Wilhem Weinberg 1862 1937 Gregor Mendel G H Hardy 1877 1947 18221884 Lectures 411 Mechanisms of Evolution Microevolution Hardy Weinberg Principle Mendelian Inheritance ID: 778718
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Slide1
Hardy Weinberg Equilibrium
Wilhem Weinberg
(1862 – 1937)
Gregor Mendel
G. H. Hardy
(1877 - 1947)
(1822-1884)
Slide2Lectures 4-11: Mechanisms of Evolution (Microevolution)
Hardy Weinberg Principle (Mendelian
Inheritance)Genetic DriftMutationSex: Recombination and Random Mating
Epigenetic InheritanceNatural SelectionThese are mechanisms acting WITHIN populations, hence called “population genetics”—EXCEPT for epigenetic modifications, which act on individuals in a Lamarckian manner
Slide3Evolution acts through
changes in allele frequency at each generation
Leads to
average change in characteristic of the population
Recall from Previous Lectures
Darwin’s Observation
Slide4HOWEVER, Darwin did not understand how genetic variation was passed on from generation to generation
Recall from Lecture on History of Evolutionary Thought
Darwin
’
s Observation
Slide5Gregor Mendel,
“Father of Modern Genetics
”
Mendel presented a mechanism for how traits got passed on
“
Individuals pass alleles on to their offspring intact”
(the idea of particulate (genes) inheritance)
Gregor Mendel
(1822-1884)
http://www.biography.com/people/gregor-mendel-39282#synopsis
Slide6Gregor Mendel,
“Father of Modern Genetics
”
Mendel’
s Laws of InheritanceLaw of Segregationonly one allele passes from each parent on to an offspringLaw of Independent Assortment
different pairs of alleles are passed to offspring independently of each other
Gregor Mendel
(1822-1884)
http://www.biography.com/people/gregor-mendel-39282#synopsis
Slide7Gregor Mendel
Slide8In cross-pollinating plants with either yellow or green peas, Mendel found that the first generation (f1) always had yellow seeds (dominance). However, the following generation (f2) consistently had a 3:1 ratio of yellow to green.
Using 29,000 pea plants, Mendel discovered the 1:3 ratio of phenotypes, due to dominant
vs.
recessive alleles
Slide9Mendel uncovered the underlying mechanism, that there are dominant and recessive alleles
Slide10Mathematical description of
Mendelian inheritance
Hardy-Weinberg Principle
Godfrey Hardy
(1877-1947)
Wilhem
Weinberg
(1862 – 1937)
Slide11Testing for Hardy-Weinberg equilibrium can be used to assess whether a population is evolving
Slide12The Hardy-Weinberg Principle
A population that is not evolving shows allele and genotypic frequencies that are in
Hardy Weinberg equilibriumIf a population is not in Hardy-Weinberg equilibrium, it can be concluded that the population is evolving
Slide13Evolutionary 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
)
Slide14Requirements of HW
Evolution
Large population size
Genetic drift
Random Mating
Inbreeding & other
No Mutations
Mutations
No Natural Selection
Natural Selection
No Migration
Migration
An evolving population is one that violates Hardy-Weinberg Assumptions
Violation
Slide15Fig. 23-5a
Porcupine
herd range
Beaufort Sea
NORTHWEST
TERRITORIES
MAP
AREA
ALASKA
CANADA
Fortymile
herd range
ALASKA
YUKON
What is a “
population
?”
A group of individuals within a species that is capable of interbreeding and producing fertile offspring
(definition for sexual species)
Slide16Patterns of inheritance should always be in “Hardy Weinberg Equilibrium”
Following the transmission rules of Mendel
In the absence of Evolution…
Slide17Hardy-Weinberg Equilibrium
According to the
Hardy-Weinberg principle,
frequencies of alleles and genotypes
in a population remain constant from generation to generation
Also, the genotype frequencies you see in a population should be the Hardy-Weinberg expectations, given the allele frequencies
Slide18“Null Model”
No Evolution: Null Model to test if no evolution is happening should simply be a population in Hardy-Weinberg EquilibriumNo Selection:
Null Model to test whether Natural Selection is occurring should have no selection, but should include Genetic DriftThis is because Genetic Drift is operating even when there is no Natural Selection
Slide19Example
: Is this population in Hardy Weinberg Equilibrium?
