Name_BS50 Exam 3 Key (Fall 2005) Page 2 of 5
|
|
- Chester Cox
- 6 years ago
- Views:
Transcription
1 Name_BS50 Exam 3 Key (Fall 2005) Page 2 of 5 Question 1. (14 points) Several Hfr strains derived from the same F + strain were crossed separately to an F - strain, giving the results indicated in the table below. Draw a chromosome map for the genes indicated. Include the location and orientation of the F factors. Define all symbols used Time elapsed before cells carrying markers were recovered Hfr strain 5min 10min 15min 20min 25min 30min 1 pro gal lac 2 pur ala leu lac 3 pro thy thr 4 pur ile thr Fertility Factor Origin (first to enter) Terminus (last to enter) Thr pro Hfr 3 Hfr 1 Thy 15 min gal lac (14 pts) Genes (order and mapping times): 8 pts; no deduction if a single mapping time in error; -1 pt. if mapping times listed, but not indicated on map. Hfr s: 6 pts; positions (2 pts); relative orientations (2 pts); details (2 pts). Position of Hfr s must be indicated as single sites: -1 pt if drawn as an extents. Details for Hfr s (any 2 of the following): absolute orientation; use of conventional symbol to indicate; precise localization (esp. for Hfr4 adjacent to ala). For aborted attempts, 1 pt if indicated as circular; if drawn as 4 linears, graded on the basis of one (2 pts for genes; 1 pt for Hfr). Question 2. (22 points). You are studying a fairly common human trait and have discovered a linked restriction fragment length polymorphism (RFLP). Use the pedigree from the following family along with the Southern blot analyzing inheritance of the RFLP to answer the following questions. Ile Hfr 2 5 min pur 5 min Hfr 4 ala leu
2 Name_BS50 Exam 3 Key (Fall 2005) Page 3 of 5 a) Just by looking at the pedigree (and ignoring the DNA data), what mode(s) of inheritance could account for the pedigree? Circle all possible answers. autosomal dominant autosomal recessive X-linked dominant X-linked recessive (4 pts) 1 pt for each correctly chosen or not b) Considering the DNA data as well, what is the mode of inheritance for this trait? (4 pts) X-linked recessive c) Give the genotypes of the following individuals are completely as possible. Define all symbols used. R 7 = 7kb RFLP A = dominant no trait R 6 = 6kb RFLP a = recessive trait R = 4kb+3kb RFLP Y = Y chromosome Individual A: R 6 a/ R 7 A Individual B: R a/ Y Individual E: R 7 A/ R a (8 pts) 2 pts for each (1 pt for each chromosome); 2 pts for correct nomenclature; -3 pts if correct linkage not indicated. d) Individual E marries a phenotypically normal male. Their son is found to have only a 4 kb and 3 kb band when analyzed with the same probe used above. Assuming the RFLP is actually 10 cm from the trait for this gene, what is the probability that their son will have this trait? Show your reasoning. (6 pts) 90 % (3 pts if no further explanation); son carries 4+3 RFLP marker from mother (3 pts; OK if implicit, deducted if misunderstood) and Y from father (1 pt; OK if implicit); 10 % chance that a recombination event occurred between the RFLP marker and the disease allele from mother (2 pts; must explicitly mention recombination, cross-over, or linkage, or show in a diagram). -2 pts if assumed E has 90% chance of carrying the disease allele; -2 to 3 pts for other extraneous factors. If answer to b/c incorrect, could receive up to 5 pts if consistent. Question 3. (18 points) a) The following lists a number of genotypes for the Lac operon (including partial diploids). In each case, say whether the genotype listed will show activity of β-galactosidase (lacz) and lactose permease (lacy) when the inducer (allolactose) is added and in the absence of inducer: Genotype β-galactosidase (lacz) lactose permease (lacy) Not induced Induced Not induced Induced I + P + O + Z + Y I s P + O + Z + Y
