Marked Assisted System

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57Iowa State University Extension and ISU Office of Biotechnology

EducatorsLesson Module II Marker Assisted Selection

Marker Assisted

Selection (MAS)

PREFACE

Marker assisted selection (MAS) is a combined productof traditional genetics and molecular biology. MASallows for the selection of genes that control traits ofinterest. Combined with traditional selection tech-niques, MAS has become a valuable tool in selectingorganisms for traits of interest, such as color, meatquality, or disease resistance.

This module examines two cases in which mutations

called single nucleotide polymorphisms or SNPs(pronounced snips) have been used for selection.Students are asked to investigate and discuss theeconomic impact that this selection technique couldhave on producers and consumers.

BACKGROUND INFORMATION

MAS Introduction

Deoxyribonucleic acid (DNA) is a molecule made up ofpairs of building blocks called nucleotides. The fourkinds of nucleotides that make up DNA are adenine(abbreviated as the single letter A), guanine (G),cytosine (C), and thymine (T). The DNA molecule hasthe shape of two intertwined spirals, referred to as adouble helix.

DNA is packaged into chromosomes that are locatedwithin the nucleus of all cells. These chromosomes arethe same in every cell of an organism and togethermake up the organisms genetic information, its

genome. Chromosomes contain stretches of DNA

called genes that code for amino acids that makeproteins. It is the proteins that are the foundation of lifefor all organisms. The interaction and structure ofproteins determine the visible characteristics or

phenotypeof an organism, while the genetic makeup ofan organism is called itsgenotype.

The sequence of nucleotides that make up a gene candiffer among individuals. The different forms of a geneare called alleles. The alleles are the result of nucle-

otide differences in a gene that affect an amino acidsequence of a protein. This can result in a change,addition, or deletion of a protein that can affectthe phenotype.

All organisms receive one copy of each gene from their

mother and one from their father. The DNA sequenceof a gene inherited from each parent may be identical,in which case the individual is said to be homozygousfor that trait. Or the sequence of the gene from one ofthe parents may be different, in which case theindividual is said to be heterozygous. Allele variationsmay differ in their DNA sequence by as little as asingle nucleotide.

Differences among alleles caused by a single nucleotide,called SNPs, can be the basis of genotyping tests.Genotyping means using laboratory methods todetermine the sequence of nucleotides in the DNA from

an individual, usually a specific gene. Genetic testsbased on SNPs utilize DNA derived from an individualto determine the nucleotide in the gene of interest.

Marker assisted selection is the process of using theresults of DNA testing in the selection of individualsto become parents for the next generations. Theinformation from the DNA testing, combined with theobserved performance records for individuals, isintended to improve the accuracy of selection andincrease the possibility of identifying organismscarrying desirable and undesirable traits at an earlier

stage of development.

Complex traits, including many of economic impor-tance, are controlled by many genes and are influencedby the environment. When an animal has a favorableperformance record for a certain trait, it means thatbased on pedigree and phenotype, the animal hasinherited a greater than average number of good allelesof each gene affecting that specific trait.

It is important to combine DNA results with perfor-mance and phenotype information to maximize theeffectiveness of selection for traits of interest. Combin-

ing information from performance records and genetictests into the selection process will be better than usingperformance, phenotype, and markers separately. Thechallenge is to determine what emphasis markerinformation should be given in the selection decision.

Molecular Markers

Until recently, researchers relied on information abouthow animals, plants, and their relatives perform to

II


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58 Iowa State University Extension and ISU Office of Biotechnology

Educators Lesson Module II Marker Assisted Selection

make observations about the genes they possess. Today,researchers can use molecular markers to find genes ofinterest that control how plants and animals perform.Some molecular markers are pieces of DNA that haveno known function or impact on animal and plantperformance. Other markers may involve the gene of

interest itself.

Linked Markers

One type of molecular marker is called a linked marker.Using well-designed experiments, scientists can findmolecular markers that are located very close to majorgenes of interest. The molecular marker is said to belinked to that gene. Linked markers are only near thegene of interest on the chromosome and are not part ofthe DNA of the gene of interest.

Suppose that scientists are trying to locate a certaingene in an animal species. Choosing animals randomly

from a population and studying them would give thescientists no clues about whether a marker is associatedwith the gene. However, if scientists studied theprogeny (offspring) of the mating of male and femaleanimals through many generations, they may determinethe presence of a useful molecular marker.

Direct Markers

A second kind of molecular marker is one that is part ofthe gene of interest. Direct markers are easier to workwith after they are found, but they often are moredifficult to find than linked markers.

Marker-Assisted Selection

Three common technologies used as molecularmarkers are: restriction fragment length polymor-phisms, simple sequence repeats, and singlenucleotide polymorphisms.

Restriction Fragment Length Polymorphisms(RFLPs)

Restriction fragment lengthpolymorphisms (RFLPs)were the first molecular markers used to diagnosegenetic variability in organisms. RFLP uses restriction

enzymesto digest (cut) the DNA molecule and identifyregions linked to a trait. The number of DNA frag-ments generated by one restriction enzyme digest canbe in the millions, with many being several thousandnucleotides long. This makes it difficult to determinespecific DNA fragments that are associated with thetrait of interest on an electrophoresis gel. To helpvisualize specific DNA fragments, a technique calledSouthern blotting was developed.

Southern blotting uses a porous membrane containingspecific radioactive DNA probes for one or more DNAfragments. Probes are very short pieces of DNA used tofind specific sequences of A, C, T, and G in very longpieces of DNA from a chromosome. The probehybridizes (attaches) to the membrane at a unique DNA

band on an electrophoresis gel. The membranecontaining the probe is developed on X-ray film andanalyzed. See Figure 1.

Simple Sequence Repeats or Microsatellites

Simple sequence repeats (SSRs), also calledmicrosatellites, are repeated units of two to six nucle-otides that occur throughout an organisms genome.The sequence ATATATAT is one example of amicrosatellite. The sequence GATGATGAT is anotherexample. SSRs are useful as molecular markers becausethey are highly polymorphic (have many forms). SSRshave been used successfully as markers in a wide rangeof analysis, particularly those involving disease diagno-sis and forensics.

1. The process begins with a

blood or cell sample from which

the DNA is extracted.

2. The DNA is cut into fragments

using a restriction enzyme. The

fragments are then separated into

bands by electrophoresis through

an agarose gel.

3. The DNA band pattern is

transferred to a nylon membrane.

4. A radioactive DNA probe is

introduced. The DNA probe binds

to specific DNA sequences on the

nylon membrane.

5. The excess probe material is

washed away leaving the unique

DNA band pattern.

6. The radioactive DNA pattern is

transferred to X-ray film by direct

exposure. When developed, the

resultant visible pattern is the

DNA FINGERPRINT.

THE PROCESS OF DNA FINGERPRINTING

Figure 1


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Single Nucleotide Polymorphisms (SNPs)

On average, SNPs will occur in an organisms DNAmore than 1% of the time. Because only about 3% to5% of an organisms DNA codes for proteins, most SNPsare found outside the regions of genes of interest. SNPsfound in a gene of interest are of particular interest to

researchers because they are directly associated with adesired trait. Because of the recent advances in technol-ogy, SNPs are playing a greater role in selection anddiagnosis of genetic traits.

Advantage of Molecular Markers

The advantage of molecular markers for researchersis that they can test for a particular trait as early as theembryo stage in animals or in the seeds of plants beforethey are planted. There is no longer a need for theorganism to develop to a stage at which the traitcan be observed, a wait that in some cases can takemany years.

The Role of PCR in MAS

Once a direct or linked marker has been located,characterized, and sequenced, a method calledpoly-

merase chain reaction (PCR) can be used to makecopies of a specific region of DNA to produce enoughDNA to conduct a test. Figure 2 on the next two pagessummarizes the PCR process. Since its conception in1983 by Kary Mullis, it has become one of the mostwidely used techniques in molecular biology. It is arapid and simple means of producing a relatively large

amount of DNA from a very small amount of DNA.

