Download Problem Set 6 Key

Problem Set 6, Bio 4181: Due Nov. 1, 2012
1. A population is polymorphic at an autosomal locus with two alleles, A and a. The probability is 0.5 for an
AA zygote surviving to adulthood, 0.4 for an Aa zygote, and 0.6 for an aa zygote. The probability is 0.6 for an
AA adult to successfully mate, 0.9 for an Aa adult, and 0.4 for an aa adult. A mated AA individual produces an
average of 9 offspring, a mated Aa individual produces 11 offspring, and a mated aa individual produces 15
offspring.
a) What are the expected numbers of offspring for AA, Aa and aa zygotes?
Genotype:
lx
c
b
w=Av. No. Off.
AA
0.5
0.6
9
2.7
Aa
0.4
0.9
11
3.96
aa
0.6
0.4
15
3.6
(3 pts)
b) Assign a fitness of 1 to aa. What are the relative fitnesses of the AA and Aa genotypes?
2.7/3.6=0.75 for AA; 1.1 for Aa (2 pts).
c) Given random mating and a frequency of the A of 0.7 in the zygotes, what are the adult genotype
frequencies?
lx
zygote freq.
Adult freq.
AA
0.5
0.49
0.525
Aa
0.4
0.42
0.360
aa
0.6
0.09
0.116
p
0.7
ave. lx= 0.467
q
0.3
(1 pt for zygote freq., 1 pt. for getting average lx, 1 pt for adult freq.)
d) What are the breeding values for the phenotype of fitness in the population described in part c.
w
zygote freq.
Genotypic D.
Average Excs
Breeding Value
AA
Aa
0.75
1.1
0.49
0.42
-0.170
0.181
A: -0.0645
-0.129
0.086
aa
p
1
0.09
0.7
0.081
ave. w= 0.9195
a: 0.1505
0.301
q
0.3
(1 pt. for average fitness, 2 pts. for average excesses; 1 pt for breeding values)
e) According to Fisher’s fundamental theorem of natural selection, is the population in part c at equilibrium?
Why or why not?
Not at equilibrium. Because the breeding values differ among the genotypes, there is a positive additive
variance of fitness. According to FFTNS, at equilibrium the additive variance must be 0, so the
population can’t be at equilibrium (2 pts).
2. The relative fitnesses in a snail population for a gene conferring cryptic coloration has been estimated to be
0.6, 1.0 and 0.8 for genotypes RR, RB, and BB, respectively. Assume random mating.
a. What is the initial change in the frequency of R starting from a gene pool in which the frequency of R
p
is 0.25? average fitness = 0.8625 (1 pt.). Average excess of R = 0.0375 (2 pts). !p = aA = 0.0109 (2 pts).
w
b. What is the initial change in the frequency of R starting from a gene pool in which the frequency of R
p
is 0.75? average fitness = 0.7625 (1 pt.). Average excess of R = -0.0625 (2 pts). !p = aA = -0.0615 (2 pts).
w
c. Calculate the intermediate equilibrium allele frequency for R. Is this equilibrium stable?
Use: ave. excess R = ave. excess B = 0 => (1-p)(wRB-wBB) = p(wRB-wRR) (2 pts), so p=1/3 (1 pt). The
equilibrium is stable because when p is below it, selection increases the allele frequency (part a), and
when p is above the equilibrium, selection decreases p (part b) (2 pts).
3. The mean and variance of fitness in a deme is 1.5 and 1 respectively. The heritability of fitness is 0.5. What
is the mean fitness in the next generation?
" a2
Use the FTNS (2 pts), !w =
, We know that w = 1.5 and that the additive variance is 0.5(1)=0.5, so
w
!w = 0.5/1.5 = 0.3333 (1 pt). Hence the average fitness in the next generation is 1.5+.3333= 1.8333 (2 pts).
4. The foraging efficiency of the individuals in a random mating population of infinite size is determined.
Under what conditions do you expect natural selection to maximize foraging efficiency given that foraging
efficiency contributes to reproductive fitness?
From FFTNS, foraging efficiency will be maximized if it has no phenotypic variance at equilibrium (2
pts) or if it contributes to fitness in a strictly linear fashion (2 pts).