Consider a locus with two alleles, A1 and A2. Mutation happens at this locus, but in a way such that A1 mutates to A2 with probability u and A2 mutates to A1 with probability v. If u = v (and neither of them = 0), the equilibrium frequency of the A1 is expected to be [Hint: set u and v to specific numbers that are equal to see what happens]:

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Biology

Wade Kelley

17 August, 23:04

Consider a locus with two alleles, A1 and A2. Mutation happens at this locus, but in a way such that A1 mutates to A2 with probability u and A2 mutates to A1 with probability v. If u = v (and neither of them = 0), the equilibrium frequency of the A1 is expected to be [Hint: set u and v to specific numbers that are equal to see what happens]:

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Nyasia Allen · Answer 1

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:

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Locus with two alleles, A1 and A2 The probability of A1 mutating to A2 is P (A1->A2) = u The probability of A2 mutating to A1 is P (A2->A1) = v V=U nor V or U is equal to cero p 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 = 0.5 q = 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.