20 POINTSSS PLEASE HURRYYY

C is the correct answer

The equilibrium frequency of the A1 (p = v/(u+v)) is expected to be equal to the equilibrium frequency of the A2 (q = 1 - p)

Explanation:

Available data:

Locus with two alleles, A1 and A2The probability of A1 mutating to A2 is P(A1->A2)=uThe probability of A2 mutating to A1 is P(A2->A1)=vV=Unor V or U is equal to cerop is the frequency for A1 and q is the frequency of A2.

The equilibrium frequency of the A1 is expected to be p = v/(u+v)

The equilibrium frequency of the A2 is expected to be q=1-p=1 - (v/(u+v))

For example, if V = U = 0.005, then

p = 0.005/(0.005+0.005) = 0.005/0.01 = 0.5q = 1-p = 1-0.5 = 0.5  

The equilibrium frequency of A1 = The equilibrium frequency of A2. The rate of mutation is equal for both alleles.