Question

2. Pn An What is the equilibrium frequency of A if a. μ-4.9 x 10-6 and v = 3.7 x 10-5? b. μ = 6.2 x 10-6 and v = 8.1 x 10-7? c. If the mutation rates are μ = 4.9 x 10-5 and v ~ 0 what is the allelic frequency pn with starting po-1.0 after 10 generations? d. 100? e. 10,000

Answer 1

When we study the population genetics we have a clear distinction to find out the allele frequencies from the change that has taken place through the Hardy-Weinberg equation. Here in this frequency dynamics we would like to find out the forward and reverse mutations rate change affects the allele frequency.

a. µ = 4.9 x 10-6, v = 3.7 x 10-5. Here the equilibrium frequency of A is

p equilibrium = v / µ + v = 3.7 x 10-5/ 4.9 x 10-6 + 3.7 x 10-5

0.000037/ 0.0000049 + 0.000037 = 0.000037/ 0.0000419 = 0.8831

b. µ = 6.2 x 10-6, v = 8.1 x 10-7, p equilibrium = v / µ + v = 8.1 x 10-7/ 6.2 x 10-6 + 8.1 x 10-7 = 0.00000081/ 0.0000062 + 0.00000081 = 0.00000081/ 0.00000701 = 0.1155

c. Two different allele, the dominant and recessive one. Here the p and q frequencies are given in the p0 generation, where p = 1 and q = 0. Taking the mutation rate into account, µ = 4.9 x 10-5, v ≈0. After 1 generation pn would be taking a formula as—p (0+1) = p0 + vq (0) - µp (0) = 1.0 + 0 - 4.9 x 10-5 = 1.0 – 0.000049 = 0.99995,

p (0+10) = 0.99951,

d. After 100 generation p (0+100) = 0.9951

e. After 100 generation p (0+1000) = 0.951