3 Name_BS50 Exam 3 Key (Fall 2005) Page 4 of 5 I + P + O c Z + Y I + P + O + Z Y + / I + P + O c Z + Y I + P O c Z Y + / I + P + O + Z + Y I + P O + Z + Y / I P + O + Z Y (15 pts) -1 pt for each incorrect; >3 incorrect scaled (eg: -11pts. given 7/15). b) For the lac operon, what would be the phenotype of a mutation that prevents CAP from binding camp. (3 pts) Genes of the lac operon would never be transcribed at high level; or basal level of transcription only, even in presence of inducer. 1-2 pts if indicate no activity, depending upon detail of explanation. Question 4. (18 points) A plant has been discovered that has two different true breeding strains, each one homozygous for a different allele of the gene "Height". One strain is homozygous for the dominant, wild-type allele H and the plants are very tall. The other strain is homozygous for the recessive, mutant h allele and the plants are all very short. Below are the DNA coding regions for the two different alleles. Both alleles have three exons, although Exon 2 of the h allele is longer and contains an in-frame stop codon (UAA). Exon 3 can encode a DNA binding motif (a homeodomain). The presence of a homeodomain is characteristic of a transcriptional regulator. A) Which allele produces a longer processed mrna? Which allele produces a longer protein? Explain. (Assume all in-frame stop codons are shown.) (8 pts) h produces longer mrna (2 pts) because exon 2 is longer (2 pts); H produces longer protein (2 pts), because h has a premature stop (2 pts). B) Based on the information given above, what do you think Height protein does in the plant? Why is the H allele dominant? (5 pts) Functions as a transcription factor (1 pt); regulates genes that affect height or growth (2 pts); is dominant because one copy of the gene is sufficient (2 pts). Partial credit for explanation of dominance: trans-acting (1 pt); produces functional product and h does not (1 pt). If additional incorrect explanation offered, -1 pt. Compensatory credit: described as haplo-sufficient (1 pt); the h allele product has no deleterious effect
4 Name_BS50 Exam 3 Key (Fall 2005) Page 5 of 5 (1 pt); if detailed explanation of transcriptional regulation, but did not address phenotype (1 pt). C) Another mutant of the H allele is discovered. It has all the Height coding DNA, but no trace of Height mrna. Suggest what the molecular nature of this mutation might be, and predict its phenotype. (5 pts) Mutation in the promoter (2 pts); phenotype (3 pts): short (2 pts), recessive (1 pt), OR equivalent to h/h (3 pts). Compensatory credit: if detailed explanation of promoter/polymerase, but did not address phenotype (1 pt). Alternative explanations, such as mutation that affects local chromatin structure, accepted if adequately explained. Question 5. (22 points) A fragment of mouse DNA digested with EcoR1 carries the gene M. This DNA fragment, which is 8 kb long, is inserted into a bacterial plasmid at an EcoR1 site. The recombinant plasmid is then digested with different restriction enzymes. The size of these fragments after digestion are described below: EcoR1: 6kb, 8kb BamH1: 3kb, 11kb EcoR1 & BamH1: 1kb, 2kb, 4kb, 7kb Pst1: 14kb Pst1 & EcoR1: 4kb, 6kb A) Draw a restriction map of this circular plasmid. PstI 3 kb 4 kb BamHI 1 kb EcoRI Gene M 4 kb EcoRI 2 kb (8 pts) Sites in correct relative order (4 pts); correct map distances (4 pts). B) If a Southern blot is prepared from EcoR1 digested DNA, BamH1 digested DNA, and EcoR1 & BamH1 digested DNA, which fragments will hybridize to a probe containing only plasmid DNA? EcoR1: _6kb BamH1: _3kb, 11kb BamHI
5 Name_BS50 Exam 3 Key (Fall 2005) Page 6 of 5 EcoR1 & BamH1: _2kb, 4kb (6 pts; 2 pts each). If one of two bands indicated, 1pt; no credit if extra bands indicated. C) What fragment sizes do you expect from a Pst1, BamH1 double digest? (8 pts) 3kb (2 pts) and 8kb (2 pts) Two different 3kb fragments are produced (not required for full credit) D) You wish to express the mouse protein in bacteria. What problem(s) might you encounter, and how could you solve it? (4 pts) If the mouse gene has introns (1 pt) they would not be spliced out (1 pt); can avoid this problem by using a cdna of the mouse gene (2 pts). OR Prokaryotic promoters and transcription differ from eukaryotic (2 pts); include a bacterial promoter in construct (2 pts). If missed the point, but described transformation process (1 pt). Question 6. (6 points) Histone acetyltransferase (HAT) is the enzyme responsible for adding acetyl groups to histones (particularly H4). What would be the consequence of having a low activity HAT enzyme? (6 pts) Would result in more tightly packed nucleosomes (4 pts) and thus lower levels of transcription (2 pts). Mention of nucleosomes required for full credit. Partial credit for more condensed chromatin structure (3 pts for first component); more heterochromatin or chromatin remodeling (2 pts for first component).
