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9/20/16 Lectures4-11:MechanismsofEvolution (Microevolution) HardyWeinbergEquilibrium • • • • • • Gregor Mendel Wilhem Weinberg (1822-1884) G. H. Hardy (1877 - 1947) (1862 – 1937) Recall from Previous Lectures Darwin’sObservation HardyWeinbergPrinciple(Mendelian Inheritance) GeneticDrift Mutation Recombination EpigeneticInheritance NaturalSelection ThesearemechanismsactingWITHINpopulations, hencecalled“populationgenetics”—EXCEPTfor epigeneticmodifications,whichactonindividuals inaLamarckianmanner Recall from Lecture on History of Evolutionary Thought Darwin’sObservation Evolutionactsthroughchangesin allelefrequency ateach generation Leadstoaverage changein characteristicofthepopulation GregorMendel,“FatherofModern Genetics” http://www.biography.com/people/gregor-mendel-39282#synopsis Gregor Mendel • Mendel presented a mechanism for how traits got passed on HOWEVER,Darwindidnot understandhowgenetic variationwaspassedonfrom generationtogeneration GregorMendel,“FatherofModern Genetics” http://www.biography.com/people/gregor-mendel-39282#synopsis Mendel’sLawsofInheritance • LawofSegregation Gregor Mendel – onlyoneallelepassesfromeach parentontoanoffspring “Individuals pass alleles on to their offspring intact” • LawofIndependentAssortment – differentpairsofallelesarepassedto offspringindependentlyofeachother (the idea of particulate (genes) inheritance) (1822-1884) (1822-1884) 1 9/20/16 Using 29,000 pea plants, Mendel discovered the 1:3 ratio of phenotypes, due to dominant vs. recessive alleles • Incross-pollinatingplantswitheitheryelloworgreenpeas, Mendelfoundthatthefirstgeneration(f1)alwayshadyellow seeds(dominance). However,thefollowinggeneration(f2) consistentlyhada3:1ratioofyellowtogreen. • Mendeluncoveredtheunderlying mechanism,thattherearedominantand recessivealleles Hardy-Weinberg Principle • Mathematical description of Mendelian inheritance Godfrey Hardy (1877-1947) TestingforHardy-Weinbergequilibriumcan beusedtoassesswhetherapopulationis evolving Wilhem Weinberg (1862 – 1937) TheHardy-WeinbergPrinciple • Apopulationthatisnotevolvingshowsalleleand genotypicfrequenciesthatareinHardyWeinberg equilibrium • IfapopulationisnotinHardy-Weinberg equilibrium,itcanbeconcludedthatthe populationisevolving Evolutionary Mechanisms (will put population out of HW Equilibrium): • • • • Genetic Drift Natural Selection Mutation Migration *Epigenetic modifications change expression of alleles but not the frequency of alleles themselves, so they won’t affect the actual inheritance of alleles However, if you count the phenotype frequencies, and not the genotype frequencies , you might see phenotypic frequencies out of HW Equilibrium due to epigenetic silencing of alleles. (epigenetic modifications can change phenotype, not genotype) 2 9/20/16 Fig.23-5a M AP Requirements of HW Violation AR EA Evolution Large population size Genetic drift Random Mating Inbreeding & other No Mutations Mutations No Natural Selection Natural Selection No