma S ( S ) (B) Sigma32( 32 ) (C) Sigma E ( E ) (D) Sigma70( 70 ) Exp. Sigma 32 (RpoH) The heat-shock signe factor is turned on when the bacteria are exposed to heat. 30. In type II splicing (A) a G-OH from outside makes a nucleophilic attack on 5'P of first base of intron (B) a free 2'O of an internal adenosine makes a nucleophilic attack on 5'P of first base of intron (C) A 3'O of an internal adenosine makes a nucleophile attack on 5'P of first base of intron (D) the hydrolysis of last base of exon is carried out by U2/U4/U6 Exp. The 2 OH of the adenosine will attack the 5 s plice site. 31. Given below are some statements about prokaryotic and eukaryotic mobile genetic elements or transposons. A. Most mobile genetic elements in bacteria transpose via an RNA intermediate. (b) Most mobile genetic elements in bacteria are DNA. (c) Mobile genetic elements in eukaryotes are only retrotransposons. (d) Both, RNA and DNA transposons, are found in eukaryotes. Choose the correct combination. (A) (a) and (c) (B) B and (c) (C) (a) and D (D) B and D Exp. In bacteria most transposaus are DNA and eucaryotes have both DNA and RNA. 32. Copying errors occurring during replication are corrected by the proof reading activity of DNA polymerases that recognize incorrect bases (A) at the 5' end of the growing chain and remove them by 5' 3' exonuclease activity (B) at the 3' end of the growing chain and remove them by 5' 3' exonuclease activity (C) at the 3' end of the growing chain and remove them by 3' 5' exonuclease activity (D) at the 5' end of the growing chain and remove them by 3' 5' exonuclease activity Exp. As the proof reading activity is carried out at the 3, growing end and it done by 3 5 exonuclease activity. 33. During each cycle of chain elongation in translation, how many conformational changes does the ribosome undergo that are coupled to GTP hydrolysis? (A) Zero (B) One (C) Two (D) Three Exp. As only elongation factors undergo for conformational changes, ribosomes do not undergo for conformational changes during the GTP hydrolysis.

121 A/UG Collagens are the most abundant component of the extracellular matrix. In order to maintain normal physiological processes like wound healing, bone development, etc., which one of the following type of enzymes is MOST important? (A) Peptidases (B) Proteases (C) Amylase (D) Lipases Exp. Procollagen peptidase helps in the procesisng of procellagen to collagen. 35. Which one of the following events NEVER activates the G-protein coupled receptor for sequestering Ca 2+ release? (A) Interaction of bindin to sperm receptors (B) Activation of Frizzled by Wnt (C) Cortical reaction blocking polyspermy (D) DNA synthesis and nuclear envelope breakdown. Exp. ALl others would result in increased concentration of Ca 2+ except for DNA synthesis and nucleur envelope breakdown which might occur due to activation of a camp dependent kinase. 36. The main difference between normal and transformed cells are (A) immortality and contact inhibition (B) shorter generation time and cell mobility (C) apoptosis and tumour suppressor gene hyperfunction (D) inactivation of oncogenes and shorter cell cycle duration Exp. Immortality and contact inhibition transformed cells have lossof tumorsuppressor activity and activation of tumor suppressor gene. 37. When bacteria growing at 20 C are warmed at 37 C, they are most likely to synthesize membrane lipids with more (A) short chain saturated fatty acids (B) short chain unsaturated fatty acids (C) long chain saturated fatty acids (D) long chain unsaturated fatty acids Exp. Long chain saturated fatty acids would be respondisble for decreasing the fluidity of the plasma membrane. As the bacteria would be shifted from lower to higher temperature therefore, it would be requireing long chain saturated fattya cids for its survival. 38. Hydra shows morphallactic regeneration and involves which one of the following signal transduction pathway in its.axis formation? (A) Wnt/ -catenin pathway P4/5 Amar Ujala Education (B) Retinoic acid pathway (C) FGF pathway (D) Delta-Notch pathway Exp. The pathway which plays an important role in axis formation during morphollactic regeneration in Hydra would be beta catenin/wnt pathway. 39. The mammalian oocyte prior to sperm entry is arrested at what stage of cell division? (A) Prophase of mitosis (B) Prophase of meiosis I (C) G 1 phase of mitotic cell cycle (D) Metaphase of meiosis II Exp. Secondary occyte is arrested at metaphase II of meiosis II till the time sperm enters the secondary ocyte which results in completion of second meiotic division of formation of ovum. 40. The pluripotency of the inner cell mass in mammals is maintained by a core of three transcription factors namely, (A) Oct 4,. Sox 2 and Nanog (B) Oct 4, Sox 2 and Cdx2 (C) Sox 2, Nanog and Cdx2 (D) Oct 4, Cdx2 and Nanog Exp. Inside cells, however, do not turn on the Cd 2 gene, and express high level sof Oct 4, Nanog, and So 2. These genes suppress Cd 2 and the inside cells maintain pluripotency generate the ICM and eventually the rest of the embryo proper. 41. Which one of the following statements about LEAFY (LFY), a regulatory gene in Arabidopsis thaliana, is correct? (A) LEAFY (LFY) is involved in floral meristem identity (B) LEAFY (LFY) is involved in leaf expansion (C) LEAFY (LFY) is involved in root meristem identity (D) LEAFY (LFY) is responsible for far-red light mediated seedling growth Exp. In Arabidopsis thaliama, LEAFY is involve in floral meristen identity. 42. Hie quantum yield of oxygen, evolution during photosynthesis drastically drops in far-red light This effect is known as : (A) Far red drop. (B) Red drop. (C) Blue drop. (D) Visible spectrum drop. Exp. Wavelengths beyond 700 nm are apperently of insufficient energy to drive any part of photosynthesis. Therefore a huge drop in efficiency has been noticed at 100 nm. It is a part of Emerson effect.

122 Amar Ujala Education 43. Dark-grown seedlings display triple response when exposed to ethylene. Which one of the following is NOT a part of triple response? (A) Decrcase in epicotyl elongation. (B) Rapid unfolding and expansion of leaves. (C) Thickening of shoot. (D) Horizontal growth of epicotyl. Exp. In triple response of seedings when exposed to ethylene, thickening and shortening of hypocotyl with prominent apical hook occurs. 44. Which one of the following compounds is generally translocated in the phloem? (A) Sucrose (B) D-Glucose (C) D-Mannose (D) D-Fructose Exp. Sucrose is translocated through the phloem in plants. 45. Nitrogen gas is reduced to ammonia by nitrogen fixation method. In order to execute the process, which one of the following compounds is usually required? (A) ATP (B) GTP (C) UDP (D) ADP Exp. The process is coupled to the hydrolysis of 16 equivalents of ATP and is accompanied by the conformation of one molecule of hydrogen. 46. The S wave of normal human ECG originates due to. (A) septal and left ventricular depolarization. (B) late depolarization of the ventricular walls moving back toward the AV junction (C) left to right septal depolarization (D) repolarization of atrium Exp. The S wave is the first negative deflection after the R wave. It represents the late ventricular depolarisation. 47. Which one of the following skeletal muscles of human body contains highest number of muscle fibre in a motor unit? (A) Muscles of hand (B) Extraocular muscles (C) Muscles of leg (D) Muscles of face Exp. Thigh muscles can have a thousand fibers in each unit. 48. In which of the following conditions is Basal Metabolic Rate (BMR) the lowest? (A) Awake and resting (B) Prolonged starvation (C) Sleep (D) Higher environmental temperature 49. Which one of the following combinations must be present in a steroid receptor that is located in the cytoplasm? (A) Nuclear export sequence (NES) leucine zipper A/UG-15 (B) NES, zinc finger motif (C) Nuclear localization sequence (NLS), zinc finger motif (D) NLS, leucine zipper Exp. Steroid receptor requires a NLSsignal and DNA binding domain is zinc fingers. 50. Segregation of alleles can occur at Anaphase I or at Anaphase II of meiosis. With reference to this statement, which one of the following organism is an ideal model system, for identifying stage of allelic segregation at meiosis? (A) Neurospora crassa (B) Saccharomyces cerevisiae (C) Drosophila melanogaster (D) Pisum sativum Exp. The ordered tetral is observed Neurospora crasa. 51. Genes A, B and C control three phenotypes which assort independently. A plant with, the genotype Aa Bb Cc is seifed. What is the probability for progeny which shows the dominant phenotype for AT LEAST ONE of the phenotypes controlled by genes A, B and C? (A) 1/64 (B) 27/64 (C) 63/64 (D) Cannot be predicted Exp. the probability of having no dominant phenotype = 1/ 64 (triple homozygous recessive) therefore the probability of having atleast one dominant = 1 (1/64) = 63/ Hybrid dysgenesis in Drosophila is caused by P-elements, Which one of the following crosses between different cytotypes will lead to dysgenesis? (A) M-cytotype o + M-cytotypeo (B) M-cytotype o + P-cytotypeo (C) P-cytotype o + M-cytotypeo (D) P-cytotype o + P-cytotype o Exp. Hybrid dysgenesis refers to the high rate of mutation in germ line cells of Drosophila strains resulting from a cross of males with autonomous P-element. 53. In an experiment, clones of a plant is grown in a field. The plants were observed to be of different heights. When a graph was plotted for frequency of plants (Y-axis) against different heights (X-axis), a belt-shaped curve was obtained. From the above it can be concluded that the observed variation in height is due to (A) it being a polygenic trait (B) environmental effect (C) variation in genotype (D) influence of environment on different genotypes Exp. Bell-shaped is a characteristics of polygenic inheritance and almost all the polygenic inheritance is effected by environment. P4/6

123 A/UG Which one of the following viruses cause acute gastrointestinal illness due to contamination of drinking water? (A) Norovirus (B) Poliovirus (C) Rotavirus (D) Filoviruses Exp. norovirus and rotarivus both cause acute gastrointestinal illness. Rotavirus is more associated with pediatric acute gastrointeritis. Whereas norovirus causes gastrointeritis in all age groups. It is more contagious as compared to rotavirus and is often called winter vomiting bug. 55. The phylum in which the animals are bilaterally symmetrical in the larval stage and radially symmetrical in the adult stage is (A) Coelenterata (B) Nematoda (C) Mollusca (D) Echinodermata 56. Which of the following fungal groups has septate hyphae and reproduces asexually by budding, conidia and fragmentation? (A) Basidiomycota (B) Zygomycetes (C) Chytrids (D) Glomeromycota Exp. Only basidiomycetes have separate hyphae which reproduce asexually by budding, conidia and fragmentation, e.g., yeast. In zygomycetes hyphae are aseptate. Chytrids is also a group of fungi which are aseptate and they reproduce through zoospores. Glomeromycota have generally coenocytic mycelia and reproduce aseuxally through blastic development of hyphal tip to produce glomerospores. 57. The most commonly used molecular tool for phylogentic analysis involves sequencing of (A) mitochondrial DN A (B) mitochondrial RNA (C) ribosomal RNA (D) nuclear DNA Exp. Ribosomal RNAs are used in resconstructing polylogeniesdue to the slow rates of evolution in this region of gene concerned. 58. The dynamics of any subpopulation within a metapopulation differs from that of a normal population in that the (A) birth rates are lower than the death rates, (B) death rates are lower than the birth rates. (C) immigration and emigration rates are significantly higher. (D) immigration and emigration rates are negligible. Exp. As compared to normal population, individuals of subpopulation within a metapopulation are dynamic in nature which is determined by their contineous immigration and emigration. P4/7 Amar Ujala Education 59. Which of the following is NOT semelparous? (A) Dracena (B) Bamboo (C) Cicada (D) Mayfly Exp. A species is considered semelparous if it is characterised by a single reproductive episode before death. 60. The general relation between generation time (T) and population growth rate (r) is described by the equation (A) lnr = lna b lnt (B) r = a bt (C) lnr = lna +blnt (D) r=a+bt Exp. As there should be inverse relation between r and T. So that can be justified only by option Which of the following is likely to contribute to the stability of an ecosystem? (A) High number of specialists (B) Fewer number of functional links (C) More omnivores (D) Linear rather than reticulate food webs 62. In eusocial insects, males develop from unfertilized eggs while females develop from fertilized eggs. The ultimate consequence of this difference is that. (A) in any colony there are always more males than females (B) a female is genetically more closely related to her sister than to her own offspring (C) females are behaviorally more dominant than the males (D) in any colony there are always more females than males Exp. Females have about 3/4 similarity with their sisters, however, have only 1/2 similarities with their daughters. 63. An extraordinary sensory ability that elephants possess is (A) emission and detection of ultra high frequency sounds (B) emission and detection of ultra low frequency sounds (C) detection of changes in earth's magnetic field (D) possession of ultraviolet vision Exp. The hearing capabilities of elephant is superior in comperison to human beings. they use infrasonic more frequently for hearing. they detect sounds as low as 14 to 15 hertz (human range low range is 20 hertz) and as high as 12,000 hertz (human being range = 20,000 hertz).

124 Amar Ujala Education 64. According to which evolutionary theory, there are long periods without significant evolutionary changes interrupted by short episodes of rapid evolution? (A) Punctuated equilibrium (B) Saltation (C) Mutation (D) Neutrality Exp. In punctuated equilbrium the species do not evolue for a long period of time and would evolve very rapidly for a short period of time. 65. The mean (μ) and standard deviation ( ) of body size in a Drosophila population are 8.5 and 2 mm, respectively. Under natural selection over many generations the μ and of body size change to 8.5 and 0.8 mm, respectively. The type of natural selection responsible for the change is called. (A) directional (B) neutral (C) disruptive (D) stabilizing Exp. As the mean remains same but variance decreases and hence this is an example of stabilizing selection. 66. A researcher would like to monitor changes in the level of a serum protein for which an antibody is available. Which one of the following methods would be best suited for the purpose? (A) Immunofluorescence microscopy (B) Fluorescence in situ hybridization (C) Enzyme linked immunosorbent assay (D) Fluorescence activated cell sorting Exp. As serum is found outside the cell and hence ELISA would be used. 67. Which one of the methods listed below is the most sensitive label-free quantification method for proteins (A) UV spectroscopy (B) Infra-red spectroscopy (C) Raman spectroscopy (D) 13 C content of protein Exp. Ultraviolet spectroscopy can be used to quantitate proteins using tyrosine and tryptophan content, rest of the methods can not be effectively used for quantitation. 68. What will happen if DNA is labeled by nick translation while doing DNA foot- printing? (A) Nick translation will facilitate better analysis because entire DNA will be labeled and proteins binding at any region of DNA can be demarcated with precision, P4/8 A/UG-15 (B) This will allow arranging the DNA fragment in the desired order, (C) Labeling by random priming may be advantageous as it generates smaller fragments which can penetrate tissue easily, (D) The linear order of fragments from 5' 3' end of DNA cannot be arranged. Exp. The protocol of DNA foot printing involves one of the ends to be labelled so that after digestion with DNA the distance from the end can be observed by autoradiography. But we incorporate label by nick translation. This will allow the label to present across the lenght of the DNA to be analysed and we can not accurately determine the length of the DNA fragments. 69. A protein has one tryptophan and one tyrosine in its sequence. Assume molar extinction coefficients at 280 nm of tryptophan and tyrosine as 3000 and 1500 M 1 cm 1, respectively. What would be the molar concentration of that protein if its absorption at 280 nm is 0.90? (A) 2 mm (B) 0.4 m M (C) 0.2 mm (D) 0.02 mm Exp. In the question taking path length as 1 cm A = ecl. The combined extinction coefficient needs to be considered which is e = e trp no. of trp + e tyr no. of tyr. In the question since both tyr and trp are one each. So the combined molar extinction coefficient is simply the sum of both i.e = Putting volues in equation, 0.90 = 4500 XC 1, C = 0.90/4500 = M or 0. mm. 70. Two groups (Control, Treated) are to be compared to test the effect of a treatment Since individual variability is high in both groups, the appropriate statistical test to use is (A) Analysis of variance. (B) Kendall s test, (C) Student s t-test (D) Mann-Whitney U-test. Exp. T-test is used in case of high variability.

125 A/UG-15 PART-C Amar Ujala Education 71. Sting of a bee causes pain, redness and swelling. Melittin is a major peptide in bee venom. Melittin is a membrane binding peptide that is involved in activating phospholipases in the membrane. The possible target phospholipases that is activated by melittin is (A) Phospholipase C to generate inositol phosphates. (B) Phospholipase A 2 to generate arachidonic acid (C) Phospholipase D to generate 1', 3'-inositol (D) Phospholipase A 1 to generate palmitic acid Exp. PLA2 are commonly found in mammalian tissues as well as insect and snake venom. [2] Venom from both snakes and insects is largely composed of melittin, which is a stimulant of PLA2. Due to the increased presence and activity of PLA2 resulting from a snake or insect bite, arachidonic acid is released from the phospholipid membrane disproportionately. As a result, inflammation and pain occur at the site. 72. The gtycolysis and citric acid cycles are important pathways to generate energy in the cell. Given below are statements regarding the production of ATP, (a) Electrons released during the oxidative steps of glycolysis and citric acid cycle produce 10 molecules of NADH and 2 molecules of FADH 2 per molecule of glucose. (b) Electrons released during the oxidative steps of glycolysis and citric acid cycle produce 20 molecules of NADH and 4 molecules of FADH 2 per molecule of glucose. (c) The coenzymes produced are oxidized by electron transfer chain. (d) The conversion of ADP and P i to ATP takes place in the intermembrane space of mitochondria. Which one of the following combinations of above statements is correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (a) and (c) Exp. Oxidation of glucose involves glycolysis during which 2 NADH are produced by the action of glyceraldehyde- 3-dehydrogenase, conversion of pyruvate to acetyl coa by PDH complex produces 2 NADH, oxidation of acetyl coa by citric acid cycle produces 6 NADH molecules and 2 FADH2 molecuels. These reduced co-enzymes are oxidized back by electron transfer chain. 73. In a 30-residue peptide, the dihedral angles, have been determined by one or more methods. When their values are examined in the Ramachandran plot, it is (A) not possible for, values to be distributed in the helical as well as beta sheet region. (B) possible that the, values are all in the helical region although circular dichroism spectral studies indicate beta sheet conformation (C) possible to conclude that the peptide is composed of entirely D-amino acids (D) not possible to conclude if the peptide is entirely helical or entirely in beta sheet conformation. 74. Hydrogen bonds in proteins occur when two electronegative atoms compete for the same hydrogen atom Donor H....Acceptor The angle ' ' between donor and acceptor of a hydrogen bond was determined from large number of X-ray structures of proteins, as shown below; No. of hydrogen bonds a b c No. of hydrogen bonds Which one of the distribution of ' ' was observed from the proteins? (A) Only b (B) Onlya (C) Onlyc (D) a and b Exp. It has been observed that 90% of N-H---O bonds in proteins lie between 140 and 180, and that they are centred around 158 C. There fore graph b seems more appropriate. 75. In the accompanying figure, reaction kinetics of three proteins (a, b, c) is presented. Protein concentrations used to obtain this data are a 1 mg/ml; b 4 mg/ml, c 2mg/ml. If catalytic efficiency is defined as k cat /K m, which of the following statements is correct? (A) b>c>a (B) a>b>c (C) a>c>b (D) c>a>b Exp. Calculate Kcat for each of the protein as Vmax/Protein concentration. Km is calculatd from the graph as substrate conc at which 1/2 Vmax is obtained, this value for all the protein comes to be 1 um. So the catalytic efficiency depends on Kcat alone. No. of hydrogen bonds P4/9

126 Amar Ujala Education 76. Lipid rafts are involved in signal transduction in cells. Rafts have composition different from rest of the membrane. Rafts were isolated and found to have cholesterol to sphingolipid ratio of 2 : 1. The estimated size of the raft is 35 nm 2. If the surface areas of cholesterol is 40 Å and sphingolipid is 60 Å, how many cholesterol and sphingolipids are present in one raft? (A) 50 cholesterol : 25 sphingolipid (B) 200 cholesterol : 100 sphingolipid (C) 40 cholesterol : 20 sphingolipid (D) 20 cholesterol : 10 sphingolipid Exp. The surface area of the raft is 35 nm2 which is equal to 3500 A2. Cholesterol to sphingolipid ratio is 2 : 1. So if we have y residues of sphingolipid there are 2y molecules of cholesterol. the equation will be 2y *40+y*60= 3500, this gives y = 25 (sphingolipids) and cholesterol as Glycolipids and sphingomyelin are produced by the addition of sugars or phosphorylcholine to ceramide on cytosolic and luminal surfaces, respectively, of the Golgi apparatus. Finally, after such modifications, these molecules are located on the outer half of the plasma membrane. What key events are responsible for such localization? (A) Membrane fusion only, (B) Action of Flippase and membrane fusion. (C) Action of only Flippase, (D) Flip flop of these molecules in the golgi membrane catalyzedbyprotonpump. Exp. According to the question, glycolipids would be facing the cytosolic side while sphingomyelin would be facing the luminal side of Golgi. Now as the vesicle would be budding from the Golgi, the luminal portion of Golgi would be becoming the extracellular portion of the plasma membrane while the cytoplasmic portion would remain the cytoplasmic portion of plasma membrane, so, sphingomyelin would be facing the extracellular side but for glycolipids to face the extracellular side, we will be requiring flippase. So, flippase along with membrane fusion between vesicle and plasma membrane would be involved. 78. In an attempt to study the transport of secretory vesicles containing insulin along microtubules in cultured pancreatic cells, how would treatment with colcemid affect the transport of these vesicles? (A) Colcemid induces polymerization of microtubules, which in turn would activate vesicular transport, (B) Polymerization of microtubules is inhibited by colcemid, which in turn would inhibit the transport of secretory vesicles. A/UG-15 (C) Colcemid inhibits the vesicular trafficking through inactivation of v-snare protein. (D) Colcemid activates t-snare proteins and in turn activates vesicular transport, Exp. Polymerization of microtubles is inhibited by colcemid, which in turn would inhibit the transport of secretory vesicles. 79. When circular plasmids having a centromere sequence are transformed into yeast cells, they replicate and segregate in each cell division. However, if a linear chromosome is generated by cutting the plasmid at a single site with a restriction endonuclease, the plasmids are quickly lost from the yeast. It is known that genes on the plasmids are lost because of the instability of the chromosome ends. What could be done so as to restore its stability and can be inherited? (A) Methylation of adenine residues of the plasmid (B) Complexing the plasmid ends with histone proteins. (C) By incorporating telomere sequences to the end of plasmid. (D) By incorporating acetylated histone proteins to the plasmid ends. Exp. Telomeres are ends of linear chromosomes. they are specific structures that protect the inside relevant DNA region from getting digested by nucleases and also prevent from chromosome fusion. Plasmid like pyac system use this technology of having telomeres at the ends to make the plasmid more stable. 80. Rec 8 is a meiosis specific cohesin that maintains centromeric cohesion between sister chromatids in meiosis I. Which of the phenotypes listed below would you predict will be manifested in a rec 8 yeast? (A) Only low viability of dyads. (B) Improper reduction division and low viability of tetrads, (C) Improper equational division and low viability of dyads, (D) Low tetrad viability with no effect on reduction division. Exp. Since cohesion is mutated, therefore sister chromatid will not be joined together and hence improper reduction division in meiosis will result into low viability of tetrads. 81. Eukaryotic DNA polymerase has tightly associated primase activity but moderate processivity. DNA polymerase and are highly processive but lack primase activity, Given below are four statements about leading and lagging strand synthesis in eukaryotes. Which one is true? P4/10

127 A/UG-15 (A) Both leading and lagging strands are synthesized by DNA polymerase. Moderate processivity is essential to maintain fidelity of replication. (B) Entire leading and tagging strands are synthesized by and. Bukaryotic replication is primer independent process. (C) Only the lagging strand synthesis needs primer and synthesized by DNA polymerase. (D) Primers for both the strands are synthesized by DNA polymerase followed by Polymerase switching with and. Exp. Once primase has created the RNA primer, Pol a strats replication elongating the primer with ~ 20 nucleotides. [18] Due to their high processivity, Pol e and Pol d take over the leading and lagging strand synthesis from Pol a respectively. 82. In context to lac operon, if two bacterial strains P 1 and P 2 with the genotypes O C I + Z and O + I Z + respectively, were used to produce mero-diploid daughter strain D, which one of the following statements correctly predicts the expression of Z gene ( galactosidase activity) in all the three strains? (O +,I + and Z + denote the wild type allelle of the respective genes). (A) P 1 No expression; P 2 constitutive expression; D inducible expression, (B) P 1 No expression; P 2 constitutive expression; D constitutive expression. (C) P 1 No expression; P 2 Inducible expression; D Inducible expression, (D) P 1 Inducible expression; P 2 constitutive expression; D Inducible expression. Exp. P1 has genotype Oc 1 + z-, this strain even though the operator carries constitutive mutation phenotype, will still not result in beta-gal production as the structural genes Z-is defective. Strain P2 phenotype is O + 1 Z +, here repressor is defective, so this will result in constitutive expression of beta-gal as there is no active repressor to block the operator region. Merodiploid will be represented as Oc 1 + Z-/O + 1 -Z +. Here, Lac 1 is a trans acting factor, so in the merodiploid strain, which has both 1 + and 1-copy,will essentially behaves as 1+. However operator is a cis-acting factors so merodiploid Oc/O+ will behave as O+ as functional Z gene is present on O+ chromosome. Therefore the expression of beta-gal will be inducible. 83. In order to study the transcription factor TFIIH, it was cloned from a large number of human subjects, Surprisingly, the subjects having mutation in TFIIH, also showed defects in their DNA repair system. Given below are the explanations : (a) DNA damage is always associated with transcription inhibition, Amar Ujala Education (b) TFIW has no role in DNA repair. (c) In mammalian system, TFIIH plays an active role in transcription coupled DNA repair process, (d) Because of mutation in TFIIH, transcription initiation is inhibited and incompletely synthesized mrnas remain attached to the template DNA leading to DNA damage. Choose the correct answer. (A) (a) and (b) (B) (c) only (C) (b) and D (D) (d) only Exp. TFIIH, have helicase and ATPase activities and help create the transcription bubble, and TFIIH is involved in nucleotide excision repair. 84. The 3' end of most eukaryotic mrnas is defined by the addition of a poly A tail a processing reaction called polyadenylation. The addition of poly A tail is carried out by the enzyme Poly (A) Polymerase, Given below are few statements about this process; (a) Poly (A) Polymerase is a template independent enzyme. (b) Poly (A) Potymerase catalyses the addition of AMP (c) from datt to the 3' end of mrna. Poly (A) Polymerase is a RNA-template dependent enzyme. (d) Poly (A) Polymerase catalyzes the addition of ADP from ATP to the 3' end of mrna. E. Poly (A) Polymerase catalyzes the addition of AMP from ATP to the 3' end of mrna, F. Poly (A) Polymerase catalyzes the addition of AMP from dadp to the 3' end of mrna. Which of the following combination is true? (A) (b) and (c) (B) (c) and D (C) (a) and E (D) (c) and F Exp. Transcription termination in eukaryotes is less understood but involves cleavage of the new transcript followed by template-independent addition of adenines at its new 3' end, in a process called polyadenylation. When the RNA is cleaved, polyadenylation starts, catalysed by polyadenylate polymerase. Polydenylate polymerase builds the poly (A) tail by adding adenosing monophosphate units from adenosine triphosphate to the RNA, cleaving off pyrophosphate. 85. With an intention to identify the genes expressed in an organism at specific stage of development, mrnas were isolated from the given organism, cdnas were synthesized. cloned in a suitable vector and sequenced. A few of the cdna sequences showed no matches with the genomic DNA sequence. Further, it was observed that these sequences were U-rich and found to be in stretches dispersed along the sequence. The following may be possible reasons for appearance of such RNA; P4/11

128 Amar Ujala Education (a) Splicing (b) Alternate splicing (c) Tram-splicing (d) Guide RNA mediated introduction of Us involving endonuclease, terminal-u- transferase and RNA ligase E. Deaminations converting C to U Which of the following is the most appropriate reason/s? (A) (a) and (c) (B) (b) and (d) (C) (c), (d) and E (D) (d) only Exp. RNA editing through the addition and deletion of uracil has been found in kinetoplasts from the mitochondria of Trypanosoma bruce the mechanism of the editosome involves an endonucleolytic cut at the mismatch point between the guide RNA and the unedited transcript. The next step is catalyzed by one of the enzymes in the complex, a terminal U-transferase, which adds Us from UTP at the 3' end of the mran. Here U is found in the stretches dispersed along the sequence, so deamination of C resulting in U will also happen. Further trans splicing followed by demaination could result in sequences which doesn t match genomic DNA. 86. When one isolates ribosomes from bacterial lysate, apart from 70S, 50S and 30S ribosomal subunits, one also finds a small population of 100S, 130S and 150S subunits. EDTA dissociates these larger ribosomal subunits into 50S and 30S, suggesting that they have both the subunits. Upon addition of cations, they reassociate into 70S, but none of the other forms could be detected. What is the reason for not obtaining the >70S forms? (A) The effects of EDTA cannot be reversed by the addition of cations. (B) 100S, 130S and 150S are modified form of ribosomes that are irreversibly damaged by EDTA. (C) 100S, 130S, etc. represent polysome that cannot be reassembled denovo without other cellular components. (D) They are obtained as an experimental artifact in preparations of ribosomes. Exp. Under such specific conditions in microorganisms during stationary phase there is presence of 100 S ribosome which do not have translational activity but formation of such ribosomes requires specific proteins. 87. Certain chemokines are known to suppress HIV infection whereas proinflammatory cytokines are known to enhance infection. In order to explain these findings, control and chemokine receptor knock-out animals were treated with proinflammatory cytokines followed by HIV administration and then infection was assessed periodically. Which one of the graphical representation given below best explain the experimental results : (A) Viral load in blood (B) Viral load in blood (C) (D) Control Chemokine receptor knockout animals Time Time A/UG-15 Exp. Viral load will be increased in the test group because of knockout of chemokine receptors. Plus due to further addition of proinflammatory cytokines and HIV viral load will further increase. In the control group, viral load will be comparitively less due to presence of chemokine receptors. 88. Of the following signaling processes, which one is NOT involved in cellular movement or cytoskeletal changes? (A) B B B P4/12

129 A/UG-15 (B) (C) (D) Ca Rock 3+ GAP Daam1 bloding site Raa-GTP Raf RhoA SOS Dsh 7 Calonin DEX ERX Exp. In picture 1. no cellular signalling is involved, it depicts cellular migration via polymerisation of actin filaments along the leading end of the cell. 89. A researcher was studying a protein X which has been observed to move across cells when an extracellular electrical stimulus is provided. An artificial peptide P was prepared which resembles the structure of connexins and competitively inhibits connexon formation. Which one of the following statements will best explain the fate of protein X if the cells are treated with peptide P and then electrical stimulus is provided. (A) X fails to move across cells due to improper formation of tight junctions (B) X fails to move across cells due to improper formation of gap junctions. (C) X moves freely across cells as before. (D) X fails to move across cells due to improper formation of desmosomes. Exp. Connexons are gap junction proteins. Gap junctions are a specialized intercellular connection between a multitude of animal cell-types. They directly connect the cytoplasm of two cells, which allows various molecules, ions and electrical impulses to directly pass through a regulated gate between cells. So when the inhibior P is added, gap junctions are not properly formed, thus the protein P can not move between the cells. Amar Ujala Education 90. A patient with breast cancer was given a dose of radiation along with chemotherapy and was apparently cured of the tumor. After five years, a tumor was noticed in the patient's lungs, but the doctors confirmed that it was derived from cells of the mammary gland. The following possibilities were suggested by the doctor. (a) Bacterial infection, after radiation, led to development of the tumors in the lungs. (b) Migration of residual chemo-resistant cells from the (c) mammary gland resulted in tumors in the lungs. Epithelial-to-mesenchymal transition had occurred in the lungs. (d) Cells in the lungs were induced to become a tumor after chemotherapy, and from factors secreted by mammary cells. Which of the following is correct? (A) (b) and D (B) Only (b) (C) (a) and (b) (D) (a) and (c) Exp. Breast cancer categorized as ER + ve are more prone to distant reccurence and are associated with dormancy. The main cause being cells that are chemo-resistant get disseminated to lungs incude the expression of Periostin (POSTN) in fibroblasts, which reciprocally increases WNT signaling and promotes tumor-initiating potential and tumor outgrowth. Similarly, lung micrometastases can also produce tenascin C (TNC), an ECM protein, that fcilitates maintenance of tumor initiating potential by activating WNT and Notch pathway in cancer cells. 91. Chromatin condensation is driven by protein complexes called condensins which are members of a family of structural maintenance of chromatin (SMC) proteins that play a key role in the organization of eukaryotic chromosomes. Condensins along with another family of SMC proteins called cohesins significantly contribute to chromosome segregation during mitosis. If the cells are treated with an inhibitor of cdk1 phosphorylation immediately before the cells enter M phase, which of the following statements is most likely to be true? (A) Sister chromatids are held together by condensins along the entire length of the chromosome. (B) Sister chromatids are held together by cohesins along the entire length of the chromosome. (C) Sister chromatids are held together by condensins and attached to each other only at the centromere. (D) Sister chromatids are held together by condensins and attached to each other only at the telomere. Exp. The cells cannot enter the M phase because of the inhibition of cdk 1 phosphorylation. At this stage sister chromatids will be held together by cohesins throught the length of the chromosome. P4/13

130 Amar Ujala Education 92. There are three substances A, B and C. Given below are the pattern of immunological responses in rabbits when (i) A is administered along with C, (ii) B is administered along with C and (iii) A is conjugated with B and administered along with C. 1. Serum Ab response Serum Ab response Serum Ab response anti A anti B anti B Time Time anti A Time Which one of the following is the correct identification? (A) A-protein, B-hapten, C-adjuvant (B) A-hapten, B-protein, C- adjuvant (C) A-protein, B- adjuvant, C-hapten (D) A-hapten, B- adjuvant, C-protein Exp. A is a protein, B is hapten and C is adjuvant Hence in first graph there is no anti A production. in the second graph the adjuvant has enhanced the immunogenicity of B (hapten) and there is increased antbody production. In the third graph there is hapten protein conjugation plus enhanced immuno-genicity when administered with an adju-vant. Hence there in production of more of anti B compared to anti A. 93. Following are certain statements regarding seed development in plants : (a) During final phase of development, embryos of orthodox seeds become tolerant to desiccation, dehydrate losing up to 90% of water (b) Dormant seeds will germinate upon rehydration while quiescent seeds require additional treatments (c) or signals for germination Precocious germination is germination of seeds without passing through the normal quiescent and/ or dormant stage of development (d) Abscisic acid is known to inhibit precocious germination Which one of the following combinations is correct? (A) (a), (b) and (c) (C) (b), (c) and (d) (B) (a), (b) and (d) (D) (a), (c) and (d) P4/14 A/UG-15 Exp. Option A is correct as orthodox seeds are long lived seeds. They can be successfully dried to moisture contents as low as 5% without injury. Option C and D are also correct as precocious germination is defined as that kind of germination wherein the seed germinates without undergoing all the four stages of germination, i.e., globular, heart shape, torpedo shape, and cotyledonary stage. ABA is known to inhibit precocious germination. Option B says that dormant seeds would germinate upon rehydration. This is incorrect. They might germinate or might not as germination of seeds is dependent not only on water but also on oxygen and temperature. 94. Development, of vulva in C, elegans is initiated by the induction of a small number of cells by short range signals from a single inducing cell. With reference to this, following statements were put forward. (a) When the anchor cell was ablated early in development, no vulva formed, (b) In a dominant negative mutant of let-23, a primary vulva formed but the secondary vulva formation did not take place. (c) A cell adopting a primary fate inhibits adjacent cells from adopting the same fate by lateral inhibition involving LIN-39 and also induces the secondary fate in these cells. (d) A constitutive signal from the hypodermis inhibits the development of both the primary and secondary fates but it is overruled by the initial signal from the anchor cell. Which of the above statements is true? (A) (a) and (b) (B) (a) and (c) (C) (a) and (d) (D) (b) and (d) Exp. Since all the vulval precursor cells are under the control of anchor cell. So if anchor cell is removed then it would result in no vulva formation. Also, signals from hypodermis would be inhibiting the development of primary and secondary vulva. However, these signals are overcomed by the signals from anchor cell (LIN 3). 95. Which one of the following about development of sea urchin embryos is TRUE? (A) Each blastomere of a 4-cell stage possesses a portion of the original animal-vegetal axis and if isolated and allowed to develop will form a complete but smaller sized larva. (B) Each blastomere of a 8-cell stage has the capacity to form a complete embryo but by the 16-cell stage, blastomeres will develop according to their presumptive fate. (C) Any blastomere isolated till the pluteus larva formationwillregulatetogoonanddevelop into a full sized embryo.

131 A/UG-15 (D) After an intricate recombination at the 16-cell stage, the resulting embryo looses its ability to form a complete larva Exp. Till 4 cell stage, since the cells, contain part of the animal and vegetal axis, therefore, if isolated they would be giving rise to the entire embryo. However, at the 8 cell stage, since the blastomeres have been formed as a result of equational division, thereby the animal and vegetal exis gets divided. So these blastomerese would not be able to give rise to the intact pluteus larvae. 96. What will happen if wingless RNAi expressed in wingless expressing cells from the stage when this gene initiates its expression in a developing Drosophila embryo? (a) The enhanced expression of wingless thus caused will broaden the area of engrailed expression. (b) Since wingless protein makes a long range gradient, its effect will not be seen in the same segment. (c) The posterior compartment of each future segment will get affected, (d) Since engrailed expression is initiated by pair rule genes, the posterior segment will not be affected, Which one of the following will most appropriately answer the question? (A) (a) and (c) (B) Only (c) (C) B and (d) (D) Only (d) Exp. Wingless is expressed in the posterior region of each segment or in the anteior portion of each parasegment. If RNAI of wingless is present then will not allow wingless to be expressed because of which the posterior compartment of each furture segment will get affected. 97. Column A Column B Column C (i) Invagination (i) Movement of epithelial (i) Hypoblast cells as a unit to enclose in birds deeper layers of the embryo (ii) Involution (ii) Splitting of one cellular (ii) Ectoderm sheet into two parallel sheets in amphibians (iii) Ingression (iii) Infolding of epithelium (iii) Mesoderm in amphibians (iv) Delamination (iv) Migration of individual (iv) Endoderm cells from surface into in sea urchin interior of the embryo (v) Epiboly (v) Inward movement of (v) Mesoderm expanding outer layer so in sea urchin that it spreads over the internal surface of remaining external cells Amar Ujala Education Which one of the following is the correct combination? (A) (a)(i), (b)(iv), (c)(ii) (B) (a)(iv), (b)(iii), (c)(i) (C) (a)(iii), (b)(iv), (c)(v) (D) (a)(v), (b)(ii), (c)(iii) 98. Formation of digits and sculpting the tetrapod limb requires death of specific cells in the limb in a programmed manner. Which one of the following interactions could explain proper limb formation? (A) (B) (C) (D) BMP Cartilage BMP Cartilage Wnt/ -Catenin FGF Wnt/ -Catenin Wnt/ -Catenin BMP Dkk-1 Apoptosis Dkk-1 FGF Apoptosis BMP Dkk-1 Apoptosis 99. Light is an important factor for plant growth and development. There are several photoreceptors in higher pl ants such as Arabidopsis thaliana involved in perception of various wavelengths of light. Some statements are given below related to photoreceptors : (a) Red light photoreceptors are represented by a gene family. (b) Phytochrome C is the most prominent photoreceptor to perceive red light. (c) Cryptochrome 1 and cryptochrome 2 have evolved from bacterial DNA photolyases. (d) Far-red light is perceived by phytochrome D. Which one of the following combinations of above statements is correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (a) and (c) Exp. Phytochrome C is not the most prominent photoreceptor of red light. In rice and Arabidopsis, PHYTOCHROME C (PHYC) requires other phytochromes for stability and function. Option D is false, as Far red light is preceived by phy A. Statements A and C are true. P4/15

132 Amar Ujala Education 100. Pyrurate dehydrogenase is subject to feed back inhibition by its products in glycolysis. Some of the chemical compounds which might be involved in the process, are listed below : (a) NADH (b) FAD (c) Acetyl-CoA (d) Acetaldehyde Which one of the following combinations of above chemicalcompounds isinvolved in feedback inhibition of pyruvate dehydrogenase? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (a) and (c) Exp. Pyruvate dehydrogenase is inhibited when the following ratios are increased : NADH/NAD + and acetyl-coa/ CoA Carbohydrates synthesized by photosynthesis are converted into sucrose and transported via phloem to other parts of the plant. The following aspects are associated with sucrose uploading in phloem and its transport : (a) Both reducing and non-reducing sugars are transported efficiently through phloem. (b) Sucrose uploading can be both symplastic and (c) apoplastic. The route of phloem uploading is mesophyll cells phloem parenchyma companion cells sieve tubes. (d) Transport in sieve tubes is as per the pressure-flow model. Which one of the following combinations is correct? (A) (a), (b) and (c) (B) (b), (c) and (d) (C) (c), (d) and (a) (D) (d), (b) and (a) Exp. Nonreducing sugars are less likely to react with other substances along the way. Sucrose can be uploaded by either symplastic and apoplastic route. Route : Mesophyll cells->phloem parenchyma- >companion->sieve Transport is via pressure flow model Symbiotic nitrogen fixation in legume nodules involves complex interaction between Rhizobium and legume roots. This complex interaction is governed by (a) Integration of sym plasmid of Rhizobium in the root nuclear genome. (b) Sensing of plant flavonoids by rhizobia. (c) Activation of nod genes in rhizobia. (d) Activation of NODULIN genes in legume roots. Which one of the following combinations is correct? (A) (a), (b) and (c) (B) (a), (c) and (d) (C) (b), (c) and (d) (D) (a), (b) and (d) Exp. As B, C and D are involved in the steps of nitogen fixation. P4/16 A/UG A Z scheme describes electron transport in O 2 -evolving phofosynthetic organisms. The direction of electron flow is presented in the following sequences : (a) P680* Pheophytin Q A Q B PC cytochrome b 6 f P700 (b) P700* Phylloquinone FeS A FeS B FeS X Fd (c) P680* Pheophytin Q A QB cytochrome b 6 f PC P700 (d) P700* Phylloquinone FeS X FeS A FeS B Fd Which one of the following combinations is correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (d) and (a) Exp. Pheophytin->QA->QB->b6f->PC->P700->P700->PQ- >FeSx->FeSA->FeSB->FD Following are certain statements related to plants exposed to dehydration stress : (a) When the water potential of the rhizosphere decreases due to water deficit, plants can continue to absorb water as long as plant water potential is lower than that of soil water. (b) The ratio of root to shoot growth increases in (c) response to water deficit. Plant cells tend to release solutes to lower water potential during periods of osmotic stress. (d) Abscisic acid is synthesized at a higher rate when leaves are dehydrated, and more ABA accumulates in the leaf apoplast. Which one of the following combinations of above statements is correct? (A) (a), (b) and (c) (B) (b), (c) and (d) (C) (a), (b) and (d) (D) (a), (c) and (d) Exp. Statement C is false. Plant would accumulate a larger amoutn of solute in it s cells rather than relasing A cross was made between Hfr met + arg + feu + str S XF met arg leu str R,inwhichleu + exconjugants are selected. If the linear organization of the genes are leu + arg + met +, which one of the following genotypes is expected to occur in the lowest frequency? (A) leu + arg met (B) leu + arg + met (C) leu + arg + met + (D) leu + arg met + Exp. The frequency depends upon the order of the genes. Most frequent Leu+ arg-met-followed by leu + arg + met - and least frequent leu + arg + met +.

133 A/UG Two homozygous individuals (P1 and P2) were genotyped using dominant DNA markers A and B, as shown below. The F 1 progeny obtained was test crossed. The frequency of progeny with which different genotypes appear, is given below : The following conclusions were made : (a) In the F 1, markers A and B are linked and in coupling phase (cis) (b) In the F 1, markers A and B are linked and in repulsion phase (trans) (c) The distance between A and B is 10 cm (d) The distance between A and B is 5 cm Which of the above conclusions are correct? (A) (a) and (c) (B) (a) and (d) (C) (b) and (c) (D) (b) and (d) Exp. As the parents have only one band therefore F1 will have the alleles in trans. The recombinants are 10 (5+5)/ Therefore distance = 10 cm The above pedigree shows the inheritance of a rare allele. The allele is : (A) X-linked recessive (B) autosomal recessive (C) dominant with incomplete penetrance (D) autosomal recessive with incomplete penetrance Exp. Could not be X linked recessive as for the trait to appear in daughter, it has to be present in could not be autosomal recessive as the trait is because of rare allele and person coming from outside the family would not have the allele (male in the second generatioe) and hence his offspring would not have the trait, which is not the case. Autosomal dominant with incomplete penetrance, can explain the pedigree. the female in the first generation Amar Ujala Education would be heterozygous and she would pass this dominant allele to her daughter, but the trait is not seen because of incomplete penetrance. In the next generation it would be again expressed A plant with red fruit is crossed to a plant with white fruit The F 1 progeny had red fruits. On selfing the F 1, two kinds of progeny were observed, plants with redfruits and those with white-fruits. To test whether it was a case of recessive epistatic interaction a chi-square test was performed, A value of was obtained (chisquare value at P 0.05 = for Degree of Freedom = 1) The following statements were made : (a) The null hypothesis was that plants with red and white fruits will occur in a 9 : 7 ratio (b) The null hypothesis was that plants with red and (c) white fruits will occur in a 1 : 1 ratio Based on the chi-square value, it is a case of recessive epistatic interaction (d) Based on the chi-square value, it is not a case of recessive epistatic interaction Which of the combination of above statements is correct? (A) (a) and (c) (B) (a) and (d) (C) (b) and (c) (D) (b) and (d) Exp. Recessive expistasis ratio is 9 : 3 : 4, more over in chi squared test we assume null hypothesis to be true.null hypothesis will be that plants with red and white fruits will occur in 1 : 1 ratio. While interpreting the results, if the calculated chi squared is less than the critical value we accept the null hypothesis. Only when this value is greater we reject it. In the question, observed Chi square valu eis less than the critical value so we accept the null hypothesis. Therefore statements B and D are ture Somatic recombination was caused by mild exposure to radiation on flies heterozygous for a given allele during specific stages of development and the individuals were allowed to develop. Such individuals are likely to have (a) clones of homozygous cells in heterozygous body. (b) site specific mutagenesis. (c) twin spots, i.e., patches of mutants cells and homozygous wild type cells in heterozygous body. (d) tissue specific expression of the given allele. Which of the following combination of answers will be most appropriate? (A) (a) and (b) (B) (b) and (c) (C) (a) and (c) (D) (c) and (d) Exp. Somatic hybridization would cause, other wise heterozygous individual to transform into homozygous and hence patches of mutants cells and homozygous wild type. Some of the tissue might end up with only one kind of alleles and hence would cause tissue specific expression of the allele. P4/17

134 Amar Ujala Education 110. Sickle cell anemia is a recessive genetic disease caused due to a point mutation In the 6th codon abolishing one of the MspII endonuclease digestion site present in the - globin gene, MspII digested DNA from a normal person gives two bands, 1150 bp and 200 bp, in gtobin, gene. A family with a proband (based on the disease phenotype) gave the following MspII digestion pattern : The following conclusions were drawn : (a) Son (I) is the proband and the given mutation is not present in Son (II). (b) The daughter is a carrier for the given mutation, (c) The gene is X-linked and thus Son (I) becomes the proband, (d) The father and daughter are affected, (e) A fife novo mutation in same site on normal allele has allowed appearance of disease phenotype in the proband. Which of the following combination of conclusions will be the most appropriate for the figure given above? (A) (a), (b) and (e) (B) (a), (b) and (c) (C) (b), (c) and (e) (D) (c) and (d) Exp. Statement A, B and E are correct. Based on the Mspll digestion profile, we can observe that the mother is normal phenotype (since she has both active restrictionsite clear from the band pattern), father is a carrier (since the digestion profile shows the presence of bands corresponding to both presence as well as absence of the site-heterozygous for the RFLP allele), son (i) is expressing the phenotype so he is the proband. Son (ii) again has normal phenotype. Daughter also is a carrier for the RFLP allele In high altitude, hypoxia induces increased number of circulating red blood cells, which can be explained by the following changes : (a) The transcription factors, HIFs, are produced (b) Erythropoietin secretion is increased (c) Myoglobin content is decreased (d) Cytochrome oxidase is decreased Which one of the following is NOT true? (A) Only (a) (B) (a) and (b) (C) (b) and (c) (D) (c) and (d) A/UG-15 Exp. Transcription factors and erythropoitin secretion and cytochrome oxidase levels and myoglobin content are elevated in hypoxia. Hence C and are wrong The mechanism of sound localization in a hprizoatal plane by the human auditory system can be explained by (a) the difference in time between the arrival of stimulus in two ears (b) the difference in phase of the sound waves on (c) two ears the difference in tuning curves of two auditory nerves (d) the activity of neurons in superior olivary nucleus, but not by the neuronal activity of auditory cortex, Which one of the following is NOT correct? (A) Only (a) (B) (a) and (b) (C) (b) and (c) (D) (c) and (d) Exp. Sound localization is the ability of an individual to first locate the origin of sound. Hence it depends on time between the arrival of stimulus in 2 ears as the ear closer to be sound source percieves it earlier compared to the other. At the same time the phase also plays a role in sound localization. When sound comes from one side the ear on the other side percieve the sound shadow. The tuning curves of audiory nerves helping sensing sound but not localization and the tuning curves of auditory nerves have many over laps. Hence option c is false. The superior olivaliry nucleus and the auditory cortex play a role in sound localization hence option D is also wrong During physical exercise, the oxygen supply to the active muscles is increased, which has been explained by the following statements : (a) Po 2 declines and Pco 2 rises in the active muscles (b) The temperature is increased and ph is decreased in active muscles (c) 2, 3-biphosphoglycerate is decreased in RBC and P 50 rises (d) Metabolites accumulating in the active muscles increase the affinity of hemo-globin to oxygen Which one of the following is NOT correct? (A) (a) only (B) (a) and (b) (C) (b) and (c) (D) (c) and (d) Exp. Changes in 2, 3 BPG concentration are not related to biochemical changes during physical exercise. It only decreases the oxygen affinity of Hb at high altitudes. Metabolites like lactic acid that accumulate after xercise they decrease the ph thus leading to decrease in affinity of Hb to oxygen (right shift). Thus the statement D is also incorrect as it says that accumulation of metabolites lead to increase in affinity of Hb to oxygen. P4/18

135 A/UG The fractional clearance of neutral and cationic dextran molecules of various sizes through kidneys of a rat is shown in the figures below. Which one of the following is correct? (A) (B) (C) (D) Farctional clearance Farctional clearance Farctional clearance Farctional clearance Effectives molecular diamoter (nm) Effectives molecular diamoter (nm) Effectives molecular diamoter (nm) Effectives molecular diamoter (nm) Exp. Since the glomerulus is negatively charged, therefore, fractional clearance of cationinc dextrans would be more as compared to the fractional clearance o neutral dextrans which should be the of the same molecule size. With increase in Stoke radius or the moleculear size, the Amar Ujala Education fractional clearance keeps on decreasing gradually. Hence, this is justified by graph Estradiol synthesis follows a 2-cell-2-gonadotropin theory, where partial synthesis occurs in the granulosa cells and the rest in the theca interna cells of the Graafian follicle. Which one of the fallowing correctly represents estradiol synthesis and secretion? (A) (B) (C) (D) Exp. Androstenedione is synthesized from cholesterol which in turn gets converted to estrone or estradiol directly or gets converted to testosterone first and then to estradiol. Since LHR is present on Theca Interna cells and FSHR on Granulosa cells. Therefore, option 1 would be correct Which one of the following is the most appropriate match for the protected areas of India? Category Protected Area (a) Biosphere Reserve (i) Chambal (b) National Park (ii) Loktak (c) Ramsar Site (iii) Nanda Devi (d) Wildlife Sanctuary (iv) Rajaji (v) Sundarbans (A) (a)-(iii) (b)-(iv) (c)-(ii) (d)-(i) (B) (a)-(ii) (b)-(iv) (c)-(iii) (d)-(i) (C) (a)-(i) (b)-(v) (c)-(iii) (d)-(i) (D) (a)-(iii) (b)-(ii) (c)-(v) (d)-(iv) 117. The following table shows selected characters used in analyzing the phylogenetic relationships of four plant taxa : Characters Taxon Xylem or Wood Seed Flowers Phloem T 1 + T T T 4 P4/19

136 Amar Ujala Education Taxa T 1,T 2,T 3 and T 4 are respectively : (A) Ferns, Oaks, Pines, Hornworts (B) Oaks, Pines, Hornworts, Ferns (C) Hornworts, Pines, Oaks, Ferns (D) Ferns, Pines, Oaks, Hornworts Exp. Only oaks have flowers and rest 3 don t have any flowers. Ferns are peteridophytes, pine is a gymnosperm and hornwort is a bryophyte Given below are the main characteristics of a few mammalian orders. Match the names of the animals with the characteristics of their orders : (a) Hooves with even number of toes on each foot, omnivorous (b) Teeth consisting of many thin tubes cemented together, eats ants and termites (c) Opposable thumbs, forward facing eyes, well developed cerebral cortex, omnivorous (d) Hooves with an odd number of toes on each foot; herbivorous (i) Tapir (ii) Lemur (iii) Aardvark (iv) Pig Choose the correct combination (A) (a)-(iii) (b)-(iv) (c)-(i) (d)-(ii) (B) (a)-(i) (b)-(iv) (c)-(ii) (d)-(iii) (C) (a)-(iv) (b)-(iii) (c)-(ii) (d)-(i) (D) (a)-(iv) (b)-(iii) (c)-(i) (d)-(ii) 119. Which of the following options match the plant tissue type with its correct function in vascular plants? Tissue Function (a) Tracheids (i) Chief water-conducting element in gymnosperms (b) Vessel elements (ii) Chief water-conducting element in angiosperms (c) Sieve tube element (iii) Food-conducting element in gymnosperms (d) Sieve cell (iv) Food-conducting element in angiosperms (A) (a)-(i) (b)-(ii) (c)-(iv) (d)-(iii) (B) (a)-(ii) (b)-(ii) (c)-(iii) (d)-(iv) (C) (a)-(i) (b)-(ii) (c)-(iii) (d)-(iv) (D) (a)-(i) (b)-(iii) (c)-(iv) (d)-(ii) Exp. Tracheids-chief water conducting element in gymnosperms, vessel elements are chief waterconducting element in angiosperms, sieve cell is the food conducting element in gymnosperms and sieve tube element is the food conducting element in angiosperms. P4/20 A/UG Which of the-following gives the correct human disease causal microbe match for each? Human Disease Causal Microbe (a) Chronic gastritis (i) Borrelia burgdorferi (b) Lyme disease (ii) Helicobacter pylori (c) Scarlet fever (iii) Rickettsia prawazekii (d) Typhus (iv) Streptococcus pyogenes (A) (a)-(ii) (b)-(i) (c)-(iv) (d)-(iii) (B) (a)-(ii) (b)-(iii) (c)-(i) (d) (iv) (C) (a)-(iv) (b)-(i) (c)-(ii) (d)-(iii) (D) (a)-(iv) (b)-(iii) (c)-(i) (d)-(ii) Exp. Lyme disease also known as Lyme boreliosis, is an infectious disease caused bybacteria of the Borrelia type. H.pylori cause chronic gastritis, Typhus fevers are caused by the rickettsiae bacteria and transmitted by arthropod. Scarlet fever from streptococcus pyogenes Match the five (A E) group of organisms with their correct taxonomic rank (i v) given below : Group Taxonomic rank (a) Crustacea (i) Order (b) Hominidae (ii) Domain (c) Dermaptera (iii) Class (d) Ctenophora (iii) Phylum (e) Archaea (iv) Family (A) (a) -(iii) (b)-(i) (c)-(v) (d)-(iv) (e)-(ii) (B) (a)-(i) (b)-(ii) (c)-(iii) (d) (iv) (e)-(v) (C) (a)-(iv) (b)-(iii) (c)-(ii) (d)-(i) (e)-(v) (D) (a) -(iii) (b)-(v) (c)-(i) (d)-(iv) (e)-(ii) Exp. Crustacea is a class of Arthropoda. Hominidae is a family. Dermaptera is an order in Arthropoda. Ctenophora is a phy-lum while archae is a domain Compared to K-selection, r-selection favours (A) rapid development, smaller body size and early, semelparous reproduction. (B) rapid development, smaller body size and early, iteroparous reproduction. (C) slow development, larger body size and late, iteroparous reproduction. (D) slow development, smaller body size and late, iteroparous reproduction. Exp. R-selected species have smaller body size, reproduce only once in their lifetime producing many offsprings (semelparous reproduction) and develop rapidly.

