Download Problem Set 9 Key

Problem Set 9, Bio 4181: Due Dec. 6, 2012
1. Consider a Levene model of a panmictic population dispersed every generation across three
environments such that the relative fitnesses for the genotypes determined by an autosomal locus
with 2 alleles are:
Genotype
AA
Aa
aa
1
2
3
1.2
0.8
0.9
1
1
1
0.5
1
1.2
Env.
Which of the following niche outputs under a soft selection model will protect the
polymorphism:
a). c1=0.2, c2=0.2, c3=0.6
b). c1=0.6, c2=0.2, c3=0.2
c). c1=0.3, c2=0.3, c3=0.4
Use the conditions:
a). 0.909090909
b). 0.638297872
c). 0.810810811
1
< 1 and
ci
!v
i
i
1
<1
ci
!w
i
i
0.923076923; protected (2 points)
1.028571429; not protected (2 points)
0.935064935; protected (2 points)
2. Same problem as above, except now consider the hard selection model and interpret all the
c’s as z’s (zygotic inputs).
Use the conditions: v = ! ziv i < 1 and w = ! zi w i < 1
i
a). 0.94
b). 1.06
c). 0.96
i
1.02; not protected (2 points)
0.74; not protected (2 points)
0.93; protected (2 points)
3. Same problems as in 1) and 2), except now assume an island model of gene flow with m=0.1.
For soft selection, first look at vi < 1 ! m = 0.9 and wi < 1 ! m = 0.9
In a, b, and c, v2= 0.8<.9, and w1=0.5<.9, so protected in all cases (4 points).
For hard selection, first look at ! i = (1" m)v i + m# ziv i $ 1" m and
i
! i = (1" m)w i + m# zi w i $ 1" m
i
a). ν 2=0.814 < 0.9, and ω 1=0.552 < 0.9, so protected (2 points)
b). ν 2=0.826 < 0.9, and ω 1=0.524 < 0.9, so protected (2 points)
c). ν 2=0.816 < 0.9, and ω 1=0.543 < 0.9, so protected (2 points)
4. A large population consists of only AA individuals at an autosomal locus with random mating
such that the population is stable in size and the number of progeny is given by a Poisson
distribution. A new mutation occurs, a, such that Aa individuals have an average of 1.05
offspring and a variance of 4 in the number of offspring.
a) What is the probability of a surviving in this large population?
Use equation 14.20 (2 pts). First calculate s=0.05 (1 pt). Second, note that the variance in
offspring number = 4 =1+s+σ s2 (1 pt), so the probability of survival = 0.1/4 = 0.025 (1 pt)
b) Another new mutation, a’, occurs such that Aa’ individuals have a Poisson number of
progeny, but with a mean of 1.04. Is a more likely than a’ to survive in this population?
First calculate s=0.04 (1 pt). Second, note that the variance in offspring number =1+s =
1.04 (from the Poisson assumption; 2 pts), so the probability of survival = 0.08/1.04 =
0.0769 (1 pt). Therefore, a’ is more likely to survive than a (1 pt).
5. A recessive allele, a, is associated with two different life history phenotypes in a population
with overlapping generations such that:
AA & Aa
lx
mxbx
0.8
0
0.6
0.5
0.5
1
0.2
1
0.1
0.2
x
0
1
2
3
4
aa
lx
mxbx
0.5
0
0.2
4
0.06
3
0.04
1
0.01
0
Calculate the fitness of both phenotypes using a) the net reproductive rate, and b) the Malthusian
parameter (use an approximation).
Use equations 15.1 (1 pt) and 15.10 (1 pt).
x
0
lx
0.8
mxbx
0
AA & Aa
lxmxbx
0
xlxmxbx
0
lx
mxbx
0.5
0
aa
lxmxbx
0
xlxmxbx
0
1
2
3
4
0.6
0.5
0.2
0.1
0.5
1
1
0.2
Sums
0.3
0.5
0.2
0.02
1.02
0.3
1
0.6
0.08
1.98
0.2
0.06
0.04
0.01
a) R0 = 1.02 for AA and Aa (1 pt).
R0 = 1.02 for aa (1 pt).
b) r ≈ 0.0101 for AA and Aa (1 pt).
r ≈ 0.0156 for aa (1 pt).
4
3
1
0
0.8
0.18
0.04
0
1.02
0.8
0.36
0.12
0
1.28
c). Which genotype(s) has the highest fitness, using the more appropriate fitness measure for
this population?
Because the population has overlapping generations, r is the more appropriate fitness
measure (2 pts). Hence, aa has the highest fitness (1 pt).