Bio6–PrinciplesofGeneticInheritanceLab

OverviewInthis laboratoryyouwill learnaboutthebasicprinciplesofgenetic inheritance,orwhatiscommonlyreferredtoas“genetics”.Atrueappreciationofthenatureofgeneticinheritancewillrequiresolvingofavarietyofgeneticsproblems,andtodosoyouwillneedtounderstandseveralrelatedconcepts,somewhichshouldbefamiliarandotherswhichmaybenewtoyou.Thusyouwillbeginthislabbyexaminingtheconceptsofgenes,gameteproductionbymeiosis,andprobability.Youwillthenusetheseconceptstoworkthroughaseriesofgeneticsproblemsaddressingvariousaspectsofgeneticinheritanceinplantsandanimals.

Part1:KEYGENETICCONCEPTSWeallknowthatwhenlivingorganismsreproduce,theiroffspringaremuchliketheirparents.Chickensdon’tgivebirthtolizardsandappletreesdon’tgiverisetopinetrees.Sowhatisthebiologicalbasisforthisobviousreality?Youprobablyalreadyknowthishastodowithgenes,genesoneinheritsfromone’sparents.Howevertheprocessofpassingongenesfromonegenerationtothenextismorecomplexthanitmayappear.Thesimplestformofgeneticinheritanceinvolvesasexualreproduction.Thisisthecasewhenasingleparentorganismpassesitsgenestooffspringwhicharebasicallyclonesoftheparent(i.e.,genetically,and for themost part, physically identical). Although thismode of reproduction is quite convenient(imagineifyoucouldsimplyhavechildrenidenticaltoyourself,nopartnernecessary),ithasoneextremelysignificantshortcoming:NOgeneticdiversity.Forsomespeciesasexualreproductionworksquitewell,however formostplantsandanimals (includinghumans) this justwon’tcut it,geneticdiversity is tooimportantforthelongtermsurvivalofspecies.Sohowisgeneticdiversityproduced?Theanswerissexualreproduction: theproductionofgametes(sperm and eggs) bymeiosis followed by the fusion of sperm and egg (fertilization) to form a new,genetically unique individual. Sexual reproduction essentially “shuffles” the genes of each parentproducingauniquecombinationofparentalgenesineachandeveryoffspring.Thisisthesortofgeneticinheritancewewillfocuson,geneticinheritancebasedonsexualreproduction.Throughsexualreproduction,eachoffspringinheritsacompletesetofgenesfromeachparent,howeverthestudyofgeneticinheritanceisgenerallylimitedtooneortwogenesatatime.Thuswhenyoubegintoworkwithgeneticsproblemsyouwillfocusinitiallyonasinglegeneattime,andthenlearnhowtofollow the inheritanceofmore thanone gene. To focuson largenumbersof geneswouldbe rathercomplicatedandisnotnecessaryforourpurposes.Beforeyoubegintoexaminegeneticinheritanceviageneticsproblems,youwillneedtounderstandsomeimportant concepts that are central to the process: the nature of chromosomes, genes and geneticalleles; the process of gamete production bymeiosis; and the concept of probability. Once theseconceptsandtheirassociatedterminologyareclear,youwillthenbereadytoimmerseyourselfintotheworldofgeneticinheritance.

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Chromosomes,GenesandAllelesAsyoulearnedinthepreviouslab,chromosomesareextremelylongpiecesofDNAinthenucleiofcellsthatcontainuptoathousandormoregeneseach.Eachspeciesoforganismhasacharacteristicnumberofchromosometypes:distinctchromosomeseachhavingauniquesetofgenesandauniquelength.Forexample, the fruit fly Drosophila, an organism used in many genetic studies, has only 4 types ofchromosomes–3autosomes(non-sexchromosomes)andtheXandYsexchromosomes.Humanbeings(Homo sapiens) on the other hand have 23 types of chromosomes – 22 autosomes and the sexchromosomes(XandY)asillustratedinthehumanmalekaryotypeshownbelow(noticetheXandYsexchromosomes):

Notice one more thing about this humankaryotype:therearetwoofeachautosomeaswellas twosexchromosomes. This isbecausehumanbeingsarediploid,whichmeanshavingtwoofeachchromosometype.Mostplantsandanimalsareinfactdiploid,andasweinvestigatetheprocessof genetic inheritancewewill onlyconcernourselveswithdiploidspecies.Howeveryoushouldbeawarethatnotallorganismsarediploid.Somearenormallyhaploid(oneofeachchromosome)suchasthefungi,andsomemayhavemorethantwoofeachchromosome(e.g.,four of each = tetraploid, eight of each =octoploid) as seen in a fair number of plantspeciesaswellasafewanimalspecies.As shown in thediagramto the left, genes arediscrete sections of chromosomal DNAresponsible for producing a specific protein orRNAmolecule.Theprocessofgeneexpression,theproductionofproteinorRNA fromagene,willbeaddressedinafuturelab.

The functional protein or RNA molecule produced from a particular gene is its gene product. It isimportant to realize that theDNAsequenceofagene,andhence itsgeneproduct, canvarywithinaspecies.Inotherwords,aparticulargeneinaspeciessuchasHomosapienscanhavedifferentversions,whatarereferredtoingeneticsasalleles.Thegeneproductsproducedfromanorganism’sgeneticallelesaccountforitsphysicalandbehavioralcharacteristics,whatwecollectivelycallanorganism’straits.Thespecific traitsan individualexhibits,whetherphysicalorbehavioral,are referred toas the individual’sphenotype.Thespecificgeneticallelesanindividualhasforaparticulargeneistheindividual’sgenotype.Asyoushallsoonsee,anindividual’sphenotypeislargelydeterminedbyitsgenotype.

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Sincediploidorganismshavetwoallelesforeachgene,anindividualcanhavetwocopiesofthesamealleleforageneortwodifferentalleles.Iftheallelesarethesame,theindividualissaidtobehomozygousforthatgene.Iftheallelesaredifferent,theindividualissaidtobeheterozygousforthatgene.Whenanindividualisheterozygousforagene,oneallelemayoverrideor“mask”theotherallelebydeterminingthephenotyperegardlessoftheotherallele.Inthissituation,theallelethatdeterminesthephenotypeissaidtobedominantwhiletheotheralleleissaidtoberecessive.Thetermsdominantandrecessivearerelativetermsjustlikethetermsbigandsmall.Somethingisonlybigorsmallinrelationtosomethingelse,andinthesamewayanalleleforageneisonlydominantorrecessiveinrelationtoanotheralleleforthesamegene.