137 A/UG Which of the following is true about the Digestion Efficiency (DE) (assimilation/consumption) and Ecological Efficiency (EE) (production/consumption) of ectotherms and endotherms? (A) Endotherms have a high DE and ectotherms have a high EE. (B) Endotherms have a low DE and ectotherms have a high EE. (C) Endotherms have a high DE and ectotherms have a low EE. (D) Endotherms have a low DE and ectotherms have a low EE. Exp. Since endotherms lose much of their energy as heat and in respiration, therefore they would be having low digestion and ecological efficiency as compared with ectotherms. This is because ectotherms do not need to maintain their body temperature thus they would be more efficient in deriving maximum benefit from the consumed food Two lakes (I and II) with a similar trophic structure of phytoplankton-zooplankton-planktivorous fish food chain were chosen, To understand the top-down effects, some piscivorous fish (those that feed on planktivorous fish) were introduced into Lake I, making it a system with four trophic levels. Lake II was enriched by adding large quantities of nitrates and phosphates to study the bottom-up effects over a period of time, Changes in the biomasses of each trophic level were measured. The expected major changes in the two lakes are (A) In Lake I zooplankton biomass increases, phytoplankton biomass decreases. In Lake II both phytoplankton and planktivorous fish biomasses increase. (B) In Lake I zooplankton biomass decreases, phytoplankton biomass increases. In Lake II both phytoplankton and planktivorous fish biomasses increase. (C) In Lake I planktivorous fish biomass and phytoplankton biomass decrease. In Lake II phytoplankton biomass increases, planktivorous fish biomass decreases. (D) In Lake I planktivorous fish and zooplankton biomasses increase. In Lake II both phytoplankton and planktivorus fish biomasses increase. Exp. In lake I planktivorous fish and zooplankton biomasses increase. In Lake II both phytoplankton and planktivorus fish biomasses increase Given below is a matrix of possible interactions (beneficial (+), harmful ( ), neutral (0) between species A and (B) The names of interactions, A, B, C and D, respectively, are Species 2 + Species C B A Amar Ujala Education (A) Predation, competition, mutualism, commensalism (B) Mutualism, competition, amensalism, commensalism (C) Competition, predation, mutualism, amensalism (D) Competition, mutualism, commensalism, predation Exp. In case of A, as both the species are negatively affected so it would be competition. In B, one of the species is benefitted another species is negatively affected so it is a type of predation. In C, both the species are positively affected, so it is mutualism. In D, one of the species is negatively affectd while other is not affected at all so this would be ammensalism Which of the following is/are NOT valid explanation(s) for the observed pattern of species richness? (a) Older communities are more species-rich. (b) Large areas support more species. (c) Natural enemies promote reduced species richness at local level. (d) Communities in climatically similar habitats may themselves be similar to species richness. (e) Greater productivity permits existence of more species. (A) (b),(c)and(d) (B) Only(c) (C) Only (d) (D) (a), (b) and (e) Exp. Species richness is highest in area of intermediate disturbances. This is because very frequent disturbance eliminates sensitive species, whereas very infrequent disturbance allows time for superior competitiors to eliminate species that cannot compete. Hence, all the other statements are correct except for statement C The following table stows the number of individuals of each species found in two communities : Community Species A B C D C C (Hint : ln values for 0.05, 0.10, 0.25 and 0.80 are , 1.4 and 0.2, respectively) The calculated Shannon diversity index (H) values for communities C1 and C2, respectively are (A) 1. 4 and 0.69 (B) 1.2 and 0.34 (C) 2.1 and 0.43 (D) 1.8 and 0.37 D P4/21

138 Amar Ujala Education Exp. Shannon diversity index = - summation of (pi) (lnpi) pi = No.of oganisms in species/total No.of organisms of all the species. Community 1 pi of A = 25/100 = 0.25 Same would be the pi of B, C and D Shannon Diversity Index for Community 1 = - 4 (0.25) (-1.4)=1.4 Community2 pi of A = 80/100 = 0.8 pi of B = 5/100 = 0.05 pi of C = 5/100 = 0.05 pi of D = 10/100 = 0.1 Shannon Diversity Index for Community 2 = - (0.8) (- 0.2) + 2 (0.05) (-3) + (0.10) (-2.3) = In a population at Hardy-Weinberg equilibrium, the genotype frequencies are : f(a 1 A 1 ) = 0.59; f(a 1 A 2 ) = 0.16; f(a 2 A 2 ) = What are the frequencies of the two alleles at this locus? (A) A 1 =0.59 A 2 =0.41 (B) A 1 =0.75 A 2 =0.25 (C) A 1 =0.67 A 2 =0.33 (D) A 1 =0.55 A 2 =0.44 Exp. Frequency of A = homozygous frequency + 1/2* heterozygous = /2 (0.16) = Frequency of 'a' = = Homing pigeons, when released at a place far away from their home, use earth's magnetic field or the sun as navigational cues and choose the right direction to fly. To test the hypotheses, two experiments were conducted. In. Expt.I, one group of pigeons (Test) were equipped with Helmholtz coils (which disrupt magnetic field detection), while the second group (Control) were not. Both groups were released on a sunny day, Expt II used the same Control and Test groups of pigeons, but they were released on a cloudy, overcast day. The expected results, if the hypotheses is true, would be (A) In Expt. I, both Control and Test groups fly in the right direction, but in Expt. II, only Control group does, (B) In both experiments, Test groups fail to choose the right direction, (C) In Expts. I and II, Test groups fly in the right direction, (D) In Expt. I both groups fly in the right direction but in Expt. II both groups fail to choose the right direction. A/UG-15 Exp. In experiment, I both the pigeons will be able to find the right direction on basis of sunlight. While in second experiment the control group will be able to find the direction by using magnetic field while the test one will not Brothers A and B have the same father but different mothers. B wants A to help him, which involves both benefits (b) and costs (c) for A. If A incurs a cost of 36 Darwinian fitness units in that act, under what condition should he help B, following Hamilton s rule? (A) only if b > 30 (B) only if b > 60 (C) only if b > 120 (D) only if b > 240 Exp. rb>c. Here C=30 and r=0.25 (for step sibligs) So B>30/0.25 or B> Following are the main types of defense employed by prey species against predators Types of defense : Chemical with aposematic coloration (a); Cryptic coloration (b); Batesian mimicry (c); Intimidation display (d) Prey Species : Grasshoppers and seahorses (i); Hoverflies and wasps (ii); Bombardier beetles, ladybird beetles, many butterflies (iii); Frilled lizard, Porcupine fish (iv) Which one of the following combinations is correct? (A) (a)-(i) (b)-(iii) (c)-(ii) (d)-(iv) (B) (a)-(iv) (b)-(ii) (c)-(i) (d)-(iii) (C) (a)-(iii) (b)-(i) (c)-(ii) (d)-(iv) (D) (a)-(ii) (b)-(iii) (c)-(i) (d)-(iv) 132. Following is the list of some important events in the history of life and the names of the epochs of Cenozoic era. Events (a) Angiosperm dominance increases; continued radiation of most present day mammalian orders (b) Major radiation of mammals, birds and pollinating insects (c) Origins of many primate groups (d) Origin of genus Homo (e) Appearance of bipedal human ancestors (f) Continued radiation of mammals is and angiosperms, earliest direct human ancestors Epochs (i) Paleocene (ii) Pleistocene (iii) Oligocene (iv) Pliocene (v) Eocene (vi) Miocene Which one of the following is the correct match of events with the epochs? (A) (a)-(v) (b)-(ii) (c)-(i) (d)-(iii) (e)-(iv) (f)-(vi) (B) (a)-(vi) (b)-(i) (c)-(ii) (d)-(iv) (e)-(iii) (f)-(v) (C) (a)-(v) (b)-(i) (c)-(iii) (d)-(ii) (e)-(iv) (f)-(vi) (D) (a)-(iv) (b)-(i) (c)-(ii) (d)-(iii) (e)-(v) (f)-(vi) P4/22

139 A/UG In the evolutionary tree given below, terms A, B, C, D and E, represent respectively Today E D C Million years ago 1 B A 2 (A) Homo erectus, Homo heidlebergensis, Neanderthal, Denisovan and Homo sapiens. (B) Homo heidelbergensis, Homo erectus, Denisovan, Neanderthal and Homo sapiens. (C) Homo erectus, Homo heidelbergensis, Denisovan, Neanderthal and Homo sapiens. (D) Homo heidelbergensis, Homo sapiens, Denisovan, Neanderthal, and Homo erectus In transgenic mice, the orientation and location of the loxp sites determine whether Cre recombinase induces a deletion, an inversion or a chromosomal translocation. If a researcher wants to put the loxp sites in such a manner that only inversion will take place, which one of the following construct best justifies their intention. Gene X is the target gene. (A) Gene X lox P lox P (B) lox P Gene X lox P (C) lox P Gene X lox P (D) Gene X lox P lox P Exp. Cre-Lox recombination is knwon as a site specific recombinase technology, and is widely used to carry out deletions, inser-tions, translocations and inversions at specific sites in the DNA of cells. The systems consists of a single enzyme, Cre recombinase, that recombines a pair of short target sequences called the Lox sequences. Placing Lox sequences appropriately allows genes to be activated, repressed, or exchanged for other genes The following genes have been genetically engineered to develop herbicide resistance in plants : (a) Resistance to glyphosate using the 5-enopyruvyl shikimate-3-phosphate synthase gene (b) Bialaphos resistance using the bar gene (c) Sulfonyl urea resistance using the acetolactate synthase gene Amar Ujala Education (d) Atrazine resistance using the glutathione S- transferase gene In which of the above two cases the mechanism is based on abolition of herbicide binding to the enzyme? (A) (a) and (d) (B) (b) and (c) (C) (c) and (d) (D) (a) and (c) Exp. Sulphonylureas inhibit ALS (acetolactate synthase) by binding to the enzyme. Glyphosate s mode of action is to inhibit a plant enzyme involved in the synthesis of the aromatic amino aicds Factor IX is essential for blood clotting. Deficiency of Factor IX could be corrected by delivering factor IX gene using viral vectors. In an experiment the gene for Factor IX was cloned appropriately into Adenovirus (AV), Adeno-associated virus (AAV) and Retrovirus (RV). Retrovirus integrate the gene into the dividing ceils. AAV integrates partially into non-dividing cells. AV does not integrate the gene but transfects both dividing and non-dividing cells, Following expression profile for Factor IX was observed when the three vectors were injected intramuscularly into three groups of experimental mice, Which one of the following outcomes is correct? a b Factor DC in sorum Time, days (A) a by RV; b by AV; c by AAV (B) a by AV; b by AAV; c by RV (C) a by AAV; b by RV; c by AV (D) a by AV; b by RV; c by AAV Exp. Since in case of Retrovirus, gene for Factor IX would be integrated in non-dividing cells, therefore, these will keep on expressing factor IX, In Adeno-associated virus, the gene for factor IX is integrated in non-dividing cells so it would be expressed for the time till non-dividing cells survive. In adenovirus, the gene for factor IX is not integrated and only transfected in both dividing and non-dividing cells, so they would be showing only a transient expression of factor IX Somatic embryogenesis is an important exercise in micropropagation and genetic engineering of plants. The following steps are considered as critical for achieving somatic embryogenesis : (a) Reducing the concentration of sucrose in the medium by half. (b) Addition of the hormone, 2, 4 dichlorophenoxyacetic acid, to induce somatic embryos. (c) Reduce agar concentration to 0.6% (w/v) (d) Use maltose in place of sucrose as a carbon source c P4/23

140 Amar Ujala Education Which one of the following combinations is correct? (A) (a) and (c) (B) (b) and (d) (C) (a) and (b) (D) (c) and (d) Exp. Auxin 2, 4-D is critical for somatic embryogenesis, maltose is more successful than sucrose For the aquaculture farming of Indian Major carps several techniques are used. Which one of the following techniques is NOT used for this purpose : (A) Induced breeding (B) Selective breeding (C) Inbreeding (D) Composite fish farming Exp. Inbreeding causes decrease in growth rate, fecundity, and survival A gene from genomic library is screened by using hybridization technique. After hybridizing the probe, usually a stringent washing step is given. The following statements are given to explain the stringent washing step : (a) Stringent washing takes care of removing unincorporated and non-specifically hybridized probe molecules (b) Stringent washing is done in solution having high salt concentration and lower temperature (c) Stringent washing is done in solution having low salt concentration keeping higher temperature (d) Salt present in washing solution supports hybrids to stay intact by shielding the interference of water molecules (e) Salt reacts with DNA molecules and allows easy dissociation of hybrids (f) Stability of hybrid is directly proportional to the temperature Which combination of the above statements is most appropriate for stringent washing step? (A) (a), (b) and (d) (B) (a), (c) and (d) (C) (a), (b) and (f) (D) (c), (e) and (f) Exp. More stringent washing will remove non-specific hybridization mainly. More stringent the wash is, the partially complementary sequence will be pulled apart. More stringency is brought about by low salt and high temperature. the role of salt is to shield the negative charges on the DNA thus stabilizing the hybrid A student, while constructing knock-out mice, isolated mouse embryonic stem cells and introduced an engineered DNA into the cells. However, none of the mice were transgenic. On checking the cells containing DNA construct, he found that he had made a mistake in constructing the DNA since the cells were resistant to gancyclovir but sensitive to G418. Which one of the following constructs had he designed? P4/24 (A) (B) (C) (D) A/UG-15 Exp. Since the cells were resistant to ganciclovir this means that the region of DNA construct between the homology regions will not carry TK gene (thymidine kinase). This is becasue ganciclovir along with TK is a suicidal combination. TK converts it into a toxic product that is generally used for elemination of TK+ cells. But the knock out mice was resistant to it so it means it did not carry a TK genes. These cells were sensitive to G418, this is an antibiotic similar to gentamicin and resistance is conferred by neo resistance genes. Since the cells were sensitive to G418, so the region between homology also doesn t have neo gene. So option 3 seems to be the construct that doesn t contain either neo or TK gene In a confirmatory test for HIV, one or more viral antigens are detected in the blood of patients. Following are the steps to be performed for the experiment : (a) Transfer of viral antigens to nitrocellulose paper (b) Incubation with the buffer containing antibodies specific for viral antigens (c) Separation of viral antigens by SDS-PAGE (d) Detection of bands by enzyme-linked secondary antibody Identify the correct sequence steps to be performed for the experiment. (A) (a)-(b)-(c)-(d) (B) (b)-(c)-(d)-(a) (C) (c)-(a)-(b)-(d) (D) (c)-(b)-(a)-(d) Exp. The procedure is for Western blotting : C-A-B-D The most important property of any microscope is its resolution (D) and can be calculated from the formula 0.61 D N sin where D is minimum distance between two distinguishable objects, is the wave length of incident light, is the angular aperture and N is the refractive index, of the medium. Given below are several suggestions to improve the resolution of a microscope :

141 A/UG-15 (a) decrease the wave length of incident light (b) increase the wave length of incident light (c) use oil which has a higher refractive index (d) use oil because of its lower refractive index Which one is the correct suggestion? (A) (a) and (c) (B) Only (b) (C) Only (d) (D) (b) and (d) Exp. Improving resolution means having a lower value of D, Which can be achieved by lower value of lambda or higher value of N The mean and standard deviation of serum cholesterol in a population of senior citizens are assumed to be 200 and 24 mg/dl, respectively. In a random sample of 36 senior citizens, what values of cholesterol (to the nearest whole number) should lead to rejection of the null hypothesis at 95% confidence level? (A) above 224 (B) above 248 (C) below 176 and above 224 (D) below 192 and above 208 Exp. The 95% of the data would be in between u + 2 sigma and u 2sigma: u = 200 and sigma = 24. Therefore null hypothesis will be rejected with 95% confidence limit above (2 * 24) = In response to a drug, changes in protein levels were examined in a cell line. A pulse chase experiment was performed using 35 S labeled methionine. In comparison to untreated samples, the following observations were made : few minutes after stimulation, protein X accumulates and this was followed by Amar Ujala Education reduction in protein Y and Z. The correct interpretation of these observations would be : (A) Protein X is a protease which degrades Y and Z (B) Protein X is a transcriptional repressor that controls expression of Y and Z (C) Expression of protein X and toss of Y and Z are unrelated (D) Information is not sufficient to distinguish between the three possibilities stated above Exp. According to the question, statement 1, 2 and 3 all can be justified. So, option 4 would be correct A protein undergoes post-translational modification. In an experiment to identify the nature of modification, following experimental results were obtained. (a) Protein moved more slowly in an SDS-PAGE. (b) Isoelectric focusing (IEF) showed that there was no change in the p1 (c) Mass spectrometric analysis showed that the modification was on serine The modification that the protein undergoes is likely to be (A) phosphorylation (B) glycosylation (C) ubiquitination (D) ADP-ribosylation Exp. PTMs on serine can be either phosphorylation or glycosylation among the 4 options given. Ubiquitination occurs on Lysine andadp-ribsoylation occurs on Arginine. Among phosphorylation or glycoslation, both will cause increase in molecule weight leading to a slowly moving protein. However, phosphorylation ac bring about change in the isoelectric point of the protein as phosphoryl group is a charged molecule and thus will affect the ionization of the protein. P4/25

142 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, December-2014 PART-A 1. We define a function f(n) = sum of digits of N, expressed as decimal number. e.g. f(137)=1+3+7=11.evaluate f(273556). (A) 10 (B) 18 (C) 28 (D) 11 Exp. f(n) = sum of the digits of N f(273556) = sum of the digits obtained by solving the expression given. So we need to obtain the product of all the terms given after expanding them can be written as 25 * 35 * 55 * 22 * 51 taking number with same powers together we get (30)5 * 20 = 35*105*2*10 = 243 * 2 * 106 = So sum of the digits of this number = = Every month the price of a particular commodity falls in this order : 1024, 640, 400, 250,... What is the next value? (A) (B) Approximately 39 (C) 64 (D) 40 Exp. 2 ^ 10 = 1024, 5 * 2 ^ 7 = 640, 5*5*2^4=400, 5 * 5 *5*2^1=250sonextis5*5*5*5*2^-2=625/ 4 = A cubical piece of wood was filed to make it into the largest possible sphere. What fraction of the original volume was removed? (A) More than 3/4 (B) 1/2 (C) Slightly less than 1/2 (D) Slightly more than 1/2 Exp. Area of cube = a ^ 3, area of new sphere = 4/3 pie (a/2) ^ 3 = 0.52a ^ 3. And hence volume removed = 0.48a3. 4. Three volumes of a Hindi book, identical in shape and size, are next to each other in a shelf, all upright, so that their spines are visible, left to right : I, II and III. A worm starts eating from the outside front cover of volume 1, and eats its way horizontally to the outside back cover of volume III. What is the distance travelled by the worm, if each volume is 6 cm thick? (A) 6cm (B) 12cm (C) 18 (D) a little more than 18 cm Exp. The thickness of the books is 3 * 6 = 18. As there is some distance between each books. So the worm travels little more than 18 cm. 5. What is the area of the triangle bounded by the lines y = 2x, y = 2xand y =6? (A) 36 (B) 18 (C) 12 (D) 24 Exp. A diagram need to be drawn. 6. A 2.2 m wide rectangular steel plate is corrugated as shown in the diagram. Each corrugation is a semi-circle in cross section having a diameter of 7 cm. What will be the width of steel sheet after it is corrugated? 2.2 m (A) 1.4 m (B) 1.6 m (C) 0.7 m (D) 1.1 m Exp. 2.2 = pie rn: n would come out to be 20. One n corresponds to 7 cm and hence 20 n = 140 cm or 1.4 m. 7. Ajay, Bunty, Chinu and Deb were agent, baker, compounder and designer, but not necessarily in that order. Deb told the baker that Chinu is on his way. Ajay is sitting across the designer and next to the compounder. The designer didn t say anything. What is each person s occupation? (A) Ajay-compounder, Bunty-designer, Chinu-baker, Deb-agent (B) Ajay-compounder, Bunty-baker, Chinu-agent, Debdesigner P5/1

143 Amar Ujala Education (C) Ajay-baker, Bunty-agent, Chinu-designer, Debcompounder (D) Ajay-baker, Bunty-designer, Chinu-agent, Debcompounder Exp. As ajay is sitting next to designer and comp, he is baker as per the options. A and B excluded. As chinu is on the way, the wont be designer or comp, as they are sitting next to Ajay. 8. Two platforms are separated horizontally by distance A and vertically by distance B. They are to be connected by a staircase having identical steps. If the minimum permissible step length is a, and the maximum permissible step height is b, the number of steps the staircase can have is : (A) > B/b (B) < A/b (C) > B/b and < A/b (D) < B/b and > A/b Exp. Number of steps should be equal to or less than the ratio of vertical distance and should be equal to or more than the ratio of horizontal distance. 9. The sum of first n natural numbers with one of them missed is 42. What is the number that was missed? (A) 1 (B) 2 (C) 3 (D) 4 Exp = 45, 45 42= A mouse has to go from point A to B without retracing any part of the path, and never moving backwards. What is the total number of distinct paths that the mouse may take to go from A to B? A (A) 11 (B) 48 (C) 72 (D) 24 Exp. 2*4*3*2= A certain day, which is x days before 17 th August, is such that 50 days prior to that day, it was 4x days since March 30 th of the same year. What is x? (A) 18 (B) 30 (C) 22 (D) 16 Exp. Let that day be y; y + x = 17th Aug; y = 17th Aug-x--- Eq1, y-50=30thmar+4x---eq2; Substituting y from eq 1, 17th aug-x-50 = 30th Mar + 4x;5x = 17th aug-30th Mar-50; # of days from 30th Mar to 17th Aug (140); 5x = ; x = 18. B P5/2 A/UG What is the next term in the following sequence? 7, 11, 13, 17, 19, 23, 29,... (A) 37 (B) 35 (C) 31 (D) Which of the following figure best shows that y is inversely proportional to x? (A) (C) Y (0, 0) Y X (B) (D) (0, 0) (0, 0) X Exp. y is exactly inversely proportional to x, asx value increases, y starts decreases. 14. What is the maximum number of whole laddoos having diameter of 6 cm that can be packed in a box whose inner dimensions are cm 3? (A) 24 (B) 30 (C) 33 (D) If N, E and T are distinct positive integers such that N E T = 2013, then which of the following is the maximum possible sum of N, E and T? (A) 39 (B) 2015 (C) 675 (D) 671 Exp = Theareasoftheinnercircleandtheshadedringare equal. The radii r 1 and r 2 are related by : Y (0, 0) (A) r 1 = r 2 (B) r 1 = r 2 2 r 1 (C) r 1 = r 2 3 (D) r 1 =2r 2 Exp. p*r1 square = 2*p*r2 square, so the ans is r1=2r2. r 2 Y X X

144 A/UG The equation m 2 33n +1=0,wherem and n are integers, has : (A) no solution (B) exactly one solution (C) exactly two solution (D) infinitely many solutions 18. Which of the following numbers is a perfect square? (A) (B) (C) (D) Exp. It is perfect square of 1011* What is the 94 th term of the following sequence? 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4... (A) 8 (B) 9 (C) 10 (D) 11 Exp. 10th no. will come in this sequence at 94th term. 20. The following graphs depict variation in the value of Dollar and Euro in terms of the Rupee over six months : Jan Feb Mar Apr May Amar Ujala Education Which of the following statements is true? (A) Values of Dollar and Euro rose steadily from January to June (B) Values of Dollar and Euro rose by equal rate between January to March (C) The rise in the value of Dollar from April to May is three times the fall Euro during the same period (D) Values of Dollar and Euro rose equally between May and June Exp. Explanation as per the graph provided. PART-B 21. Chirality of DNA is due to : (A) the bases (B) base stacking (C) hydrogen bonds between bases (D) deoxyribose Exp. Deoxyribose, only deoxyribose exhibits chirality and is present naturally as D-deoxyribose is right handed helical DNA. 22. Proton motive force during oxidative phosphorylation is generated in mitochondria by : (A) exchanging protons for sodium ions (B) pumping protons out into intermembrane space (C) pumping hydroxyl ions into the mitochondria (D) hydrolysis of ATP Exp. As electrons are transferred through the complexes in mitchondrial mem, proton are pumnped out in intermembranous space generaing PMF. 23. In proteins, hydrogen bonds form as follows : Donor (D) H Acceptor(A). Hydrogen bond is more favorable if the angle between D H and A is : (A) < 90 (B) 180 (C) > 180 (D) 120 Exp. 180 degrees, the most stable hydrogen bonds are close to linear (D to H to A angle of 180 ). [hydrogen bonds tend to form with a geometry in which the hydrogen bond donor, the hydrogen and the hydrogen-bond acceptor are arranged in a straight line]. 24. Reaction products inhibit catalysis in enzymes by : (A) covalently binding to the enzyme (B) altering the enzyme structure (C) occupying the active site (D) form a complex with the substrate Exp. Altering the enzyme structure. Inhibition by product is an example of feedback inhibition that allosterically regulates the activity of the enzyme. 25. Which of the following statements regarding membrane transport is false? (A) Polar and charged solutes will not cross cell membranes effectively without specific protein carriers (B) Each protein carrier will only bind and transport one (or a few very similar) type of solute (C) Sugars such as glucose are always transported by active transport rather than by facilitated diffusion carriers (D) Ions are typically transported by special proteins that form membrane channels Exp. Glucose can transport itself on both ways, for example, the sugar glucose is transported by active transport from the gut into intestinal epithelial cells, but by facilitated diffusion across the membrane of red blood cells. P5/3

145 Amar Ujala Education 26. What will happen if histones are depleted from a metaphase chromosome and viewed under a transmission electron microscope? (A) 30 nm chromatin fibres will be observed (B) 10 nm chromatin fibres will be observed (C) A scaffold and a huge number of loops of DNA fibres will be observed (D) A huge number of loops of DNA fibres without scaffold will be observed Exp. If histone proteins are removed then still non-histone proteins such as scaffolding proteins will persist which will be responsible for scaffold and loops of the DNA. 27. Inchloroplast, the site of coupled oxidation-reduction reaction is the : (A) outer membrane (B) inner membrane (C) thylakoid space (D) stromal space Exp. Thylakoid membranes are the site of coupled oxidationreduction reactions that generate the proton-motive force. 28. Which of the following statements about meiosis is not true? (A) Kinetochores of sister chromatids attach to opposite poles in Meiosis I (B) Kinetochores of sister chromatids attach to opposite poles in Meiosis II (C) Chiasma is formed inprophase I (D) Homologous chromosomes are segregated in Meiosis I Exp. In meiosis I, homologous chromosomes separate from each other not the sister chromatids. So kinetochores of sister chromatids will not be attached to opposite poles in meiosis I. This will be observed in Meiosis II. 29. Leader sequence in some of the protozoan parasites is transcribed elsewhere in the parasite genome and gets joined with several transcripts to make the functional RNA. The joining of the two transcripts occur by the process of : (A) alternate splicing (B) trans splicing (C) ligation (D) RNA editing Exp. Trans splicing, int rans-splicing, a common spliced leader (SL) is donated to all mrnas from a small RNA molecule, the SL RNA. It was first discovered in the primitive eukaryotes, the trypanosomatids and later shown to occur also in C. elegans and other nematodes. 30. Small nucleolar RNAs used to process and chemically modify rrnas are called : (A) ScaRNAs (B) SiRNAs (C) SnoRNAs (D) SnRNAs A/UG-15 Exp. Small nucleolar RNA. The modifictions of rrnas are determined by approximately 150 different small nucleolus restircted RNA species, called small nucleolar RNAs (snornas), which hybridize transiently to prerrna molecules. Like snrnas, snornas associate with proteins, forming snornps. 31. During replication, the RNA primer is degraded by the 5' 3' exonuclease activity of : (A) RNase H1(ribonuclease H1) (B) FEN-1 (flap endonuclease 1) (C) Topoisomerase II B (D) DNA polymerase Exp. RNase H1 reconises and removes most of RNA primer. 32. Which one of the following statements about eukaryotic translation is NOT true? In eukaryotic translation : (A) ribosome binding site on mrna is called Kozak consensus sequences (B) initiator trna is RNA f t met t (C) initiator amino acid is methionine (D) translocation factor is eef2 Ans. (A & B) Exp. Both options (A) and (B) are wrong. Kozak sequence is not the ribosome binding site, rather it signifies the translational start site. Initiator trna in eukaryotes is charged with methionine but is not formylated. 33. Some T lymphocytes respond to antigenic stimulation by synthesizing a growth factor that causes T cell proliferation thereby increasing the responsive T lymphocytes resulting in amplification of the immune response. This is an example of : (A) endocrine signaling (B) paracrine signaling (C) autocrine signaling (D) cyclic signaling Exp. The cells secrete interleukines which has their receptors on the same. The cells which further activate the cell for division and repeating the same action. 34. Glycosaminoglycans are usually linked to proteins to form proteoglycans. Which of the following is not a proteoglycan? (A) Hyaluronan (B) Aggrecan (C) Betaglycan (D) Syndecan-1 Exp. Proteoglycans Are composed of GAG chains covalently linked to a Core protein, Except for hyaluronan, all GaGs are found covalently attahced to protein int he form of proteoglycans, which are made by most animals cells. 35. A patient with ER + /PR + breast cancer was cured with a drug T, whereas a second patient did not respond to T. Which one of the following is the best therapy that you should suggest for the second patient? P5/4

146 A/UG-15 (A) Surgery, followed by HER-2/neu targeted drugs (B) A drug that targets triple negative (ER /PR /HER- 2 ) breast cancer (C) Radiation, followed by drug T (D) Surgery, followed by radiation only Exp. Some breast cnacers--estimates range between 10% and 17%--are known as triple negative because they lack estrogen and progesterone receptrs and do not over express the HER2 protein. the majority of breast cancers associated with the breast cancer gene known as BRCA1 are triple negative. 36. If you run a pentavalent IgM through SDSpolyacrylamide gel electrophoresis, how many bands you are supposed to get by Western blotting using alkaline phosphatase conjugated secondary antibody? (A) Five (B) Four (C) Three (D) One Exp. IgM will form three bands one of 10 light chains (identical), one band of heavy chain (identical) and one of j chain. 37. Lens formation requires sequential events whereby the anterior neural plate signals the anterior ectoderm to promote secretion of Pax 6, which renders the anterior ectoderm more receptive to secretions from the optic vesicle. The above can be best explained by which of the following phenomenon? (A) Instructive interactions only (B) Epithetial-Mesenchymal interactions (C) Permissive-interactions (D) Induction and competence Exp. It can be explained by induction and copetences as first the neural plate is making lens ectoderm competent by expression of Pax 6 and then only lens ectoderm can be induced by optic vesicle. 38. The splitting or migration or one sheet of cells into two sheets as seen during hypoblast for-mation in bird embryogenesis is termed as : (A) delamination (B) ingression (C) involution (D) invagination Exp. Delamination is the splitting of one sheet into two sheets. 39. The group of cells of amphibian blastula capable of inducing the organizer is called as : (A) Hensen s node (B) Nieuwkoop centre (C) Dorsal blastopore lip (D) Hypoblast Exp. Nieuwkoop Center is responsible for induction of organizer by accumulation of beta catenin in the organizer region. Amar Ujala Education 40. Which one of the following statements regarding seed germination of a wild type plant is not correct? (A) Low ABA and high bioactive GA can break seed dormancy (B) Light accompanied with high temperature can break seed dormancy (C) GA induces synthesis of hydrolytic enzymes in cereal grains (D) Degradation of carbohydrates and storage proteins provide nourishment and energy to support seedling growth Exp. Dormancy period can be maintained also in high temperature. 41. Light is the dominant environmental signal that controls stomatal movement in leaves of well-watered plants grown in natural environment. Which one of the following wavelengths of light is responsible for such regulation? (A) Red light (B) Blue light (C) Green light (D) Far-red light Exp. Blue light activates hydrogen pump, which increases ions uptake inside guard cells and leads to opening of stomata. 42. Which one of the following is NOT the main factor that contributes to water potential during plant growth under normal conditions? (A) Solute potential (B) Hydrostatic pressure (C) Gravity (D) Temperature Exp. Water potential is dependent on osmotic potential, pressure potential and gravity potential, but least one temperature, as water atmosphere continum is dependent on temperature and humidity. 43. Which one of the following is NOT a characteristic property of carotenoids? (A) They posses complex porphyrin ring (B) They are integral constituent of thylakoid membrane (C) They are also called accessory pigments (D) They protect plants from damages caused by light Exp. Carotenoids does nt have porphyrin. 44. The plant hormone indole-3-acetic acid (IAA) is present in most plants. The structure of this hormone is related to which one of the following amino acids? (A) Glutamic acid (B) Aspartic acid (C) Lysine (D) Tryptophan Exp. IAA is derived from tryptophan. P5/5

147 Amar Ujala Education 45. The type-i glomus cells present in the carotid bodies contain granules which release some substances during hypoxia. Which one of the following is released in hypoxia? (A) Serotonin (B) GABA (C) Dopamine (D) IL 8 Exp. The signalling within the chemoreceptors is thought to be mediated by the release of neurotransmitters by the glomus cells, including dopamine, noradrenaline, acetylcholine, substancep, vasoactive intestinal peptide and enkephalins. 46. Which one of the following cells in the renal corpuscle can influence glomerular filtration by its contraction? (A) Podocytes (B) Endothelial cells of glomerular capillaries (C) Parietal epithelial cells of Bowman s capsule (D) Mesangial cells Exp. Mesangial excitability enables a homeo-static intraglomeruar stretch reflex that integrates an increase in filtration pressure with a reduction in capillary surface area. 47. Production of excessive amount of corticotropin (ACTH) occurs in which one of the following : (A) Grave s disease (B) Cushing s syndrome (C) Grieg s syndrome (D) Alport s syndrome Exp. Cushings syndrome is caused due to hypersecretion of ACTH (Hyperadrenalism). 48. Which one of the following functions is NOT served by the plasma proteins? (A) Blood clotting (B) O 2 transport (C) Hormone binding and transport (D) Buffering capacity of blood Exp. Plasma proteins are not involved in oxygen transport whereas they are involved in other mentioned functions. 49. The following pedigree shows the inheritance of a common phenotype controlled by an autosomal recessive allele. The probability of carriers in the population is 1/3 : I 1 2 II ? What is the probability that a child from parents II-3 and II-4 will show the phenotype? (A) 1/16 (B) 1/18 (C) 1/36 (D) 3/16 P5/6 A/UG Two plants with white flowers are crossed. White flowers arise due to recessive mutation. All F 1 progeny have red flowers. When the F 1 plants are selfed, both red and white flowered progeny are observed. In what ratio will red-flowered plants and white-flowered plants occur? (A) 1 : 1 (B) 3 : 1 (C) 9 : 7 (D) 15 : 1 Exp. Since the cross of both white flowers is giving rise to red flowers, means the mutation in the parents are in different genes and the presence of both the genes is required for red color formation. So, the gene interaction working here is complementary gene action. In that case the cross will be aabb AAbb; F1 : AaBb; F2 : 9 red (A_B_) : 7 white (aab_, A_bb, aabb) Bromouracil is a base analog that can cause mutation when incorporated into DNA. Which of the following is the most likely change that 5-Bromouracil induces : (A) T:AtoC:G (B) T:AtoA:T (C) G:CtoT:A (D) C:GtoA:T Exp. 5-Bromouracil is bae analogue for Thymine. Instead of binding to Adenine it pairs up with guanine. Base pair mutation will be T : A to C : G. 52. An interrupted mating experiment was performed between Hfr Str x a + b + c + and F Str r ab c strains. the genotype of majority of streptomycin resistant (Str r ) exconjugant after 10, 20 and 30 minutes of interrupted mating is given below : 10 min 20 min 30 min a b c a b c a b c Themostprobablegeneorderwouldbe: (A) abc (B) cab (C) bac (D) acb Exp. When the mating is interrupted at 10 min, only a + has passed. At 20 mins a + and c + has passed (meaning a + wasfollowedbyc+)andat30minsa+,b+andc+all have passed. So the gene order : a c b. 53. Individuals with greater mass have a smaller surface area to volume ratio, which helps to conserve heat. This is known as : (A) Leibig s rule (B) Cope s rule (C) Gloger s rule (D) Bergmann s rule Exp. According to Bergmann s rule :-larger animals have a lower surface area to volume ratio than smaller animals, so they radiate less body heat per unit of mass, and therefore stay warmer in cold climates.

148 A/UG Schizocoelous coelom formation, mouth formation from embryonic blastopore, spiral and determinate cleavage are characteristic of : (A) deuterostomes (B) pseudocoelomates (C) protists (D) protostomes Exp. Protostomers : Cleavage : Spiral, deter-minate, Coelom formation : schizocoelous, Blastopore : becomes mouth. 55. Which of the following is a correct match of the animal with its taxonomic group? (A) Hirudinea-Leech, Chelicerata-Horse shoe crab, Cestoda-Tapeworm Echinoidea-Sea Urchins Cephalopoda-Earthworm; Oligochaeta-Leech (B) Hirudinea-Earthworm, Chelicerata-Horse shoe crab, Cestoda-Octapus, Echinoidea-Tapeworm, Cephalopoda-Earthworm, Oligochaeta-Leech (C) Hirudinea-Tapeworm, Chelicerata-Leech, Cestoda- Tapeworm, Echinoidea-Horse shoe crab, cephalopoda-earthworm, Oligochaeta-Octopus (D) Hirudinea-Leech, Chelicerata-Tapeworm, Cestoda- Earthworm, Echinoidea-Sea Urchins, Cephalopoda- Octopus, Oligochaeta- Horse shoe crab 56. During which geological period was there an explosive increase in the number of many marine invertebrate phyla? (A) Ordovician (B) Devonian (C) Permian (D) Cambrian Exp. Cambrian period show a massive increase in number of marine invertebrates. 57. In which ecosystem is the detrial pathway of energy flow most important? (A) Lakes (B) Grasslands (C) Tropical rain forests (D) Oceans 58. What parameter, plotted on Y-axis against generation time, would yield the curve shown in the figure? High? Low Low Generation time High (A) Survivorship (B) Body size (C) Lifespan (D) Intrinsic rate of increase Exp. Intrinsic rate of increase is inversely proportional to generation time. Amar Ujala Education 59. An example of the species interaction called commensalism is : (A) nitrogen-fixing bacteria in association with legume plant roots (B) microbes in living human gut (C) female mosquito deriving nourishment from human blood (D) orchid plant growing on the trunk of a mango tree 60. The population density of an insect species increases from 40 to 46 in one month. If the birth rate during that period is 0.4 What is the death rate? (A) 0.25 (B) 0.15 (C) 0.87 (D) 0.40 Exp. No. of Birth = per capita birth Rate X No. of individuals (.4 40) = No.of death in one year = = Now No. of death = per capita deaht rate X no. of individuals (10 = d 40) = Worker bess, instead of themselves reproducing, help the queen reproduce. this behaviour is explained as an example of : (A) kin selection (B) group selection (C) sexual selection (D) natural selection 62. The degree of genetic relatedness between the offspring and their parents is : (A) higher than that between sister and brother (B) lower than that between sister and brother (C) the same as that between sister and brother (D) dependent on the number of siblings Exp. Genetic relatedness to parents and the offsprings is Which species concept utilizes morphological and molecular characters to distinguish between species? (A) Evolutionary (B) Ecological (C) Biological (D) Phylogenetic Exp. Phylogenetic is thestudy of evolutionary relationships among groups of organisms (e.g. species, populations), which are discovered through molecular sequencing data and morphological data matrices. 64. The wings of insects and the wings of bats represent a case of : (A) divergent evolution (B) convergent evolution (C) parallel evolution (D) natural evolution Exp. Wings of insects and bats are analogous organs, therefore, they are an example of convergent evolution. P5/7

149 Amar Ujala Education 65. You want to purify a recombinant protein of your interest. You can use affinity chromotography to purify as you have nickel columns available in the laboratory. With what molecule will you tag the protein to purify using those columns? (A) GST (B) Histidine (C) Histamine (D) Proline Exp. the polyhistidine binds strongly to divalent metal ions such as nickel and cobalt during Affinity chromatography. 66. Lower limits of detection by sensors is important. Which method of detection is more sensitive than glass electrode used for ph measurement? (A) Absorption spectroscopy (B) Refractive index (C) Circular dichroism (D) Fluorescence spectroscopy 67. Which of the following statement is incorrect for fluorescence in situ hybridization (FISH) technique? (A) A fluorescence or confocal microscope is used for detection of signal (B) A labeled sequence of nucleotides is used (C) Specific fluorescence tagged antibodies are used (D) A stringent washing step is essential to remove appearance of non specific signal Exp. FISH (fluorescent in situ hybridization) is a cytogenetic technqiue developed by biomedical researchers in the early that is used to detect and localize the presence or absence of specific DNA sequences on chromosomes. So no requirements of antibody, as no protein detection. A/UG In an experiment to detect a new protein in fixed cells, no secondary antibody tagged with fluoresence dye is available. What should be the best choice out of the following to detect the protein? (A) Protein A-FITC (B) Protein A-Sepharose (C) Biotin-FITC (D) Avidin-FITC Exp. Option (A) and (B) involves antibodies while option (D) reacts with biotin while biotin FITc is the only which can be used to detect biotin-binding sites or the concentration of avidin or streptavidin through the strong quenching associated when this molecule binds to avidin. 69. If a researcher intends to identify a specific brain area activity linked to a cognitive function in human subjects, which one of the following techniques should be used? (A) CAT (B) MRI (C) fmri (D) Patch-clamp Exp. fmri, functional MRI its a techniques for measuring brain activity 70. Two 18-residue helical peptides A and B are enantiomers. They can be distinguished by : (A) recording their MALDI mass spectrum (B) hydrolysis followed by amino acid analysis (C) sequencing by Edman s method (D) examining their circular dichroism spectra Exp. CD spectroscopy. Chiral molecules exist as pairs of mirror-image isomers. These mirror image isomers are not super-imposable and are known as enantiomers. they physical and chemical properties of pair of enantiomers are identical with two exceptions : the way that they interact with polarised light and the way that they interact with other chiral molecules. 71. The lifetime of a peptide bond in protein is very large (~ 1000 years). Which statement below is incorrect with respect to stability of the peptide bond? (A) The free energy of hydrolysis if negative (B) the free energy of hydrolysis is positive and large (C) The energy barrier to be crossed to go to the hydrolyzed state is large (D) The peptide bond can be hydrolyzed by 6N HCl at 100 C Exp. Free energy of hydrolysis is of a peptide bond is negative releasing around 10 KJ/mol of energy, so free energy change is negative and not positive. 72. Acetyl-(Ala) 18 CONH 2 exists in -helical conformation in solution. Most of the backbone dihedral angles (, ) will be : PART-C P5/8 (A) 60, 30 (B) 60, 30 (C) 60, 30 (50%) and 60, 60 (50%) (D) 80, 120 Exp. The average phi and psi values for alpha-helices and beta-sheets is around -57, -47 and -80, +150, respectively. 73. DNA is not hydrolyzed by alkali whereas RNA is readily hydrolyzed. This is due to : (A) The double helical structure of DNA (B) The presence of uridine in RNA (C) Due to features observed in RNA such as stem-loop structures (D) The presence of 2'-OH group in RNA Exp. Presence of 2'OH in Ribose of RNA.

150 A/UG Two homologous proteins were isolated from a psychrophile (P) and a thermophile (T). The purified proteins were subjected to denaturation, protease digestion and circular dichroism (CD). Following obsrvations were made : (1) The CD spectra of P and T proteins are identical (2) Their amino acid composition is 95% identical (3) T and P are equally susceptible to proteolysis in the presence or absence of reducing agent (4) T has higher midpoint of thermal denaturation than P The reason for enhanced stability in T is due to : (A) Altered secondary structure (B) Increased number of disulfides in T (C) Increase in water of hydration (D) Increase in number of salt bridges Exp. Increase in no. of salt bridges. It is observed that salt bridges and side chain-side chain hydrogen bonds interactions are increased in the majority of the thermophilic proteins. 75. Binding of two ligands to their binding proteins were investigated. Following binding isotherms were obtained : Bound Bound/Total Which of the following statements is correct? (A) A is obtained with an oligomeric protein and B is obtained with a monomeric protein (B) B is obtained with protein with positive cooperativity (C) A and B were obtained by the same protein at two different temperatures (D) The profile B is not possible Exp. The plot is of Scatchard plot which describes ligandproton binding. Profile B describes positives cooperative bidning. 76. Enzyme parameters of four isozymes is given below : Isozyme K m micromolar V max A B C D These isozymes are localized in different tissues. In liver the substrate concentration is 0.2 micromolar. The liver isozyme is likely to be : (A) A (B) B (C) C (D) D A B Amar Ujala Education Exp. Since, the km value for Enzyme D is loest which is an indication about the high affinity of the enzyme towards the substrate and the high. affinity enzyme always work at low substrate concentration. 77. A null mutation is created in a gene which is responsible for specific phosphorylation at 6 th carbon position of mannose on acid hydrolases occuring in cis-golgi. The following statements are given towards explaining the effect of this mutations : (1) The lysosomes will be devoid of lysosomal enzymes (2) Lysosomal enzyme will be secreted out (3) Lysosomal enzyme will get localized in cytoplasm Which statement or combination of statements will explain the effect of mutation if the acid hydrolases in the mutant do not get degraded? (A) (1) and (3) (B) (2) and (3) (C) (3) only (D) (1) and (2) Exp. If the mannose group is not phosphorylated at 6th carbon residue then lysosomal enzymes will not be largeted to the lysosomes as this is the sorting signal. So lysosomes will be devoid of lysosomal enzyme, also since they will be present in the Golgi so these enzymes will be secreted out of the cell. 78. In order to prove that liposome can serve as a model membrane (mimicking cellular plasma membrane) and can be used as a target for complement-mediated immunolysis, an experiment as below is designed. To initiate such experiment, hapten-conjugated liposomes are made and loaded with umbelliferyl phosphate (UMP, hydrolysed product of UMP is umbelliferone and is fluorescent). Such loaded, hapten-conjugated liposomes in 10 mm Tris buffered saline, ph 7.4 were mixed with anti-hapten antibodies and fresh guinea pig serum (as a source of complement) to induce immunolysis of liposomal membrane. To quantify only the membrane lysis component which of the assay sequences below is MOST appropriate? (A) Mixture is ultra centrifuged and the supernatant reacted with alkaline phosphatase and fluorescence measured (B) Mixture is sequentially reacted with phospholipase and alkaline phosphatase followed by fluorescence measurements (C) Mixture is directly subjected to fluorescence measurement (D) Mixture is treated with Triton X-100 and fluorescence measured Exp. If membrane is lysed by complement system, UMP loaded in the liposomes will be released in the supernatant. This can be detected by treating with alkaline phosphatase and mesuring fluorescence. P5/9

151 Amar Ujala Education 79. A gene producing red pigment was placed near centromeres of fission yeast and thus subjected to position effect variegation and produced white colonies. A screen for mutants that increased the red pigment production was undertaken. Which of the following genes, when mutated, is likely to produce this genotype? (A) Histone Deacetylase (B) Histone acetylase (C) RNA polymerase II (D) TATA binding factor Exp. Histone deacetylase, this enzyme removes the acetyl group from Histones and makes the histone-dna binding more tight. So mutation in the gene encoding for this enzyme will allow for acetylation of histone and hence opening up of chromatin structure. This will enable the red gene to express. 80. A newly identified sequence was experimentally tested by in vitro transport assay using a radiolabelled protein containing the sequence to test import into mitochondria. Transport assay was done for a short time with or without membrane potential and after the assay, the mitochondria were either treated or not treated with proteinase K. At the end of the assay the mitochondria were pelleted and total protein of the pellet was isolated and separted on SDS-PAGE and autoradiographed. A representative autoradiogram is shown below. Based on this experiment data, which of the following statements is not correct? A/UG-15 (B) Arrest cells in G2 phase (C) Arrest cells in Anaphase (D) Block chromatin condensation Exp. If proteasome inhibitor is added in G2 phase then cells will be arrested in Anaphase as APC will not be able to cause degradation of securin which is the key step for metaphase to anaphase transition. Also, cell requires ubiquitinylation to end M-phase. So in both the cases, cell ought to remain arrested in anaphase. 82. Which of the following statements best describe archaebacteria? (A) Mostly autotrophic, cell wall contains peptidoglycan, 60S ribosomes, live in extreme environment (B) Divide by fission, not susceptible to lysozyme, live in extreme environments mostly autotrophic (C) Not susceptible to lysozyme, contain Golgi and linear chromosomes (D) Chitinous cell wall, obligate aerobic, circular chromosomes Exp. Archaebacteria they are capableof surviving in extreme conditions, are not susceptible to lysosomal enzymes and are mostly autotrophic. 83. The following graph represents the expression of tryptophan synthetase (TS) in E. coli cells in absence ( ) or presence ( ) of tryptophan in the medium : 5 Activity of TS (A) The protein goes into the matrix (B) Not all of the added protein was imported (C) The protein requires membrane potential for import (D) The protein is associted with the outer mitochondrial membrane Exp. As is depicted in the diagram, protein requires membrane potential to be transported into the matrix. Since it goes into the matrix it can not be cleaved by proteinase K once it is transported. So option 4 is wrong according to which it is an outer mitochondrial protein. 81. If a proteasome inhibitor is added to synchronously cycling human cells in G2 phase which one of the following events is likely to happen? (A) Induce re-replication of DNA If the two trp codons in the leader sequence of trp operon is mutated to ala, which of the following graphs will best represent activity of TS in E. coli cells grown in the absence ( ) or presence ( ) of tryptophan? (A) (C) (B) 1 Activity of TS (D) 1 Activity of TS Activity of TS Activity of TS5 P5/10

152 A/UG-15 Exp. In case with tryptophan present, there is no need to synthesise the tryptophan genes ie tryp A an tryp B. But in case with absent it has to synthesised so from the picture, it is clear that in the presence of tryptophan tryptophan synthase is less and absent it is high. Because of mutation again it is going to be the same case. Tryp codons got mutated in leader region 1 will in turn lead to 3 and 4 pairing, because it undergoes a normal pause in the stop codon if trypophan is present. And coming to the tryptophan absent, due to mutation in tryp codons in leader region 1 will make the ribosome to proceed in a similar fashion i.e. 2 and 3 pairing, hence tryptophan synthase will be synthesised. Moreover when comparing to all the other options given in the picture in the absence of tryptophan same level of plain box (tryptophan synthase) is given but only in 1st option in the presence of tryptophan very less synthesis of tryptophan is given with shaded colour. 84. Puromycin is an antibiotic used to inhibit protein synthesis. Given below are few statement about the antibiotic : (1) It enters the E-site of the ribosome where it prevents the release of deacetylated trna after the action of peptidyl transferase (2) It blocks the translocation process by binding to the translocation factor EF-G (3) Puromycin resembles the initiatior trna, -met trna f and binds exclusively to the P-site i (4) It resembles the aminoacyl trna and binds to the A-site of the ribosome (5) Puromycin inhibits only prokaryotic protein synthesis (6) Puromycin inhibits both prokaryotic and eukaryotic protein synthesis Which of the above statement (s) is/are true? (A) (1) and (5) (B) (2) only (C) (4) and (6) (D) (3) and (5) Exp. Puromycin is an aminonucleoside antibiotic, derived from the streptomyces alboniger-bacterium, [1] that cuses premature chain termination during translation taking place in the ribosome. Part of the molecule resembles the 3' end of the aminoacylated trna.it enters the A site and transfers to the growing chain, causing the formation of a puromycylated nascent chain and premature chain release. Puromycin is used in cell biology as selective agent in cell culture system. It is toxic to prokaryotic and eukaryotic cells. 85. Following are certain statements related to eukaryotic DNA replication : (1) The genome of multicellular animals contain many potential origins of replication P5/11 Amar Ujala Education (2) During early development, when embryos are undergoing rapid cell divisions, origin sites are uniformly activated (3) Pulse-chase technique is used to label sites of DNA replication (4) The rate of elongation of different DNA chains during genome replication varies drastically Which one of of the following combinations of above statement is correct? (A) (1), (2) and (3) (B) (1), (3) and (4) (C) (2), (3) and (4) (D) (1), (2) and (4) Exp. Option D is wrong as rate of elongation will be similar forallthednastrandsassamednapolymeraseisused to replicate the entire genome. 86. The expression of a hypothetical gene was analyzed by Northern and Western blot hybridization under control and induced condition.the results are summarized below : Control Induced Control Induced Northern Blot Western Blot Expression of genes can be regulated by : (1) control at transcription initiation (2) alternative splicing (3) control of translation initiation (4) protein stability Which of the above regulatory mechanisms can explain the observations shown in the figures? (A) only (2) (B) only (1) and (2) (C) only (2) and (3) (D) (1), (2), (3) and (4) Exp. As is clear from the picture of northern blot, the size of the transcript in induced condition is lower than of control indicating likelihood of alternative splicing in case of western blot picture band is visible only in case of induced condition indiacting a requirement of a regulation at initiation of translation. 87. Total RNA was isolated separately from cytosol and nuclei of human cells growing in a cell culture. Each sample was mixed with a purified denatured fragment of a DNA corresponding to a large intron of a house keeping gene and incubated under renaturating condition. Given below are the statements made about the outcome of the experiment : (1) RNA isolated from nuclei will form RNA-DNA duplexes because of the presence of introns in the primary RNA (2) Cytosolic RNAs usually will not form RNA-DNA duplexes

153 Amar Ujala Education (3) Both cytosolic and nuclear RNA will not form RNA- DNA duplexes as transcription and splicing occur simultaneously (4) Cytosolic RNA will form RNA-DNA duplexes because unspliced cytosolic RNAs are exceptionally stable. Which of the above statement (s) is/are true? (A) only (3) (B) (1) and (4) (C) (1) and (2) (D) only (4) 88. A promoter deletion study was done in order to detemrine the binding sites for a transcription factor on the promoter, which is activated on treatment with the drug X. the following constructs were made : LUC LUC LUC LUC Luciferase assay revealed the following results : Luciferase activity Control Treated with drug X The following statements can be made : (1) Region between 1800 and 1210 contains a binding site for the activator. (2) Region between 868 and 1210 contains a binding site for a repressor (3) Region between 868 and 432 contains a binding site for a repressor (4) Region between 1210 and 868 contains a binding site for the activator. Which of the above is/are true? (A) (1) and (3) (B) (2) and (3) (C) (1) and (4) (D) (2) only Exp. The luciferase activity is seen to decrease when there is a deletion in the region and and hence suggests that at these sites activator binds a deletion after 868 does not change the acticity of luciferase. P5/12 A/UG A researcher wanted to immunize individuals of a particular area with viral infection. The researcher developed two different vaccine type (A and B) with the following properties : (1) When vaccine type A specific for a viral strain is administer to individuals, they develop strong neutralizing antibody response with very poor immunological memory. Hence it has to be administered in repetitive doses (2) When vaccine type B specific for a viral strain is administered to individuals, they fail to develop circulating antibody response at the time of infection but they develop strong immunological memory If two viral strains V1 (incubation period-2 days) and V2 (incubation period-15 days) are likely to infect the area, which of the following vaccine combination would provide maximum immunization? (A) V1 specific type A and V1 specific type B (B) V1 specific type A and V2 specific type B (C) V2 specific type A and V1 specific type B (D) V2 specific type A and V2 specific type B Exp. Since V1 and 2 days of incubation period then it need quick response against it that can be provided by vaccine A and V2 has incubation period of 15 days then though the antibodies are not formed but the memory created can help in treatment against it. 90. A pharmacy student designed a drug to specifically target the receptors for retinoic acid in order to prevent stem cell differentiation. After in vitro trial, the investigator found that the cells underwent differentiation and the drug seemed to be ineffective. The following reasons were given by the student : (1) The size of the drug exceeded the size of molecules that could cross the membrane (2) The drug was small in size but hydrophobic in nature (3) The drug did not bind to its receptors Which of the above could be the probable reason for drug ineffectiveness? (A) only (3) (B) (1) and (3) (C) (1), (2) and (3) (D) only (2) Exp. The most probable reasons can be a and C only as if the drug is small and hydrophobic then it should easily cross the membrane os in that case drug should be effective. 91. Which one of the following statements about cell-cell interactions is not true? (1) Cadherins are transmembrane linker proteins which carry out Ca 2+ -mediated adhesion between adjacent cells.