BS 50 Genetics and Genomics Week of Nov 29
BS 50 Genetics and Genomics Week of Nov 29 Additional Practice Problems for Section Problem 1. A linear piece of DNA is digested with restriction enzymes EcoRI and HinDIII, and the products are separated
More informationR1 12 kb R1 4 kb R1. R1 10 kb R1 2 kb R1 4 kb R1
Bcor101 Sample questions Midterm 3 1. The maps of the sites for restriction enzyme EcoR1 (R1) in the wild type and mutated cystic fibrosis genes are shown below: Wild Type R1 12 kb R1 4 kb R1 _ _ CF probe
More informationDO NOT OPEN UNTIL TOLD TO START
DO NOT OPEN UNTIL TOLD TO START BIO 312, Section 1, Spring 2011 February 21, 2011 Exam 1 Name (print neatly) Instructor 7 digit student ID INSTRUCTIONS: 1. There are 11 pages to the exam. Make sure you
More informationCHAPTER 13 LECTURE SLIDES
CHAPTER 13 LECTURE SLIDES Prepared by Brenda Leady University of Toledo To run the animations you must be in Slideshow View. Use the buttons on the animation to play, pause, and turn audio/text on or off.
More informationBio 311 Learning Objectives
Bio 311 Learning Objectives This document outlines the learning objectives for Biol 311 (Principles of Genetics). Biol 311 is part of the BioCore within the Department of Biological Sciences; therefore,
More informationConcepts: What are RFLPs and how do they act like genetic marker loci?
Restriction Fragment Length Polymorphisms (RFLPs) -1 Readings: Griffiths et al: 7th Edition: Ch. 12 pp. 384-386; Ch.13 pp404-407 8th Edition: pp. 364-366 Assigned Problems: 8th Ch. 11: 32, 34, 38-39 7th
More informationDO NOT OPEN UNTIL TOLD TO START
DO NOT OPEN UNTIL TOLD TO START BIO 312, Section 1: Fall 2012 December 4 th, 2012 Exam 3 Name (print neatly) Signature 7 digit student ID INSTRUCTIONS: 1. There are 12 pages to the exam. Make sure you
More informationChapter 18: Regulation of Gene Expression. 1. Gene Regulation in Bacteria 2. Gene Regulation in Eukaryotes 3. Gene Regulation & Cancer
Chapter 18: Regulation of Gene Expression 1. Gene Regulation in Bacteria 2. Gene Regulation in Eukaryotes 3. Gene Regulation & Cancer Gene Regulation Gene regulation refers to all aspects of controlling
More informationGenomes summary. Bacterial genome sizes
Genomes summary 1. >930 bacterial genomes sequenced. 2. Circular. Genes densely packed. 3. 2-10 Mbases, 470-7,000 genes 4. Genomes of >200 eukaryotes (45 higher ) sequenced. 5. Linear chromosomes 6. On
More information7 Gene Isolation and Analysis of Multiple
Genetic Techniques for Biological Research Corinne A. Michels Copyright q 2002 John Wiley & Sons, Ltd ISBNs: 0-471-89921-6 (Hardback); 0-470-84662-3 (Electronic) 7 Gene Isolation and Analysis of Multiple
More information7.1 The lac Operon 7-1
7.1 The lac Operon The lac operon was the first operon discovered It contains 3 genes coding for E. coli proteins that permit the bacteria to use the sugar lactose Galactoside permease (lacy) which transports
More informationM I C R O B I O L O G Y WITH DISEASES BY TAXONOMY, THIRD EDITION
M I C R O B I O L O G Y WITH DISEASES BY TAXONOMY, THIRD EDITION Chapter 7 Microbial Genetics Lecture prepared by Mindy Miller-Kittrell, University of Tennessee, Knoxville The Structure and Replication
More informationAnswers to additional linkage problems.
Spring 2013 Biology 321 Answers to Assignment Set 8 Chapter 4 http://fire.biol.wwu.edu/trent/trent/iga_10e_sm_chapter_04.pdf Answers to additional linkage problems. Problem -1 In this cell, there two copies
More informationBIOLOGY. Chapter 16 GenesExpression
BIOLOGY Chapter 16 GenesExpression CAMPBELL BIOLOGY TENTH EDITION Reece Urry Cain Wasserman Minorsky Jackson 18 Gene Expression 2014 Pearson Education, Inc. Figure 16.1 Differential Gene Expression results
More informationRegulation of Gene Expression
Slide 1 Chapter 18 Regulation of Gene Expression PowerPoint Lecture Presentations for Biology Eighth Edition Neil Campbell and Jane Reece Lectures by Chris Romero, updated by Erin Barley with contributions
More informationGENE REGULATION IN PROKARYOTES
GENE REGULATION IN PROKARYOTES Prepared by Brenda Leady, University of Toledo Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Gene regulation refers to
More informationGene mutation and DNA polymorphism
Gene mutation and DNA polymorphism Outline of this chapter Gene Mutation DNA Polymorphism Gene Mutation Definition Major Types Definition A gene mutation is a change in the nucleotide sequence that composes
More informationRegulation of Gene Expression in Eukaryotes
12 Regulation of Gene Expression in Eukaryotes WORKING WITH THE FIGURES 1. In Figure 12-4, certain mutations decrease the relative transcription rate of the -globin gene. Where are these mutations located,
More informationBio 101 Sample questions: Chapter 10
Bio 101 Sample questions: Chapter 10 1. Which of the following is NOT needed for DNA replication? A. nucleotides B. ribosomes C. Enzymes (like polymerases) D. DNA E. all of the above are needed 2 The information
More informationGENETICS. I. Review of DNA/RNA A. Basic Structure DNA 3 parts that make up a nucleotide chains wrap around each other to form a
GENETICS I. Review of DNA/RNA A. Basic Structure DNA 3 parts that make up a nucleotide 1. 2. 3. chains wrap around each other to form a Chains run in opposite direction known as Type of bond between the
More informationQ1 (1 point): Explain why a lettuce leaf wilts when it is placed in a concentrated salt solution.