MeiosisandtheProductionofGametesSexualreproductionindiploidplantandanimalspeciesrequirestheproductionofhaploidgametesbytheprocessofmeiosiswhichisillustratedbelow.

Sincemeiosiswascoveredinthepreviouslab,wewon’treviewtheprocessinmuchdetailotherthantoremindyouofseveralkeypointsthatpertaintogeneticinheritance:

1) Diploid organisms have two of each chromosome type, one haploid set of chromosomesinheritedfromthemother(maternalchromosomes)andanotherhaploidsetinheritedfromthefather(paternalchromosomes),andthustwoallelesforeachgene,onematernalalleleandonepaternalallele.

2) Thehaploidgametesproducedbymeiosis(eggorsperm)containoneofeachchromosometype(e.g.,oneofeachautosomeandonesexchromosome),andthuscontainonlyonealleleforeachgene.

3) Itiscompletelyrandomwhichchromosomeofagiventype,maternalorpaternal,endsupinagameteproducedbymeiosis.Thusitiscompletelyrandomwhichalleleforaparticulargene(maternalorpaternal)endsupinagamete.

4) Sinceanindividualhasonlytwoallelesforeachgene,thereisa50%chancethateitherallele(maternalorpaternal)willendupinagivengamete.

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Thesepointsarecentral to thestudyofgeneticsas itpertains tosexual reproduction. Diploidparentorganisms produce large numbers of haploid gametes, any of which may fuse at the moment offertilization toproduceadiploidzygote, ageneticallyuniquenew individual. It is thecombinationofallelesinheritedfromeachparentthatdeterminethegenotypeandphenotypeofeachnewoffspring.Giventhevarietyofpossiblegenotypesandphenotypesfortheoffspringofanytwoparents,geneticsmustalsoaddresstheprobabilityofeachpossibility.Sobeforeyoubeginworkingwithgeneticcrossesitisimportantthatyouunderstandbasicconceptsofprobability.

Probability

Tounderstandgeneticinheritance,youneedtohaveabasicunderstandingofprobability. Probabilityreferstothelikelihoodthatsomethingwillhappenasopposedtowhatactuallyhappens.Forexample,weallknowthatasinglecoinfliphasa50%chanceofbeing“heads”or“tails”,thustheprobabilityofheadsis50%or0.5or½asistheprobabilityoftails.Howeverwealsoknowthatwecannotknowwhattheoutcomeofasinglecoinflipwillbe. Allwecando ispredictthe likelihoodorprobabilityofeachpossibleoutcome,inthiscaseheadsortails.Whenanalyzingtheinheritanceofgenesyouwillalsobedealingwithprobabilities.Whentwoorganismsmateandproduceoffspringwecannotknowwhatgeneticalleleswillbeinheritedbyagivenindividual(genotype) or its physical characteristics (phenotype). We can only predict the likelihood of variouscharacteristicsbasedontheprobabilityofinheritingparticulargeneticallelesfromeachparent.Thenexttwo exercises will help illustrate the nature of probability and help prepare you to solve geneticsproblems.

Exercise1A–ProbabilityandsamplesizeInthisexerciseyouwillperformseveralsetsofcoinflips.Thiswillbedonetocomparethepredictedoutcomebasedonprobabilitytotheactualoutcome,andhowthisrelatestosamplesize(i.e.,thenumberofrepetitions):

1. Onyourworksheet,determinetheexpectednumbersofheadsandtailsforsamplesizesof10and100coinflips(theprobabilityofheadsis50%or0.50,asistheprobabilityoftails).

2. Perform10setsof10coinflips,recordingtheresultsonyourworksheet.

3. Combinetheresultsforall10setsofcoinflipsonyourworksheettogetatotalnumberofheadsandtailsoutof100coinflips.

4. Onyourworksheet,comparetheactualresultstoexpectedresultsforeachsetof10coinflipsaswellasthecombinedsetof100coinflipsandanswerthecorrespondingquestions.

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Whenflippingacointhereareonlytwopossibleoutcomes, each with the same probability.Sometimes, however, youmust dealwithmorethantwopossibleoutcomes,eachwithadifferentprobability. This is thecase, forexample,whenyourollapairofdice,forwhichthereareelevenpossibleoutcomesintermsofthetotalsum:2,3,4,5,6,7,8,9,10,11and12.

Theprobabilitiesofeachpossiblesumofthedicearenotthesame.Ifyouhaveeverplayedaboardgameorcrapsyouknowforexamplethatarolltotaling7ismuchmorelikelythanarolltotaling2.Sowhyaretheprobabilitiesdifferent?Ithastodowithhowmanydifferentcombinationsadduptoagivensum.Forexample,torollapairofdiceandgetasumof2,bothdicemustshowa“1”.Thereisnootherwaytorollatotalof2withapairofdice.Therearesixdifferentways,however,torolla7: Reddie Blackdie total 1 6 7 6 1 7 2 5 7 5 2 7 3 4 7 4 3 7Sohowdoyoudeterminetheprobabilitiesofeachpossiblesumof thedice? Asshown inthematrixbelow,thereare6x6=36differentcombinationsyielding11possiblesums:

black1 black2 black3 black4 black5 black6red1 2 3 4 5 6 7red2 3 4 5 6 7 8red3 4 5 6 7 8 9red4 5 6 7 8 9 10red5 6 7 8 9 10 11red6 7 8 9 10 11 12

Eachdiehassixpossibleoutcomes(1through6),eachofwhichareequallylikelyandeachofwhichcanbe paired with any roll on the other die. Looking carefully at thematrix, you can see there is onecombinationoutof36thatyieldsasumof2(probability=1/36or0.028or2.8%),twocombinationsthatyieldasumof3(2/36or0.056or5.6%),threecombinationsthatyieldasumof4(3/36or0.083or8.3%),andsoon.If you think about it, the rolling of a pair of dice is actually two independent events in combination.Whetheryourollthereddieorblackfirst,orbothatthesametime,isirrelevant.Theprobabilityofrollingasumof2(“snakeeyes”)forexampleisstill1/36,regardlessofthetiming.Thisleadstoanimportantruleregardingprobabilitycommonlyreferredtoasthe“productrule”:

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Theprobabilityofaspecificcombinationoftwoormoreoutcomesisthe

productoftheprobabilitiesofeachindividualoutcome.