AA Aa
aa Generation 1
0.25 0.50 0.25
Generation 2 0.20 0.60 0.20
Generation 3
0.10 0.80 0.10
Slide20Hardy-Weinberg Theorem
In a non-evolving population, frequency of alleles and genotypes remain
constant
over generations
You should be able to predict the genotype frequencies, given the allele frequencies
Slide21important concepts
gene: A region of genome sequence (DNA or RNA), that is the unit of inheritance , the product of which contributes to phenotypelocus
: Location in a genome (used interchangeably with “gene,” if the location is at a gene… but, locus can be anywhere, so meaning is broader than gene)loci
: Plural of locusallele: Variant forms of a gene (e.g. alleles for different eye colors, BRCA1 breast cancer allele, etc.)genotype: The combination of alleles at a locus (gene)
phenotype: The expression of a trait, as a result of the genotype and regulation of genes (green eyes, brown hair, body size, finger length, cystic fibrosis, etc.)
Slide22important concepts
allele: Variant forms of a gene (e.g. alleles for different eye colors, BRCA1 breast cancer allele, etc.)
We are diploid (2 chromosomes), so we have 2 alleles at a locus (any location in the genome)
However, there can be many alleles at a locus in a population.For example, you might have inherited a blue eye allele from your mom and a brown eye allele from your dad… you can’t have more alleles than that (only 2 chromosomes, one from each parent)BUT, there could be many alleles at this locus in the population, blue, green, grey, brown, etc.
Slide23Alleles
in a population of diploid organisms
A1
A2
A3
A4
A1
A1
A2
Sperm
Eggs
Genotypes
Random Mating (Sex)
Zygotes
A1A3
A1A1
A1A1
A2A4
A3A1
A1A1
A1
A2
A1
A1
A3
A4
Slide24So then can we predict the % of alleles and genotypes in the population at each generation?
A1
A2
A3
A4
A1
A1
A2
Sperm
Eggs
Zygotes
A1A3
A1A1
A1A1
A2A4
A3A1
A1A1
A1
A2
A1
A1
A3
A4
Slide25Hardy-Weinberg Theorem
In a non-evolving population, frequency of alleles and genotypes remain
constant
over generations
Slide26Fig. 23-6
Frequencies of alleles
Alleles in the population
Gametes produced
Each egg:
Each sperm:
80%
chance
80%
chance
20%
chance
20%
chance
q
= frequency of
p
= frequency of
C
R
allele = 0.8
C
W
allele = 0.2
Hardy-Weinberg proportions indicate the expected allele and genotype frequencies, given the starting frequencies
Slide27By convention, if there are 2 alleles at a locus, p and q are used to represent their frequencies
The frequency of all alleles in a population will add up to 1For example, p
+ q = 1
Slide28If p and
q represent the relative frequencies of the only two possible alleles in a population at a particular locus, then for a diploid organism (2 chromosomes),
(p +
q) 2 = 1 = p
2 + 2pq + q2 = 1
where p2
and q2 represent the frequencies of the homozygous genotypes and 2pq represents the frequency of the heterozygous genotype
Slide29What about for a triploid organism?
Slide30What about for a triploid organism?
(p +
q)3 = 1
= p3 + 3p
2q + 3pq2
+ q3 = 1
Potential offspring: ppp,
ppq
,
pqp
,
qpp
,
qqp
,
pqq
,
qpq
,
qqq
How about
tetraploid? You work it out.
