The M allele is at a frequency of 0.2 in a population of 500 west-slope cutthroa
ID: 37061 • Letter: T
Question
The M allele is at a frequency of 0.2 in a population of 500 west-slope cutthroat trout from Rock Creek that is in Hardy-Weinberg proportions. An outbreak of whirling disease kills 30% of the individuals with the MM genotype, but none of the other fish (i.e., mm or Mm fish)
a. What is the frequency of each genotype before the disease outbreak?
b. Based on these data, what are the fitnesses of each genotype and what is the mean fitness of the population?
c. What is the frequency of each genotype after the disease outbreak?
d. After the outbreak described above, the surviving fish mate at random to produce the next generation. What are the expected allele and expected genotype frequencies in the next generation?
Explanation / Answer
Based on the given data, the total population = 500
a)
The genotypic frequencies for this population before disease:
p = f(A) = 0.2
q = f(a) = 0.8
f(AA) = p2 = f(A) × f(A) = 0. 04
f(Aa) = 2pq = 2 × [f(A) × f(a)] = 0.32
f(aa) = q2 = f(a) × f(a) = 0.64
Thus, p2 + 2pq + q2 = 0.04 + 0.32 + 0.64 = 1 (It is in Hardy-Weinberg equilibrium)
Thus, the frequency of each genotype before the disease outbreak is MM = 0.04, Mm = 0.32 and mm = 0.64.
Therefore, the number of individuals with frequency MM are 20, Mm are 160 and mm are 320 before the outbreak of the disease.
b)
The fitnesses of each genotype and the mean fitness of the population are calculated as:
w = fitness thus, wMM represents the relative fitness of genotype MM, wMm the relative fitness of genotype Mm, and wmm the relative fitness of genotype mm.
The mean fitness of the population at generation 0 is:
Mean fitness = p2(wMM) + 2pq(wMm) + q2(wmm) 0.04 + 0.32 + 0.64 = 1
c)
The genotypic frequencies after the disease outbreak are given as:
The individuals with genotypic frequency MM (p2) is 0.04
Therefore, the number of individuals with p2 frequency is
0.04 × 500 = 20
From the given data, 30% of the MM genotypic fishes are eliminated.
30/100 × 20 = 6 individuals died of the disease
Therefore, the number of individuals becomes 500 – 6 = 494
Hence, the frequencies after disease outbreak are calculated as:
p2 frequency (MM) = 14/494 = 0.028
2pq frequency (Mm) = 320/494 = 0.648
q2 frequency (mm) = 160/494 = 0.324
d)
Allelic frequency after random mating is given as:
p = p2 + 1/2 pq
= 0.028 + 0.162
= 0.19
q = q2 + 1/2 pq
= 0.648 + 0.162
= 0.81
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