DNA replication in natural systems requires:

a source of the nucleotides adenine (A), cytosine (C),thymine (T), and guanine (G);

the DNA polymerase (DNA synthesis enzyme); a short RNA molecule (primer); a DNA strand to be copied; and proper reaction conditions (pH, temperature).

The DNA is unwound enzymatically, the RNA moleculeis synthesized, the DNA polymerase attaches to the

RNA, and a complementary DNA strand is synthesized.

Use of PCR in the laboratory involves the same compo-nents and mechanisms of the natural system, but thereare three primary differences:

(1) DNA primers are used instead of the RNA primerfound in the natural system. DNA primers areusually 18-25 nucleotide bases long and are

designed so that they attach to both sides of theregion of DNA to be copied.

(2) Magnesium ions that play a role in DNA replicationare added to the reaction mixture.

(3) A DNA polymerase enzyme that can withstandhigh temperatures, such as Taq, is used.

(4) A reaction buffer is used to establish the correctconditions for the DNA polymerase to work.

The DNA primers are complementary (match up) toopposite strands of the DNA to be copied, so that bothstrands can be synthesized at the same time. A and Tmatch, and C and G match. Because the reactionmixture contains primers complementary to bothstrands of DNA, the products of the DNA synthesis canthemselves be copied with the opposite primer.

The length of the DNA to be copied is determined bythe position of the two primers relative to the targeted

DNA region. The DNA copies are a defined length andat a specific location on the original DNA. BecauseDNA replication starts from the primers, the newstrands of DNA include the sequence of the primers.This provides a sequence on the new strands to whichthe primers can attach to make additional DNA copies.

Over the years, the PCR procedure has been simplifiedand the results made uniform as a result of two impor-tant developments. The first was the isolation of a heat-stable DNA polymerase, Taq polymerase. This enzymegets its name from the bacteria from which it was

isolated, Thermus aquaticus. This bacteria was discov-ered living in the boiling water of hot springs. Until Taqpolymerase was discovered, the DNA polymerasesavailable to researchers were destroyed at 65C. TheTaq enzyme is not destroyed by the high temperaturerequired to denature the DNA template (pattern).Therefore, using this enzyme eliminates the need to addnew enzyme to the tube for each new cycle of copying,commonly done before Taqs discovery.

The PCR procedure involves three steps that make up acycle of copying. Each step allows the temperature ofthe mixture to change to optimize the reaction. The

cycles are repeated as many times as necessary to obtainthe desired amount of DNA.

STEP 1 DENATURATION

The double-stranded DNA that is to be copied is heatedto ~95C so that the hydrogen bonds between thecomplementary bases are broken. This creates two,single stranded pieces of DNA.


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Figure 2

The Polymerase Chain Reaction Process

(continued on next page)

1. Denaturation

The double-stranded DNA

containing the area of interest

(target DNA) is heated to about

95 C.

The hydrogen bonds between

the bases on the strand are

broken. This results in two

single-stranded pieces of DNA.

2. Annealing

The single-stranded pieces of

DNA are cooled to about 58 C.

The primers form hydrogen

bonds to attach themselves to

their complementary bases on

the single-stranded pieces of

DNA.

3. DNA Synthesis

The DNA pieces resulting from

step 2 are heated to about 72 C.

Polymerase enzyme, Taq,

attaches at each priming site

and extends by adding As, Ts,

Cs, and Gs, forming a new

DNA strand.

Cycle two begins by again

raising the temperature toabout 95 C. to denature the

DNA made in cycle 1. The

entire PCR cycle begins again.

95 C.

58 C.

72 C.

Cycle One

Target DNA

Taq

Taq

Primers

(4bp)


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4. Denaturation

Heating separates the DNA

strands from cycle one. Theoriginal strands and the strands

made in cycle one each contain

the target DNA.

5. AnnealingThe primers attach themselves

to the two original strands of

DNA and the two strands

produced in cycle one.

6. DNA Synthesis

Four new DNA strands are

synthesized. Millions of copies

of the target DNA can be

produced within hours.

95 C.

58 C.

Cycle Two

72 C.

Target DNA

Taq

Taq

Taq

Taq

Original DNA

Copied DNA

Copied DNA

Original DNA


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STEP 2 ANNEALING or HYBRIDIZATIONThe temperature is lowered to ~58C so the DNAprimers can bind to the complementary sequence onthe single-stranded DNA by forming hydrogen bondsbetween the bases of the template and the primers.

STEP 3 DNA SYNTHESIS or EXTENSIONDuring the replication step, the reaction solution isheated to ~72C so the DNA polymerase incorporatesthe nucleotide bases A, C, T, and G into the new copyof DNA. The new DNA strand is formed by connectingbases that are complementary to the template until itcomes to the end of the region to be copied.

To view simulations of the PCR process, go towww.biotech.iastate.edu/publications/ed_resources/Laboratory_protocols.html and find the PCR activityand simulation links.

To view an animation of the PCR process, visitwww.dnalc.org/resources/BiologyAnimationLibrary.htmand view or download the polymerase chain reaction.

DNA Sequencing

A technology used to detect molecular markers of DNAis called DNA sequencing. DNA sequencing is theprocess of determining the exact order of the bases A, T,C, and G in a piece of DNA. The DNA to be sequencedis used to generate a set of fragments that differ inlength from each other by one base pair. The fragments

are separated by size using electro-phoresis. By reading the gel from thebottom up, the sequence of DNA canbe determined.

The most commonly used method ofsequencing DNA was developed byFrederick Sanger in 1977. He modifiedthe chemical structure of the normalnucleotides used in PCR by replacing ahydroxyl group (OH) with a hydrogen(H) on the 3 carbon. The modifiedmolecule is referred to as a

dideoxynucleotide (ddNTP). Thischemical change in the nucleotidecauses the replication of the DNAstrand to terminate during the PCRprocess. Four PCR reactions, eachcontaining a different ddNTP alongwith the normal dNTPs, are conducted.This generates many different sizes offragments in the reaction solution, eachending with a specific nucleotide. ddC ddT ddA ddG

Actual Fragment Sizes

CATTCGAATGCA

CATTCGAATGC

CATTCGAATG

CATTCGAAT

CATTCGAA

CATTCGA

CATTCG

CATTC

CATT

CAT

CA

C

Figure 3

The four reaction solutions are loaded into side-by-sidewells and electrophoresed in one of several gel matrixes.The distance the fragment migrates is inversely propor-tional to its size. The smallest fragment travels fartherand faster through the gel matrix than the largerfragments, thus creating a ladder or pattern of bands

that can be read from the bottom to the top of the gel.In the gel pictured in Figure 3, the size of the fragmentincreases by one base pair relative to its position on thegel. The DNA sequence for the gel is read asCATTCGAATGCA.

To view a sequencing simulation, go to www.dnalc.org/shockwave/cycseq.html or www.pbs.org/wgbh/nova/genome/sequencer.html.

Part I

Sire Osborndale Ivanhoe:The Story of Bovine Leukocyte Adhesion

Deficiency (BLAD)

By the year 1988, a genetic disease specific to Holsteincattle was claiming an ever-increasing number ofanimals. Because Holsteins are a major breed in milkproduction throughout the world, the disease wascausing serious economic loss to the milk industry. Thedisease, called bovine leukocyteadhesion deficiency orBLAD, is characterized in young calves by their inability


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to fight off common bacterial infections like pneumo-nia. Death usually occurs at an early age.

BLAD is caused by a hereditary genetic mutation thatdisrupts the function of a protein on white blood cellscalled leukocytes. Leukocytes are part of the immune

system and help cattle fight disease. When a cow isexposed to an infectious agent called an antigen, leuko-cytes are attracted to the site of the infection by mole-cules that appear on the walls of blood vessels closest tothe infected area. When the leukocytes reach theinfected area, they attach to the vessel walls, go throughthe walls into the infected tissue, and destroy theantigen. The mutation associated with BLAD changesthe leukocyte so it cannot attach to the vessel wall andreach infected tissue.

