1 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 genomic libraries and how they are generated Understand how cdna libraries are constructed and the use of reverse transcriptase for their construction Understand the rationale for library screening Understand the method of plaque hybridization Understand the four methods for library screening and when they are put into use
2 Molecular cloning in bacterial cells. This strategy can be applied to genomic DNA as well as cdna
3 Library construction two types of libraries a genomic library contains fragments of genomic DNA (genes) a cdna library contains DNA copies of cellular mrnas both types are usually cloned in bacteriophage vectors Construction of a genomic library Bam HI sites vector DNA (bacteriophage lambda) lambda has a linear doublestranded DNA genome the left and right arms are essential for the phage replication cycle the internal fragment is dispensable left arm right arm internal fragment (dispensable for phage growth)
4 Bam HI sites: NNG NNCCTAG GATCCNN GNN human genomic DNA (isolated from many cells) cut with Bam HI (6-base cutter) cut with Sau 3A (4-base cutter) which has ends compatible with Bam HI: left arm internal fragment remove internal fragment right arm NNN GATCNNN NNNCTAG isolate ~20 kb fragments NNN
5 left arm right arm combine and treat with DNA ligase left arm right arm package into bacteriophage and infect E. coli genomic library of human DNA fragments in which each phage contains a different human DNA sequence
6 Partial restriction enzyme digestion allows cloning of overlapping fragments isolation of ~20 kb fragments provides optimally sized DNAs for cloning in bacteriophage partial digestion with a frequent-cutter (4-base cutter) allows production of overlapping fragments, since not every site is cut overlapping fragments insures that all sequences in the genome are cloned overlapping fragments allows larger physical maps to be constructed as contiguous chromosomal regions (contigs) are put together from the sequence data number of clones needed to fully represent the human genome (3 X 10 9 bp) assuming ~20 kb fragments theoretical minimum = ~150,000 99% probability that every sequence is represented = ~800,000 a contig
7 All possible sites: Results of a partial digestion: = uncut = cut
8 Genomic Library making The partial digest is one of the most important steps. Why??? Due to the production of overlapping DNA fragments
9 The production of a cdna library
10 Construction of a cdna library reverse transcriptase makes a DNA copy of an RNA The life cycle of a retrovirus depends on reverse transcriptase retrovirus 2. the capsid is uncoated, releasing genomic RNA and reverse transcriptase 1. virus enters cell and looses envelope 3. reverse transcriptase makes a DNA copy new viruses 6. it is translated into viral proteins, and assembled into new virus particles 4. then copies the DNA strand to make it double-stranded DNA, removing the RNA with RNase H 5. the DNA is then integrated into the host cell genome where it is transcribed by host RNA polymerase II
11 cdna library construction AAAAA 5 3 mrna (all mrnas in cell) anneal oligo(dt) primers of bases in length 5 AAAAA 3 3 TTTTT 5 add reverse transcriptase and dntps AAAAA TTTTT 3 5 cdna add RNaseH (specific for the RNA strand of an RNA-DNA hybrid) and carry out a partial digestion AA TTTTT short RNA fragments serve as primers for second strand synthesis using DNA polymerase I
12 AAAAA TTTTT short RNA fragments serve as primers for second strand synthesis using DNA polymerase I AAA TTTTT DNA polymerase I removes the remaining RNA with its 5 to 3 exonuclease activity and continues synthesis DNA ligase seals the gaps AAA TTTTT 5 3 AAAAA TTTTT double-stranded cdna
13 AATTCNNNNNNNN GNNNNNNNN AAAAA TTTTT NNNNNNNNG EcoRI linkers are ligated to both ends NNNNNNNNCTTAA using DNA ligase AAAAANNNNNNNNG TTTTTNNNNNNNNCTTAA double-stranded cdna copies of mrna with EcoRI cohesive ends are now ready to ligate into a bacteriophage lambda vector cut with EcoRI
14 left arm EcoRI sites right arm cdnas combine cdnas with lambda arms and treat with DNA ligase left arm right arm package into bacteriophage and infect E. coli cdna library in which each phage contains a different human cdna
15 DNA libraries Genomic DNA libraries contains both introns and exons and promoters etc Usually made with 4 base cutters that cut frequently ( every 275 bases or so). The production of overlapping sequences is due to partial digestion. Libra sqry complexity is important to make sure that the sequence you are looking for is found in the DNA that has been sampled. N = ln (1-P) / ln (1-f) where N = number of clones, P = probability that the DNA fragment is found in your library and f = the frequency of the DNA in your library.