Short questions 1 point per question. Q1 (1 point): Explain why a lettuce leaf wilts when it is placed in a concentrated salt solution. Answer: Water is sucked out of the cells by osmosis (this reduces
More informationEinführung in die Genetik
Einführung in die Genetik Prof. Dr. Kay Schneitz (EBio Pflanzen) http://plantdev.bio.wzw.tum.de schneitz@wzw.tum.de Prof. Dr. Claus Schwechheimer (PlaSysBiol) http://wzw.tum.de/sysbiol claus.schwechheimer@wzw.tum.de
More informationRegulation of gene expression. (Lehninger pg )
Regulation of gene expression (Lehninger pg. 1072-1085) Today s lecture Gene expression Constitutive, inducible, repressible genes Specificity factors, activators, repressors Negative and positive gene
More informationGen e e n t e i t c c V a V ri r abi b li l ty Biolo l gy g Lec e tur u e e 9 : 9 Gen e et e ic I n I her e itan a ce
Genetic Variability Biology 102 Lecture 9: Genetic Inheritance Asexual reproduction = daughter cells genetically identical to parent (clones) Sexual reproduction = offspring are genetic hybrids Tendency
More informationGENE REGULATION slide shows by Kim Foglia modified Slides with blue edges are Kim s
GENE REGULATION slide shows by Kim Foglia modified Slides with blue edges are Kim s 2007-2008 Bacterial metabolism Bacteria need to respond quickly to changes in their environment STOP GO if they have
More informationDifferential Gene Expression
Biology 4361 Developmental Biology Differential Gene Expression September 28, 2006 Chromatin Structure ~140 bp ~60 bp Transcriptional Regulation: 1. Packing prevents access CH 3 2. Acetylation ( C O )
More informationEinführung in die Genetik
Einführung in die Genetik Prof. Dr. Kay Schneitz (EBio Pflanzen) http://plantdev.bio.wzw.tum.de schneitz@wzw.tum.de Twitter: @PlantDevTUM, #genetiktum FB: Plant Development TUM Prof. Dr. Claus Schwechheimer
More informationLecture 9 Controlling gene expression
Lecture 9 Controlling gene expression BIOLOGY Campbell, Reece and Mitchell Chapter 18 334- (352-356) Every cell in your body contains the same number of genes approximately 35, 000 DNA is wound around
More informationfour chromosomes ` four chromosomes correct markers (sister chromatids identical!)
Name KEY total=107 pts 1. Genes G and H are on one chromosome; gene F is on another chromosome. Assume the organism is diploid and that there is no crossing over in this species. You are examining the
More informationGenetics Final Exam Summer 2012 VERSION B. Multiple Choice (50 pts. possible) IF you completed the in-class workshop put a CHECK MARK HERE --->
enetics Final Exam Summer 2012 VERSION B Name Multiple hoice (50 pts. possible) Problems (50 points possible) Total (100 points possible) KEY IF you completed the in-class workshop put a HEK MRK HERE --->
More informationAP Biology Gene Expression/Biotechnology REVIEW
AP Biology Gene Expression/Biotechnology REVIEW Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Gene expression can be a. regulated before transcription.