BLAD is an autosomal (non-sex chromosome) recessivedisease. To suffer from the disease, calves must have

two defective alleles for the trait, one donated byeach parent.

Through an investigation of pedigrees of affected calves,a common sire was determined. Osborndale Ivanhoe, aHolstein bull, is now known to have had the largestimpact of any bull on the Holstein breed. It is estimatedthat he sired over 79,000 daughters and over 1,200 sonsthat produced additional female cows. By the timeBLAD was understood and a molecular test developedin 1991, some estimates are that 28% of the Holsteinpopulation tested positive as BLAD carriers, and an

estimated 16,000-20,000 calves were born with BLADeach year in the United States.

How could one bull be responsible for a genetic diseasethat spread through a large segment of the Holsteinbreed? The answer lies in the way the dairy industrybreeds its cows for milk production. Bulls are selectedfor breeding by evaluating the milk production of theirfemale offspring. When a bull has female offspring with

superior milk production, its sperm are collected foruse in artificial insemination (AI). The benefit of AI isthat one bull of superior genetics can improve theperformance of herds on many farms. One of the risksof AI is that if a sire is a heterozygous carrier of anundesirable recessive allele, that allele can be spreadundetected to many progeny.

Because bulls and cows with heterozygous alleles forthe trait are healthy, a recessive allele can spreadundetected for many generations. See the BLADpedigree, Figure 4.

When a heterozygous bull is crossed with a heterozy-gous cow, there is a 25% chance the calf will be inflictedwith BLAD, and a 50% chance the calves will becarriers. See Figure 5. Breeders needed a reliable test toidentify cattle that were heterozygous carriers. The testthat was developed was marker assisted selection.

Scientists found that the deleterious recessive allele forBLAD had two mutations in the CD18 gene. One of themutations did not affect the amino acid sequence, butthe second mutation caused an incorrect amino acid tobe produced. In that second mutation, the nucleotide

guanine (G) replaced adenine (A) so the amino acidglycine is produced instead of aspartic acid. See Figure6. Figure 7 illustrates the DNA strands from each allele

Figure 4

BLAD Pedigree

II

I

III

IV

V

inbreeding

Osborndale Ivanhoe


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B b

B BB bB

b bB bb

Figure 5

containing the site of the BLAD mutation during thePCR process. When these PCR products are treatedwith the restriction (cutting) enzyme TaqI, the enzyme

recognizes the TCGA sequence and cuts between the Tand C nucleotides. Each strand will generate DNAfragments consistent with the presence or absence ofthe restriction site TCGA on the strand. A normal DNAsequence will contain a TaqI restriction site andgenerate two fragments, one of 26 base pairs (bp) and

Cattle that have the Bb alleles are carriers of BLAD.

Affected cattle have two copies of the b allele.

the other 32 bp. In the case of the BLAD mutation, therestriction site for TaqI is lost. Since there is no restric-tion site for TaqI on the mutation, a single fragment of58 bp (the size of the PCR product) remains. Thepresence of the 26, 32 and 58 bp fragments indicate thecarrier. See Figure 8.

The protein with the amino acid change preventsleukocytes from reaching and destroying the invadingantigen by interfering with their ability to adhere to theblood vessel walls at the area of infection. This is whycalves with BLAD cannot fight infections and die earlyin life.

Using molecular marker technology, it has beenpossible to identify the heterozygous carriers of BLADand remove those individuals from the breeding stock.As a result, the disease has been virtually eliminatedfrom the Holstein cattle breed.

The defective allele causing BLAD has not been foundin breeds other than Holsteins. However, a similar formof the genetic disorder has been described in humans.

Figure 6

Protein Synthesis from the Normal CD18 Gene

DNA Strand 5ggc tac ccc atc gac ctg tac tac ctg 3

Amino Acids gly tyr pro lle asp leu tyr try leu

Protein Synthesis from the BLAD Mutation CD18 Gene

DNA Strand 5ggc tac ccc atc ggc ctg tac tac ctg 3

Amino Acids gly tyr pro lle gly leu tyr try leu

When the nucleotide adenine (a) is replaced by guanine (g) in the DNA strand, the amino acid glycine (gly) is

produced instead of the correct amino acid aspartic acid (asp). The result is the BLAD condition in cattle.


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Figure 7

Figure 7 shows the DNA fragments produced by normal cattle, heterozygous BLAD carriers, and

homozygous BLAD affected cattle when their DNA is mixed with the Taq1 enzyme. The enzyme

recognizes thetcga sequence and cuts (t / cga)between the t and c nucleotides. When guanine

(g) replaces adenine (a), the tcgasequence is replaced by tcgg and the enzyme does not cut

the strand.

32 bp 26 bp

DNA Strands Involved in Diagnosis of BLAD

Normal: 32 and 26 bp segments produced

5gtgaccttccggagggccaagggctaccccat / cgacctgtactacctgatggacctct 3


BLAD Carrier: 32, 26, and 58 bp segments produced


5gtgaccttccggagggccaagggctaccccatcggcctgtactacctgatggacctct 3

BLAD Affected: 58 bp segment produced



32 bp 26 bp

nucleotide change

58 bp

nucleotide change

58 bpnucleotide change

58 bp


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Credit Notes

Atherly, Alan G.; Girton, Jack R.; and McDonald, John F.The Science of Genetics. Saunders College Publishing.1999

Basics of Marker Assisted Selection (BMAS). Julius vander Werf, Department of Animal Science, and BrianKinghorn, Twynam Chair of Animal Breeding Technolo-gies, University of New England.

Campbell, Neil A. and Reece, Jane B. Biology. Seventh

edition. Pearson-Benjamin Cummings Publications.San Francisco, California. 2005

Doggy DNA: The Power of PCR. 2000 Summer BiologyInstitute: Biodiversity. The Woodrow Wilson Founda-tion Leadership Program for Teachers.www.woodrow.org/teachers/bi/2000/Doggy_DNA/background_for_polymerase_chai.html

Figure 8

Kehrli, Marcus E., Jr. Bovine Leukoctye AdhesionDeficiency (BLAD) in Holstein Cattle. Virus and PrionDiseases of Livestock Research Unit, National AnimalDisease Center, USDA-ARS, Ames, Iowa.

Marker-Assisted Selection: Applications to AnimalProduction. The Agbiotech Infosource. AG-WESTBIOTECH INC. Issue 39. October 1998

Olson, Tim. Automated DNA Sequencing and PrimerDesign. Department of Animal Sciences, University of

Florida.

Olson, Tim. New Genes: Good and Bad. Departmentof Animal Sciences, University of Florida.

Polking, Gary F. The Polymerase Chain Reaction:Introduction. Introduction to Molecular BiologyTechniques Gen 542A. Iowa State Universitys Officeof Biotechnology. Summer 2003

The BLAD mutation results in the loss of the TaqI restriction site. See Figure 7. A normal cow will

display two fragments, 26 and 32 bp. A carrier will display three fragments, 26, 32, and 58 bp. A BLAD

infected animal has only a single fragment of 58 bp. Based on a gel photo provided by Marcus E. Kehrli, Jr., DVM,PhD, National Animal Disease Center, USDA-ARS.

Homozygous

Normal

Heterozygous

Carrier

Homozygous

BLAD

Base Pairs (bp)

26 bp

32 bp

26 bp

32 bp

58 bp 58 bp

Agarose Gelof BLAD

PCR ProductDigested with Taq1


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Suszkiw, Jan. Mapping the Way to Bovine Bounty. ARSNational Program Publication.

Terminology based on Campbell, Neil A. and Reece,Jane B. Biology. Seventh edition. Pearson-BenjaminCummings Publications. San Francisco, California.