Migration Migration BeaufortSea •What is a “population?” A group of individuals within a species that is capable of interbreeding and producing fertile offspring Porcupine herdrange (definition for sexual species) Fortymile herdrange An evolving population is one that violates Hardy-Weinberg Assumptions In the absence of Evolution… Patternsofinheritanceshouldalwaysbein “HardyWeinbergEquilibrium” FollowingthetransmissionrulesofMendel Hardy-WeinbergEquilibrium • AccordingtotheHardy-Weinbergprinciple, frequenciesofalleles andgenotypes ina populationremainconstantfromgenerationto generation • Also,thegenotypefrequenciesyouseeina populationshouldbetheHardy-Weinberg expectations,giventheallelefrequencies “NullModel” • NoEvolution: NullModeltotestifno evolutionishappeningshouldsimplybea populationinHardy-WeinbergEquilibrium • NoSelection:NullModeltotestwhether NaturalSelectionisoccurringshouldhaveno selection,butshouldincludeGeneticDrift Example: Is this population in Hardy Weinberg Equilibrium? Generation 1 Generation 2 Generation 3 AA 0.25 0.20 0.10 Aa 0.50 0.60 0.80 aa 0.25 0.20 0.10 – ThisisbecauseGeneticDriftisoperatingeven whenthereisnoNaturalSelection 3 9/20/16 importantconcepts • gene: Hardy-WeinbergTheorem Inanon-evolvingpopulation, frequencyofallelesandgenotypes remainconstant overgenerations You should be able to predict the genotype frequencies, given the allele frequencies Aregionofgenomesequence(DNAorRNA),thatis theunitofinheritance,theproductofwhichcontributesto phenotype • locus: Locationinagenome(usedinterchangeablywith “gene,”ifthelocationisatagene…but,locuscanbeanywhere, someaningisbroaderthangene) • loci: • allele: Pluraloflocus Variantformsofagene(e.g.allelesfordifferenteye colors,BRCA1breastcancerallele,etc.) • genotype: Thecombinationofallelesatalocus(gene) • phenotype: Theexpressionofatrait,asaresultofthe genotypeandregulationofgenes(greeneyes,brownhair,body size,fingerlength,cysticfibrosis,etc.) importantconcepts • allele: Variantformsofagene(e.g.allelesfordifferenteye colors,BRCA1breastcancerallele,etc.) • Wearediploid(2chromosomes),sowehave2alleles atalocus(anylocationinthegenome) • However,therecanbemanyallelesatalocusina population. – Forexample,youmighthaveinheritedablueeyeallelefrom yourmomandabrowneyeallelefromyourdad…youcan’t havemoreallelesthanthat(only2chromosomes,onefrom eachparent) – BUT,therecouldbemanyallelesatthislocusinthe population,blue,green,grey,brown,etc. A2 Eggs A1 Sothencanwe predictthe%of allelesandgenotypes inthepopulationat eachgeneration? A1 A1 A1 A3 A2 A4 A2 Sperm A1 A3 A4 A1 A1 RandomMating(Sex) Zygotes A1A1 A1A3 • Genotypes A1A1 A1A1 A2A4 A3A1 Hardy-WeinbergTheorem A1 A3 A4 A1 A1 A1A1 A1A3 A1 A2 A2 Zygotes A2 Eggs A1 A3 A4 Sperm • Allelesina populationof diploidorganisms Inanon-evolvingpopulation, frequencyofallelesandgenotypes remainconstant overgenerations A1A1 A1A1 A2A4 A3A1 4 9/20/16 Fig.23-6 • Byconvention,ifthereare2allelesatalocus,p and q areusedtorepresenttheirfrequencies Allelesinthepopulation Frequenciesofalleles