Inour“snakeeyes”example,theprobabilityofrolling1onthereddieis1/6andtheprobabilityofrolling1ontheblackdieis1/6,thustheprobabilityofrolling1onboththereddieANDtheblackdieis1/6x1/6whichequals1/36.Anotherexamplewouldbetheprobabilityofflippingtwocoinsandhavingbothturnup“heads”,orinotherwordstheprobabilityofcoin1ANDcoin2bothyielding“heads”.Usingtheruleabovetheprobabilitywouldbe1/2x1/2whichis1/4.Anotherimportantrulecommonlycalledthe“sumrule”or“additionrule”addressesprobabilitywhentherearemultiplewaystoarriveataparticularoutcome:

Theprobabilityofanoutcomethatcanoccurinmultiplewaysisthesumoftheprobabilitiesofeachindividualoutcome.

Forexample,wehavealreadydeterminedthattheprobabilityofrollingasumof3withapairofdiceis2/36sincethereare2differentwaystoobtainasumof3:red1plusblack2(1/36)ORred2plusblack1(1/36).Eachoutcomeyieldsarollof3andbyaddingbothprobabilitiesyouarriveatanoverallprobabilityof2/36.Anotherexamplewouldbetheflippingofacoin.TheprobabilityofflippingheadsORtailsonagiventossisclearly100%: 50%heads+50%tails=100%. Whatabouttheprobabilityof flippingtwocoinsandgettingheadsforoneandtailsfortheother?Thereareactuallytwowaysthiscanoccur:

headsfromcoin1andtailsfromcoin2(50%x50%=25%)

OR

tailsfromcoin1andheadsfromcoin2(50%x50%=25%)

Inthiscasetheoverallprobabilityofgettingoneheadsandonetailsfromtwocoinflips is25%+25%whichequals50%.Youarenowreadyforthenextsetofexercisesinprobability,andkeepinmindthattheprobabilityofXANDYoccurringmeansyoumultiply,andtheprobabilityofXORYoccurringmeansyouadd.Exercise1B–DiceBaseballInthisexerciseyouwillplayagameofbaseballusingapairofdice.Thiswillrequirethatyouconsidertheprobabilitiesofvariousoutcomesforan“atbat”andmatchthesewithprobabilitiesinrollingapairofdice:

1. Onyourworksheet,determinetheprobabilityofeachpossiblesumforapairofdice.

2. Considertheprobabilitiesofeachsumanduseyourworksheettodesignyourowndicebaseballgameinwhichagivensumofthediceequalsaparticularoutcomeofan“atbat”(e.g.,7=out,12=homerun).Doyourbesttoberealistic–yourinstructorwillexplainthisfurther.

3. Arrangeteamsof1to2peopleandplayanineinninggamekeepingscoreonyourworksheet(you

cangotoextrainningsifnecessary).210

Part2:SOLVINGGENETICSPROBLEMS

ImportantToolsforSolvingGeneticsProblemsBeforeyoubegintosolvegeneticsproblemsitisimportantthatyouarefamiliarwiththetypesofsymbolsusedtorepresentgeneticallelesaswellastwoimportanttoolsyouwillusetosolvegeneticproblems.Let’sfirstaddressthesymbolsforgeneticalleles:

SymbolizingAllelesGeneticallelesarecommonlyrepresentedbyasingleletter,withthedominantallelebeinguppercaseandtherecessiveallelebeinglowercase:

dominantallele–Arecessiveallele–a

Whenworkingwithmorethanonegeneinageneticproblem,theallelesofeachgenewillberepresentedusingadifferentletter:

Gene1 Gene2dominantallele–A dominantallele–Brecessiveallele–a recessiveallele–b

Whenwritingthegenotypeforanindividualitisbesttokeeptheallelesforaparticulargenenexttoeachotherandtolistadominantallelebeforearecessiveallele:

AaBb

Ifanalleleisunknown,indicatethiswithanunderscore:

A_bb

Whentherearemorethantwoallelesforagene,differentsuperscriptsonasingleletterarecommonlyusedtodistinguishalleles:

alleles–Cx,Cy,Cz

genotype–CxCyWhenageneislocatedontheXchromosome(“X-linkedgene”),allelesarerepresentedbysuperscriptsonacapitalX.Inmalesa“Y”withoutasuperscriptisusedtosymbolizetheYchromosomewhichwouldnothaveanallelefortheX-linkedgene:

femalegenotypeforanX-linkedgene–XAXamalegenotypeforanX-linkedgene–XAY

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PedigreesForsomeproblemsitwillbehelpfultodiagramapedigreeofallfamilymembersrelevanttotheproblemasshownhereforthreegenerationsofahypotheticalfamily.As you can see, males are represented bysquares and females by circles. Individualsexhibiting the phenotype of concern have afilled-insymbol.Pedigreesareusefulinthattheyallowyou to representall familial relationshipsvisuallysothatyoucanfillinanddeduceasmanygenotypes as possible. In addition, they alsoallowyoutodeducethemodeofinheritanceofthe condition in question (i.e., dominant vsrecessive condition, sex-linked vs autosomalgene).Onmany occasions youmay find it simpler todiagram your pedigree without shapes andcolors, and simply represent individuals in thepedigreebytheirgenotypes(orasmuchofthegenotypeasyouknow)asshowntotheright.

AaxAa

A_,A_,aa

PunnettsquaresAnotherextremelyusefultoolisthePunnettsquare,whichisbasicallyagridormatrixshowingallpossiblecombinationsofgametesfromeachparenttoproduceoffspring.APunnettsquareisusefulifyouknowthegenotypesoftheparentsbeingcrossed.OnceyoudetermineallpossiblehaploidgametesforeachparentyousimplyfilloutthePunnettsquareasshownbelow:

cross:Aa(female)xAa(male)

gametes:A&a(female);A&a(male)

ByplacingthedifferentgametesforeachparentalongeithersideofaPunnettsquare,itbecomeseasytofillinallpossiblecombinationsofeggandsperm.Allthat’sleftistodeterminetheprobabilitiesofeachgenotypeand/orphenotype.Thisgetsabitmorecomplicatedwhendealingwithtwogenes,howevertheprincipleisthesameasshowninthenextexample:

A a

A AA Aa

a Aa aa

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cross:AaBbxAabb

gametes:AB,Ab,aB&ab;Ab&ab

AB Ab aB ab

Ab AABb AAbb AaBb Aabb

ab AaBb Aabb aaBb aabb

NoticethatthePunnettsquaredoesn’thavetobeasquareatall, itsimplyneedstoaccommodateallparentalgametesandtheirpossiblecombinations,nothingmore.