154 A/UG-15 (2) Integrins are transmembrane adhesion proteins that mediate homophilic adhesion through actin and intermediate filaments. (3) Selectins are cell surface lectins that mediate a variety of transient, cell-cell adhesion interactions in the bloodstream. (4) ICAMs (intracellular cell adhesion mole-cules) and VCAMs (vascular cell adhesion molecules) are members of immunoglobulin (Ig) superfamily. (A) only (1) (B) only (2) (C) both (3) and (4) (D) both (1) and (2) Exp. Integrins are transmembrane proteins which that mediate heterophilic adhesion. Hence, option B is not true. 92. You have isolated a short chain polypeptide from a bacteria which upon administering to animal cells in culture is demonstrated to get transported to mitochondria and to cause mitochondrial membrane disruption. This peptide is being considered for its therapeutic effects on cancer and is to be tested for its anticancer properties in the following models of the disease : (1) Cancer with Apaf-1 mutation (2) Cancer where Bax genes are inactivated (3) Cancer where Bak genes are inactivated Which of the following is likely to provide best effect (i.e. induce apoptosis) : (A) only (1) (B) (2) and (3) (C) (1) and (2) (D) (1) and (3) Exp. As Bax and Bak are the Pro-apoptotic genes which get localised on the mitchondrial membrane and induce apoptosis. While Apaf 1 is present in cytosol of the cell andbindstocytreleasedbythemitochondriaafter localisation of Bax andbak to forapoptosome. 93. A technician wanted to make rabbit antiserum specific for mouse IgG.The technician injected rabbit with purified mouse IgG but obtained antiserum which reacted strongly with each of the other mouse isotypes. Which of the procedures mentioned below will allow him to make antiserum specific for IgG only? (A) Injecting rabbit with purified F(ab) 2 region of the IgG antibody (B) Injecting rabbit with purified heavy chain of the IgG antibody (C) Injecting rabbit with purified light chain of the IgG antibody (D) Injecting rabbit with purified F(ab)' region of the IgG antibody 94. In order to precipitate a particular protein by its specific antiserum, it was found that the protein formed crosslinked lattice with specific polyclonal antiserum but failed to precipitate with specific monoclonal antiserum. Amar Ujala Education Which of the following would accurately justify the reason for this behavior? (A) The protein has multiple copies of the same epitope specific for the monoclonal antibody (B) The protein has multiple distinct epitopes but each has a single copy (C) There is total absence of epitopes in the protein (D) The protein has multiple copies of different epitopes 95. Activation of the Wnt signal transduction pathway is extremely important during early development. Of the various pathways, which one of the following is most likely to induce cytoskeletal changes, like cell shape and movement? (A) Wht (B) (C) (D) Dsh Daam1 RhoaA ROCK Wht Dsh Axin GSK3 APC PLC PKC Adaptor protein Wht Wht SOS Ca 2+ Rac CaMK11 Active Ras Raf MEK Exp. Activation of Rho-GTPases is responsible for the different cytoskeletal changes such as changes in cell shape or cell movement. P5/13

155 Amar Ujala Education 96. For successful fertilization in sea urchin, interaction between the surface of the egg and acrosomal proteins, specifically a 30.5 kda protein called bindin, is necessary. The following factors could affect this interaction and prevent fertilization : (1) Removal of egg jelly polysaccharides. (2) Removal of bindin receptors on the eggs vitelline membrane. (3) Removal of bindin receptors from the egg jelly. (4) Removal of bindin receptors from a single cluster on the vitelline membrane. Which one or the combination of the above statements is correct? (A) (1) and (4) (B) Only (2) (C) (1) and (2) (D) Only (3) Exp. Bindin receptors are present on the vitelline membrane. 97. A two-celled embryo is made of blastomeres A and B. If the two blastomeres are experimentally separated, the A blastomere generates all the cells it would normally make. However, the B blastomere in isolation makes only a small fraction of cells it would normally make. Based on the above observations only, which one of the following conclusions is correct? (A) A blastomere is autonomously specified while B blastomere is conditionally specified (B) A blastomere is conditionally specified, while B blastomere is autonomously specified (C) Descendents of A blastomere are autonomously specified (D) Descendents of B blastomere can either be autonomously specified or conditionally specified Exp. A blastomere is able to form all the cells according to its fate therefore it is autonomously specified whereas B blastomere will be conditionally specified. 98. A mutant embryo of Drosophila in which one of the major sex determining gene, sex lethal, can only undergo default splicing, was allowed to develop. The following statements are towards explaining the determination of sex of the embryo : (1) Theembryowilldevelopintoamalefly (2) The embryo will develop into a female fly (3) Sex lathal gene product direct regulates sex specific alternate splicing of double sex RNA (4) Sex lethal gene product regulates sex specific splicing of transformer RNA which inturn regulates splicing of double sex DNA The correct combination of above statements to explain sex determination of the given embryo is : (A) (1) and (3) (B) (1) and (4) (C) (2) and (4) (D) (2) and (3) A/UG A mutant was experimentally generated which had wings reduced to haltere like structure. The following statements are put forward regarding this phenotype : (1) ultrabithorax gene extopically expressed in second thoracic segment (2) antennapedia gene ectopically expressed in second thoracic segment (3) a homeotic mutation (4) a mutation in gap gene The following combination of statement will be most appropriate explaining the molecular basis of mutant phenotype : (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (3) Exp. Ultrabiothorax is responsible for formation of Hlateres so halteres can be observed in the 2nd thoracic segment if ultrabithorax is expressed ectopically in the 2nd throacic segment. this a homeotic gene mutation Following are certain statement regarding the activities of homeotic genes of classes A, B and C invilved in floral organ identity : (1) Activity of A alone specifies sepals (2) Activity of B alone specifies petals (3) Activities of B and C form stamens (4) Activity of C alone specifies carpels Which one of the following combination of above statements is correct? (A) (1), (2) and (3) (B) (1), (2) and (4) (C) (2), (3) and (4) (D) (1), (3) and (4) Exp. As petal formation is regulated by A and B both A transgenic lettuce plant was generated by overexpressing isopentenyl transferase (IPT) gene under the control of the promoter of senescence activator gene (SAG12). Following are some statements regarding this transgenic plant. The transgenic plants : (1) exhibit delayed senescence (2) exhibit fast senescence (3) have higher amount of cytokinin during senescence (4) have higher amount of gibberellins during senescence Which one of the following combination of above statement is correct? (A) (1), (2) and (3) (B) (1), (2) and (4) (C) (2), (3) and (4) (D) (1), (3) and (4) Exp. As IPT, is responsible for cytokinin synthe-sis, as cytokinin is growth hormone, so it delays senescence. P5/14

156 A/UG Following are certain statements regarding secondary metabolites found in plants : (1) All terpenes are derived from a six-carbon element (2) Alkaloids are nitrogen containing compounds (3) Pyrethroids, a monoterpene easter found in the leaves and flower of Chrysanthemum species, show insecticidal activity (4) Limonoids are groups of alkaloids and have antiherbivoral activity. Which one of the following combinations of above statements is correct? (A) (1) and (2) (B) (1) and (4) (C) (2) and (3) (D) (3) and (4) Exp. Alkaloids are nitrogen containing compounds and pyrethroids, pyrethroid is an organic compound similar to the natural pyrethrins produced by the flowers of pyrethrums (Chrysanthemum cinerariae-folium and C occineum). Pyrethroids now constitute the majority of commercial household insecticides [1] in the concentrations used in such products they may also have insect repellent properties and are generally harmless to human beings in low doses but can harm sensitive individuals Light is crucial for plant growth and development. Following are certain statements related to photoreceptors in model plant Arabidopsis thaliana : (1) Among the five phytochrome genes, representing a gene family, PHYB plays a predominant role in redlight perception (2) Cryptochromes are involved in the regulation of flowering time and hypocotyl length (3) phy A photoreceptor is predominantly involved in far-red light perception (4) The LOV domains of phytochrome C (PHYC) is an important domain for signal transmission Which one of the following combinations of above statements is correct? (A) (1), (2) and (3) (B) (1), (3) and (4) (C) (2), (3) and (4) (D) (1), (2) and (4) Exp. Statement B : incorrect. Cryptochromes not involved in hypocotyl length One of the important functions of program cell death (PCD) in plants is protection against pathogens. PCD also appears to occur during the differentiation of xylem tracheary elements that leads to nuclei and chromatin degradation. These changes results from the activation of certain genes. Following are certain genes encoding : (1) Topoisomerase (2) Nuclease (3) RNA polymerase (4) Protease Amar Ujala Education Which one of the following combination of the above is involved in differentiation of xylem tracheary elements? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (4) and (1) 105. Following are certain statemnets regarding C 3,C 4 and CAM plants? (1) The ratio of water loss to CO 2 uptake is higher in CAMplantsthatitisineitherC 3 and C 4 plants (2) The rate of photosynthesis attains maximum rate at lower intracellular CO 2 partial pressure in C 4 plants than in C 3 plants (3) The compensation point in C 3 plants are always lower than C 4 plants (4) Plants with C 4 metabolism need less rubisco than C 3 plants to achieve a given rate of photosynthesis Which one of the following combination of above statement is correct? (A) (1) and (2) (B) (1) and (3) (C) (3) and (4) (D) (2) and (4) 106. Following are some statements related to osmotic stress in plants : (1) The accumulation of ions during osmotic adjustement is predominantly restricted to the vacuoles (2) In order to maintain the water potential equilibrium within the cell, other solutes called as compatible solutes or compatible osmolytes accumulate in the cytoplasm (3) Galactose is one of the compatible osmolytes involved in osmotic stress in plants (4) There are mainly four groups of molecules that frequently serve as compatible solutes Which one of the following combinations of above statement is correct? (A) (1), (2) and (3) (B) (2), (3) and (4) (C) (1), (2) and (4) (D) (1), (3) and (4) Exp. Galactose is not working as compatible solute or osmoprotectant, main osmoprotectant are mannitol, sorbitol, glycine betaine, proline The changes in left ventricular stroke work (LVSW) according to the different left ventricular end-diastolic pressures (LVEDP, which indicates the initial myocardial fiber length) in a dog, under control conditions, were recorded, which follows Starling s law of the heart. This LVSW-LVEDP relationship was investigated in the same dog after constant infusion of norepinephrine, and these two data sets were plotted. Which one of the following graphs correctly represents the results obtained? P5/15

157 Amar Ujala Education (A) (B) (C) (D) LVSW LVSW LVSW LVSW Control LVEDP LVEDP LVEDP LVEDP Noreplanephrine Exp. In the presence of norepinephrine SW will increase due to increase in end diastolic volume Different frequencies of sound were presented on the ear and the movement of basilar membrane was experimentally determined. The characteris-tics of movement of basilar membrane after presentation of 100 Hz sound are described in the following statements : (1) The base of basilar membrane showed resonance (2) The apex of basilar membrane showed resonance (3) A wave travelled from the base to apex of basilar membrane but the maximum displacement was noted near the apex (4) A wave travelled from the base to apex of basilar membrane but the maximum displacement was noted near the base A/UG-15 Which one of the following is correct? (A) (1) and (3) (B) (2) and (4) (C) (3) only (D) (4) only Exp. For the lowest frequency (60 Hz) the maximum displacement is near the apical end for the highest frequency (2000 Hz) the maximum displacement near the base, while the intermediate frequency has maximal displacement between the two. Therefore, highfrequency sounds cause a small region of the basilar membrane near the stapes to move, while low frequencies cause almost the entire membrane to move. However, the peak displacement of the membrane is located near the apex. This shows that the travelling wave always travels from base to apex, and how far towards the apex it travels depends on the frequency of stimulation : lower frequencies travel further When a nerve fiber is stimulated with increasing strength of stimulus, the action potential fails to generate even though thethreshold level may be passed. The following statement may explain this accommodation of nerve fiber : (1) The critical number of open sodium channels required to trigger the action potential may never be attained due to slow depolarization (2) Potassium channels open in response to slow depolarization, which makes the nerve fiber refractory to depolarization (3) The low threshold sodium channels remain open, which increases the threshold of firing of action potential (4) The efflux of sodium and influx of potassium due to operation of Na +,K + -ATPase oppose the depolarization Which one of the following is correct? (A) (1) only (B) (1) and (2) (C) (3) only (D) (3) and (4) 110. The heart rate shows variation during respiratory rhythm in most human subjects. Which one of the following statements describing the changes of heart rate during respiratory phases is true? (A) The heart rate is accelerated during expiration, but no change occurs during inspiration (B) The heart rate is accelerated during inspiration and decelerated during expiration (C) The heart rate is accelerated during expiration and decelerated during inspiration (D) The heart rate is accelerated during inspiration and no change occurs during expiration P5/16

158 A/UG-15 Exp. Respiratory sinus arrhythmia (RSA) is a naturally occurring variation in heart rate that occurs during breathing cycle. Heart rate increase during inspiration and decreases during expiration. Heart rate is normally controlled by centers in the medulla oblongata. One of these centers, the nucleus ambiguus, increases parasympathetic nervous system input to the heart via the vagus nerve. The vagus nerve decreases heart rate by decreasing the rate of SA node firing. Upon expiration the cells in the nucleus ambiguus are activated and heart rate is slowed down. In contrast, inspiration triggers inhibitory signals to the nucleus ambiguus and consequently the vagus nerve remains unstimulated Given below are few statements with reference to blood clot formation which results from triggered chain of reactions : (1) Conversion of fibrinogen to fibrin (2) Activation of factor XIII, which stabilizes fibrin mesh work (3) Activation of factor XII, which promotes plasmin activation (4) Enhancement of platelet aggregation Which one of the following combination of statement is correct with reference to roles of thrombin in hemostasis? (A) (2), (3) and (4) (B) (1), (2) and (4) (C) (1), (3) and (4) (D) (1), (2) and (3) Exp. If plasma will be activated then it will dissolve the clot hence, option B will be wrong At 17 years, a 7 feet tall human was diagnosed with gigantism caused by pituitary tumor. The condition was surgically corrected by removal of the person s pituitary gland. Doctors advised hormonal therapy. The possible hormonal therapies that would be required for survival are : (1) Thyroid hormone (2) Glucocorticoids (3) Glucagon (4) Growth hormone (5) Insulin Which one of the following combination can be used? (A) (1) and (2) only (B) (2) and (4) only (C) (1), (2) and (4) (D) (1), (3) and (5) Exp. He would require only Thyroid hormone and Glucocorticoids for survival. If thyroid hormones will be absent then basic metabolic rate of the body will decrease and if glucocorticoids will be absent then that individual will not be able to survive in stressful conditions Autotetrapoloids arise by the doubling of 2n complement to 4n. There are three different pairing possibilities at meiosis in tetraploids as given below : (1) Two bivalents P5/17 Amar Ujala Education (2) one quadrivalent (3) One univalent + one trivalent Which of the above pairings can lead to production of diploid gamete? (A) only (1) (B) (2) and (3) (C) (1) and (3) (D) (1) and (2) Exp. If 2 bivalents or one quadrivalent is present in a cell only then diploid gametes can be produced due to distribution of bivalents between the two cells. Whereas, unequal distribution of chromosomes will be observed in case of univalents and trivalents and trivalents, therefore, diploid gametes cannot be produced The following is the amino acid sequence of a part of a protein encoded by gene X.... Phe Leu Val Pro Ser Tyr Cys... A mutant for gene X is isolated following treatment with a mutagen. the amino acid sequence of the same region encoded by the mutant gene is as follows :... Phe Leu Phe Arg Arg Ile... Which of the following mutagenes is most likely to have been used? (A) 5-bromouracil (B) 2-amino-purine (C) Ethyl methanesulfonate (D) Acridine orange Exp. After leuicine the amino acids sequence is changed and that reflect an indel which is caused by acridine orange In Neurospora, The mutant stp exhibits erratic stopand-start growth. When a female of stp strain is crossed with a normal strain acting as a male, all progeny individuals showed stp mutant phenotype. However, the reciprocal cross resulted in all normal progeny individuals. These results can be explained on the basis of : (1) Maternal inheritance (2) sex limited inheritance (3) sex influenced inheritance (4) stp mutation may be located in mitochondrial DNA The most appropriate statement or combination of the above statements for explaining the experimental results is : (A) (1) and (3) (B) (3) only (C) (1) and (4) (D) (2) and (4) Exp. When mother is stp, all offsprings are stp. When mother is normal, all offspring are normal. this shows that the character is getting transfered from mother to the offspring directly. This can happen when the stp mutation is present in extranuclear DNA i.e. mitochondrial DNA. Statements A and D, both are correct.

159 Amar Ujala Education 116. A co-transduction experiment was performed to decipher the linear order of 4 genes : a, b, c and d. Three sets of experiments were done where transductants were selected for a (Set-1) or b (Set-2) or c (Set-3) and screened for co-transduction of the other markers : Set-1 Selected for Co-transduction Frequency a b 31 a c 3 a d 89 Set-2 Selected for Co-transduction Frequency b a 22 b c 78 b d 68 Set-3 Selected for Co-transduction Frequency c a 0 c b 69 c d 43 Based on the frequencies shown above, identify the most likely order in the genome : (A) abcd (B) bcda (C) cdab (D) adbc Exp. Co-transduction frequency between a and c is least. So a and c should be farthest A hypothetical biochemical pathway for the formation of eye color in insect is given below : Substrate X Substrate Y (Colorless) (Colorless) Colorless Intermediate mutant a x Brown pigment Colorless Intermediate xmutant b organge pigment Orange-Brown Eye color (wild type) Two autosomal recessive mutants a and b are identified which block the pathway as shown above. Considering that the mutants are not linked, what will be the phenotype of the F 2 progeny if crosses were made between parents of the genotype aabb AAbb and the F1 progeny are intercrossed? P5/18 A/UG-15 (A) 9 orange-brown : 3 orange : 3 brown : 1 colorless (B) 9 orange-brown : 7 colorless (C) 1 orange : 2 colorless (D) 15 orange-brown : 1 colorless Exp. Parental cross : aabbx AAbb; F1 : AaBb; F1 self cross : AaBb AaBb, since genes are unlinked, they will get assorted independently F2 : A_B_(Organge brown) 9 : A_bb(Brown) 3 : aab_(orange) 3 : aabb (white) 1. Answer An analysis of four microsatellite markers was carried out in a family showing a genetic disorder. The results are summarized below : F1 F F1 F M1 M2 M3 M4 Based on the above, which of the markers shows linkages to the disorder? (A) M1 (B) M2 (C) M3 (D) M4 Exp. P1, 1, 2, 5, 6-all affected individual share M2 as their common marker. this means M2 is linked with the genetic disorder In group I are given 4 orders of class insecta. Match each one with a common name (Group II) and its diagnostic characters (Group III) : Group I Dermaptera (A) Ephemeroptera (B) Odonata (C) Plecoptera (D) Group II Ant (E) Mayfly (F) Grasshopper (G) Damselfly (H) Stonefly (I) Earwig (J) Group III (i) Elongate, membranous wings with net like venation, abdomen long and slender, compound eyes occupy most of head, hemimetabolous metamorphosis

160 A/UG-15 (ii) Elongate chewing mouthparts, thread-like antennae, abdomen with unsegmented forceps-like cerci, hemimetabolous metamorphosis (iii) Forewing long, narrow and leathery, hindwing broad and membranous, chewing mouthparts, hemimetabolous metamor-phosis (iv) Elongate abdomen with two or three tail filaments, two pairs of membranous wings with many veins, forewings traingular, short, bristle-like antennae, hemimetabolous metamorphosis (v) Adults with reduced mouthparts, elongate antennae, long cerci, nymphs aquatic with gills, hemimetabolous metamorphosis (vi) Wings membranous with few veins, well developed ovipositiors, sometimes modified into a sting, mouth parts modified for biting and lapping, holometabolous metamorphosis (A) A-J-(ii) (B) B-I-(vi) (C) C-H-(v) (D) D-F-(i) Exp. Damselfly belong to Odonata order of insecta nymphs are aquatic with gills Which of the following is a correct match of the animal with its attribute? Animal Attribute (1) Rotifer (i) Nauplius larval stage (2) Sea anemone (ii) Radial symmetry (3) Barnacle (iii) Pacudocoelomate body cavity (4) Sea urchin (iv) Water vascular system (A) (1) (iii), (2) (ii), (3) (i), (4) (iv) (B) (1) (ii), (2) (iii), (3) (i), (4) (iv) (C) (1) (iii), (2) (iv), (3) (i), (4) (ii) (D) (1) (iv), (2) (iii), (3) (ii), (4) (i) Exp. Rotifer-are pseudocoelomate animals. Sea animonebelong to cnideria phylum they have radial symmetry. Sea urchin-belong to Echinodemata phylum. they possess water vascular system Which of the following characteristics make Amborella the most basal living angiosperm? (A) Carpels fused by tissue connection and absence of vessel elements (B) Absence of carpels and presence of vessel element (C) Carpels free and presence of vessel elements (D) Presence of carpels and absence of vessel elements P5/19 Amar Ujala Education Exp. The carpels are sealed by a secretion of stick fluid, rather than developmentally fused as in most angiosperms and vessels are absent Which of the following statements is not correct? (A) Stomata are present in mosses and hornworts but absent in liverworts (B) Only the lycophytes have microphylls and almost all other vascular plants have megaphylls (C) Monocot pollen grains have three openings whereas eudioot pollen grains have one opening (D) Monocots have fibrous root system whereas eudicots have taproot Exp. Monocot pollen have one opening whereas dicot have three Following is a table showing selected characte-ristics of important fungal groups : Fungal group Characteristic a No regularly occuring septa in thallus b Perforated septa c Forms arbuscular mycorhizae on plant roots d Have zoospores with flagella In the above table, the fungal groups A, B, C and D, are respectively : (A) Chytridiomycetes, Ascomycetes, Glomero-mycetes, Zygomycetes (B) Zygomycetes, Ascomycetes, Glomero-mycetes, Chytridiomycetes, (C) Ascomycetes, Zygomycetes, Glomero-mycetes, Chytridiomycetes (D) Chytridiomycetes, Zygomycetes, Ascomy-cetes, Glomeromycetes 124. Which of the following is a correct statement? (A) Euglenids have a spiral or crystalline rod inside flagella (B) Pheophytes have a spiral or crystalline rod inside flagella (C) Euglenids have a hairy and smooth flagella (D) Euglenids and pheophytes both have a spiral of crystalline rod inside flagella Exp. Euglenids are the one having spiral or crystalline rod inside flagellas. And pheophytes are otherwise called as brown algae and they are having hairy and smooth flagella kindly find it from the reference where it is tabulated.

161 Amar Ujala Education 125. In life history evolution there is generally a trade-off between the size and number of offspring produced. Some conditions are listed below : (1) Scarcity of food during the early stages of life (2) Provision of parental care (3) High mortality during early stages of life (4) Predator s prefernece for large sized prey What are the above two conditions that would favour production of a small number of large-sized offspring? (A) (2) and (3) (B) (2) and (4) (C) (1) and (2) (D) (1) and (3) Exp. Small sized offsprings are produced in large No. and they dont get any parental care usually. As the population of offspring is large that will lead to the scarcity of the food The possible relationship between level of disturbance and species diversity in a biological community are that species diversity : (i) is unaffected by disturbance (ii) is highest at intermediate levels of disturbance (iii) decreases exponentially with increasing levels of disturbance (iv) starts decreasing only at higher levels of disturbance (1) High Species diversity Low Low (2) High Species diversity Low Low (3) High Species diversity Low Low Disturbance Disturbance Disturbance High High High (4) High Species diversity Low Low Disturbance High A/UG-15 Match each graph with its corresponding state-ments above : (A) (1) (iv), (2) (iii), (3) (ii), (4) (iv) (B) (1) (iii), (2) (iv), (3) (ii), (4) (i) (C) (1) (i), (2) (ii), (3) (iii), (4) (iv) (D) (1) (iii), (2) (i), (3) (ii), (4) (iv) Exp. As it is clear from the figure that in case second is showing the independent pattern and option 4th is the only optoin The diagram represents competition between species 1 and species 2 according to Lotka-Volterra model of competition : Sp 2 K 2 K 1/ K 2/ Sp 1 Given the conditions in the diagram, the predicted outcome of competition is : (A) Unstable coexistence between species 1 and 2 because K 1 >K 2 / and K 2 >K 1 / (B) Unstable coexistence between species 1 and 2 because K 1 <K 2 / and K 2 <K 1 / (C) Stable coexistence between species 1 and 2 because K 1 >K 2 / and K 2 >K 1 / (D) Stable coexistence between species 1 and 2 because K 1 <K 2 / and K 2 <K 1 / Exp. The isoclines are intersecting at the stable equilibrium point Following four types of species were observed in a community : (1) Species A has a large effect on community because of its abundance. (2) Species B has a large role in community out of proportion to its abundance. K 1 P5/20

162 A/UG-15 (3) Status of species C provides information on the overall health of an ecosystem. (4) Significant conservation resources are allocated to species D which is single, large and instantly recognizable According to above description, species A, B C and D are called respectively : (A) Dominant, Keystone, Indicator and Flagship (B) Keystone, Flagship, Dominant and Indicator (C) Keystone, Dominant, Indicator and Flagship (D) Flagship, Dominant, Keystone and Indicator Exp. Keystone species effects the community largely irrespective to its abundance as per this B is keystone species. Thus option 1 is correct Complete the following hypothetical life table of a species to calculate the net reproductive rate R 0 : Age Number Number Age Age l x m x Class (X) Alive of dying Specific Specific (n x ) (d x ) survivor Fertility -ship The calculated R 0 will be : (A) 0.75 (B) 1.00 (C) 0.65 (D) 0.15 Exp. Explanation : dx is age specific death is 200 and 300 (missing one), missing nx are 500 and 200, Now lx of each class are 1, 0.8, 0.5, 0.3, 0.2. Then, lxbx foreachclassare(1 0=0)+(0.8 0=0)+ ( = 0.25) + (0.3 1 =.3) + (0.2 1 = 2) =.75 (R0) Which of the following is the correct decreasing order for the rate of decomposition of litter constituents? (A) Hemicellulose, cellulose, lignin, phenol (B) Cellulocse, hemicellulose, phenol, lignin (C) Hemicellulose, cellulose, phenol, lignin (D) Lignin, phenol, hemicellulose, cellulose 131. Which of the following is not a benefit for the female adopting polyandry? (A) Greater probability of getting all her eggs fertilized (B) Ability to receive more resources from the males (C) Ability to produce more offspring than normal (D) Improved chances of genetic compatbility with her own DNA Amar Ujala Education 132. Assume that individual A wants to do an altruistic act to individual B and that the benefit and cost of doing this act are, in fitness units, 40 and 12, respectively. According to Hamilton s Rule, A should perform the altruistic act only if B is his : (A) nephew (B) niece (C) grandson or granddaughter (D) daughter or son Exp. Polyandry increases the probability of getting all the eggs fertilised, it also increases genetic compatibility and female receives more rewards in this. But it does not result in production of more offspring than normal rather increases the survival rate of those offsprings due to the parental care received Individual A performs to another individual a behavioural act which has a fitness consequences. Match the behavioural acts (a) to (e) with the correct fitness consequence ((i) to (iv)). Behavioral act Cooperation (a) Adaptive altruism (b) Spite (c) Deceit and manipulation (d) Reciprocity (e) Fitness consequence to A Gains direct fitness but after delay (i) Loses inclusive fitness (ii) Gain indirect fitness (iii) Gains direct fitness but immediately (iv) (A) (a) (iv), (b) (iii), (c) (ii), (d) (i), (e) (i) (B) (a) (i), (b) (ii), (c) (ii), (d) (iii), (e) (iv) (C) (a) (i), (b) (iii), (c) (ii), (d) (ii), (e) (iv) (D) (a) (ii), (b) (ii), (c) (iii), (d) (i), (e) (iv) Exp. In adaptive Alturism there is the indirect gain of fitness (inclusive fitness) and in Spite, deciet and manipulation No gain of inclusive fitness is there, these are the selfish behaviours In a random sample of 400 individuals from a population with alleles of a trait in Hardy-Weinberg equilibrium 36 individuals are homozygous for allele a. How many individuals in the sample are expected to carry at lest one allele A? (A) 36 (B) 168 (C) 364 (D) 196 Exp. # of individuals with At lest 1 A = # of AA + # of Aa; = 364 will be the number of people with at least one A. P5/21

163 Amar Ujala Education 135. Two kinds of natural selection (A and B) acting on a trait are shown in the figure below. In each, the top graph shows the trait frequency before and the bottom graph frequency after the action of natural selection : F R E Q U E N C Y A TRAIT F R E Q U E N C Y B TRAIT The kind of natural selection in A and B are : (A) A Directional, B Disruptive (B) A Netural, B Disruptive (C) A Stabilizing, B Disruptive (D) A Disruptive, B Stabilizing Exp. In stabilising selection intermediate forms are selected by nature while in disrupted both the extreme forms of the population are selected Which of the following statements is not correct regarding effect of genetic drift? (A) It alters allele freuency substantially only in small population (B) It can cause allele frequencies to change at random (C) It can lead to a loss of genetic variation within population (D) It can cause harmful alleles to become eliminated Exp. All other statements are correct moreover genetic drift is a random process. It can not change the genetic structure in any specific direction During the production of alcohol by fermentation using budding yeast, oxygen supply is kept limited. Why? (A) Budding yeasts are obligate anaerobes and cannot tolerate oxygen (B) Budding yeats lose mitochondria in the absence of oxygen (C) Budding yeast are facultative anaerobes (D) Alcohol is oxidized further in the presence of oxygen Exp. In presence of oxygen, pyruvate is completely oxidized into carbon dioxide and water. However htere is no oxidation of Alcohol. Majority of yeast species, prefer fermention or respiration. i.e. they are facultative anaerobes and they will produce alcohol in absence of oxgyen. A/UG A student was asked to design a knockout cassette for specifically deleting the p53 gene from the prostate gland of mice. Which one of the following pairs of cassettes will ensure deletion of the gene? TSP Cre p53 (A) Cre p53 p53 (B) TSP Cre p53 (C) TSP Cre p53 (D) TSP = Tissue specific parameter Cre = Cre recombiness = kap sites Exp. Cre should be under TSP and the gene p53 should be surrounded by direct repeats of lox as direct repeats causes deletion Following are certain statement regarding somatic hybridization, a technique used for plant improvement : (1) Protoplasts of only sexually compatible plant species can be fused (2) Hybrids are produced with variable and asymmetric amounts of genetic material of parental species (3) Protoplast fusion permits transfer of gene block or chromosomes (4) Genes to be transferred need to be identified and isolated Which one of the following combinations of the above statements is correct? (A) (1) and (3) (B) (2) and (3) (C) (1) and (4) (D) (2) and (4) Exp. Protoplast of sexually incompatible can also be fused (POMATO) and genes are not be isolated and identified for transfer Small molecular weight ocmpounds affect the activity of luciferase differently. the oil/water solubility of various compounds is one property important for its effect on luciferase. the straight line in the graph was obtained by plotting the activity data with 50 different compounds. the luciferase activity, of two new derivatives of Benzene (A and B) are shown below : Luciferase activity B A Oil/water partitioning coefficient P5/22

164 A/UG-15 Which of the following statement is correct? (A) A is phosphate and B is amine (B) A is methyl and B is amine (C) A methyl and B is propyl (D) A is amine and B is phosphate Exp. A partition coefficient is the ratio of concentrations of a compoundinthetwophasesofamixtureoftwo immiscible liquid at equilibrium. Higher values indicate that compound is more hydrophobic. According to graph derivative A is hydrophobic and derivative B is hydrophilic. So a correct combination of the functional groups is option 2 (methyl is hydrophobic and amine is hydrophilic) A researcher was repeating a FACS experiment but somehow got confused with the labeling of the tubes. There are four tubes, one control, C (with no fluorescent label), one standard 1, S1 (with FITC label), one standard 2, S2 (with PE label) and the last one test, T (which should be FITC positive). Given below is the result of the FACS experiment : (1) PE (2) PE (3) PE (4) PE FITC FITC FITC FITC Amar Ujala Education What should be the correct labeling? (A) a,s2;b,s1;c,t;d,c (B) a, S1; b, T; c. C; d, S2 (C) a,s2;b,s1/t;c,c;d,s1/t (D) a, S1/T, b, S2; c, S1/T; d, C Exp. Control-zero fluorescence figure d that has minimum fluorescence, S1-has only FITC label so could either be figure a or c, S2 has onlype label sois figure b, test also is FITC positive so can either be figure a or c. (corresponding labels will show more fluorescnece with respect to their region towards X-axis for FITC and towards Y-axis for PE) Nichrome coated stainless steel electrodes were implanted in a rat brain for chronically recording the electrical activity of deep brain structures. During a study of 3 months the intensity of electrical signals gradually decreased. The following statements may explain the cause of this observation : (1) The depositions of metallic iron from the electrode tips caused degeneration of some neurons (2) The gradual accumulation of microglia at the electrode tips increased the resistnace of electrodes (3) The neurons at the electrode tips were hyperpolarized gradually (4) The threshold for firing action potential in the neurons at the electorde tips was increased due to prolonged presence of electordes Which one of the following is correct? (A) (1) only (B) (1) and (2) (C) (3) only (D) (3) and (4) Exp. Under evaluation 143. In an attempt to detect protein expression profile in a cell, Western blot technique is employed. Expression of two new proteins is to be followed by probing with respective high affinity antibodies (raised in rabbit). Unfortunately, the two proteins were found to co-migrate in SDS-PAGE profile. Under this situations, using one dimensional SDS-PAGE and by Western blot, which one of the following is the best way to demonstrate the presence of both the proteins? (A) Develop Western blots with their antibodies in the same gel (B) Prior to doing SDS-PAGE/Western blot, one protein could be removed by immuno precipitating in the cell extracts (C) Silencing the expression of one protein at a time by sirna and performing Western blotting P5/23

165 Amar Ujala Education (D) Subjecting the technique of stripping/reprobing of the gel after transferring to nitrocellulose membrane while doing Western blotting Exp. Researchers have used been using stripping or reprobing techniques to screen a single blot with multiple antibodies. Protocols have been specifically formulated to dissociate and remove all antibodies from the membrane immobilized proteins without destroying the antigenic binding sites or removing the protein A chromatin immuno precipitation (ChIP) assay was performed to determine specific transcription factor binding sites on the promoter of a gene. Pull down was done using either IgG or antibodies against c-myc. A DNA containing c-myc binding regions was used as a control for PCR amplification (input). Which one of the following PCR representations of DNA is correct? (A) (B) Input IgG Anti-C-myc (C) (D) A/UG the number of seeds in the fruit of a plant species, H0 : μ = 30. A random sample of 9 fruits gives the mean number ofseedsas24withastandarddeviationof6.12(a)what are the confidence limits for the sample mean? (b) Would you reject or accept the null hypothesis at 95% confidence level? (A) (a) 18 and 30, (b) reject the hypothesis (B) (a) 20 and 28, (b) reject the hypothesis (C) (a) 20 and 28, (b) reject the hypothesis (D) (a) 18 and 30, (b) reject the hypothesis Exp. The sigma M = sigma/(n)^-1/2. Sigma = 6.12 and N = 9. Sigma M would come out to be Since the Confidence interval = 95 percent, hence range of C.I. can be calculated as 1.96*2.04 = 4. Therefore actual CL = 24-4 to = 20 to 28. And since mu = 30 does not fall in the CI and hence will reject the null hypothesis. P5/24

166 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, June-2014 PART-A 1. B A D One a semi-circle of diameter 10 m drawn on a horizontal ground are standing 4 boys A, B, C and D with distance AB = BC = CD. The length of line-segment joining A and B is : (A) 5 m (B) 6 m (C) 7 m (D) 5 m 3 Exp. B 60 C C A D Length of the line segment AB = 5 m. 2. The following table shows the price of diamond crystals of a particular quality : Wt. of a diamond Price per carat Crystal (in (in lakh carat) Rs.) What will be the price (in lakh Rs.) of a 2.5 carat diamond crystal? (A) 10 (B) 20 (C) 25 (D) 50 Exp. Cost of 2.5 carat diamond crystal = = 25 lakhs. 3. A man on the equator moves along 0 longitude up to 45 N. He then turns east and moves up to 90 E, and returns to the equator along 90 E. The distance covered in multiples of Earth s radius R is : 3 (A) R 4 (B) R (C) R Exp. Distance covered = (D) R R cos 45 2 R sin = R = R What is the next number in the following sequence? 2, 3, 5, 6, 3, 4, 7, 12, 4, 5, 9,... (A) 10 (B) 20 (C) 13 (D) 6 Exp. 2,3 2+3,2 3 3, 4 3+4,3 4 4, 5 4+5, 4 5 =20 P6/1

167 Amar Ujala Education 5. To go from the engine to the last coach of his train of length 200 m, a man jumped from his train to another train moving on a parallel track in the opposite direction,waited till the last coach of his original train appeared and then jumped back. In how much time did he reach the last coach if the speed of each train was 60 km/hr? (A) 5 s (B) 6 s Exp. (C) 10 s (D) 12 s 200 m Total time taken = m/sec 5 (60 60) 18 = 3600 sec = 6 sec How many digits are there in : ? (A) 14 (B) 15 (C) 16 (D) 17 Exp = = which has 17 digits. 7. The following sum is : =? (A) 10 (B) 10 (C) 11 (D) 9 Exp. 1 + (1 2) + (3 4) + (5 6) (19 20) = 1 + ( 1) + ( 1) ( 1) = You are given 100 verbs using which you have to form sentences containing at least one verb, without repeating the verbs, under the condition that the number of verbs (from this set of 100) in any two sentences should not be equal. The maximum number of sentences you can form is : (A) 10 (B) 13 (C) 14 (D) 100 Exp. Let, there be maximum of n sentenes, then n < 100 nn ( 1) n(n +1)<200 n < 13 n = 13 (maximum). 9. After giving 20% discount on the marked price to a customer, the seller s profit was 20%. Which of the following is true? A/UG-15 Marked price + Cost price (A) Sale price = 2 Marked price + Cost price (B) Sale price < 2 2 (C) (Marked price + Cost price) > Sale price > 3 Marked price + Cost price 2 (D) Sale price = 2 (Marked price + Cost 3 price) Exp. Let C = cost price; M = marked price and s = sale price =0.8MandS=1.2C S S M+C = S = (0.8) (1.2) = S 0.96 M+C S November 9, 1994 was a Wednesday. Then which of the following is true? (A) November 9, 1965 is a Wednesday and November 9, 1970 is a Wednesday (B) November 9, 1965 is not a Wednesday and November 9, 1970 is a Wednesday (C) November 9, 1965 is a Wednesday and November 9, 1970 is not a Wednesday (D) November 9, 1965 is not a Wednesday and November 9, 1970 is not a Wednesday Exp. November 9, 1994 is = 36 days. i.e. 1 day ahead from November 9, So, November 9, 1965is Tuesday and November, 9, 1970 is 542 = 7 days means no days ahead from November 9, So it also falls on Tuesday. 11. A 4 4 magic square is given below : How many 2 2 squares are there in it whose elements addupto34? (A) 6 (B) 9 (C) 4 (D) 5 Exp. There are 5 such squares whose 4 elements sum to 34. P6/2

168 A/UG Three years ago, the difference in the ages of two brothers was 2 years. The sum of their present ages will double in 10 years. What is the present age of the elder brother? (A) 6 (B) 11 (C) 7 (D) 9 Exp. If elder brother is x-years old, then younger brother will be x 2yearsold. x + x 2+20=2(x + x 2) 2x +18=4x 4 2x =22 x = (20) (2) = 3 (20) % Amar Ujala Education = = 99.9% 16. What is the length of the longest rod that can be put in a hemispherical bowl of radius 10 cm such that no end of the rod is outside the bowl? (Assume that the rod has negligible thickness) : (A) 10 2cm (B) 10 3cm (C) 10 4cm (D) 10 5cm Exp. Two identical cylinders of the same height as a bigger hollow cylinder were put vertically into the latter to fit exactly into it. The volume of the bigger cylinder is V. The volume of each of the smaller cylinders is : (A) V/8 (B) V/4 (C) V/2 (D) V Exp. If V = pr 2 H is volume of larger cylinder, then the volume of smaller cylinder : 2 R = 2 RH H = = V You get 20% returns on your investment annually but also pay a 20% tax on the gain. At the end of 5 years, the net gain made by you (as percentage of the capital) is approximately : (A) 0 (B) 16 (C) 80 (D) 100 Exp. Net profit per annum = 20% 80% = 16% So, net profit percentage in 5 years =16 5= A cubic cavity of edge 20 μm is filled with a fluid with a cubic solid of edge 2 μm. What percentage of the cavity volume is occupied by the fluid? (A) 10.0 (B) 20.0 (C) 90.0 (D) 99.9 Exp. Total volume cavity = (20) 3 μm Volume covered by the fluid = (2) 3 μm % of volume unoccupied P6/3 The length of largest rod is equal to diameter of the hemisphere =10 2= Marks obtained by two students S1 and S2 in a four semester course are plotted in the following graph : S2 S1 I II III IV Semester Which of the following statements is true? (A) S2 got higher marks than S1 in all four semesters (B) Over four semesters, S1 improved by a higher percentage compared to S2 (C) Total makrs of S1 and S2 are equal (D) S1 and S2 did not get the same marks in any semester Exp. According to graph option (2) is correct. 18. Find the missing numebr in the sequence : 61, 52, 63, 94, 18, 001, 121. (A) 46 (B) 70 (C) 66 (D) 44 S2 S1

169 Amar Ujala Education Exp. 4 2 = 16 whose, reverse is = 25 whose, reverse is = 36 whose, reverse is = 49 whose, reverse is = 64 whose, reverse is = 81 whose, reverse is = 100 whose, reverse is = 121 whose, reverse is If a 4 digit year (e.g. 1927) is chosen randomly, what is the probability that it is NOT a leap year? (A) 3 4 (B) 1 4 A/UG Coordinates of a point in x, y, z space is (1, 2, 3). What would be the coordinates of its relfection in a mirror along the x, z plane? (A) ( 1, 2, 3) (B) (1, 2, 3) (C) (1, 2, 3) (D) ( 1, 2, 3) Exp. Reflection of (1, 2, 3) in X-2 plane is : (1, 2, 3) Y (1, 2, 3) (C) < 1 (D) > Exp. AsEveryfourthyearneednotbealeapyear.So, probability of not a leap year is > 3. 4 PART-B Z X (1, 2, 3) 21. Which one of the following is unfavorable for protein folding? (A) Hydrophobic interaction (B) van der Waals interaction (C) Conformational entropy (D) Hydrogen bonding Exp. Conformational entropy is unfavourable for protein folding. Conformational entropy is the entropy associated with the number of molecule. 22. The maximum number of hydrogen bonds that can form between H 2 N NH 2 (hydrazine) and water is : (A) 2 (B) 1 (C) 3 (D) the peptide unit (C C'O NH C ) is planer due to : (A) restriction around C C' bond (B) restriction around C' C bond (C) restriction around N C bond (D) hydrogen bonding between carbonyl oxygen and imino hyrogen of the peptide backbone Exp. The linking of two amino acids is accompanied by the loss of a molecule of water. A series of amino acids joined by peptide bonds form a polypeptide chain. 24. Which one of the following is the most appropriate statement regarding folded proteins? (A) Charged amino acid side chains are always buried (B) Charged amino acid side chains are seldom buried P6/4 (C) Non-polar amino acid side chains are seldom buried (D) Tyrosine residues are always buried 25. Molecules primarily responsible for the formation of lipid raft are : (A) phosphatidyl serine and phosphatidyl choline (B) phosphatidyl inostol and cholesterol (C) glycosylphosphatidyi inositol and cholesterol (D) sphingolipids and cholesterol 26. Membrane-bound and free ribosomes, are structurally identical, but differ only at a given time in terms of association with (A) acetylated proteins (B) glycosylated proteins (C) phospholipids (D) nascent proteins Exp. A protein as it is being formed by a ribosome before it folds into its active shape is called nascent protein. 27. Histone acetylase and chromatin remodeling complexes are recruited to specific regions of chromatin by (A) gene activator proteins (B) specific promoter sequence (C) phosphorylation of histone acetylase (D) dephosphorylation of chromatin remodeling complexes 28. The dye used in Gram staining is : (A) Rhodamine (B) Methylene blue (C) Giemsa (D) Crystal violet

170 A/UG Each origin of replication is activated only once. This is achieved because : (A) pre-replicativecomplexcanonlyforming1and replication can only be initiated when prereplication complexes is disassembled at the beginning of S-phase (B) replication can only be initiated when prereplication complex is intact (C) replication can only be initiated when unphosphorylated Rb is present (D) pre-replicative complex can only from in S-phase 30. The homologous genetic recombination is a DNA repair process referred to as recombination repair. Which one of the following statements is INCORRECT for recombination repair? (A) DNA polymerase III stalls at the site of the damage (B) DNA polymerase III leaves a gap in the daughter strand (C) The gap is filled by recombination between complementary parent strand homologous to daughter strand and the gapped daughter strand (D) Homologous recombination process is catalyzed by topoisomerase II 31. During protein synthesis in prokaryotes, the peptidyl transferase activity required for peptide bond formation is due to : (A) ribosomal protein L26 (B) 16S ribosomal RNA (C) 23S ribosomal RNA (D) aminoacyl trna 32. After activation of a promoter by the DNA binding activity of a transcription factor, a co-activator is recruited at the region targeted for transcription which in turn creates a binding site for a chromatin remodeling complex. Which one of the following activities of the co-activator is responsible for the recruitment of chromatin remodeling complex? (A) Histone deacetylase activity (B) Histone methyl transferase activity (C) Histone acetyl transferase activity (D) DNA methyl transferase activity 33. Prior to transcription, chromatin changes from an inactive state to an active state by various factors in a stepwise manner. Which one of the following is involved in the initial step during activation of a chromatin loop? Amar Ujala Education (A) HMG 14 (B) Single stranded DNA-binding protein (C) DNA polymerase III (D) Topoisomerase I Exp. A pathogen like. 34. A pathogen like Mycobacterium, which colonizes inside the cells of the host, is likely to be least affected by which one of the following host immune defense mechanisms? (A) Cell-mediated immune response (B) CD4 + T lymphocytes (C) Cytokines (D) Humoral immune response Exp. The humoral immune response denotes immunologic responses that are mediated by antibodies. 35. Ethylene signaling pathway is important for fruit ripening. Which one of the following responses is routinely used to identify ethylene signaling pathway components? (A) Cotyledon expansion response (B) Lateral root formation response (C) Triple response (D) Flowering time response 36. Which one of the following is a type of inter- cellular junction in animal cells? (A) Middle lamella (B) Plasmodesmata (C) Desmosomes (D) Glycocalyx 37. Which one of the following is not related to immediate hypersensitivity reactions? (A) Mast cell degranulation results in histaminemediated allergic reactions (B) Reaginic antibodies trigger allergic reactions (C) Granulomatous reaction is a key to contain infection (D) Anaphylactic reaction is triggered primarily by IgE Exp. Granulomas are complex multicellular tissue reactions formed after a cascade of interactive systems that encompass adhesion molecule activity. 38. Which one of the following developed pro-cesses in animal is more dependent on cellular movements? (A) pattern formation (B) morphogenesis (C) cell differentiation (D) growth P6/5

171 Amar Ujala Education 39. Engrailed expression in Drosophila melanogas-ter defines : (A) anterior margin of the segment (B) anterior compartment of each segment (C) posterior margin of each parasegment (D) posterior compartment of each segment 40. During which one of the following stages of Arabidopsis embryogenesis, cell elongation throughout the embryonic axis and further development of the cotyledons occur? (A) Globular stage (B) Torpedo stage (C) Heart stage (D) Mature stage 41. Which one of the following graphs represents the relative expression of proteins of Wnt-activated signaling cascades involved during development of an embryo? (A) (C) Dsh ROCK (B) (D) PLC Ca Virus induced gene silencing (VIGS) is a process that takes advantage of the RNAi-mediated antiviral defence mechanism. Which one of the following ultimately guides SiRNA to degrade the target transcript (mrna)? (A) dsrna (B) ssrna (C) RNA Induced Silencing (RIS) complex (D) dsrna binding protein 43. During the operation of C 2 oxidative photosynthetic cycle, which of the following metabolites is transported from chloroplast to peroxisome? (A) Glycerate (B) Serine (C) Glycine (D) Glycolate P6/6 A/UG Arabidopsis thaliana seeds were planted on Murashige Skoog (MS) plates with or without a hormone added to the medium. Seeds were found to germinate late in the hormone containing MS plates as compared to MS plates without hormone. Identify the hormone : (A) Jasmonic acid (B) Cytokinin (C) Auxin (D) Abscisic acid 45. Which one of the following molecular marker types uses combination of both restriction enzyme and PCR technique? (A) SSR (B) AFLP (C) SNP (D) RAPD Exp. The acronym of AFLP is amplified fragment length polymorphism 46. One hemoglobin molecule containing four heme groups can bind with four O 2. The reactions of Hb and O 2 are shown below. Which one of the following reactions is fastest? (A) Hb 4 +O 2 Hb 4 O 2 (B) Hb 4 O 2 +O 2 Hb 4 O 4 (C) Hb 4 O 4 +O 2 Hb 4 O 6 (D) Hb 4 O 6 +O 2 Hb 4 O Which one of the following is required to anchor the synaptic vesicles to the cytoskeletal proteins in the presynaptic nerve terminals? (A) Syntaxin (B) Synaptobrevin (C) Synaptotagmin (D) Synapsin 48. Vasopressin acts on blood vessels leading to their constriction. Which one of the following signaling cascades will apply to the effect of vasopressin? (A) V 1 R Phosphatidylinositol hydrolysis Ca 2+ increase (B) V 1 R Adenylate cyclase camp (C) V 2 R Phosphatidylinositol hydrolysis Ca 2+ increase (D) V 2 R Adenylate cyclase camp 49. In an experiment involving mapping of 3 genes (a, b and c) indrosophila, a three point test cross is carried out. The parental cross was AAbbCC aabbcc. The genotypes of the double crossovers are : Aabbcc and aabbcc. Based on this, determine the order of the genes : (A) acb (B) cab (C) abc (D) bac

172 A/UG In the following example, 3 independently assorting genes are known to govern coat color in mice. The genotype of few of the coat colors is given below : Agouti : A B C Black : aa B C Albino : c c What will be the expected frequency of albinos in the F 2 progeny from crosses of pure black with albino of the genotype AAbbcc? (A) 1/4 (B) 1/16 (C) 1/64 (D) 9/ The distance between bacterial genes as determined from interrupted mating experiments are measured in units of : (A) cm (B) minutes (C) bp (D) micrometers 52. A black Labrador homozygous for the dominant alleles (BBEE) is crossed with a yellow Labrador homozygous for the recessive alleles (bbee). On intercrossing the F 1, the F 2 progeny was obtained in the following ratio : 9black:3brown:4yellow This is an example of : (A) recessive epistasis where allele e is epistatic to B and b (B) dominant epistasis where allele E is epistatic to B and b (C) recessive epistasis where allele e is epistatic to E (D) complementary epistasis where allele b is epistatic Exp. Because both genes are required for the correct phenotype, this epistatic interaction is called complementary gene. 53. The organs radula and clitellum are found in : (A) Coelenterata and Echinodermata, respecti-vely (B) Echinodermata and Coelenterata, respecti-vely (C) Annelida and Mollusca, respectively (D) Mollusca and Annelida, respectively 54. Which of the following is a correct hierarchial sequence or classifying a living organism? (A) Domain-Kingdom-Phylum-Class-Order-Family- Genus-Species (B) Kingdom-Domain-Phylum-Class-Order-Family- Genus-Species (C) Domain-Kingdom-Phylum-Order-Class-Family- Genus-Species (D) Kingdom-Domain-Phylum-Order-Class-Family- Genus-Species P6/7 Amar Ujala Education 55. Branchiostoma is a : (A) deuterostome and schizocoelomate (B) protostome and schizocoelomate (C) deuterostome and enterocoelomate (D) protostome and enterocoelomate 56. The population size of a bird increased from 600 to 645 in one year. If the per capita birth rate of this population in 0.125, what is its per capita death rate? (A) 0.25 (B) 0.15 (C) 0.05 (D) The present global warming trend is expected to result in an increased incidence of malaria in temperate countries. the supposed underlying mechanism is that : (A) higher temperatures make temperate country people more vulnerable to diseases (B) malarial parasite grows better at higher temperatures (C) the vector mosquito species requires warmer temperatures for reproduction (D) anti-malaria drugs are less effective in temperate countries 58. The most important reproductive strategies of big trees in a forest are : (A) earlier age at first reproduction and produc-tion of a large number of small seeds (B) earlier age at first reproduction and produc-tion of a small number of large seeds (C) later age at first reproduction and produc-tion of a large number of small seeds (D) later age at first reproduction and produc-tion of a small number of large seeds 59. In rats, after the delivery of the offspring, mother shows the following behaviors is NOT mater- nal? (A) Licking the pups (B) Huddling above the pups to access the ventrum (C) Lordosis of mother rat (D) Bringing back to the nest pups that wander away from it Exp. Lordosis is a reflexive behaviour that is triggered by a touch on the lower back, flanks or geninal organs. 60. In very smallpopulations, genetic variation is often lost through genetic drift. If the population size of a mammal on an isolated island is 50, what percentage of its genetic variation is lost every generation? (A) 0.01 (B) 0.5 (C) 0.1 (D) 0.05

173 Amar Ujala Education 61. The long feather train of a peacock is quoted as an example supporting : (A) Hamilton s rule (B) Zahavi s handicap principle (C) The Red Queen hypothesis (D) Haldane s rule 62. An ant moving straight, upon encountering an obstacle, may turn either right or left and continue moving. The test the hypothesis that the direction chosen by the ant is random, the most appropriate statistical test is : (A) Student s t-test (B) 2 -test of independence (C) 2 -test of goodness of fit (D) correlation test Exp. A correlation coefficient is a coefficient that illustrates a quantitative measure of some type of correlation and dependence. 63. Among the following antigens specific to a pathogen, which one is most likely to be ineligible as a vaccine with long lasting host protective effect? (A) A cell surface protein (B) An enzyme involved in pathogen metabo-lism (C) A signalng intermediate, which is a kinase (D) A long chain fatty acid 64. Which one of the following techniques is generally used to produce transgenic animals? (A) processed mrna containing only exons are introduced into blastocyst stage embryo (B) Entire foreign nucleus is introduced in enulceated unfertilized egg (C) Desired DNA is microinjected into fertilized eggs followedbyimplantationofembryoinafoster mother (D) cdna of desired gene is introduced into animal embryos and implanted in a foster mother 65. Electron microscopes have much higher resolution than any type of light microscope because : (A) of their higher magnification (B) the lenses used are of much higher quality (C) of very short wave length of electrons (D) the images are viewed on screen rather than directly using an eye-piece or occular lens A/UG Which one of the following statements is NOT correct for propagation and maintenance of mammalian cells is vitro? (A) Transformed cell lines do need external supply of serum to grow (B) The cells that are obtained directly from the organism is a primary culture (C) Trypsin is added to cell culture media to maintain cell s health (D) HEPES buffer is generally used to maintain ph of the culture media 67. A bioinformatics tool used to find the sequence similarity in the subunits of hemoglobin is : (A) FASTA (B) BLAST (C) HUMMER (D) PSI : PLOT 68. The electro spray ionization spectrum of a highly purified protein shows multiple closely spaced peaks. This most likely arises due to : (A) degradation of the protein during recording of the spectrum (B) the presence of multiple conformations (C) multiple charged species of the protein (D) extensive aggregation of the protein 69. The average energy absorbed by 10 gm of tissue from 32 P radiation is 14.9 Jkg 1. The average dose in rads is : (A) 1490 (B) 1.49 (C) (D) In isoelectric focussing experiments, proteins are separated on the basis of their : (A) relative content of only positively charged residues (B) relative content of only charged residues (C) relative content of positively and negatively charged residues (D) mass to charge ratios P6/8

174 A/UG-15 Amar Ujala Education mm acetate buffer (ph 4.00) is diluted one million times with distilled water (ph 7.00). ph of this diluted buffer is : (A) 4.00 (B) 7.04 (C) 8.00 (D) 6.96 [Help : log x =x;log = 0.04, log = 0.004] Exp. Distilled water ph ph of 10 mm acetate buffer = = Consider the structureless oligopeptide, R-G-P-S-T-K- M-P-E-Y-G-S-T-D-Q-S-N-W-H-F-R. The number of bonds that will be cleaved by trypsin and chymotrypsin treatments separately, are : (A) 1,2 (B) 2,2 (C) 2,3 (D) 1,3 73. The internal energy of a gas increases by 1J when it is compressed by a force of 1 Newton through 2 metres. The heat change of the system is : (A) 1 J (B) 1 J (C) 2 J (D) 2 J 74. G for the base pairing of oligonucleotides (n = 5) at 300 K is 18 kj mol 1. What would be the approximate value of the equilibrium constant K? (A) 100 (B) 10 (C) 1000 (D) the reaction of glutamate and ammonia to glutamine and water has a value of + 14 kj mol 1 for G. This is coupled with the ATP reaction (ATP + H 2 O ADP + phosphate). The G for this reactions is 30 kj mol 1. The G (kj mol 1 ). for the coupled reaction Glutamate +NH 3 +ATP Glutamine + ADP + Phosphate under equilibrium condition is : (A) 16 (B) 44 (C) 16 (D) A protein is composed of leucine, isoleucine, alanine, glycine, proline, one lysine, one arginine and two cysteines connected by a disulfide bond. Conformational analysis indicates that the protein has elements of helix and beta structure. The protein is most likely : (A) a non-specific protease (B) not an enzyme (C) a lipase (D) a flippase PART-C P6/9 77. You are studying a protein that inserts itself into a model membrane (liposomes) during a reconstitution process. The protein has an N-terminal, 18-amino acid hydrophilic segment that is located on the outside of the membrane, a 19-amino acid hydrophobic transmembrane segment flankedbynegativelyand positively charged amino acid and a C-terminal domain that resides inside the lumen (as depicted below in the form of a cartoon) : N + C Lumen Outer surface (+) Inner surface ( ) For proper reconstitution of the protein, which of the following strategies will be appropriate? (A) Increase the number of negatively charged amino acids in the N-terminal (B) Increase the number of positively charged amino acids in the C-terminal (C) Removal of positively charged amino acids from the C-terminal (D) Increase the length of hydrophobic segment I-labelled diaminofluorene (DAF) is a well known photoactivable hydrophobic probe of plasma membrane integral protein. To determine the approximate length and number of hydrophobic domain in any integral membrane protein, a controlled experiment (following standard protocol) is carried out. In order to ascertain the aforesaid aspects indicate the correct combination of experimental protocols from the following choices : (A) Intact membrane was allowed to interact with DAF and unicorporated DAF measured (B) Intact membrane was allowed to interact with DAF, lysed and total protein precipitated with TCA and amount of radioactivity incorporated in the total proteins in the TCA precipitated fraction measured (C) Intact membrane was allowed to interact with DAF, then membrane was solubilized with detergent, digested with protease (limited proteolysis) run on SDS-PAGE followed by autoradiography (D) Intact membrane was allowed to interact with DAF followed by complete proteolysis, SDS-PAGE and finally autoradiography Exp. Diaminofluorene is more sensitive than benzidine for detecting hemoglobin in erythroprotein responsive J2E cells.

