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.