2005

Van Eenennaam, Alison. Marker-assisted selectionbackgrounder. Agriculture and Natural ResourcesResearch and Extension Centers. University ofCalifornia-Davis. 2004

A Holstein dairy cow. Keith Weller, ARS-USDA


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Marker Assisted

Selection

Part I

TEACHING RESOURCES

Laboratory Lesson Plan:

Student Exercise on Polymerase

Chain Reaction (PCR)

Science Education Standards

Science as Inquiry, Content Standard A

Abilities necessary to do scientific inquiry(p. 175)

Understanding about scientific inquiry (p. 176)

Life Science, Content Standard C

The cell (p. 184) Molecular basis of heredity (p. 185) Matter, energy, and organization in living systems

(p. 186)

Science and Technology, Content Standard E

Understandings about science and technology(p. 192)

Source: National Science Education Standards, National Academy ofSciences, 1996. Used with permission. Page numbers refer to theseventh printing, November 1999 also available on the Internet athttp://books.nap.edu/html/nses/pdf/index.html.

Science Process Skills

Comparing and measuring Observing Ordering

Relating Inferring

Life Skills

Learning to learn Science processing Problem solving Decision making Communicating

IITimePreparation: Ten minutes to photocopy studenthandouts MAS-1 and MAS-2 on p. 73-84.Activity: One 30-minute block of class time.

Materials

Educators should make enough copies of the studenthandouts MAS-1, Learning More About Marker AssistedSelection, and MAS-2, Student Exercise on PolymeraseChain Reaction, so that each student has a copy.

Procedure

The background information contained in studenthandout MAS-1, Learning More About Marker AssistedSelection, should be presented before the class period in

which educators want to do the polymerase chainreaction exercise. Overhead transparencies MAS a-mon p. 91-115 may be helpful. Give students the studenthandout MAS-2, See for Yourself: Student Exercise onPolymerase Chain Reaction.

The answer key for the exercise appears on the next fewpages. Student answers are in bold print. Educatorsalso may wish to use the PowerPoint or animatedversions of the exercise that can be viewed or down-loaded from www.biotech.iastate.edu/publications/ed_resources/Laboratory_protocols.html. Scroll downto the Polymerase Chain Reaction section. The Internetversions have two additional parts that focus on themathematical aspects of PCR.


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Student Exercise on Polymerase Chain Reaction (PCR)

Prepared by the Office of Biotechnology, Iowa State University

Part I

Original-1 3' T C G G C T A C A G C A G C A G A T G G T A C G T A 5'

Original-2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

1. The purpose of PCR is to make copies of the target DNA, such as the one

above. In our exercise, one strand of the double helix of DNA will be

designated Original-1. Write the nucleotide sequence of the complementary

strand in the blanks designated Original-2 above.

A G C C G A T G T C G T C G T C T A C C A T G C A T

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Target DNA


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Part II


5' _ _ _ _ _

(Primer-1)

Original-2 5' A G C C G A T G T C G T C G T C T A C C A T G C A T 3'

3' _ _ _ _ _ 5'

(Primer-2)

2. A piece of DNA known as the primer is artificially made that has a

nucleotide sequence complementary to the bases adjacent to the target

DNA on the 3' end of Original-1. Write the nucleotide sequence of the

five bases of Primer-1 in the blanks above.

3. A primer is artificially made that has a nucleotide sequence

complementary to the bases adjacent to the target DNA on the 3' end of

Original-2. Write the nucleotide sequence of the five bases of Primer-2

in the blanks above.

Part III


Copy-1 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer-1)


Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A T G G T 5'

(Primer-2)

4. In cycle 1 and all subsequent cycles of the PCR reaction, a copy of each

of the two original strands will be made beginning at the 3' end of the

primer and continuing to the 5' end of the original strand. Write the

sequence of the copies that are made from the strands of Original-1 and

Original-2 in the blanks above.

C C G A T

A T G G T

G T C G T C G T C T A C C A T G C A T

T C G G C T A C A G C A G C A G

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _


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Part IV


Copy-1 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer-1)

Copy-1 5' C C G A T G T C G T C G T C T A C C A T G C A T 3'

Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

(Primer)


Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A T G G T 5'

(Primer-2)

Copy-2 3' T C G G C T A C A G C A G C A G A T G G T 5'

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer)

5. During the second cycle of PCR, a copy is made of each of the strands of

Original-1, Copy-1, Original-2, and Copy-2 obtained in cycle 1. In the

blanks above, write the sequence of Copy-1 formed during the replication

of Original-1 in cycle 2. Does the sequence differ from that of Copy-1

made in the first cycle? No. Write the sequence of Copy-2 formed duringthe replication of Original-2 in the second cycle. Does the sequence

differ from that of Copy-2 made in the first cycle? No.

6. To make a copy of the Copy-1 strand, a primer attaches to appropri-

ate sequences on the strand. Note that only one of the two primers will

be appropriate. Write the sequence of the primer and complete the

sequence of Copy-C1 in the blanks above.

7. To make a copy of the Copy-2 strand, a primer attaches to appropriate

sequences on the strand. Write the sequence of the primer and complete

the sequence of Copy-C2 in the blanks above.

8. How many strands of each of the following are present after the second

cycle?

Original-1 ___ Original-2 ___

Copy-1 ___ Copy-2 ___

Copy-C1 ___ Copy-C2 ___


T C G G C T A C A G C A G C A G

G G C T A C A G C A G C A G A T G G T

C C G A T G T C G T C G T C T A C C A

1

2

1

1

2

1

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _


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Part V


Copy-1 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer-1)


Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

(Primer)

Copy-1 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

(Primer)

Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer)

Original-2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

(Primer-2)

Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

(Primer)

Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _(Primer)

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

(Primer)

9. For the third cycle of PCR, each of the eight strands produced by Cycle

2 are replicated. Write the sequence of the primers and all the new

strands that are formed during copying. You may refer to part IV of the

exercise for assistance.

11. How many strands of each of the following types are present after the

third cycle?

Total number _____ Original-1 ___ Original-2 ___

Copy-1 ___ Copy-2 ___


16 1

3

4

1

3

4



_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

C C G A T G T C G T C G T C T A C C A T G C A T 3'

G G C T A C A G C A G C A G A T G G T 5'




T C G G C T A C A G C A G C A G A T G G T 5'


C C G A T G T C G T C G T C T A C C A 3'


C C G A T G T C G T C G T C T A C C A 3'




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73

Learning more about . . .

Iowa State University Extension and ISU Office of Biotechnology

Student Handout

Marker Assisted Selection (MAS)

Lesson Module II Marker Assisted SelectionMAS-1

BACKGROUND INFORMATION

MAS Introduction

Deoxyribonucleic acid (DNA) is a molecule made up ofpairs of building blocks called nucleotides. The fourkinds of nucleotides that make up DNA are adenine(abbreviated as the single letter A), guanine (G),cytosine (C), and thymine (T). The DNA molecule has

the shape of two intertwined spirals, referred to as adouble helix.

DNA is packaged into chromosomes that are locatedwithin the nucleus of all cells. These chromosomes arethe same in every cell of an organism and togethermake up the organisms genetic information, its

genome. Chromosomes contain stretches of DNAcalled genes that code for amino acids that makeproteins. It is the proteins that are the foundation of lifefor all organisms. The interaction and structure ofproteins determine the visible characteristics or

phenotypeof an organism, while the genetic makeup of

an organism is called itsgenotype.

The sequence of nucleotides that make up a gene candiffer among individuals. The different forms of a geneare called alleles. The alleles are the result of nucle-otide differences in a gene that affect an amino acidsequence of a protein. This can result in a change,addition, or deletion of a protein that can affectthe phenotype.

All organisms receive one copy of each gene from theirmother and one from their father. The DNA sequenceof a gene inherited from each parent may be identical,

in which case the individual is said to be homozygousfor that trait. Or the sequence of the gene from one ofthe parents may be different, in which case the indi-vidual is said to be heterozygous. Allele variations maydiffer in their DNA sequence by as little as asingle nucleotide.

Differences among alleles caused by a single nucleotide,called SNPs, can be the basis of genotyping tests.