Gametesproduced p =frequencyof CR allele=0.8 Eachegg: q =frequencyof CW allele=0.2 80% chance 20% chance Eachsperm: 80% chance 20% chance Hardy-Weinbergproportionsindicatethe expectedalleleandgenotypefrequencies, giventhestartingfrequencies Ifp andq representtherelativefrequenciesofthe onlytwopossibleallelesinapopulationata particularlocus,thenforadiploidorganism(2 chromosomes), • Thefrequencyofallallelesinapopulationwilladd upto1 – Forexample,p +q =1 Whataboutforatriploidorganism? (p +q) 2 =1 =p2 +2pq +q2 =1 – wherep2 andq2 representthefrequenciesofthe homozygousgenotypesand2pq representsthe frequencyoftheheterozygousgenotype Whataboutforatriploidorganism? • (p +q)3 =1 =p3 +3p2q +3pq2+q3=1 Potentialoffspring:ppp,ppq,pqp,qpp, qqp,pqq,qpq,qqq Howabouttetraploid?Youworkitout. Hardy Weinberg Theorem ALLELES ProbabilityofA =p Probabilityofa =q GENOTYPES AA: p x p = p +q =1 p2 Aa: p x q +q x p = 2pq aa: q x q = q2 p2 +2pq +q2 =1 5 9/20/16 ALLELE Frequencies Frequency of A = p = 0.8 Frequency of a = q = 0.2 p+q=1 MoreGeneralHWEquations • Onelocusthreealleles:(p +q +r)2=p2 +q2+r2+2pq+2pr+ 2qr Expected GENOTYPE Frequencies AA: p x p = p2 = 0.8 x 0.8 = 0.64 Aa: p x q + q x p = 2pq = 2 x (0.8 x 0.2) = 0.32 aa: q x q = q2 = 0.2 x 0.2 = 0.04 • Onelocusn #alleles:(p1 +p2 +p3 +p4……+pn)2=p12 +p22+ p32 +p42……+pn2 +2p1p2 +2p1p3 +2p2p3 +2p1p4 +2p1p5 +… …+2pn-1pn • Forapolyploid (morethantwochromosomes): (p +q)c,wherec =numberofchromosomes • Ifmultipleloci(genes)codeforatrait,eachlocusfollowsthe HWprincipleindependently,andthentheallelesateachloci interacttoinfluencethetrait Hardy Weinberg Theorem ALLELEFrequency FrequencyofA=p =0.8 Frequencyofa=q =0.2 p2 + 2pq + q2 = 0.64 + 0.32 + 0.04 = 1 Expected Allele Frequencies at 2nd Generation p = AA + Aa/2 = 0.64 + (0.32/2) = 0.8 q = aa + Aa/2 = 0.04 + (0.32/2) = 0.2 Allelefrequenciesremainthesameat nextgeneration Similarexample, Butwithdifferentstartingallelefrequencies p +q =1 ExpectedGENOTYPEFrequency AA: Aa: aa : p x p = p2 =0.8x 0.8=0.64 p x q +q x p = 2pq =2x (0.8x 0.2)=0.32 q x q = q2 =0.2x 0.2=0.04 p2 +2pq+q2 =0.64+0.32+0.04=1 p ExpectedAlleleFrequencyat2ndGeneration q p =AA+Aa/2=0.64+(0.32/2)=0.8 q =aa +Aa/2=0.04+(0.32/2)=0.2 p2 2pq q2 6 9/20/16 Calculating Allele Frequencies from # of Individuals • Thefrequencyofanalleleinapopulationcanbe calculatedfrom#ofindividuals: – Fordiploidorganisms,thetotalnumberofallelesat alocusisthetotalnumberofindividualsx2 – Thetotalnumberofdominantallelesatalocusis2 allelesforeachhomozygousdominantindividual – plus1alleleforeachheterozygousindividual;the samelogicappliesforrecessivealleles Calculating Allele and Genotype Frequencies from # of Individuals AA 120 Aa 60 aa 35(#ofindividuals) #A=(2xAA)+Aa=240+60=300 #a=(2xaa)+Aa=70+60=130 ProportionA=300/total=300/430=0.70 Proportiona=130/total=130/430=0.30 A+a=0.70+0.30=1 ProportionAA=120/215=0.56 ProportionAa=60/215=0.28 Proportionaa=35/215=0.16 AA+Aa+aa=0.56+0.28+0.16=1 ApplyingtheHardy-WeinbergPrinciple • Example:estimatefrequencyofadiseaseallelein apopulation • Phenylketonuria(PKU)isametabolicdisorderthat resultsfromhomozygosityforarecessiveallele • TheoccurrenceofPKUis1per10,000births • Howmanycarriersofthisdiseaseinthe population? • Individualsthatarehomozygousforthedeleterious recessiveallelecannotbreakdownphenylalanine,results inbuildupà mentalretardation – Raredeleteriousrecessivesoftenremainina populationbecausetheyarehiddeninthe heterozygousstate(the“carriers”) – Naturalselectioncanonlyactonthehomozygous individualswherethephenotypeisexposed (individualswhoshowsymptomsofPKU) So, let’s calculate HW frequencies • TheoccurrenceofPKUis1per10,000births (frequencyofthediseaseallele): q2 =0.0001 q =sqrt(q2)=sqrt(0.0001)=0.01 • Thefrequencyofnormalallelesis: – WecanassumeHWequilibriumif: • Thereisnomigrationfromapopulationwithdifferent allelefrequency • Randommating • Nogeneticdrift • Etc p =1– q =1– 0.01=0.99 • Thefrequencyofcarriers(heterozygotes)ofthe deleteriousalleleis: 2pq =2x 0.99x 0.01=0.0198 orapproximately2%oftheU.S.population 7 9/20/16 ConditionsforHardy-WeinbergEquilibrium • TheHardy-Weinbergtheoremdescribesa hypotheticalpopulation • Thefiveconditionsfornonevolvingpopulations arerarelymetinnature: – – – – – Nomutations Randommating Nonaturalselection Extremelylargepopulationsize Nogeneflow DEVIATION from Hardy-Weinberg Equilibrium Indicates that EVOLUTION Is happening • So,inrealpopulations,alleleandgenotype frequenciesdochangeovertime Hardy-Weinberg across a Genome AlleleA1Demo • Innaturalpopulations,somelocimightbeoutof HWequilibrium,whilebeinginHardy-Weinberg equilibriumatotherloci • Forexample,somelocimightbeundergoing naturalselectionandbecomeoutofHW equilibrium,whiletherestofthegenomeremains inHWequilibrium Howcanyoutellwhetherapopulation isoutofHWEquilibrium? • PerformHWcalculationstoseeifitlookslike thepopulationisoutofHWequilibrium • Thenapplystatisticalteststoseeifthe deviationissignificantlydifferentfromwhat youwouldexpectbyrandomchance 8 9/20/16 Generation 1 Generation 2 Generation 3 Example: Does this population remain in Hardy Weinberg Equilibrium across Generations? Generation 1 Generation 2 Generation 3 AA 0.25 0.20 0.10 Aa 0.50 0.60 0.80 aa 0.25 0.20 0.10 Howcanyoutellwhetherapopulation isoutofHWEquilibrium? 1. When allele frequencies are changing across generations 2. When you cannot predict genotype frequencies from allele frequencies (means there is an excess or deficit of genotypes than what would be expected given the allele frequencies) Example • GenotypeCount:AA 30Aa 55 aa 15 • Calculatethec2value: Genotype AA Aa aa Total ObservedExpected (O-E)2/E 30 33 0.27 55 49 0.73 15 18 0.50 100 100 1.50 • Sincec2 =1.50<3.841(fromChi-squaretable,alpha=0.05), weconcludethatthegenotypefrequenciesinthispopulation arenotsignificantlydifferentthanwhatwouldbeexpectedif thepopulationisinHardy-Weinbergequilibrium. AA 0.25 0.20 0.10 Aa 0.50 0.60 0.80 aa 0.25 0.20 0.10 ■ In this case, allele frequencies (of A and a) did not change. ■ ***However, the population did go out of HW equilibrium because you can no