TypesofGeneticsProblemsSolving geneticsproblems involvesusing the knowncharacteristicsof some individuals todeduce theprobability of unknown characteristics of other individuals. The characteristics given will relate tophenotype(physicaldescription),genotype(geneticalleles),orboth.Ingeneral,thegeneticsproblemsyouwillbeaskedtosolveareoftwobasictypes:

1) Characteristicsoftheparentsaregivenandyoumustdeterminetheprobabilitiesofeachpossibletypeofoffspring.

2) Characteristicsoftheoffspringaregivenandyoumustdeducecharacteristicsoftheparents.Regardlessofwhichtypeofproblemyouaresolving,youwillneedtousetheprincipleswehaveaddressedinPart1(genes,alleles,meiosis,gametes,probability),thetoolsreviewedintheprevioussection,andabitofcommonsense.Theapproachtosolvingeachtypeofproblem,however,isslightlydifferent.

SolvingType1problemsInType1problems,youwillbegivencharacteristicsoftheparentsbeingcrossedandaskedtofigureouttheprobabilitiesofallpossiblegenotypesand/orphenotypesintheoffspring.Solvingsuchproblemswillrequire4basicsteps:

1) Determinethegenotypesofbothparentsandwriteoutthecross.• genotypesmaybegivenintheproblemoryoumayhavetodeducethem

2) Determinethegenotypesofallpossiblegametesproducedbyeachparent,andwritethemonadjacent

sidesofaPunnettsquare.• gametesarehaploidandthushaveonealleleforeachgeneofinterest

3) FillinallpossibleoffspringgenotypesinthePunnettsquare.

• allpossiblecombinationsofgeneticallydifferenteggsandspermmustbeaccountedfor

4) Determinetheprobabilitiesofeachpossiblegenotype/phenotypeintheoffspring.

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Toillustratethesestepslet’ssolvethesampleproblembelow: In pea plants, purple flower color is determined by a dominant allele P and white flower color is determinedbyarecessiveallelep.Iftwoheterozygouspeaplantswithpurpleflowersarecrossed,what aretheprobabilitiesofeachpossiblegenotypeandphenotypeintheoffspring?Step1:Writeoutthegenotypesofeachparentinthecross.

Due to the phrase “two heterozygous pea plants with purple flowers are crossed” we know that the genotypesoftheparentsbeingcrossedmustbe:

PpxPp

Step2:Determinethegenotypesofalldifferentgametesproducedbyeachparent,andarrangethemonadjacentsidesofaPunnettsquare.

Duetomeiosis,eachparentwillproducehaploidgametescontainingeitherthePalleleorthepallele, eachinequalproportions(50%ofeach):

genotypesofmalegametes–P,p

genotypesoffemalegametes–P,p

P pP

p

Step3:Determineallpossiblegenotypesintheoffspring.

SimplyfillinthePunnettsquarewiththegenotypesofallpossibleunionsofgametes:

P(1/2) p(1/2)P(1/2)

PP(1/4) Pp(1/4)

p(1/2)

Pp(1/4) pp(1/4)

NOTE:Theprobabilitiesofeachgameteandeachresultingoffspringgenotypeareshowninparentheses,

thoughthisnormallywouldnotbenecessary.

Step4:Summarizetheprobabilitiesofeachpossiblegenotype/phenotypeintheoffspring.

ThePunnettsquareaboverevealsthatthereareonly4differentcombinationsofspermandeggyielding 3possiblegenotypesand2possiblephenotypes.Probabilitiesaredeterminedbyfollowingtheproduct rule(e.g.,PandPinagamete=½x½=¼)andthesumrule(e.g.,PporPp=¼+¼=½):

Genotypes Phenotypes¼PP ¾purpleflowers(¼+¼+¼)

½Pp(¼+¼) ¼whiteflowers ¼pp

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Onceyoucompletestep4youwillhaveyouranswer.Whetheryourepresenttheprobabilitiesasfractions,decimalsorpercentagesisuptoyou.Allthatmattersisthatthevalueforeachprobability,regardlessofhowitispresented,iscorrect.Nowyouarereadyforthenextexerciseinwhichyouwillsolvesimilarproblemsonyourown…Exercise2A–Type1geneticsproblemsForthefollowingtwoproblems,tallplantheight isdeterminedbyadominantallele(T)andshortplantheight isdeterminedbyarecessiveallele(t).Determinetheprobabilitiesforallpossiblegenotypesandphenotypesoftheoffspringresultingfromeachcross.

1. homozygousdominant(TT)xheterozygous(Tt)

2. heterozygous(Tt)xhomozygousrecessive(tt)

SolvingType2problemsInType2problemsyouwillbegiventhecharacteristicsofoffspringandmustfigureoutthecharacteristicsofoneorbothparents.Tosolvesuchproblemssimplyfollowthesetwosteps:

1) Diagramapedigree containingall individuals in theproblem,and indicate thegenotypesofasmanyindividualsaspossibleusingtheinformationprovided.

2) Usethepedigreetodeducetheinformationtheproblemasksfor.

Let’sdoasampleproblemofthistype: In cats, black fur color is determined by a dominant allele B and white fur color is determined by a recessivealleleb.Awhitecatgivesbirthto3whitekittensand2blackkittens.Whatisthegenotypeand phenotypeofthefather?Step1:Diagramapedigreeshowingthegenotypes.

bbx__

3bb,2B_

Thephenotypesgivenforthemothercatandherkittensindicatethegenotypesabove(forthisproblemafullpedigreeisnotnecessary,asimplediagramsuchasthiswillsuffice).

Step2:Deducemissinggenotypesinpedigreetoarriveattheanswer.