175 Amar Ujala Education 79. The tetrapeptide "KDEL" is well known as a retrieval signal of several newly synthesized proteins. This process is mediated through specific receptor - KDEL interaction. Any single amino acid change in this tetrapeptide is not allowed in terms of its binding with its receptors and its subsequent retention in specific organelle whereas, secretory proteins are devoid of such tetrapeptide. From this observation indicate the localization of the receptor of this tetrapeptide: (A) Plasma membrane (B) Golgi (C) Endoplasmic reticulum (D) Mitochondria 80. Cyclins are the regulatory subunits and cyclindependent kinases (CDKs) are the catalytic subunits. Following diagram represents the involvement of cyclins and CDKs in various stages of cell cycle: Cyclin A CDK2 Cyclin B CDK1 S G 1 M Cyclin E CDK2 G 2 Cyclin D CDK4,5 If we knock down cyclin D in a cell by shrna, which one of the following graphs correctly represents the level of CDK2 activity? (A) (C) CDK2 Activity CDK2 Activity Time Time (B) (D) CDK2 Activity CDK2 Activity Time Time P6/10 A/UG Chromosome organization becomes clearer from a series of biochemical, electron microscopic and X-ray crystallographic studies. When interphase chromatin is isolated in low salt buffer and observed under EM, 11 nm bead on string organization is seen. Interphase chromatin directly observed under EM shows 30 nm fibre. When histones are depleted from metaphase cromosome and visualized under EM, it shows a huge number of very large loops associated with scaffold. Following interpretations can be made from these: (1) 11 nm fibre is formed when nucleosomes are brought closer by scaffold. (2) 30 nm interphase chromatin is formed by zig-zag organization of nucleosomes of 11 nm fibre. (3) 30 nm fibre makes a solenoid packing to from the metaphase chromosome. (4) 30 nm fibre gets organized into loops due to SARs getting associated with scaffold proteins and coming closer. The correct combination of interpretations is: (A) (1) and (4) (B) (1) and (3) (C) (1) and (2) (D) (2) and (4) 82. To test whether bacteria with enhanced toluene degradative abilities could be created for cold environment, a TOL (toluenedegrading) plasmid from a mesophilic bacterial strain was transferred by conjugation into a facultative psychrophile. The psychrophile was able to degrade salicylate (SAL) but not toluene. The recombinant strain carried the introduced TOL plasmid and its own SAL plasmid. The results are as follows : Optimum growth in presence of X Y Temp ( C) A B C A B C 37 C + 25 C + 10 C C Identify A, B, C, X and Y : (A) A. mesophile B. psychrophile C. transfor-mant X. Toluene Y. Salicylate (B) A. mesophile B. psychrophile C. transfor-mant X. Salicylate Y. Toluene (C) A. psychrophile B. transformant C. mesophile X. Salicylate Y. toluene (D) A. transformant B. psychrophile C. meso-phile X. Toluene Y. Salicylate

176 A/UG Aminoacyl-tRNA synthetases are very specific for aminoacylation of trnas with the correct cognate amino acids. However, there is a possibility of a mismatch between the trna and its cognate amino acid.this error is corrected by the inherent proof-reading activity of the aminoacyl-trna synthetase. In case of two very similar amino acids, namely valine and isoleucine, isoleucyl-trna synthetase employs the following possible approaches for an error-free aminoacylation : (1) It removes an incorrect amino acid by hydrolyzing the aminoacyl-amp linkage following the first reaction step (2) It is activated for proof-reading activity, leading to breakage of the bond between the wrong amino acid and trna (3) It has an intrinsic ability to recognize the structural difference between amino acids leading to abortive elimination of the non-cognate amino acid (4) It gets sequestered in the second step with the wrong amino acid, and that freezes the aminoacylation process. Which of the following combinations is correct? (A) (1) and (2) (B) (1) and (4) (C) (2) and (4) (D) (3) and (4) 84. Attenuation is a mechanism involved in the regulation of tryptophan operon in E. coli. When tryptophan levels are high in the cell, region 2 of the trpl is blocked from pairing with region 3. This allows the pairing of region 3 and 4 leading to the formation a rho-independent terminator. What would be the structure of the trpl region in E. coli cells where protein synthesis has been inhibited? (A) Region 2 pairs with region 3 allowing transcription of the structural genes (B) Region1and2willpairs,allowing3and4topair leading to attenuation (C) There is no pairing in the trpl region and transcription of structural gene occurs (D) Region 2 and 3 will pair leading to attenuation 85. MicroRNAs (mirnas) have recently been shown to play a significant role in the fine tuning of gene expression. Some mirnas induce gene silencing by binding to mrnas and inducing inhibition of translation. On the other hand, there are mirnas that bind to mrnas and activate their degradation. The following characteristics can be applicable to the mirnas that inhibit mrna translation : (1) mirna is partially complementary to a region of target mrna in the 3' UTR (2) mirna always base pairs with mrna around a AUrich sequence Amar Ujala Education (3) mirna base pairs with mrna through 6-7 nucleotides at its 5' end referred to as seed sequence as well as few additional bases elsewhere (4) mirna is always partially complementary to the 5' UTR of the target mrna Choose the correct option from the following : (A) (1) and (2) (B) (1) and (3) (C) (3) and (4) (D) (1) and (4) 86. A 6.4 Kb plasmid DNA has two restriction endonuclease sites, HindIII and EcoRI. Complete double digestion of the plasmid with both the enzymes yields two fragments of 3.1 and 3.3 Kb. Inorder to study DNA repair process, a G : T mismatch was introduced in one strand of HindIII site and the damaged plasmid was incubated in a reconstituted repair system containing all the factors and enzymes required for repair. If the efficiency of the repair system is 50%, which one of the following band patterns on agarose gel will be obtained after treating the repaired plasmid with both HindIII and EcoRI? (A) (B) (C) 6.4 Kb 3.3 Kb 6.4 Kb 3.3 Kb 3.1 Kb (D) 3.3 Kb 3.1 Kb 6.4 Kb 3.3 Kb 3.1 Kb Exp. A plasmid is a small DNA molecule with in a cell that is physically separated from a chromosomal DNA and can replicate indipently. 87. The following statements are made on DNA replication : (1) Replication fork is a branch point in a replication eye or bubble (2) A replication bubble contains two replication forks (3) DNA replication is continuous according to the interpretation made by Okazaki (4) Multiple priming events are required for both leading and lagging strands to initiate DNA synthesis Which one of the following is the correct combination? (A) (1) and (2) (B) (1) and (3) (C) (3) and (4) (D) (1) and (3) P6/11

177 Amar Ujala Education A/UG A Promoter Activity 100% 20% (A) -LDL bound pmale/10 cells Normal -LDL bound pmale/10 cells 1-LDL only 1-LDL only + excess unlabeled LDL D C B 90% 140% 105% In order to identify the regulatory regions of a novel promoter sequence shown above, four 150 bp deletion constructs were made in a luciferase reporter system as indicated above in boxes A to D. After transfection, the observed level of promoter activity (%) as analyzed by luciferase assay of all the constructs is indicated in the right of the figure. Identify the best correct combination of regions in the options given below that indicate the presence of a positive and a negative regulatory element, respectively : (A) (2) and (4) (B) (1) and (3) (C) (1) and (4) (D) (1) and (2) 89. Following sets of Plasmodium falciparum sporozoites : (i) normal sporozoites (ii) sporozoites with mutation in the C-termianl of circumsporozoite (CS) antigen (iii) sporozoites with mutation in the N-terminal region of circumsporozoite (CS) antigen are injected into 2 groups of mice, one normal (Group A) and one (Group B) where localized known down of heparan sulfate receptor for CS antigen in liver (cells) is achieved by injecting specific shrna expressing lentiviral particles in the liver prior to infection. 15 days post infection, parasitemia is measured by counting infected RBC through Giemsa staining. Which of the following groups will show maximum level of parasitemia? (A) Group B with set (i) (B) Group A with set (ii) (C) Group A with set (iii) (D) Group B with set (ii) Exp. Plasmodim falciparum is a protozoan parasite, one of the species of Plasmodium that cause malaria in humans. 90. Patients suffering from familial hypercholestero-lemia (FH) are mostly homozygous for the defective gene and have profoundly elevated levels of serum cholesterol. The reason may be that the gene for highly specific receptor for LDL is either defective or missing in these patients. In an experiment, cells were taken from both normal individual and homozygote (FH) subjects, incubated in buffer with 125 I-labeled LDL in presence or absence of excess unlabeled LDL for various time periods and then 125 I-labeled LDL bound to cells was measured. Which of the following is the bestfit graph for the above experiment? (B) (C) (D) -LDL bound pmale/10 cells -LDL bound pmale/10 cells -LDL bound pmale/10 cells Time Time Time Time -LDL bound pmale/10 cells -LDL bound pmale/10 cells -LDL bound pmale/10 cells Time Time Time Time 91. Collagen is the most prevalent extracellular matrix protein. Which of the following is NOT true for collagen? (A) Collagen is composed of triple helix consisting of two polypeptide chain and one polypeptide chain wound around one another in a rope-like structure (B) Glycine accounts for almost one third of the amino acids within collagen molecule (C) Ascorbate is essential for collagen formation required for hydroxylation of proline (D) Individual collagen polypeptide chains are synthesized on membrane-bound ribo-somes with N-terminal signal sequences for directing them to ER lumen 92. The following statements have been proposed for a cancer cell : (1) Binding of p53 with MDM2, a ubiquitin E3 ligase, is precondition for cancer progre-ssion (2) Phosphorylation of a tyrosine residue in the C- terminus of human c-src is essential for cell invasion and motility (3) Loss of function of both alleles of a tumor suppressor gene prevents metastasis (4) Dimerization of c-myc-max leads to enhanced cell proliferation Which of the combinations of the above statements is correct? (A) (1) and (2) (B) (3) and (4) (C) (1) and (4) (D) (2) and (3) P6/12

178 A/UG CD4 + T cells are co-cultured with macrophages in the presence of immobilized anti-cd3 antibody under four different conditions : (1) Interleukin (IL)-4 plus anti-ifn antibody (2) IL-12 and anti-il-4 antibody (3) Transforming growth factor (TGF)- (4) TGF- and IL-6 For three rounds to induce T-helper cell differentiation identifiable by the cytokines they express predominantly Which one of the following is the most likely combination of predominant cytokine expression in these cultures? (A) (A) IL-4, (B) IFN-, (C) IL-10, (D) IL-17 (B) (A) IFN-, (B) IL-4, (C) IL-17, (D) IL-10 (C) (A) IL-17, (B) IL-4, (C) IFN-,(D)IL-10 (D) (A) IL-17, (B) IL-10, (C) IL-4, (D) IFN- 94. Four groups of one-day old female BALB/c neonates had received the following treatments : (1) Epidermal cells from C57BL/6 male plus anti- 2 microglobulin antibody (2) Epidermal cells from C57BL/6 male plus antibodies to 2 microglobulin, CD40 ligand, CD80 and CD86 (3) Epidermal cells from C57BL/6 female plus anti- CD80 antibody (4) Epidermal cells from C57BL/6 female mice When these BALB/c neonates grew six weeks old, they received skin transplant from C57BL/6 male mice. Transplantation rejection time varied between these four groups. Starting from the fastest to the slowest rejection, which one of the following is the most likely order? (A) D > C>A>B (B) A>B>C>D (C) B>A>D>C (D) D>B>C>A 95. At the 2-celled stage of Caenorabditis elegans development the blastomeres were experimentally separated and allowed to proceed in development. One of the blastomere (P1) developed generating all types of cells it would normally make while the other blastomere (AB) made only a fraction of the cell types it would normally make. The following conclusion could be drawn : (1) The determination of both P1 and AB was autonomous (2) The determination of both P1 and AB was conditional (3) The determination of P1 was autonomous and AB was conditional (4) Both asymmetric division and cell-cell interactions specify cell fate in early development Which of the above combination is correct? (A) (1) and (3) (B) (2) and (4) (C) (1) and (4) (D) (3) and (4) Amar Ujala Education 96. The proximal distal growth and differentiation of the tetrapod limb bud are made possibel by a series of interactions between the apical ectodermal ridge (AER) and limb bud mesenchyme directly beneath it. Some of the interactions performed in chick demonstrated the following results : (1) When the AER was removed at any time of development further development of distal limb skeletal elements ceased (2) When leg mesenchyme was placed directly beneath the wing AER, distal hindlimb structures developed at the end of the wing (3) When limb mesenchyme was replaced by a nonlimb mesenchyme beneath the AER, the limb still developed (4) When an extra AER was grafted onto an existing limb bud, the development of the limb ceased Which of the above combination is correct? (A) (1) and (2) (B) (1) and (3) (C) (2) and (4) (D) (2) and (3) 97. Torpedo is a trans-membrane receptor on follicle cells that binds with Gurken protein located in the presumptive dorsal surface of the oocytes and inhibits a cascade leading to nuclear localization of the Dorsal protein. In an experiment, Drosophila germ line chimeras were made by interchanging pole cells (germ line precursors) between wild type embryos and embryos from mother homozygous for a mutation of torpedo gene. These transplants produced : (i) Wild type females whose eggs came from mutant mother, and (ii) torpedo deficient females whose egg came from wild type mother The possible outcome of thie experiment can be : (1) Torpedo deficient eggs developed in wild type ovary produced normal embryos (2) Wild type eggs developed in Torpedo deficient ovary produced ventralized embryos (3) Torpedo deficient eggs developed in wild type ovary produced ventralized embryos (4) Dorsal protein enters in the nuclei of dorsal side of embryos which came from wild type eggs developed in Torpedo deficient ovary (5) Dorsal protein remains cytoplasmic in the dorsal side of the embryos which came from wild type eggs developed in Torpedo deficient ovary. Which of the above combinations is correct? (A) (1), (2) and (4) (B) (2), (3) and (5) (C) (2), (4) and (5) (D) (1), (3) and (5) P6/13

179 Amar Ujala Education 98. During lens formation in the Xenopus, the following statements have been proposed : (1) Lens induction can be achieved in the absence of optic vesicle after priming of head ectoderm by the anterior neural plate (2) The optic vesicle can induce the presump-tive trunk ectoderm to form the lens (3) Only the head ectoderm can respond to direct signals from the optic vesicle to form the lens (4) The anterior neural plate primes the head ectoderm via BMP4 and Fgf8 prior to signals from the optic vesicle Which of the above combinations is correct? (A) (3) and (4) (B) (2) and (4) (C) (1) and (4) (D) (1) and (3) 99. Mutation in a gene x in Arabidopsis thaliana results in more number of lateral root formation. Which one of the following is the correct statement? (A) The gene product acts as a positive regulator of lateral root formation (B) The gene product acts as a negative regulator of lateral root formation (C) the gene product is not likely to be involved in lateral root formation (D) The gene product promotes replication for lateral root development Exp. Mutation is a permanent change of the nucleotide sequence of the genome of an organism (e.g., Arabdiopsis thaliana) virus, or extra chramosomal DNA or other genetic elements Nitrogen fixation is basically a process of converting nitrogen gas into ammonia NH 3. One of the key enzymes in the process is "nitro-genase". The production and activity of nitro-genase is very highly regulated as highlighted below. (1) Nitrogen fixation through nitrogenase is an energetically expensive process (2) Nitrogenase encoding gene is under a constitutive promoter (3) Nitrogenase is highly sensitive to oxygen (4) Endogenous availability of the cofactor of nitogenase enzyme is very low Which one of the following combinations of above statements is correct? (A) (1) and (2) (B) (1) and (3) (C) (2) and (3) (D) (2) and (4) A/UG The amount of each enzyme present in the chloroplast stroma is regulated by mechanisms that control the concerted expression of nuclear and chloroplast genomes. Following are certain statements regarding the regulation of chloro-plast enzymes: (1) Nucleus-encoded enzymes are translated on 70S ribosomes in the cytosol and subse-quently transported into the plastid. (2) Plastid encoded enzymes are translocated in the stroma on prokaryote-like 70S rebosomes (3) Light modulates the expression of stromal enzymes encoded by the nuclear genome via specific photoreceptors. (4) The eight small subunits of rubisco is encoded in plastid. Which one of the following combinations of above statements is correct? (A) (1) and (2) (B) (1) and (3) (C) (2) and (3) (D) (3) and (4) 102. Aquaporins are a class of proteins that are relatively abundant in plant membranes. Following are certain statements regarding the properties of aquaporins: (1) Aquaporins form water channels in membrane. (2) Some aquaporins also transport uncharged molecules such as NH 3. (3) The activity of aquaporins is not regulated by phosphorylation. (4) The activity of aquaporin is regulated by calcium concentration and reactive oxygen species. Which one of the following combinations of above statements is correct? (A) (1), (2) and (4) (B) (2), (3) and (4) (C) (1), (3) and (4) (D) (1), (2) and (3) 103. In terpene biosynthesis pathways, three acetyl- CoA are joined together stepwise to form mevalonic acid. Which one of the following three steps is required by mevalonic acid to form isopentenyl pyrophosphate (IPP)? (A) Pyrophosphorylation, decarboxylation, and dehydration. (B) Alkylation, pyrophosphorylation and decarboxylation (C) Methylation, dehydration and alkylation (D) Phosphorylation, carboxylation and methylation. P6/14

180 A/UG Following statements are related oxidative phosphorylation. (1) Redox reactions of electron transport chain coupled with ATP synthesis are collectively called oxidative phosphorylation. (2) Three major processes: glycolysis, oxidative pentose phosphate pathway and citric acid cycle are related to oxidative phosphory-lation. (3) Electron transport proteins are bound to outer of the two mitochondrial membranes. (4) In the electron transport chain electrons are transferred to oxygen from NADH. Which one of the following combinations of above statements is correct? (A) (1) and (4) (B) (2) and (3) (C) (3) and (4) (D) (1) and (3) 105. Satellite RNAs (sat-rnas) are species of RNA associated with specific strains of some plant RNA viruses, although it is not necessary for their replication. Few statements are given below on sat-rna (1) Presence of sat-rna leads to reduction in severity of disease symptoms. (2) Presence of sat-rna leads to increase in severity of disease symptoms. (3) sat-rna is constitutively expressed like coat proteins and is independent of virus infection. (4) sat-rna is not constitutively expressed like coat proteins but is expressed only after virus infection. Which one of the following combinations of above statements regarding sat-rna is correct? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) 106. The following statements are related to excretion in invertebrates: (1) Flame cells are found in molluscs and jelly fish. (2) Nephridia and Malpighian tubules convert ammonia to urea for water conservation. (3) Green glands are found in flatworms and help in the excreta elimination. (4) Excretory canals in nematodes carry waste materials to excretory pores in the body wall. Choose the correct answer (A) Only (3) (B) (1) and (3) (C) Only (2) (D) (2) and (4) 107. Respiration can be inhibited voluntarily for some time. The point at which respiration cannot be voluntarily inhibited is known as breaking point. Following explanations are offered for the breaking point: Amar Ujala Education (1) J-receptors stimulate respiratory centers (2) Hering-Breuer reflex operates (3) The rise of arterial pco 2 stimulates the respiratory centre (4) The fall of arterial po 2 stimulates the respiratory centre Which of the above combination is correct? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) 108. A mechanical pressure was exerted on a specific location of a peripheral nerve of a mammal. The touch or pain rececptors were stimulated from the skin surface innervated by the same nerve. The action potential generated by touch receptor stimulation was blocked beyond the point of mechanical pressure. But the action pontential generated by pain receptor stimulation passes through the point of mechanical pressure. Following explanation were offered for these observations: (1) The large diameter 'A' fibres were affected by mechanical pressure (2) The small diameter 'C' fibres were not affected by mechanical pressure (3) The intermediate diameter 'B' fibres were affected by mechanical pressure (4) The large diameter 'A' fibres were not affected by mechanical pressure Which of the above combinations is correct? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) 109. When a skeletal muscle with an intact nerve supply is stretched, the muscle contracts and the tone increases. The following explanations are offered for this observation: (1) Golgi tendon organ was stimulated by the stretching for muscle (2) y-motor neurons were excited by the stimulated afferent nerve fibres from the stretched muscle (3) Muscle spindle was stimulated by stretching of muscle (4) -motor neurons were excited by the stimulated afferent nerve fibres from the stretched muscle Which of the combination is correct? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) P6/15

181 Amar Ujala Education 110. Upon prolonged illumination, activated rhodopsin does not activate transducin, hence the vision is impaired. This could be because of the following explanations: (1) Most of the activated rhodopsin gets phosphorylated and is unable to activate transducin. (2) Most of the activated rhodopsin gets dephosphorylated and is unable to activate transducin. (3) Arrestin further interacts with phosphorylated rhodopsin. (4) Arrestin further interacts with dephosphorylated rhodopsin. Which of the above combinations is correct? (A) (1) and (3) (B) (2) and (4) (C) (1) and (4) (D) (2) and (3) 111. An elderly person suffering from calcium deficiency was advised to take calcium rich food and to supplement the diet with vitamin D. The absorption of calcium in theintestine wasincreasedwith the supplementation of vitamin D. Following explanations were offered for this increased calcium absorption by vitamin D: (1) The synthesis of calbindin-d 9k and calbindin-d 28k in enterocytes was stimulated. (2) The number of Ca 2+_ ATPase molecules in enterocytes was increased. (3) The synthesis of divalent metal transporters 1 (DMT1) in the enterocytes was stimulated. (4) The number of hephaestin in the entero-cytes was increased. Which of the above combinations is correct? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) 112. A series of cell lines was created by fusing mouse and human somatic cell hybrids, human chromosomes tend to get lost before becoming a stable cell line. Some hybrid cell lines may carry human chromosome deletions. each cell line was examined for the presence of chromosomes and for the production of an enzyme. the following results were obtained : Cell Gene Chromosomal segments line Pro- 1p 1q 2p 2q 3p 3q 4p 4q 5p 5q duct A B C D E A/UG-15 Which segment of the chromosome has the gene encoding for the enzyme? (A) 1p (B) 5p (C) 5q (D) 4p 113. The following table summarizes the result of a cross between two strains of Neurospora having the alleles D and d, respectively. The table shows the different patterns of octad arrangement and the number of ascus observed of each type. Octads D d D d D d D d D d D d D d d D d D D d d D d D d D D d d D d D D d d D d D d D D d d D d D D d Number of ascus observed Total = 300 Based on the above, fill in the blanks from the options given below, The first two columns are from meiosis with no crossover between locus D and...[1]... The patterns for these two columns represent...[2]... segregation pattern. The distance between the locus 4 and the centromere is...[3]... map units (A) d allele first division 10 (B) centromere first division 10 (C) d allele second division 20 (D) centromere second division The inheritance pattern of a common trait which shows complete penetrance is shown below: I II III ? Based on the above pedigree, fill in the blanks from the options given below : "The trait is...[1]... The probability that a child from the marriage of individual III-1 and III-2 will show the trait is...[2]... considering that the individual III-1 is a carrier of the trait." 2 P6/16

182 A/UG (A) Y-linked 0 (B) Y-linked 1/2 (C) Autosomal 1/8 (D) Autosomal 1/6 Exp. A pedigree chart is a diagram that shows the occurrence and apprearance or phenotypes of a particular gene or organism and its ancestors from one generation to the next A hypothetical gene encodes a protein with the following amino acid sequence: Phe- Pro- Thr- Ala- Val- Arg- Ser A mutation of single nucleotide alters the amino acid sequence to Phe- Leu- Leu- Leu- Leu- Val A second single nucleotide mutation occurs in same gene restoring back the amino acid sequence to the original. The following statements were made regarding the nature and location of the first mutation and that of the intragenic suppressor mutation : (1) The first mutation is a deletion in the second codon (2) The first mutation is an insertion in the second codon (3) The intragenic suppressor mutation is an insertion in the second codon (4) The intragenic suppressor mutation is a deletion in the third codon Which combination of the above statements is correct? (A) (1) and (3) (B) (1) and (4) (C) (2) and (3) (D) (2) and (4) 116. The following statements were made regarding chromosome pairing (shown in the figure below) and subsequent segregation during meiosis I in the reciprocal translocation heterozygote : G F E G F E A B C D A B C D P6/17 Amar Ujala Education (1) Three ways of segregation in Anaphase I would be: adjacent 1 (vertically in the above figure), adjacent 2 (horizontally) and alternate. (2) Gametes resulting from adjacent 1 and adjacent 2 segregation will be nonviable because of deletions and duplications of several genes. (3) All gametes resulting from alternate segregation will be viable as they will carry both normal chromosomes or both chromosomes having translocations in the two poles, respectively. (4) A dicentric and an acentric chromosome will be generated following alternate segrega-tion. Which of the following combination of statements will most appropriately explain the consequence? (A) (1), (2) and (4) (B) (1), (2) and (3) (C) Only (1) and (2) (D) Only (1) and (3) 117. Three Indian animals - cormorant, lion-tailed macaque and gerbil are to be matched with the ecosystem they inhabit - Wetland (A), Desert (B) Deciduous forest (C), or Rain forest (D). Which of the following is the correct match of each animal with its habitat? (1) Cormorant - D; Lion-tailed macaque - C; Gerbil - B (2) Cormorant - A; Lion-tailed macaque - C; Gerbil - D (3) Cormorant - A; Lion-tailed macaque - D; Gerbil - B (4) Cormorant - B; Lion-tailed macaque - C; Gerbil - D 118. The list below includes names of animal phyla and classes. 1. Echinodermata 2. Cephalopoda 3. Annelida 4. Mollusca 5. Hirudinea 6. Asteroidea 7. Arthropoda 8. Crustacea. For a leech and lobster, the correct classification of phylum and class, respectively, is (A) Leech : Phylum - 4, Class - 2; Lobster : Phylum - 1, Class - 8 (B) Leech : Phylum - 3, Class - 2; Lobster : Phylum - 4, Class - 3 (C) Leech : Phylum - 3, Class - 5; Lobster : Phylum - 7, Class - 8; (D) Leech : Phylum - 1, Class - 7; Lobster : Phylum - 3, Class - 6

183 Amar Ujala Education 119. Some key characteristics of the four classes of phylum Mollusca are listed below : (1) They have two lateral (left and right) shells (valves) hinged together dorsally; they do not have distinct head or radula; they disperse from place to place largely as larvae (2) they generally creep on their foot; the heads of most of this group have a pair of tentacles with eyes at the end; during embryological development, they undergo torsion (3) They have oval bodies with overlapping calcareous plates underneath the plates, the body is not segmented; they creep along using a broad, flat foot surrounded by a groove or mantle cavity in which the gills are arranged (4) They have highly developed nervous system most members of this class have closed circulatory system The correct match of the above charactersitics with the classes of Mollusca is : (A) A Polyplascophora B Bivalvia C Gastropoda D Cephalopoda (B) A Cephalopoda B Polyplacophora C Bivalvia D Gastropoda (C) A Bivalvia B Gastropoda C Polyplacophora D Cephalopoda (D) A Gastropoda B Bivalvia C Cephalopoda D Polyplacophora 120. The diagram below depicts a simplified tree of Life with three domains and one of the domains including Whittaker s three major kingdoms : Which of the following is the correct naming of the numbered boxes? (A) 1 Bacteria 2 Archaea 3 Eukarya 4 Fungi (B) 1 Archaea 2 Bacteria 3 Eukarya 4 Plants (C) 1 Eukarya 2 Bacteria 3 Archaea 4 Plants (D) 1 Archaea 2 Bacteria 3 Eukarya 4 Fungi P6/18 A/UG Given below are some of the methods used to assess evolutionary phylogenetic relationships among plant taxa : (1) 16s rrna sequence (2) Mitochondrial microsatellite (3) Biochemical characterization (4) Morphology Which two of the above methods can best reveal the evolutionary phylogenetic relationships? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) 122. Sleeping sickness is a disease caused by a protozoan parasite and the following statements pertain to that disease : (1) The vector for this disease is tsetse fly. (2) The vector for this disease is Trypanosoma brucei. (3) The parasite s body is covered by a dense coat of variable surface glycoprotein (VSG) (4) There are several thousands of VSG genes,only one of them expressing at a time, which helps the parasite in evading host s immune response (5) Several thousands copies of VSG genes express concurrently, paralyzing the host immune system Which of the following is the correct combination of the statement given above? (A) (2), (3) and (4) (B) (1), (3) and (4) (C) (1), (3) and (5) (D) (2), (3) and (5) 123. There are three species of frogs A, B and C species A does not provide parental care for its eggs and larvae. Species B is subjected to predation by a predator that selectively feeds only on small-sized larvae. Species C faces progressively decreasing opportunities for breeding with increasing age. Assuming that resources available for reproduction are similar for A, B and C, which of the following strategies would have been favored? (A) A should produce large number of small-sized offspring; B should produce a small number of largesized offspring; C should breed earlier in life (B) Species A and B should produce a small number of large-sized offspring and C should breed earlier in life (C) Both species A and B should produce a large number of small-sized offspring and C should breed later in life but increase its clutch size (D) Number of large-sized offspring; B should produce a large number of small-sized offspring and C should breed earlier in life with a small clutch size

184 A/UG Fish species X and Y feed on mayfly nymphs in their stream habital. In a laboratory experiment, the predation intensity of X and Y on their prey was tested under dark (D) and light (L) conditions. Thus, the experimental protocol included four aquaria-lx, LY, DX and DY. In each aquarium containing 100 mayfly nymphs, one fish was introduced and allowed to feed for 30 minutes. Then the fish was removed and the number of mayfly nymphs left uneaten in each aquarium was counted. The results are shown graphically below : No. Of left uneaten LX LY DX DY The most significant conclusion from the results is : (A) X is a visual predator, but has less predation impact on the prey than Y (B) X is a visual predator and has greater predation impact on the prey than Y (C) Y is a visual predator and has greater predation impact on the prey than X (D) Y is a visual predator but has less predation impact on the prey than X 125. The three graphs (A, B, C) show population growht (N) patterns in relation to N or time (t) : A?? t B N Which of the following is correct with reference to the Y-axis label and the type of population growth? (A) A : Y-axis : N, exponential growth B : Y-axis : dn/dt, logistic growth C : Y-axis : ln(n), exponential growth (B) A : Y-axis : dn/dt, exponential growth B : Y-axis : ln(n), logistic growth C : Y-axis : N, exponential growth (C) A : Y-axis : ln(n), exponential growth B : Y-axis : dn/dt, logistic growth C : Y-axis : N, exponential growth (D) A : Y-axis : dn/dt, exponential growth B : Y-axis : ln(n), logistic growth C : Y-axis : N, exponential growth? C t Amar Ujala Education 126. Which of the following set of observations is true with reference to a comparison of aquatic (A) and terrestrial (T) ecosystems? (A) Number of trophic levels is more in A than in T Productivity/Biomass ratio is higher in T than in A Herbivore assimilation efficiency is higher in A than in T (B) Number of trophic levels is more in T than in A Productivity/Biomass ratio is greater in A than in T Herbivore assimilation efficiency is higher in T than in A (C) Number of trophic levels is more in T than in A Productivity/Biomass ratio is higher in T than in A Herbivore assimilation efficiency is higher in T than in A (D) Number of trophic levels is more in A than in T Productivity/Biomass ratio is greater in A than in T Herbivore assimilation efficiency is higher in A than in T 127. A small lake has three trophic levels-phytoplankton (autotrophs), Zooplankton (hervivore) and planktivorous fish (primary carnivore). Into this lake, a population of piscivorous fish (secondary carnivore) was introduced to study the top-down effects. What is the expected long-term consequence of such an introduction to phytoplankton and zooplankton trophic levels? (A) Zooplankton biomass will increase and phytoplankton biomass will decrease (B) Zooplankton biomass will decrease and phytoplankton biomass will increase (C) The biomasses of both zooplankton and phytoplankton will increase (D) The biomasses of both zooplankton and phytoplankton will decrease 128. Following is the diagram of three idealized survivorship curve of animals : Survivorship Type III Age Type II Type I Find the correct match between the group of animals and the respective survivorship curves : (A) Marine pelagic fish and large mammals-iii and I, respectively P6/19

185 Amar Ujala Education (B) Margine pelagic fish and large mammals-i and II, respectively (C) Some birds and large mammals-i and III respectively (D) Maring pelagic fish and some birds-i and III, respectively 129. Following are the characteristics of spices that make themmore or less prone to extinction : Rare -a Common-b Good dispersal rate-c poor dispersal rate-d Low specialization-e high specialization-f High variability-g low variability-h Low trophic status-i high trophic status-j Long life span-k High reproductive output-m short life span-i Low reproductive output-n Which of the following is the correct combination of characteristics that makes the species more prone to extinction? (A) adfgjln (B) acfhikm (C) bd cgiln (D) befhjkm 130. Assume that in terms of genetic fitness the benefit of performing an altruistic act to a relative is 500 units and the cost involved is 150 units. Following Hamilton s Rule, the act should be performed if the relative is a : (A) only brother (B) nephew or niece (C) brother or step-sister (D) only step-sister 131. The following situations might lead to the evolution of monogyny in birds : (1) Male has to assist the female in rearing the offspring (2) Male guards the female against other males trying to mate with her (3) one male may not produce enough sperm required to fertilize all the eggs produced by the female Which of the above is/are correct? (A) Only (1) (B) Only (2) (C) (1) and (2) (D) (1) and (3) 132. The body weight of adult female of a strain of Drosophila is mg (mean + standard deviation). In a laboratory experiment, each of the 3 groups (A, B, C) of this strain was subjected to a different type of selection pressure having influence on the female body weight. After many generations of experimental selection pressure, the body weight changed as follows : Group A : Body weight distribution- Normal, mg Group B : Body weight distribution- Bimodal at 1.4 and 2.2 mg Group C : Body weight distribution- Normal, mg P6/20 A/UG-15 Which of the following correctly gives the types of selection that have occured in the three groups? (A) Group A : Directional; Group B : Stabilizing; Group C : Disruptive (B) Group A : Disruptive; Group B : Directional; Group C : Stabilizing (C) Group A : Stabilizing; Group B : Disruptive; Group C : Directional (D) Group A : Directional; Group B : Disruptive; Group C : Stabilizing 133. A few in the history of life on earth are given below : (1) Radiation of mammals and birds Flourishing of insects and angiosperms (2) Primitive plants and fungi colonize land; Diversification of echinoderms (3) Seed plants appear; Fishes and Trilobites abundant; earliest amphibians and insects (4) Earliest birds and Angiosperms appear; Gymnosperms dominant (5) Invasion of land by primitive land plants and Arthropods (6) Mass marine extinctions; reptiles radiate; Amphibians decline Which of the following is a correct match of the above events with the geological period during which they hadoccurred? (A) A : Ordovician; B : Tertiary; C : Permian; D : Silurian; E : Devonian; F : Jurassic (B) A : Permian; B : Devonian; C : Silurian; D : Ordovician; E : Tertiary; F : Jurassic (C) A : Tertiary; B : Ordovocian; C : Devonian; D : Jurassic; E : Silurian; F : Permian (D) A : Permian; B : Devonian; C : Jurassic; D : Tertiary; E : Silurian; F : Ordovician 134. Four different species concepts are given below : (1) Species separate based on their use of different ecological niches and their presence in different habitats and environments. (2) Differences in physical characteristics or molecular characteristics are used to distinguish species (3) Species are distinct if they are reproductively isolated (4) Phylogenetic trees and analyses of ancestry serve to differentiate species

186 A/UG-15 Which of the following gives the correct names of the above concepts? (A) A : Biological; B : Phylogenetic; C : Evolutionary; D : Ecological (B) A : Ecological; B : Phylogenetic; C : Biological; D : Evolutionary (C) A : Evolutionary; B : Ecological; C : Biological; D : Phylogenetic (D) A : Phylogenetic; B : Evolutionary; C : Ecological; D : Biological 135. Following are certain statemnets regarding the use of Agrobacterium in plant transformation : (1) A tumerfaciens causes crown gall disease and A. rhizogenes causes hairy root disease (2) Region A in Ti plasmid is responsible for replication (3) Region D in Ti plasmid is responsible for virulence (4) Oncogenic (onc) region in T-DNA is responsible or unusual amino acid synthesis. Which one of the following combinations of above statements is correct? (A) (1) and (2) (B) (3) and (4) (C) (1) and (3) (D) (2) and (4) 136. Marker-assisted selection (MAS) defined as selection based on molecular markers should have some important criterion for plant breeding activities. Some statements about these criteria are mentioned below : (1) Marker should co-segregate with the desired trait of interest (2) Marker should not co-segregate with the desired trait of interest (3) Marker should be un-linked with the desired trait of interest (4) Marker is used for indirect selection of a genetic dete rminant or determinants of a trait interest Which one of the above combinations is correct? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) 137. In order to develop a vaccine against regulatory T cellpromoting but Th1 suppressing viral infection, four groups (A-D) of mice were primed with either killed virus (A) or a virus-derived immune dominant peptide (B), or the same peptide but with two substitutions (C) or were left unprimed (D). Upon infection challenge, the order of increasing severity of infection was observed to be B > A > D > C. To explain the contrasting effects of these two peptides (B) and (C), the MHCbinding affinities were assessed but no difference was found. Which of the following possibilities most likely to explain their contrasting effects? (A) The wild-type peptide (B)elicits T cell expansion but the mutant peptide (C) fails Amar Ujala Education (B) The wild-type peptide induces T cell deletion but the mutant peptide does not (C) The wild-type peptide-mhc complex binds T cell receptor with significantly higher affinity than the mutant peptide-mhc complex (D) The wild-type peptide induces deletion of T-Reg cells but increases IFN production by T cells, whereas the mutant peptide fails to induce these effects 138. In at ransgenic mice line, lox P sites are introduced in the target gene A in the following manner : Promoter Exon1 loxp Exon2 loxp This transgenic mice line was mated with another transgenic mice line where Cre recombinase is expressed only in B cells What will be expression profile of gene A in Cre/lox recombinant mice? (A) Gene will not be expressed in B cells, as orientation of exon 1 will be inverted by Cre (B) Gene will not be expressed in B cells, as exon 2 will be deleted by Cre (C)GenewillonlybeexpressedinBcellsofthe recombinant mice where Cre removes the two lox P sites (D) Gene will not be expressed in B cells as orientation of exon 2 will be inverted 139. Cultured animal cells were transfected with expression vector encoding either -galactosidase ( -gal) alone or expressing a fusion protein of -gal and glucocorticoid receptor (GR). After transfection, cells were kept in presence or absence of Dexamethasone. Immunofluorescence with a labeled antibody specific for -gal was used to detect the expressed protein in cytoplasm or nucleus of transfected cells. Possible results of the experiments are : (1) Expression of -gal alone in the cytoplasm in both absence or presence of Dexamethasone (2) Expression of -gal-gr in the cytoplasm in the absence of Dexamethasone (3) Expression of -gal alone in the nucleus both in the presence or absence of Dexamethasone (4) Expression of -gal-gr in the nucleus in presence of Dexamethasone (5) Expression of -gal alone in both cytoplasm and nucleus in presence or absence of Dexamethasone (6) Expression of -gal-gr in both cytoplasm and nucleus in presence of Dexamethasone Choose the correct combination of results from the following options : (A) (2), (3) and (4) (B) (1), (2) and (4) (C) (2), (4) and (5) (D) (1), (2) and (6) P6/21

187 Amar Ujala Education 140. A small fraction of clear cellular lysate was run on an isoelectric focusing gel (IEF) to purify a particular protein, which showed a number of sharp bands corresponding to different pi values. The protein of interest has a pi of 5.2. Therefore, the band corresponding to pi 5.2 was cut, eluted with appropriate buffer and subjected to SDS-PAGE, which showed 3 distinct bands. Which one of the following inferences CANNOT be drawn from the above observations? (A) Several different proteins having same PI may be present at the single band on IEF gel (B) SDS-PAGE showed 3 distinct bands which may represent molecular mass of different proteins (C) The protein of interest may be composed of 3 subunits (D) As the IEF gel showed a distinct band corresponding topi 5.2, which is the PI of the protein of interest, the protein is composed of a single subunit 141. In a plaque-forming cell assay, antigen specific B-cell numbers are assessed. In this assay, antigen coated sheep red blood cells (SRBCs) are lysed by the hapten-specific B cells by complement-mediated cyotoxicity. In an assay that tried to enumerate the TNP-specific B-cells elicited in TNP-KLH. Primed mice, no plaques were formed despite the presence of antigen-specific antibody producing B cells. Which of the following is not the reason for the absence of plaques? (A) The source of complement has anti-tnp antibody (B) The SRBC were stored for too long (C) The B cells wre contaminated with LPS or lectins (D) The complement had anti-klh antibody 142. Cytochrome-c has only one tryptophan residue (W) which is burid. The protein in cacodylate buffer (ph 6.0) is excited at 280 nm, and its emission spectrum measured in the ragne of nm. the same measurement was repeated on the protein in the buffer containing 6M guanidine hydrochloride. It was observed that there is an increase in the intensity of the emission spectrum of the guanidine hydrochloride treated cytochrome-c. The most probable reason for this increase is : (A) W is near a hydrophobic patch present in the unfolded protein (B) W is near the heme in the native protein (C) W is near carboxylate amino acid side chains in the native protein (D) W is in a polar pocket in the native protein A/UG The hydrogen atoms in the (delta) methylene group of lysine will give the following splitting pattern in the 1 HNMR spectra of lysine. (A) Triplet of triplets (B) Quintet (C) Doublet of triplets (D) Triplet of a doublet Exp. Nuclear magnetic resonance spectroscopy, most commonly known as NMR spectroscopy, is a research technique that exploits the magnetic property of certain atomic nuclei Hind III Bam H1 0.8 Kb 1.2 Kb 0.9 Kb Bam H1 2Kb Eco R1 Vector 3.8 Kb 2.9 Kb The above diagram represents a 2 Kb insert successfully introduced in between two BamHI sites of a 3.8 Kb vector in desired orientation. The HindIII site on the insert and EcoRI site on the vector is also indicated. If the insert was introduced in the opposite orientation, which one of the following statements is incorrect? (A) Digestion with EcoRI will linearize the 5.8 Kb plasmid (B) Digestion with BamHI will yield one 3.8 and one 2.0 Kb fragment (C) Double digestion with EcoRI and HindIII will produce 1.7 Kb and 4.1 Kb fragments (D) Double digestion with EcoRI and HindIII will produce 2.1 Kb and 3.7 Kb fragments 145. Combination of molecular markers with their classification based on either dominant or co-dominant types are shown below : (1) SSR and RFLP : co-dominant (2) SSR and RAPD : co-dominant (3) RAPD and RFLP : co-dominant (4) AFLP and RAPD : co-dominant Which one of the following is the correct combination? (A) (1) and (2) (B) (2) and (3) (C) (3) and (4) (D) (1) and (4) P6/22

188 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, December-2013 PART-A 1. A cart wheel rolls along a straight line. If the distance covered is equal to the diameter of the wheel, what is the angle through which the wheel has turned? (A) 90 (B) between 90 and 120 (C) between 120 and 150 (D) between 150 and In a class of 10 students, 3 failed in History, 6 failed in Geographyandtotalnumberofpapers.2failedinboth. Howmanypassedinboththesubjects? (A) 1 (B) 2 (C) 3 (D) 0 C 3m A 4m 3. As shown in the diagram above, a sphere is placed on the top of an incline. It rolls down the incline without slipping in exactly 50 turns. The radius of the sphere is (A) (5/ )cm (B) (5/ )m (C) (10/ )cm (D) 10cm B 2 2 Exp. BC = AB AC 2 2 = = 25 = 5 5m = 50 2 r r= = 5 cm 4. A set of concentric circles of integer radii 1,2,... N is shown in the figure above. An ant starts at point A, goes round the first circle, returns to A, moves to A, goes round the second circle, returns to A, moves to A, and repeats this until it reaches A. The distance covered by the ant is (A) N (N + 1) p (B) 2N p + N (C) p (N + l)n + N - 1 (D) p(n-l)n+n-1 Exp. 2p 1 + 2p 2 + 2p p N + (N 1) = 2 ( N) + N 1 N( N 1) = 2 N 1 2 = (N)(N +1)+N 1 φ r 1 A 1 A 2 A 3 A N φ r 2 θ θ θ φ r 3 r 4 P7/1

189 Amar Ujala Education 5. The figure above shows an infinite series of triangles, in which r 1 >r 2 >r 3... What is the total length of the solid line segments in the figure? (A) r 1 /r 2 +r 2 /r (B) r 2 /r -r (C) r 2 /r +r (D) r 1 -r 2 /r If a i, b i and C i are distinct, how many terms will the expansion of the product (a 1 +a 2 +a 3 )(b 1 +b 2 +b 3 +b 4 ) (C 1 +C 2 +C 3 +C 4 +C 5 ) contain? (A) 12 (B) 30 (C) 23 (D) 60 number 4 8 papers citations 8 6 papers citations Exp. a 1 + a 2 + a 3 =3 b 1 + b 2 + b 3 + b 4 =4 c 1 + c 2 + c 3 + c 4 + c 5 =5 Hence =60 7. The above plot depicts the number of research publications of a scientist along with the number of citations. Which of the following statements is not correct? (A) In the year 2012, 50% more papers were published but citations decreased by 25%. (B) In the year 2012, 100% more papers were published but citations were 75% of the number of the papers in that year. (C) The papers published in year 2011 is only 33.33% of the total. The papers published in year 2011 is only 33.33% of the total (D) The total number of citations for both years is 16.66% more than the total number of papers 8. The next number of the sequence 1, 5, 14, 30, 55,... is (A) 85 (B) 90 (C) 91 (D) 95 height Weight A/UG The distribution of heights and weights in a 91population is shown The distribution of heights and weights in a 91population is shown proportional to the number of persons having a particular combination of weight and height. Which statement best describes the trend in the population? (A) Height and weight are strongly correlated. (B) Height and weight are anti-correlated. (C) Large heights do not imply proportionately large weights. (D) Height and weight are independent characteristics. 10. What is the maximum sum of the numbers of Saturdays and Sundays in a leap year? (A) 104 (B) 105 (C) 106 (D) Two trains of length 150 m a nd 250 m pass each other with Two trains of length 150 m a nd 250 m pass each other with and guards are at the extremities of the trains. The time gap between the drivers passing each other and first driver-guard pair passing each other is 30 s. How much later will the other driver-guard pair pass by? (A) 10 s (B) 20 s (C) 30s (D) 50s 12. In a room, we have one grandfather, two fathers, two sons, and a grandson. The age of one father is seven times the age of his son. The age of the other father is twice his son s a ge. Assuming that there are age of the other father is twice his son s a ge. Assuming that there are is the grandson? (A) 1 (B) 2 (C) 5 (D) Cannot be determined P7/2

190 A/UG A hemispherical bowl is being filled with water at a constant volumetric rate. The level of water in the bowl increases (A) indirectproportiontotime (B) in inverse proportion to time (C) fasterthandirectproportiontotime (D) slowerthandirectproportiontotime 14. Equal masses of two liquids of densities 6 kg/m 3 and 4 kg/m 3 are mixed thoroughly. The density of the mixture is (A) 4.8 km/ 3. (B) 5.0 km/ 3. (C) 5.2 km/ 3. (D) 5.4 km/ Two points A and B on the surface of the Earth have the following latitude and longitude co-ordinates. (A) 30 0N, 45 0 H (B) N, 135 W If R is the radius of the Earth, the length of the shortest path from A to B is (A) v3/2pr (B) p. R /3 (C) PR/6 (D) 2p R/3 16. Amoebae are known to double in 3 min. Two identical vessels A & B, respectively contain one and two amoebae to start with. The vessel B gets filled in 3 hours. When will Agetfilled? Amar Ujala Education (A) 3 hours (B) 2 hours 57 min (C) 3 hours 3 min (D) 6 hours 17. Students of a school are divided into 4 groups. What is the probability that three friends get into the same group? (A) 3/4 (B) 1/64 (C) 1/16 (D) 1/3 18. A fruit vendor buys 120 Shimla apples at 4 for Rs. 100, and 120 Golden apples at 6 for Rs She decides to mix them and sell at 10 for Rs She will make (A) no profit, no los (B) a loss of 4% (C) a gain of 4% (D) a loss of 10% /2 +4-1/2 (A) 4º (B) (C) 19 9/16 (D) 22 9/ In an enclosure there were both crows and cows. If there were 30 heads and 100 legs, what fraction of them are crows? (A) 1/3 (B) 1/4 (C) 1/10 (D) 3/ The interaction energy between atom A and B is 400 k/ mol. The type of interaction between them is (A) pi - pi (B) covalent. (C) ion - dipole. (D) hydrogen bond. 22. Which one of the following bases has the largest hydrogen bonding possibility? (A) adenine (B) Guanine (C) cytosine (D) Uracil Exp. In the given options, Guanine has the largest hydrogen bonding possibility. 23. Enzymes help to lower the activation energies of reactions by (A) covalent interaction with substrates. (B) binding only with the solvent molecules. (C) changing reaction equilibria. (D) forming weak interactions with substrates. Exp. Enzymes help to lower the activation energies of reactions by forming weak interactions with substrates. 24. Glucose residues in amylose are linked by (A) ß1 4 (B) 1 4 (C) 1 6 (D) ß1 6 PART-B P7/3 Exp. Glucose residues in amylose are linked by 1 4. Amylose is a linear polymer of several thousand glucose residues, (1,4) glycosidically linked. 25. Whe n a membrane is depolarized to a voltage value more positive than the threshold voltage, it leads to the generation of (A) Donnan potential. (B) Action potential. (C) Resting potential. (D) Electrochemical potential. Exp. Whe n a membrane is depolarized to a voltage value more positive than the threshold voltage, it leads to the generation of Action potential. 26. E. coli proliferates faster on glucose than it does on lactose because lactose is (A) taken up more slowly than glucose (B) not hydrolyzed by E. coli (C) take n up fa ster than glucose. (D) toxic to the cells. Exp. E. coli proliferates faster on glucose than it does on lactose because lactose is taken up more slowly than glucose.

191 Amar Ujala Education 27. Out of the list given below, which is the correct order of increasing lipid bilayer permeability? (A) N>Ethanol>H 2 O> Glucose> Ca + 2 >RNA (B) H 2 O > Glucose > Ethanol> N 2 >Ca>RNA (C) Ca + 2 > RNA > N 2 >Ethanol>H 2 O > Glucose + 2 (D) Ethanol>RNA>Ca +>H 2 O > Glucose >N2 28. In the lysogenic -phage (A) Both CI and Cro are on. (B) Both CI and Cro are off (C) CIisonwhileCroisoff. (D) CI is off while Cro is on. Exp. In the lysogenic -phage CI is on while Cro is off. 29. In bacteria chromosomal DNA replication stops at (A) one specific locus. (B) several specific loci. (C) a single locus, randomly. (D) from several loci, randomly. Exp. In bacteria chromosomal DNA replication stops at one specific locus. 30. The Uvr ABC repair mechanism is involved in repairing (A) missing bases. (B) missing bases. (C) cross linked strands. (D) DNA damage caused by bulky chemical adducts. Exp. The Uvr ABC repair mechanism is involved in repairing DNA damage caused by bulky chemical adducts. 31. Which of the following phenomena is observed in compatible plant-pathogen interactions? (A) Virulence in pathogen. (B) Hypersensitive response in host. (C) Resistance in host. (D) Avirulence in pathogen. Exp. Virulence in pathogen phenomena is observed in compatible plant-pathogen interactions. 32. Which one of the following is NOT an extracellular matrix protein? (A) Fibronectin (B) Vitronectin (C) Laminin (D) Cyclin Exp. Cyclin is not an extracellular matrix protein. 33. The cylindrical channels in gap junctions are made of (A) connexin. (B) collagen. (C) fibronectin. (D) N-CAM. Exp. The cylindrical channels in gap junctions are made of connexin. P7/4 A/UG A tumour suppressor protein (A) is one whose function brings about regression of a tumour (B) is one whose function brings about regression of a tumour tumours. (C) tumours. (D) inhibits the progression of the cell cycle by phosphorylating cyclins. Exp. A tumour suppressor protein is one whose function brings about regression of a tumour. 35. Immediate hypersensitivity reactions are associated with (A) IgG (B) IgE (C) 1gM (D) 1gA Exp. Immediate hypersensitivity reactions are associated with IgE. IgE is made by a small proportion of B cells and is present in the blood in low concentrations. Each molecule of IgE consists of one four-chain unit and so has two antigen-binding sites, like the IgG molecule; however, each of its H chains has an extra constant domain (CH4), which confers on IgE the special property of binding to the surface of basophils and mast cells. 36. The change in the state of specification of imaginal disc of Drosophila to that of a different disc type is known as (A) transdetermination (B) transdifferentiation (C) transformation (D) transduction Exp. The change in the state of specification of imaginal disc of Drosophila to that of a different disc type is known as transdetermination. 37. During double fertilization in plants, one sperm fuses with the egg cell and the other sperm fuses with (A) synergid cell. (B) central cell. (C) antipodal cell (D) nucellar cell. Exp. During double fertilization in plants, one sperm fuses with the egg cell and the other sperm fuses with central cell. 38. Over-expression of a dominant negative FGF receptor during amphibian development would prevent formation of (A) trunk and tail. (B) head and trunks. (C) trunk and fore limbs. (D) head and forelimbs. Exp. Over-expression of a dominant negative FGF receptor during amphibian development would prevent formation of trunk and tail.

192 A/UG The cell death pathway in C. elegans can be schematically represented as : ced9 ced 4 ced3 Based on the above, which one of the following statements is TRUE? (A) A loss of function allele of ced9 would lead to survival of cells that normally die. (B) A loss of function allele of ced9 would lead to excessive cell death. (C) A gain of function allele of ced9 would lead to excessive cell death. (D) Neither loss or gain-of-function of ced9 would make any change to the cell death pathway. 40. Which of the following mechanisms is NOT involved in providing photo protection to plants? (A) Degradation of D1 protein. (B) Zeaxanthin formation. (C) Zeaxanthin formation. (D) Thermal dissipation. Exp. Zeaxanthin formation. mechanisms is not involved in providing photo protection to plants. 41. Which of the following plant hormones can mimic the det1 mutation, causing de-etiolation and chloroplast development in dark? (A) Cytokinin (B) Cytokinin (C) Auxin (D) Ethylene Exp. Cytokinin plant hormones can mimic the det1 mutation, causing de-etiolation and chloroplast development in dark. 42. Under which conditions do members of the family Gramineae synthesize and release phytosiderophores? (A) Iron deficiency. (B) Phosphorus deficiency. (C) Availability of iron complexes in rhizosphere. (D) Availability of phosphorus complexes in rhizosphere. Exp. Under Iron deficiency conditions members of the family Gramineae synthesize and release phytosiderophores. 43. The membranes of chilling-sensitive plants is characterized by (A) high proportion of saturated fatty acids (B) lower transition temperature (C) lower proportion of saturated fatty acids (D) lower transition temperature and higher proportion of unsaturated fatty acids Exp. The membranes of chilling-sensitive plants is characterized by high proportion of saturated fatty acids. Amar Ujala Education 44. The only bone marrow cell that never appears in peripheral blood is (A) myeloblast (B) myelocyte. (C) lymphoblast. (D) megaloblast. Exp. The only bone marrow cell that never appears in peripheral blood is myeloblast. 45. Various types of excitable tissues when stimulated showed response as shown in the above figures. Which one of them is an example of fast adapting tissue? (A) A (B) B (C) C (D) D (A) (B) Response (C) Response Time Time Response (D) Response Time Time 46. After gull nestlings hatch, the parents remove the eggshells from the nest. This behaviour is to (A) clean the area. (B) reduce infection (C) makemorespaceinthenest. (D) minimise nest detection by predators. Exp. After gull nestlings hatch, the parents remove the eggshells from the nest. This behaviour is to minimise nest detection by predators. 47. Thyroxin releasing hormone (TRH) receptor belongs to (A) nuclear receptor family. (B) receptor tyrosin kinase family (C) G-protein - coupled receptor family. (D) guanyla te cyclase receptor family. Exp. Thyroxin releasing hormone (TRH) receptor belongs to G-protein - coupled receptor family. 48. A cross is made between a pure breeding plant having red coloured flowers with a pure breeding plant having white coloured flowers. Such a cross is called as (A) test cross. (B) monohybrid cross. (C) dihybrid cross. (D) back cross. Exp. A cross is made between a pure breeding plant having red coloured flowers with a pure breeding plant having white coloured flowers. Such a cross is called as monohybrid cross. P7/5

193 Amar Ujala Education 49. A cis-trans complementation test is carried out to identify (A) if two mutations are allelic in nature. (B) if two genes interact with one another. (C) the number of genes influencing phenotype. (D) to understand the dominance/recessive relationships between alleles. Exp. A cis-trans complementation test is carried out to identify if two mutations are allelic in nature. 50. The following is the inheritance pattern of a trait under observation: (i) The trait often skips a generation (ii) The number of affected males and females is almost equal (iii) The trait is often found in pedigrees with consanguineous marriages. The trait is likely to be (A) autosomal recessive. (B) autosomal dominant. (C) sex-linked recessive. (D) sex-linked dominant. Eumetezoe Multicellulerity Bilateria Parazoe B A Ecdysozoans Lophotrochozoans Radiata Chordata X Arthropoda Nematoda Annelids Y phatyhelminthes Z Porifera Exp. Autosomal recessive is one of several ways that a trait, disorder, or disease can be passed down through families. An autosomal recessive disorder means two copies of an abnormal gene must be present in order for the disease or trait to develop. 51. Given below is an evolutionary tree Based on the above, which one of the following combinations is correct? (A) A - Protostome; B - Deuterostome; X - Mollusca; Y - Cnidaria Z - Protozoa (B) A- Protostome; B - Deuterostome; X - Echinodermata; Y - Mollusca; Z - Cnidaria (C) A - Deuterostome; B - Protostome; X - Crustacea; Y - Mollusca; Z-Cnidaria (D) A - Deuterostome; B - Protostome; X - Echinodermata Y-Roundworm; Z -Ctenophora 52. The fungal group presently classified under protists is (A) Zygomycetes. (B) Oomycetes. (C) Deuteromycetes. (D) Discomycetes. Exp. The fungal group presently classified under protists is Oomycetes. P7/6 A/UG Name the common Indian bird that is ge nerally seen in groups (aggregation) (A) Bulbul (B) Warbler (C) Babbler (D) Sun bird Exp. Babbler is the common Indian bird that is ge nerally seen in groups (aggregation). 54. The fungus associated with human oral or vaginal infection is (A) Fusarium (B) Aspergillus (C) Candida (D) Pneumocytis Exp. The fungus associated with human oral or vaginal infection is Candida. 55. The most common vegetation in the Western Ghats of India is tropical moist deciduous forest but that in Deccan plateu is depleted thorn forest. The possible reason is (A) richer soil of Western Ghats compare to Deccan plateau. (B) extensive deforestation in Deccan plateau compared to Western Ghats. (C) higher rainfall in Western Ghat compared to Deccan plateau. (D) higher rainfall in Western Ghat compared to Deccan plateau. 56. The following graph is for a logistically growing population, with N plotted on the X-axis. What is the parameter plotted on the Y-axis?? N 1 (A) dn/dt (B) N (C) dn/dt.1/n (D) K 57. Annual weeds of arable lands are classified as (A) phonerophytes. (B) therophytes. (C) chamaephytes. (D) geophytes. Exp. Annual weeds of arable lands are classified as therophytes.