Slide31Hardy Weinberg Theorem
ALLELES
Probability of
A
= p p
+ q
= 1Probability of
a
=
q
GENOTYPES
AA
:
p
x
p
=
p2Aa: p x q + q
x
p
= 2
p
q
aa
:
q
x
q
=
q
2
p
2
+ 2
p
q
+
q
2
= 1
Slide32More General HW Equations
One locus three alleles: (p + q + r)
2 = p2 + q2 + r
2 + 2pq +2pr + 2qrOne locus n # alleles: (p1 + p2 + p3 + p4
… …+ pn)2 = p12 + p22 + p
32 + p42… …+ p
n2 + 2p1p2 + 2p1p
3
+ 2p
2
p
3
+ 2p
1
p
4
+ 2p
1
p
5
+ … … + 2pn-1pnFor a polyploid (more than two chromosomes):
(p + q)c, where c = number of chromosomesIf multiple loci (genes) code for a trait, each locus follows the HW principle independently, and then the alleles at each loci interact to influence the trait
Slide33ALLELE Frequencies
Frequency of A =
p
= 0.8
Frequency of a =
q
= 0.2
p
+
q
= 1
Expected GENOTYPE Frequencies
AA:
p
x
p
= p
2
= 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
= q
2
= 0.2
x
0.2 = 0.04
p
2
+ 2pq + q
2
= 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
Allele frequencies remain the same at next generation
Slide34Hardy Weinberg Theorem
ALLELE Frequency
Frequency of A =
p = 0.8
p + q
= 1Frequency of a =
q = 0.2
Expected GENOTYPE Frequency
AA:
p
x
p
= p
2
= 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
= q
2
= 0.2
x
0.2 = 0.04
p
2
+ 2pq + q
2
= 0.64 + 0.32 + 0.04 = 1
Expected Allele Frequency 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
Slide35Similar example,
But with different starting allele frequencies
p
q
Slide36Slide37p
2
2
pq
q2
Slide38The frequency of an allele in a population can be calculated from # of individuals:
For diploid organisms, the total number of alleles at a locus is the total number of individuals x 2 The total number of dominant alleles at a locus is 2 alleles for each homozygous dominant individual
plus 1 allele for each heterozygous individual; the same logic applies for recessive alleles
Calculating Allele Frequencies from # of Individuals
Slide39AA Aa aa
120 60 35 (# of individuals)
#A = (2 x AA) + Aa = 240 + 60 = 300#a = (2 x aa) + Aa = 70 + 60 = 130
Proportion A = 300/total = 300/430 = 0.70Proportion a = 130/total = 130/430 = 0.30
A + a = 0.70 + 0.30 = 1Proportion AA = 120/215 = 0.56
Proportion Aa = 60/215 = 0.28
Proportion aa = 35/215 = 0.16 AA + Aa + aa = 0.56 + 0.28 +0.16 = 1
Calculating Allele and Genotype Frequencies from # of Individuals
Slide40Applying the Hardy-Weinberg Principle
Example: estimate frequency of a disease allele in a population
Phenylketonuria (PKU) is a metabolic disorder that results from homozygosity for a recessive allele
Individuals that are homozygous for the deleterious recessive allele cannot break down phenylalanine, results in build up mental retardation
Slide41The occurrence of PKU is 1 per 10,000 birthsHow many carriers of this disease in the population?
Slide42Rare deleterious recessives often remain in a population because they are hidden in the heterozygous state (the “carriers”)
Natural selection can only act on the homozygous individuals where the phenotype is exposed (individuals who show symptoms of PKU)
We can assume HW equilibrium if:
There is no migration from a population with different allele frequencyRandom matingNo genetic drift
Etc
Slide43The occurrence of PKU is 1 per 10,000 births
(frequency of the disease allele):
q2 = 0.0001
q = sqrt(q2 ) = sqrt(0.0001) = 0.01
The frequency of normal alleles is: p
= 1 – q = 1 – 0.01 = 0.99
The frequency of carriers (heterozygotes) of the deleterious allele is:
2
pq
= 2
x
0.99
x
0.01 = 0.0198
or approximately 2% of the U.S. population
So, let’s calculate HW frequencies
Slide44Conditions for Hardy-Weinberg Equilibrium
The Hardy-Weinberg theorem describes a hypothetical population
The five conditions for nonevolving populations are rarely met in nature:No mutations Random mating
No natural selection Extremely large population sizeNo gene flowSo, in real populations, allele and genotype frequencies do change over time
Slide45DEVIATION
from
Hardy-Weinberg Equilibrium
Indicates that EVOLUTION
Is happening
Slide46In natural populations, some loci might be out of HW equilibrium, while being in Hardy-Weinberg equilibrium at other loci
For example, some loci might be undergoing natural selection and become out of HW equilibrium, while the rest of the genome remains in HW equilibrium
Hardy-Weinberg across a Genome
Slide47Allele A1 Demo
Slide48How can you tell whether a population is out of HW Equilibrium?
Slide49Perform HW calculations to see if it looks like the population is out of HW equilibrium
Then apply statistical tests to see if the deviation is significantly different from what you would expect by random chance
Slide50Example
: Does this population remain in Hardy Weinberg Equilibrium across Generations?
AA Aa
aa Generation 1
0.25 0.50 0.25
Generation 2 0.20 0.60 0.20
Generation 3
0.10 0.80 0.10
Slide51AA Aa
aa
Generation 1 0.25 0.50 0.25 Generation 2 0.20 0.60 0.20
Generation 3 0.10 0.80 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, p
2
= 0.25, but in Generation 3, the observe p
2
= 0.10
Slide52How can you tell whether a population is out of HW Equilibrium?
