Question

The relative fitness of three genotypes at a particular locus are W₁₁=0.65, W₁₂=0.75, W₂₂=0.9. Four populations have the following set of allele frequencies. In which population will the change in allele frequencies (due to selection) be the most rapid? (Show work) CLEAR typing explanation please. (A, B, C, and D are answer choices not givens in the problem so explain and select one answer choice)

A.    p = 0.86 & q = 0.14

B.    p = 1.0 & q = 0.0

C.   p = 0.05 & q = 0.95

D. p = 0.49 & q = 0.51

Answer 1

Answer :)

We have values of relative fitness. Therefore, we can estimate the allele frequency by using the following method.

Allele frequency of allele w1 (p) = Total relative fitness of w1 allele / Total fitness of the population

Total relative fitness of w1 allele = 0.65 + 0.75/2

We divide the fitness of w12 by 2 because it includes the fitness of both alleles.

Total relative fitness of w1 allele = 1.025

Total fitness of the population = 0.65 + 0.75 + 0.9

Total fitness of the population = 2.3

Therefore, the Allele frequency of allele w1 (p) = 0.45

If the population is in equilibrium, then the allele frequency of the second allele will be 1-p due to the Hardy Weinberg principle.

Allele frequency of w2 allele (q) = 1 – p

Allele frequency of w2 allele (q) = 1 – 0.45

Allele frequency of w2 allele (q) = 0.55

Option D is nearer to the calculated values. Therefore, option D is the correct answer.

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