longer predict genotypic frequencies from allele frequencies ■ For example, p = 0.5, p2 = 0.25, but in Generation 3, the observe p2 = 0.10 TestingforDeviaton fromHardyWeinbergExpectations • Ac2 goodness-of-fittest canbeusedtodetermineifa populationissignificantlydifferentfromtheexpections of Hardy-Weinbergequilibrium. • Ifwehaveaseriesofgenotypecountsfromapopulation, thenwecancomparethesecountstotheonespredictedby theHardy-Weinbergmodel. • O =observedcounts,E =expectedcounts,sumacross genotypes TestingforDeviaton fromHardyWeinbergExpectations • Ac2 goodness-of-fittest canbeusedtodetermineifa populationissignificantlydifferentfromtheexpections of Hardy-Weinbergequilibrium. • Ifwehaveaseriesofgenotypecountsfromapopulation, thenwecancomparethesecountstotheonespredictedby theHardy-Weinbergmodel. • O =observedcounts,E =expectedcounts,sumacross genotypes 54 9 9/20/16 TestingforDeviaton fromHardy-WeinbergExpectations • O =observedcounts,E =expectedcounts,sumacross genotypes • Wetestourc2 valueagainsttheChi-squaredistribution(sum ofsquareofanormaldistribution),whichrepresentsthe theoreticaldistributionofsamplevaluesunderHW equilibrium à Less likely to get these values by chance • Anddeterminehowlikelyitistogetourresultsimplyby chance(e.g.duetosamplingerror);i.e.,doourObserved valuesdifferfromourExpectedvaluesmorethanwhatwe wouldexpectbychance (=significantlydifferent)? 55 TestforDeviationfromHWequilibrium • GenotypeCountGeneration4: AA 65Aa 31aa 4 • Calculatethec2value: Genotype AA Aa aa Total ObservedExpected (O-E)2/E 65 64.8 0.00062 31 31.4 0.0051 4 3.8 0.0105 100 100 0.016 • Sincec2 =0.016<3.841(fromChi-squaretableforcriticalvalues, alpha=0.05),weconcludethatthegenotypefrequenciesinthis populationarenotsignificantlydifferentthanwhatwouldbe 56 expectedifthepopulationwereinHardy-Weinbergequilibrium. • ? • Thechi-squareddistributionisusedbecauseitisthesumofsquared normaldistributions • • • • CalculateChi-squaredteststatistic Figureoutdegreesoffreedom Selectconfidenceinterval(P-value) CompareyourChi-squaredvaluetothetheoretical distribution(fromthetable),andacceptorrejectthenull hypothesis. TestforSignificanceofDeviationfromHWEquilibrium DegreesofFreedom isn– 1 =2alleles(p,q)-1= 1 – Iftheteststatistic>thanthecriticalvalue,thenullhypothesis(H0 =thereisnodifferencebetweenthedistributions)canberejected withtheselectedlevelofconfidence,andthealternative hypothesis(H1=thereisadifferencebetweenthedistributions) canbeaccepted. – Iftheteststatistic<thanthecriticalvalue,thenullhypothesis 57 cannotberejected Testingforsignificance • TheresultscomeoutnotsignificantlydifferentfromHW equilibrium • Thisdoesnotnecessarilymeanthatgeneticdriftisnot happening,butthatwecannotconcludethatgenetic driftishappening • Eitherwedonothaveenoughpower(notenoughdata, smallsamplesize),orgeneticdriftisnothappening • Sometimesitisdifficulttotestwhetherevolutionis happening,evenwhenitishappening...Thesignal needstobesufficientlylargetobesurethatyoucan’t gettheresultsbychance(likebysamplingerror) 59 58 