Question

ay between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the mean realtive fitness within this population? Please give your answer to two decimal places.
Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the expected allele frequency change for A after one generation with selection? Please give your answer to two decimal places.

Answer 1

Answer:

Question 1.

The mean relative fitness is 0.85

Question 2.

The expected allele frequency change of A, p = 0.63: AA, p2 =0.63 x 0.63 = 0.39.

Explanation:

Mean relative fitness is given as

p2W(AA) + 2pqW(Aa) + q2W(aa).

Where W(AA)= 1.00, W(Aa) = 0.80, W(aa) = 0.60

Firstly, find p2, 2pq, q2 which are the frequencies of AA, Aa and aa respectively.

AA= 500,Aa=250 ,aa=250. Total is 1000

Frequency of A, p=

250/2)+500= 625.

625/1000= 0.625.

Therefore A, p is 0.625

p2 = 0.625x0.625= 0.39

For q, since p+q =1

q= 1 - 0.625 = 0.375

q2= 0.375x0.375=0.14.

2pq= 2x0.625x0.375= 0.47.

Mean relative fitness is

p2W(AA) + 2pqW(Aa) + q2W(aa).

0.39x1+ (0.47x0.80) + (0.14x0.60)=0.85.

Question 2.

From Hardy-Weinberg equilibrium

F = p2 + 2pq + q2.

Frequency of A, p= 0.625 and AA, p2 is 0.625 x 0. 0625= 0.39.