Genotyping means using laboratory methods todetermine the sequence of nucleotides in the DNA froman individual, usually a specific gene. Genetic testsbased on SNPs utilize DNA derived from an individualto determine the nucleotide in the gene of interest.

Marker assisted selection is the process of using theresults of DNA testing in the selection of individualsto become parents for the next generations. Theinformation from the DNA testing, combined with the

observed performance records for individuals, isintended to improve the accuracy of selection andincrease the possibility of identifying organismscarrying desirable and undesirable traits at an earlierstage of development.

Complex traits, including many of economic impor-tance, are controlled by many genes and are influencedby the environment. When an animal has a favorableperformance record for a certain trait, it means thatbased on pedigree and phenotype, the animal hasinherited a greater than average number of good alleles

of each gene affecting that specific trait.

It is important to combine DNA results with perfor-mance and phenotype information to maximize theeffectiveness of selection for traits of interest. Combin-ing information from performance records and genetictests into the selection process will be better than usingperformance, phenotype, and markers separately. Thechallenge is to determine what emphasis markerinformation should be given in the selection decision.

Molecular Markers

Until recently, researchers relied on information abouthow animals, plants, and their relatives perform tomake observations about the genes they possess. Today,researchers can usemolecular markers to find genes ofinterest that control how plants and animals perform.Some molecular markers are pieces of DNA that haveno known function or impact on animal and plantperformance. Other markers may involve the gene ofinterest itself.


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Student Handout


Lesson Module II Marker Assisted Selection

MAS-1

Linked Markers

One type of molecular marker is called a linked marker.Using well-designed experiments, scientists can findmolecular markers that are located very close to majorgenes of interest. The molecular marker is said to belinked to that gene. Linked markers are only near the

gene of interest on the chromosome and are not part ofthe DNA of the gene of interest.

Suppose that scientists are trying to locate a certaingene in an animal species. Choosing animals randomlyfrom a population and studying them would give thescientists no clues about whether a marker is associatedwith the gene. However, if scientists studied theprogeny (offspring) of the mating of male and femaleanimals through many generations, they may determinethe presence of a useful molecular marker.

Direct Markers

A second kind of molecular marker is one that is part ofthe gene of interest. Direct markers are easier to workwith after they are found, but they often are moredifficult to find than linked markers.

Marker-Assisted Selection

Three common technologies used as molecularmarkers are: restriction fragment length polymor-phisms, simple sequence repeats, and singlenucleotide polymorphisms.

Restriction Fragment Length Polymorphisms(RFLPs)

Restriction fragment lengthpolymorphisms (RFLPs)were the first molecular markers used to diagnosegenetic variability in organisms. RFLP uses restrictionenzymes to digest (cut) the DNA molecule and identifyregions linked to a trait. The number of DNA frag-ments generated by one restriction enzyme digest canbe in the millions, with many being several thousandnucleotides long. This makes it difficult to determinespecific DNA fragments that are associated with thetrait of interest on an electrophoresis gel. To helpvisualize specific DNA fragments, a technique called

Southern blotting was developed.

Southern blotting uses a porous membrane containingspecific radioactive DNA probes for one or more DNAfragments. Probes are very short pieces of DNA used tofind specific sequences of A, C, T, and G in very longpieces of DNA from a chromosome. The probehybridizes (attaches) to the membrane at a unique DNAband on an electrophoresis gel. The membrane

containing the probe is developed on X-ray film andanalyzed. See Figure 1.

Simple Sequence Repeats or Microsatellites

Simple sequence repeats (SSRs), also calledmicrosatellites, are repeated units of two to six nucle-otides that occur throughout an organisms genome.The sequence ATATATAT is one example of amicrosatellite. The sequence GATGATGAT is anotherexample. SSRs are useful as molecular markers becausethey are highly polymorphic (have many forms). SSRs

have been used successfully as markers in a wide rangeof analysis, particularly those involving disease diagno-sis and forensics.

Single Nucleotide Polymorphisms (SNPs)

On average, SNPs will occur in an organisms DNAmore than 1% of the time. Because only about 3% to5% of an organisms DNA codes for proteins, most SNPsare found outside the regions of genes of interest. SNPsfound in a gene of interest are of particular interest to








an agarose gel.






nylon membrane.



DNA band pattern.





DNA FINGERPRINT.


Figure 1


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Student HandoutLesson Module II Marker Assisted SelectionMAS-1

researchers because they are directly associated with adesired trait. Because of the recent advances in technol-ogy, SNPs are playing a greater role in selection anddiagnosis of genetic traits.

Advantage of Molecular Markers

The advantage of molecular markers for researchersis that they can test for a particular trait as early as theembryo stage in animals or in the seeds of plants beforethey are planted. There is no longer a need for theorganism to develop to a stage at which the traitcan be observed, a wait that in some cases can takemany years.

The Role of PCR in MAS

Once a direct or linked marker has been located,characterized, and sequenced, a method calledpoly-

merase chain reaction (PCR) can be used to makecopies of a specific region of DNA to produce enoughDNA to conduct a test. Figure 2 on the next two pagessummarizes the PCR process. Since its conception in1983 by Kary Mullis, it has become one of the mostwidely used techniques in molecular biology. It is arapid and simple means of producing a relatively largeamount of DNA from a very small amount of DNA.

DNA replication in natural systems requires:

a source of the nucleotides adenine (A), cytosine (C),thymine (T), and guanine (G);

the DNA polymerase (DNA synthesis enzyme); a short RNA molecule (primer); a DNA strand to be copied; and proper reaction conditions (pH, temperature).

The DNA is unwound enzymatically, the RNA moleculeis synthesized,the DNA polymerase attaches to theRNA, and a complementary DNA strand is synthesized.

Use of PCR in the laboratory involves the same compo-nents and mechanisms of the natural system, but thereare three primary differences:

(1) DNA primers are used instead of the RNA primerfound in the natural system. DNA primers areusually 18-25 nucleotide bases long and aredesigned so that they attach to both sides of theregion of DNA to be copied.

(2) Magnesium ions that play a role in DNA replicationare added to the reaction mixture.

(3) A DNA polymerase enzyme that can withstandhigh temperatures, such as Taq, is used.

(4) A reaction buffer is used to establish the correctconditions for the DNA polymerase to work.

The DNA primers are complementary (match up) toopposite strands of the DNA to be copied, so that bothstrands can be synthesized at the same time. A and

T match, and C and G match. Because the reactionmixture contains primers complementary to bothstrands of DNA, the products of the DNA synthesis canthemselves be copied with the opposite primer.

The length of the DNA to be copied is determined bythe position of the two primers relative to the targetedDNA region. The DNA copies are a defined length andat a specific location on the original DNA. BecauseDNA replication starts from the primers, the newstrands of DNA include the sequence of the primers.This provides a sequence on the new strands to whichthe primers can attach to make additional DNA copies.

Over the years, the PCR procedure has been simplifiedand the results made uniform as a result of two impor-tant developments. The first was the isolation of a heat-stable DNA polymerase, Taq polymerase. This enzymegets its name from the bacteria from which it wasisolated, Thermus aquaticus. This bacteria was discov-ered living in the boiling water of hot springs. Until Taqpolymerase was discovered, the DNA polymerasesavailable to researchers were destroyed at 65C. TheTaq enzyme is not destroyed by the high temperaturerequired to denature the DNA template (pattern).

Therefore, using this enzyme eliminates the need to addnew enzyme to the tube for each new cycle of copying,commonly done before Taqs discovery.

The PCR procedure involves three steps that make up acycle of copying. Each step allows the temperature ofthe mixture to change to optimize the reaction. Thecycles are repeated as many times as necessary to obtainthe desired amount of DNA.

STEP 1 DENATURATION

The double-stranded DNA that is to be copied is heatedto ~95C so that the hydrogen bonds between the

complementary bases are broken. This creates two,single stranded pieces of DNA.

STEP 2 ANNEALING or HYBRIDIZATION

The temperature is lowered to ~58C so the DNAprimers can bind to the complementary sequence onthe single-stranded DNA by forming hydrogen bondsbetween the bases of the template and the primers.