TestforDeviationfromHWequilibrium • GenotypeCountGeneration4à increasesamplesize AA 65000Aa 31000aa 4000 • Calculatethec2value: Genotype AA Aa aa Total ObservedExpected 65000 64800 31000 31400 4000 3800 100,000 100,000 (O-E)2/E 0.617 5.10 10.32 16.04 • Sincec2 =16.04>3.841(fromChi-squaretableforcriticalvalues, alpha=0.05),weconcludethatthegenotypefrequenciesinthis populationAREsignificantlydifferentthanwhatwouldbe 60 expectedifthepopulationwereinHardy-Weinbergequilibrium. 10 9/20/16 TestforSignificanceofDeviationfromHWEquilibrium • OnegenerationofRandomMatingcouldputa populationbackintoHardyWeinberg Equilibrium DegreesofFreedom isn– 1 =2alleles(p,q)-1= 1 61 WhatwouldGeneticDriftlook like? ExamplesofDeviationfrom Hardy-WeinbergEquilibrium ExamplesofDeviationfrom Hardy-WeinbergEquilibrium Generation1 Generation2 Generation3 Generation4 AA 0.64 0.63 0.64 0.65 Aa 0.32 0.33 0.315 0.31 ExamplesofDeviationfrom Hardy-WeinbergEquilibrium aa 0.04 0.04 0.045 0.04 IsthispopulationinHWequilibrium? Ifnot,howdoesitdeviate? Whatcouldbethereason? • Mostpopulationsareexperiencingsomelevel ofgeneticdrift,unlesstheyareincredibly large Generation1 Generation2 Generation3 Generation4 AA 0.64 0.63 0.64 0.65 Aa 0.32 0.33 0.315 0.31 aa 0.04 0.04 0.045 0.04 ThisisacaseofGeneticDrift,where allelefrequenciesarefluctuating randomlyacrossgenerations 11 9/20/16 ExamplesofDeviationfrom Hardy-WeinbergEquilibrium AA 0.64 Aa 0.32 aa 0 IsthispopulationinHWequilibrium? Ifnot,howdoesitdeviate? Whatcouldbethereason? ExamplesofDeviationfrom Hardy-WeinbergEquilibrium AA 0.64 Aa 0.70 aa 0.05 IsthispopulationinHWequilibrium? Ifnot,howdoesitdeviate? aa 0 HerethisappearstobeDirectional Selection favoringAA Or…NegativeSelection disfavoringaa ExamplesofDeviationfrom Hardy-WeinbergEquilibrium AA 0.25 Aa 0.32 ExamplesofDeviationfrom Hardy-WeinbergEquilibrium AA 0.25 Aa 0.70 aa 0.05 ThisappearstobeacaseofHeterozygote Advantage(orOverdominance) Whatcouldbethereason? ExamplesofDeviationfrom Hardy-WeinbergEquilibrium AA 0.10 Aa 0.10 aa 0.80 IsthispopulationinHWequilibrium? Ifnot,howdoesitdeviate? ExamplesofDeviationfrom Hardy-WeinbergEquilibrium AA 0.10 Aa 0.10 aa 0.80 Selectionappearstobefavoringaa Whatcouldbethereason? 12 9/20/16 Summary HardyWeinbergEquilibrium (1)Anonevolving populationisinHW Equilibrium Gregor Mendel (2)Evolutionoccurswhentherequirementsfor HWEquilibriumarenotmet (3)HWEquilibriumisviolatedwhenthereis GeneticDrift,Migration,Mutations,Natural Selection,andNonrandomMating Fig.23-7-4 80%CR ( p =0.8) Fig.23-7-1 20%CW (q =0.2) Sperm CR (80%) CW (20%) 64%(p2) CR CR (1862 – 1937) G. H. Hardy (1877 - 1947) 80%CR (p =0.8) Perform the same calculations using percentages 16%( pq) CR CW 16%(qp) CR CW Wilhem Weinberg (1822-1884) 20%CW (q =0.2) Sperm CR (80%) CW (20%) 64%(p 2) C RC R 16%(pq) C RC W 4%(q2) CW CW 64%CR CR, 32%CR CW,and 4%CW CW Gametesofthisgeneration: 64%CR + 16%CR = 80%CR = 0.8=p 4%CW + 16%CW = 20%CW = 0.2=q Genotypesinthenextgeneration: 16%(qp) C RC W 64%CR CR, 32%CR CW,and 4%CW CW plants Fig.23-7-2 4%(q 2) CW