16 Genomic DNA complexity To screen for a clone in a library usually want a 99% probability that your clone is found there. Frequency is the size of the DNA fragment in the library/the size of the haploid genome. For a lambda library 17 kb (1.7 x 10 4 ) is the average size of library. The size of the genome is 3 x 10 9 bp F = 1.7 x 10 4 / 3 x 10 9 bp N = ln (1-.99) / ln (1- [1.7 x 10 4 / 3 x 10 9 ]) N = ln.01 / ln ( x 10-5 ) N = / N = 822,351 clones
17 Genome equivalents How many genome equivalents are there in this library? How do you calculate this? 822,351 x 1.7 x 10 4 bps = 1.40 x bps Divide by the genome size 3.0 x 10 9 bps = 4.67 times the genome equivalent How many positives will you get if you screen for a single copy gene?
18 Insertional mutagensis In all of the vectors that are currently used to date there is a system that can either identify or select for vectors containing clones. This is the backbone of recombinant DNA technology. Initial vectors involved the cloning into a antibiotic resistance gene making a bacteria containing a vector with a DNA fragment sensitive to the antibiotic. This is not the best situation, Why?
19 Insertional mutagenesis II The use of the beta-galactosidase gene for an insertional mutagenesis target allowed the screening of all clones for those that contained inserts by a simple blue white color assay. This gene cleaves X gal (chromagen) to give rise to a blue dye that colors the bacteria or phage plaque. This allows the screening those plasmids or phage particles that contain DNA disrupting the target gene.
20 Insertional mutagenesis III In addition suppressor trna genes can be used to identify YAC that contain an insert. The suppressor trna can suppress the effects of a Ade2 ochre mutation. This gives a white yeast colony. When the trna gene is disrupted the colonies are pink due to the accumulation of a precursor of Adenine. Pink colonies are what is desired. See Figure 4.16
21 Clones are usually characterized first by restriction digestion. This DNA fragment was digest with various enzymes giving rise to specific sizes. These can be used to generate a restriction map
22 Vectors for library construction Plasmid vectors Small circles of DNA that contain a selection marker like antibiotic resistance. Insertional mutagenesis target with a multi cloning site. A variety DNA replicons. Bacterial, Yeast. Maximum size of insert is about 10 kb.
23 Lambda and Cosmid vectors Bacteriophage lambda can be used as a cloning vector. It has a genome of about 50 kb of linear DNA. Its life cycle is condusive to the use as a cloning vector The lytic cycle can be supported by only a portion of the genes found in the lambda genome.
24 Lambda life cycle. The lytic life cycle produces phage particles immediately The lysogenic life cycle requires genes in the middle of the genome, which can be replaced
26 Lambda insertion and replacement vectors Only 37 to 52 kb DNA fragments can be packaged into the lambda head. This can be done in vitro.because the middle portion of the lambda genome can be replace if the lytic life cycle is used up to 23 kb DNA can be inserted in lambda genome. These are used for genomic DNA libraries. Insertion vectors can hold up to 7 kb of cdna.
27 Lambda genome
28 In vitro Packaging of ligated lambda DNA.
29 Cosmid vectors A cosmid is a hybrid between a lambda vector and a plasmid. The COS sites are the only thing that is necessary for lambda DNA packaging. Therefore if one can ligate COS sites about 50 kb apart then the ligation products can be in vitro packaged. Therefore cosmid vectors can contain 33 to 45 kb.
30 Cosmid vector ligation
31 Making a Genomic Library with Cosmids Partial digest Tet R 21.5 kb cos Ligation into site EcoRI EcoRI
32 Final Steps of a Genomic Library Package into heads and plug with tails Transduce E. coli receptor cell Select white colonies with tet R Check for plasmid Screen in your mutant for phenotype restoration
33 Things You Should Remember Some plasmids are used as vectors or cloning vehicles but they are limited to the amount of DNA that can be cloned. A cosmid is a plasmid that has at least one COS (cohesive end site). COS comes from a bacteriophage. A genomic library contains at least one copy of every gene in an organism.