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76

Student Handout


Figure 2

The Polymerase Chain Reaction Process


MAS-1

1. Denaturation




95 C.



broken. This results in twosingle-stranded pieces of DNA.

2. Annealing







DNA.

3. DNA Synthesis







DNA strand.

Cycle two begins by againraising the temperature to

about 95 C. to denature the



95 C.

58 C.

72 C.

Cycle One

Primers

(4 bp)

Target DNA

Taq

Taq

(continued on next page)


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Student HandoutLesson Module II Marker Assisted Selection

MAS-1

4. Denaturation


strands from cycle one. The

original strands and the strands


the target DNA.

5. Annealing

The primers attach themselves




6. DNA Synthesis





95 C.

58 C.

Cycle Two

72 C.

Target DNA

Taq

Taq

Taq

Taq

Original DNA

Copied DNA

Copied DNA

Original DNA


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78

Student Handout



MAS-1

Figure 3

STEP 3 DNA SYNTHESIS or EXTENSIONDuring the replication step, the reaction solution isheated to ~72C so the DNA polymerase incorporatesthe nucleotide bases A, C, T, and G into the new copyof DNA. The new DNA strand is formed by connectingbases that are complementary to the template until it

comes to the end of the region to be copied.

To view simulations of the PCR process, go towww.biotech.iastate.edu/publications/ed_resources/Laboratory_protocols.html and find the PCR activityand simulation links. To view an animation of the PCRprocess, visit www.dnalc.org/resources/BiologyAnimationLibrary.htm and view or download thepolymerase chain reaction.

DNA Sequencing

A technology used to detect molecular markers of DNAis called DNA sequencing. DNA sequencing is theprocess of determining the exact order of the bases A, T,C, and G in a piece of DNA. The DNA to be sequencedis used to generate a set of fragments that differ inlength from each other by one base pair. The fragmentsare separated by size using electrophoresis. By readingthe gel from the bottom up, the sequence of DNA canbe determined.

The most commonly used method of sequencing DNAwas developed by Frederick Sanger in 1977. Hemodified the chemical structure of the normal nucle-

otides used in PCR by replacing a hydroxyl group (OH)with a hydrogen (H) on the 3 carbon. The modifiedmolecule is referred to as a dideoxynucleotide (ddNTP).This chemical change in the nucleotide causes thereplication of the DNA strand to terminate during thePCR process. Four PCR reactions, each containing a

different ddNTP along with the normal dNTPs, areconducted. This generates many different sizes offragments in the reaction solution, each ending with aspecific nucleotide.

The four reaction solutions are loaded into side-by-sidewells and electrophoresed in one of several gel matrixes.The distance the fragment migrates is inversely propor-tional to its size. The smallest fragment travels fartherand faster through the gel matrix than the largerfragments, thus creating a ladder or pattern of bandsthat can be read from the bottom to the top of the gel.In the gel pictured in Figure 3, the size of the fragment

increases by one base pair relative to its position on thegel. The DNA sequence for the gel is read asCATTCGAATGCA.

To view a sequencing simulation, go to www.dnalc.org/shockwave/cycseq.html or www.pbs.org/wgbh/nova/genome/sequencer.html.

Credit NotesAtherly, Alan G.; Girton, Jack R.; and McDonald, John F.The Science of Genetics. Saunders College Publishing.

1999.

Basics of Marker Assisted Selection(BMAS). Julius van der Werf,Department of Animal Science, andBrian Kinghorn, Twynam Chair ofAnimal Breeding Technologies,University of New England.

Campbell, Neil A. and Reece, Jane B.Biology. Seventh edition. Pearson-Benjamin Cummings Publications.San Francisco, California. 2005

Doggy DNA: The Power of PCR.2000 Summer Biology Institute:Biodiversity. The Woodrow WilsonFoundation Leadership Program forTeachers. www.woodrow.org/teachers/bi/2000/Doggy_DNA/background_for_ polymerase_chai.htmlddC ddT ddA ddG


CATTCGAATGCA

CATTCGAATGC

CATTCGAATG

CATTCGAAT

CATTCGAA

CATTCGA

CATTCG

CATTC

CATT

CAT

CA

C


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MAS-1

Learn the Language

Alleles

Different forms of the same gene

Deoxyribonucleic acid (DNA)

A molecule made up of four bases; adenine (A),cytosine (C), thymine (T), and guanine (G); thatcarries the genetic information in the nucleus of

all cells

Direct marker

A sequence of DNA located within a geneof interest

DNA sequencing

The process of determining the exact order of thebases adenine (A), cytosine (C), thymine (T), andguanine (G) in a piece of DNA


Marker-Assisted Selection: Applications to Animal

Production. The Agbiotech Infosource. AG-WESTBIOTECH INC. Issue 39. October 1998

Olson, Tim. Automated DNA Sequencing and PrimerDesign. Department of Animal Sciences, Universityof Florida.


Polking, Gary F. The Polymerase Chain Reaction:Introduction. Introduction to Molecular BiologyTechniques Gen 542A. Iowa State Universitys Office

of Biotechnology. Summer 2003


Terminology based on Campbell, Neil A. and Reece,Jane B. Biology. Seventh edition. Pearson-BenjaminCummings Publications. San Francisco, California.2005

Van Eenennaam, Alison. Marker-assisted selectionbackgrounder. Agriculture and Natural Resources

Research and Extension Centers. University ofCalifornia-Davis. 2004

Genome

The complete genetic code of an organism

Genotype

The genetic makeup of an individual, typicallyexpressed in alphabetical letters

Heterozygous

Having two different alleles of a gene (Rr)

Homozygous

Having two identical alleles of a gene (RR or rr)

Linked marker

A sequence of DNA located near, but outside of, agene of interest

Marker assisted selection

Use of molecular markers to selectdesirable individuals

Molecular marker

A piece of DNA linked to or part of a geneof interest

Phenotype

Description of an observable trait

Polymerase chain reaction (PCR)

A method used to make enough copies of a specific

region of DNA

Polymorphism

Having many forms

Restriction enzyme

A class of enzyme that was originally discovered inbacteria and is extensively used in genetic engineer-ing to recognize a specific target nucleotidesequence in DNA and break the DNA chain atthe target

and justice for allThe U.S. Department of Agriculture (USDA) prohibits discrimination in all itsprograms and activities on the basis of race, color, national origin, gender, religion,age, disability, political beliefs, sexual orientation, and marital or family status. (Notall prohibited bases apply to all programs.) Many materials can be made availablein alternative formats for ADA clients. To file a complaint of discrimination, writeUSDA, Office of Civil Rights, Room 326-W, Whitten Building, 14th and IndependenceAvenue, SW, Washington, DC 20250-9410 or call 202-720-5964.

Issued in furtherance of Cooperative Extension work, Acts of May 8 and June 30,1914 in cooperation with the U.S. Department of Agriculture. Stanley R. Johnson,director, Cooperative Extension Service, Iowa State University of Science andTechnology, Ames, Iowa.


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Iowa State University Extension and ISU Office of Biotechnology 81

See for yourself . . .

Student Handout

Polymerase Chain Reaction (PCR)

Lesson Module II Marker Assisted SelectionMAS-2

and justice for allThe U.S. Department of Agriculture (USDA) prohibits discrimination in all its programs and activities on the basis of race, color, national origin, gender, religion, age, disability,political beliefs, sexual orientation, and marital or family status. (Not all prohibited bases apply to all programs.) Many materials can be made available in alternative formats forADA clients. To file a complaint of discrimination, write USDA, Office of Civil Rights, Room 326-W, Whitten Building, 14th and Independence Avenue, SW, Washington, DC 20250-9410 or call 202-720-5964.

Issued in furtherance of Cooperative Extension work, Acts of May 8 and June 30, 1914 in cooperation with the U.S. Department of Agriculture. Stanley R. Johnson, director,Cooperative Extension Service, Iowa State University of Science and Technology, Ames, Iowa.