CW Fig.23-7-3 64%CR CR , 32%CR CW ,and 4%CW CW Gametesofthisgeneration: 64%CR CR , 32%CR CW ,and 4%CW CW 64%C + 16%CR = 80%CR = 0.8=p 4%CW = 20%CW = 0.2=q R Gametesofthisgeneration: 64%C + 16%CR = 80%CR = 0.8=p 4%CW = 20%CW = 0.2=q R + 16%CW + 16%CW Genotypesinthenextgeneration: 64%CR CR , 32%CR CW ,and 4%CW CW plants 13 9/20/16 Gregor Mendel 1. Nabila is a Saudi Princess who is arranged to marry her first cousin. Many in her family have died of a rare blood disease, which sometimes skips generations, and thus appears to be recessive. Nabila thinks that she is a carrier of this disease. If her fiancé is also a carrier, what is the probability that her offspring will have (be afflicted with) the disease? (A) 1/4 (B) 1/3 (C) 1/2 (D) 3/4 (E) zero The following are numbers of pink and white flowers in a population. Generation 1: Pink 901 White 302 Generation 2: 1204 403 Generation 3: 1510 504 2. Which of the following is most likely to be TRUE? (A) The heterozygotes are probably pink (B) The recessive allele here (probably white) is clearly deleterious (C) Evolution is occurring, as allele frequencies are changing greatly over time (D) Clearly there is a heterozygote advantage (E) The frequencies above violate Hardy-Weinberg expectations The following are numbers of purple and white peas in a population. (A1A1) (A1A2) (A2A2) Purple Purple White Generation 1: 360 480 160 Generation 2: 100 200 200 Generation 3: 0 100 300 3. What are the genotype frequencies at each generation? (A) Generation 1: 0.30, 0.50, 0.20 Generation 2: 0.20, 0.40, 0.40 Generation 3: 0, 0.333, 0.666 (B) Generation 1: 0.36, 0.48, 0.16 Generation 2: 0.10, 0.20, 0.20 Generation 3: 0, 0.10, 0.30 (C) Generation 1: 0.36, 0.48, 0.16 Generation 2: 0.20, 0.40, 0.40 Generation 3: 0, 0.25, 0.75 (D) Generation 1: 0.36, 0.48, 0.16 Generation 2: 0.36, 0.48, 0.16 Generation 3: 0.36, 0.48, 0.16 4. From the example on the previous slide, what are the frequencies of alleles at each generation? (A) Generation1: Dominant allele (A1) = 0.6, Recessive allele (A2) = 0.4 Generation2: Dominant allele = 0.4, Recessive allele = 0.6 Generation3: Dominant allele = 0.125, Recessive allele = 0.875 (B) Generation1: Dominant allele = 0.6, Recessive allele = 0.4 Generation2: Dominant allele = 0.6, Recessive allele = 0.4 Generation3: Dominant allele = 0.6, Recessive allele = 0.4 (C) Generation1: Dominant allele = 0.6, Recessive allele = 0.4 Generation2: Dominant allele = 0.5, Recessive allele = 0.5 Generation3: Dominant allele = 0.25, Recessive allele = 0.75 5. From the example two slides ago, which evolutionary mechanism might be operating across generations? (A) Mutation (B) Selection favoring A1 (C) Heterozygote advantage (D) Selection favoring A2 (E) Inbreeding (D) Generation1: Dominant allele = 0.4, Recessive allele = 0.6 Generation2: Dominant allele = 0.5, Recessive allele = 0.5 Generation3: Dominant allele = 0.25, Recessive allele = 0.75 14 9/20/16 Answers: 1.Parents:AaxAa=Offspring:AA(25%),Aa(50%),aa(25%) Answer=A 2.A 3.C 4.A 5.D 15