What is the coefficient(s) for individuals homozygous for normal globin gene? Wh

Question

What is the coefficient(s) for individuals homozygous for normal globin gene?
What is the selection coefficient (t) for individuals homozygous for mutated gene?
If the initial frequency of the normal gene in the population (p) is 80% and the initial frequency of the mutated gene in the population (q) is 20%, what will the change in frequency of the normal gene (^p) after one generation? delta p
Predict the stable internal equilibrium frequency of the mutated globin gene in that population


PLEASE EXPLAIN

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. 2.

Explanation / Answer

relative fitness of homozygotesfor normal globin gene HbAHbA = 0.95 = Rep rate of HbAHbA /highest rep rate

highest rep. rate = Rep rate of HbAHbA/ 0.95

relative fitness of mutated globin gene HbsHbs= 0.25

highest rep. rate = Rep. rate of HbsHbs / 0.25

Relative fitness of heterozygote = HbAHbs = reproductive rate of heterozygote/ highest reproductive rate. So, we need to know the individual reproductive rate and /or survival rates for calculating this problem.

We can assume relative fitness of heterozygote to be 1, because it is given that the heterozygote has highest relative fitness

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