194 A/UG Which one of the following advancements in on the animal classification is correct? (A) Protostomes Pseudocoelomates Deuterostomes Eucoelomates (B) Acoelomates Protostomes Eucoelomates Deuterostomes (C) Pseudocoelomates Encoclomates Protostomes Deuterostomes (D) Protostomes Deuterostome Acoelomates Eucoelomates 59. Which of the following is the most appropriate spectral bands for vegetation analysis using remote sensing platforms? (A) Red, Near Infrared (B) Infrared, Visible (C) Red, Microwave (D) Visible, Microwave 60. Which of the following diseases doe s not leave any paleontological evidence? (A) Tuberculosis (B) Arthritis. (C) Rickets (D) Cholera Exp. Cholera does not leave any paleontological evidence. 61. Greater male investment in the care of offspring is most likely to lead to (A) a lek system. (B) stronger female choice. (C) Reverse sexual dimorphism (D) run-away selection. Exp. Greater male investment in the care of offspring is most likely to lead to Reverse sexual dimorphism. 62. A neuron that fires when an individual is eating by hand, also fires when he sees someone else eating with hand. Such neurons are called (A) mirror neurons. (B) mimicry neurons. (C) motor neurons. (D) reward neurons. Exp. A neuron that fires when an individual is eating by hand, also fires when he sees someone else eating with hand. Such neurons are called mirror neurons. 63. Which of the following microbial fermentations are anaerobic? (A) Ethanol and acetone-butanol. (B) Citric acid and propionic acid. (C) Penicillin and vitamin B (D) Streptomycin and rifampicin. Exp. Ethanol and acetone-butanol microbial fermentations are anaerobic. Amar Ujala Education 64. Encasing of which of the following planteel in a gelatinous matrix is referred as artificial seed? (A) Microcalli (B) Somaticembryos (C) Roottips (D) Shoottips Exp. Encasing somatic embryos planteel in a gelatinous matrix is referred as artificial seed. 65. Which of the following tra nsgenic crops have been approved for commercial cultivation in India? (A) Cotton (B) Brinjal (C) Cotton and Brinjal (D) Cotton, Brassica,Brinjal Exp. Transgenic crop Cotton has been approved for commercial cultivation in India. Transgenic plants are plants that have been genetically engineered, a breeding approach that uses recombinant DNA techniques to create plants with new characteristics.they are identified as a class of genetically modified organism (GMO). 66. Chinese Brake fern (Pteris vittata) is hyperaccumulator of : (A) Cadmium. (B) Arsenic. (C) Lead. (D) Chromium. Exp. Chinese Brake fern (Pteris vittata) is hyperaccumulator of Arsenic. 67. Which of the following is NOT a post -translational modification in a (A) Palmitoylation (B) glycosylation (C) peptidylation (D) phosphorylation 68. Haemoglobin has characteristic circular dichroism (CD) peaks C in (A) aromatic amino acid residues. (B) heme group. (C) heme and aromatic amino acid residues. (D) peptide bonds and aromatic amino acid residues. 69. A weed is assumed to be dispersed randomly in a meadow. What statistical distribution will describe the dispersion correctly? (A) Binomial (C) Negative Binomial (B) Poisson (D) Normal Exp. A weed is assumed to be dispersed randomly in a meadow. Poisson statistical distribution will describe the dispersion correctly. 70. Co-localization of two fluorescently labeled proteins in an organelle in cells is usually visualised by (A) interference-contrast microscopy. (B) scanning electron microscopy. (C) confocal microscopy. (D) atomic force microscopy. P7/7

195 Amar Ujala Education PART-C A/UG Tryptic digest of a heptapeptide (built from 3 lysine (K), 2 alanine (A), 1 tyrosine (Y) and 1 phenylalanine (F) yielded tri and tetrapeptide. Which of the following is the correct sequence of the heptapeptide? (A) KAYAKFK (B) YKAAFKK (C) KYKAAKF (D) KYAAKFK ml of 0.1 M sodium acetate solution has a ph of To this solution 1000 μl of 1M acetic acid (pka = 4.76) of ph 2.80 is added. The ph of this mixture will be: (A) 8.90 (B) 4.76 (C) 2.80 (D) What are A, B and C in the following reactions? Glucose Glucose 6-phosphate Glucose1-phoshateFructose6-phosphate6-phosphogluconate A B C (A) Pyruvate, ribose 5-phosphate, glycogen. (B) Ribose 5-phosphate, glycogen, pyruvate. (C) Glycogen, pyruvate, ribose 5-phosphate. (D) Glycogen, citrate, ribose 5-phosphate. 74. Michaelis-Menten enzyme kinetics for a simple reaction involving an enzyme (E) and substrate (S) is given by the scheme: Km Kcat E+S ES E+P. description of K m,k cat and their relationship is provid in the following statements. (A) Km represents association constant of the ES complex. (B) Km represents dissociation constant of the ES complex. (C) the rate constant for the chemical conversion of the ES complex to substra te bound enzy me and product. (D) k cat /K m is a rate constant that refers to the properties and reactions of the free enzyme and free substrate. Which of the combinations of above statements is true? (A) AandC (B) BandD (C) AandD (D) BandC 75. A 26-residue peptide composed of alanine and leucine shows a circular dichroism (CD) spectrum characteristic of a-helix at 50 C in 5 mm phosphate buffer at ph 7.4. Deconvolution of the spectrum peptide solution is cooled gradually to 25 C, and the CD spectra are that (A) the % helical content will decrease and % random conformation will increase. (B) the % helical content will increase and % random conformation will decrease. (C) there will be transition from a-helix to ß- sheet. (D) here will be transition from a-helix to ß-hairpin. 76. It has been observed that for the DNA double helix melting, the value of H (enthalpy change of dena turation) are 80 and 90 kcal/mole at 70 and 80, respectively. Assuming that Cp (constant-pressure heat capacity change) is independent of temperature, estimate H associated with the denaturation of DNA at 37 C. This estimated value of H (kcal/ mole) is (A) 27 (B) 37 (C) 47 (D) 57 Ans. (?) 77. Using FRA P (Fluorescence Recovery After Photobleaching) techniques, diffusion coefficient of three inte gral membrane s proteins M 1,M 2 and M 3 in a kidney cell is calculated as 1 μm/s, 0.05 μm/s and0.005 μm/s, receptively. Considering fluid-mosaic nature of biological membrane and relationship of structural organization of integral membrane and relationship of structural organization of integral membrane and relationship of structural organization of integral (A) M 2 andm 3 2. M (B) M 2 only (C) M 3 only (D) M 1 and M One highly pathogenic DNA virus enters into the host cells by endocytosis replicates in the nucleus followed by cell lysis. You have drugs at you disposal that bloks. A. acidification of vesicles. B. mitochondrial transport. C. nuclear export. D. exocytosis. Identify the right combination to prevent the infection. (A) AandB (B) BandD (C) AandC (D) AandD Exp. One highly pathogenic DNA virus enters into the host cells by endocytosis replicates in the nucleus followed by cell lysis. We have drugs at you disposal that bloks : acidification of vesicles and nuclear export. 79. When cells enter mitosis, their ex isting array of cytoplasmic microtubules has to be rapidly broken down and replaced with the mitotic spindle, which pulls the chromosomes into the daug hter cells.the enzyme Katanin is activated during the onset of mitosis and chops microtubules into short pieces. The possible fate of the microtubule fragments created by Katanin will be P7/8

196 A/UG-15 (A) depolymerization (B) aggregation. (C) degradation. (D) translocation. Exp. When cells enter mitosis, their ex isting array of cytoplasmic microtubules has to be rapidly broken down and replaced with the mitotic spindle, which pulls the chromosomes into the daug hter cells.the enzyme Katanin is activated during the onset of mitosis and chops microtubules into short pieces. The possible fate of the microtubule fragments created by Katanin will be depolymerization. 80. A bacterial strain can use carbohydrates and hydrocarbons as growth substrates. The strain uses glucose following a minimal lag period after inoculation, regardless of the other carbohydrates and hydrocarbons in the growth medium. The following observations were also made. (a) In the absence of glucose, lactose is used after a lag period of about three times as long as the lag period for glucose utilization. (b) The presence of hydrocarbons does not affect the lag period for the utilization of lactose. (c) The utilization pattern for all hydrocarbons is similar to that of lactose. (d) Branched hydrocarbons are not immediately utilized if straight chain hydrocarbons are initially present. Which one of the following specific regulatory mechanisms is consistent with the above observations related to carbohydrate and hydrocarbon utilization? (A) Diauxie (B) End point repression (C) Catabolite repression (D) Transcription attenuation 81. Cell cycle is regulated by various cyclins and cyclin dependent kinases (CDK). On receiving mitotic stimuli, cyclin D, the first cyclin expressed, binds with existing CDK4 to form the active cyclin D-CDK4 complex. This in turn phosphorylates retinoblastoma protein (Rb) which activates E2f to further activate the transcription of various downstream cyclins. In a particular cell type there is a mutation in Rb such that it cannot be phosphorylated. What will be the correct expression pattern of cyclin E in these cells after mitotic stimulation? (A) (C) Expreselon of cycin E Expreselon of cycin E TIME TIME (B) (D) Expreselon of cycin E Expreselon of cycin E Amar Ujala Education TIME TIME 82. A bacterial culture was in log phase in the following figure. At time x, an antibacterial compound was added to the culture. Number of cells (logs scale) Time Which of the following lines in the growth curve represents the antibacterial activities of the compound? (A) (C) Number of cells (log scale) Number of cells (log scale) X Time Bacteriostatic Bactericidal Bacteriostatic X Time Bactericidal (B) (D) Number of cells (log scale) Number of cells (log scale) X Bacteriostatic Bactericidal X Time X Time Bacteriostatic P7/9

197 Amar Ujala Education 83. In a cell free extract containing DNA poly merase I, Mg dgtp, dctp and dttp (3H), the following DNA molecules were added: a. Single stranded closed circular DNA molecule containing 824 nucleotides. b. Single stranded closed circular DNA molecule having 1578 nucleotides base paired with a linear single standard DNA molecule of 824 nucleotides having a free-3'-oh group. c. Double stranded linear DNA molecule containing 1578 nucleotides having free-3' - OH group at both ends. d. Double stranded closed circular DNA molecule having 824 nucleotides. The rate of DNA synthesis was measured by incorporation of thymidine in the DNA molecule and expressed as the percentage of DNA synthesis relative to total DNA input. Which one of the following graphs represents the correct result? (A) 100% DNA Synthesis (C) 100% DNA Synthesis Incubation time (min) Incubation time (min) (B) 100% DNA Synthesis (D) 100% DNA Synthesis Incubation time (min) Incubation time (min) 84. The complex responses to different types of DNA damage in both prokaryotes and eukaryotes fall into three main categories: (i) damage bypass (ii) damage reversal (iii) damage removal Many repair proteins are isolated like (a) DNA methyl transferase (b) DNA glycosylase (c) DNA polymerase IV Which one of the following represents the correct combination? (A) (i) - (a), (ii) - (b), (iii) - (c) (B) (i) - (b), (ii) - (c), (iii) - (a) (C) (i) - (c), (ii) - (a), (iii) - (b) (D) (i) - (c), (ii) - (b), (iii) - (a) P7/10 A/UG In recent years, genome-wide transcription study using high throughput sequence analysis has revealed some novel results that include: (i) presenceofrnapolymeraseinbothintra- and intergenic regions of the genome (ii) existence of non-coding R NAs generated from mrna coding genes. (iii) existence of sense and antisense transcripts generated from the promoter and untra-nslated region of many annotated genes. Possible interpretation of the above results are: A. RNA polymerase can loosely bind to any part of the genome but its affinity becomes strong only when it reache s the promoter. B. Binding of RNA polymerase to non-promoter regions of the genome leads to the generation of various noncoding regulatory RNAs. C. Non-coding RNAs are generated from mrna coding genes due to aberrant transcription initiation and termination. D. Sense and antisense transcripts are generated from the promoter and untranslated regions of protein coding genes by a novel mechanism of bidirectional transcription. Identify the correct combination of the above interpretations: (A) A and B (B) B and D (C) A and D (D) B and C 86. Aminacyl trna synthetases face two important challenges: (i) They must recognize the correct set of trnas for a particular amino acid. (ii) they must charge all of these isoaccepting trnas with the correct amino acid. Both of these processes are carried out with high fidelity by the following possible mechanisms: A. The discrimination ability resides predominantly at the acceptor stem of the trnas. B. The specificity is contributed by the anticodon loop in trnas. C. The specificity is e mbedded in the amino acyl synthetase at the N Terminus D. The specificity is contributed by the variable loop of the trna. Which of the following is correct? (A) A and B (B) A and C (C) B and C (D) A and D

198 A/UG During heat shock, mammalian cells shut down global protein During heat shock, mammalian cells shut down global protein molecular regulation(s) tha t could explain the phenomenon are: A. mrna of all proteins, except those of Hsps, undergoes degradation during heat shock. B. Cap-dependent translation of most mr NAs is affected during heat shock due to denaturation of cap binding protein, eif-4e. C. Translation initiation of Hsp mrnas takes place through their internal ribosome entry sites (IRES) D. Hsp mrnas are abundant during heat shock and thus they compete out other mrnas for ribosome binding and translation. Which of the following sets is correct? (A) AandD (B) BandC (C) CandD (D) AandD 88. Bacteriophage is a temperate bacteriophage and has two modes in its life cycle, lysogenic and lytic. Several genes are involved in these two processes like N, ci, cii, ciii,q,int,xis, etc.which one of the following diagrams represents the control mechanism correctly? (A) N (B) N (C) (D) N CIII CII CIII Q (+)ve (Lytic) CII CI CIII XIS.Q (Lytic) CI CII CIII XIS.Q CII CI (Lysogeny) INT INT (Lysogeny) (Lysogeny) INT (Lytic) CI INT (Lysogeny) XIS N Q (Lytic) 89. Clearance of phagocytosed intracellular parasite like Leishmania requires the involvement of reactive oxygen species (ROS) and reactive nitrogen species (RNS). Administration of IFN- to macrophages harbouring an intracellular pathogen leads to the production of ROS and RNS by JAK/STAT pathway. A macrophage cell line J774 infected with Leishmania is given the following treatments. A. IFN- B. IFN- +AMT, a potent inos inhibitor. C. IFN- + a pocyanin, a NADPH oxidase inhibitor. D. IFN- + NMM (N-monomethyl arginine), an arginine analogue. Amar Ujala Education What will be the most appropriate graph showing the survival of parasites after these treatments? (A) 100 %survtvalofparasltes 50 (C) 100 %survtval of parasltes 50 A B A B C D C D (B) 100 % survtval of parasltes %survtval of parasltes 50 (D) A B A B C D C D 90. A ligand recognizes two different cell surface receptors, A and B, on the same cell type. Receptor A, after binding with the ligand is internalized along with the ligand whereas receptor B, after binding with the ligand, initiates tyrosine kinase activity of the intracellular domain. One particular disease is associated with the loss of receptor- mediated signal transduction of the ligand. Different observers inferred that the disease may be resulted due to A. loss of binding affinity of receptor A due to mutation in the extracellular domain. B. loss of binding affinity of receptor B due to mutation in the extracellular domain. C. mutation in the tyrosine kinase domain rendering it inactive. D. mutation in the intracellular domain rendering it incapable of endocytosis. Which combination of the above inferences do you think is appropriate for the cause of the disease? (A) AandB (B) BandC (C) CandD (D) AandD 91. Vascular endothelial (VE)-cadherin is an important cell adhesion molecule for endothelial cells. Endothelial cells that are unable to express VE-cadherin still can adhere to one another via N-cadherin (neural cadherin), but these cells do not survive. Which of the following is the most appropriate reason for this? (A) N-cadherinusesVE-cadherinasco-receptorfor adhesion. (B) VE-cadherin acts as co-receptor for VEGF (vascular endothelial growth factor) mediated signal transduct ion in endothelial cells. (C) VE-cadherin is importa nt for desmosome formation and interaction of intermediate filaments. (D) Loss of VE-cadherin impairs Cahomeostasis of vascular endothelial cells leading to their death. P7/11

199 Amar Ujala Education Exp. Vascular endothelial (VE)-cadherin is an important cell adhesion molecule for endothelial cells. Endothelial cells that are unable to express VE-cadherin still can adhere to one another via N-cadherin (neural cadherin), but these cells do not survive. VE-cadherin acts as coreceptor for VEGF (vascular endothelial growth factor) mediated signal transduct ion in endothelial cells is the most appropriate resons for this fact. 92. An important role of Fas and me diate elimination of tumor cells by killer lymphocytes. In a study of 35 primary lung and colon tumors, half the tumors were found to have amplified and overexpressed a gene for a secreted protein that binds to Fas ligand. The main reason for survival of these tumor cells by this secreted Fasligand bindingprotein may be attributed to its (A) decoy receptor activity. (B) anti-proliferative activity. (C) cellular defense activity against cytotoxic killing. (D) anti-contact inhibition activity. Exp. An important role of Fas and me diate elimination of tumor cells by killer lymphocytes. In a study of 35 primary lung and colon tumors, half the tumors were found to have amplified and overexpressed a gene for a secreted protein that binds to Fas ligand. The main reason for survival of these tumor cells by this secreted Fasligand bindingprotein may be attributed to its decoy receptor activity. 93. A BALB/c mouse was thymectomized on the first day after birth (mouse 1) whereas a nother was thymectomized on day 7 after birth (mouse 2). A third mouse underwent the same operation on day 21 after birth. After 56 days, sera were prepared from these mice and also from control mice, which had sham operation. The sera were checked for anti-dna antibodies. Which one of the following observations is the most plausible? (A) 1 Both mouse 1 and mouse 2 had anti-dna antibodies but mouse 3 did not have anti-dna antibodies. (B) Only mouse 1 had anti-dna antibodies. (C) Only mouse 3 had anti-dna antibodies. (D) Only the control mice had anti-dna antibodies. 94. A potentially valuable therapeutic approach for killing tumour cells without affecting normal cells is the use of immunotoxins. Immunotoxins consist of particular cellspecific monoclonal antibodies coupled to lethal toxins. Which of the following molecular approaches is NOT appropriate for killing tumor-cells? (A) Cell surface receptor binding polypeptide chain of toxin molecules should be replaced by monoclonal antibodies which are specific for a particular tumor cell. P7/12 A/UG-15 (B) Constant region Fc domain of tumor cell-specific monoclonal antibody should be replaced by toxin molecules. (C) Variable region F(ab) domain of tumor c specific monoclonal antibody should be replaced b toxin molecules. (D) Inhibitor polypeptide chain of toxin molecules should be conjugated to the F(ab) domain of monoclonal antibody tumor-specif monoclonal antibodies. Exp. A potentially valuable therapeutic approach for killing tumour cells without affecting normal cells is the use of immunotoxins. Immunotoxins consist of particular cellspecific monoclonal antibodies coupled to lethal toxins. Variable region F(ab) domain of tumor c specific monoclonal antibody should be replaced b toxin molecules molecular approaches is NOT appropriate for killing tumor-cells. 95. A set of experiments that were carried out to demonstrate the effect of Apical Ectodermal Ridge (AER) of the chick limb bud on the underlying mesenchy meare enlisted below, along with their expected outcomes: A. Removal of the AER of forelimb leads to cessation of limb development. B. If an extra AER is placed in the forelimb bud, duplication of the distal region of the wing takes place. C. If an extra AER is placed in the forelimb bud, a leg develops instead of a wing. D. If AER of forelimb bud is replaced with beads soaked in FGF2, a normal wing develops. E. If a non limb mesenchyme is placed below ane AER, the AER directs the mesenchyme to form a normal wing. Which of the above statements are correct? (A) A, C and E (B) C, D and E (C) B, D and E (D) A, B and D 96. Hensen s node is established as the avian equivalent of the amphibian dorsal blastopore lip. The following observations are presumed to be support of the same. A. It is the re gion whose cells induce and pattern a second embryonic axis when transplanted into other locations of the gastrula. B. It is equivale nt in terms of tissue structure. C. It expresses the same marker genes as the Spemann s organizer in Amphibians. D. The same micro RNA can interfere with the formation of pre-chordal plate in both Hensen s node and Spemann s organizer. Choose the correct set among the following : (A) AandD (B) AandC (C) BandC (D) AandB

200 A/UG In Amphibians, when due to some injury, the eye lens is damaged, the fully differentiated iris cells can regenerate the lens. It is achieved through the possible processes: A. Iris cells through some signaling undergo dedifferentiation and transdifferentiation into lens cells to regenerate the lens. B. Iris cells transform into lens cells spontaneously. C. Iris cells induce in a stepwise manner, specific genes responsible for their dedifferentiation and then conversion to lens cells. D. Stem cells present in iris tissue differentia te into lens cells. Which of the following is correct? (A) AandB (B) AandC (C) BandD (D) BandC 98. The control of flowering is a complex process inv olving several key regulatory genes. Some statements on flower development are given below: a. Two major types of genes regulate floral development: meristem identity genes and floral organ ide ntity genes. b. The important genes in Arabidopsis that play key regulatory roles in meristem identity are: APETALA1, LEAFY and SUPPRESSOR OF CONSTANS1. c. The genes that determine floral organ identity were discovered as floral homeotic mutants. d. Most plant homeotic genes belong to a class of related sequences known as FAD box genes. Which one of the following combinations of the above statements is correct? (A) a,bandc (B) b,candd (C) a,candd (D) a,bandd 99. A few state ments on early developmental stages in plants are given below: a. The cells of flower are diploid in nature. b. Only some specialized cells in reproductive organs undergo meiosis produce haploid cells. c. The haploid cells produced in (b) above, undergo a few normal mitotic cell divisions. d. All the progeny cells produced in (b) above, differentiate either into haploid egg cells or into haploid sperm cells. Which one of the following combinations of t above statements is correct? (A) a,bandc (B) b,candd (C) a,candd (D) a,bandd 100. Cells from an early frog blastula were remove from the animal pole and used to replace cells fror the vegetal pole of the blastula. The following eve nts may be expected. a. Transplanted cells would develop normally as part of the cells of the vegetal pole. Amar Ujala Education b. Transplanted cells would develop as cells of the animal pole of the adult on the vegetal pole. c. Region of the animal pole from where the cells were removedwouldbemissingintheadult. d. Remaining cells in the animal pole would compensate for the cells that were removed. Which of the following are true? (A) b,candd (B) a,bandd (C) a,bandc (D) a,candd 101. Capacitation of mammalian sperms allows them to be activ ated within the uterus and facilitate fertilization. The following sta tements were ma de regarding events occurring during capacitation: a. Removal of cholesterol from sperm head. b. Removal of non-covalently bound glycoproteins. c. Increased expression of fibronectin. d. Decreased permeability of calcium ions. Identify the correct statements: (A) b,candd (B) a,bandd (C) a,bandc (D) a,candd 102. Asada-Halliwell pathway protects plants against oxidative stress during unfavorable environmental growth regimes. The following are some statements related to the stress-tolerance mechanism through this pathway in plants. a. Oxygen accepts electrons as an alternative electron acceptor. b. Hydrogen peroxide is reduced by catalase to form water. c. Ascorbate is oxidized and re generated. d. Glutathione is oxidized and reduced. Which one of the following combinations of the above statements is true? (A) b,candd (B) a,candd (C) a,bandd (D) a,c andd 103. The following are certain facts regarding biological cnitrogen fixation in plants: a. Oxygen irreversibly inactiv ates nitrogenase enzy me involved in nitrogen fixation. b. The nod genes that code for nodulation proteins are activated by NodD c. The two components of nitrogenase enzyme complex, the Fe protein and MoFe protein, can show catalytic activity independently. d. During the reaction cataly zed by nitrogenase enzyme, the Fe protein reduces the MoFe protein while the MoFe protein reduces N Which one of the following combination of the above statements is correct? (A) a,bandc (B) b,candd (C) a,candd (D) a,b andd P7/13

201 Amar Ujala Education 104. Secondary metabolites are div erse array of organic compounds in plants. The following are certain statements about secondary metabolites: a. They protect plants against being eaten by herbivores and againstbeinginfectedbymicrobialpathogens. b. Terpenes, the largest class of secondary metabolites are synthesized by methyl erythritol phosphate(mep) pathway and shikimic acid pathway. c. The most a bundant classes of phenolic compounds in plants are derived from phenylalanine. d. Alkaloids are nitroge n containing secondary metabolites in plants. Which one of the following combinations of the above statements is correct? (A) a,bandc (B) b,candd (C) a,candd (D) a,bandd 105. The following are some statements about long distance translocation of photoassimilates in higher plants: a. Sugars are translocated in the phloem by mass transfer along a hydrostatic pressure. b. Gibberellic acid stimulates the unloading of sugars from phloem tissue into apoplasts. c. Munch pressure-flow hypothesis is crucial to drive translocation in the phloem. d. Allocation and partition of carbon within a source leaf determine the phloem loading phenomenon. Which one of the following combinations of the above statements is true? (A) a,bandd (B) a,bandc (C) a,candd (D) b,candd 106. Directional growth of plants induced by lig ht is called phototropism. Some statements on phototropism are given below: a. Phototropism is a photomorphogene tic response. b. PHOT1 and PHOT2 genes mediate phototropism. c. CRY1 and CRY2 genes although help to perceive blue light are not involved in phototropism. d. Perception of blue light by phya photoreceptor initiates phototropism. Which one of the following combination of the above statements is correct? (A) a, b and c (B) b, c and d (C) a, c and d (D) a, b and d 107. Plants make several hormones that are important for growth and development. Some statements on plant hormones are given below: a. Auxin is produced primarily in the root apices b. Cytokinins are a smaller group of related compounds. P7/14 A/UG-15 c. Gibberellins are a large group of related compounds defined not by their biological functions but by their structures. d. Brassinosteroids are a n important class of plant hormones, which control a broad spectrum of developmental reposes including pollen tube growth. Which one of the following combination of the above statements is correct? (A) a,bandc (B) b,candd (C) a,candd (D) a,bandd 108. An action potential was generated on a nerve fibre by a threshold electrical stimulus. When a second stimulus was applied, no matter how strong it was, during the absolute refractory period of the action potential, the nerve fibre was unable to generate second action potential. This observation was explained in the following statements: a. A large fraction of potassium channel was voltage inactivated b. The critical number of sodium channels required to produce an action potential could not be recruited. c. A large fraction of sodium channel was voltage inactivated. d. The critical number of potassium channels required to produce an action potential could not be recruited. Which one of the following is true? (A) Only a (B) aandb (C) Onlyc (D) candd 109. Water and electrolytes like Na + and clare lost from the body indiarrhoea. Oral administration of NaCl solution in this condition does not improve the situation. When glucose is a dministered with normal NaCl solution through oral route, the absorption of electrolytes along with water occurs and the patient recovers. a. Glucose enhances ATP production in the mucosal cells of small intestine and thus facilitates sodium absorption. b. Glucose inhibits the diarrheal toxin induced camp production in the mucosal cells of small intestine. c. Na + is co-transported with glucose on the apical surface of the mucosal cells of small intestine. d. The epithelial sodium channels (ENaC) are activated by glucose in colon. Which one of the following is true? (A) Only a (B) a and b (C) Only c (D) c and d

202 A/UG A patient has episodes of painful spontaneous muscles contraction, followed by periods of paralysis of the affected muscles. It was identified as primary hyperkalemic paralysis, an inherited disorder. The possible causes of the paralysis are a. The elevation of extracellular K + causes hyperpolarization of skeletal muscle cells. b. The hyperpolarization of the muscle cell membrane inactivates sodium channels. c. The elevation of extracellular K + causes depolarization of skeletal muscle cells. d. The sodium channels are voltage inactivated in depolarized state Which one of the following is true? (A) Only a (B) a and b (C) Only c (D) c and d 111. My oglobin (Mb) in muscles, Hemoglobin A (HbA) in adult RaC, Hemoglobin C (HbC) in patients with thinner RBC and He moglobin S (HbS) in sickle cell disease are four different hemoproteins. Oxygen saturation kinetics of these four proteins is different. Which of the following profile is most plausible? (A) (C) (B) Pertial pressure of oxygen [mm Hg] (D) 100 Amar Ujala Education 112. Maintaining the salt concentration and volume plasma are two key parameters for physiological processes achieved by kidney. Which of following structural and functional combines the most efficient renal regulatory system of mammals? Combination Structural Functional No. (A) Large glomerulus, long Trans-epithelial proximal and distal in proximal tubules, long Henle s countercurrent loop multiplier, ADH responsiveness of distal tubule (B) Small glomerulus, short Transepithelial proximal and distal potential in distal tubule short Henle s loop tubule, very high ADH concentration in circulation (C) Very large glomerulus, short Very efficient proximal tubule, glomerular very long distal filtration, prevention tubule, of solute loss long Henle s loop (D) Small glomerulus, long Preventing water proximal and distal and solute tubules, Henle s loop filtration, e xcreting solute, lowering ADH responsiveness 113. Which one of the following graphs best repre-sents the hormone profile in a rat right after matting? (A) Expreselon of cycin E TIME (B) Expreselon of cycin E TIME Pertial pressure of oxygen [mm Hg] Mb HbA Hbc HbS (C) Expreselon of cycin E TIME (D) Expreselon of cycin E TIME P7/15

203 Amar Ujala Education 114. Which of the following representations of chromosomal arrangement in meiotic metaphase I best explains the Law of Independent Assortment? % of cells with arrangement (A) (B) (C) (D) 50% 50% 50% AND AND AND OR 50% 50% 50% 100% 100% 115. Which of the following statements are true Robertsonian translocations? a. The size of the homologous chromosome involved in translocation will differ. b. Genes on the chromosome involved in translocation will show linkage with genes with which it normal independently assorts. c. There will be change in the physical map, but no change in the genetic map. d. It can be identified by G-banding chromosomes. e. It can be identified by C-banding chromosomes. f. ItcanleadtoDownsyndrome Which one of the following combination is correct? (A) a,c andd (B) a,dandf (C) a,b,dandf (D) a,c,eandf 116. A mes test is used to evaluate mutagens in the environment. Which of the following statements, about A mes test are true? a. The mutagenic effect of a compound tested using an auxotrophic strain of b. The mutagenic effect of a compound is teste d using His strain of Escherichia coli c. Using appropriate strains, compounds causing base substitutions and frame shift d. Liver enzymes are important as they are activated by test compound to ev aluate its mutagenicity potential. e. Many compounds may have to be converted to bioactive metabolites, which is carried out by the enzymes from the liver A/UG-15 (A) a,c andd (B) a,bandd (C) a, c, and e (D) a, and e only 117. The genetic map of three genes in Drosophila melanogaster given below: a b c b + 10 C M 5 C M A cross, as given below individuals of the genotype: a + b + c a b X a + b + c a b The female F1 progeny are test-crossed a nd 1000 progeny are obtained. Assuming that there has been no double crossover, what is the expected number of progeny withthegenotypes: Select the set which shows the correct number of expected progeny. (a) (b) (c) a + b c + a + b + c + a + c a b c a b c a b c IFAS Jodhpur Select the set which shows the correct number of expected progeny. Set (a) (b) (c) (A) (B) (C) (D) The following pedigree shows the inheritance pattern of a rare recessive disorder with complete penetrance. A child from marriage between indiv iduals II- disorder only if the parents carry the recessive allele. What is the 2 & II-3 will show the probability that the child will show the disorder? (A) 1/9, and the probability of the parents carrythe recessive allele is 2/3 (B) 1/4, and the probability of the parents carry the recessive allele is 3/4. (C) 1/16, and the probability of the parents carry the recessive allele is 2/3. (D) 1/64, and the probability of the parents carry the recessive allele is 3/4. P7/16

204 A/UG DNA from a strain of bacteria with genotype b + c + d + e + was isolated and used to tra nsform a strain of bacteria that was a - b - c - d - e -.The transformed eel were tested for the presence of donat ed genes. The following genes are found be co-transformed i. a+andd+ ii. b+ande+ iii. c + and d + iv. c + and e + The order of genes on the bacterial chromosome is (A) a-b-c-d-e (B) a-d-c-e-b (C) a-c-d-e-b (D) a-d-b-e-c 120. Coelomates have a. fluidfilledbodycavity. b. a complete lining called peritoneum, derived from mesoderm covering the body cavity. c. a complete lining called peritoneum, derived from ectoderm covering the body cavity. d. Round worm as representative of this group. e. Flat worm as representative of this group. Select the correct combination (A) a,candd (B) a,cande (C) a,bandd (D) aandb 121. Most biologists agree that seaweeds are protists. Some biologists think that at least some seaweeds should be considered plants, not protists. Which of the following would support the latter one? (A) Certain seaweeds contain several kinds of specialised cells (B) Certain seaweeds have multicellular organization. (C) Certain seaweeds are found to be prokaryotic. (D) Certain seaweeds undergo sexual and asexual reproduction. Exp. Most biologists agree that seaweeds are protists. Some biologists think that at least some seaweeds should be considered plants, not protists. Certain seaweeds contain several kinds of specialised cells that would support the latter one Which of following shows the correct systematic hierarchy? (A) Kingdom Phylum Subphylum Superclass Class Subclass Cohort Order Suborder Super family Family Subfamily Genus Subgenus Species Subspecies (B) Kingdom Phylum Subphylum Cohort Superclass Classo btained Assuming that there has been no double crossover, what is Subclass Superfamily Family Subfamily Order Suborder Genus Subgenus Species Subspecies (C) Kingdom Phylum Class Order Cohort Family Genus Species (D) Kingdom Phylum Class Cohort Family Order Genus Species P7/ Match the following : Disease Amar Ujala Education Pathogen/Causative factor A. Creutzfeldt Jakob (i) Fungi B. Pneumocystis (ii) Virus C. Legionnaires diseases (iii) Prion D. Rabies (iv) Bacteria (A) A-iv, B- iii, C- ii, D 1 (B) A-iii, B-i, C-,iv, D-ii (C) A-i, B-ii, C - iii, D iv (D) A-ii, B-iv, C-i, D iii 124. In the following diagram, two models succession are represented. In this diagram A, C and D are species and arrows indicate replaced by Based on the above, which statement is correct. (i) A B C D A B (ii) C D (A) Fig.(i) represents facilitation and Fig.(ii) represents tolerance model (B) Fig.(i) represents tolerance model Fig. (ii) represents facilitation model (C) Fig.(i) represents facilitation m and Fig.(ii) represents inhibition model (D) Fig.(i) represents tolerance model Fig.(ii) represents inhibition model 125. Lindeman s efficiency between trophic levels is depicted by the formula : Efficiency = A/B Where, A and B, respectively are : (A) assimilation at trophic level n and assimilation at trophic level n - 1' (B) intake at trophic level n and assimilation at trophic level n-1 (C) assimilation at trophic level n and net productivity at trophic level n - 1' (D) intake at trophic level n and productivity at trophic level n - 1' 126. The following matrix shows the relationship between probability of death and duration of species association. Duration of Association Low High A C B D Low High Probability of Death In the above, A, B, C and D are (A) A - Parasites, B = Parasitoids, C - Grazers, D-Predators (B) A - Carnivores, B = Herbivores, C- Parasites, D - Parasitoids (C) A - Grazers, B = Parasites, C - Herbivores, D -arasitoids (D) A-Predators, B=Parasitoids,C-Parasites, D-Carnivores:

205 Amar Ujala Education 127. An observation was made on a species experiencing three factors A, B and C in order to infer a density dependent population regulation by a factor. The following graph shows the relationship between the adverse effect of the factors in terms of number and population density.based on the above observation, which of the following is correct? populatlon Density A B (A) A - Density independent; B = Density de pendent; C - Inv ersely density dependent (B) A - Inversely density dependent; B- Density independent; C - Density dependent (C) A - Density dependent; B = Inversely density independent; C - Density independent (D) A - Density dependent; B = Density independent; C - Inversely density dependent 128. Which of the following biotic provinces are part of Deccan Peninsula biogeographic zone of India? (A) Malabar Coast, Western Plateau, Eastern Plateau (B) Western Ghats, Ce ntral Plateau, Eastern Plateau (C) Central Plateau, Eastern Plateau, chhota Nagapur (D) Central Plateau, Malabar Coast, Western Ghats 129. Which one of the following plants group combinations reflects the correct increasing order of the number of species it has? (A) Gymnosperms, Bryophytes, Algae, Angiosperms (B) Angiosperms, Algae, Fungi (C) Algae, Bryophytes, Gy mnosperms, Angiosperms (D) Angiosperms, Gymnosperms, Bryophytes, Algae 130. Micro-evolution is the term used for change s in allele frequencies that occur over time. a. Within a population at species level b. within a community at genus level c. due to appearance of new genes infections d. due to mutation, natural selection, flow and genetic drift Which of the following combinations is NOT appropriate? (A) aandc (B) aandd (C) bandc (D) band d C P7/18 A/UG The following genotypes were observed in a population Genotype Number HH 90 Hh 60 hh 50 Which of the following is the correct frequency of H allele and what will be the expected number of HH in the given population? (A) 0.60 and 72 (B) 0.80 and 96 (C) 0.50 and 32 (D) 0.30 and The first vertebrate animal appeared in which of the following geological ages? a. Paleozoic era b. Mesozoic era c. Ordovician period d. Cretaceous period e. Mississippian epoch f. Paleocene epoch Which of the following combinations give the best answer? (A) a,candf (B) aandf (C) b,dande (D) aandc 133. Fore limb of huma n and flippers of whale are embryologically homologous structures. What does the study of homologous structures tell us about evolution? A. This is the example of adaptive radiation, occurred due to similar group of organisms inhabiting different environments B. This is the example of divergent evolution, occurred due to similar group of organisms inhabiting different environments C. Similar group of organisms with mutations and variations getting naturally selected in different environments D. This is the example of convergent evolution, occurred due to similar group of organisms inhabiting different environments Which of the following is the correct combination? (A) A,BandC (B) AandB (C) BandD (D) OnlyD 134. Foster pups were presented to a primiparous rat at the mid gestation period. Which of the following behaviour will be found in the pregnant rat? (A) The rat shows maternal behaviour after few days of presentation of pups. (B) The rat attacks and kills the pups time they are presented. (C) The ra t rejects the pups after repeated presentation. (D) The rat shows fear response for a days Exp. Foster pups were presented to a primiparous rat at the mid gestation period. The rat shows maternal behaviour after few days of presentation of pups. behaviour will be found in the pregnant rat.

206 A/UG Electrons are transferred from reduce d enzy mes like NADH, NADPH to pyruvic acid its derivatives during fermentation. Those electron acceptors are reduced to the end-product for example latic acid, propionic acid, etc. end products depend on the particular microorganism and the substrate Organism End product (s) A Streptococcus a. Butane diol, formic acid B Clostridium b. Lactic acid C Salmonella c. Butyric acid and iso proanol D Enterobacter d. Succinic acid, acetic acid The correct match is (A) A-a,B-b,C-c,D d (B) A-d,B-c,C-a,D-c (C) A-b,B-c,C-d,D-a (D) A-c,B-d,C-a,D b 136. In resting cells, proteins X and Y are localized in the cytosol. Upon stimulation with lipopolysaccharide (LPS), both of them are phosphorylated and translocate to the nucleus. You have used antibodies against phosphorylated forms of proteins X and Y which are conjugated to either red, or green or blue dye. Keeping optical aberration of light in mind, which one of the following will be the best for visualizing X and Y In the nucleus by fluorescence microscopy? (A) Anti green X and anti red Y (B) Anti red X and anti green Y (C) Anti red X and anti blue Y (D) Anti blue X and anti green Y 137. A transgenic plant is developed with the following T- DNA construct RB Gene A Probe A EcoRI Gene B Probe B LB 8kb 6kb In order to analyze the nature of integration, genomic DNA digested with EcoRI was use d for Southern hybridization using either probe A or B. The result obtained is as Probe A Probe B kb P7/19 Amar Ujala Education The following conclusions were made: a. There are two copies of the T-DNA cassette integrated at one loci and a third copy at another loci. b. There are two copies of the T-DNA cassette integrated at one loci. c. Complete T-DNA cassette has been integra ted in all cases. d. In one T-DNA cassette there is a truncation towards the RB e. In one T-DNA cassette there is a towards the LB f. The arrangement of the T-DNA cassettes integrated at the same loci is RB LB LB RB g. The arrangement of the T-DNA cassette integrated at thesamelociis RB LBRB LB Which of the above are correct? (A) b,eandg (B) b,dandf (C) a,dandf (D) b,candf 138. While designing an experiment for Agrobacterium mediated plant transformation, a student noted down the following points: a. Ti and Ri plasmids induce crown gall and hairy root disease, respectively b. Enzymes octopine syntha se and nopaline synthase involved in the synthesis of octopine and nopaline, respectively are encoded by T-DNA. c. All the six v ir genes, vir A, vir B, vir C, v ir D, vir E and vir G are absolutely required for virulence. d. Almost perfect 25 bp direct repeat sequences fla nking all Ti and Ri plasmids in the T-DNA region is essential for T-DNA transfer. Which one of the following combinations of the above statement is correct? (A) a,bandc (B) b,candd (C) a,candd (D) a,bandd 139. The following are certain facts regarding bioremediation: a. Biodegradable plastics are made using polyhy droxy alkanoates (PHAs) such as polyhydroxybutyrate (PHB). b. Pseudomonas putida F1 bacterial strain is involved in degradation of aromatic hydrocarbon. c. The bacterium Deinococcus radiodurans consume and digest toluene and ionic mercury from highly radioactive nuclear waste. d. Bioaugmentation is a process of improving the microorganisms already existing in the system for degradation of xenobiotic compound. Which one of the following combination of above statements is correct? (A) a,bandc (B) a,bandb (C) a,candd (D) b,candd

207 Amar Ujala Education 140. An unknown pe ptide was isolated from the leaf of a medicinal plant and purified to homogeneity. The peptide did not yield any sequence when subjected to Edman degradation. However, tryptic digest of the peptide produced a unique sequence. The mass of the intact peptide was found to be 18 Da less than that obta ined from the trypsin treated sample. The possible interpretation of the above experimental results could be that the a. the N-terminus of the peptide wa s blocked by acetylation or methylation. b. the peptide was cyclic and contained at least one internal arginine or lysine residue. c. the peptide was cy clic and contained a lysine or arginine residue at the C terminus. d. the peptide was cyclized by peptide bond formation between a - amino group and a-carboxyl group. Which of the above statements is true? (A) aandd (B) aandb (C) bandc (D) band d 141. A gene is regulated by a novel transcription factor. The following technique s may be used to identify the cisregulatory element in the 1 kb promoter sequence of the gene where the novel transcription factor binds: a. Bioinformatics analysis. b. Cell based reporter assay. c. S 1 nuclease assay d. Electrophoretic mobility shift assay. e. DNAse-I foot-printing analysis Which one of the following can help to identify the cis element? (A) aandb (B) cande (C) d only (D) e only 142. A protein D is encoded by a gene, which is 5 Kb long and has three Hind III restriction enzy me sites. The first one is 0.5 Kb from thetranscription start site, the second one is 2.5 Kb from the first site and the third one is 0.5 Kb internal to the stop codon. The second site is poly morphic. In order to find out whether fetal cells contain the normalor the mutated gene, total genomic DNA from fetal cells was isolated,completely digested with Hind III, separated in an agarose gel transferred to membrane and detected by a probe against the region between the second and third restriction site. Which one of the following band patterns will be obtained if the fetal cell is heterozygous? 4kb 2.5 kb 1.5 kb 4kb 1.5 kb 2.5 kb 2.5 kb 2kb 1.5 kb 0.5 kb P7/20 A/UG Given below are the ex perimental protocols to find out the exact location of repetitive DNA sequence in mitotic chromosome by FISH (fluorescence in situ hybridization). Which one of the protocols will give the correct result? (A) Mitotic chromosomes were fixe d on glass slide incubated with biotinylated telomeric DNA denatured incubated with fluorescently labeled avidin localization observed under fluorescence microscope. (B) Mitotic chromosomes were fixed on glass slide denatured incubated with FITC labeled unrelated non- repetitive DNA sequence counterstained with propidium iodide localization observed under fluorescence microscope. (C) Mitotic chromosomes were fixed on glass slide denatured incubated with biotinylated satellite DNA incubated with fluorescently labeled avidin localization observed under fluorescence microscope. (D) mitotic chromosomes were fixed on glass slide incubated withrepetitivednasequence binding protein denatured FITC labeled antibody against the protein localization observed under fluorescence microscope t25 Cvaluesof[ ], the mean residue ellipticity at 222 nm, are - 33,000 and -3,000 deg cm dmol for a poly peptide existing in a- 2-1 helical ( ) and ß-structure (ß), respectively. If this polypeptide undergoes a two-state heat-induced ß transition, and a value of [ ]= - 18,000 deg cm dmol is observ ed at 60 C, then this observation leads to the conclusion that the a helix conversion to ß- structure is : (A) 40% (B) 50% (C) 55% (D) 60% 145. An EEG was recorded and its power spectrum analyses were done in rats with implant electrode for a long time. The power of the EEG waves decreased two months after electrode impla- ntation. This observation may be due to the following: a. Glial cells accumulate surrounding exposed tips of electrodes. b. Degeneration of neurons occur surrounding the electrode tips due metal ion deposition. c. Coating of electrodes are destroyed with time. d. The microsocket becomes loose time. Which one of the following is true? (A) Onlya (B) aandb (C) Only c (D) candd Exp. An EEG was recorded and its power spectrum analyses were done in rats with implant electrode for a long time. The power of the EEG waves decreased two months after electrode impla- ntation. This observation may be due to Glial cells accumulate surrounding exposed tips of electrodes.

208 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, June-2013 PART-A 1. A 16.2 m long wooden log has a uniform deameter of 2m. To what length the log should be cut to obtain a piece of 22m 3 volume? (A) 3.5 m (B) 7.0 m (C) 14.0 m (D) 22.0 m Exp. Cross section area of base = p (1) 2 Let require length to be cut = x m. Hence volume of price = ( x)m 3 Given that x =22 = 22 =7m. 2. What is the last digit of 7 73? (A) 7 (B) 9 (C) 3 (D) 1 Exp. Last digit 7 1 =7 7 2 =9 7=3 7 4 =1 7 5 =7 Unit digit value is get repeated in following exponent values on 7 1, 5, 9, 13, =1+(n 1)4 (n 1)4=72 n 1=18,n =19 3. A lucky man finds 6 pots of gold coins. He counts the coins in the first pots to be 60, 30, 20 and 15, respectively. If there is a difinite progression, what would be the numbers of coins in the next two pots? (A) 10 and 5 (B) 4 and 2 (C) 15 and 15 (D) 12 and 10 Exp. Coins in first four unit value 7. 60, 30, 20, 15, x, y = 2 1, = 3 2, = 4 3 similarly, 15 x = 5 4 x = 15 4 =12 5 Now, 12 y = 6 5 y= 12 5 =10 6 Hence, No. of coins in 6 pots will be 60, 30, 20, 15, 12, A bee leaves its hive in the morning and after flying for 30 minutes due south reaches a garden and spends 5 minutes collecting honey. Then it flies for 40 minutes due west and cellecting honey in another garden for 10 minutes. Then it returns to the hive taking the shortest route. How long was the bee away from its hive? (Asume that the bee flies at constant speed) (A) 85 min (B) 155 min (C) 135 min (D) Less than 1 hour Exp. Let bee leaves its hive at point A. After flying 30 minutes reaches point B. Finally its fly for 40 minute and reaches at point C. Shortest distance from C to A is along line AC. Time required to cover distance AC 2 2 = ( 30) + ( 40) = 50 minutes. Total time consumed to reach at Point A (including its stay time at point B and Point C) = 50 = = 5 = 135 minutes. P8/1

209 Amar Ujala Education 5. A bird perched at the top of a 12 m high tree sees a centipede moving towards the base of the tree from a distance equal to twice the height of the tree. The bird flies along a straight line to catch the centipede. If both move at the same speed, at what distance from the base of the tree will the centipede be picked up by the bird? (A) 16m (B) 9m (C) 12m (D) 14m Exp. Let the bird catch the centipede at point D. such that CD = x m. A/UG-15 AE EB BF FC AH HD DG GC AE EI IG GC AH HI IF FC AH HI IG GC AE EI IF FC 7. A point is chosen at random from a cirular disc shown below. What is the probability that the point lies the sector OAB? A O x B As the speed of bird and centipede are same. So in equal time t, both will cover equal distance. therefore AD = CD = x m. From right angle triangle ABD, (AB) 2 +(BD) 2 = (AD) 2 (12) 2 +(24 x) 2 =(x) (576 + x 2 48x)=x x =0 48x = 720 x = = 15m. Hence, required distance = =9m. 6. An at goes from A to C in the figure crawling only on the lines and taking the least length of path. The number of ways in which it can do so is A B C D (A) 2 (B) 4 (C) 5 (D) 6 Exp. There are 6 possible ways (where AOB = x radians) 2x (A) (B) (C) x 2 (D) x x 4 Exp. Area of whole circular disc =p(oa) 2 8. Area of sector OAB = x 360º ( OA) ( OA) x = 360º 2 x 2. A ray of light, after getting reflected twice from a hemispherical mirror of radius R (see the above figure), emerges parallel to the incident ray. The separation of the original incident ray and the final reflected ray is (A) R (B) R 2 (C) 2 R (D) R 3 Exp. Ray AB CD 2 P8/2

210 A/UG-15 So ABC + BCD = 180º From law of reflection, ABO = OBC = 45º R BE = R.cos 45º = 2 Hence, BC = 2BE = 2R. 9. A king ordered that a golden crown be made for him from 8 kg of gold and 2 kg of silver. The goldsmith took away some amount of gold and replaced it by an equal amount of silver and the crown when made, weighed 10 kg. Archemedes knew that underwater gold lost 1/20th of its weight, while silver lost 1/10th. When the crown wes weighed underwater, it was 9.25 kg. How much gold was stolen by the goldmith? (A) 0.5 kg. (B) 1 kg. (C) 2 kg. (D) 3 kg. Exp. Let goldsmith took away x kg gold and replace it by same weight of silver. According to Archimedes, Under water calculaton of weight, (8 x) (2+x) 9 10 =9.25 or, x x = x = 185 x =3kg Hence, goldsmith took 3 kg. of gold and replace it by same weight of silver. 10. What is the angle between the minute and hour hands of a clock at 7 : 35? (A) 0º (B) 17.5º (C) 19.5º (D) 20º Exp. In 60 minutes, minute hand of any watch rotated by 360º Angle rotated by minute hand in 35 minutes. = º = 210º 60 Amar Ujala Education 7 hour 35 minutes = hours. Now, in 12 hours, hour hand of any watch rotated by 360º. Angle rotated by hour hand in hours. = 360º = 227.5º Angle between hour hand & minute hand of watch at 7.35 = 227.5º 210º = 17.5º. 11. A stream of ants go from point A to point B and return to A long the same path. All the ants move at a constant speed and from any given point 2 ants pass per second one way. It takes 1 minute for an ant to go from A to B. How many returning ants will an ant meet in its journey from A to B? (A) 120 (B) 60 (C) 240 (D) The capacity of the coincal vessel shown above is V. It is filled with water upto half its height. The volume of water in the vessel is : (A) V 2 (B) (C) V 8 (D) Exp. Volume of cone (v)= 1 3 p r2 h here, v h and v r 2 V 4 V 16 When, h becomes 2 h,r becomes 2 r Volume of water (v)= r 2. H 2 = 1 r 2. h 3 8 Hence, volume of water (v)= 8 v. P8/3

211 Amar Ujala Education 13. A large tank filled with water is to be emptied by removing half of the water present in it everyday. After how many days will there be closest to 10% water left in the tank? (A) One (B) Two (C) Three (D) Four Exp. 10% of water left = 0.1v of water left. here v = total volume of water in the tank. In first day water left out in tank = 0.5v. In second day water left in the tank =0.25v. In the third day water left iin the tank = 0.125v Which is almost equal to 0.1v. 14. n is a natural number. If n 5 is odd, which of the following is true? 1. n is odd 2. n 3 is odd 3. n 4 is even (A) 1 only (B) 2 only (C) 3 only (D) 1 and 2 only Exp. Product of odd numbers is always a odd number. Product of odd number with any even number is always an even number. Given that (n) 5 is odd number. n 5 = n n n n n Let n is an even number So, n =2k where, k = 1, 2, 3... Now, (n) 5 =(2k) 5 =32 k 5 32(K) 5 is always even number whether k is even or odd. Which contracdict the given question. Hence our assumption is wrong and n is an odd number. 15. Suppose you expand the product (x 1 + y 1 )(x 2 + y 2 )... (x 20 + y 20 ). How many terms will have only one x and rest y's? (A) 1 (B) 5 (C) 10 (D) 20 Exp. (x 1 + y 1 )(x 2 + y 2 )(x 2 + y 3 )...(x 20 + y 20 ) When x 1 = x 2 = x 3 =...= x 20 = x (say) y 1 = y 2 = y 3 =...= y 20 = y (say) =(x+ y) 20 From binomial expension, (x + y) 20 = 20 c 0 x c 1 x 19 y c 19 xy c 20 y 20 Coeficient of xy 19 = 20 c 19 = 20. P8/4 A/UG In the figure below the numbers of circles in the blank rows must be?? (A) 12 and 20 (B) 13 and 20 (C) 13 and 21 (D) 10 and 11 Exp. Number of circles in the nth row = sum of the circles in the (n 1) th and (n 2) th row. Number of circle in the 7 th row =sumofcirclesin6 th and 5 th row. =8+5 =13 Again, number of circles in the 8th row = Sum of circles in 7th and 6th row. =13+8= If we plot the weight (w) versus age (t) of a child in a graph, the one that will never be obtained from amongst the four graphs given below is : (A) w w (C) t 18. Find the missing number : t (B) (D) w w ? (A) 1 (B) 0 (C) 2 (D) 3 Exp = = x + 7 = 100 x = x =3 t t

212 A/UG In solving a quadratic equation of the form x 2 + ax + b = 0, one student took the wrong value of a and got the roots as 6 and 2; while another student took the wrong value of b and got the roots as 6 and 1. What are the correct values of a and b, respectively? (A) 7 and 12 (B) 3 and 4 (C) 7 and 12 (D) 8 and 12 Exp. First student took wrong value of a, but correct value of b. so, product of root (b)= Next student took wrong value of b, but correct value of a so, sum of roots ( a)=6+1=7 Hence, correct value of a and b are 7 and 12. Amar Ujala Education 20. The distance between two oil rigs is 6 km. What will be the distance between these rigs in maps of 1 : and 1 : 5000 scales, respectively? (A) 12cmand1.2cm (B) 2cmand12cm (C) 120 cm and 12 cm (D) 12 cm and 120 cm Exp. 6 km = cm Distance between two oil rigs on the given scale = =12cm and = 120 cm. PART-B 21. Which one of the following non covalent interactions between two non bonded atoms A and B is most sensitive to the distance between them? (A) A and B are permanent dipoles and are in volved in hydrogen bonding. (B) A and B are fully ionized and are involved in salt bridge formation. (C) A and B are uncharged and repel each other. (D) A and B are uncharged and attract each other. Exp. Non covalent interactions between two non bonded atoms A and B is most sensitive to the distance between them because A and B are uncharged and repel each other. 22. Which statement best describes the pka of amino groups in proteins? (A) pka of amino group is higher than the pka of amino group. (B) pka of amino group is lower than the pka of amino group. (C) pka of amino groip is same as the pka of amino group. (D) pka of amino group is higher than the pka of guanidine side chain of ariginine. Exp. Lysine. an essential amino acid, has a positively charged e-amino group (a primary amine). Lysine is basically alanine with a propylamine substituent on the b-carbon. The e-amino group has a significantly higher pk a (about 10.5 in polypeptides) than does the a-amino group. The amino group is highly reactive and often participates in a reactions at the active centers of enzymes, Proteins only have one amino group, but numerous amin so groups. However, the higher pk a renders the lysyl side chains effectively less necleophilic. Specific environmental effects in enzymes active centers can lower the pk a of the lysyl side chain such that it becomes reactive. Note that theside chain has three methylene groups. So that even though the terminal amino group will be charged under physiological conditions, the side chain does have significant hydrophobic character. Lysines are often found buried with only the amino group exposed to solvent pk R = NH3 CH2 CH2 CH2 CH2 +H N CH C O 3 pk 2 =9.0 O pk 1 = What is the effect of 2, 4 dinitrohphenol on mitochondria? (A) Blocks ATP syntheses without inhibiting electron transport by dissipating the proton gradient. (B) Blocks electron transport and ATP synthesis by in hibiting ATP ADP exchange across the inner mi tochondrial membrane. (C) Blocks electron transport and proton pumping at complexes I, II and III. (D) Interacts directly with ATP synthase and inhibits its activity. Exp. 2, 4-dinitrophenol (DNP) uncouples the mitochondria by shuttling H+ ions across the inner membrane, by passing ATP synthase. DNP is a mobile ionophore, this means that it is a lipid soluble molecule that is capable of transporting ions across a biological membrane. P8/5

213 Amar Ujala Education The way it does this is by shielding the charge of the ion within its hydrophobic exterior, thereby facilitating the transport of H+ ions across a biological membrane. The mode of action provides less resistence to the H+ ion than as if it were to move through ATP synthase and thus it becomes the preferred route. Because of this, less ATP is produced inversely proportional to the concentration of DNP. Subsequently the potential energy that was stored in the concentration gradient is released as heat energy. If the concentration of DNP becomes too great, the cell will be unable to produce ATP and eventually die. 24. A protein has 30% alanine. If all the alanines are replaced by glycines. (A) Helical content will increase. (B) sheel content will increase. (C) There will be no change in conformation. (D) The alanine substituted protein will be less struc tured than the parent protein. Exp. In internal helical positions, alanine is regarded as the most stabilizing residue, whereas gylcine. after proline, is the more destrabilizing. 25. The gel to liquid crystalline transition temperature (Tm)of phospholipids is dependent on the fatty acid composition. Considering this, Tm of (A) All the phospholipids will be indentical. (B) DPPC will be lowest and DOPC will be highest. (C) POPC and DOPC will be indentical and lower than DMPC or DPPC. (D) DOPC will be lowest and DPPC will be highest. 26. You have created a fusion between the trp operon, which encodes the enzymes for tryprophan biosynthesis, under the regulatory control of the lac operatir. Under which of the following condetions will tryptophan synthase be induced in the strain that carries the chimeric operator fused operons? (A) Only when both lactose and gluse are absent. (B) Only when both lactose and glucose are present. (C) Only when lactose is absent and glucose is present. (D) Only when lactose is present and glucose is absent. Exp. Lac repressor protein always binds to the operator there by blocking transcription. In the presence of glucose and lactose, the bacterium will utilize glucose as the primary Carbon source - a transcription off Lac Operon off. In the presence of lactose alone, Lactose binds to the lac repressor-transcription on. 27. Which of the following pairs of subcellular compartments is likely to have same ph and electrolyte composition? A/UG-15 (A) Cytosol and lysosomes. (B) Cytosol and mitochondrial inter membrane space. (C) Cytosol and endosome. (D) Mitochondrial matrix and inter membrane space. 28. Regarding microtubule assembly and disassembly during cell division, which will be the most appropriate answer? (A) Once formed, kinetochore microtubules depolymer ize at the plus ends throughout mitosis. (B) Once formed, kinetochore microtubules depolymerize at the plus ends throughout mitosis. (C) Kinetochore microtubules polymerize at their plus ends up to anaphase, at which point they begin to depolymerize. (D) Kinetochore microtubules polymerize at their mi nus ends up to cytokinesis, at which point they depolymerize. 29. Origin of replication usually contains : (A) GC rich sequences (B) Both AT and GC rich sequences (C) No particular stretch of sequences (D) AT rich sequences Exp. Repeated sequences are commonly present in the sites for DNA replication initiation in bacterial, archaeal, and eukaryotic replicons. Those motifs are usually the binding places for replication initiation proteins or replication regulatory factors. In prokaryotic replication origins, the most abundant repeated sequences are DnaA boxes which are the binding sites for chromosomal replication initiation protein DnaA, iterons which bind plasmid or phage DNA replication initiators, defined motifs for site-specific DNA methylation, and 13- nucleotide-long motifs of a not too well-characterized function, which are present within a specific region of replication origin containing higher than average content of adenine and thymine residues. In this review, we specify methods allowing identification of a replication origin, basing on the localization of an AT-rich region and the arrangement of the origin's structural elements. We describe the regularity of the position and structure of the AT-rich regions in bacterial chromosomes and plasmids. The importance of 13-nucleotide-long repreats present at the AT-rich region, as well as other motifs overlapping them, was pointed out to be essential for DNA replication initiation including origin opening, helicase loading and replication complex assembly. We also summarize the role of AT-rich region repeated sequences for DNA replication regulation. P8/6

214 A/UG csubunit of E. coli RNA polymerase does not : (A) Initiate transcription and fall off diring elongation (B) Increase affinity of the core enzyme to the promoter (C) Binds to DNA, independent of the core enzyme (D) Ensures specificity of transcription by inter acting with the core enzyme Exp. Every molecule of RNA polymerase holoenzyme contains exactly one sigma factor subunit, which in the model bacterium Escherichia coli is one of those listed below. The number of sigma factors varies between bacterial species. E. coli has seven sigma factors. RNA polymerase holoenzyme complex consists of core RNA polymerase and a sigma factor executes transcription of a DNA template strand. Once initiation of RNA transcription is complete, the sigma factor can leave the complex. It long has been thought that the factor obligatorily leaves the core enzyme and initiate transcription allowing the free to link to another core enzyme and initiate transcrption at another site. Thus, the cycles from one core to another. However, does not obligatorily leave the core. Instead, the changes its binding with the core during initiation and elongation. Therefore, the cycles between a strongly bound state during initiation and a weakly bound state during elongation. 31. The cap binding protein (eif4e), which is involved in the global regulation of translation, is highly regurlated in eukaryotic cells. In an experiment, a researcher transfected mammalian cells with (eif4e) gene for its overxpression. Due to this, the cells will undergo : (A) Apoptosis (B) Neoplastic transformation (C) No change (D) Differentiation 32. Bacteriophage T4 infects E. coil and injects its DNA inside the cell. The transcription of viral genes occurs in three stages : immediate early, early and late. All the promotrs on viral genome are available, but the control takes place at the level of : (A) Promotes strength (B) Modification of host RNA polymerase (C) Synthesis of new RNA merases (D) Turn over rate of RNA synthesis Exp. T4 early promoter strength probed in vivo with unribosylated and ADP-ribosylated Excherichia coli RNA polymerase : a mutation analysis. 33. Gram negative bacteria, Klebsiella pneumoniae, upon infecting humans, results in severe septic shock after a few hours of infection.which of the following is not true for this type of infection? P8/7 Amar Ujala Education (A) Cell wall endotoxins cause overproduction of cytokinnes. (B) Septic shock can be treated by anti TNF antibodies. (C) Recombinant bacterial proteins can be used for the treatment of septic shock (D) Recombinant TNF receptor antagonist can be used for the treatment of septic shock 34. Which of the following is NOT associated with insulin action? (A) Increased glucose transport (B) Increased glycogen formation (C) Enhanced lipolysis in adipose tissue (D) Decreased rate of gluconeogenesis Exp. Insulin inhibited glycerol release by adipose tissue in all groups of rats with fructose in the medium. in fed rats a barely significant insulin inhibition occured also in the presence of glucose, and in fasted-refed rats it was indepedent of whether a hexose was present in the medium or not. The following effects of insulin on tissue of fasted-refed rats incubated in the absence of hexose were observed: Inhibition of glycerol release with a dose effect relationship between 10 and 250 units insulin per milliliter; partial inhibition of glycogen breakdown during incubation; diminished lactic and pyruvic acid ratio. These same effects were also exerted by the antilipolytic drug 5-methylpyarazole 3-carboxylic acid, which does not stimulate glucose uptake. 35. When adenoma is converted to metastatic adenocarcinoma, which of the following compination of proteins is almost certainly to be degraded? (A) Type IV collagen and laminin (B) Fibranectin and 2 integrin (C) Metalloprotease and serine protease (D) Elastin and selectin 36. Which of the following is considered to be a combined B and T cell deficiency? (A) Ataxia telanglectasia (B) Swiss type agammaglobulinemia (C) Wiskott Aldrich syndrome (D) Bruton's agammaglobulinemia Exp. Combined B-cell and T-cell immuno deficiencies, or SCID, is a group of medical disorders that are the result of genetic defects in both cellular and humoral immunity. The defects in humoral and cellular immunity have an early clinical presentation and, if untreated, result in a fatal outcome in the first few years of life. This article focuses only on SCID disorders and outlines recent advances in therapeutics options for patients.