34 Cloning large DNA fragments Due to the large size of the human genome and the fact that many genes are very large and some DNA fragments cannot be replicated in lambda other vector systems needed to be developed. Bacterial Artificial chromosomes (BAC) vectors These vectors are based on the E. coli F factor. These vectors are maintained at 1-2 copies per cell and can hold > 300 kb of insert DNA. Problems are low DNA yield from host cells. (due to low copy number when compared to 300 copies per cell with a plasmid vector like puc19.
35 Cloning large DNA fragments II Bacteriophage P1 These vectors are like lambda and can hold up to 110 to 115 kb of DNA. This DNA can then be packaged by the P1 phage protein coat. The use of T4 in vitro packaging systems can enable the recovery of 122 kb inserts. See Figure 4.15
36 Bacteriophage P1 vector system.
37 Cloning large DNA fragments III Yeast Artificial Chromosomes Many DNA fragments cannot be propagated in bacterial cells. Therefore yeast artificial chromosomes can be built with a few specific components. 1. Centromere 2. Telomere 3. Autonomously replicating sequence (ARS) Genomic DNA is ligated between two telomeres and the ligation products are transformed into yeast cells using the spheroplast method.
38 YAC cloning system
39 Cloning systems Vector systems that can be used to clone DNA
40 Plaque hybridization This is a general technique required for a number of specific approaches for isolating cdna or genomic clones Generally, one starts by 1). Isolating a cdna sequence from a cdna library, then 2). The gene from a genomic library using the cdna as a probe Information gained from cdna and genomic clones 1). cdna clones provide the amino acid sequence of the full-length protein, unencumbered by intron sequences 2). Genomic clones provide the control regions and are required for searching for mutations Library screening: four experimental approaches Starting with a protein 1). Synthetic oligonucleotide - plaque hybridization 2). Antibody - variation of plaque hybridization Starting with mrna 1). Differential cdna library screening - yet another variation 2). Expression screening - does not utilize plaque hybridization
41 Library screening plaque hybridization plate phage library on lawn of E. coli (bacteria >>> phage) plaques form as a consequence of a spreading lytic infection starting with a single phage-infected bacterial cell each phage plaque is a clone of identical recombinant phage prepare replica of phage plaques and hybridize DNA with probe E. coli lawn is grown on agar plate and then overlayered with the recombinant phage library. Wherever a single bacteriophage particle infects a bacteria cell, a plaque will form. This is a clear area caused by the lysis of bacteria on the lawn of E. coli. A replica of the agar plate is made on a nitrocellulose sheet - the DNA is denatured and adheres to the nitrocellulose. X-ray film The nitrocellulose is hybridized with a labeled DNA probe (such as an oligonucleotide) and the nitrocellulose is exposed to X-ray film. spot on film indicates a plaque containing DNA of interest
42 32 P label Labeled probe in solution e.g. an oligo probe Hybridization of probe to immobilized DNA Probe hybridized to immobilized DNA forming double-stranded region
43 1 X 2 X 2 X 2 X 3 X 2 X 2 X 2 = 192-fold degenerate How does one isolate a gene for an inherited disorder? Start with a candidate protein DNA protein If a protein candidate has been identified for a genetic disease it can be used to make a probe to screen for the gene 1. oligonucleotide probe purify the protein of interest partially sequence the protein find a region having amino acids with the fewest possible codons predict a DNA sequence that could represent a gene region encoding a portion of the protein synthesize a set of degenerate oligonucleotides for that region hybridize the labeled oligonucleotide to the phage library MET.GLU.PHE.TYR.ILE.CYS.GLN.LYS amino acids } AUG.GAA.UUU.UAU.AUU.UGU.CAA.AAA G C C C C G G all possible A oligonucleotides
44 2. antibody probe purify the protein of interest make an antibody to that protein construct cdna library to express recombinant proteins in E. coli use the antibody to detect the protein being made from the cloned cdna encoded by the recombinant phage in E. coli using plaque hybridization method modified for antibody probes bacterial promoter and Shine-Dalgarno sequence left arm right arm human cdna insert