Sincesomeofthekittensarewhite(bb),bothparentsmustcarrythewhiteallele(b).Sincesomeofthekittensareblack(B_),atleastoneparentmusthaveablackallele(B)whichinthiscasecanonlybethefather. Thus the father catmust beblackwith the genotypeBb,which is the answer to theproblem.Thoughnotaskedintheproblem,youcanalsodeducethattheblackkittensallhavethegenotypeBbsince

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allkittensreceiveawhiteallele(b)fromthemother.Keepinmind,itisalwaysagoodideatoconfirmyouranswerwithaPunnettsquarebasedontheparentalgenotypesyouhavededuced.

bbxBb

3bb,2Bb

Exercise2B–Type2geneticproblemsAlbinisminhumanbeings(lackofskinpigment)isduetoarecessiveallele.UseAtorepresentthedominant“normal”alleleandatorepresenttherecessivealbinismalleleinsolvingthetwoproblemsbelow:

1. Analbinochildisborntotwoparentswithnormalskinpigmentation.Whatarethegenotypesoftheparents?

2. Analbinowomanhas8childrenwiththesameman,3withnormalpigmentationand5thatarealbino.Whatcanyouconcludeaboutthefather?

IncompleteDominance,Codominance,andMulti-AllelicInheritance

Theproblemsyouhavesolvedsofararerelativelystraightforwardasfarasgeneticsproblemsgo.Ineachcasetheprobleminvolvedonlyonegene,twoalleles,andthealleleswereeitherdominantorrecessive.Thenext fewproblems youwill solve involve somenewconcepts that are frequently encountered ingenetics.The first issue is incomplete dominance, which occurs when a heterozygous genotype results in aphenotypethatisintermediatebetweenthetwohomozygousphenotypes.Forexample,twodifferentalleles forasinglegeneresult inthreedifferentcolorsofcarnationflowers.TheRalleleresults inredpigment production (red) and the r allele results in a lack of pigment production (white). The threepossiblegenotypesandphenotypesareshownbelow:

Genotype Phenotype

RR redcarnations

Rr pinkcarnations

rr whitecarnationsThisclearlyisnotcompletedominance,otherwisetheheterozygotes(Rr)wouldhaveredflowers.Thewhitealleleisstillconsideredrecessivesinceitresultsinthelackofpigment.Howeverthereislessredpigmentwithoneredallele(Rr)thanwithtworedalleles(RR).Thusinthiscasewewouldsaythattheredalleleisincompletelydominantoverthewhiteallele.Thenicethingaboutincompletedominanceisthatthephenotyperevealsthegenotype,somethingyouwillappreciatewhenyousolveproblemssuchasthoseinthenextexercise.

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Exercise2C–IncompletedominanceInsnapdragons,redflowersareproducedbyhomozygousRRplantsandwhiteflowersareproducedbyhomozygousrrplants.Heterozygousplants(Rr)producepinkflowers.Solvethefollowingproblemsonyourworksheet,beingsuretoshowallyourwork:

1. Indicatetheprobabilitiesofeachgenotypeandphenotyperesultingfromtheacrossbetweentwopinksnapdragons.

2. Acrossbetweentwosnapdragonsproduced15offspringwithredflowersand17offspringwithpinkflowers.Whatarethegenotypesandphenotypesoftheparents?

Sometraitsinvolveagenethathasmorethantwoalleles.OnesuchexampleishumanABObloodtypewhichisdeterminedbyasinglegenewiththreedifferentalleles.The“A”and“B”alleleseachresultinadifferentglycoproteinonthesurfaceofredbloodcells.Whenpresentinthesameindividual,bothallelesareexpressedequallyresultingintheABbloodtype.Thisisanexampleofcodominance.The“O”alleleproduces no glycoprotein and thus is recessive to both the “A” and “B” alleles. Based on theserelationshipsthethreeallelesaresymbolizedIA,IB,andi,withthevariousABOgenotypesandphenotypessummarizedbelow:

Genotype BloodType

IAIA A IAi A IBIB B IBi B IAIB AB ii O

RefertotheinformationaboveasyousolvegeneticproblemsinvolvinghumanABObloodtypeinthenextexercise…Exercise2D–Multi-allelicinheritanceandcodominanceSolvethefollowingproblemsonyourworksheet,beingsuretoshowallyourwork:

1. IndicatetheprobabilitiesofeachgenotypeandphenotypeinthechildrenofawomanwithbloodtypeOandamanwithbloodtypeAB.

2. AwomanwithbloodtypeAandamanwithbloodtypeBhave3children,oneeachwithbloodtypesA,BandO.Whatarethegenotypesoftheparents?

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SexLinkageAll of the geneticproblems youhave solvedup to this point involve genesonanautosome, i.e., anychromosomethatisnotasexchromosome.Forgenesonautosomes,inheritancepatternsaregenerallyno different for males and females. This is not the case, however, for genes on the X and Y sexchromosomes. Females have two X chromosomes and thus two alleles for each gene on the Xchromosome.MaleshaveonlyoneXchromosomealongwithasingleYchromosome. Withveryfewexceptions,genesontheXchromosomearenotfoundontheYchromosomeandviceversa.ThusmaleswillhaveonlyonealleleforeachgeneontheXchromosome.As a result of this disparity, the inheritancepatterns for geneson the sex chromosomes are typicallydifferent for males versus females. This phenomenon is called sex linkage, which refers to geneticinheritancethatdiffersdependingonthesexoftheindividual.GenesontheXchromosome,referredtoasX-linked,andgenesontheYchromosome,referredtoasY-linked,exhibitsexlinkageorsex-linkedinheritance.BecausetheYchromosomehassofewgenes,wewilllimitourfocustoX-linkedgenes.

TheapproachtosolvinggeneticproblemsinvolvingX-linkedgenesisbasicallythesameasforautosomalgenes.TheallelesforX-linkedgenesaresymbolizedusingan“X”withasuperscriptrepresentingtheallele(e.g.,XA).Inplaceofasecondalleleinmalesthesymbol“Y”isused,withnosuperscript,torepresenttheYchromosome.Let’snowlookatasampleproblem:

Inhumans,ageneticalleleresponsibleforcolor-blindnessisrecessiveandX-linked.Ifacolor-blindwoman andamanwithnormalcolorvisionplantohavechildren,whatsortofcolorvisionwouldyoupredict in theirchildren?Step1:Writeoutthegenotypesofeachparentinthecross.

UsingXBandXbtorepresentthedominantnormalandrecessivecolor-blindnessalleles,respectively,the onlypossiblegenotypesfor“acolor-blindwomanandamanwithnormalcolorvision”are:

XbXbxXBY

Step2:Determinethegenotypesofalldifferentgametesproducedbyeachparent.

Bymeiosis,thewomanproduceseggsallwiththegenotypeXb.ThemanproducesspermcontaininghisX chromosome(XB)orhisYchromosome:

genotypeoffemalegametes–Xbgenotypesofmalegametes–XBorY

Step3:Determineallpossiblegenotypesintheoffspring.

SimplydrawaPunnettsquareaccountingforallparentalgametesandfillinallpossiblegenotypesforthechildren:

XB YXb

XBXb XbY

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Step4:Summarizetheprobabilitiesofeachpossiblegenotype/phenotypeintheoffspring.