215 Amar Ujala Education The profound degree of immune compromise in SCID leads to infections with bacterial, viral, and fungal pathogens that cause significant morabidity and eventually mortality in patients. Swiss type agammaglobulenemia is a type of combined immuno-deficiency syndromes. 37. The part of the embryo from which the ectoderm, mesoderm and endoderm are formed in chick is known as (A) Primitive streak (B) Hypoblast (C) Epiblast (D) Cytotrophoblast Exp. The epiblast is capable of forming all three germ layers (ectoderm, mesoderm and endoderm) during gastrulation. Epiblast cells migrate to the primitive streak and invaginate into a space between the epiblast and the hypoblast. It is a cross section through the cranial region of the primitive streak at 15 days, illustrating the invagination of epiblast cells. The invaginating epiblast cells displace the hypoblast to create the definitive endoderm. Once the definitive endoderm is established, migrating epiblast cells also form the intraem-bryonic mesoderm. The remaining epiblast cells, which do not migrate through the primitive streak. remain in the epiblast to form the ectoderm. Hence, the epiblast gives rise to all three layers in the embryo. 38. Which protein secreted by the amphibian orgainizer indced neural tissue formation by inhibiting Bone Morphogenetic Protein? (A) catenin (B) Noggin (C) Armadillo (D) Cubitus interruptus 39. The homologue of catenin in Drosophila is (A) Fushi tatazu (B) Engrailed (C) Armadillo (D) Cubitus interruptus Exp. The armadillo repeat family of proteins is defined by the existence in length (42 amino acids) and spacing (typically end-to0end with no intervening sequences), though not necessarily highly conserved in sequence. Rather, it is the structure of the arm motif that is conserved; repeats of this domain together form a positively charged groove that mediates protein-protein interactions. This is so-called 'arm motif' was originally identified in the protein product of the Drosophila segment polarity gene known as armadillo. It was subsequently determined that the vertebrate homologues of armadillo, -catenin and plakoglobin and the Src tyrosine kinase substrate p120 catenin, also contain multiple arm repeats. Together, these four proteins initially defined the armadillo repeat family of proteins. Although the function of these four proteins all coincidentally relate P8/8 A/UG-15 to cell-cell adhesion, it is now clear that members of the armadillo repeat family have widely varied roles in cellular physiology. 40. Which of the floral whorls is affected in apetala 3/ pistillata (ap3/pt) mutants? (A) Sepals and petals (B) Petals and stamens (C) Stamens and carpels (D) Sepals and stamens 41. Which one of the following statements is INCORRECT about the role of oxidative pentose phosphate pathway in plant metabolism? (A) Generation of NADPH required to drive biosynthetic reactions (B) Production of pentose phosphate for the synthesis of nucleic acids (C) Formation of erythrose 4 phosphate for biosynthe sis of aromatic amino acids (D) Production of NADH to generate ATP 42. During photosynthetic carbon reduction cycle in green leaves, net production of one molecule of glyceraldehyde 3 phosphate requires one of the following combinations of energy equivalents : (A) 9 NADPH and 6 ATP (B) 3 NADPH and 9 ATP (C) 2 NADPH and 3 ATP (D) 6 NADPH and 9 ATP 43. Which one of the following essential micronutrients is associated with urease enzyme found in higher plants? (A) Nickel (B) Molybdenum (C) Zinc (D) Copper Exp. Nickel (Ni) is one of the essential micronutrients for higher plants and its known function is being the metal component of urease. The effects of various Ni levels on urease activity in maize plants grown in two nutrient media containing urea or ammonium nitrate as two separate nitrogen sources were investigated. Ni, the most recently discovered essential element, is unique among plant nutrients in that its metabolic function was determined well before it was determined that its deficiency could disrupt plant growth. Subsequent to the discovery of its essentiality in the laboratory, Ni deficiency has now been observed under field condition in several perennial species. 44. Plants and able to perceive light through various photoreceptors and downstream genes. Which one of the following genes is NOT involved in light perception? (A) PIF3 (B) NPR 1 (C) PHYE (D) CRY 3

216 A/UG In the dark, rods show a large inward "dark" current which is suppressed by a flash of light. Which one of the following statements, explaining the effect of light, is true? (A) Sodium channels in the outer segment of rods are closed (B) Cytoplasmic cgmp concentraction increases (C) Sodium channels in the inner segment of rods are closed (D) Transducin dissociates from beta arrestin 46. Four group of mice were studied for the factor required for mast cell generation : IL 3 deficient, GM CSF deficient, G CSF deficient and erythropoietin deficient. In which mice, mast cell generation is most likely to be deficient? (A) IL 3 deficient (B) GM CSF dificient (C) G CSF dificient (D) Erythropoietin deficient 47. What would be the outcome if the theca interna cells were destroyed in a Graafian follicle? (A) Immediate formation of corpus albicans (B) Incereased progesterone synthesis in the granulosa cells (C) Decreased estrigen synthesis in the granlosa cells (D) Formation of corpus hemorrhagicum 48. The size of red blood cells (RBC) in venous blood is greater than that of arterial blood. This increased size of red blood cell in the venous blood is due to (A) the increased permeability of red blood cell (RBC) membrane (B) The decreased osmotic pressure in plasma (C) The increased osmotlc pressure in RBC (D) The dissociation of cytoskeletal proteins in RBC 49. A chromosome aberration leads to change in the order of genes in a genetic map but does not alter its linkage group. This is due to (A) Translocation (B) Recombination (C) Transposition (D) Inversion Exp. Single crossovers within the inversion loop lead to recombinant gametes that are imbalanced for genetic material OUTSIDE the loop. One recombination product is duplicated for the region at outside the loop at one end of the chromosome and simultaneously deleted for the material outside the loop at the other end of the chromosome. The reciprocal recombinant gamete has the reciprocal imbalance, All games have the expected amount of DNA for eveything within the inversion loop. If the centromere Amar Ujala Education is in the inversion (pericentric) then each meiotic product has a centromere. If the centromere is outside of the inversion loop then the meiotic products that result from a single cross over with in the loop are imbalanced for the centromere as well as the surrounding genes - one recombinant product will be dicentric and the other will be acentric. 50. The concept of recon was proposed by Seymour Benzer by studying recombination between : (A) Lysis mutants of bacteriophage T4 (B) White eye mutants of Drosophila melanogaster (C) Biochemical mutants of Neurospora crassa (D) Auxotrophic mutants of Escherichai coli. 51. Aspartic acid (Asp) is specified by the codon GAU and GAC. After mutation, Asp is changed to Alanine represented by GCX, where X may be A, U, C or G. The reversion of the mutation could only be done with reactive oxygen species. The nature of the mutation is considered to be (A) Transition (B) Transversion (C) Either A or B (D) Deputrination 52. A cross is made between two plants with white flowers. All the F1 progeny had red coloured flower. This is because of (A) Complementation (B) Recombination (C) Translocation (D) Reversion 53. Cladistic classification is based on: (A) Sequential order in which branches arise from a phylogenetic tree (B) The order of sequence divergence (C) Morphological features and skeleton of in dividuals (D) Cellular organization and cystoskeleton 54. Tautonym is an informal taxonomic designation used for animals referring to: (A) Same name for genus and species (B) Same name for species and subspecies (C) Trinomial nomenclature (D) The name of the author for the species 55. A marine biologist dug up a small animal from the ocean floor. The animal was uniformly segmented with short, stiff appendages and soft, flexible skin. It had a complete digestive system and an open circulatory system but no exoskeleton. Based on this description, the animal appears to be a : (A) Lancelet (B) Roundworm (C) Mollusc (D) Crustacean P8/9

217 Amar Ujala Education 56. Which of these programs is used to conserve a species facing extinction? (A) Captive breedings (B) Natural resources (C) Sustainble use (D) Edgeeffects Exp. This process ocuvers before species arereintroduced into an area where they once lived. 57. A grasshopper population is being assessed by capturemark-release-recapture method. On the first day, 100 grasshoppers were captured from a given area in 1 hour time, marked and released. on the next day during recapture, 10 marked and 90 unmarked grasshoppers could be found in the same time period from same area. What will be the estimated population size in the given area? (A) 80 (B) 100 (C) 1,000 (D) 10, Free-living nitrogen fixers can survive in different ecological niches. Identify the incorrect combination from the following list: (A) Azotobacter - acidic (B) Deraxia - alkaline soil (C) Beijernckia - acid soil (D) Frankia - neutral soil 59. A plot of soil contaminated with diesel oil was inoculated with oyster mushrooms. After 4 weeks, more than 95% of the polycyclic aromatic hydrocarbons had been reduced to non-toxic compounds. This process is called: (A) Phytoremediation (B) Chemoremediation (C) Mycoremediation (D) Zooremediation Exp. Bioremediation is the process of cleaning up or degrading environmental contaminants using biological organisms especially microbes. Here mushrooms are used. Mushrooms belong to fungi. Therefore myco (fungus) remediation. 60. In pre-industrial period in England, peppered moths had light coloration which effectively camouflaged them against light coloured trees and lichens. During industrial revolution, many lichens died out and trees became blackened by soot from factories and interestingly, dark coloured moths were predominantly seen. This happened due to (A) Natural selection of dark coloured moths which were initially present in fewer numbers (B) New mutation which arose due to environmental pollution (C) Macroevolution occuring due to environmental change (D) Natural selection of the camouflaging mechanism of the moths P8/10 A/UG The speciation in which a population splits into two geographically isolated populations experience dissimilar selective pressure and genetic drift is known as (A) Sympatric specification (B) Parapatric specification (C) Peripatric specification (D) Allopatric specification Exp. In allopatric specification, a population splits into two geographically isolated populations due to formation of a barrier between portions of a population, for example, because of mountain building as depicted in the animal to the left. The isolated populations then experience differentiating genotypic and phenotypic divergence as a result of different selective pressures in their differing environments. Additionally, genetic driff will occur differently, eventually differentiating the population's genotypes and phenotypes. 62. Evolution of multi-gene family occurs by (A) Only gene duplication (B) Only unequal crossing-over (C) Random mutations (D) BothAandB Exp. Evolution of multigene families by gene duplication and subsequent diversification is analyzed assuming a haploid model without interchromosomal crossing over. Chromosomes with more different genes are assumed to have higher fitness. Advantageous and deleterious mutation and duplication/deletion also affect the evolution, as in previous studies. In addition, negative selection on the total number of genes (copy number selection) is incorporated in the model. 63. One aims to find out the role of a gene product in macrophages by using a transgenic mouse expressing the genes under a promotor. Which of the following is the most appropriate promoter? (A) Actin promoter (B) MHC Class II promoter (C) MAC-1/cd l 1b promoter (D) IL-2 promoter 64. Which of the following genes was engineered in the "Flavr Savr" transgenic tomato variety? (A) 1-Amino cyclopropane-1-carboxylic acid synthase (B) 1-Amino cyclopropane-1-carboxylic acid oxidase (C) Expansin (D) Polygalacturonase Exp. 'Flavr Savr' tomato has increased shelf life by silencing the polygalactouronase gene involved in ripening by anti sense technology.

218 A/UG For developing transgenic mice, embryonic stem cells are engineered to express the transgene. These cells are selected by (A) Novobiocin (B) Neomycin (C) Tetracycline (D) Penicillin 66. Microbial leaching involves the process of dissolution of metals from ore breaking rocks using mircoorganisms. Which one of the following bacteria helps in leaching copper from its ore? (A) Acidithiobacillus ferroxidans (B) Pseudomonas putida (C) Deinococcus radiodurans (D) Rhodopseudomonas capsulate 67. Molar absorption coefficient of phenylalanine is 200 M-1 cm-1 at 257 nm. What concentration (g/l) of this amino acid will give an absorption of 1 in a cell of 0.5- cm path length at 257 nm? (A) 3.30 (B) 0.33 (C) 1.65 (D) Which of the following atomic nuclei cannot be probed by nuclear magnetic resonance spectroscopy? (A) 1 H (B) 31 P (C) 18 O (D) 15 N Exp. 1 H : most commonly used. 15 N, 31 p : used in biochemical studies. 18 O : not used Amar Ujala Education 69. t 1/2 of an irreversible first order reaction, S - P is 1 hour. The time (in hours) required to reach 75% completion is: (A) 1.5 (B) 2.0 (C) 2.5 (D) In the case of monoclonal antibody production by hybridoma technology, myeloma cells used lack the enzyme hypoxanthine-guanine phosphoriboxyl transferase (HGPRT) such that fused cells can only survive when selected on hypoxanthine-aminopterinthymidine (HAT). What is the role of aminopterin in this medium? (A) To be used as cell cycle inhibitor of myeloma cells (B) To block the pathway for nucleotide synthesis (C) To facilitate fusion of myeloid B cells and antibody producing B cells (D) To facilitate production of antibody producing b cells Exp. To block the pathway for nucleotide synthesis monoclonal antibody production by hybridoma technology, myeloma cells used lack the enzyme hypoxanthine-guanine phosphoriboxyl transferase (HGPRT) such that fused cells can only survive when selected on hypoxanthine-aminopterin-thymidine (HAT). PART-C 71. The amino acid alanine has high propensity to occur in helical conformation. The circular dichroism spectrum of an equimolar mixture of two 20-residue peptides, one composed of only L-alanine and the other only D- alanine is recorded in the region of nm. Which one of the following will be observed? (A) No signal: as the chiroptical properties of the two peptides will cancel out (B) Bands with only negative ellipticity: as helix formed by the D-Ala peptide will be unstable (C) Bands with only positive ellipticity: as both the peptides will form right-handed helices (D) Bands with identical negative and positive ellipticity 72. The following small peptide substrates are used for determining elastase activity and the following data have been recorded Substrate K M (mm) k cat (s 1 ) PAPA G PAPA A PAPA F The arrow indicates the cleavage site. From the above observations, it appears that 1. PAPAF is digested most rapidly 2. PAPAG is digested most rapidly 3. A hydrophobic residue at the C-terminus seems to be favored 4. A smaller residue at the C-terminus seems to be favoured 5. Elastase always requires a smaller residue at the N- terminus of the cleavage site. Which of the following is true? (A) 1, 3, 5 (B) 2, 4, 5, (C) 5 only (D) 4, 5 only 73. The apparent ph of a fluid is 7.45, where bicarbonate buffer is involved for maintaining its ph. Values of pka of carbonic acid are 6.15 and The molar ratio of [conjugate base] : [acid] is (A) 1:20 (B) 20:1 (C) 1 : 1000 (D) 1000 : 1 (Hint : antilog 1.3 = 20.0, and antilog 10-3 = 1000) P8/11

219 Amar Ujala Education 74. A segment of B-DNA encodes an enzyme of molecular mass 50 kd. The estimated length of this segment in m would be (A) (B) x 10-3 (C) (D) x In order to determine the primary structure of an octapeptide, amino acid composition was determined by acid hydrolysis (A). The intact oligopeptide was treated with carboxyepeptidase (B), chymotrypsin (C), trypsin (D) and CNBr (E). The peptides were separated in each case and acid hydrolysis was carried out for B-E. Following results were obtained (the brackets represent mixtures of amino acids in each fragment): 1. (2Ala, Arg, Lys, Met, Phe, 2Ser) 2. (Ala, Arg, Lys, Met, Phe, 2Ser) and Ala 3. (Ala, Arg, Phe, Ser), (Ala, Lys, Met, Ser) 4. (Ala, Arg), (Lys, Phe, Ser), (Ala, Met, Ser) 5. (Ala, Arg, Lys, Met, Phe, Ser), (Ala, Ser) Which one is the correct sequence of the oligopeptide? (A) Arg-Ala-Ser-Lys-Met-Phe-Ser-Ala (B) Arg-Ala-Ser-Lys-Phe-Met-Ser-Ala (C) Ala-Arg-Ser-Phe-Lys-Met-Ser-Ala (D) Ala-Arg-Phe-Ser-Lys-Met-Ser-Ala 76. You are following the intercellular sorting of an integral plasma membrane protein in a living cell, in culture. You have decided to probe this protein by metabolic labelling technique with 35 S-methionine (pulse-chase technique). After one cycle of division, the cells were treated with a potent inhibitor of protein biosynthesis and processed for subcellular fractionation. In which of the following fractions will you expect the presenc of this protein upon immunoprecipitation with a specific antibody? (A) Only cytoplasm (B) Only plasma membrane (C) Both endoplasmic reticulum and plasma membrane (D) Only secretory vesicles and endoplasmic reticulum 77. The principal pathway for transport of lysosomal hydrolases from the trans Golgi network (ph 6.6) to the late endosomes (ph 6.0) and the recycling of M6P (mannose 6 phosphate) receptors back to the Golgi depends on the ph difference between those two compartments. From what you know about M6P receptor binding and recycling and the pathways for delivery of material to lysosomes, predict what would happen if the ph in late endosomes was raised to 6.6? (A) M6P will bind to hydrolases but will not re lease the hydrolases in the late endosomes (B) M6P will bind to hydrolases and will release the hydrolases in the late endosomes P8/12 A/UG-15 (C) At higher endosomal ph, the receptor would not release the hydrolase and could not be recycled back to the trans Golgi network (D) M6P will be degraded at higher ph 78. The diploid genome of a species comprises 6.4 x 10 9 bp and fits into a nucleus that is 6 m in diameter. If base pairs occur at intervals of 0.34 nm along the DNA helix, what is the total length of DNA in a resting cell? (A) 3.0 m (B) 3.5 m (C) 2.2 m (D) 4.0 m 79. Phosphotylation of serines as well as methylation and acetylation of lysines in histone tails affect the stability of chromatin structure above the nucleosome level and have important consequences for gene expression. The resulting changes in charge are expected to affect the ability of the tails to interact with DNA because (A) DNA is negatively charged (B) DNA-histone interaction is independent of net charge (C) Phosphorylation of serine increases DNA-histone interaction (D) Methylation and acetylation of lysine increases DNa-histone interaction 80. Cells that grow and divide in a medium containing radioactive thymidine covalently incorporate the thymidine into their DNA during S phase. Consider a simple experiment in which cells are labelled by a brief (30 minutes) exposure to radioactive thymidine. The medium is then replaced with one containing unlabelled thymidine, an the cells grow and divide for some additional time. At different time points after replacement of the medium, cells are examined under a microscope. Cells in mitosis are easy to recognize by their condensed chomosomes and the fraction of mitotic cells that have radioactive DNA can be estimated by autoradiography and plotted as a function of time after the thymidine labelling as in the figure below

220 A/UG-15 The rise and fall of the curve is because (A) Initial rise of the curve corresponds to cells that were just finishing DNA replication when radioactive thymidine was added (S phase) (B) The peak of the curve corresponds to cells in M phase (C) The rise in curve after 20 min corresponds to cells in apoptotic (D) The fall in curve after 10 min indicates the cells exiting M phase 81. A rapidly growing bacterial species such as E. coli exhibits a typical phase of growth cycle in liquid nutrient broth (lag phase log phase stationary phase death phase). If a bacterial culture has a starting density of 10 3 cells/ml has a lag time of 10 minutes and a generation time of 10 minutes, what will the cell density be at (cells/ml) 30 minutes? (A) 6.0 x 10 3 (B) 2.0 x 10 3 (C) 3.0 x 10 3 (D) 4.0 x in order to study the role of telomeres in DNA replication, genetically engineered mice were prepared, where the gene for telomerase RNA was knocked out. when cells from these knock out mice were taken and cultured in vitro, they proliferated even after 100 cell divisions which is quitw unlikely in the case of human cells. Which of the following is the correct reason? (A) Human and mice are fundametally different with respect to their requirements for telemerase enzyme in the context of DNA replication (B) In vitro, mice DNA becomes circular due to end to end chromosome fusion and does not require telomerase for DNA end replication (C) Mice have very long stretch of telomere DNA sequence compared to that of human (D) In vitro, mice DNA replication does not require the removal of RNA primers 83. You are working with an in vitro eukaryotic transcription system, which produced both capped and uncapped mrnas. You incubated these mrnas with mammalian cell nuclear extract and then quantified the different products as shown below. Which of the following graphs correctly represents the expected result? (A) Level of mrna Capped Uncapped (B) (C) (D) Level of mrna Level of mrna Level of mrna Capped Capped Capped Uncapped Uncapped Uncapped Amar Ujala Education 84. A non-enzymatic viral protein X was found to be inducing a cellular gene promoter activity. Although no in vitro DNA binding activity could be co-recruited on the cellular promoter along with a cellular transcription factor in vivo. Which one of the following statements seems to be the best interpretation of the above findings? (A) X is a DNA-binding protein (B) X physically interacts with the transcription factor (C) X modifies the chromatin for transcription activation (D) X is a chaperone 85. During elongation step of protein synthesis, translocation moves the mrna and rhe peptidyl t-rna by one codon through the ribosome. Translocation in E. coli involves GTP and EF-G. However, in vitro translocation can take place indepedent of GTP and EF-G. Based on these observations, the following hypotheses can be made: 1. The molecular mechanism of translocation in vitro is completely different from that in vivo. 2. Translocation activity is independent of GTP hydrolysis 3. Translocation activity is completely dependent on GTP and EF-G 4. Translocation activity is inherent in ribosomes, however, the rate of translocation in vivo is enhanced significantly in presence of GTP and EF-G Which one of the following combination is correct? (A) only 4 (B) 1 and 3 (C) 1 and 2 (D) 3 and 4 P8/13

221 Amar Ujala Education 86. DNA methylation plays an important role in transcription regulation in vertbrates. There is an inverse correlation between the level of DNA methylation in the vicinity of a gene and its transcription rate, whereas there is a direct correlation between histone acetylation and increased transcription. thalassemia is a common genetic impairment of haemoglobin chain in its place, they would be notably benefitted. Administration of 5- azacytidine to thalassemia patients increases haemoglobin-f level in erythrocytes and thus benefit the patients. Which one of the following statements about 5- azacytidine in NOT correct? (A) Cells exposed to 5-zacytidine incorporate it into DNa in place of cytidine (B) 5-azacytidine decreases DNA methylation (C) 5-zacytidine promotes histone acetylation (D) 5-zacytidine does not promote gene expression 87. In cells having G protein coupled receptor, inhibition of protein kinase A by sirna technology led to diminished transcription of androgen binding protein (ABP) and CREB protein. Addition of camp, which is a second messenger, will lead to (A) Increased transcription of ABP (B) Increased phosphorylation of CREB protein (C) No change in transcription level (D) Increased GTPase activity of G subunit 88. Binding of a ligand to a cell-surface receptor activates an intracellular signal transduction pathway through thr sequential activation of four protein kinase. In the human cell line A, by a scaffolding protein whereas in another cell line B, these kinase are freely diffusible. Which one of the following possibilities do you think is NOT correct? (A) Speed of signal transduction will be higher in cell A (B) Possibility of cross-linking with other signal transduction pathways will be lesser in cell A (C) Possibility of signal amplification will be higher in cell A (D) Potency of spreading signal through other signaling pathways will be higher in cell B 89. Mouse erythroleukemia (MEL) cells are used as an in vitro cell culture model for understanding erythropoiesis. These cells are arrested at the stage of pro-erythroblast due to transformation. These cells could be induced by heme to differentiate further so as to synthesize haemoglobin. The most probable molecule mechanism for this could be that heme may suppress and/or downregulate an endogenous heme-regulated inhibitor (HRI) kinase, an inhibitor of globin synthesis. This P8/14 A/UG-15 downregulation in turn promotes differentiation. To validate this hypothesis which of the following approaches is NOT appropriate? (A) Transfect MEL cells with HRI kinase gene (B) Knock down HRI kinase gene in MEL cells (C) Determine the rate of protein synthesis in situ as a function of differentiation. (D) Measure HRI kinase activity as a function of differentiation 90. Cells undergo apoptosis by two distinct and interconnected pathways: extrinsic and intrinsic. Extrinsic pathway is activated by extracellular ligand binding to cell surface death receptors. Whenever an apoptotic stimulus activates intrinsic pathway, the pro-apoptotic BAx and Bak proteins become activated and induce the release of cytochrome C from mitochondria leading to capase cascade activation resulting in apoptosis. In cell A, cythchrome C is introduced by mocroinjection but BAX and BAk are inactivated. What will be the most appropriate apoptotic response type in both cells? (A) (C) (B) (D) 91. Dendritic cells (DC) from BALB/c mice were treated with 1L-10 or IFN-. These cells were co-cultured with CD8 + T cells from hen egg lysozyme (HEL)-specific T cell receptor transgenic mice in presence of the HEL peptide. Five days later, CD8 + T cells were assayed for target cells lysis. Which one of the following combinations will have the highest target cytotoxicity? (A) DC (BALB/c) 1L-10 xcd8 + T (B) DC (BALB/c) 1FN- xcd8 + T (C) DC ( 2-microglobulin - deficient) 1L-10 xcd8 + T (D) DC ( 2-microglobulin - deficient) 1FN- xcd8 + T 92. Polyspermy results when two or more sperms fertiliuze an egg. It is usually lethal since it results in blastomers with different numbers and types of chromosomes. Many species therefore, have two blocks to ployspermy: the fast block and the slow block. In the case of sea urchins: 1. The fast block is immediate and causes the egg membrane resting potential to rise which does not allow the sperm to fuse with the egg and is mediated by an influx of sodium ions.

222 A/UG The fast block is immediate and causes the egg membrane resting potential to rise which does not allow the sperm to fuse with the egg ans is mediated by an efflux of sodium ions. 3. The slow block or cortical granule reaction is mediated by calcium ions. 4. The slow block or cortical granule reaction is mediated by potassium ions. (A) 1 and 3 (B) 1 and 4 (C) 2 and 3 (D) 2 and In an experiment, the cells that would normally become the middle segment of a Drosophila leg were removed from the leg forming area of the larva and were placed in the tip of the fly's antenna. Based on the "French flag" analogy for the operation of a gradient of positional information, which of the following statements is true? (A) Thetransplanted cellsretain their committed status as leg cells, but respond to the positional information of their environment by becoming leg tip cells - i.e., claws (B) The transplanted cells are determined as legcells and therefore would form a complete limb instead of an antenna (C) The transplanted cells would intermingle with the cells present in the new environment and develop accordingly to give rise to an antenna. (D) The transplanted cells retain committed status as leg cells and would develop to form a chimeric structure having proximal region made of antenna and the distal region ending in a complete leg (D) 2 and Which of the inferences (A - D) given below would you draw from the following tissue transplantation experiments performed with the early and late gastrula stages of the newt? Host regions Donar Differentiation regions of donor tisuue EARLY GASTRULA (i) Prospective neurons Prospective Epidermis epidemis (ii) Prospective Prospective neurons Neurons epidermis LATE GASTULA (i) Prospective neurons Prospective Neurons epidermis (ii) Prospective Prospective neurons Epidermis epidermis 1. Cells of early newt gastrula exhibit conditional development 2. Cells of early newt gastrula exhibit autonomous development Amar Ujala Education 3. Cells of late newt gastrula exhibit conditional development 4. Cells of late gastrula exhibit autonomous development The correct inferences are (A) 1 and 4 (B) 2 and 3 (C) 1 only (D) 4 only 95. Segmentation genes in Drosophilla are divided into three groups (gap, pair rule and segment polarity) based on their mutant phenotype. Below are some of the major genes expressed in a sequential manner (with respect to the groups) affecting segmentation pattern. 1. Hairy paired tailles patched 2. Hunchback even-skipped fushi tarazu wingless 3. Odd-skipped giant paired wing less 4. Tailless hairy fushi tarazu goose berry Which of the above sequence(s) of genes expressed from early to late embryo is/are correct? (A) 4 only (B) 1 and 2 (C) 2 and 3 (D) 2 and Human chrionic gonadotropin (hcg) is known to facilitate attachment of blastocyst to uterus. In women with mutation in hcg gene, biologically inactive hcg was formed but implantation occured. When hcg was immunoneutralized in the uterus of normal woman, implantation failed. This suggests that for implantation in humans: (A) Biologically active circulating hcg is not required (B) Blastocyst can produce the required hcg, which helps locally in uterine attachment (C) Trophoblastic cells do not require hcg for the invansion of uterus (D) Extra-embryonic tissue is not responsible for the attachment of embryo to uterus 97. During reproductive development in plants: 1. Male and female gametes are produced as a result of two mitotic divisions after meiosis. 2. Vegetative cells in pollen contribute to pollen development. 3. Antipodals provide nourishment to developing embryos. 4. Pollen tube ruptures and releases both the male gametes in one of the degenerated synergids. Which of the above statements are true? (A) 1 and 2 (B) 2 and 4 (C) 2 and 3 (D) 1 and 4 P8/15

223 Amar Ujala Education 98. During fertilization in mammals, sperm-egg interaction is mediated by zona pellucida (ZP) membrane proteins and their receptors present in sperm membrane. ZP3 has been identified to be the principle ZP protein whose post-translational modification is important for spermegg interaction. In a competitve inhibition assay the sperm is saturated with the either active ZP3 ir its modified forms, before studying sperm-egg-interaction. Which of the following experiments will NOT inhibit sperm egg-interaction? (A) Saturate sperm with ZP3 protein prior to use (B) Deglycosylate the ZP3 protein and use it for saturation of sperm (C) Phosphorylate the ZP3 protein and use it for saturation of sperm (D) Dephosphorylate the ZP3 protein and use it for saturation of sperm Exp. If an Arabidopsis plant, mutated in lycopene biosynthetic pathway is grown in sunny tropical climate in the presence of oxygen, It would show reduced biomass due to photo oxidative damage 99. If an Arabidopsis plant, mutated in lycopene biosynthetic pathway is grown in sunny tropical climate in the presence of oxygen: (A) It would accumulate higher biomass due to higher rate of photosynthesis (B) There will not be any influence of this mutation one the rate of photosynthesis and plant growth (C) It would show reduced biomass due to photo oxidative damage (D) The leaves would be bluish purple in colour because of higher accumulation of xanthophylls 100. According to the current model of alternative oxidase regulation, the following factors cause induction of alternative oxidase: 1. Significant increase in the ubiquitin pool in the cytosol 2. Presence of -keto acids (like pyruvate and glyoxylate) 3. Cold stress 4. Increase in cytosolic ATP concentrantion Which one of the following combinations of above statements is true? (A) 1 and 4 (B) 2 and 3 (C) 1 and 2 (D) 1 and The oxidative pentose phosphate pathway provides the reducing equivalents for nitrite reduction in plastids (leucoplasts) of non-green tissues. Which one of the following statements would be correct for the above mentioned pathway? P8/16 A/UG-15 (A) Glutamate synthesized from NH 4+ is translo cated from cytosol to leucoplast. (B) -ketoglutarate is translocated from cytosol to leucoplast (C) Glucose-6-phosphate is translocated and moves from leucoplast to cytosol (D) Triose phosphate is translocated from cytosol to leucoplast 102. Perception of blue light in plants causes (A) Inhinition of cell elongation and stimulation of stomatalopening (B) Stimulation of cell elongation and inhibition of stomatal opening (C) Inhibition of stomatal opening (D) Inhibition of cell elongation Exp. Blue light is involved in many plant responses, including phototropism initiation, stomata opening, and hypocotyl elongation when the seedling emerges from the soil. Although Charles Darwin first documented plant responses to blue light in the 1800s, it was not until the 1980s that research began to identify the pigment responsible. Arabidopsis was used extensiively in the study of the genetic basis of phototropism, chloroplast alignment, and stomatal aperture and other blue light-influenced processes. These traits respond to blue light, which is perceived by three different blue light receptors: cryptochromes, phototropin light receptors, and zeaxanthin. Cryptochromes are necessary for seedling elongation just after emergence. Phototropin is a protein kinase involved in plant curvature during phototropic responses, and zeaxanthin is a carotenoid, which appears to be involved in stomatal openings in response to blue light Following are few statements regarding water potential of soil. 1. The osmotic potential ( s ) of soil water is generally negligible, except in saline soils. 2. The osmotic potential ( s ) of saline soil is always more than zero. 3. In dry soils the hydrostatic pressure ( p ) of soil water potential is always positive. 4. Gravitational potential ( g ) of soil water is always proportional to height of the tree. Which one of the following combinations of above statements is true? (A) 1 and 3 (B) 2 and 4 (C) 2 and 3 (D) 1 and 4

224 A/UG Which one of the following pairs of precursor amino acid and alkaloid is correct? (A) 'Ornithine aspartate-nicotine' and 'tryptophanquinine' (B) 'Ornithine-nicotine' and 'tyrosine-orphine' (C) 'Tyrosine-quinine' and 'tryptophan-orphine' (D) 'Ornithine-quinine' and 'ornithine aspartatenicotine' 105. Typical morphological defects are routinely used in genetic screens to identify novel genes in signal transduction pathways. which one of the following morphology has been used to decipher the ethylene signaling pathways? (A) Light grown morphology of seedling (B) Triple response morphology of seedling (C) Dark grown morphology of seedling (D) Morphology of true leaves 106. In bone marrow, stem cells are commited to different lineages. Factors that stimulate the colonies of these different lineages are interleukin-3 (multi-csf), granulocyte-macrophage colony stimulating factor (GM-CSF) and granulocyte or macrophage colony stimulating factor (G-CSF or M-CSF). In a mouse deficient in GM-CSF, the number of hematopoietic cells will be altered. Which one of the following is correct? (A) Mast cells will be normal in number while granulocytes and macrophages will be deficient in number (B) Granulocytes count will be normal but not of macrophages (C) Macrophages number will remain unaltered (D) Mice will be deficient in all the three cell types 107. An individual was suffering from digestive complications. It was observed that the individual had dehydrated gastrointestinal tract. When advanced investigation was done, the person was found to have defects in the following: 1. Cystic fibrosis transmembrane conductance regular protein 2. Glucose transporter protein 3. Na + /K + ATPase 4. Ca 2+ ATPase Which one of the above could be the cause for such a digestive disorder? (A) 1 only (B) 2 and 3 (C) 3 and 4 (D) 4 only 108. the action potential was recorded intracellularly from a squid giant axon bathed in two types of fluid such as sea water and artificial sea water having lower concentration of sodium ions while choline chloride. Amar Ujala Education The nature of action potential was different in the two bathing fluids. Which of the following results is most likely? (A) The resting transmembrane potential was not changed but the amplitude of action potiential was increased with lower sodium concentration in the bathing fluid (B) The amplitude of action potential was gradu ally decreased with reduction of sodium concentration in bathing fluid but the duration of action potential was prolonged (C) The resting transmembrane potential was decreased and the amplitude of action potential was also decreased with lower sodium concentration in the bathing fluid (D) The amplitude of action potential was not changed with reduction of sodium concen tration in the bathing fluid but the duration of action potential was prolonged 109. Three forms of dextrans namely neutral, polyanionic and polycationic having different molecular radii were injected separately in three groups of rats. The concentrations of dextrans in glomerular filtrate were measured to determine the filterability of the dextrans. The possible outcomes could be as follows: 1. The dextrans having smaller diameter have greater filterability than larger dextrans 2. Neutral dextrans were filtered more than polycationic and polyanionic dextrans 3. Polycationic dextrans were filtered more than neutral and polyanionic dextrans 4. Polyanionic dextrans were filtered more than neutral and polycationic dextrans Which one of the following combination is correct? (A) 1 only (B) 2 only (C) 1 and 3 (D) 2 and A novel enzyme was identified in humans. The following approaches are available to identify the chromosome on which the gene encoding the enzyme is present: 1. Suppress the activity of enzyme by RNAi. 2. Identify polymorphism in the population and carry out pedigree analysis to study its inheristance. 3. Purify the enzyme, decipher its amino acid sequence, predict its DNA sequence and search for its presence in the available human genome sequence. 4. Create chromosomes addition lines by making somatic hybrids between human and mouse cells, identify lines showing the enzyme activity and the human. Which one of the above approaches can be used? (A) 1 or 2 (B) 2 or 3 (C) 3 or 4 (D) 1 or 3 P8/17

225 Amar Ujala Education 111. In an experiment on trasposition in an eukaryotic system, an atom intron was cloned within a transposablr element and allowed to transpose from a plasmid to genomic DNA. The nitron was found to be absent in the transposable element in its new location. it is: (A) Not a case of transposition (B) A case of replicative mode of transposition (C) A case of conservation mode of transposition (D) A retroposon Exp. Retroposon : A transposition of DNA sequences that does not occur in DNA itself but rather when mrna is transcribed back into the genomic DNA. Or it is A class of genetic elements that includes retroviruses and transposons that have an intermediate RNA stage. A transposon that was created by reverse transcription of an RNA molecule In a plant species, a segregating line (one that contains both homozygotes and heterozygotes at a locus) can be made homozygous by repeated selfing for several generations. What is the level of remaining heterozygosity after three generations of selfing, if the level of heterozygosity in generation '0' is denoted as 1? (A) 0.5 (B) 0.25 (C) (D) Given below is the result of a complementation test for six independent mutants (1 to 6). '+' represents complementation; '0' represents noncomplementation Based on the above, which one of the following conclusions is correct? (A) The mutations can be ordered in a single cistron as (B) All mutations belong to a single cistron, but their order cannot be determined (C) There are three cistrons, mutations 1, 3 and 6 represent one cistron. 4 and 5 represent the second cistron and 2 represents the third cistron (D) There are three linkage groups, mutations 1, 3 and 6 represent linkage group A, 4 and 5 represent linkage group B, and 6 represents linkage group C A/UG In a hospital three babies were mixed up. The blood group of the babies were A, B and AB. In order to identify the parents of the babies the blood groups of the parents were determined. The results obtained were: Parent set 1 - A and AB Parent set 2 - AB and O Parent set 3 - B and AB Which of the following conclusions can be definitively made? (A) The baby with blood group A is the child of the parent set 2 (B) The baby with blood gropu AB is the child of the parent set 1 (C) The baby with the blood group B is the child of the parent set 3 (D) The parentage of none of the babies can be determined from the given information 115. There are two mutants plants. One shows taller phenotype than wild type, whereas the other has the same height as the wild type. When these two mutations were brought in together by genetic crosses, the double mutant displayed even taller phenotype than the tall mutant plants. This genetic interaction is called (A) Antagonistic interaction (B) Additive interaction (C) Synergistic interaction (D) Suppressive interaction Exp. Synergy occurs when the contribution of two mutations to the phenotype of a double mutant exceeds the expectations from the additive effects of the individual mutations. The molecular characterization of genes involved in synergistic interactions has revealed that synergy mainly results from mutations in functionally related genes. Recent reseach in Arabidopsis thaliana has advanced out understanding of the scenarios resulting in synergistic phenotypes. Those involving homologous loci usually result from various levels of functional redundancy. Some of these loci fail to complement each other or become dosedependent in certain multiple mutant combinations, suggesting that the effects of haploinsufficienct and redundancy can combine. Synergy involving non-homologous genes probably reflects the topology of the regulatory or metabolic networks and can arise when pathways that converge at a node are disrupted. The Hub genes provide a remarkable example, these genes have an extraordinary number of connections and mutations that interact with many unrelated pathways. P8/18

226 A/UG The following tables gives vascular tissue characteristics of four divisions of Tracheophyta. Divisions Vascular tissue characteristics 1. Psilophyta i. Well-developed tracheid and pits in lateral wall 2. Lycopodiophyta ii. Tracheids 3. Sphenophyta iii. Tracheids, vessels and well-developed phloem 4. Pteridophyta iv. Primitive tracheids and pits in lateral wall Identify the correct combinations:- (A) 1-i, 2-ii, 3-iii, 4-iv (B) 1-ii, 2-i, 3-iv, 4-iii (C) 1-iv, 2-iii, 3-ii, 4-i (D) 1-iii, 2-iv, 3-i, 4-ii 117. Which one of the following is NOT an advantage to seed-based reproduction? (A) Reserve food material is provided for the developing embryo (B) Seed coat protects the embryo and allows it to remain dormant until favourable environmental conditions are available (C) The amount of energy spent per female gametophyte is less than that spent on making a spore (D) The female gametophyte remains on the sporophyto which provides protection and nourishment 118. In a study of sexual isolation in a species of salamander, scientists brought together males and females from different populations and from the same population. They observed the frequency of mating and calculated a sexual isolation index. One graph shows the relationship between mating frequency and genetic distance, and the other shows the relationship between sexual isolation index and geographic isolation. Choose the appropriate terms for X 1,Y 1,X 2 and Y 2 in the figure, above. (A) X 1 = Geographic distance, Y 1 = Sexual isolation index; X 2 = Genetic distance Y 2 = Mating frequency (B) X 1 = Geographic distance Y 1 = Mating frequency X 2 = Genetic distance Y 2 = Sexual isolation index (C) X 1 = Genetic distance Y 1 = Mating frequency X 2 = Sexual isolation index Y 2 = Geographic distance (D) X 1 = Genetic distance Y 1 = Geographic distance X 2 = Sexual isolation index Y 2 = Mating frequency Amar Ujala Education 119. As per the of International Code of Botanical Nomenclature, 2006 (Vienna Code), which of the following is a Nothospecies? (A) Polypodium vulgare subsp. prionodes (Asch.) Rothm. (B) Polypogon Monspeliensis (L.) Desf. (C) Agrostis Stolonofera L. (D) Agrostis stolonofera L. x Polypogon monspeliensis (L.) Desf. Exp. As per the of International Code of Botanical Nomenclature, 2006 (Vienna Code), Agrostis stolonofera L. x Polypogon monspeliensis (L.) Desf. is a Nothospecies Which of the following groups have only two wings? (A) Honey bee, beetle, ant (B) Butterfly, housefly, fruitfly (C) Dragonfly, butterfly, fruitfly (D) Housefly, fruitfly, mosquito Exp. Flies and mosquitoes are classified as order Diptera, which mean two wings. The insects in this order have only one pair of membranous flying wings. The second pair of wings are reduced to small knobs, called halteres, for the purpose of balancing. Their body is relatively soft and hairy. They have a pair of large compound eyes, a pair of very short antennae and a sucking mouth. Example of such flies are Housefly, fruitfly, mosquito India has currently 17 biosphere reserves representing different ecosystems. These conservation area significantly differ from the conventional protected areas significantly differ from the conventional protected areas of the country. Identify the correct combination of attributes (A to G) that best explains the concept of biosphere reserve. 1. Conservation, 2. Education, 3. Human habitation allowed, 4. Human habitation not allowed, 5. Strong legal back-up, 6. No supporting act, 7. Research. (A)1,2,3,6,7 (B)1,2,4,6,7 (C)1,2,3,5,7 (D)1,4,5, Followings are the niche characteristics of the constituent species and resource partitioning pattern in different ecosystems. Which of these would lead to competitive exclusion of species? (A) P8/19

227 Amar Ujala Education (B) (C) (D) A/UG /km 2. On the third island, both A and B co-exist with densities 297/km 2 and 150/km 2, respectively. Which of the following can be inferred from this? (A) The two species do not compete with each other (B) The intra-species competition is more intense than inter-species competition (C) The intra-species competition is more intense than inter-species competition (D) The inter and intra species competition are of the same intensity A few males and females of a species were introduces to a new island. Their population was monitored over sevral generations and followed a pattern shown in the figure : 123. Environmental conditions can influence accumulation of species in successional communities. Curves representing changes in forest species over time are given in the figure below. which of the following keys is correct for the curves? (A) A = xeric, B = mesic, C = intermediate (B) A = intermediate, B = xeric, C = mesic (C) A = intermediate, B = mesic, C = xeric (D) A = mesic, B = intermediate, C = xeric 124. A plant with blue-coloured flowers was observed to attract a large number of pollinators. However, these flowers were not producing any nector. Which of the following can be logical explanation to the observation? (A) There could be another species in the vicnity that has blue flowers and is rich in nectar (B) There is not other species with blut flowers in the vicnity so pollunitors are compelled to visit this species (C) Pollinators may not have blue-colour vision (D) Pollinators may be able to see only blue colour 125. Three islands have identical habitat characteristics. on first island rodent species A is present at a density 325/ km 2. Second island has only species B at a density of P8/20 Which of the characteristics of the species does NOT explain the pattern? (A) Skewed sex ratio (more females than males) (B) Large litter size (C) Delayed sexual maturity (D) Effects of intra-uterine development on fecundity 127. Plasmids are self replicating small circular DNA elements in bacterial cells that can be said to have a stable symbiotic existence with the host cell. They often carry genes useful to the host. Which of the following is a potential threat to the evolution and stability of the symbiotic coexistence with the host cell. They often carry genes useful to the host. Which of the following is a potential threat to the evolution and stability of the symbiotic coexistence? (A) 'Copy-up' mutations that increase the rate of plasmid replication per host cell cycle (B) Reversible integration of plasmid DNA into the host DNA (C) Transfer of plasmids to new cells by conjugation (D) Spontaneous curing of plasmids in a small proportion of host cells Complex eukaryotic cells may have evolved from simpler prokaryotic cells because complexity of organization increases the (A) Growth rate (B) Efficiency of energy untilization (C) Tolerance to starvation (D) Ability to attain larger size

228 A/UG-15 Exp. Prokaryotic cells are known to be much less complex than eukaryotic cells since eukaryotic cells are usually considered to be present at a later time of evolution. It is likely that Eukaryotic cells have evolved from Prokaryotic cells. These differences in complexity can be seen ast the cellular level. The DNA of prokaryotes is circullar and attached to the plasma membrane, while eukaryotic DNA is packed into chromosome bundles. Prokaryote's DNA is contained in nucleoid which is not surrounded by nuclear membrane. Eukaryotic DNA is more complex where it has histone (protein that winds the DNA into a more compact form), and non-histone proteins in chromosomes. Chromosomes are contained in the nucleus with a nuclear envelope (a defining feature of eukaryotic cells) Knox genes code transcriptional factors important for the regulation of indeterminate growth in plant shoots. These genes also regulate patterns of development of plant organs such as leaves and flowers. The figure represents a phylogenetic tree of the multigene family in some land plants. The circles represent genes that act to maintain shoot apical meristem (equivalent to stem cells). Orthologues are genes that duplicate due to speciation and paralogues are genes that duplicate within a species. From the figure, the following inferences were made. 1. Multiple gene duplication occured in vascular plants. 2. Gene duplications may have enabled shoot diversification in vascular plants 3. Shoot apical meristems are regulated by orthologous genes in vascular plants 4. Shoot apical meristems are regulated by paralogues genes in vascular plants Which of the following represents a combination of correct inferences? (A) 1, 2 and 4 (B) 1,2,and3 (C) 2 and 3 only (D) 2 and 4 only 130. In an experiment that has continued for more than 50 years, corn has been propagated by breeding only from plants with the highest amount of oil in the seeds. The average oil content is now much greater than any of the plants in the original population. P8/21 Amar Ujala Education The following hypotheses were proposed as explanations for this observation. 1. Mutations occured that increased the oil content in seeds. 2. Plants with high oil content were stimulated to produce offspring with more oil in their seeds. 3. The breeding led to increased frequency of alleles at multiple loci, so that new combinations of genes for even higher oil content were formed. Which of the following represents a combination of correct statements? (A) 1 and 2 only (B) 1 and 3 only (C) 2 and 3 only (D) 1, 2 and The Galapagos finches were an important clue to Darwin's thinking about the origin of species. These finches are believed to have descended from a single ancestral species that colonized the Galapagos archipelago, America, over a short period of time. The Galapagos finches differ in their beak shape and size. Different species feed on seeds that vary in size and hardness. Which of the following is the most likely explanation for these patterns? (A) The finches represent an example of directional trend in beak size from small to big (B) Beak shapes changed in response to different seed types and these changes were in herited by subsequent generations (C) The ancestral finch already has all the beak variations and different lineages formed that were specialized to eat different seed types (D) The finches represent an example of adaptive radiation in which beak variation was generated by mutation followed by selection by different seed types 132. In order to demonstrate that the long tails of males attracted females in a bird species, experiments captured and cut the tails of 'n' number of males and monitored the number of females mated by each male. They had two types of controls in the experiment. (i) 'n' males that were not captured (ii) 'n' males that were captured, had their tails cut and then stiched back to attain the original size. The males with cut tails mated with a significantly smaller number of females than both the controls. Which of the following alternative explanations is NOT ruled out by the experiment? (A) The stress of cutting tails affected the performance of males (B) The time wasted in the capture reduced mating opportunities of males (C) Females avoided any deviation from normal (D) Females chose males randomly

229 Amar Ujala Education 133. In the phylogenetic tree above, branch-lengths are drawn proportional to the number of changes along a lineage. The following inferences were made from this tree. 1. Bacteria and Archea are more similar to each other than either is to Eukarya. 2. Bacteria and Archea are more similar to each other than either is to Eukarya. 3. Archea and Eukrya diverged from each other after their common ancestor diverged from bacteria. Which of the following represents a combination of correct inferences? (A) 1, 2 and 3 (B) 1 and 3 only (C) 2 and 3 only (D) 1 and 3 only 134. While attempting to create a disease model of poliomyelitis in mice, it was found that mice can not be infected with the said virus. Since human beings are suseptible to this viral infection, which kind of transgenic mice should be generated to have a transgenic mouse model that can be infected with polio virus? Select the right approach from below (A) A mouse expressing surface protein of polio virus (B) A mouse human receptor gene which makes cells surface protein for docking and internatlization of polio virus (C) A mouse expressing human MHC class II invariant chain (D) A mouse expressing human receptor gene which makes cell surface protein for docking and interlization of polio virus along with a gene designed to express surface protein of this virus at puberty Protoplast fusion is used in plant tissue culture for various applications. In protoplast fusion: 1. Naked plant 2. Transfer of organelles is not possible 3. Partial genome transfer is involve 4. Cells from two different plants can be mixed together and forced to fuse Which one of the following combinations of the above statements is correct? (A) 1, 2 and 3 (B) 1, 3 and 4 (C) 1, 2 and 4 (D) 2, 3 and 4 P8/22 A/UG Which of the following is a mismatch between the plant drug and its source? (A) Codeine - papaver somniferum (B) Vinblastine - Catharanthus roseus (C) Quinine - Cinchona ledgeriana (D) Digitalin - Artemisia annua 137. Which of the following curves correctly represents the process of ethanol production by yeast? (A) (C) (B) (D) 138. Inbreading for 5 generations led to production of homozygrous transgentic mice. However, these homozygous males or females were infertile. Which of the following approach is most preferable and economical to obtain heterozygous transgentic animals continuously? (A) More transgenic founder (1 st animal) should be generated (B) Crossing (breeding) of transgenic mice with wild type mice in earlier generations should be done for continued production of transgenic heterozygous offspring (C) Inbreeding should be avoided after 5th generation (D) Homozygous transgenic mice should be mated with wild type mice for continued production of transgenic heterozygous off spring 139. Following are few statements for regeneration of plants from explant/tissues. 1. Cytokinin is required for shoot development. 2. Auxin is required for shoot development. 3. Auxin to xytokinin ratio is very important. 4. Jasmonic acid is required for both root and shoot development Which of the following combinations of above statements is true? (A) 1 and 3 (B) 2 and 4 (C) 1 and 4 (D) 2 and 3

230 A/UG A set of neonatal mice are divided into four groups. Group 1 neonates were not primed with any antigen. Group 2 neonates were primed with KLH. Group 3 neonates were primed with KLH but thymectomized, but reconstituted with KLH bu thymectomized. Group 4 neonates were KLH-primed, thymectomized, but reconstituted with KLH-specific CD4 + T cells. All these mice, when grown adult, were challenged with KLH and the anti-klh IgG antibody was measured in sera. Which of the following is the correct order of magnitude of antibody response? [> = greater than, > = greater than or equal to] (A) Group 1 > Group 2 > Group 3 > Group 4 (B) Group 2 > Group 1 > Group 3 > Group 4 (C) Group 2 > Group 3 > Group 1 > Group 4 (D) Group 4 > Group 1 > Group 2 > Group Choose the correct sequences of events in a next generation sequencing technology-based on whole genome sequencing project. (A) DNA extraction shearing library preparation sequencing assembly finishing annotation submission to Genbank (B) DNA extraction library preparation sequencing assembly annotation finishing submission to Genbank (C) DNA extraction shearing adapter ligation library amplification sequencing assembly finishing annotation submission to Genbank (D) DNA extraction adapter ligation library amplification shearing sequencing finishing assembly annotation submission to Genbank 142. An investigator discovers a new receptor for a known ligand and wanted to identify the binding partner of the receptor i.e. is coreceptor. The antireceptor antibody is not available but anti-gfp antibody is available. Which one of the following strategies is most likely to identify the co-receptor? (A) The GFP-receptor fusion protein is expressed in a cell line and analyzed by LC-MS/MS (B) The GFP-receptor fusion protein is expressed in a cell line and the cells positive for GFP were sorted out, lysed and run on a poly acrylamide gel (C) The GFP-receptor protein is coated on ELISA plate, followed by ELISA with anti- GFP an tibody (D) The receptor is cloned as a fusion protein of GFP and expressed in stimulated cells. The immunoprecipitated complex obtained by anti-gfp antibody was analyzed by LC-MS/MS. Amar Ujala Education Exp. An investigator discovers a new receptor for a known ligand and wanted to identify the binding partner of the receptor i.e. is coreceptor. The antireceptor antibody is not available but anti-gfp antibody is available. This strategy is most likely to identify the co-receptor.the receptor is cloned as a fusion protein of GFP and expressed in stimulated cells. The immunoprecipitated complex obtained by anti-gfp antibody was analyzed by LC-MS/MS The frequency distribution of tree heights in two forest areaa with different annual rainfall are given Which of the following statistical analysis will you choose to test whether rainfall has an effect on tree heights? (A) t test for comparison of means (B) A non parametric comparison of the two groups (C) Correlation analysis of rainfall and mean tree heights (D) Regression of tree heights on rainfall 144. Two species of plants were sampled in 32 quadrats in a forest. The mean and variance for the occurence of species 1 were 16.2 and 48 and species 2 were 3.6 and 3.2 respectively. Which of the following statements about the distribution of the two species in these quadrats is supported by these findings? (A) Both species are distributed randomly (B) Species 1 is distributed randomly and species 2 is clustered (C) Species 1 is clustered and species 2 is distributed randomly (D) Both species are clustered 145. Poly-L-lysine exists in pure -helix, -sheet and random coiled conformations depending upon the solvent conditions. The values of mean residue ellipticity at 220nm ([ ] 220 ) are -35,700, -13,800 and +3,900 deg cm 2 dmol -1 for -helix, -sheet and random coil conformations of this polypeptide, respectively. The polypeptide exists in -helix conformation at ph 10.8 and 25 O C. Addition of urea leads to a two state transition between -helix and random coil conformation. It has been observed that [ ] 222 of the polypeptide is deg cm 2 dmol -1 inthepresenceof6 Murea.The percentage of the polypeptide in -helix conformation is (A) 37 (B) 41 (C) 47 (D) 50 P8/23

231 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, December-2012 PART-A 1. Agraniteblockof2mx5mx3msizeiscutinto5cm thick slabs of 2 m x 5 m size. These slabs are laid over a 2 m wide pavement. What is the length of the pavement that can be covered with these slabs? (A) 100 m (B) 200 m (C) 300 m (D) 500 m Exp. Volume of granite block = =30m 3. Volume of slab = =0.5m 3. Number of slab = = 60. Now, let the length of the pavement = x m. Area of pavement = 2 x And base area occupied by single slab =2 5 =10m 2. Area occupied by 60 slabs = = 600 m 2. According to the problem, 600 = 2 x or, x = = 300 m. 2. Which is the least among the following? , ,,e -1/e (A) (B) (C) (D) e -1/e Exp. (0.33) 0.33 = (0.44) 0.44 = ( ) 1 = 1 ( ) 1 1 = (3.14) 1 = = (e) 1 1 e = 1 (2.718) = = What is the next number in this "see and tell" sequence? (A) (B) (C) (D) A vertical pole of length a stands at the centre of a horizontal regular hexagonal ground of side a. A rope that is fixed taut in between a vertex on the ground and the tip of the pole has a length: (A) A (B) (C) 2a 3a (D) 6a Exp. Every sector of a hexagonal Ground is equilateral triangle. So, distance of a vertex from the center of haxagon BC = side length of Pole length = AC = a From DABC, AB = 2 2 a + a = 2a. P9/1

232 Amar Ujala Education 5. A peacock perched on the top of a 12 m high tree spots a snake moving towards its hole at the base of the tree from a distance equal to thrice the height of the tree. The peacock flies towards the snake in a straight line and they both move at the same speed. At what distance from the base of the tree will the peacock catch the snake? (A) 16 m (B) 18 m (C) 14 m (D) 12 m Exp. Height of tree AB = 12 m. Distance of smake from its hole BC = 36 m. Let the snake & Peacock meets at point D such that BD = x m. (A) 1 : 1 (B) 1 : 2 (C) 1 : 3 (D) 1 : 4 Exp. Let BC = h and BE = DE = AD = x A/UG-15 Area of ADC =Areaof ABC Area of BCD = 1 2 (3x)h = 1 2 (2x)hx = 1 2 xh As speed of two are same, so in equal time, distance AD = CD = (36 x)m. Again, from rigth triagle ABD AB = 12 m BD = x m AD = (36 x)m (AD) 2 =(AB) 2 +(BD) 2 (36 x) 2 = (12) 2 +(x) x 2 72x = x 2 72x = x = 1152 x = =16m. 6. The cities of a country are connected by intercity roads. If a city is directly connected to an odd number of other cities, it is called an odd city. If a city is directly connected to an even number of other cities, it is called an even city. Then which of the following is impossible? A. There are an even number of odd cities B. There are an odd number of odd cities C. There are an even number of even cities D. There are an odd number of even cities 7. In the figure ABC = /2 AD = DE = EB What is the ratio of the area of triangle ADC to that of triangle CDB? Area of BCD = 1 2 (2x) h 1 xh 2 Ratio of two areas = 1 ( 2 x) h =1: A rectangular sheet ABCD is folded in such a way that vertex C, thereby forming a line PQ. Assuming AB = 3 and BC = 4, find PQ. Note that AP = PC and AQ = QC. (A) 13/4 (B) 15/4 (C) 17/4 (D) 19/4 9. A string of diameter 1 mm is kept on a table in the shape of a close flat spiral i.e. a spiral with no gap between the turns. The area of the table occupied by the spiral is 1 m 2. Then the length of the string is: (A) 10m (B) 10 2 m (C) 10 3 m (D) 10 6 m P9/2

233 A/UG % of 25% of a quantity is x% of the quantity where x is (A) 6.25% (B) 12.5% (C) 25% (D) 50 % Exp. Let quantity is P. According to the question, P =P x or, x = x = In sequence {a n } every term is equal to the sum of all its previous terms. If a 0 = 3, then lim (A) 3 (B) 2 (C) 1 (D) e Exp. {an}=a 0 + a 1 + a a n 1 {a n+1 }=a 0 + a 1 + a a n lim n n a a + 1 n =lim n a0+ a1+ a a a + a + a a n 1 n Amar Ujala Education 2 BC Area of semicircle II = 2 2 = (BC) AB Area of semicirle III = 2 2 = (BC) 2 8 In given rigth triangle ABC (AC) 2 =+(AB) 2 +(BC) 2 or, 8 (AC)2 = 8 (AB)2 + 8 (BC) 2 II + III = I. 13. What is the minimum number of days between one Friday the 13 th and the next Friday the 13 th? (Assume that the year is a leap year). (A) 28 (B) 56 (C) 91 (D) Suppose a person A is at the North-East corner of a square (see the figure below). From that point he moves along the diagonal and after covering 1/3rd portion of the diagonal, he goes to his left and after sometime he stops, rotates 90 O clockwise and moves straight. After a few minutes he stops, rotates 180º anticlockwise. Towards which direction he is facing now? = lim n an 1+ a + a + a a n 1 = lim n a + a + a a 1+ a + a + a a = lim n (1 + 1) = n n In the figure below, angle ABC = /2. I, II, III are the areas of semicircles on the sides opposite angles B, A and C respectively. Which of the following is always true? (A) II 2 +III 2 =I 2 (B) II + III = I (C) II 2 +III 1 >I 2 (D) II + III < I 2 Exp. Area of semicircle I = 2. AC 2 = 8 (AC) 2 (A) North-East (B) North-West (C) South-East (D) South-West 15. Cucumber contains 99% water. Ramesh buys 100 kg of cucumbers. After 30 days of storing, the cucumbers lose some water. They now contain 98% water. What is the total weight of cucumbers now? (A) 99 kg (B) 50 kg (C) 75 kg (D) 2 kg 16. In a museum there were old coins with their respective years engraved on them, as follows (1) 1837 AD (2) 1907 AD (3) 1947 AD (4) 200 BC Identify the fake coin(s) (A) coin 1 (B) coin 4 (C) coins 1 and 2 (D) coin 3 P9/3

234 Amar Ujala Education 17. A student observes the movement of four snails and plots the graphs of distance moved as a function of time as given in figures (1), (2), (3) and (4). Which of the following is not correct? (A) Graph (1) (B) Graph (2) (C) Graph (3) (D) Graph (4) Exp. From graph (C0, snails exist at two different places at same time. Which is practicaly not acceptable. 18. Find the missing letter : A/UG Consider the following equation x 2 +4y 2 +9z 2 = 14x + 28y + 42z = 147 where x, y and z are real numbers. Then the value of x + 2y + 3z is (A) 7 (B) 14 (C) 21 (D) Not unique Exp. x 2 +4y 2 +9z 2 =14x +28y +42z 147 or, x 2 14x +4y 2 28y +9z = 0 { x 2 2(x)(7) +(7) 2 } + {(24) 2 2. (24).(7) + (7) 2 }= {(3z) 2 2.(3z)(7) + (7) 2 }=0 {(x 7) 2 +(2y 7) 2 +(3z 7) 2 }= 0 or, x 7=0,2y 7=0and 3z 7= 0 x +2y +3z 21= 0 x +2y +3z The map given below shows a meandering river following a semi-circular path, along which two villages are located at A and B. The distance between A and B along the east-west direction in the map is 7 cm. What is the length of the river between A and B in the ground? (A) H (C) Z (B) L (D) Y (A) 1.1km (C) 5.5 km (B) 3.5km (D) 11.0 km PART-B 21. Out of the following hydrogen bonding schemes shown by..., which one corresponds to the weakest hydrogen bond in a given solvent condition? (A) O - H... O < (B) N -H... O < (C) O - H... N < (D) N - H... N < 22. Which peptide bonds(s) marked as a, b, c, d and e will be broken when the following oligopeptide is treated with trypsin at ph 7.0? (A) a,b,d,e (B) b,d,e, (C) d, e (D) d 23. In cellular respiration, which of the following processes occur only inside mitochondria and not in the cytoplasm? (A) Glycolysis and the pentose-phosphate path way. P9/4 (B) Glycolysis and the crtitic acid cycle. (C) The citric acid cycle and oxidative phospho rylation. (D) Glycolysis and oxidative phosphorylation. Exp. The second stage of cellular respiration occurs in the mitochondrial matrix of cell and takes part through the citric acid cycle. Oxygen is essential in this stage. Acetyl Co A is the substrate of citric acid cycle. Two molecules of Acettyl Co A are produced in first stage of cellular respiration by breakdown of one molecule of glucose.in second stage, these two molecules of Acetyl Co A are oxidized to carbon dioxide with release of two molecules of ATP. six molecules of NADH and two molecules of FADH2. One NADH provides three ATP and one FADH2 provides two ATP via the electron transport chain. Therefore, in secondstage of cellular respiration, two molecules of Acetyl Co A provide total 24 molecules of ATP.