When allele frequencies are changing across generations
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)
Slide53Testing for Deviaton from Hardy-Weinberg Expectations
A c2
goodness-of-fit test can be used to determine if a population is significantly different from the expections of Hardy-Weinberg equilibrium.
If we have a series of genotype counts from a population, then we can compare these counts to the ones predicted by the Hardy-Weinberg model. O = observed counts, E = expected counts, sum across genotypes
Slide54Example
Genotype Count: AA 30 Aa 55 aa 15
Calculate the c2 value:
Genotype Observed Expected (O-E)2/E AA 30 33 0.27
Aa 55 49 0.73
aa 15 18 0.50 Total 100 100 1.50
Since c2 = 1.50 < 3.841 (from Chi-square table, alpha = 0.05), we conclude that the genotype frequencies in this population are not significantly different than what would be expected if the population is in Hardy-Weinberg equilibrium.
Slide55Testing for Deviaton from Hardy-Weinberg Expectations
A c2 goodness-of-fit test
can be used to determine if a population is significantly different from the expections of Hardy-Weinberg equilibrium. If we have a series of genotype counts from a population, then we can compare these counts to the ones predicted by the Hardy-Weinberg model.
O = observed counts, E = expected counts, sum across genotypes
55
Slide56Testing for
Deviaton from Hardy-Weinberg Expectations
O
= observed counts, E = expected counts, sum across genotypes We test our c2 value against the Chi-square distribution (sum of square of a normal distribution), which represents the theoretical distribution of sample values under HW equilibrium
And determine how likely it is to get our result simply by chance (e.g. due to sampling error); i.e., do our Observed values differ from our Expected values more than what we would expect by chance (= significantly different)?
?
Less likely to get these values by chance
56
Slide57Test for Deviation from HW equilibrium
Genotype Count Generation 4: AA 65 Aa 31
aa 4 Calculate the c2
value: Genotype Observed Expected (O-E)2/E AA
65 64.8 0.00062 Aa 31 31.4 0.0051
aa 4 3.8 0.0105
Total 100 100 0.016Since c2 = 0.016 < 3.841 (from Chi-square table for critical values, alpha = 0.05), we conclude that the genotype frequencies in this population are not significantly different than what would be expected if the population were in Hardy-Weinberg equilibrium.
57
Slide58The chi-squared distribution is used because it is the sum of squared normal distributionsCalculate Chi-squared test statisticFigure out degrees of freedom
Select confidence interval (P-value)Compare your Chi-squared value to the theoretical distribution (from the table), and accept or reject the null hypothesis.If the test statistic > than the critical value, the null hypothesis (H0 = there is no difference between the distributions) can be rejected with the selected level of confidence, and the alternative hypothesis (H1 = there is a difference between the distributions) can be accepted.
If the test statistic < than the critical value, the null hypothesis cannot be rejected
58
Slide59Test for Significance of Deviation from HW Equilibrium
Degrees of Freedom is n – 1
= 2 alleles (p, q) -1 = 1
59
Slide60Testing for significance
The results come out not significantly different from HW equilibriumThis does not necessarily mean that genetic drift is not happening, but that we cannot conclude that genetic drift is happening
Either we do not have enough power (not enough data, small sample size), or genetic drift is not happeningSometimes it is difficult to test whether evolution is happening, even when it is happening... The signal needs to be sufficiently large to be sure that you can’t get the results by chance (like by sampling error)
60
Slide61Test for Deviation from HW equilibrium
Genotype Count Generation 4 increase sample size
AA 65000 Aa 31000 aa 4000
Calculate the c2 value: Genotype Observed Expected (O-E)2
/E AA 65000 64800 0.617 Aa
31000 31400 5.10 aa
4000 3800 10.32 Total 100,000 100,000 16.04Since
c
2
= 16.04 > 3.841 (from Chi-square table for critical values, alpha = 0.05), we conclude that the genotype frequencies in this population ARE significantly different than what would be expected if the population were in Hardy-Weinberg equilibrium.