45 How does one isolate a gene for an inherited disorder? Start with a candidate mrna DNA mrna mrna candidates can be identified by comparing mrna populations between normal and abnormal tissues, or by looking for a specific function encoded by the mrna 1. differential cdna library screening prepare duplicate plaque replica plates hybridize one with a labeled cdna probe made to all the mrnas in the normal cell and hybridize the other (duplicate) with the corresponding probe to the abnormal cell differences in cell function should be reflected by differences in the mrna populations any plaques showing differential hybridization are candidates hybridization with cdna probe differentially expressed clone no hybridization to this plaque hybridization with cdna probe from abnormal cells
46 2. expression screening develop a cell-based functional assay for the abnormality (e.g., a transport assay) construct cdna library in a way that will allow expression of protein in mammalian cells inject groups of cdna clones into cells and assay function narrow down cdna clones using smaller groups of clones until the function is observed with a single cdna species for example, inject clones and test cells for transport activity left arm inject groups of cdna clones if the function being assayed is observed, divide the group of clones into smaller groups and retest continue process of testing smaller groups until the function being assayed is obtained with one clone right arm mammalian promoter human cdna insert
48 Expression Cloning Certain vector systems can be used to produce specific products. The type of expression product RNA Riboprobes Protein product The type of environment In vitro cell free In vivo mammalian or prokaryotic cells Purpose of the expression system To produce large quantities of proteins for protein studies or antibody production.
49 cdna expression libraries The gene for a specific protein can be cloned from an expression cdna library if an antibody to the protein is available. A variety vectors can be used to produce fusion proteins which can be detected with Ab in question. See Figure 4.18
50 Expression for Ab detection
51 Expression in Eukaryotic cells Many proteins need specific modifications to work properly expression in bacterial cells is not sufficient Plasmid based Eukaryotic expression systems which work after transient transfection into mammalian cell lines have been produced. Viral based system are also popular.
Molecular Genetics Techniques BIT 220 Chapter 20 What is Cloning? Recombinant DNA technologies 1. Producing Recombinant DNA molecule Incorporate gene of interest into plasmid (cloning vector) 2. Recombinant
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
Molecular Biology: Gene cloning Author: Prof Marinda Oosthuizen Licensed under a Creative Commons Attribution license. CLONING VECTORS The central component of a gene cloning experiment is the vector or
Chapter 20 Recombinant DNA Technology Copyright 2009 Pearson Education, Inc. 20.1 Recombinant DNA Technology Began with Two Key Tools: Restriction Enzymes and DNA Cloning Vectors Recombinant DNA refers
Lecture 8 Reading Lecture 8: 96-110 Lecture 9: 111-120 DNA Libraries Definition Types Construction 142 DNA Libraries A DNA library is a collection of clones of genomic fragments or cdnas from a certain
Cloning Vectors A M I R A A. T. A L - H O S A R Y L E C T U R E R O F I N F E C T I O U S D I S E A S E S F A C U L T Y O F V E T. M E D I C I N E A S S I U T U N I V E R S I T Y - E G Y P T DNA Cloning
PLNT2530 (2018) Unit 6b Sequence Libraries Molecular Biotechnology (Ch 4) Analysis of Genes and Genomes (Ch 5) Unless otherwise cited or referenced, all content of this presenataion is licensed under the
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
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
Molecular Cloning Methods Mohammad Keramatipour MD, PhD email@example.com Outline DNA recombinant technology DNA cloning co Cell based PCR PCR-based Some application of DNA cloning Genomic libraries
Student Name: All questions are worth 5 pts. each. GENETICS EXAM 3 FALL 2004 1. a) is a technique that allows you to separate nucleic acids (DNA or RNA) by size. b) Name one of the materials (of the two
Lecture Four. Molecular Approaches I: Nucleic Acids I. Recombinant DNA and Gene Cloning Recombinant DNA is DNA that has been created artificially. DNA from two or more sources is incorporated into a single