235 A/UG An ezyme catalysed reaction was measured in the presence and absence of an inhibitor. For an uncompetitive inhibition, (A) Only K m is increased (B) Both K m and V max are decreased (C) Only V max is decreased (D) Both K m and V max are not affected Exp. On rearranging the equation K max =k m /(1+I/K II )andv max =V m /(1+I/K ii ) we see that the apparent K m and V m do decrease as we predicted. K ii is the inhibitor dissociation constant in which the inhibitor affects the intercept of the double reciprocal plot. Note that if I is zero, K m and V max unchanged. 25. KCl (100nM) was entrapped inside large unilamellar vesicles. A diffusions potential across the bilayer can be generated by diluting with buffer containing: (A) 100 nm KCl and a protonophore (B) 100 nm NaCl and a protophore (C) 100 nm KCl and a K + - specific ionophore (D) 100 nm NaCl and a K + -specific ionophore 26. Acetylcholine receptor is an archetype for: (A) Ligand-gated ion channel (B) ATPase dependent voltage-gated ion channel (C) ATPase dependent Ca 2+ - gated ion channel (D) ATPase independent voltage gated ion channel Exp. The nicotinic acetylcholine receptor (nachr) is the archetypal ligand gated ion channel. A model of the alpha 7 homopentameric nachr is described in which the pore lining M2 helix bundle is treated atomistically and the remainder of the molecule is treated as a "low resolution" cylinder. The surface charge on the cylinder is derived from the distribution of charged amino acids in the amino acid sequence (excluding the M2 segments). 27. With reference to lac operon, what will be the phenotype of an E. coli strain having a genotype i - o + z + y - /F' i + o' z - y +? (A) Constitutive for both -galactosidase and lac permease. (B) Inducible for both -galactosidase and lac permease. (C) Inducible for -galactosidase and constitutive for lac permease. (D) Constitutive for -galactosidase and inducible for lac permease. 28. An organism that has peroxidase and superoxide dismutase but lacks catalase is most likely an: (A) Aerotolerant anaerobe (B) Aerotolerant aerobe (C) Obligate anaerobe (D) Facultative anaerobe P9/5 Amar Ujala Education 29. During DNA replication, events at the replication fork require different types of enzymes having specialized functions except: (A) DNA polymerase III (B) DNA gyrase (C) DNA ligase (D) DNA glycosylase 30. Which of the following names is appropriate for the sequence 5' -G/ANNAUG - 3' in a mammalian mrna? (A) Shine-Dalgamo sequence (B) Kozak sequence (C) Internal ribosome entry sites (D) Translation termination site 31. The specificity of trna recognition by a aminoacy trna synthetase that is intrinsic to the trna molecule lies on: (A) Acceptor stem (B) Acceptor stem and anticodon loop (C) Anticodon loop (D) D-arm 32. Viral gene expression after T3 bacteriophage infection is controlled by: (A) Repressor molecule (B) Slow injection of nucleic acid (C) Modification of RNA polymerase (D) DNA polymerase 33. Which of the following factors id NOT true for the low levels of immune response in Plasmodium infection? (A) Different types pf antigens are expressed at various stages of Plasmodium life cycle. (B) Most of the phases in the life cycle of Plas modium are intracellular. (C) Sporozoites are rapidly cleared from blood circulation. (D) Plasmodium infection primarily destroys macrophages and dendritic cells. 34. Presence of the nuclear localization signal (NLS) in a steroid recepto indicates that the receptor resides: (A) On the nuclear membrane (B) Within the nucleus (C) On the cell membrane (D) In the cytosol Exp. The activated GR complex up regulates the expression of anti inflammatory proteins in the nucleus or represses the expression of proinflammatory proteins in the cytosol.

236 Amar Ujala Education 35. Which of the following is an intracellular anchor protein? (A) Vitronectin (B) Vinculin (C) Integrin (D) Elastin Exp. As intracellular protein of 130,000 molecular weight was recently isolated in this laboratory from chicken gizzard smooth muscle. By immunofluorescence observations of cultured chicken fibroblasts, it was shown to be concentrated on the ventral surfaces of the cells where they formed focal adhesions to the substratum [Geiger, B. (1979) Cell 18, ]. Focal adhesions are sites where, inside the fibroblast, microfilament bundles are known to terminate at the cell membrane. 36. Out of the following matches of oncogenes with the proteins that each specifies, which one is incorrect? (A) Erb A - thyroid hormone receptor (B) Erb B - epidermal growth factor receptor (C) Ras - guanine - nucleotide binding protein with GTPase activity (D) Fos - platelet - derived growth factor Exp. Fructooligosaccharides (FOS) also sometimes called oligofructose or oligofructan, are oligosaccharide fructans, used as an alternative sweetener. FOS exhibits sweetness levels between 30 and 50 percent of suger in commerically reapared syrups. 37. Capacitation of sperms in humans: (A) Occurs during copulation (B) Occurs after the acrosome reaction (C) Takes place in the ampulla of the oviduct (D) Takes place in the epididymis of testis 38. With respect to development of any organism, "autonomous specification" would result in which type of development? (A) Regulative (B) Mosaic (C) Syncytial (D) Definitive 39. The group of cells which generates the vascular tissues including the pericycle in roots of higher plants are called: (A) Procambium (B) Protoderm (C) Ground meristem (D) Apical meristem 40. If an embryo undergoes 13 cleavages divisions during embryogenesis, then the size of the embryo compared to zygote: (A) Increases 13 times (B) Increases in an exponential fashion (C) Increases only 6-7 times (D) Remains almost the same A/UG The chlorosis (yellowing) symptom of iron deficiency is influenced by: (A) Sodium and Potassium (B) Sodium and Phosphorufdfdds (C) Calcium and Nitrogen (D) Potassium and Phosphorus Exp. Excesses of potassium, magnesium and phosphorus also contribute to chlorosis. When present in excess, these elements cause some trees, particularly oaks and maples, to take up inadequate amounts of the nicronutrients iron and manganese. If iron or manganese deficiency is suspected, there are both longterm and short term treatment strategies, but a soil test will determine the ph as well as the availability of nutruents that cause chlorosis. Stressors, such as temperature extremes, drought, poor drainage (which limits soil aeration) or restricted root growth, further limit nutrient uptake in plants sensitive to chlorosis. 42. A plant hormone that promotes the acquisition of desiccation tolerance in developing seed is: (A) ABA (B) Ethylene (C) IAA (D) GA3 Exp. ABA plays important roles in many aspects of seed development, including accumulation of storage compounds, acquisition of desiccation tolerance, induction of seed dormancy and suppression of precocious germinaton. 43. Change in Ca 2+ concentration can initiate various responses in plants. Which one of the following responses is NOT known to be initiated by change in Ca 2+ concentration? (A) Closure of stomata (B) Reorientation of growth in pollen tubes (C) Thickening of cell walls in young tobacc seedlings in response to wind (D) Lateral root formation 44. Water can move through the soil-plant-atmosphere continuum, only if water potential ( w ) Along that path (A) Decreases (B) Increases (C) Remains unchanged (D) Fluctuates rapidly in either direction 45. Which one of the following is responsible for the ejection of milk from mammary glands in mammals? (A) Oxytocin (B) Prolactin (C) Serotonin (D) Melatonin P9/6

237 A/UG-15 Exp. There are two horones involved in the synthesis and release of milk: prolactin and oxytocin. Both are produced in a pea sized gland attached to the base of the brain called the pituitary gland, and carried to the breasts by the blood.prolactin is produced by the anterior (front part) of the pituitary gland. It stimulates the cells in the breasts to synthesise milk. Prolactin synthesis is stimulated by several minutes of the infant sucking at the breast. oxytrocin is synthesised by the posterior (rea part) of the pitutary gland. It stimulates the release of the milk from the breast (also called milk 'let down'). Oxytocin production is also stimulated by suckling at the breast. 46. A nerve fibre cannot be stimulated during the absolute refractory period of a previous stimulus because (A) Sodium permeability remains high (B) Sodium-potassium pump does not operate (C) Voltage -gated calcium channels remain closed (D) Potassium conductance remains low 47. the T-wave of ECG indicates (A) Atrial depolarization (B) Ventricular depolarization (C) Ventricular repolarization (D) Atrial repolarization Exp. The T wave represents ventricular repolarization and is longer in duration than depolarizaton and is longer in duration than depolatization (i.e., conduction of the repolarization). Sometimes a small positive U wave may be seen following the T wave (not shown in figure at top of page). This wave represents the last remnants of ventricular represents the last remnants of ventricular repolarization. Inverted or prominent U waves indicates underlying pathology or conditions affecting repolazation. 48. Blood group type A antigen is a complex oligosaccharide which differs from H antigen present in type O individual by the presence of terminal (A) Glucose (B) Galactose (C) N-acetyl galactosamine (D) Fucose 49. A cross was made between pure wild type males and brown eyed, curled winged females of D. melanogaster. The F 1 females were test crossed. The F 2 progency obtained was a follows: Wild type 200 Brown eyes, curled wings 150 Brown eyes, normal wings 30 Normal eyes, curled wings 20 Total 400 Amar Ujala Education The genetic distance (cm) between brown eye and curled wing loci is: (A) 12.5 (B) 50 (C) 150 (D) The effect of nonsense mutation could be nullified by reversion as well as suppression. Which of the following processes will help to distinguish between the two kinds of revertants? (A) Complementation (B) Transgenesis (C) Test for allelism (D) Recombination Aminopurine induces mutation by: (A) Base pair change (B) Frameshift (C) Duplication (D) Insertion 52. In a transformation experiment, donor DNA from an E. coli strain with the genotype Z + Y + was used to transform a strain of genotype Z - Y -. The frequencies of transformed classes were: Z + Y Z + Y Z - Y Total 1000 What is the frequency (%) with which Y locus is cotransformed with the Z locus? (A) 1 (B) 20 (C) 33.3 (D) The 'Tribe' refers to a taxonomic group recognized between the ranks (A) Genus and species (B) Family and genus (C) Order and family (D) Class and order Exp. In biology, a tribe is a taxonomic rank between family and genus. It lies below ranks with names derived from family, such as subfamily. It is sometimes subdivided into subtribes. 54. A plant species has been described for the first time by author 'x'. Later, the species has been transferred to some other genus by author 'y'. Then the author citation for the new combination will be (A) x et y (B) x ex y (B) (x) y (D) (y) x 55. The group which is no longer considered under fungi is : (A) Ascomycetes (B) Basidiomycetes (C) Chytridiomycetes (D) Omycetes Exp. Fungi and Oomycetes are the two most important groups of eukaryotic plant pathogens. Fungi form a separate kingdom and are evolutionarily related to animals. Oomycetes are classified in the kingdom Protoctista and are related to heterokont, biflagellate, golden brown algae. P9/7

238 Amar Ujala Education 56. Character similarly that can be misinterpreted as common descent is called: (A) Symplesiomorphy (B) Synapomorphy (C) Homology (D) Homoplasy Exp. In cladistice, a homoplasy or homoplastic character is a trait (genetic, morphological etc.) that is shared by two or more taxa because of convergence, parallelism or reversion. 57. The following table shows survival and fertility data for a seasonally breeding species. Season Proportion Surviving Fertility Based on above data, net reproductive rate (R 0 )ofthe species will be 58. Which of the following is NOT a characteristic of late successional forest plant species? (A) Large seed size, high root to shoot ratio. (B) Long seed dispersal distance, long seed viability. (C) Slow growth rate, long maximum life span. (D) Low light saturation intensity, high efficiency at low light. 59. Which of the following organisms do not posses the ability to fix nitrogen? (A) Organisms specialized for high altitude (B) Marine plankton (C) Eukaryotic organisms (D) Acidophilic organisms 60. Which of the following greenhouse gases has got highest atmospheric lifetime? (A) CO 2 (B) CH 4 (C) N 2 O (D) CFCs Exp. The longest lifetime in the atmosphere would have to be CFC 13, CCIF3 which has a half life of 640 years. One you didn't ask about Sulphur hexafluoride, SF6, is actually the longest at 3,000 years half life. 61. Which of the following evolutionary processes played an important role in the evolution of complex immune system? (A) Reproductive isolation (B) Adaptive radiation (C) Neutral evolution (D) Co-evolution P9/8 A/UG In some species of new world monkeys, only one female redpoduces in a group. One or more younger females have suppressed reproduction and assist the reproductive female. This is an example of: (A) Sexual selection (B) Group selection (C) Kin selection (D) Reciprocal altruism 63. In bird species where both parent contribute equally to parental care, generally: (A) Males are larger than females (B) Females are more colourful than males (C) Females are larger than males (D) Both sexes are morphologically similar 64. The idea that an altruistis gene will be favoured if r > C/ B, where r is the coefficient of the altruism, and C is the cost incurred to the donor, is known as : (A) Red queen hypothesis (B) Handicap principle (C) Hamilton's rule (D) Competitive exclusion principle 65. Use of doubled haploid in plant breeding helps to : (A) Reduce generation time while interogressing recessive traits (B) Reduce generation time while interogressing dominant traits (C) Develop somatic hybrids (D) Interogress transgenic traits 66. For sustained expression of a transgene in the successive generation of a cell line in culture, the ideal gene transfer can be obtained using: (A) Lentiviral vector (B) Adenoviral vector (C) Plasmid DNA containing the transgene (D) Only transgenic DNA Exp. Cystic fibrosis (CF) is a chronic autosomic recessive syndrom, caused by mutations in the CF Transmembrane Conductance Regulator (CFTR) gene, a chloride channel expressed on the apical side of the airway epithelial cells. The lack of CFTR activity brings a dysregulated exchange of ions and water through the airway epithelium, one of the main aspects of CF lung disease pathophysiology. Lentiviral (LV) vectors. of the Retroviridae family, show interesting properties for CF gene therapy, since they integrate into the host genome and allow long lasting gene expression. Proof of principls that LV vectors can transduce the airway epithelium and correct the basic eletrophysiological defect in CF mice has been given.

239 A/UG Desulphovibrio desulfuricans (A) and Pseudomonas species (B) are involved in mercury bioremediation. Which of the statements below is correct? (A) A converts methyl mercury to mercuric ion, B converts mercury to methyl mercury (B) A converts mercury to methyl mercury, B converts to mercury to mercuric ion (C) A converts mercury to methyl mercury, B converts methyl mercury to mercuric (D) A converts methyl mercury to mercuric ion, B converts to mercuric ion 68. Optical density of a 400 base pair long 1 ml DNA solution was found to be How many DNA molecules are present in the solution? (1 base pair = 650 dalton, optical density of 1.0 D corresponds to 50 g DNA/ml) (A) x (B) x (C) x (D) 5.2x In which of the following techniques does molecular fragmentation offer clues to the covalent chemical structures of biochemicals? (A) MALDI-TOF MS mass spectrometry (B) MALDI-TOF MS/MS mass spectrometry Amar Ujala Education (C) ESI-TOF MS mass spectrometry (D) LC-coupled ESI-TOF MS mass spectrometry Exp. Maxtrix-assisted laser desorption/ionisation-time of flight mass spectrometry (MALDI-TOF M0S) isa relatively novel technique in which a co-precipitate of n UV-light absorbing matrix and a biomolecule is irradiated by a nanosecond laser pulse. Most of the laser energy is absorbed by the maxtrix, which precents unwanted fragmentation of the biothe laser energy is absorbed by the matrix, which pervents unwanted fragmentation of the biomlecule. The mass accuracy of MALDI-TOF MS will be sufficient to characterise proteins (after tryptic digestion) from completely sequenced genomes (e.g.methanogens, yeast). The use of MALDI-TOF MS for typing of single nucleotide polymorphisms, using single nucleotide primer extension has made important progress. 70. The movement of a single cell was required to be coutinually monitored during development. This cell was marked with a reporter gene. To visualize this movement one would use: (A) Phase contrast microscopy (B) Bright field microscopy (C) Fluorescence microscopy (D) Atomic force microscopy PART-C 71. The Gibbs free energy of binding of a ligand with a protein is determined using alorimetric measurements at 25 o C. The value of G o thus determined is 1.36 kcal/ mole. the binding constant for the liagant-protein association is: (A) 1.30x10-12 (B) 0.10 (C) 1.00 (D) A is converted to E by enzymes E A,E B,E C,D d.thek m (M) values of the enzymes are 10-2,0-4,10-5 and 10-4 respectively. If all the subtrates and products are present at a concentration of 10-4 M, and the enzymes have approximately the same V max,..., the rate limiting step will be : (A) (B) (C) (D) 73. The molecular mass of a protein determined by gel filtration is 120 kda. When its mass is determined by SDS-PAGE with ans without -mercaptoethanol, it is only 60 kda. What is the most probable explanation for these observations? (A) Protein is a dimer in which two identical chains are cross-linked by disulphide bond(s). (B) Protein is a monomer of molecular mass 60 kda but it is excluded from the gel matrix due to dtring repulsion between the gel matrix and the protein. (C) Protein is most likely to be composed of two subunits having identical molecular mass. (D) Protein is a monomer but it is nicked into half its size by SDS. 74. Mouse IgG is left either intact (left lane in A, B, C, D) or digested with papain or pepsin or treated with mercaptoe thanol ( -ME) and run on non-reducing SDS- PAGE and stained with Coosmassie blue. In a separate experiment, papain-digested products are immunoblotted with an anti-idiotypic monoclonal antibody. Following four profiles are attributed to each of these treatments. P9/9

240 Amar Ujala Education Which one of the following possibilities is correct? (A) A(pepsin), B(papain), C( -ME), D(papain, followed by anti-idiotype immunoblot) (B) A(papain), B(pepsin), C)papain, followed b antiidiotype immunoblot, F( -ME) (C) A (papain, followed by anti-idiotype immunoblot), B(papain), C(pepsin), D( -ME) (D) A ( -ME), B(papain), C(pepsin), D(papain, followed by anti-idiotype immunoblot ) 75. The citric acid cycle in respiration yields: (A) 1 GTP, 3 NADH, 1 FADH, 2CO 2 (B) 2 GTP, 2 FADH 2, 6NADH, 2CO 2 (C) 4 GTP, 6 NADH, 4 FADH 2,CO 2 (D) 32 GTP, 2 NADH, 4 FADH 2,4CO 2 Exp. Citric Acid Cycle : The citric acid cycle can be summarized by eigher of the diagrams below. The diagram below occurs twice, once for each acetyl CoA. Coenzyme A is removed when the two-carbon compound is attached to a four-carbon compound producing a sixcarbon compound (citrate). Each citrate molecule undergoesa series of reactions that removes 2 carbon atoms which are released as CO 2. Inaddition, 3 NADH, 1ATPand1FADH 2 are produced. In addition, the fourcarbon compund that began the cyle is regenerated. 76. Phosphatidyl serine (PS) is mostly locatd in the inner bilayer of plasma membrane of red blood cells (RBCs). You have to prove this fact about PS by an experiment. A/UG-15 You are provided with PS-specific lytic enzymes (PSE) and other reagents needed. Identify the correct sequence of experiments to be carried out to settle this issue. (A) RBCs - inside out vesicles - PSE Thin LAyer Chromatography (TLC) (B) RBCs - right side out vesicles - TLC - PSE (C) RBCs - PSE - Inside out vesicles - TLC (D) RBCs - PSE - TLC - Inside out vesicles 77. ATP-driven pumps hydrolyze ATP to ADP and phosphate and use the energy released to pump ions or solutes across a membrane. There are many classes of these pumps and representatives of each are found in all prokaryotic and eukaryotic cells. Which of the following statements about these pumps is NOT correct? (A) P-type pumps are multipass transmembrane proteins which phosphorylate themselves during pumping and involve in ion transport. (B) F-type pumps normally use the H + gradient across the membrane to drive the synhesis of ATP. (C) V-type pumps normally use voltage gradient for transport of small molecules. (D) ABC transporters primarily pum small molecules across cell membrane. 78. Following are statements related to the organization of the four major protein complexes of thylakoid membrane. (a) Photosystem Ii is located predominantly in the stacked regions of the thylakoid membrane. (b) Photosystem I is found in the unstacked regions protruding into stroma. (c) Cytochrome B 6 complex is confined to stroma only. (d) ATP synthase is located in the unstacked regions ptotruding into stroma. Which one of the following combinations of above statements is correct? (A) (a) and (c) (B) (a), (b) and (d) (C) (b), (c) and (d) (D) (c),(d) and (a) Exp. The light reactions of photosynthesis in green plants are mediated by four large protein pants compexes, embedded in the thylakoid membrane of the chloroplast. Photosystem I (PSI) and Photosystem II (PSII) are both organized into large supercomplexes with variable amounts of membrane-bound peripheral antenna complexes. with PSI consists of a monomeral core complex with single copis of four different LHCI proteins and has binding sites for additional LHCIand/ or LHCII complexes. PSII supercomplexes are dimeric and contain usally two to four copies of trimeric LHCII complexes. These supercomplexes have a further P9/10

241 A/UG-15 tendency to associate into mega complexes or into crystalline domains, of which several types have been characterized. Together with the specific lipid composition, the strutural features of the main protein complexes of the thylakoid membranes from the main trigger for the segregation of PSII and LHCII from PSI and ATPase into stacked grana membranes. We suggest that the margins, the strongly folded regions of the membranes that connct the grana, are essentially proteinfree, and that protein-protein interactions in the lumen also determine the shape of the grana. We also discuss which mechanisms determine the stacking of the thylakoid membranes and how the supramolecular organization of the pigment-protein complexes in the thylakoid membrane and their flexbility may play roles in various regulatatory mehanisms of green plant phtosynthesis. 79. A bacterial population has a plasmid with copy number 'n'. It was observed that on an average in one out of 2 (n- 1) cell divisions, there was spontaneous plasmid curing. It was inferred from the observation that: (a) Each cell division does not have equal probability of plasmid curing. (b) There is no evidence for any mechanism of plasmid segregation in the two daughter cells. (c) Plasmid distribution to daughter cells is random. (d) Each plasmid has an equal chance of being in either of the two daughter cells. Which of the combinations of above statements is true? (A) (a) and (b) (B) (b) and (d) (B) only (a) (D) (b), (c) and (d) Mitotic cells (% Labeled) Time (hours) In given experiment the cells were labelled for 30 minutes with radioactive thymidine. The medium was then replaced with that containing unlabelled thynidine and the cells were grown for additional time. At different time points after replacement of medium the fraction of mitotic cells were analysed. Based on the results obtained, the above figure was drawn which shows the percentage of mitotic cells that are labelled as function Amar Ujala Education of time after brief incubation with radioactive thymidine. Considering the above experiment, the following statements were made: (a) Cells in the S-phase of the cell cycle during the 30 minute labelling period contain radioactive DNA. (b) It takes about 3 hours before the first labelled mitotic cells appear. (c) The cells enter the second round of mitosis at t 30 hours. (d) The total length of the cell cycle is about 27 hours with G 1 being more than 15 hours. Which of the combinations of the above statements is correct? (A) (a) and (b) (B) (b) and (c) (C) (c)y - - and (d) (D) (a) and (d) 81. Mutants of lac Y (Y ) gene of E. coli do not synthesize the lactose permease protein. The following statements refer to the behaviour of lac Y mutant under different experimental conditions. (a) No synthesis of -galactosidase when Y - cells are induced with lactose. (b) Synthesis of -galactosidase when cells are induced with lactose. (c) No synthesis of -galactosidase when cells are induced with IPTG. (d) Synthesis of -galactosidase when cells are induced with IPTG. (e) The cells induced with IPTG cannot grow in the presence of TONPG (TONPG is a compound, whose uptake is mediated by lac tose permease and cleaved by -galactosi dase to release a toxic compound). (f) Cells induced with IPTG can grow in the pres ence of TONPG. Which combination of the above statements is correct? (A) (a), (d) and (f) (B) (b), (c) and (e) (C) (a), (c) and (f) (D) (a), (c) and (e) 82. The semiconservative nature of DNA replication was established by Meselson and Stahl in their classic experiment with bacteria. They grow bacteria in N 15 - NH 4 Cl containing medium, washed and then incubated in fresh medium with N 14 - containing compounds and allowed to grow for three generations. CsCl density gradient centrifugation of isolated DNA established the nature of semiconservations DNA replication. The pictorial representation below shows the position of differentially labelled DNA in CsCl density gradient. 14 N DNA N /N N /DNA DNA P9/11

242 Amar Ujala Education Had the DNA replication been conservative, what would have been the patter? (A) (B) (C) (D) 83. HeLa cell extract was used to study transcription of gene X having six introns. RNA Pol II complex containing all associated proteins was isolated from actively transcribing system and subjected to proteome analysis. Results showed the presence of both splicing and capping enzymes. When phosphorylation of the CTD domain of Pol II was inhibited by a kinase inhibitor, the complex contained neither splicing not capping enzymes. From these results, following conclusions were made: (a) Transcription of gene X is coupled to mrna capping. (b) Transcription elongation is coupled to splicing. (c) Phosphotylation of CTD is required for the recruitment of capping and splicing enzymes. (d) Both capping and splicing of mrnas ocurs simultaneously. Identify the correct set of conclusions: (A) (a), (b) and (c) (C) (c), (d) and (a) (B) (b), (c) and (d) (D) (d), (a) and (b) 84. In bacteria, N-formyl methionine is the first amino acid to be incorporated into a polypeptide chain. Accordingly, one would think that all bacterial proteins have a formyl group at their amino terminous and the first amino acid is methionine. However, this is not the case, because of the following possible reasons. (a) Deformylase removes the formyl group only during or after the synthesis of the polypep tides. (b) Aminopeptidase removes only the amino terminal (c) methionine. Aminopeptidase removes the amio terminal methionine as well as one or two additional amino acids. (d) Deformylase removes the formyl group as well as amino terminal methionine and adds one or two amino acids to it. Choose the combination of correct answers from the following: (A) (b) and (c) (C) (a) and (c) (B) (a) and (b) (D) (a) and (d) 85. Bacteriophage is a temperate phage. Immediately after infection, viral specific mrnas for N and Cro proteins are expressed followed by early mrnas. At the P9/12 A/UG-15 commitment phase, either lytic cycle starts with the expression of genes for head tail, and lytic proteins or lysogenisation cycle begins with the expression of repressor and integrase genes. During induction of lysogens both INT and XIS proteins are needed along with host factors. Out of the four processes below, some govern integration of viral genome and its excision? (a) Repression of transcription (b) Retroregulation (c) Rearrangement of viral genome (d) Repression of translation Identify the correct set of combination: (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (d) and (a) 86. In E. coli, reca gene is involved in rcombination as well as repair and dnab gene in involved in unwinding of DNA double strands during replication. Which of the following statement is/are correct about Rec. A and dnab? (a) Mutation in E. coli reca gene is lethal. (b) E. coli with mutated dnab gene does not survive. (c) Dna B after uncoiling DNA double strands, prevents further reannealing at the eparated strands. (d) Rec A gene is involved in SoS response and helps DNA repair. The correct option are : (A) (b) and (c) (B) (a) and (b) (C) (b) and (d) (D) (a) and (c) Exp. The dnab mutation in Escherichia coil K.12 is an amber mutation such that strains carrying this mutation are not viable in a sup+ strain. With five different R plasmids, it has been possible to construct viable R+ derivatives of this amber mutant and show that the plasmide themselves do not carry amber supressors. The reca gene product is a multifunctional enzyme that plays a role in homologous recombinatiion, DNA repair and induction of the SOS response [PMID: ]. In homologous recombination, the protein functions as a DNAdependent ATPase, promoting synapsis, heteroduplex formation and strand exchange between homologous DNAs. 87. The challenges faced by aminoacyl trna synthetases in selecting the correct amino acid is more daunting than its recognition of the appropriate trna. In case of amino acids with similar structures like valne and isoleucine, this challenge is met by the enzyme possibly through its. (a) Catalytic pocket (b) Editing pocket (c) Anticodon loop (d) Acceptor arm Choose the correct set from the following : (A) (a) and (b) (B) (a) and (c) (C) (b) and (d) (D) (b) and (c)

243 A/UG p24 is an important core protein of HIV. This protein is abundant during active replication of the virus. The serum of an HIV patient was examined for the presence of p24 and antibody against p24 for proper diagnosis of the infection stage. Match the clinical observations in column A with the inferences in column B. Column A Column B (a) p24 is present in the serum 1. viral latency (b) Anti-p24 antibody is high in 2. progression of the serum HIV latnecy to lytic stage (c) Anti-p24 antibody begins to 3. early stage of decline with corresponding infection increase in p24. Choose the correct matching (a) (b) (c) A C B D Epidermal growth factor (EGF) is needed for growth of almost all cells. EGF receptor is a transmembrane protein having an extracellular ligand-binding domain, a transmembrane domain and a cytosolic domain of protein tyrosine kinase (PTK). Binding of EGF to the receptor activates PTK resulting in activation of transcription factors through intracellular transducers. In cell type A, much of the extracellular ligand-binding domain is deleted by proteases such that cytosolic domain of PTK becomes constitutively active whereas cell type B is having normal EGF receptor. What will be the best-fit graph for the growth of the cultures of cell type A and B in complete medium in presence (+) and absence (-)ofegf? (A) (B) (C) No. of cells No. of cells No. of cells Time A B Time B Time A A A B A B B +EGF EGF +EGF EGF +EGF +EGF EGF EGF +EGF EGF +EGF EGF P9/13 (D) No. of cells B B A A Time +EGF EGF +EGF EGF Amar Ujala Education 90. A particular type of cancer cell undergoes apoptosis by both extrinsic and intrinsic pathways when treated with a chemotherapeutic agent X. Caspase 8 and Caspase 9 are the initiator capases associated with extrinsic and intrinsic pathways respectively. Now, if caspase 9 is silenced in the cancer cell by shrna transfection, what will be the best-fit graph for apoptosis scenario in the cancer cell when treated with agent X? (A) (C) % Apoptosis % Apoptosis (B) (D) 91. After successive surgery and chemotherapy, the tumour of a breast cancer patient subsided. However, after almost 5 years, the tumour relapsed in a more aggressive manner and did not respond to the conventional chemotherapy delivered earlier. The following postulations were made. (a) Chemoresistant cells were persisting within the tumour even after therapy. (b) A population of quiescent cells existed, which under favourable conditions, transformed to new tumour (c) cells. High ABC (ATO-Binding Cassette) - transporter expressing cells persisted in the breast during chemotherapy. (d) Breast tumour cells which may have migrated to other tissues, returned to the breast immediately after chemotherapy was terminated. Which of the above combination of statements is true? (A) (a) and (d) (C) only (b) % Apoptosis % Apoptosis (B) (a), (b) and (c) (D) (b) and (d)

244 Amar Ujala Education 92. Following are the experimental observations made in treatment of B cells: (a) Anti-immunoglobulin (anti-ig) antibody treatment results in B cell apoptosis. (b) Anti-Ig plus CD40 ligand treatment results in B cell proliferation. (c) Anti-Ig plus CD40 ligand plus 1l-4 treatment results in B cell proliferation and switching toigg1. (d) Anti-Ig plus L4 treatment resultsin less B cell Proliferation but switching to IgE. From the above observation, which one of the following is the correct interpretation for the role CD40 in B cell function? (A) Induce death of B cells. (B) Reascue B cells from death and Ig class switch to IgG1. (C) Inducing Ig class switch to 1gE. (D) InduceIg class switching to both IgG1and IgE and inhibit B cell proliferation. 93. A potentially valueble therapeutic approach for killing tumour cells without affecting normal cells is the use of immunotoxins. Immunotoxins constitute monoclonal antibodies against tumour cells conjugated to lethal toxins. Which of the following molecular approaches do you think is NOT appropriate for generating tumour cell-specific immunotoxin that will not kill normal cells? (A) Cell surface receptor binding polypeptide chain of toxin molecule shouldbe replaced by monoclonal antibodiesagainst a particular tumour cell type. (B) Constant region Fe domain of tumour cell specific monoclonal antibody should be replaced by ligation of toxin. (C) Variable region F(ab) 2 domain of tumour cellspecific monoclonal antibody should be replaced by ligation of toxins. (D) Inhibitor polypeptide chain of toxin should be conjugated to F(ab) 2 domain of tumour cell specific monoclonal antibody. 94. Flowers represent a complex array of functionally specialized structures that differ substantially from the vegetative plant body in from and cell types. Following are statements made regarding floral meristems by their larger size. (a) Floral meristems can usually be distinguished from vegetative meristems by their larger size. (b) The increase in the size of the meristems is largely a result of increased rate of cell division in central cells. (c) The increase in the size of the meristem is due to larger size of the cells, which in turn result from rapid cell expansion only. (d) A network of genes control floral morphogenesis in plants. A/UG-15 Which combination of the above statements is ture? (A) (a), (b) and (d) (B) (a), (b) and (c) (C) (b), (c) and (d) (D) (a), (c) and (d) 95. Three embroys, X (wild type),y (mutant for bico id) and Z (mutant for nanos) were injected with bicoid mrna in their posterior pole at early cleavage stage. What would be the phenotypes of the resulting embryos? (A) Embryos X will deelope head on both anterior and posterior side, while embroysy and Z will develop head on posterior side only. (B) Embroys X and Z will develop head on both anterior and posterior side, while embroys Y and Z will develop head on posterior end only. (C) Embroys X, Y and Z will develop head on both anterior as well as posterior side. (D) Embroys X will develop head on anterior side, embryo Z will develop head on anterior as well as posterior side. 96. In C. elegans during embryogenesis, an anchor cell and 6 hypodermal vulval precursor cells (VPCs) get involved in forming the vulva. If 3 of the hypodermal VPCs get are killed by a laser beam, a normal vulva is still formed. This could be due to the following possible reasons. (a) Six hypodermal VPCs form equivalence group of cells, out of which only 3 participate in vulva formation and 3 cells remain as reserve cells. (b) When 3 hypodermal VPCs are killed, the 3 neighbouring hypodermal non -VPCs get freshly recuited. (c) Anchor cell function as an inducer which can induce epithelial cells of the gonad to get recruited to compensate for the loss. (d) Anchor cell acts as an inducer which can spatially induce only 3 hypodermal cells to form the vulva. Which combination of the above statements is correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (a) and (d) 97. In tadpoles, if the tail is amputated it can regenerate. However, if the tail is amputated and then exposed to retinoic acid, it develops limbs instead of regenerating the tail. This could be due to the following reasons: (A) Retinoic acid is a morphogen and induces genes responsible for limb formation. (B) Retinoic acid raises the positional values in that reason for limb development to take place. (C) This is a random phenomenon and is not well understood. (D) Retinoic acid possibly acts as a mutagen and the phenotype observed is a result of several mutations. P9/14

245 A/UG In sea urchins, a group of cells at the vegetal pole become specified as the large micromere cells. These cells are determined to become skeletogenic mesenchyme cells that will leave the blastula epihthelium to ingress into the blastocoel. This specification is controlled by the expression of Pmarl which is a repressor of HesC. HesC represses the genes encoding transcription factors activating skeleton forming genes. The gene regulatory network is given below. (b) Amar Ujala Education (c) (d) Below, column I lists the experiments carried with nrna/ antisense RNA of different genes injected into singlecelled sea urchin embroy while column II lists the developmental outcomes. Match the following : Column I Column II (Injection of) (Developmental outcomes) (a) mrna of Pmar1 1. All cells will start ingressing into the blastocoel (b) mrna of HesC 2. Skeleton mesen chyme will not be formed (c) Antisense of Pmar1 (d) Antisense of HesC Which of the following combinations is correct? (a) (b) (c) (d) A C B D Which of the following cellular communications shown below will override the process of normal development and lead to cancer? (a) (A) (b) and (c) (B) (a) and (c) (C) (a) and (d) (D) (b) and (d) 100. In plants, the energy of sunlight is first absorbed by the pigments present in their leaf cells followed by the fixation of carbon through photosynthesis. Consider the following statements. (a) Chlorophylls a and b are abundant in green plants. (b) Chlorophylls c and d are found in some protists and cyanobacteria. (c) Out of different types of bacteriochloro phyll, type a is the most widely distributed. (d) Out of different types of bacteriochloro phyll, type b is the most widely distributed. Which of the following combination of above statements is correct? (A) (a), (b) and (c) (B) (a), (c) and (d) (C) (b), (c) and (d) (D) (a), (b) and (d) 101. Nitrate reductase is an important enzyme for nitrate assimilation. Given below are some statements on nitrate reductase enzyme: (a) Nitrate reductase of higher plants is composed of two identical subunits. P9/15

246 Amar Ujala Education (b) One subunit of nitrate reductase contains three prosthetic groups. (c) One of the prosthetic groups attached to both subunits is heme. (d) One of the prosthetic groups complexed with pterin is magnesium. Which of the following combination of statements on nitrate reductase mentioned above is correct? (A) (a), (b) and (c) (B) (a), (c) and (d) (C) (b), (c) and (d) (D) (a), (b) and (d) 102. A farmer growing a particular variety of grape plants in vineyard, observes the following: (a) Fruit size normally remained small. (b) Natural seed abortion. (c) Development of fungal infection as the pedicels are small in size due to which moistures is retained in the bunches of grapes. Experts suggested spraying gibberellic acid during the fruit development. This treatment would help in getting rid of: (A) (a), (b) and (c) (B) only (a) and (b) (C) only (a) and (c) (D) only (b) and (c) 103. Light is perceived by various photoreseptors in plants. The photoreceptors predominantly work at specific wavelenghts of light. Some of the following statements are related to the functions of plant photoreceptors. (a) Phytochrome A predominantly perceives the red and far-red light. (b) Phytochrome B predominantly perceives red light. (c) Cryptochromes regulate plant development. (d) Phototropins are involved in blue light perception and chloroplast movements. Which of the following combinations based on above statements is correct? (A) (a), (b) and (c) (B) (b), (c) and (d) (C) (c), (d) and (a) (D) (a), (b) and (d) 104. From the following statements: (a) Triose phosphate is utilized for the synthesis of both starch and sucrose. (b) Triose phosphate is translocated to cytosol from chloroplast. (c) Triose phosphate is confined to chloroplast and is utilized for synthesis of starch only. (d) Triose phosphate is translocated from cytosol to chloroplast. (A) (a) and (b) (C) (c) and (d) (B) (b) and (c) (D) (d) and (a) P9/ A/UG-15 Shown above, is a graph representing the growth of different plant species subjected to salinity relative to that of unsalinized control. Which of the following statements is NOT true? (A) Plants in group IA are extreme halophytes while very salt sensitive species will be part of group III. (B) Plants in group IA are very salt sensitive and extreme halophytes will be part of group III. (C) Halophytes, which can tolerate salt but their growth is retarded will be of part of group IB. (D) Non-halophytes, which are salt tolerant but lacks salt glands will be a part of group II An experimentalist stimulates a nerve fibre in the middle of an axon and records the following observations. Which one of the observation is correct? (A) Nerve impulse is travelling in a direction towards cell body. (B) Nerve impulse is travelling in a direction towards telodendrons. (C) Nerve impulses are travelling in both the directions opposite to each other. (D) Nerve impulse is not moving in either direction. Exp. A nerve electrical impulse onlytravels in one direction. There are a few reason why nerve impulses only travel in one direction, the most important being synaptic transport. In order for a depolarization wave, or "nerve impulse" to pass from cell to cell, there are what we call synaptic junctions. This means that the nerve cells are lined up head to tail all the way down a nerve track, and are not connected, but have tiny gaps between them and next cell. These tiny gaps are called synapses Desert animals have longer loop of Henle compared to that of humans. It may be due to the following reasons: (a) Long loop of Henle is associated with greater amount of vasopressin secretion (b) In long loop of Henle, the counter-current exchanger is more effective. (c) Long loop of Henle conserves more water. (d) Long loop of Henle stimulates production of angiotensin II. Which of the above reason(s) is/are correct? (A) (a) and (b) (C) (c) and (d) (B) (b) and (c) (D) only (d)

247 A/UG A boy eats a large serving of cheese having high amount of sodium. He hardly drinks any fluid. Inspite of this, the water and electrolyte balance was mantained. Which one of the following explanation is correct? (A) His aldosterone was decreased and ADH alcohol dehydrogenase (ADH) was increase. (B) His aldosterone was increased and ADH was decreased. (C) There was no change in either of the hormones. (D) His sympathoadrenal system was stimu lated. Exp. Aldosterone acts mainly at the renal tubules and stimulates the reabsorptoin of Na+ and water and excretion of K+ and phosphate ions. Thus aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure and (ADH) or vasopressin acts mainly at the kidney and stimulates reabsorption of water and thereby reduces loss of water through urine (diuresis) The blood volume decreased when a mammal was bled rapidly. However, the cardiovascular changes resulting from haemorrhage could be minimized by the following compensatory mechanisms: (a) Increased cerebral blood flow. (b) Reduction of baroreceptor activity and timulation of chemoreceptors.' (c) Reabsorption of tissue fluid in blood. (d) Increased release of enkephalins and betaendorphins. Which of the above is/are correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) only (d) 110. The stomach of a person was partially removed during surgery of a gastric tumour.despite taking a balanced diet, the person developed anaemia. following possible explanations were offered: (a) Lower gastric secretion inhibits folic acid absorption. (b) Protein digestion was disturbed in partial (c) gastrectomy. Lower HCl secretion from stomch reduced iron absorption. (d) Lower secretion of intrinsic protein factor from stomach reduced Vit B12 absorption. Which of the above explanations were correct? (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (a) and (d) P9/17 Amar Ujala Education 111. The RFLP pattern observed for two pure parental lines (P1ndP2)andtheirF 1 progeny is represented below. Further, the P1 plant had red flowers while the P2 had white flowers. The F 1 progeny was backcrossed to P2. The result obtained, showing the number of progeny with red and white flowers and their RFLP patterns is also represented below. P1 P2 F 1 Progeny of the cross between F 1 and P2 Phenotype Red White Red Red White RELP pattern Which one of the following conclusion made is correct? (A) The DNA marker and the gene for the flower colour are 10cM apart. (B) The market and the gene for the flower colour are 5cM apart. (C) The marker and the phenotype are indepedently assorting. (D) The marker and the gene for the colour segregate from one another Wild type T4 bacteriophage can grow on B and K strains of E. coli forming small palques. rii mutants of T4 bacteriophage cannot grow on E.coli strain K (nonpermissive host), but form large plaques on E.coli strain B (permissive host). The following two experiments were carried out: Experiment I: E. coli K cells were simultaneously infected with two rii mutants (a - and b - ) Several plaques with wild type morpholgy were formed. Experiment II: E. coli B cells were simultaneously infected with the same mutants as above. T4 phages were isolated from the resulting plaques and used to infect E. coli K cells. Few plaques with wild type morphology were formed. Which one is the correct conclusion made regarding the rii mutants, a - and b - from the above experiments? (A) The mutations a - and b - belong to two different cistrons (experiment I) and there is norecombination between them (experiment II). (B) The mutants a - and b - belong to two different cistrons (experiment I) and they recom bined (experiment II). (C) The mutants a - and b - belong to two different cistrons (experiment II) and they recom bined (experiment I). (D) The mutants a - and b - belong to the same cistron (experiment I) and they did not recombine (experiment II).

248 Amar Ujala Education 113. The following pedigree represents inheritance of a trait in an extended family: I II III IV What is probable mode of inheritance and which individuals conclusively demonstrate the mode of inheritance. (A) Autosomal recessive, III-2, 3 and IV-1, 2 con clusively demonstrate the mode of inheritance. (B) Autosomal recessive I-1, 2 and II-2 conclusively demonstrate the mode of inheritance. (C) Autosomal dominant, III-2, 3 and IV-1, 2 conclusively demonstrate the mode of inheritance. (D) X-linked recessive, II-3, 4 and 5 conclusively demon-strate the node of inheritance, 114. Following is the diagram of a paracentric inversion heterozygote ABCDEFG/ABFEDCG involved in recombination during meiosis I : The consequence of this recombination will be the formation of: (a) A dicentric and an acentric chromosome in meiosis I as the chiasmata gets terminated. (b) No dicentric or acentric chromosome but appearance of deletion and duplication in both the chromosomes. (c) All non-viable gametes. (d) Non-viable gametes from crossover products. Which of the above statements are correct? (A) (a) and (b) (B) (a) and (c) (C) (a) and (d) (D) (b) and (c) 115. An E. coli strain has metb1 (90 min) and leua5 (2 min) mutations. It also has stra7 (73 min) mutation and Tu5 transposon which confers streptomycin and kanamycin resistance, respectively, inserted in its chromosome. The mutant strain was crossed with an Hfr strain that is P9/18 A/UG-15 streptomycin sensitive and has a hisg2 mutation (44 min) that makes it require histidine. After incubation for 100 min, the cells were plated on minimal plate supplemented with leucine, histidine and streptomycin to select the metb marker. After purifying 100 of the Met + transconjugants, one finds that 15 are His -, 2 are leu + and 12 are kanamycin sensitive. The unselected markers are: (a) MetB1 and leua5 mutation. (b) LeuA5 and Tn5 insertion mutation. Which of the above statement is correct and what is the position of transposon insertion? (A) (a) and before 73 min (B) (b) and before 44 min (C) (c) and before 73 min (D) (d) and before 44 min 116. A chemist synthesized three new chemical compounds, M1, M2 and M3. The compounds were tested for their mutagenic potential and were found to be highly mutagenic. Tests were made to characterize the nature of mutations by allowing the reversion with other mutagens. The following results were obtained : Mutations Reversed by produced by 2 Aminopurine Nitrous Hydroxyl Acridine acid amine orange M 1 No No No No M 2 Yes Yes No No M 3 No No No Yes Which one of the following conclusions drawn regarding the nature of mutations by the compounds is correct? (A) M1-transversion, M2- insertion, M3- deletion (B) M1-transition, M2- transversion, M3- insertion (C) M1-insertion, M2- transition, M3- transversion (D) M1- transversion, M2- transition, M3- insertion 117. Four Cnidarians with the following characteristics were observed: (a) Asexual polyps and sexual medusae; solitary or colonial; both freshwater and marine. (b) Polyp stage reduced or absent, medusae with velum; (c) solitary; all marine. Polyp stage reduced, bell shaped medusae; solitary; all marine. (d) All polyps, no medusae; solitary or colonial; all marine. They can be identified to their respective classes: (A) (a)- Scyphozoa, (b)- Anthozoa, (c)- Cubozoa, (d)- Hydrozoa (B) (a)- Hydrozoa, (b)- Scyphozoa, (c)- Cubozoa, (d)- Anthozoa (C) (a)- Anthozoa, (b)- Cubozoa, (c)- Hydrozoa, (d)- Scyphozoa (D) (a)- Cubozoa, (b)- Scyphozoa, (c)- Anthozoa, (d)- hydrozoa

249 A/UG The following table shows the summary of characters between two taxa based on presence (1) and absence (0) data (C) Amar Ujala Education Which of the following represents Jaccard's coefficient and simple matching coefficient respectively? (A) 0.8, 0.5 (B) 0.6, 0.5 (C) 0.8, 0.6 (D) 0.5, Identify the proteobacteria based on the key given below: i. Cause disease in humans (ii) i. Do not cause disease in humans (iii) ii. An obligate intracellular parasite (a) ii. Not an obligate intracellular parasite (b) iii. Live in insects (c) iii. Do not live in insects (iv) iv. Chemautotrophic (d) iv. Not Chemautotrophic (v) v. Plant pathogen (e) v. Not a plant pathogen (vi) vi. Fix nitrogen (vii) vi. Do not fix nitrogen (f) vii. Associated with legumes (g) vii. Not associated with legumes (h) (A) (a) - Rickettsia; (b) - Brucella; (c) - Wolbachia; (d) - Nitrobacter; (e) - Agrobacterium; (f) - Acetobacter; (g) - Rhizobium; (h) -Azospirillum (B) (a) - Rickettsia; (b) - Wolbachi; (c) - Brucella; (d) - Nitrobacter; (e) - Acetobacter; (f) - Agrobacterium; (g) - Rhizobium; (h) - Azospirillum (C) (a) - Rickettsia; (b) - Brucella; (c) - Wolbachia; (d) - Nitrobacter; (e) - Agrobacterium; (f) - Acetobacter; (g) - Azospirillum; (h) - Rhizobium (D) (a) - Rickettsia; (b) - Brucella; (c) - Wolbachia; (d) - Nitrobacter; (e) - Acetobacter; (f) - Agrobacter; (g) - Azospirillum; (h) - Rhizobium Which of the following phylogenetic trees appropriately uses principle of parismony? (A) (B) P9/19 (D) 121. Identity the characters shown in the diagram depicting phylogenetic relationships among major groups of ferns and fern allies. (A) (a) Roots absent. (b) Sporangiophores, (c) Vertical, interrupted annulus, (d) Heterospory, (e) Leaves scale like, (f) Elaters. (B) (a) Roots absent, (b) Leaves scale like, (c) Sporangiophores, (d) Elaters, (e) Heterospory, (f) Vertical, interrupted annulus. (C) (a) Leaves scale like, (b) Sporangiophores, (c) Elaters, (d) Heterospory, (e) Roots absent, (f) Vertical, interrupted annulus. (D) (a) Heterspory, (b) Roots absent, (c) Elaters, (d) Sporangiophores, (e) Leaves scale like, (f) Vertical, interrupted annulus Associate the forest/vegetation type with the plants: (a) Grass land (b) Subalpine forest (c) Shola forest (d) Subtropical pine forest (e) Tropical thorn forest (f) Tropical dry deciduous forest (g) Tropical semievergreen forest (h) Tropical wer evergreen forest (A) (a)-ilex, (b)-dichanthium, (c)-abies, (d)- Pinus, (e)-acacia, (f)-anogeissus, (g)- cinnamomum, (h)-dipterocarpus (B) (a)-dichanthium, (b)-abies, (c)-ilex, (d)- Pinus, (e)-acacia, (f)-anogeissus, (g)- Cinnamomum, (h)-dipterocarpus (C) (a)-dichanthium, (b)-abies, (c)-ilex, (d)- Pinus, (e)-dipterocarpus, (f)-cinnamomum, (g)-acacia, (h)-anogeissus (D) (a)-anogeissus, (b)-dichanthium, (c)-ilex, (d)-pinus, (e)-acacia, (f)-abies, (g)-cinnamomum, (h)-dipterocarpus

250 Amar Ujala Education 123. Possible explanations for the age related decline in primary productivity of trees are: (a) As trees grow larger with age, they have more tissues that respire and loose energy and proportionately less leaf area to photosynthesize. (b) Nutrient limitation by nitrogen due to reduced rate of woody litter decomposition as forest ages. (c) As trees become larger, water transport to the top conopy leaves become limited because of increased hydraulic resistance. This results in reduced stomatal conductance and reductionin photosynthetic rate. Which of the above is/are correct? (A) (a), (b) and (c) (B) only (a) (C) only (a) and (c) (D) only (b) and (c) 124. Species characteristics that make them more prone to extinction are listed below: (a) High degree of specialization (b) High sexual dimorphism (c) High trophic status (d) Short life span Which of the following is the correct combination? (A) (a), (b) and (c) (B) (a), (c) and (d) (C) (a), (b) and (d) (D) (b), (c) and (d) 125. Based on the information given in the table below, which combination is correct? Biographic zone Plant Animal A 1 Mediterranean B 1 Rhododendron C 1 Gibbon A 2 Indio-Chinese B 2 Dipterocarpus C 2 Jungle Fowl A 3 Indo-Malayan B 3 Euphorbia C 3 Takin A 4 Peninsular India B 4 Deodar C 4 Ibex (A) A 1 -B 3 -C 3 (B) A 2 -B 1 -C 4 (C) A 3 -B 2 -C 1 (D) A 4 -B 4 -C In a census for a lake fish, 10 individuals were marked and released. In second sampling after a few days 15 individuals were caught, of which 5 individuals were found marked. The estimated population of the fish in the ake will be: (A) 20 (B) 30 (C) 25 (D) Identify the pollinators for the flowers with following pollination syndromes: (a) Flowers dull coloured, located away from foliage, floral parts turgid. (b) Flowers bright red, crowded, turgid, nectar watery and sucrose rich. (c) Flowers white pleasant odour, corolla tube long, night blooming. P9/20 A/UG-15 (A) (a) Bird; (b) Bat; (c) Butterfly (B) (a) Bat; (b) Bird; (c) Moth (C) (a) Bat; (b) Bird; (c) Bee (D) (a) Bird; (b) Bat: (c) Carrion fly 128. Which of the following combinations is good for setting up a nature reserve: (i) (ii) (iii) (iv) (A) (i), (ii), (iii) (B) (ii), (iii), (iv) (C) (i), (iii), (iv) (D) (i), (ii), (iv) 129. If the number of new species evolving is directly proportional to the number of existing species and the probability of extinction of any species is inversly proportional to the number of existing species, the number of species present at a time during evolution will follow a curve given by: (A) (B) (C) (D)

251 A/UG If the relationship between life time reproductive success and body size for males and females of a species as shown in figure below: The species is most likely to evolve: (A) Sexual dimorphism (B) Asexual reproduction (C) Polyandry (D) Obligate monogamy 131. Following tree represents pholygenetic relationships among species of a moth family. Circles represent species having eye spots on the wings. Other species do not have eye spots. The following inferences were made by different researchers: (a) Eye spots were present in the ancestors and some species lost them. (b) Eye spots were not present in the ancestors. (c) Eye spots were lost more than one in evolution of the family. (d) Eye spots were gained only once while evolving from ancestor without them. Which of the inferences are correct? (A) (a) and (b) (C) (a) and (c) (B) (c) and (d) (D) (b) and (d) 132. Wolbachia are obligate intracellular bacteria, many different strains of which are abundantly present in insects. They induce mating incompatibility in host, i.e. males infected with one strain can only fertilize females infected with the same strain. No other pathological effects are observed in host. A possible evolutionary consequence of this phenomenon would be: (A) Extinction of many insects species. (B) Termination of sexual reproduction in many insect species. (C) Co-extinction of host and parasite. P9/21 Amar Ujala Education (D) Reproductive isolation leading to rapid specification in insects Twenty small populations of a species, each polymorphic for a given locus (T, t) were bred in captivity. In 10 of them the population size was kept constant by random removal of individuals, while other 10 were allowed to increase their population size. After several generations it was observed that in 7 of the size restricted populations only T was present, in the remaining 3 only t was present. in the growing populations 8 retained their polymorphism and in 2 only t was observed. The experiment illustrates: (A) Genetic drift which is more likely in large populations. (B) Genetic drift which is more likely in populations. (C) Density depedent selection against T. (D) Density depedent selection against t Some important events in the history of life on Earth are given below. (a) First vertebrates (jawless fishes); first plants. (b) Forest of ferns and conifers; amphibians arise; insects radiate. (c) Conifers dominant; dinosaurs arise; insects radiate (d) Flowering plants appear; climax of dinosaurs followed by extinction. (e) Radiation of flowering plants, most modern mammalian orders represented. (f) Ice Ages, Modern humans appear. Match the above with the geological time periods and choose the correct combination, (A) (a)- Silurian; (b)- Permian; (c)- Triassic; (d)-jurassic; (e)- Cretaceous; (f)- Tertiary (B) (a)- Ordovician; (b)- Carboniferous;, (c)-triassic, (d)- Cretaceous, (e)- Terriary; (f)-quaternary (C) (a)- Cambrian; (b)- Ordovician; (c)- Silurian; (d)- Devonian; (e)- Permian; (f)- Tertiary (D) (a)- Devonian; (a)- Permian; (c)- Triassic; (d)- Cretaceous; (e)- Tertiary; (f)- Quaternary 135. Microbes produce either primary or secondary metabolites during fermentation. A metabolite production curve is shown below:

252 Amar Ujala Education The following statements refer to the above figure : (a) A primary metabolite has a production curve that lags behind the line showing cell growth. (b) A primary metabolite is produced after the Trophophase is completed. (c) A secondary metabolite is produced mainly during Idiophase. (d) The curve shows the production of Penicillin from mold. Which of the above statements are correct? (A) (a) and (b) (B) (a) and (d) (C) (a) and (c) (D) (b) and (d) 136. During transgenesis, the location of the genes and their number integrated into the genome of the transgenic animal are random. It is often necessary to determine the copy number of genes and their tissue-specific transcription. The following are the possible methods used for the determination. (a) Polymerase Chain Reaction (PCR) (b) Southern blot hybridization (c) Reverse Transcriptase PCR (d) Western blot Choose the correct set of combinations. (A) (a) and (b) (B) (b) and (c) (C) (b) and (d) (D) (a) and (d) 137. Agrobacterium tumefaciens, also known as natural genetic engineers, causes crown-gall disease in plants. However,whenthesamebacteriaareusedtoraise transgenic plants with improved agronomic traits, no such tumour (disease) is observed. This is due to: (a) Vir D 2 gene is mutated in Ti plasmid. (b) Disarmed Ti plasmid is generally used. (c) Heat-shock during transformation destroys virulence. (d) Oncegenes have been removed. Which one of the following combination of above statements is correct? (A) (a) and (c) (C) (b) and (c) (B) (a) and (d) (D) (b) and (d) 138. Locus control region (LCR) lies far upstream from the gene cluster and is required for the appropriate expression of each gene in the cluster. LCR regulates expression of globin genes in the cluster through the following ways. (a) LCR interacts with promotors of individual genes by DNA looping through DNA-binding proteins. (b) The LCR-bound proteins attract chromatinremodelling complexes including histrone modifying enzymes and components of the transcription machinery. P9/22 A/UG-15 (c) LCR acts as an enhancer for global regulation of gene cluster and does not regulate individual genes. (d) LCR participates in converting inactive chromatin to active chromatin around the gene cluster. Choose the correct set of combinations. (A) (a) and (b) (B) (a) and (c) (C) (b) and (c) (D) (b) and (d) 139. A student wrote following statements regarding comparison of Restriction Fragment Length Polymorphism (RFLP), Random Amplified Polymorphic DNA (RAPD), Amplified Fragment Length Polymorphism (AFLP) and Simple Sequence Repeats (SSRs) techniques used for generating molecular markers in plants: (a) All these techniques can be used for finger printing. (b) Detection of allelic variation can be achieved only (c) by RFLP and SSRs. Use of radioisotopes is required in RFLP and RAPD only. (d) Polymerase chain reaction is required for all the techniques. Which one of the following combination of above statements is correct? (A) (a) and (b) (C) (c) and (d) (B) (b) and (c) (D) (d) and (a) 140. In order to clone an eukaryotic gene in pbr322 plasmid vector, the desired DNA fragment was produced by PstI cleavage and incubated with PstI digested prr322 (PstI cleavage site lies within thr ampicillin resistant gene) and ligated. Mixture of ligated cells were used to transform E. coli and plasmid containing bacteria were selected by their growth in tetracycline containing medium. Which type of plasmid/s will be found? (A) Circular pbr322 plasmid containing the target gene and resistant to only tetracycline. (B) Circular pbr322 plasmid containing the target gene and resistant to tetracycline only and recircularised pbr322 plasmid resistant to both ampicillin and tetracycline. (C) Circular pbr322 plasmid containing the target gene and resistant to only tetracycline, recircularised pbr322 resistant to both ampicillin and tetracycline and concatemerized pbr322 resistant to both ampicillin and tetracycline. (D) Circular pbr322 plasmid containing the tar get gene and resistant to both ampicillin and tetracycline.

