Download Lecture PDF - Carol Eunmi LEE

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