Whenevertheresultsofacrossdifferformalesandfemales,youshouldindicatetheprobabilitiesfor maleandfemalegenotypes/phenotypesseparately.Theanswertothisproblemthereforeisbest expressedasfollows:

100%oftheirgirlswillhavenormalcolorvision(thoughallwillbecarriers)

100%oftheirboyswillbecolor-blindHopefully it isclear that themethodofsolvinganX-linkedgeneticproblem is reallynodifferent thansolvinganyotherproblem.SomegeneralfeaturesofX-linkedgenesyoumayhaverealizedbysolvingthisproblemandshouldkeepinmindare:

• sonsalwaysreceivetheirXchromosomefromtheirmother(dadcontributesY)

• daughtersalwaysreceivetheirfather’sXchromosome

• femalesarediploidforX-linkedgenesandcanbecarriersofrecessivealleles

• malesarehemizygousforX-linkedgenes(haveonlyoneallele)andthusshowthephenotypeofthesinglealleleinheritedfromthemother

Exercise2E–X-linkedinheritanceHemophilia,likecolor-blindness,isanX-linkedrecessivecondition.Solvethefollowingproblemsonyourworksheet,beingsuretoshowallyourwork:

1. Anormalwomanwhosefatherisahemophiliacmarriesanormalman.Whataretheoddsthathemophiliawillafflictanygivenmalechild?anygivenfemalechild?

2. Acouplehasthreechildren,anormalboyandaboyandgirleachwithhemophilia.Whatcanyousayabouttheparents?

TheInheritanceofMultipleGenesWhenfollowingtheinheritanceofmorethanonegene,theapproachissimilartosinglegeneinheritance.Thetrickistocorrectlydeterminealldifferentparentalgametesandkeeptrackoftheallelesforeachgene.Toillustratehowthisisdone,let’sdoasampleproblem: Inpeaplants,flowercolorisdeterminedbyadominantpurpleallelePandarecessivewhiteallelep,and plantheightisdeterminedbyadominanttallalleleTandarecessiveshortallelet.Atallpeaplantwith purpleflowersthatisheterozygousforbothgenesiscrossedwithashortplantwithpurpleflowersthatis heterozygousforflowercolor.Whataretheprobabilitiesofeachpossiblephenotypeintheoffspring?

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Step1:Writeoutthegenotypesofeachparentinthecross.

TtPp(tallpurpleplant)xttPp(shortpurpleplant)

Thesearetheonlygenotypesconsistentwiththeinformationgivenintheproblem.

Step2:Determinethegenotypesofalldifferentgametesproducedbyeachparent.

Bymeiosis,eachparentwillproducehaploidgametescontainingonealleleforeachgene.Thedifferent gametesproducedbyeachparentare:

TP,Tp,tP,tp(tallpurpleplant)

tP,tp(shortpurpleplant)

Step3:Determineallpossiblegenotypesintheoffspring.

DrawaPunnettsquareshowingthegametesforeachparentandfillinthegenotypesofallpossibleoffspring:

TP Tp tP tptP

TtPP TtPp ttPP ttPp

tp

TtPp Ttpp ttPp ttpp

Step4:Summarizetheprobabilitiesofeachpossiblephenotypeintheoffspring.

BasedonthePunnettsquare,3outof8genotypesgivea“tallpurple”phenotype,3of8“shortpurple”,1of8“tallwhite”,and1of8“shortwhite”.Thustheprobabilitiesofallpossiblephenotypesintheoffspringare:

3/8–tallwithpurpleflowers3/8–shortwithpurpleflowers1/8–tallwithwhiteflowers1/8–shortwhiteflowers

Problemssuchasthiscangetrathercomplicated,howeverifyoukeeptrackofallallelesproperlyateachstep,youcaneasilysolveanysuchproblem.

Exercise2F–GeneticsproblemsinvolvingtwogenesSolvethefollowingproblemsonyourworksheet,beingsuretoshowallyourwork:

1. Indicatetheprobabilitiesofallphenotypesintheoffspringoftwotallpeaplantswithpurpleflowers,bothofwhichareheterozygousforbothgenes.

2. Anormalwomanwithanalbinofatherandnofamilyhistoryofhemophiliaplanstohavechildrenwithahemophiliacmanwithnormalskinpigmentationwhosemotherisalbino.Determinetheprobabilitiesofallpossiblephenotypesintheirchildren.

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InterpretingPedigrees

Asmentionedearlier,pedigreescanbeaveryusefultooltovisualizephenotypesinafamilylineage(seepage8ofthislab).Onceapedigreehasbeenproducedonecandeducethelikelymodeofinheritanceofthegeneticcondition(i.e.,dominantvsrecessive,autosomalvsX-linked)andasaresultdeterminethegenotypesofvariousindividualsinthepedigree.Toillustratethislet’sconsidersomeexamples:

Thispedigreerevealsanautosomal recessivemodeofinheritance, i.e., the gene responsible for the geneticconditionisrecessiveandlocatedonanautosome(non-sex chromosome). The condition is clearly recessivesince unaffected couples have affected children. Thiswouldnotbepossiblewithadominantallele.Youwillalsonoticethatthecondition“skipsageneration”whichalso is only possible with a recessive allele. It also

appears tobeautosomalsincemalesandfemalesareaffectedequally (recall thatmalesare farmorelikelytobeaffectedbyanX-linkedrecessivecondition).Youshouldalsoexamineaffectedfemalessuchastheoneatthetopofthepedigree.IfaconditionisactuallyX-linkedthenallsonsofaffectedfemaleswouldhavethecondition.Ifthisisnotthecase,asinthispedigree,thegenemustbeonanautosome.Oncethemodeofinheritancehasbeendetermined,genotypescanthenbededucedforeveryoneinthepedigree. For example, all affected individuals abovemust be homozygous recessive (aa) and if twounaffectedparentshaveaffectedchildren,theybothmustbeheterozygous(Aa).Forsomeunaffectedindividualsthesecondallelecannotbededucedbasedonthepedigreeandthusisunknown(A?).Here is an example of apedigree revealing a geneticcondition to be X-linkedrecessive. This condition isrecessive for the samereasons as the previouspedigree. It ismost likelyX-linked since only males areaffected. However thiscannot be said with 100%certaintysincethereisaslightchancethegeneresponsibleforthisconditionisautosomalandbyrandomchanceonlymalesareaffectedinthisfamily.Thenextexample showsanautosomaldominant condition. Noticethat,consistentwithaconditioncausedbyadominantallele,thereareaffectedindividualswhoshowupeverygeneration.Also,therearenounaffectedcoupleshavingaffectedchildrenasyouwouldseewitharecessivecondition.ThisconditioncannotbeX-linkedsince,ifitwas,all fathers would pass the condition to their daughters since alldaughters receive their fathersXchromosome. Ifyou lookcarefullyyouwillseeoneaffectedfatherhavinganunaffecteddaughtermakingitcertainthatthelocusofthismutantalleleisanautosome.