Lecture 3 Reading Lecture 3: 24-25, 45, 55-66 Lecture 4: 66-71, 75-79 Vectors Definition Properties Types Transformation 56 VECTORS- Definition Vectors are carriers of a DNA fragment of interest Insert
1 Selected Techniques Part I Gel Electrophoresis Can be both qualitative and quantitative Qualitative About what size is the fragment? How many fragments are present? Is there in insert or not? Quantitative
Chapter 8 Recombinant DNA and Genetic Engineering Genetic manipulation Ways this technology touches us Criminal justice The Justice Project, started by law students to advocate for DNA testing of Death
XXII DNA cloning and sequencing 1) Deriving DNA for cloning Outline 2) Vectors; forming recombinant DNA; cloning DNA; and screening for clones containing recombinant DNA [replica plating and autoradiography;
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
Computational Biology 2 Pawan Dhar BII Lecture 1 Introduction to terms, techniques and concepts in molecular biology Molecular biology - a primer Human body has 100 trillion cells each containing 3 billion
Chapter 20 Biotechnology Manipulation of DNA In 2007, the first entire human genome had been sequenced. The ability to sequence an organisms genomes were made possible by advances in biotechnology, (the
Lecture 22: Molecular techniques DNA cloning and DNA libraries DNA cloning: general strategy -> to prepare large quantities of identical DNA Vector + DNA fragment Recombinant DNA (any piece of DNA derived
Chapter 20 Biotechnology 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 Copyright
Chapter 6 - Molecular Genetic Techniques Two objects of molecular & genetic technologies For analysis For generation Molecular genetic technologies! For analysis DNA gel electrophoresis Southern blotting
MODULE 4-LECTURE 4 CONSTRUCTION OF GENOMIC LIBRARY 4-4.1. Introduction A genomic library is an organism specific collection of DNA covering the entire genome of an organism. It contains all DNA sequences
Basics of Recombinant DNA Technology Biochemistry 302 March 5, 2004 Bob Kelm Applications of recombinant DNA technology Mapping and identifying genes (DNA cloning) Propagating genes (DNA subcloning) Modifying
Chapter 15 The Biotechnology Toolbox Cutting and Pasting DNA Cutting DNA Restriction endonuclease or restriction enzymes Cellular protection mechanism for infected foreign DNA Recognition and cutting specific
Your Name: Your UID# 1. (20 points) Match following mutations with corresponding mutagens (X-RAY, Ds transposon excision, UV, EMS, Proflavin) a) Thymidine dimmers b) Breakage of DNA backbone c) Frameshift
Bi 8 Lecture 4 DNA approaches: How we know what we know Ellen Rothenberg 14 January 2016 Reading: from Alberts Ch. 8 Central concept: DNA or RNA polymer length as an identifying feature RNA has intrinsically
Design Construction Characterization DNA mrna (messenger) A C C transcription translation C A C protein His A T G C T A C G Plasmids replicon copy number incompatibility selection marker origin of replication
MOLECULAR BIOLOGY 2003-4 Topic B Recombinant DNA -principles and tools Construct a library - what for, how Major techniques +principles Bioinformatics - in brief Chapter 7 (MCB) 1 Motivation From Protein
STANDARD CLONING PROCEDURES Shotgun cloning (using a plasmid vector and E coli as a host). 1) Digest donor DNA and plasmid DNA with the same restriction endonuclease 2) Mix the fragments together and treat
Recitation CHAPTER 9 DNA Technologies DNA Cloning: General Scheme A cloning vector and eukaryotic chromosomes are separately cleaved with the same restriction endonuclease. (A single chromosome is shown
Molecular Cell Biology - Problem Drill 11: Recombinant DNA Question No. 1 of 10 1. Which of the following statements about the sources of DNA used for molecular cloning is correct? Question #1 (A) cdna
Enzymes Restriction Enzymes (Site-Specific Endonuclease) Enzymes that recognize and cleave dsdna in a highly sequence specific manner. Generally recognize an inverted repeat sequence 4, 6, or 8 base pairs
Tools in Genetic engineering The science of using living systems to benefit humankind is called biotechnology. Technically speaking, the domestication of plants and animals through farming and breeding
Name: AP Biology Biology, Campbell and Reece, 7th Edition Adapted from chapter reading guides originally created by Lynn Miriello Chapter 18 The Genetics of Viruses and Bacteria Unit 8: Genomics Guided
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.