More informationSummary 12 1 DNA RNA and Protein Synthesis Chromosomes and DNA Replication. Name Class Date
Chapter 12 Summary DNA and RNA 12 1 DNA To understand genetics, biologists had to learn the chemical structure of the gene. Frederick Griffith first learned that some factor from dead, disease-causing
More informationDO NOT OPEN UNTIL TOLD TO START
DO NOT OPEN UNTIL TOLD TO START BIO 312, Section 1: Fall 2012 September 25 th, 2012 Exam 1 Name (print neatly) Signature 7 digit student ID INSTRUCTIONS: 1. There are 14 pages to the exam. Make sure you
More information3. human genomics clone genes associated with genetic disorders. 4. many projects generate ordered clones that cover genome
Lectures 30 and 31 Genome analysis I. Genome analysis A. two general areas 1. structural 2. functional B. genome projects a status report 1. 1 st sequenced: several viral genomes 2. mitochondria and chloroplasts
More informationCHAPTER 18 LECTURE NOTES: CONTROL OF GENE EXPRESSION PART B: CONTROL IN EUKARYOTES
CHAPTER 18 LECTURE NOTES: CONTROL OF GENE EXPRESSION PART B: CONTROL IN EUKARYOTES I. Introduction A. No operon structures in eukaryotes B. Regulation of gene expression is frequently tissue specific.
More information(b) Draw a genetic linkage map showing map distances between met, thi, and pur.
Botany 132 Final exam 2002 Name Please show all of your work in answering the questions below. 1. F- bacterial cells of genotype met - thi - pur - were conjugated with F+ cells with the genotype met +
More informationREGULATION OF GENE EXPRESSION
REGULATION OF GENE EXPRESSION Each cell of a living organism contains thousands of genes. But all genes do not function at a time. Genes function according to requirements of the cell. Genes control the
More informationThe Regulation of Bacterial Gene Expression
The Regulation of Bacterial Gene Expression Constitutive genes are expressed at a fixed rate Other genes are expressed only as needed Inducible genes Repressible genes Catabolite repression Pre-transcriptional
More informationDNA Structure and Properties Basic Properties Predicting Melting Temperature. Dinesh Yadav
DNA Structure and Properties Basic Properties Predicting Melting Temperature Dinesh Yadav Nucleic Acid Structure Question: Is this RNA or DNA? Molecules of Life, pp. 15 2 Nucleic Acid Bases Molecules of
More informationRNA and PROTEIN SYNTHESIS. Chapter 13
RNA and PROTEIN SYNTHESIS Chapter 13 DNA Double stranded Thymine Sugar is RNA Single stranded Uracil Sugar is Ribose Deoxyribose Types of RNA 1. Messenger RNA (mrna) Carries copies of instructions from
More informationMultiple choice questions (numbers in brackets indicate the number of correct answers)
1 Multiple choice questions (numbers in brackets indicate the number of correct answers) February 1, 2013 1. Ribose is found in Nucleic acids Proteins Lipids RNA DNA (2) 2. Most RNA in cells is transfer
More informationBasic Concepts of Human Genetics
Basic Concepts of Human Genetics The genetic information of an individual is contained in 23 pairs of chromosomes. Every human cell contains the 23 pair of chromosomes. One pair is called sex chromosomes
More informationCHAPTERS , 17: Eukaryotic Genetics
CHAPTERS 14.1 14.6, 17: Eukaryotic Genetics 1. Review the levels of DNA packing within the eukaryote nucleus. Label each level. (A similar diagram is on pg 188 of your textbook.) 2. How do the coding regions
More informationUnit 6: Gene Activity and Biotechnology
Chapter 16 Outline The Molecular Basis of Inheritance Level 1 Items students should be able to: 1. Recognize scientists and the experiments that lead to the understanding of the molecular basis of inheritance.
More informationProkaryotic Transcription
Prokaryotic Transcription Contents 1 The Lactose Intolerance of Bacteria 2 The Lac Operon 3 Lac Operon Simulation 4 LacZ as a reporter gene The Lactose Intolerance of Bacteria The standard growth kinetics
More informationI. Prokaryotic Gene Regulation. Figure 1: Operon. Operon:
I. Prokaryotic Gene Regulation Figure 1: Operon Operon: a) Regulatory Elements consist of an Operator that serves as the on-off switch for the genes of the operon. Also contains a promoter for the Structural
More informationGenetics Lecture Notes Lectures 17 19
Genetics Lecture Notes 7.03 2005 Lectures 17 19 Lecture 17 Gene Regulation We are now going to look at ways that genetics can be used to study gene regulation. The issue is how cells adjust the expression
More informationBio 121 Practice Exam 3
The material covered on Exam 3 includes lecture since the last exam and text chapters 13-21. Be sure that you read chapter 19, which was not represented in the notes. 1. Which of the following is an enveloped