Student Exercise on Polymerase Chain Reaction (PCR)

Prepared by the Office of Biotechnology, Iowa State University

Part I


Original-2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

1. The purpose of PCR is to make copies of the target DNA, such as the one

above. In our exercise, one strand of the double helix of DNA will be

designated Original-1. Write the nucleotide sequence of the complementary

strand in the blanks designated Original-2 above.

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Target DNA


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Iowa State University Extension and ISU Office of Biotechnology82

Student Handout Lesson Module II Marker Assisted Selection

MAS-2

Part II


5' _ _ _ _ _

(Primer-1)


3' _ _ _ _ _ 5'

(Primer-2)

2. A piece of DNA known as the primer is artificially made that has a

nucleotide sequence complementary to the bases adjacent to the target

DNA on the 3' end of Original-1. Write the nucleotide sequence of the

five bases of Primer-1 in the blanks above.

3. A primer is artificially made that has a nucleotide sequence

complementary to the bases adjacent to the target DNA on the 3' end of

Original-2. Write the nucleotide sequence of the five bases of Primer-2

in the blanks above.

Part III


Copy-1 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer-1)


Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A T G G T 5'

(Primer-2)

4. In cycle 1 and all subsequent cycles of the PCR reaction, a copy of each

of the two original strands will be made beginning at the 3' end of the

primer and continuing to the 5' end of the original strand. Write the

sequence of the copies that are made from the strands of Original-1 and

Original-2 in the blanks above.

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _


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Iowa State University Extension and ISU Office of Biotechnology 83


MAS-2

Part IV


Copy-1 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer-1)


Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

(Primer)


Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A T G G T 5'

(Primer-2)

Copy-2 3' T C G G C T A C A G C A G C A G A T G G T 5'

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer)

5. During the second cycle of PCR, a copy is made of each of the strands of

Original-1, Copy-1, Original-2, and Copy-2 obtained in cycle 1. In the

blanks above, write the sequence of Copy-1 formed during the replication

of Original-1 in cycle 2. Does the sequence differ from that of Copy-1

made in the first cycle?_____ Write the sequence of Copy-2 formed during

the replication of Original-2 in the second cycle. Does the sequencediffer from that of Copy-2 made in the first cycle?_____

6. To make a copy of the Copy-1 strand, a primer attaches to appropri-

ate sequences on the strand. Note that only one of the two primers will

be appropriate. Write the sequence of the primer and complete the

sequence of Copy-C1 in the blanks above.

7. To make a copy of the Copy-2 strand, a primer attaches to appropriate

sequences on the strand. Write the sequence of the primer and complete

the sequence of Copy-C2 in the blanks above.

8. How many strands of each of the following are present after the second

cycle?

Original-1 ___ Original-2 ___

Copy-1 ___ Copy-2 ___


_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _


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Iowa State University Extension and ISU Office of Biotechnology84

Student Handout Lesson Module II Marker Assisted Selection

MAS-2

Part V


Copy-1 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer-1)


Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

(Primer)

Copy-1 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

(Primer)

Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

(Primer)

Original-2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

(Primer-2)

Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

(Primer)

Copy-2 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _(Primer)

Copy-C2 5' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'

Copy-C1 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5'

(Primer)

9. For the third cycle of PCR, each of the eight strands produced by Cycle

2 are replicated. Write the sequence of the primers and all the new

strands that are formed during copying. You may refer to part IV of the

exercise for assistance.

11. How many strands of each of the following types are present after the

third cycle?

Total number _____ Original-1 ___ Original-2 ___

Copy-1 ___ Copy-2 ___


_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

C C G A T G T C G T C G T C T A C C A T G C A T 3'






A T G G T


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85

Learning more about . . .


Student HandoutLesson Module II Marker Assisted SelectionMAS-3

Sire Osborndale Ivanhoe:The Story of Bovine Leukocyte Adhesion

Deficiency (BLAD)

By the year 1988, a genetic disease specific to Holstein

cattle was claiming an ever-increasing number ofanimals. Because Holsteins are a major breed in milkproduction throughout the world, the disease was

causing serious economic loss to the milk industry. The

disease, called bovine leukocyteadhesion deficiency orBLAD, is characterized in young calves by their inabilityto fight off common bacterial infections like pneumo-

nia. Death usually occurs at an early age.

BLAD is caused by a hereditary genetic mutation that

disrupts the function of a protein on white blood cellscalled leukocytes. Leukocytes are part of the immunesystem and help cattle fight disease. When a cow isexposed to an infectious agent called an antigen, leuko-cytes are attracted to the site of the infection by mole-cules that appear on the walls of blood vessels closest tothe infected area. When the leukocytes reach the

infected area, they attach to the vessel walls, go throughthe walls into the infected tissue, and destroy the

antigen. The mutation associated with BLAD changesthe leukocyte so it cannot attach to the vessel wall and

reach infected tissue.

BLAD is an autosomal (non-sex chromosome) recessivedisease. To suffer from the disease, calves must havetwo defective alleles for the trait, one donated by

each parent.

Through an investigation of pedigrees of affected calves,

a common sire was determined. Osborndale Ivanhoe, aHolstein bull, is now known to have had the largest

impact of any bull on the Holstein breed. It is estimatedthat he sired over 79,000 daughters and over 1,200 sons

that produced additional female cows. By the timeBLAD was understood and a molecular test developed

in 1991, some estimates are that 28% of the Holsteinpopulation tested positive as BLAD carriers, and anestimated 16,000-20,000 calves were born with BLAD

each year in the United States.

BLAD and Sire Osborndale Ivanhoe

How could one bull be responsible for a genetic diseasethat spread through a large segment of the Holstein

breed? The answer lies in the way the dairy industrybreeds its cows for milk production. Bulls are selected

for breeding by evaluating the milk production of theirfemale offspring. When a bull has female offspring with

superior milk production, its sperm are collected foruse in artificial insemination (AI). The benefit of AI isthat one bull of superior genetics can improve the

performance of herds on many farms. One of the risksof AI is that if a sire is a heterozygous carrier of an

undesirable recessive allele, that allele can be spreadundetected to many progeny.

Because bulls and cows with heterozygous alleles forthe trait are healthy, a recessive allele can spread

undetected for many generations. BLAD can onlymanifest itself when heterozygous male and female

descendants of Ivanhoe are bred and two recessive

alleles, one from each parent, combine to producecalves with a homozygous recessive condition. See theBLAD pedigree, Figure 1, on the next page.

When a heterozygous bull is crossed with a heterozy-gous cow, there is a 25% chance the calf will be inflicted

with BLAD, and a 50% chance the calves will becarriers. See Figure 2. Breeders needed a reliable test to

identify cattle that were heterozygous carriers. The testthat was developed was marker assisted selection.

A Holstein dairy cow. Keith Weller, ARS-USDA


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86

Student Handout


B b

B BB bB

b bB bb

Figure 2

Figure 1

BLAD Pedigree

II

I

III

IV

V

inbreeding

Osborndale Ivanhoe

Scientists found that the deleterious recessive allele forBLAD had two mutations in the CD18 gene. One of themutations did not affect the amino acid sequence, butthe second mutation caused an incorrect amino acidto be produced. In that second mutation, thenucleotide guanine (G) replaced adenine (A) so the

amino acid glycine is produced instead of asparticacid. See Figure 3. Figure 4 on p. 88 illustrates theDNA strands from each allele containing the site ofthe BLAD mutation during the PCR process. Whenthese PCR products are treated with the restriction(cutting) enzyme TaqI, the enzyme recognizes theTCGA sequence and cuts between the T and C nucle-otides. Each strand will generate DNA fragmentsconsistent with the presence or absence of the restric-tion site TCGA on the strand. A normal DNA sequencewill contain a TaqI restriction site and generate twofragments, one of 26 base pairs (bp) and the other 32bp. In the case of the BLAD mutation, the restriction

site for TaqI is lost. Since there is no restriction site forTaqI on the mutation, a single fragment of 58 bp (thesize of the PCR product) remains. The presence of the26, 32 and 58 bp fragments indicate the carrier. SeeFigure 5 on p. 89.

