Question

What is the equilibrium frequency of a recessive gene arising with a mutation rate of 4×10−6 and a reproductive fitness in homozygotes of 0.8? What would it be if the gene were partially recessive with h = 0.05? How different are these two equilibrium frequencies? Explain the difference. Calculate and compare the genetic loads under these two conditions (use the full equation for ¯w).

Answer 1

The following is the information provided: Mutation rate (H) 4x106 Relative (reproductive) fitness in homozygotes Equilibrium frequency q- 0.8 where s-selection coefficient-1-relative fitness-1-0.8-0.2 A 4x10-6 Therefore, equilibrium frequency q = V20 x 10-6-10.0045 or 4.5 x 10-3 0.2 Given that the heterozygous effect (h)- 0.05, the equilibrium frequency is then calculated as given below 4x10-6 4x10-6 ^ 00k02 0010.0004 or 4-10 hs 0.05×0.2 0.01 When the heterozygous effect is 5%, the equilibrium frequency is decreased by more than one- tenth or one order in magnitude.

(since p = 1-ql-q is substituted for p) 1-(1-qs) The geneticload (L) (wa is relative fitness or reproductive fitness) Therefore, the genetic loads are calcul ated as follows: * case 1: qxs = 0.0045×0.2 =0.009= 9x10-3 * case 2: qxs= 0.0004×0.2= 8x10-5