More informationCHAPTER 9 DNA Technologies
CHAPTER 9 DNA Technologies Recombinant DNA Artificially created DNA that combines sequences that do not occur together in the nature Basis of much of the modern molecular biology Molecular cloning of genes
More informationDifferential Gene Expression
Biology 4361 Developmental Biology Differential Gene Expression June 19, 2008 Differential Gene Expression Overview Chromatin structure Gene anatomy RNA processing and protein production Initiating transcription:
More informationAP Biology. Chapter 20. Biotechnology: DNA Technology & Genomics. Biotechnology. The BIG Questions. Evolution & breeding of food plants
What do you notice about these phrases? radar racecar Madam I m Adam Able was I ere I saw Elba a man, a plan, a canal, Panama Was it a bar or a bat I saw? Chapter 20. Biotechnology: DNA Technology & enomics
More informationencodes a sigma factor to modify the recognition of the E.coli RNA polymerase (Several other answers would also be acceptable for each phage)
Name Student ID# Bacterial Genetics, BIO 4443/6443 Spring Semester 2001 Final Exam 1.) Different bacteriophage utilize different mechanisms to ensure that their own genes (and not host s genes) are transcribed
More informationConfirming the Phenotypes of E. coli Strains
Confirming the Phenotypes of E. coli Strains INTRODUCTION Before undertaking any experiments, we need to confirm that the phenotypes of the E. coli strains we intend to use in the planned experiments correspond
More informationLearning Objectives :
Learning Objectives : Understand the basic differences between genomic and cdna libraries Understand how genomic libraries are constructed Understand the purpose for having overlapping DNA fragments in
More informationCHAPTER 20 DNA TECHNOLOGY AND GENOMICS. Section A: DNA Cloning
Section A: DNA Cloning 1. DNA technology makes it possible to clone genes for basic research and commercial applications: an overview 2. Restriction enzymes are used to make recombinant DNA 3. Genes can
More informationBC2004 Review Sheet for Lab Exercises 7-11 Spring Semester 2005
BC2004 Review Sheet for Lab Exercises 7-11 Spring Semester 2005 Lab Exercise 7 Drosophila crosses, three weeks Vocabulary: phenotype, genotype, gene, allele, locus (loci), sex chromosomes, autosomes, homozygous,
More informationSolutions to Quiz II
MIT Department of Biology 7.014 Introductory Biology, Spring 2005 Solutions to 7.014 Quiz II Class Average = 79 Median = 82 Grade Range % A 90-100 27 B 75-89 37 C 59 74 25 D 41 58 7 F 0 40 2 Question 1
More informationGENE EXPRESSION AT THE MOLECULAR LEVEL. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
GENE EXPRESSION AT THE MOLECULAR LEVEL Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Gene expression Gene function at the level of traits Gene function
More informationBiology Lecture 2 Genes
Genes Definitions o Gene: DNA that codes for a single polypeptide/mrna/rrna/trna o Euchromatin: region of DNA containing genes being actively transcribed o Heterochromatin: region of DNA containing genes
More informationSept 2. Structure and Organization of Genomes. Today: Genetic and Physical Mapping. Sept 9. Forward and Reverse Genetics. Genetic and Physical Mapping
Sept 2. Structure and Organization of Genomes Today: Genetic and Physical Mapping Assignments: Gibson & Muse, pp.4-10 Brown, pp. 126-160 Olson et al., Science 245: 1434 New homework:due, before class,
More informationEUKARYOTIC REGULATION C H A P T E R 1 3
EUKARYOTIC REGULATION C H A P T E R 1 3 EUKARYOTIC REGULATION Every cell in an organism contains a complete set of DNA. But it doesn t use all of the DNA it receives Each cell chooses different DNA sequences
More informationProblem set questions from Exam 1 Unit Basic Genetic Tests, Setting up and Analyzing Crosses, and Genetic Mapping
Problem set questions from Exam 1 Unit Basic Genetic Tests, Setting up and Analyzing Crosses, and Genetic Mapping Basic genetic tests for complementation and/or dominance 1. You have isolated 20 new mutant
More informationPELLISSIPPI STATE COMMUNITY COLLEGE MASTER SYLLABUS
PELLISSIPPI STATE COMMUNITY COLLEGE MASTER SYLLABUS GENERAL GENETICS BIOL 2120 Class Hours: 3.0 Credit Hours: 4.0 Laboratory Hours: 3.0 Revised Spring 2017 Catalog Course Description Prerequisites Corequisites
More informationRapid Learning Center Presents. Teach Yourself AP Biology in 24 Hours
Rapid Learning Center Chemistry :: Biology :: Physics :: Math Rapid Learning Center Presents Teach Yourself AP Biology in 24 Hours 1/35 *AP is a registered trademark of the College Board, which does not
More information1a. What is the ratio of feathered to unfeathered shanks in the offspring of the above cross?