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ThislastexamplebelowshowsanX-linkeddominantcondition.Ithasallthecharacteristicsofadominantconditionasindicatedforthepreviouspedigree.YoucantelliftheconditionisX-linkedbylookingattheoffspring of an affectedman and an unaffected woman (there are 3 such couples in this pedigree).RememberthatfathersgivetheirXchromosometoalldaughtersandtheirYchromosometoallsons.ThusiftheconditionisX-linkedanddominant,alldaughtersofthesemenshouldbeaffectedandalloftheirsonsshouldbeunaffected.All7daughtersofthesemenhavethecondition,whereasnoneoftheir5sonsareaffected.Theoddsofthisoccurringforanautosomaldominantconditionwouldbe2-12or1in4096sincetherewouldbea50%chanceforeachdaughtertoinherittheconditionanda50%chanceforeachsontobeunaffected.ThereforeitisalmostcertainthatthisgeneticconditionisdominantandX-linked.

Inthefinalexerciseforthislabyouwillpracticediagrammingandinterpretingpedigrees.Exercise2G–InterpretingpedigreesSolvethefollowingproblemsonyourworksheet,beingsuretoshowallyourwork:

1. Examinethepedigreeonyourworksheetanddeterminethemodeofinheritance(dominantorrecessive,autosomalorX-linked).Basedonthemodeofinheritance,determinethegenotypesofeachindividualinthepedigreerepresentinganyunknownalleleswithaquestionmark(?).

2. Ahypotheticalmutationinasinglegeneresultsina“zombie”phenotype.Anormalmanandwomanhave3children–azombiegirl,anormalgirlandanormalboy.Themanhasanormalsisterandmotherbuthisfatherisazombie.Thewomanhasanormalbrotherandbothherparentsarenormal.Diagramapedigreeofthisfamilyanddeterminethemodeofinheritanceforthezombiecondition.Onceyouhavedoneso,determinethegenotypesofallfamilymembersandindicateanyunknownalleleswithaquestionmark(?).

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GeneticInheritanceLabWorksheet Name________________________Exercise1A–Probabilityandsamplesize

Probabilityofheads=______ Probabilityoftails=______Fill in the table to the right with the expectednumbers of heads and tails based on theprobabilitiesabove:

10flips 50flips wholeclassheads

tails

Recordthenumbersofheads&tailsandthepercentagesforyouractualcoinflips:

set> 1 % 2 % 3 % 4 % 5 % all %heads % % % % % %

tails % % % % % %

TOTAL 10 100% 10 100% 10 100% 10 100% 10 100% 50 100%Usethetablebelowtorecordtheresultsfortheentireclass(whichshouldbeonthewhiteboard):

Comparetheresultsforsamplesizesof10,50andthewholeclassandcommentonwhichsamplesizecomesclosesttotheexpectednumberofheadsandtailsyoudeterminedabove.

class percentheads %

tails %

TOTAL 100%

Exercise1B–DiceBaseball

Refertothematrixonpage5ofthelabtofillintheprobabilitiesforeachsumofthedice:

sum> 2 3 4 5 6 7 8 9 10 11 12 totalprobability

(x/36)

36/36Determinethefollowing“atbat”outcomesforeachsumofthedice–single,double,triple,homerun,out,walk,doubleplay(ifrunneronfirst)–doingyourbesttobeconsistentwithrealbaseball:

sum> 2 3 4 5 6 7 8 9 10 11 12outcomefor“atbat”

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Keepscoreforyourbaseballgameinthetablebelow.Asyouplay,keeptrackofyourbaserunnersusingthepropsprovidedinyourlabkit:

team 1 2 3 4 5 6 7 8 9 10 11 12 total

Exercise2A–Type1geneticsproblems

Solvetheproblemsbelowinthespaceprovided.Besuretoindicatetheprobabilitiesofeachpossiblephenotypeandgenotypeandtoshowyourwork:1.homozygousdominant(TT)xheterozygous(Tt)

2.heterozygous(Tt)xhomozygousrecessive(tt)

Exercise2B–Type2geneticsproblems

Solvetheproblemsbelowinthespaceprovided,andbesuretoshowyourwork:1.Analbinochildisborntotwoparentswithnormalskinpigmentation.Whatarethegenotypesoftheparents?2.Analbinowomanhas8childrenwiththesameman,5ofwhicharealbino.Whatcanyouconcludeaboutthefather?

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Exercise2C–Incompletedominance

Solvetheproblemsbelowinthespaceprovided,andbesuretoshowyourwork:1.Indicatetheprobabilitiesofeachgenotypeandphenotyperesultingfromtheacrossbetweentwopinksnapdragons.2.Acrossbetweentwosnapdragonsproduced15offspringwithredflowersand17offspringwithpinkflowers.Whatarethegenotypesandphenotypesoftheparents?Exercise2D–Multi-allelicinheritanceandcodominance

Solvetheproblemsbelowinthespaceprovided,andbesuretoshowyourwork:1.IndicatetheprobabilitiesofeachgenotypeandphenotypeinthechildrenofawomanwithbloodtypeOandamanwithbloodtypeAB.2.AwomanwithbloodtypeAandamanwithbloodtypeBhave3children,oneeachwithbloodtypesA,BandO.Whatarethegenotypesoftheparents?

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Exercise2E–Multi-allelicinheritanceandcodominance

Solvetheproblemsbelowinthespaceprovided,andbesuretoshowyourwork:1.Anormalwomanwhosefatherisahemophiliacmarriesanormalman.Whataretheoddsthathemophiliawillafflictanygivenmalechild?anygivenfemalechild?2.Acouplehasthreechildren,anormalboyandaboyandgirleachwithhemophilia.Whatcanyousayabouttheparents?Exercise2F–Geneticsproblemsinvolvingtwogenes

Solvetheproblemsbelowinthespaceprovided,andbesuretoshowyourwork:1.Indicatetheprobabilitiesofallphenotypesintheoffspringoftwotallpeaplantswithpurpleflowers,bothofwhichareheterozygousforbothgenes.2.Anormalwomanwithanalbinofatherandnofamilyhistoryofhemophiliaplanstohavechildrenwithahemophiliacmanwithnormalskinpigmentationwhosemotherisalbino.Determinetheprobabilitiesofallpossiblephenotypesintheirchildren.

