Imagine a population living in part of the world where malaria is very common. I
ID: 58806 • Letter: I
Question
Imagine a population living in part of the world where malaria is very common. In that population, most people are homozygous for normal globin genes; they are vulnerable to Malaria, so their relative fitness is 0. 95. People that are homozygous for a mutated globin gene have a relative fitness of 0. 25 because of anemia. Heterozygotes have intrinsic resistance to malaria, but lack anemia; they have the highest relative fitness. What is the relative fitness of the heterozygote? What is the selection coefficient (s) for individuals homozygous for the normal globin gene? What is the selection coefficient (t) for individuals homozygous for the mutated gene? If the initial frequency of the normal gene in the population (p) is 80% and the initialirequcncy of the, muitated gene in the population (q) is 20%, what will be the change in frequency of the normal gene (Dleta p) after one generation? Predict the stable, internal equilibrium frequency of the mutated globin gene in that population.Explanation / Answer
Relative fitness of the heterozygote can be determined by using below formula
The maximum relative fitness will be more than the mutant genes.
As given,
Relative fitness for an individual with homozygous normal globin genes (WAA) = 0.95
Relative fitness for an individual with homozygous mutated globin genes (Waa) = 0.25
Relative fitness for an individual with Heterozygous normal globin genes (WAa) = ?
The relative fitness can be determined either by 0 or 1. Here, 1 represents the healthy indvidual and 0 represents mutated individual.
WAA + 2(WAa )+ Waa
Here,
= 0.95 + 2(0.95 x 0.25) + 0.25
= 0.95 + 0.475 + 0.25
So, the relative fitness of the heterozygote (WAa ) = 0.475
2) Selection coefficient = s = 1 - WAA
Insert the values in the given formula
= 0.95 (1-s)/1-WAA
= 0.95 (1-S)/97
= 0.98
3) Selection coefficient for individuals homozygous for mutated gene
Waa (1-t)/s
=0.25 (1-T)/s
= 0.5