Problem Set 5 answers 1. Whether or not the shanks of chickens contains feathers is due to two independently assorting genes. Individuals have unfeathered shanks when they are homozygous for recessive
More information1. DNA replication. (a) Why is DNA replication an essential process?
ame Section 7.014 Problem Set 3 Please print out this problem set and record your answers on the printed copy. Answers to this problem set are to be turned in to the box outside 68120 by 5:00pm on Friday
More informationDNA and RNA. Chapter 12
DNA and RNA Chapter 12 Warm Up Exercise Test Corrections Make sure to indicate your new answer and provide an explanation for why this is the correct answer. Do this with a red pen in the margins of your
More information2054, Chap. 14, page 1
2054, Chap. 14, page 1 I. Recombinant DNA technology (Chapter 14) A. recombinant DNA technology = collection of methods used to perform genetic engineering 1. genetic engineering = deliberate modification
More informationExam 1 Answers Biology 210 Sept. 20, 2006
Exam Answers Biology 20 Sept. 20, 2006 Name: Section:. (5 points) Circle the answer that gives the maximum number of different alleles that might exist for any one locus in a normal mammalian cell. A.
More informationDNA Structure and Analysis. Chapter 4: Background
DNA Structure and Analysis Chapter 4: Background Molecular Biology Three main disciplines of biotechnology Biochemistry Genetics Molecular Biology # Biotechnology: A Laboratory Skills Course explorer.bio-rad.com
More informationUsing mutants to clone genes
Using mutants to clone genes Objectives 1. What is positional cloning? 2. What is insertional tagging? 3. How can one confirm that the gene cloned is the same one that is mutated to give the phenotype
More informationToday s lecture: Types of mutations and their impact on protein function
Today s lecture: Types of mutations and their impact on protein function Mutations can be classified by their effect on the DNA sequence OR the encoded protein 1 From my Lecture 4 (10/1): Classification
More information8/21/2014. From Gene to Protein
From Gene to Protein Chapter 17 Objectives Describe the contributions made by Garrod, Beadle, and Tatum to our understanding of the relationship between genes and enzymes Briefly explain how information
More informationLecture 25 (11/15/17)
Lecture 25 (11/15/17) Reading: Ch9; 328-332 Ch25; 990-995, 1005-1012 Problems: Ch9 (study-guide: applying); 1,2 Ch9 (study-guide: facts); 7,8 Ch25 (text); 1-3,5-7,9,10,13-15 Ch25 (study-guide: applying);
More informationMicrobiology 微生物学 Spring-Summer
Microbiology 微生物学 2017 Spring-Summer Relevant Information and Resources Course slides can be found at http://mypage.zju.edu.cn/haichun 教学工作 Course-related questions will be answered through emails. Textbook:
More informationBasic Concepts of Human Genetics
Basic oncepts of Human enetics The genetic information of an individual is contained in 23 pairs of chromosomes. Every human cell contains the 23 pair of chromosomes. ne pair is called sex chromosomes
More informationMIT Department of Biology 7.013: Introductory Biology - Spring 2005 Instructors: Professor Hazel Sive, Professor Tyler Jacks, Dr.
MIT Department of Biology 7.01: Introductory Biology - Spring 2005 Instructors: Professor Hazel Sive, Professor Tyler Jacks, Dr. Claudette Gardel iv) Would Xba I be useful for cloning? Why or why not?
More informationBiological information flow
BCMB 3100 Chapters 36-38 Transcription & RNA Processing Biological information flow Definition of gene RNA Polymerase Gene coding vs template strand Promoter Transcription in E. coli Transcription factors
More informationThe information provided below may be useful in answering some questions.