PowerPoint Lecture Presentations prepared by Bradley W. Christian, McLennan Community College C H A P T E R 9 Biotechnolog y and DNA Technology Introduction to Biotechnology Biotechnology: the use of microorganisms,
Chapter 9 Genetic Engineering Biotechnology: use of microbes to make a protein product Recombinant DNA Technology: Insertion or modification of genes to produce desired proteins Genetic engineering: manipulation
Chapter Review 1. Explain why the brewing of beer is considered to be biotechnology. The United Nations defines biotechnology as any technological application that uses biological system, living organism,
Chapter 20 DNA Technology & Genomics If we can, should we? Biotechnology Genetic manipulation of organisms or their components to make useful products Humans have been doing this for 1,000s of years plant
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
11/27/2017 PowerPoint Lecture Presentations prepared by Bradley W. Christian, McLennan Community College CHAPTER 9 Biotechnology and DNA Technology Introduction to Biotechnology Learning Objectives Compare
7.03, 2005, Lecture 20 EUKARYOTIC GENES AND GENOMES I For the last several lectures we have been looking at how one can manipulate prokaryotic genomes and how prokaryotic genes are regulated. In the next
Recombinant DNA Technology in Today s Medicine Shelley M. Martineau Jessica A. Matthews Catherine C. Miller Carol D. Riley Institute of TIP Productions, Inc. Part A Overview rdna Objectives Steps in rdna
!! www.clutchprep.com CONCEPT: DNA CLONING DNA cloning is a technique that inserts a foreign gene into a living host to replicate the gene and produce gene products. Transformation the process by which
Genetics - Problem Drill 15: The Techniques in Molecular Genetics No. 1 of 10 1. Which of the following is not part of the normal process of cloning recombinant DNA in bacteria? (A) Restriction endonuclease
BIOTECHNOLOGY RECOMBINANT DNA TECHNOLOGY Recombinant DNA technology involves sticking together bits of DNA from different sources. Made possible because DNA & the genetic code are universal. 2004 Biology
Name KEY Section Biology 201 (Genetics) Exam #3 120 points 20 November 2006 Read the question carefully before answering. Think before you write. You will have up to 50 minutes to take this exam. After
Central Dogma Genes are units perpetuating themselves and functioning through their expression in the form of proteins 1 DNA RNA Protein 2 3 1. Replication 2. Transcription 3. Translation Spring 2002 21
MMG 301, Lec. 25 Mutations and Bacteriophage Questions for today: 1. What are mutations and how do they form? 2. How are mutant bacteria used in research? 3. What are the general properties of bacteriophage
Molecular Cell Biology - Problem Drill 06: Genes and Chromosomes Question No. 1 of 10 1. Which of the following statements about genes is correct? Question #1 (A) Genes carry the information for protein
Regulation of gene expression a. Expression of most genes can be turned off and on, usually by controlling the initiation of transcription. b. Lactose degradation in E. coli (Negative Control) Lac Operon
Chapter 19 Viral Genomes Genomes may consist of: 1. Double Stranded DNA 2. Double Stranded RNA 3. Single-stranded RNA 4. Single-stranded DNA Genome is usually organized as a single linear or circular molecule
VOLUME 2 Molecular Clonin g A LABORATORY MANUA L THIRD EDITIO N Joseph Sambrook David W. Russell Chapter 8 In Vitro Amplification of DNA by the Polymerase 8. 1 Chain Reaction 1 The Basic Polymerase Chain
Biol/ MBios 301 (General Genetics) Spring 2003 Second Midterm Examination A (100 points possible) Key April 1, 2003 10 Multiple Choice Questions-4 pts. each (Choose the best answer) 1. Transcription involves:
Genetics (see text pages 257-259, 267-298) Remember what it is we want to address: How is it that prokaryotes gain new genetic ability? The cells are haploid and reproduce by fission...so how does an genetic
Molecular Cloning Genomic DNA Library: Contains DNA fragments that represent an entire genome. cdna Library: Made from mrna, and represents only protein-coding genes expressed by a cell at a given time.