The protein with the amino acid change preventsleukocytes from reaching and destroying the invadingantigen by interfering with their ability to adhere to theblood vessel walls at the area of infection. This is whycalves with BLAD cannot fight infections and die earlyin life.

Using molecular marker technology, it has beenpossible to identify the heterozygous carriers of BLAD

and remove those individuals from the breeding stock.

As a result, the disease has been virtually eliminatedfrom the Holstein cattle breed.

The defective allele causing BLAD has not been foundin breeds other than Holsteins. However, a similar formof the genetic disorder has been described in humans.


MAS-3

Cattle that have the Bb alleles are carriers of BLAD.

Affected cattle have two copies of the b allele.


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MAS-3

Figure 3







When the nucleotide adenine (a) is replaced by guanine (g) in the DNA strand, the amino acid glycine (gly) is

produced instead of the correct amino acid aspartic acid (asp). The result is the BLAD condition in cattle.


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88

Student Handout



MAS-3

Figure 4

Figure 4 shows the DNA fragments produced by normal cattle, heterozygous BLAD carriers, and


recognizes the tcga sequence and cuts (t / cga) between the t and c nucleotides. When guanine

(g) replaces adenine (a), the tcga sequence is replaced by tcgg and the enzyme does not cut

the strand.

32 bp 26 bp





BLAD Carrier: 32, 26, and 58 bp segments produced



BLAD Affected: 58 bp segment produced



32 bp 26 bp

nucleotide change

58 bp

nucleotide change

58 bpnucleotide change

58 bp


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MAS-3

Figure 5

The BLAD mutation results in the loss of the TaqI restriction site. See Figure 4. A normal cow will

display two fragments, 26 and 32 bp. A carrier will display three fragments, 26, 32, and 58 bp. A BLAD

infected animal has only a single fragment of 58 bp. Based on a gel photo provided by Marcus E. Kehrli, Jr., DVM,PhD, National Animal Disease Center, USDA-ARS.

Homozygous

Normal

Heterozygous

Carrier

Homozygous

BLAD

Base Pairs (bp)

26 bp

32 bp

26 bp

32 bp

58 bp 58 bp

Agarose Gelof BLAD

PCR ProductDigested with Taq1


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90

Student Handout



MAS-3

Credit Notes

Atherly, Alan G.; Girton, Jack R.; and McDonald, John F.The Science of Genetics. Saunders College Publishing.1999

Basics of Marker Assisted Selection (BMAS). Julius vander Werf, Department of Animal Science, and BrianKinghorn, Twynam Chair of Animal Breeding Technolo-gies, University of New England.

Campbell, Neil A. and Reece, Jane B. Biology. Seventhedition. Pearson-Benjamin Cummings Publications.San Francisco, California. 2005

Doggy DNA: The Power of PCR. 2000 Summer BiologyInstitute: Biodiversity. The Woodrow Wilson Founda-tion Leadership Program for Teachers.www.woodrow.org/teachers/bi/2000/Doggy_DNA/background_for_polymerase_chai.html


Marker-Assisted Selection: Applications to AnimalProduction. The Agbiotech Infosource. AG-WESTBIOTECH INC. Issue 39. October 1998

Olson, Tim. Automated DNA Sequencing and Primer

Design. Department of Animal Sciences, Universityof Florida.


Polking, Gary F. The Polymerase Chain Reaction:Introduction. Introduction to Molecular BiologyTechniques Gen 542A. Iowa State Universitys Officeof Biotechnology. Summer 2003


Terminology based on Campbell, Neil A. and Reece,Jane B. Biology. Seventh edition. Pearson-BenjaminCummings Publications. San Francisco, California.2005

Van Eenennaam, Alison. Marker-assisted selectionbackgrounder. Agriculture and Natural ResourcesResearch and Extension Centers. University ofCalifornia-Davis. 2004

Learn the Language

Antigen

An infectious agent, such as a virus

Autosomal chromosome

All chromosomes, except the sex chromosomes. Adiploid cell has two copies of each chromosome.

Leukocytes

White blood cells

Recessive

An allele (r) that expresses itself in a phenotypeonly in homozygous individuals (rr)

and justice for allThe U.S. Department of Agriculture (USDA) prohibits discrimination in all itsprograms and activities on the basis of race, color, national origin, gender, religion,age, disability, political beliefs, sexual orientation, and marital or family status. (Notall prohibited bases apply to all programs.) Many materials can be made availablein alternative formats for ADA clients. To file a complaint of discrimination, writeUSDA, Office of Civil Rights, Room 326-W, Whitten Building, 14th and IndependenceAvenue, SW, Washington, DC 20250-9410 or call 202-720-5964.

Issued in furtherance of Cooperative Extension work, Acts of May 8 and June 30,1914 in cooperation with the U.S. Department of Agriculture. Stanley R. Johnson,director, Cooperative Extension Service, Iowa State University of Science andTechnology, Ames, Iowa.


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DNA Fingerprinting








an agarose gel.






nylon membrane.



DNA band pattern.





DNA FINGERPRINT.


Overhead Master: MAS-a


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Lesson Module II Marker Assisted Selection Overhead Master: MAS-b

Polymerase Chain Reaction Process

1. Denaturation




95 C.



broken. This results in two

single-stranded pieces of DNA.

2. Annealing







DNA.

3. DNA Synthesis







DNA strand.

Cycle two begins by again

raising the temperature toabout 95 C. to denature the



95 C.

58 C.

72 C.

Cycle One

Primers

(4 bp)

Target DNA

Taq

Taq


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Polymerase Chain Reaction Process

4. Denaturation


strands from cycle one. The

original strands and the strands


the target DNA.

5. Annealing

The primers attach themselves




6. DNA Synthesis





95 C.

58 C.

Cycle Two

72 C.

Target DNA

Taq

Taq

Taq

Taq

Original DNA

Copied DNA

Copied DNAOriginal DNA

Overhead Master: MAS-c


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Lesson Module II Marker Assisted Selection Overhead Master: MAS-d

In the gel pictured, the size of the

fragment increases by one base pair

relative to its position on the gel. The

DNA sequence for the gel is read as

CATTCGAATGCA.

DNA Sequencing

ddC ddT ddA ddG


CATTCGAATGCA

CATTCGAATGC

CATTCGAATG

CATTCGAAT

CATTCGAA

CATTCGA

CATTCG

CATTC

CATT

CAT

CA

C


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Lesson Module II Marker Assisted Selection Overhead Master: MAS-e

BLAD Pedigree

II

I

III

IV

V

inbreeding

Osborndale Ivanhoe

Bovine leukocyte adhesion deficiency

(BLAD) can only manifest itself when

heterozygous male and female

descendants of Ivanhoe are bred and

two recessive alleles, one from each

parent, combine to produce calves

with a homozygous recessive

condition.

BLAD Pedigree


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Lesson Module II Marker Assisted Selection Overhead Master: MAS-f

BLAD Punnett Square

B b

B BB bB

b bB bb

Cattle that have the Bb alleles arecarriers of BLAD. Affected cattle have

two copies of the b allele.


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Lesson Module II Marker Assisted Selection Overhead Master: MAS-g







Protein Synthesis and BLAD

When the nucleotide adenine (a) is replaced

by guanine (g) in the DNA strand, the amino

acid glycine (gly) is produced instead of the

correct amino acid aspartic acid (asp).

The result is the BLAD condition in cattle.


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Lesson Module II Marker Assisted Selection Overhead Master: MAS-h


This figure shows the DNA fragments produced by normal cattle, heterozygous BLAD carriers, and


recognizes the tcga sequence and cuts (t / cga) between the t and c nucleotides. When guanine

(g) replaces adenine (a), the tcga sequence is replaced by tcgg and the enzyme does not cut

the strand.

32 bp 26 bp


5gtgaccttccggagggccaagggctaccccat / cgac

Documents

Marked Assisted System