Molecular Exam 1 More Tutorial at www.dumblittledoctor.com The information provided below may be useful in answering some questions. INFORMATION ON COMPONENTS OF RIBOSOMES I. Prokaryotes (e.g. E. coli)
More informationTable of Contents. Chapter: Heredity. Section 1: Genetics. Section 2: Genetics Since Mendel. Section 3: Biotechnology
Table of Contents Chapter: Heredity Section 1: Genetics Section 2: Genetics Since Mendel Section 3: Biotechnology 1 Genetics Inheriting Traits Eye color, nose shape, and many other physical features are
More informationRegulatory Dynamics in Engineered Gene Networks
Regulatory Dynamics in Engineered Gene Networks The Physico-chemical Foundation of Transcriptional Regulation with Applications to Systems Biology Mads Kærn Boston University Center for BioDynamics Center
More information14 Gene Expression: From Gene to Protein
CMPBELL BIOLOY IN FOCS rry Cain Wasserman Minorsky Jackson Reece 14 ene Expression: From ene to Protein Lecture Presentations by Kathleen Fitzpatrick and Nicole Tunbridge Overview: The Flow of enetic Information
More informationMOLECULAR BASIS OF INHERITANCE
CHAPTER 6 MOLECULAR BASIS OF INHERITANCE POINTS TO REMEMBER Anticodon : A sequence of three nitrogenous bases on trna which is complementary to the codon on mrna. Transformation : The phenomenon by which
More informationBiological information flow
BCMB 3100 Chapters 36-38 Transcription & RNA Processing Definition of gene RNA Polymerase Gene coding vs template strand Promoter Transcription in E. coli Transcription factors mrna processing Biological
More informationHuman disease genes summary. Some examples of single-gene diseases
Human disease genes summary 1. Goals: discover the basis for disease, understand key processes, and develop diagnostics and cures. 2. Finding human disease genes -- OMIM 3. Sickle Cell Anemia 4. Inheritance
More information7.013 Practice Quiz
MIT Department of Biology 7.013: Introductory Biology - Spring 2005 Instructors: Professor Hazel Sive, Professor Tyler Jacks, Dr. Claudette Gardel 7.013 Practice Quiz 2 2004 1 Question 1 A. The primer
More information7.012 Final Exam
7.012 Final Exam 2006 You have 180 minutes to complete this exam. There are 19 pages including this cover page, the AMINO AID page, and the GENETI ODE page at the end of the exam. Please write your name
More informationFrom Gene to Protein. Chapter 17. Biology Eighth Edition Neil Campbell and Jane Reece. PowerPoint Lecture Presentations for
Chapter 17 From Gene to Protein PowerPoint Lecture Presentations for Biology Eighth Edition Neil Campbell and Jane Reece Lectures by Chris Romero, updated by Erin Barley with contributions from Joan Sharp
More informationTrasposable elements: Uses of P elements Problem set B at the end
Trasposable elements: Uses of P elements Problem set B at the end P-elements have revolutionized the way Drosophila geneticists conduct their research. Here, we will discuss just a few of the approaches
More information7.013 Problem Set 3 FRIDAY October 8th, 2004
MIT Biology Department 7.012: Introductory Biology - Fall 2004 Instructors: Professor Eric Lander, Professor Robert. Weinberg, Dr. laudette ardel Name: T: 7.013 Problem Set 3 FRIDY October 8th, 2004 Problem
More informationProtein Synthesis
HEBISD Student Expectations: Identify that RNA Is a nucleic acid with a single strand of nucleotides Contains the 5-carbon sugar ribose Contains the nitrogen bases A, G, C and U instead of T. The U is
More informationProblem Set 2B Name and Lab Section:
Problem Set 2B 9-26-06 Name and Lab Section: 1. Define each of the following rearrangements (mutations) (use one phrase or sentence for each). Then describe what kind of chromosomal structure you might
More informationRegents Biology REVIEW 5: GENETICS
Period Date REVIEW 5: GENETICS 1. Chromosomes: a. Humans have chromosomes, or homologous pairs. Homologous: b. Chromosome pairs carry genes for the same traits. Most organisms have two copies of the gene
More informationDNA RNA PROTEIN SYNTHESIS -NOTES-
DNA RNA PROTEIN SYNTHESIS -NOTES- THE COMPONENTS AND STRUCTURE OF DNA DNA is made up of units called nucleotides. Nucleotides are made up of three basic components:, called deoxyribose in DNA In DNA, there
More informationBioinformatics of Transcriptional Regulation
Bioinformatics of Transcriptional Regulation Carl Herrmann IPMB & DKFZ c.herrmann@dkfz.de Wechselwirkung von Maßnahmen und Auswirkungen Einflussmöglichkeiten in einem Dialog From genes to active compounds
More informationDivision Ave. High School AP Biology
Control of Eukaryotic Genes 2007-2008 The BIG Questions n How are genes turned on & off in eukaryotes? n How do cells with the same genes differentiate to perform completely different, specialized functions?
More informationCS273B: Deep Learning in Genomics and Biomedicine. Recitation 1 30/9/2016
CS273B: Deep Learning in Genomics and Biomedicine. Recitation 1 30/9/2016 Topics Genetic variation Population structure Linkage disequilibrium Natural disease variants Genome Wide Association Studies Gene
More informationZool 3200: Cell Biology Exam 3 3/6/15
Name: Trask Zool 3200: Cell Biology Exam 3 3/6/15 Answer each of the following questions in the space provided; circle the correct answer or answers for each multiple choice question and circle either
More informationChapter 8 DNA Recognition in Prokaryotes by Helix-Turn-Helix Motifs
Chapter 8 DNA Recognition in Prokaryotes by Helix-Turn-Helix Motifs 1. Helix-turn-helix proteins 2. Zinc finger proteins 3. Leucine zipper proteins 4. Beta-scaffold factors 5. Others λ-repressor AND CRO
More information