Key Terms Chapter 32: Genetic Engineering Cloning describes propagation of a DNA sequence by incorporating it into a hybrid construct that can be replicated in a host cell. A cloning vector is a plasmid
It s All in the Hands Genetic Engineering Genetic Engineering Genetic Engineering is the technique of modifying the genome of an organism by using recombinant DNA technology. Recombinant DNA (rdna) technology
Genetics and Genomics in Medicine Chapter 3 Multiple Choice Questions Questions & Answers Question 3.1 Which of the following statements, if any, is false? a) Amplifying DNA means making many identical
Understanding the Cellular Mechanism of the Excess Microsporocytes I (EMSI) Gene Andrew ElBardissi, The Pennsylvania State University Abstract: Hong Ma, The Pennsylvania State University The Excess Microsporocytes
PowerPoint Lecture Presentations prepared by Mindy Miller-Kittrell, North Carolina State University C H A P T E R 8 Recombinant DNA Technology The Role of Recombinant DNA Technology in Biotechnology Biotechnology?
Bi 8 Lecture 5 MORE ON HOW WE KNOW WHAT WE KNOW and intro to the protein code Ellen Rothenberg 19 January 2016 SIZE AND PURIFICATION BY SYNTHESIS: BASIS OF EARLY SEQUENCING complex mixture of aborted DNA
Regulation of enzyme synthesis The lac operon is an example of an inducible operon - it is normally off, but when a molecule called an inducer is present, the operon turns on. The trp operon is an example
Big Idea 3C Basic Review 1. A gene is a. A sequence of DNA that codes for a protein. b. A sequence of amino acids that codes for a protein. c. A sequence of codons that code for nucleic acids. d. The end
Experimental genetics - I Examples of diseases with genetic-links Hemophilia (complete loss or altered form of factor VIII): bleeding disorder Duchenne muscular dystrophy (altered form of dystrophin) muscle
Page 1 of 5 Exam 2 Key - Spring 2008 A#: Please see us if you have any questions! 1. A mutation in which parts of two nonhomologous chromosomes change places is called a(n) A. translocation. B. transition.
.. 4. Analysing s II Isolate mutants* Using the mutant to isolate the classify mutants by complementation analysis wild type study phenotype of mutants mutant 1 - use mutant to isolate sequence put individual
Introduction to Plant Genomics and Online Resources Manish Raizada University of Guelph Genomics Glossary http://www.genomenewsnetwork.org/articles/06_00/sequence_primer.shtml Annotation Adding pertinent
AGRO/ANSC/BIOL/GENE/HORT 305 Fall, 2017 Recombinant DNA Technology (Chpt 20, Genetics by Brooker) Lecture outline: (#14) - RECOMBINANT DNA TECHNOLOGY is the use of in vitro molecular techniques to isolate
BIOLOGY 205 Midterm II - 19 February 1999 Name Multiple choice questions 4 points each (Best 12 out of 13). 1. Each of the following statements are correct regarding Eukaryotic genes and genomes EXCEPT?
Chapter 19 Viruses 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 Copyright
11/21/2017 PowerPoint Lecture Presentations prepared by Bradley W. Christian, McLennan Community College CHAPTER 13 Viruses, Viroids, and Prions General Characteristics of Viruses Learning Objective Differentiate
Unit 2: Metabolism and Survival Sub-Topic (2.7) Genetic Control of Metabolism (2.8) Ethical considerations in the use of microorganisms Duncanrig Secondary JHM&MHC 2015 Page 1 of 18 On completion of this
Recombinant protein production in Eukaryotic cells Dr. W. McLaughlin BC35C Recombinant protein production in Eukaryotic cells! rhuman protein must be identical to the natural protein! Prokaryotes are generally
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
Warm-Up Plasmids are circular pieces of DNA which bacterial cells are able to take up from the environment, then replicate and transcribe. Eukaryotic cells, by contrast, contain large, linear (non-circular)
Genetic Engineering & Recombinant DNA Chapter 10 Copyright The McGraw-Hill Companies, Inc) Permission required for reproduction or display. Applications of Genetic Engineering Basic science vs. Applied
Chapter 13A: Viral Basics 1. Viral Structure 2. The Viral Life Cycle 3. Bacteriophages 1. Viral Structure What exactly is a Virus? Viruses are extremely small entities that are obligate intracellular parasites