61
Slide62Test for Significance of Deviation from HW Equilibrium
Degrees of Freedom is n – 1
= 2 alleles (p, q) -1 = 1
62
Slide63One generation of Random Mating could put a population back into Hardy Weinberg Equilibrium
Slide64Examples of Deviation from Hardy-Weinberg Equilibrium
Slide65What would Genetic Drift look like?Most populations are experiencing some level of genetic drift, unless they are incredibly large
Slide66Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa aa
Generation 1 0.64 0.32 0.04
Generation 2 0.63 0.33 0.04Generation 3 0.64 0.315 0.045Generation 4 0.65 0.31 0.04
Is this population in HW equilibrium?
If not, how does it deviate?What could be the reason?
Slide67Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa aa
Generation 1 0.64 0.32 0.04
Generation 2 0.63 0.33 0.04Generation 3 0.64 0.315 0.045Generation 4 0.65 0.31 0.04
This is a case of Genetic Drift
, where allele frequencies are fluctuating randomly across generations
Slide68Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa aa
0.64 0.36
0Is this population in HW equilibrium?
If not, how does it deviate?What could be the reason?
Slide69Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa
aa0.64
0.36 0Here this appears to be
Directional Selection favoring AA
Or… Negative Selection disfavoring
aa
Slide70Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa aa
0.25 0.70 0.05
Is this population in HW equilibrium?If not, how does it deviate?What could be the reason?
Slide71Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa
aa0.25 0.70 0.05
This appears to be a case of Heterozygote Advantage (or Overdominance
)
Slide72Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa aa
0.10 0.10 0.80
Is this population in HW equilibrium?If not, how does it deviate?What could be the reason?
Slide73Examples of Deviation from Hardy-Weinberg Equilibrium
AA Aa
aa0.10 0.10 0.80
Selection appears to be favoring aa
Slide74(1) A
nonevolving population is in HW Equilibrium
(2) Evolution occurs when the requirements for HW Equilibrium are not met
(3) HW Equilibrium is violated when there is Genetic Drift, Migration, Mutations, Natural Selection, and Nonrandom Mating
Summary
Slide75Hardy Weinberg Equilibrium
Wilhem Weinberg
(1862 – 1937)
Gregor Mendel
G. H. Hardy
(1877 - 1947)
(1822-1884)
Slide76Fig. 23-7-4
Gametes of this generation:
64%
C
R CR
, 32%
CR C
W
, and
4%
C
W
C
W
64%
C
R
+
16% CR = 80% CR = 0.8 = p
4%
C
W
+
16%
C
W
=
2
0%
C
W
=
0.2 =
q
64%
C
R
C
R
,
32%
C
R
C
W
, and
4%
C
W
C
W
plants
Genotypes in the next generation:
Sperm
C
R
(80%)
C
W
(20%)
80%
C
R
(
p
= 0.8)
C
W
(20%)
20%
C
W
(
q
= 0.2)
16% (
pq
)
C
R
C
W
4% (
q
2
)
C
W
C
W
C
R
(80%)
64% (
p2) CR CR
16% (qp) C
R CWEggsPerform the same calculations using percentages
Slide77Fig. 23-7-1
Sperm
C
R
(80%)
C
W
(20%)
80%
C
R
(
p
= 0.8)
C
W
(20%)
20% CW (
q
= 0.2)
16% (
pq
)
C
R
C
W
4% (
q
2
)
C
W
C
W
C
R
(80%)
64% (
p
2
)
C
R
C
R
16% (
qp
)
C
R
C
W
Eggs
Slide78Fig. 23-7-2
Gametes of this generation:
64%
C
RC
R,
32% CR
C
W
, and
4%
C
W
C
W
64%
C
R
+
16% CR = 80% CR =
0.8 =
p
4%
C
W
+
16%
C
W
=
2
0%
C
W
=
0.2 =
q
Slide79Fig. 23-7-3
Gametes of this generation:
64%
C
RC
R,
32% CR
C
W
, and
4%
C
W
C
W
64%
C
R
+
16% CR = 80% CR =
0.8 =
p
4%
C
W
+
16%
C
W
=
2
0%
C
W
=
0.2 =
q
64%
C
R
C
R
,
32%
C
R
C
W
, and
4%
C
W
C
W
plants
Genotypes in the next generation:
Slide801. 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
Slide81The following are numbers of pink and white flowers in a population.
Pink White
Generation 1: 901 302Generation 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
Slide82The 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
Slide834. 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
(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
Slide845. 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
Slide85Answers:
1. Parents: Aa x Aa = Offspring: AA (25%), Aa (50%), aa (25%)Answer = A2. A
3. C4. A5. D