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Exercise2G–Interpretingpedigrees

1. Examinethepedigreeshownanddeterminethemodeofinheritance(dominantorrecessive,autosomalorX-linked).

Basedonthemodeofinheritance,determinethegenotypesofeachindividualinthepedigreerepresentinganyunknownalleleswithaquestionmark(?).

2. Ahypotheticalmutationinasinglegeneresultsina“zombie”phenotype.Anormalmanand

womanhave3children–azombiegirl,anormalgirlandanormalboy.Themanhasanormalsisterandmotherbuthisfatherisazombie.Thewomanhasanormalbrotherandbothherparentsarenormal.Diagramapedigreeofthisfamilyanddeterminethemodeofinheritanceforthezombiecondition.Onceyouhavedoneso,determinethegenotypesofallfamilymembersandindicateanyunknownalleleswithaquestionmark(?).

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SupplementalProblemsToensureyouunderstandthekeyaspectsofgeneticinheritanceyouwillneedtopracticesolvinggeneticproblemsmuchliketheonesyou’vesolvedinthislaboratory.Whatfollowsaresetsofgeneticproblemsgroupedbycategory.Solvetheseproblemsonyourowninordertocompletethislaboratory.

SINGLEGENEPROBLEMSInDrosophila,atypeoffruitfly,thelong-wingallele(L)isdominantoverthevestigial-wingallele(l):1. Ifaheterozygous long-winged fly ismatedwithahomozygous long-winged fly,whatpercentageof the

offspringwouldbeexpectedtobehomozygouslong-winged?Wouldtherebeanyvestigial-wingedflies?2. Along-wingedflymatedwithaflyhavingvestigialwingsproduces35long-wingedand33vestigial-winged

offspring.Whatarethegenotypesoftheparentsandtheoffspring?PROBLEMSINVOLVINGTWOGENES3. Tworough,blackguineapigsproducetwooffspring,oneroughwhiteandtheothersmoothblack.Ifthese

sameparentsweretohaveadditionaloffspring,whatproportionofphenotypeswouldyouexpect?4. InDrosophila,graybodycolorisdominantoverebony,andstraightwingsaredominantovercurved. A

gray-bodiedfemalewithcurvedwingsismatedwithagray-bodiedmalewithstraightwings,yieldingsomeebony,curved-wingoffspring.Whatothertypesofoffspringcouldbeproducedandinwhatproportions?

Forthenextproblem,assumethatbrowneyes(B)aredominantoverblueeyes(b),andright-handedness(R)isdominantoverleft-handedness(r).(Inreality,modifyingfactorscomplicatetheinheritanceofthesetraits)5. Aright-handed,blue-eyedmanmarriesaright-handed,brown-eyedwoman.Theyhavetwochildren,one

left-handedandbrown-eyedandtheotherright-handedandblue-eyed.Byalatermarriagewithanotherwomanwhoisalsoright-handedandbrown-eyed,themanhasninechildren,allofwhomareright-handedandbrown-eyed.Whatarethegenotypesofthismanandhistwowives?

X-LINKEDPROBLEMS6. YellowbodycolorinDrosophilaisanX-linkedcharacteristicthatisrecessivetograybodycolor(Drosophila

femalesareXXandmalesXYasinhumans).Agrayfemalematedwithanunknownmaleandproducedsomeyellowandsomegrayoffspringofbothsexes.Whatisthegenotypeoftheoriginalfemale?Whatisthephenotypeofthemalewithwhichshemated?

Fortheseproblemskeepinmindthatcolor-blindnessandhemophiliaarebothX-linkedrecessiveconditions.7. Inhumans,migraine isduetoanautosomaldominantallele. Anormal-visionedwomanwhohasnever

sufferedfrommigraineheadachestakesherdaughtertoadoctorforanexamination.Intheexaminationthedoctordiscoversthatthegirliscolor-blindandsuffersfrommigraineheadaches.Whatcanthedoctorconcludeaboutthegirl’sfather?

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8. Inhumans,aniridia(atypeofblindness)isduetoanautosomaldominantallele,andopticatrophy(anothertypeofblindness)isduetoarecessiveX-linkedallele.Amanblindfromopticatrophymarriesawomanblindfromaniridia.Assumingthatthewomanishomozygousforbothgenes,wouldanyoftheirchildrenbeexpectedtobeblind?Ifso,whattypeofblindnesswouldtheyhave?

9. Anon-hemophiliacmanwhoisblindfromaniridia(seepreviousproblem)andwhosemotherisnotblind

marries a non-hemophiliacwomanwho is not blind andwhose father has hemophilia. What kinds ofchildrenmighttheyhaveandinwhatproportions?

10. A normal-visionedmanmarries a normal-visionedwomanwhose father is color-blind. They have two

daughterswho growup andmarry. The first daughter has five sons, all normal-visioned. The seconddaughterhastwonormal-visioneddaughtersandacolor-blindson.Diagramthefamilyhistoryinapedigreeandindicatethegenotypesofallfamilymembers.

MULTIPLEALLELEPROBLEMSInrabbits,fourallelesforasinglegeneaffectingcoatcolorhavethefollowingrelationships:

C>cch>ch>cThustherearefourpossiblephenotypeswhichareassociatedwiththefollowinggenotypes:

fullcolor–C_(Cwithanyotherallele)

chinchilla–cchcchorcchchorcchc

Himalayan–chchorchc

albino–cc 11. Would it bepossible for a crossbetween two chinchilla rabbits to result inbothHimalayanandalbino

offspring?(besuretoqualifyyouranswer) Inmice,threeallelesforasinglegeneaffectingcoatcolorhavethefollowingrelationships:

AY>A>aTherearefourpossiblephenotypeswhichareassociatedwiththefollowinggenotypes: yellowcoat–AYAorAYa lethal(micedieinutero)–AYAY

agouti(gray)–AAorAa black–aa12. For the following crosses indicate the phenotypes of the parents and the expected proportions of all

possibleoffspring: a)AYaxAa b)AYaxAYa c)AAxAa d)AYAxAYa e)AYAxAYA

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