253 A/UG During apoptosis, Phosphatidyl serine (PS) usually present in the inner leaflet of the plasma membrane flips to the outer membrane. Annexin V is a protein that binds to PS Using this as a tool, we identify the apoptotic cells from necrotic and normal cell populations by FACS using FITC-tagged Annexin V. Propidium iodide (PI) is used to stain the nucleus which generally identifies nectrotic and late apoptotic cells. In which area of the plot you should get early apoptotic cells by FACS analysis? (A) Quadrant I (B) Quadrant II (C) Quadrant III (D) Quadrant IV 142. The muscle tone was increased after electrolytic lesion of the caudate nucleus in a cat. The muscle tone decreased within seven days. The following explanations were given by the researcher. (a) The functional recovery was due to plastic changes of nervous system. (b) The brain tissue surrounding the lesioned area was non-functionaldue to circulatory insufficiency immediately after surgery which led to greater (c) functional loss. The circulatory status in surrounding tissue recovered with time resulting in partial functional recovery. (d) The degenerating nerve fibres were regenerated which underlie functional recovery. Which one of the following is correct? (A) (a) and (b) (C) (c) and (d) (B) (b) and (c) (D) (a) and (d) Amar Ujala Education 143. A fluorophore when transferred from solvent A to solvent B results in an increase in the number of vibrational states in the ground state without any change in the mean energies of either the ground or excited state. What would be the change seen in the fluorophore's emission spectrum? (A) An increase in emission intensity (B) An increase in emission bandwidth (C) An increase in emission wavelength (D) A decrease in emission wavelength 144. You wish to localize a given gene product at subcellular levels following immunolfuorescence staining. Routing microscopy could not resolve whether the gene product is localized inside the nucleus or on the nuclear membrane. Which of the following will resolve this unambiguously? (a) Sectioning of cell followed by phase contrast microscopy. (b) A simulation of 3D picture following confocal microscopy. (c) Optical sectioning and observing each section. (d) FreezefracturingfollowedbyscanningElectron Microscopy. (A) (a) and (b) (B) (b) and (c) (C) (c) and (d) (D) (a) and (c) 145. The Triver-willard hypothesis state that the physiological state of a female can bias the sex ratio of offspring. In an experiment in the bird species a group of females were fed a diet 30% lower in calories than the control females. After allowing both the groups to mate and breed freely, the offspring of control 1 group were 22 males and 18 females. The diet restricted females laid a total of 40 eggs. What should be the minimum deviation from the control to conclude that they have significantly female biased offspring sex ratio. (Chi sq [0.05] df = 1is 3.84) (A) 18 males 22 females (B) 20 males 20 females (C) 15 males 25 females (D) 10 males 30 females P9/23

254 A/UG-15 Amar Ujala Education CSIR-UGS (NET/JRF) LIFE SCIENCES Solved Paper, June-2012 PART-A 1. The area of the shaded region in cm 2 is : 3. AB is the diameter of the semicircle as shown in the diagram. If AQ = 2AP then which of the following is correct? (A) 2 (B) 2 2 (C) 4 2 Exp. Area of circle = 4p (D) 2 Area of 1 circle = 4 Area of triangle = 2 Shaded Region = ( 2) 2. The angles of a right-angleds triangle shaped garden are in arithmetic progression and the smallest side is m. The total length of the fencing of the garden in m is (A) (B) (C) (D) Exp. The angle should be 30º, 60º & 90º. AC BC = tan 60º = 3 AC = 10 3 BC = = 400 =20 Perimeter = 10 = = (A) 1 APB AQB 2 (B) APB 2 AQB (C) APB AQB (D) 1 APB AQB 4 Exp. Both 90º angle made on diameter are 90º. 4. The rabbit population in community A increases at 25% per year while that in B increases at 50% per year. If the present population of A and B are equal, the ratio of the number of the rabbits in B to that in A after 2 years will be (A) 1.44 (B) 1.72 (C) 1.90 (D) 1.25 Exp. The present population of A & B are equal. A B i.e., After 1 year year = = = 225 Ratio B : A = = =1.44 P10/1

255 Amar Ujala Education 5. Two moles each of O 2 and H 2 are in two separate containers, each of the volume V 0 and at 150 o Cand1 atmosphere. The two are made to react in a third container to form water vapour until H2 is exhausted. When the temperature of the mixture in the third container was restored to 150 o C, its pressure became 1 atmosphere. The volume of the third container must be (A) V 0 (B) 5V 0 /4 (C) 3V 0 /2 (D) 2V 0 6. Helium and argon gases in two separate containers are at the same temperature and so have different root-meansquare (r.m.s) velocities. The two are mixed in a third container keeping the same temperature. The r.m.s. velocity of the helium atoms in the mixture is (A) more than what it was before mixing (B) less than what it was before mixing (C) equal to what is was before mixing (D) equal to that of argon atoms in the mixture Exp. Since Root mean square (r.m.s.) velocity = V 2 7. The mineral talc is used in the manufacture of soap because it 1. Gives bulk to the product 2. Kills bacteria 3. Gives fragrance 4. Is soft and does not scratch the skin Which of the above statements is/are correct? (A) 4 (B) 1 and 3 (C) 1 and 2 (D) 1 and g of an inorganic compound X-5H 2 O containing a volatile impurity was kept in an oven at 150 o C for 60 minutes. The weight of the residue after heating is 8 g. The percentage of impurity in X was (A) 10 (B) 8 (C) 20 (D) On a certain night the moon in its waning phase was a half-moon. At midnight the moon will be (A) On the easter horizon (B) At 45 o angular height above the eastern horizon (C) At the zenith (D) On the western horizon 10. A gemstone is irradiated in a nuclear reactor for 5 days. Ten days after irradiation, the activity of the chromium radioisotope in the gemstone is 600 disintegrations per hour. What is the activity of chromium radioisotope 5 days after irradiation if its half life is 5 days? (A) 300 (B) 150 (C) 2400 (D) A/UG-15 Exp. It is decreasing at logerithmic scale. After 2nd half it becomes 600. So, after 1st half it would be Displacement versus time curve for a body is shown in the figure. Select the graph that correctly shows the variation of the velocity with time (A) (C) 12. (B) (D) The spring balance in Fig. A reads 0.5 kg and the pan balance in Fig. B reads 3.0 kg. The iron block suspended from the spring balance is partially immersed in the water in the beaker (Fig.C). The spring balance now reads 0.4 kg. The reading on the pan balance in Fig. C is (A) 3.0 kg (B) 2.9 kg (C) 3.1 kg (D) 3.5 kg Exp. Total weight will be constant = The ends of a rope are fixed to two pegs, such that the rope and moved, such that the rope always remains taut. The shape of the curve traced by the pencil would be a part of (A) Acircle (B) An ellipse (C) A square (D) A triangle P10/2

256 A/UG During ice skating, the blades of the ice skater's shoes exertpressure on the ice. Ice skater can efficiently skate because (A) Ice gets converted to water as the pressure exerted on it increases (B) Ice gets converted to water as the pressure exerted on it decreases (C) The density of ice in contact with the blades decreases (D) Blades do not penetrate into ice Exp. The ice skater can skate over ice because of decrease in pressure on ice. 15. Four sedimentary rocks A, B, C and D are intruded by an igneous rock R as shown in the cross-section diagram. Which of the following is correct about their ages? Ground Suface (A) A is the youngest followed by B, C, D and R (B) R is the youngest followed by A, B, C and D (C) D is the youngest followed by C, B, A and R (D) A is the youngest followed by R, B, C and D Exp. The R represent volcanic rock which is formed by eruption of material from inside the earth; following by A, B, C and D. 16. The strain in a solid subjected to continuous stress is plotted. Which of the following statements is true? (A) The solid deforms elastically till the point of failure (B) The solid deforms plastically till the point of failure (C) The solid comes back to original shape and size on failure (D) The solid is permanantly deformed on failure Exp. The strain on a solid material due to external stress represent that the solid permanently 17. Growth of an organism was monitored at regular intervals of time, and is shown in the graph below. Around which time is the rate of growth zero? Amar Ujala Education (C) Between days 20 and 30 (D) Between days 20 and A Tall plant with Red seeds (both dominant traits) was crossed with a dwarf plant with white seeds. If the segregating progeny produced equal number of tall red and dwarf white plants, what would be the genotype of the parents? (A) TtRrxTtRR (B) TtRrxttrr (C) TTRR x ttrr (D) TTRR x TtRr Exp. 19. Three sunflower plants were places in conditions as indicated below. Plant A : still air Plant B : moderately turbulent air Plant C : still air in the dark Plant D : still air in the dark Which of the following statements is correct? (A) Transpiration rate of plant B > that of plant A (B) Transpiration rate of plant A > that of plant B (C) Transpiration rate of plant C = that of plant A (D) Transpiration rate of plant C > that of plant A > that of plant B Exp. Windy conditions results in increased transpiration rates, the increase being more pronounced at low wind speeds (breeze). High wind results in closing of stomata which may stop transpiration. When there is no breeze then the air surrounding a leaf becomes increasingly humid thus reducing rate of transpiration. 20. Which of the following is indicated by the accompanying diagram? a b a b (A) a+ab+ab = a/(1 b)for b <1 (B) a > b implies a 3 > b 2 (C) (a+b) 2 = a 2 +2ab + b 2 (D) a > b implies a < b Exp. a b a (a 2 ) (A) Close to day 10 (B) On day 20 b (a+b) (b 2 ) The diagram indicate a 2 +(a + b)+(a + b) +b 2 = a 2 + b 2 +2(a + b) P10/3

257 Amar Ujala Education PART-B A/UG Which nitrogen of adenosine gets protonated if the ph of the nucleoside is lowered from 7 to 3? (A) N1 (B) N3 (C) N7 (D) N9 Exp. N1 will get protonated if the PH is lowered from 7 to 3 as N1 is more basic site second protonation will be at N The oligopeptide, F-A-R-P-M-T-S-R-P-G-F, is treated with trypsin, chymotrypsin and carboxypeptidase-b. Apart from the original peptide, the number of fragments obtained will be (A) 4 (B) 3 (C) 2 (D) Which one of the following interaction plays a major role in stabilizing B-DNA? (A) Hydrogen bond (B) Hydrophobic interaction (C) Van der Waal's interaction (D) Ionic interaction Exp. B-DNA is most common DNA mfound in living organism. This contain Purins (A and G) and Pyrimidines (C and T), which bind itself by double and triple H-Bonds/ 24. Phosphatidyl serine, an important component of biological membrane, is located in (A) The outer leaflet but flipflops to inner leaflet under specific conditions (B) Both the leaflets (C) The middle of the bilayer (D) The inner leaflet but flipflops to outer leaflet under specific conditions Exp. Phosphpatidyl serine a phospholipids is present preferentially in inner leaflet (cytosolic side). Negatively charged phospholipids are kept on inner leaflet by flippase. It is flipped to outer leaflet upon cell damage. Scrambalase is the enzyme that will transport negatively charged phospholipid from inner leaflet to outer leaflet. 25. Major disadvantage of using liposome as a targeted drug delivery vehicle is that (A) It gets internalized by phagocylosis inside lysosomes (B) It is very unstable and has low shelf-life (C) It gets intercalated in cell membranes (D) It's drug entrapment efficiency is very low 26. Major stimulus for spore formation in bacteria is (A) Nutrition limitation (B) Heat stress (C) Cold stress (D) phstress 27. ATP-bonding cassette (ABC) transporters (A) Are all P-glycoproteins (B) Are found only in eukaryotes (C) Are both: a membrance-spanning domain that recognizes the substrate and an ATP-binding domain (D) Affect translocation by forming channels Exp. ABC transporter consist of 2 distinct domains TMD (Transmembrane domain) and NBD (nucleoside binding domain). ATP-binding cassete transportes (ABCtransporter) are members of a protein superfamily that is one of the largest and most ancient families with representatives in all extant phyla from prokaryotes to humans. 28. All cytosolic proteins have nuclear export signals that allow them to be removed from the nucleus when it reassembles after (A) Meiosis (B) Mitosis (C) Both A and B (D) DNA replication 29. Site-specific recombination results in precise DNa rearrangements, which is limited to specific sequences. The enzymes that are important to carry out the process are (A) Restriction endonuclease and DNA polymerase (B) Nuclease and ligase (C) DNA polymerase and ligase (D) DNA polymerase and DNA gyrase Exp. DNS ligase an enzyme commonly used in molecular biology laboratories to join together DNA fragments.a DNA polymerase is an enzyme that helps catalyze the polymerization of DNA bases (deoxyribonucleotides) into a DNA strand. 30. Which of the following statements is NOT true about small interfering RNA (sirna)? (A) SiRNA has a nucleotide sequence with 2nucleotides overhanging at the 3' end (B) SiRNA is processed by the RNA-protein complex RISC (C) SiRNA is often induced by virus (D) SiRNA does not generally act at the level of transcription 31. Which of the following statements is INCORRECT in relation to treatment of pre-b cells with phorbal esters? (A) Phorbol esters activate NFkB for translocation into the nucleus (B) Phorbal esters activate protein kinase C (C) Phorbal easters lead to phosphorylation of NFkB (D) Phorbal esters remove the inhibitor from in active NFkB complex in the cytoplasm P10/4

258 A/UG Presence of an internal ribosome entry site (IRES) in mrna (A) Inhibits its translation (B) Promotes its post-transcriptional processing (C) Has no impact on its translation (D) Promotes its translation under adverse conditions 33. Regulatory elements for expression of ribosomal RNA genes reside in the (A) Transcribed spacer region (B) Non-transcribed spacer region (C) 5' flanking region of individual ribosomal RNA genes (D) Internal regions within the genes 34. Mycobacterium tuberculosis is an intra-cellular bacterium. It prefers to infect (A) Macrophages (B) B-cells (C) T-cells (D) Neutrophils Exp. M. tuberculosis is an intracellular pathogen that infects phagocytic APCs (antigen presenting cells) in lungs, including alveolar macrophages, lung macrophages and dendritic cells. 35. Integrin molecules link extracellular matrix (ECM) to the actin cytoskeleton of cell. Integrin binds to which of the following ECM macromolecules? (A) Laminin (B) Collagen (C) Fibronectin (D) Vitronectin Exp. Inter binds to the Fibronectin ECM molecules. Fibronetin is a high-molecular extracellular matrix that binds to membranespanning receptor protiens called integrins. 36. CD19 is a marker for (A) B-cells (B) T-cells (C) Macrophages (D) NK cells Exp. CD19 is a marker for B-lymphocytes. B-lymphocyte antigencd19alsoknownascd19(clusterof Differentiation 19), is a protein that in humans is encoded by the CD19 gene. It is found on the surface of B-cells, a type of white blood cell. 37. Given below are fate maps of two organisms and the patterns by which embryos undergo cleavage. Which of the following is/are the right combination(s)? Amar Ujala Education (A) 2 only (B) 2 and 1 (C) 1 and 3 (D) 2 and Which of the following matches of oncogeneprotein product is NOT correct? (A) erba - thyroid hormone receptor (B) erbb - epidermal growth factor receptor (C) ras - guanine nucleotide binding protein with GTPase activity (D) fos - platelet-derived growth factor receptor Exp. c-fosis a cellular proto-oncogene belonging to the immediate early gene family of transcription factors. c- Fos has aleucine-zipper DNA binding domain, and a transcription of c-fos is upregulated in response to many extracellular signals, e.g., growth factors. 39. Spermatohonial stem cell undergoes extensive metamorphosis to become a spermatozoan. Meiosis leads to the formation of spermatid containing 22 autosomes and one sex chromosome. A male mouse was found in a colony which always produced only female pups upon mating. Which one of the following is a possible reason? (A) Spermiogenesis was defective (B) All spermatogonial stem cells contained only X and no Y chromosome (C) Activation of Y chromosome linked postmeiotic death related gene may lead to such a situation. (D) Activation of X chromosome linked postmeiotic death related gene may lead to such a situation Exp. A male mouse always produce female puff on mating. The reason is activation of lethal gene associated with Y-chromosome. 40. In case of Xenopus laevis, which cells make up the Nieuwkoop centre and Spemann's organizer? (A) Endodermal and mesodermal, respectively (B) Mesodermal and endodermal, respectively (C) Endodermal and ectodermal, respectively (D) Ectodermal and endodermal, respectively 41. The ced-9 gene to be a binary switch the regulates cellular survival and apoptosis in nematodes. Considering that CED-9 protein can bind to an inactive CED-4, which of the following would lead to apoptosis? P10/5

259 Amar Ujala Education (A) Activation of ced-9 gene (B) Loss of function of CED-3 (C) Loss of function ced-9 gene (D) Loss of function of CED Photosystem II functions as a light-depedent waterplastoquinone oxidoreductase. What are the names of two reaction center proteins that bind electron transfer prosthetic groups, such as P680, pheophytin and plastoquinone? (A) CP43 and CP47 (B) D1andD2 (C) 33kDaand23kDa (D) F A and F B Exp. D1 and D2 proteins of PSII binds to P680 and primary accetor pheophytin. 43. Plants have evolved with multiple photoreceptors, which can perceive specific wavelengths og light. Which one of the following statements is correct about the photoreceptors? (A) Phytochrome A can perceive far red and blue light (B) Phytochrome C can perceive far red light (C) Cryptochrome 1 and phytochrome B are responsible for perceiving blue light (D) Phytochrome B can predominantly perceive far red light 44. Which one of the following statement describes the process of phloem loading? (A) Triose phosphate is transported from the chloroplast to cytosol (B) Sugars are transported into the sieve elements and companion cells (C) Sugars are transported from producing cells in the mesophyll to cells in the vicinity of the sieve elements (D) Solutes are transported from roots to the shoots 45. Which one of the following combinations of secondary metabolite biosynthetic pathways result in the biosynthesis of terpenes? (A) Mevalonic acid and MEP pathways (B) Malonic acid and MEP pathways (C) Shikimic acid and Malonic acid pathways (D) Shikimic acid and Mevalonic acid pathways Exp. The major biosynthetis pathways for terpenes synthesis are - Mevalonic acid and MEP pathways. 46. Which one of the following changes will occur in the cell membrane of nodal tissue of heart, which results in an increased heart rate due to stimulation of sympathetic nerves? (A) Opening of sodium channels is facilitated (B) Potassium conductance is increased A/UG-15 (C) Opening of L-calcium channels are facilitated (D) 'h' channels are inhibited Exp. Sympathetic system increase heart rate by regulating T type calcium channel. 47. A person takes 1.0 ml of insulin injection daily at 8.00 AM. His son gave him 1.5 ml insulin at 8.00 AM considering the father will go to party and eat more during lunch. The father also avoided breakfast, as he planned to eat more during lunch. Which one of the following events will occur? (A) Father will be normaglycemic (B) Father will be in hypoglycemic condition before lunch (C) Father will be in hyperglycemic condition before lunch (D) Blood glucose of father will be low after aking lunch Exp. Because he skipped breakfast too, which is essential for sustainable maintenance of gluose level. 48. How many genetically different gametes can be made by an individual of genotyoe AaBbccDDEe, assuming they are indepedently assorting? (A) 3 (B) 5 (C) 8 (D) 32 Exp. Three genes can have 2 different types of alleles.the genes that show only 1 type of allele need not to be considered. Hence, 2*2*2= Mutation at two different loci of the same gene X results in altered functions. These two mutated versions of the gene X are called (A) Alleles (B) Complementation groups (C) Interrupted genes (D) Linkage groups Exp. A change on one locus another locus of the same gene results in mutation. The two copies of mutated same gene termed as allele. 50. A gene encoding trna undergoes a mutational event in its anticodon region that enables it to recognize a mutant nonsense codon and permit completion of translation. Such a mutation is known as (A) Silent mutation (B) Neutral mutation (C) Reversion (D) Nonsense suppressor Exp. A trna gene is mutated at anticodon region to `reognize stop codon and allow translation beyond pre mature stop codon in mrna transcribed from mutated protien coding gene, suh that functional protein is produced. Such a phenomenon is termed as Nonsense suppressive mutation. P10/6

260 A/UG Two pure lines of corn have mean cob length of 9 and 3 inches, respectively. The polygenes involved in this trait exhibit additive gene action. Crossing these two lines is excepted to produce a progeny population with mean cob length (in inches) of (A) 12.0 (B) 7.5 (C) 6.0 (D) Which of the following organism is widely used as a biocontrol agent in organic farming? (A) Rhizobium tropicii (B) Trichoderma virdis (C) Fusarium oxysporum (D) Nostac muscorum Exp. Trichoderma viridis, is used as a biocontrol agent in organic farming. 53. A paraphyletic group (A) Contains unrelated organisms (B) Includes the most recent common ancestor but not all of its descendents (C) Includes all the representatives of a clade but not the most recent common ancestor (D) Contains all the representatives of a clade and the most recent common ancestor Exp. A paraphyletic group contains the most recent common ancestor but not all of its descendants, e.g.-reptilia contains the last common ancestor of descendants of that ancestor but not birds and mammals. 54. Which of the following is NOT an adaptive modification in a xerophytic plant? (A) Strongly developed sclerenchyma (B) Sunken stomata (C) Sparse stomata (D) Presence of lucnar tissues Exp. Xerophytic plants are those which grown in arid condition,where evaporation is more than precipitation and hence plant has sunken stomata. 55. If milk is left open, lactose is fermented first to produce acid. This is followed by proteolytic bacterial activity which increases the ph. Ultimately milk fats degraded to produce rancidity. This is an example of (A) Ecological succession (B) Microbial antagonism (C) Interference competition (D) Microevolution Exp. Ecological succession is the phenomenon or process by which an ecological community undergoes more or less orderly and predictable changes following disturbance or initial colonization of new habitat. P10/7 Amar Ujala Education 56. Symbiotic biological nitrogen fixation takes place with the association between a plant and a nitrogen fixing prokaryote as shown in the following table: List of plants Nitrogen fixing 1. Soybean (i) Frankia 2. Casuarina (ii) Bradyrhizobium 3. Gunnera (iii) Anabaena 4. Azolla (iv) Nostac The correct combination is: (A) 1 - (i), 2 - (ii), 3 - (iii), 4 - (iv) (B) 1 - (ii), 2 - (i), 3 - (iv), 4 - (iii) (C) 1 - (iii), 2 - (ii), 3 - (i), 4 - (iv) (D) 1 - (iv), 2 - (iii), 3 - (ii), 4 - (i) 57. Secondary sewage treatment involves (A) Physical removal of solids from polluted water by filteration and sedimentation (B) Removal of chemical remains by precipitation (C) Removal of dissolved organic compounds by activated sludge ot trickling filter (D) Removal of microbial pathogens by chlorination or ozonization 58. Based on per molecule, which of the following gas has the most powerful greenhouse effect? (A) CO 2 (B) CH 4 (C) N 2 O (D) CFCs Exp. Carbon dioxide equivalent is highest for CFCs. (amongst given option.it is 140~11,700) 59. Sexual selection results in variation in the reproductive success of males, often due to female choice with particular phenotypes. This type of sexual selection occurs because (A) Males cannot compete with other males (B) Cost of breeding is higher for females as compared to males (C) Inappropriate mating results in a similar reduction in fitness of females and males (D) Males are limiting resource for females Exp. Cost of breeding is higher for females as compared to males. 60. Among the following events in the history of life 1. prokaryotic cell 2. eukaryotic cell 3. natural selection 4. organic molecules 5. self-replicating molecules Which is the correct chronological order? (A) 4,5,3,1,2 (B) 4,5,1,2,3 (C) 5,4,1,3,2 (D) 4,5,1,3,2

261 Amar Ujala Education 61. The Hardy-Weinberg principle comes from considering what happens when Mendelian genes act in a population. The model predicts that there will be no change in allele frequencies when (A) Migration into the population norms at a steady rate (B) The population suffers a bottleneck (C) A rare new mutation is associated with a sharp increase in fitness (D) No evolutionary process is at work Exp. Hardy-Weinberg law states that both alleles and gesnotypic frequencies in a population remains constant from generation to generation until some specific disturbing influences are introduced like non random mating, mutation, selection, random genetic drift, gene flow and meiotic drive. 62. Which one of the following is responsible for initiation of maternal behaviour in the first time pregnant rats after parturition? (A) Higher prolactin levels in blood (B) Srimulation of sensory receptors during delivery (C) Changes of uterine volume (D) Presence of males rats Exp. Prolactin is inducer for lactation and metalnal behavior. 63. To replace animal use in testing hepatic toxicity of a drug on trial, which one of the following would be used in vitro to be closet to the in vivo scenario? (A) Liver cells (B) Hepatic cell lines (C) Liver slices (D) Co-culture of liver parenchymal cells and Kupfer cells 64. Which of the following does not represent a strategy for phytoremediation? (A) Phytodegradation (B) Phytomining (C) Continuous removal through hyper accumulators (D) Chelate-mediated extraction of pollutants Exp. Phytomining is use of plants to extract metal compound of high economic value. 65. The word "fermentation" is used in biochemistry and microbial technology to denote different phenomena. If the former is called C and latter is called T, which of the following statements is true? (A) All C is T but all T is not C (B) All T is C but all C is not T (C) T is always a product of genetic engineering while C is not (D) C is always an anaerobic process, while T can be aerobic or anaerobic P10/8 A/UG Which of the following statements is NOT true during the infection of plant cells with Agrpbacterium? (A) The protein products of virulence genes vira and virg perceive acetosyringone (B) The virb protein forms a connection between Agrobacterium and the plant cell and facilitates T- DNA transfer into the plant (C) The T-DNA is excised and bound to VirD2 protein (D) The T-DNA, after becoming coated with VirF binds to phosphorylated VIP1, which, allows the complex to enter the plant's nucleus Exp. Agrobacterium is a genus o Gram-negative bacteria established by H.J.Conn that uses horizontal gene transfer to cause tumors in plants. 67. among exixting technologies, which of the following vector systems would you prefer to use for generating a library for 140 kb eukaryotic genomic DNA fragments, while giving due consideration to size as well as stability of the insert? (A) Phage (B) Cosmid (C) Bacterial artificial chromosome (BAC) (D) Yeast artificial chromosome (YAC) Exp. COSMID-37 to 52kb BAC kb YAC kb Lamda-below 100 kb 68. If r denotes the correlation coefficient and m denotes the slope of regression line, interchanging X and Y axes would (A) Change m but not r (B) Change r but not m (C) Change both r or m (D) Not change r or m 69. The use of biotinylated secondary antibody in ELISA (A) Increases the sensitivity of the assay but compromises the specificity (B) Increases the sensitivity of the assay without compromising the specificity (C) Does not alter either sensitivity or specificity (D) Decreases both sensitivity and specificity Exp. Biotin labeled Antibodies (Abs) increase sensitivity. They reduce cross reactivity and steric hindrance. It results in higher sensitivity and better specificity. 70. Which is the best method for checking mycoplasma contamination in a mammalian cell line? (A) Southern hybridization (B) ELISA (C) PCR (D) Western hybridization

262 A/UG-15 PART-C Amar Ujala Education 71. Phosphorylation of ADP to ATP occurs through energy metabolism, comprising oxidative phosphorylation or substrate-level phosphorylation or photophosphorylation (in plants). ATP can also be formed from ADP through the action of adenylate kinase. Crystal structure determination of adenylate kinase shows that the C-terminal region has the sequence -Val-Asp-Asp-Val-Phe-Ser-Gln-Val-Cys-Thr- His-Leu-Asp-Thr-Leu-Lys What can be as possible conformation of the sequence? (A) A helix that is not amphipathic (B) Amphipathic helix (C) Leucine zipper helix (D) Beta hairpin Exp. In an amphipathic a helix, one site of the helix contains mainly hydrophilic amino acids and the other site contains mainly hydrophobic amino acids. The amino acid sequence of amphipathic a helix alternates between hydrophilic and hydrophobic residues every 3 to 4 residues,since the a helix makes a turn for every 3.6 residues. 72. Consider a 51-residue long protein containing only 100 bonds about which rotation can occur. Assume that 3 orientations per bond are possible. Based on these assumptions, how many conformation will be possible for this protein? (A) (B) (C) 3 51 (D) 51 x 100 x 3 Exp. Conformation for protein is possible =no. of orientation no. of bonds. Therefore, answer is (A). 73. A plot of V/[S] versus V is generated for an enzyme catalyzed reaction, and a straight line is obtained. Indicate the information that can be obtained from the plot. (A) V max and turnover number K m can be obtained only from a plot of I/V versus I/[S] (B) K m /V max from the slope (C) V max K m and turnover number (D) Only K m and turnover number 74. The following reactions are part of the citric acid cycle. The numbers in parenthesis indicate the number of carbon atoms in each molecule. isocitrate (6) A -ketogluratate (5) B succinyl CoA (4) C succinate (4) D fumarate (4) Which one of the following sequences of reaction systems A D is correct? (A) NAD + CO NADH + H +,NAD + CO2 2 NADH +H +, GDP P GTP, FAD FADH 1 2 P10/9 (B) NAD + CO NADH + H +,NAD + CO2 2 NADH +H +, ADP P ATP, FAD FADH 1 2 (C) NAD + CO NADH + H +,PAD + FADH 2 2 +ADP P 1 ATP, NAD + CO2 FADH + H + (D) NAD + CO NADH + H +,FAD + FADH 2 2 +GDP P 1 GTP, NAD + CO2 NADH + H + Exp. The Krebs CYcle is the central metabolic pathway in all aerobic organisms. The cycle is a series of eight reactions that occur in the mitochondrion. These reactions take a two carbon molecule (acetate) and completely oxidize it to carbon dioxide. The cycle is summarized in the following chemical equation: 2 acetyl CoA+3NAD+FAD+ADP+HPO 4 2 CO 2 +CoA+3 NADH + +FADH + +ATP 75. Phosphoglucomutase is added to 0.1 M glucose-1- phosphate (G-6-P). The standard free energy change of the reaction, G-6-P = G-1-P is 1.8 kcal/mole at 25 o C. The equilibrium concentrations of G-6-P and G-1-P respectively, etc. (A) 96 mm, 45mM (B) 100 mm, 0 mm (C) 45 mm, 96mM (D) 0 mm, 100 mm Exp. The phosphoglucomutase is an enzyme that transfers a phosphate group on an a-d-glucose monomer from the 1' to the 6' position in the forward direction or the 6' to the 1' position in the reverse direction. 76. Differential scanning calorimetric study of calf thymus DNA was carried out to measure midpoint of thermal denaturation (T m ), H m (enthalpy change at T m )and C p (constant - pressure heat capacity change). It has been observed that C p =0,T m = 75.5 o Cand Hm kcal/mole. The Gibbs free energy change at 37 o Cis (A) 25.5 kcal/mole (B) 2.6 kcal/mole (C) 0.6 kcal/mole (D) 5.6 kcal/mole Exp. Several Formulas are used to calculate T m values. Some formulas are more accurate in predicting melting temperatures of DNA duplexes. One problem in nucleic acid thermodynamics is to determine the the thermodynamic parameters for forming double-stranded nucleic acid A B from single-stranded nucleic acids A and B. AB=A + B The equilibrium constant for this reaction is K= A B AB. According to thermodynamic, the relation

263 Amar Ujala Education between free energy, G, and K is G = RTln K, where R is the ideal gas law constant, and T is the kelvin temperature of the reaction. This gives, for the nucleic A B acid system, Gº = RTln AB. the melting temperature, T m, occurs when half of the double-stranded nucleic acid has dissociated.if no additional nucleic acids are present, then [A], [B], and [AB] will be equal, and equal to half the initial concentration of double-stranded nucleic acid, [AB] initial.this gives an expression for the melting point of a nucleic acid duplex of Gº Tm = AB Rln initial 2 Because Gº = Hº T Sº, T m is also given by Hº Tm = AB Sº R ln initian Cystic fibrosis (CF) transmembrane conductance regulator (CFTR) protein is known to be a eampdependent Cl- channel. CF patients (with mutant CFTR proteins) show reduced Cl- permeability and as a result exhibit elevated Cl level in sweat. To prove this, CFTR proteins (both wild type and mutant) are inserted in a model membrance liposome) and Cl- transport is followed with radioactive Cl-. It is known that topology of CFTR in membrane is very important for its function. Despite not proteolytic degradation or denaturation of CFTR proteins, wild type CFTR failed to transport Clin liposome. Which of the following is the correct explanation of this? (A) CFTR protein gets mutated during insertion in liposomes (B) CFTR protein loses affinity with Cl- ions (C) CFTR protein gets wrongly inserted in liposomes (D) CFTR protein loses channel forming property in liposomes Exp. Cystic fibrosis is an autosomal recessive genetic disorder affecting most critically the lungs, and also the pancreas, liver, and intestine. It is characterized by abnormal transport of chloride and sodium across anepithelium, leading to think, viscous secretions. 78. The respiratory chain is relatively inaccessible to experimental manipulation in intact mitochondria. Upon disputing mitochondria with ultrasound, however, it is possible to isolate functional submitochondrial particles, which consist of broken cristae that have resealed inside out into small closed vesicles. In these P10/10 A/UG-15 vesicles the components thatoriginally faced the matrix are now exposed to the surrounding medium. This arrangement helps in studying of electron transport and ATP synthesis because: (A) It is difficult to manipulate the concentration of small molecules (NADH, ATP, ADP, Pi) in the matrix of intact mitochondria (B) In broken cristae, the enzyme and other molecules responsible for electron transport are more active (C) Intact mitochondria are more unstable than broken cristae (D) Purification of intact mitochondria is not possible Exp. The microsome obtained from inner membrane of mitochondria is preferred to investigate, because it is easy to maintain the different concentration of NAD, FAD and ADP with isolated microsomes. 79. Assuming that the histone octamer forms a cylinder 9 nm in diameter and 5 nm in height and that the human genome forms 32 million nucleosomes, what fraction (approximately) of the volume of nucleus (6 m diameter) is occupies by histone octamers? (A) 1/21 (B) 1/11 (C) 10/21 (D) 10/ Hoechst is a membrane-permeant dye that fluoresces when it binds to DNA through a intercalating process. If a population of cells is incubated briefly with Hoechst dye and sorted in a flow cytomer, the cells display various levels of fluorescence in different phases of cell cycles as shown in figure below (marked as X, Y and Z) Which of the following is correct? (A) X is G 1,YisG 2 + M and Z is S (B) X is G 1,YisSandZisG 2 +M (C) X is S, Y is G 2 +MandZisG1 (D) X is S, Y is G 1 and Z is G 2 +M Exp. Flow cytometry is a laser based, biophysical technology employed in cell counting, sorting, biomarker detection and protein engineering, by suspending cells in a stream of fluid and passing them by an electronic derection appratus. Flow cytometry is routinely used in the diagnosis o health disorders,especially blood cancers,but has many other applications in basic research, clinical practice and clinical trials.

264 A/UG The scatter plot of growth rate and growth yield for 100 random environmental isolates of bacteria is shown below: Which of the following can be inferred from the data? (A) The two parameters are not related (B) Growth rate is inversely proportional to growth yield (C) Growth yield is negatively correlated with growth rate (D) High growth cannot be accompanied by high growth yield 82. Double stranded DNA replicates in a semi-conservative manner. In an in vitro DNA synthesis reaction, dideoxy CTP in excess (in separate reaction tubes) in addition to dntps and other necessary reagents. Rate of DNA synthesis was measured by incorporation of 3 H- thymidine. The four graphs drawn below represent the rate of DNA synthesis in two separate reaction tubes. Which of the following graphs represents the expected data? Amar Ujala Education DNA synthesis proceeds in opposite direction, while the double helix is progressively unwinding and replicating in only one direction. One of the DNA strands is continuously synthesized in the same direction as the advancing replication fork and is called strand is synthesized discontinuously in segments and is fragments made as lagging strand. These short fragments made discontinuously are labeled as Okazaki fragments. These Okazaki fragments need to be matured into continuous DNA strand by which one of the following combination of enzymes? (A) DNA Pol III and DNA ligase (B) DNA Pol I and DNA ligase (C) DNA Pol II and DNA ligase (D) DNA gyrase and DNA ligase Exp. During replication, removal of Okazakki fragments (Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging template strand during DNA replication) and gap filling is done by DNA polymerase I and DNA ligase. (A) (C) (B) (D) Exp. In vitro DNA synthesis is a process of synthesis of DNA in laboratory condition i.e.in test tube by polymerase chain reaction process. The polymerase chain reation(pcr) is a biochemical technology in molecular biology to amplify a single or a few copies of a piece of DNA across several orders of magnitude, generating thousands to millions of copies of a particular DNA sequence. In the process the concentration of ddctp will decrease, whereas ddmp will increase which is visible in choise no In semi-conservative mode of DNA replication, two parental strands unwind and are used for synthesis of new strands following the rule of complementary base pairing. Synthesis of complementary strands require that P10/ a reporter cell line with stably integrated retroviral promoter-luciferase construct was transfected with an expression vector for a cellular protein. The protein seems to regulate the activation of the retroviral promoter as analyzed by luciferase activity assay. Which one of the following techniques will you use to show "in vivo" recruitment of the cellular protein on the integrated retroviral promoter? (A) Electrophoretic mobility shift assay (B) RNAse protection assay (C) DNAse hypersensitivity assay (D) Chromatin immunoprecipitation assay Exp. Chromatin immunopreciptation technique can be used for study of DNA (Promoter) - Protein interaction (Regulatory protein) under in-vivo condition. Chromatin Immunoprecipitation (ChIP) is a type of immunopreipitation experimental technique used to investigate the interaction between proteins and DNA in the cell. It aims to determine whether specific genomic regions, such as transcription factors on promoters or other DNA binding sites, and possibly defining cistromes. ChIP also aims to determine the specific location in the genome that various histono modifications are associated with, indicating the target of the histone modifiers.

265 Amar Ujala Education 85. A synthetically prepared mrna contains repetitive AU sequences. The mrna was incubated with mammalian cell extract which contains ribosomes, trnas and all the factors required for protein synthesis, which of the following peptides will most likely be synthesized? (A) A single peptide composed of the same amino acid sequece (B) A single peptide with alternating sequence of two amino acids (C) A single peptide with alternating sequence of three amino acid (D) Three different peptides each sequence composed of a single amino acid Exp. If synthetic RNA poly of alternating AU sequences is used for in-vitro translation, then peptide sequence have repetitive amino acid sequence of two different amino acids. 86. During cell cycle-regulation in eukaryotes, there are post-translational modifications of protein factors, which act as switches for different phases of cell cycle. A cell population of yeast was transfected with gene for Weel kinase (modifies Cde2 protein). Assuming that the transfection efficiency was 50% only, which of the following graphical representation of the results is most appropriate? (A) (C) (B) (D) 87. The lac operon in E.coli, is controlled by both the lac repressor and the catabolite activation protein CAP. In an in vivo experiment with lac operon, the following observations were made: 1. camp levels are high 2. Repressor is bound with allolactose. 3. CAP is interacting with RNA polymerase. Which one of the following conclusions is most appropriate based on the above observations? (A) Glucose and lactose are present (B) Glucose is present and lactose is absent (C) Both are absent (D) Glucose is absent and lactose is present Exp. In an experiment medium when only lactose is present, the lac operaon A/UG-15 Allow lactose to bind to the repressor camp concentration is high in the cell CRP/CAP was interacting with RNA polymerase 88. Upon ligand binding, cell surface receptors move laterally to be capped and internalized. Leishmania, a protozoan parasite, can use several receptors on macrophages to get internalized. One of them is Tolllike receptor 2 (TLR 2) that binds lipophosphoglycan on Leishmania. Once internalized, the parasite is destroyed in the phagolysosome. Which one of the following treatments of Leishmania-infected macrophages will result in lowest parasite number in macrophages? (A) Membrane cholesterol-depleting drug, -methyl cyclodextrin ( -MCD) (B) Ammonium chloride that increases lysosomal ph (C) Both -MCD and ammonium chloride (D) Anti-TLR2 antibody 89. Level of follicle stimulating hormone (FSH) during infancy and adulthood is the same but spermatogenesis is seen only during adulthood. mrna levels coding for FSH receptor are also found to be the same in testis of both age groups. Which of the following investigations will clarify this paradox a little more? (A) Culture testicular cells and add LH to see testosterone production (B) Culture testicular cells and add testosterone to see comparative rise in FSH mrna from both age groups (C) Culture testicular cells and add FSH to see comparative rise in camp production by both age groups (D) Add both LH and FSH to testicular cells and evaluate camp production Exp. The level of FSH during infacy and adult condition are almost same, but sperm production is observed only in adults. It is due to the fact that culture leydig cells and add FSH bring comparative rise in c-amp in both age group. 90. In a tissue, cells are bound together by physical attachment between cell to cell or between cell to extracellular matrix. Following are some of the characteristics of cell junctions: 1. Adhesens junctions are cell-cell anchoring junctions connecting actin filament in one cell with that in the next cell. 2. Desmosomes are cell-matrix anchoring junctions connecting actin filament in one cell to extracellular matrix. 3. Gap junctions are channel forming junctions allowing passage of small water soluble molecules from cell to cell P10/12

266 A/UG Tight junctions are occluding junctions, which seal gap between two cells. 5. Hemidesmosomes are cell-matrix anchoring junctions connecting intermediate filament in one cell to extracellular matrix. Which of the following combination of statements is NOT correct? (A) 1 and 2 (B) 1 and 3 (C) 3 and 4 (D) 4 and 6 Exp. A desmosome is a cell structure specialized for cell-tocell adhesion. Atype of junctional complex, they are localized spot-like adhesions randomly arranged on the lateral sides of plasma membranes. So, Desmosomes cannot join cell matrix and cannot actin to cell-matrix. 91. Oncogenes and tumor suppressor genes are termed as cancer-criti cal genes. Increasingly powerful tools are now available for systematically searching the DNA or mrnas of cancer cells for either significant mutations or altered expression. To identify independently an oncogene or a tumor suppressor gene, which of the following would be the most convincing tests to use? (A) Transgenic mice that overexpress the candidate oncogene and knockout mice that lack candidate tumor suppressor gene (B) Transgenic mice that overexpress the can didate tumor suppressor gene and the knockout mice that lack candidate oncogene (C) Transgenic mice that overexpress the candidate oncogene and tumor suppressor gene (D) Knockout mice that lack the candidate oncogene and tumor suppressor gene Exp. In an experimental condition, use of transgenic mouse over expressing oncogene and loss of function in tumor suppressor gene will help in assessing the role of onogene and tumor suppressor gene in cancer. 92. A large protein of a pathogenic bacterium has been enzymatically digested to generate a mixture of peptides ranginginsizefdrom3to8amino acidsinlength. Peptide mixtures were then administered in experimental animals to generate peptide-specific antibodies. In order to develop diagnostics for the bacteria, the antisera were used for Western blotting failed despite the use of a wide range of antisera concentrations. What is the most probable cause of the problem? (A) Peptide-specific antibody mixture is unstable (B) Peptide-specific antibodies were not generated as adjuvant was not administered (C) Peptide-specific antibodies were not generated as they were not coupled to a protein carrier (D) Peptide-specific antibodies could not recognize the bacterial antigen P10/13 Amar Ujala Education Exp. The most probable cause may be that, peptide specific antibodies were not raised as peptide were not linked to carrier. 93. A mouse was primed with trinitrophenyl- lipopolysaccharide (TNP-LPS) whereas another mouse was primed with TNP-Keyhole limpet hemocyanin (TNP- KLH). After three weeks, these mice were sacrificed and splenic cells were fractionated to B cells and T cells. B cells from TNP-LPS primed mice were co-cultured with T cells from TNP-LPS or TNP-KLH primed mice. So, we have four co-cultures: (A) B TNP-LPS xt TNP-LPS (B) B TNP-LPS xt TNP-KLH (C) B TNP-KLH xt TNP-LPS (D) B TNP-KLH xt TNP-KLH 94. The functionality of the pax6 gene in the formation of optic and nasal structures may be attrubuted to the following: 1. Pax6 makes the optic vesicle competent and allows lens formation. 2. The optic vesicle can induce any part of the head ectoderm to form the nasal and optic structures, due to the presence of Pax6. 3. Pax6 renders the head ectoderm competent to receive signals from the optic vesicle. 4. Apart from the optic vesicle, the head ecto derm may also be induced by BMP4 and FGF8, so Pax6 is not exclusive for lens formation Which of the above attributes are true? (A) 1 and 4 (B) 3 and 4 (C) 2 and 3 (D) 3 only 95. In an experiment, sperm removed from epididymis of a male mouse was added in a dish containing appropriate media and oocyte. No fertilization was seen. However, when sperm from epididymis were directly placed in uterus of an ovulated female, she became pregnant. These observations suggest that: (A) The sperm needs to travel some distance to attain fertilizing ability (B) The oocyte secrets some biochemicals or factors which help sperm to fertilize (C) The hormones in body help sperm to attain fertilizing ability (D) The contents of female reproductive tractinteract with sperm and activate it for fertilization Exp. Under experimental conditions, when sperm and egg are kept together in test tube, they fail to fertilize the egg. But when the similar sperms are directly placed in uterrus of model organism fertilization is observed. The reason for this contrast results is that the content of female reproductive tract interacts with sperm and activates fertilizing activity.

267 Amar Ujala Education 96. The following statements have been proposed for plant vegetative development 1. Lateral roots develop from epidermal cells. 2. Shoot axillary meristem develops from shoot apical meristem during differentiation of leaf primordia 3. Root cap is made up of dead cells. 4. Lateral meristems and cylindrical meristems found in roots and shoots result in secondary growth Which of the above statement are true? (A) 1 and 2 (B) 2 and 4 (C) 1, 2 and 4 (D) 3 and 4 Exp. Shoot apical meristem gives shoot axillary meristmom and lateral meristem and radial meristem are responsible for secondary growth in roots. 97. The pattern of embryonic cleavage specific to a species is determined by two major parameters. 1. The amount and distribution of yolk protein within the cytoplasm 2. The factors in the egg cytoplasm that influence the angle of mitotic spindles and the timing of its formation. Which of the following statement is true? (A) Species having telolecithal egg follow a holoblastic cleavage (B) Species having isolectithal egg follow a holoblastic cleavage (C) Species having centrolecithal egg follow a holoblastic cleavage (D) Species having isolecithal egg follow a meroblastic cleavage Exp. In the absence of large concentration of yolk, four major cleavege types can be observe in isolecithal cells (cells with a small even distribution of yolk ) or in mesolecithalcells (moderate amount of yolk in a gradient) bilateral holoblastic, radial holoblastic, rotational holoblastic, and spiral holoblastic, cleavage. 98. The fate of a cell or a tissue is "specified" when it is capable of differentiating autonomously on being placed in a neutral environment with respect to the development pathway. An embryo will show a development pattern based on its type of specification: Based on the above facts it can be said that the potency of a cell is: 1. equal to its normal fate in regulative development. 2. greater than its normal fate in regulative development. 3. equal to its normal fate in mosaic development. 4. greater than its normal fate in mosaic development Which of the above statements are true? (A) 2 and 3 (B) 1 and 4 (C) 1 and 3 (D) 2 and 4 A/UG-15 Exp. Based on the fact in question, the potency of a cell is equal to normal fate in mosaic development and geater in regulative development. Mosaic and regulative development: The molecles that underline these concepts are becoming more defined and understood. To oversimlify: mosaic development depends on agents, such as transcription factors, being placed locally in the egg by the mother. Regulative development depends in part on long-range gradients of positional information, such as that provided by the Hedgehog protein, that can pattern many cells at once. Regulative development can also be diven by shortrange signals that trigger changes in cell identity in nearby neighbours. 99. In the context of the proximal-distal growth and differentiation of a tetrapod limb following experiments were visualized: 1. If the apical ectodermal ridge (AER) is development, further development of distal limb skeletal elements ceases. 2. If leg mesenchyme is placed directly beneath the wing AER, distal hindlimb structures develop at the end of the limb. 3. If an extra AER is grafted onto an existing limb bud, supernumerary structures are formed usually at the distal end of the limb. 4. If leg mesenchyme is placed directly beneath the wing AER, proximal hindlimb structures develop at the end of the limb. Which of the above experiments would show the possible interactions between the AER and the limb bud mesenchyme directly beneath it during limb development? (A) 1 and 2 only (B) 2 abd 3 only (C) 3 and 4 only (D) 1, 2 and Following are some facts regarding localization of photosynthetic supramolecular complexes on plastid lamellae: 1. PSII is preferentially located on granal lamellae. 2. ATP synthase and PSI are preferentially located on stromal lamellae. 3. PSI and PSII are located adjacent to each other on stromal lamellae. 4. Cyt b 6/f complex is not membrane-bound complex Which one of the following combinations of the above statements is true? (A) 1 and 2 (B) 3 and 4 (C) 2 and 4 (D) 2 and 3 P10/14

268 A/UG Upon absorption of a photon, a chlorophyll molecule gets converted to its excited state when the energy of the photon is (A) More than that of the ground state of the pigment molecule (B) Equal to that of the pigment molecule's excited state (C) More than that of the ground state but less than that of the excited state of the pigment molecule (D) Equal to the energy gap between ground state energy and the excited state energy 102. Following are certain facts about the effect of abscisic acid (ABA) on the development and physiology effect of plants: 1. ABA promotes leaf senescence independent of ethylene. 2. ABA promotes shoot growth and inhibits root growth at low water potentials. 3. ABA inhibits gibberellin induced enzyme production. 4. Seed dormancy is controlled by the ratio of ABA and gibberellin. Which one of the following combinations of the above statements is true? (A) 1, 2 and 3 (B) 2, 3 and 4 (C) 1, 2 and 4 (D) 1, 3 and Red and far-red lights are perceived by plants through various photoreceptors including phytochromes. The activation of phytochrome is caused by: (A) Conversion of Pr to Pfr from through the effect of red light (B) Repression of Pr form through the effect of far-ref light. (C) Equal proportion of red and far-red lights at same fluence rates (D) Presence of red and far-red light at different fluence rate 104. While studying the primary effects of different abiotic stress on plants, a researcher observed water potetial ( p ) reduction and cellular dehydrations. Which of the following combination of abiotic stresses may cause the observed effect? (A) Water, deficit, salinity and chilling (B) Salinity, high-temperature and flooding (C) Freezing, salinity and water deficit (D) Freezing, chilling and flooding P10/15 Amar Ujala Education 105. Phenylalanine ammonia-lyase (PAL) and chalcone synthase (CHS) are involved in biosynthase of phenolic compounds in plants. Following are some statements regarding the actions of PAL and CHS: 1. Substrates for PAL and CHS are phenylalanine and chalcone, respectively. 2. PAL converts phenylalanine to transcinnamic acid. 3. PAL converts phenylalanine to p-coumeric acid. 4. p-coumaroyl-coa is converted to chalcones CHS. Which one of the following combination of the above statements is true? (A) 1 and 2 (B) 1 and 3 (C) 2 and 3 (D) 2 and The intestinal absorption of glucose is im paired by the use of ouabain, an inhibitor of Na + /K + ATPase. Indicate the correct explanation. (A) The inhibitor hsd blocked the transport of Na + from intestinal lumen to epithelial cells (B) The inhibitor has blocked the transport of Na + from epithelial cells to the intestinal lumen (C) The inhibitor has blocked Na + transport from epithelial cells to the interstitial space (D) The inhibitor ha blocked Na + transport from the interstitial space to epithelial cells 107. The stereocilia of auditory hair cells are arranged in rows but the heights of stereocilia are not the same in all the rows. Though the heights of stereocilia is the same withina particular row, the heights increase in subsequent rows. When the steriocilia of shorter rows are mechaniclly pushed towards the taller rows, the hair cells are depolarized but a push on opposite direction hyperpolarizes them.the significance of this grade height of stereocilia is: 1. Each row of stereocilia may be displaced independent of other rows in physiological conditions. 2. The tip of the talller stereocilia will show greater displacements as compared to shorter ones when all the rows are moving in the same axis. 3. The hair cells will be depolarized or hyper polarized in different grades when the axis of displacement is changed. 4. The taller stereocilia are involved with depolarization and shorter ones are responsible for hyperpolarization. Which one of the following is correct? (A) 1 only (B) 2 only (C) 2 and 3 (D) 2 and 4

269 Amar Ujala Education 108. GnRH is secreted during infancy (0-6 months) and puberty onwards (4 years and above) in monkeys. However, i.v. injection of GnRH during pre-pubertal period (about 2 years of age) led to elevated LH and FSH in blood compared to untreated 2 years old monkey. This suggests that: 1. hypothalamus is active during pre-pubertal period. 2. GnRH action on pituitary is age dependent. 3. pituitary matures during afulthood. 4. pituitary is active in all the stages of development in minkey. Which one of the following is true? (A) 1 and 2 (B) 2 and 3 (C) 3 only (D) 4 only 109. A person has been suffering fron night blindness. On consultation, the doctor advised the person to eat carrots and/or cod fish oil. After some time having seen no improvement, the doctor gave the person Vitamin A injection. Still no marked improvement was seen. The doctor mooted several suggestion indicating lack of the following enzymes for the failure of treatment: 1. Retinol dehydrogenase. 2. Retinal reductase. 3. Retinal isomerase. 4. Retinal synthase. Which one of the following is correct? (A) 1 only (B) 2 only (C) 2 and 3 (D) 3 and 4 Exp. Night blindness may also be due to the results of Zn deficiency which reduces the activity of an enzyme that helps vitamin A. This enzyme is retinol dehydrogenase A person suffering from thyrotoxicosis has extremely high level of thyroid hormone in blood. There is a failure of feed back regulation in hypothalamic-pituitarythyroid axis. The detailed blood investigation exhibited high level of the following: 1. Thyroid stimutaling hormone (TSH). 2. Thyroid stimulating immunoglobulin(tsi). 3. Thyrotropin releasing hormone (TRH). 4. Parathyroid hormone (PTH). In your opinion, which one of the following is the reason for such thyrotoxicosis? (A) 1 only (B) 2 only (C) 2 and 3 (D) 3 and Which of the following statements regarding aquaporins or water channel NOT correct? (A) Aquaporins are found in both plant and animal membranes P10/16 A/UG-15 (B) Aquaporins cannot transport unchanged molecules like NH 3. (C) Phosphorylation and calcium concentration regulate aquaporin is regulated by ph and reactive oxygen species. (D) Activity of aquaporin is regulated by ph and reactive oxygen species. Exp. Aquaporins can transport unchanged molecules like NH 3 (aquaglyceroporins) When two indepedent pure lines of pea with white flowers are crossed, the F 1 progeny has purple flowers. The F 2 progeny obtained from selfing shows both purple and white flower in a ratio of 9 : 7. The following conclusions were made: 1. Two different genes are involved, mutations in which lead to formation of white flowers. 2. These two genes show independent assortment. 3. This is an example of complementary geneaction. 4. This is an example of duplicate genes. Which of the above conclusions are correct? (A) 1 and 3 only (B) 1 and 4 only (C) 1, 2 and 4 (D) 1, 2 and A cell undergoing meiosis produces four duaghter cells, two of which are ancuploids, while two are haploid. This can occur due to: (A) Non-disjunction during first meiotic division only. (B) Non-disjunction during second meiotic division only. (C) Non-disjunction during either first or second meiotic divisions (D) Non-disjunction during both first and second meiotic divisions. Exp. Nondisjunction during meiosis 2 will produce aneupoids as n + 1 and n - 1 and haploids as n and n. Thus half gametes are affected by non dis junction during meiosis 2 while all gametes are affetcted during meiosis Three E. coli mutants are isolated which require compound 'A' for their growth. The compounds B, C and D are known to be involved in the biosynthetic pathway to A. In order to determine the pathway, the mutants were grown in a medium supplemented with ONE OF THE COMPOUNDS, A TO D. The results obtained are summarized below: Mutant Medium supplemented with compund A B C D

270 A/UG-15 '+' indicates growth: '0' indicates lack of growth Which of the following equations represents the biosynthetic pathway of A? (A) B-C-D-A (B) C-D-B-A (C) B-D-C-A (D) A-C-D-B 115. Following are four modes of inheritance 1. X-linked recessive 2. X-linked dominant 3. Autosomal recessive 4. Autosomal dominant Which of the above modes of inheritance can explain the pedigree shown below? (A) 1 and 3 (B) 2 and 3 (C) 3 and 4 (D) 4 only 116. The auxotrophic strains of E. coli:a(met -- bio -- thr + leu + thi + ) and B (met + bio + th -- leu -- thi -- ) were incubated together for 18 hours in a liquid complete medium and then ~ 10 8 cells were plated on a minimal medium. Prototrophs were observed at a frequency of 1 x 10-7 cells. This may have happened by a process of genetic recombination between the two strains or by mutation of the strains. Which of the following control experiments would help rule out the possibility of mutation? (A) Plating strains A and B directly on minimal medium. (B) Growing the mixture of strains A and B for 18 hours and then plating on complete medium (C) Growing strains A and B individually in a liquid complete medium for 18 hours and then plating them on a minimal medium. (D) Growing the obtained prototrophs in a liquid complete medium for 18 hours and then plating them on a minimal medium Four different mutant lines showing similar phenotype were identified from a genetic screen. When genetc crosses among these mutants were carried out, the first mutant was found to complement the second, third and fourth mutant lines. However, no other complementation groups do the four mutant lines belong to? (A) One (B) Two (C) Three (D) Four 118. A species has the following population characteristics: 1. Reduction in population size > 90% over the last 10 years or 3 generations. Amar Ujala Education 2. Geographic range : Extent of occurance: < 100 km 2 and Area of occupancy : < 10 km Population size less than 50 matured individuals. 4. Probability of extinction in the wild is at least 50% within the next 10 years or 3 generation. To which of the following categories the species will be assigned according to IUCN categorozation of threatened species (version 3.1)? (A) Endangered (B) Vulnerable (C) Critically (D) Extinct in the wild 119. Which of the following hypothesis best explains the occurence of Himalayan floral elements in Western Ghats in India? (A) Continental drift theory (B) Deccan trap theory (C) Himalayan glaciation theory (D) Coromandel coast hypothesis 120. Why lysogenic cycle is more beneficial to a virus than lytic cycle under certain circumstances? (A) The lysogenic cycle prevents local extinction of the host while still retaining its infection potential. (B) By integrating with the bacterial chromosomes, the genetic instructions for the virus become refreshed after one or more replication events during binary fission. (C) Lysogenic infection cycles do not harm their host cells, so they can produce virus particles indefinitely (D) Lysogeny causes more mutations to occur in the virus, creating more variants upon which natureal selection can operate 121. Identify a, b and c in the figure: (A) a = mitochondria; b = multicellularity; c = chloroplast (B) a = mitochondria; b = chloroplast; c = multicellularity (C) a = chloroplast; b = multicellularity, c = mitochondria (D) a = chloroplast; b = nucleus; c = multicellularity P10/17

271 Amar Ujala Education 122. In which of the following classes of vertebrates there are groups of animals without limbs? (A) Fish, reptiles, and mammals (B) Reptiles only (C) Reptiles and amphibians (D) Amphibians only 123. the schematic section given below of an animal indicates that the animal is: A/UG-15 The number of individuals under each species was listed as mentioned in the following Table. In which of the following communities Pielou's Evenness Index (e) will be 1? Exp. As each species has 10 individuals so it is more even Average annual precipitation and temperature are two important determinants of world's major biomes. Which of the following combination is correct? (A) Tripoblastic, coelomic, invertebrate (B) Tripoblastic, acoelomic, invertebrate (C) Diploblastic, coelomic, invertibrate (D) Triploblastic, coelomic, vertebrate Exp. AS VENTRAL nerve cord is there and coelom is also there along with mesoderm so it is triploblastic, coelomate, invertebrate At a given time, the age class distribution of a population was as shown in the figure: Which of the following can be inferred from the figure? (A) Age class 2 has maximum fecundity (B) Age class 2 has maximum survival (C) Age class distribution is at equilibrium (D) age class distribution is not at equilibrium 125. While studying the diversity of 4 communities, 5 species and 50 individuals were recorded from each community ºC temperature and 255 cmprecipitation 2. 15ºC temperature and 300 cmprecipitation 3. 15ºC temperature and 100 cmprecipitation 4. 25ºC temperature and 255 cmprecipitation (i) Temperate forest (ii) Savannah (iii) Temperate rain forest (iv) Tropical rain forest (A) 1 - (iv), 2 - (iii), 3 - (i), 4 - (ii) (B) 1 - (iii), 2 - (ii), 3 - (iv), 4 - (i) (C) 1 - (ii), 2 - (i), 3 - (iii), 4 - (iv) (D) 1 - (i), 2 - (iv), 3 - (ii), 4 - (iii) 127. A researcher collected information from four forest areas using a sensor to assess their green cover. Observed average spectral values for each of the forests are given in the table below: Spectral value Forest NIR VIS (A) (B) (C) (D) The forest green cover in the order of highest to lowest is (A) A>C>B>D (B) A>D>C>B (C) B>C>D>A (D) D>A>B>C Exp. NORMALIZED differential vegetation index =NIR-VIS/ NIR + VIS NDVI is highest for A. P10/18

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