A human population in Africa living under a severe malaria selection has the fol

Question

A human population in Africa living under a severe malaria selection has the following relative fitnesses ( w ) of homozygous genotypes in beta - hemoglobin locus: w 11 = 0.75 for A 1 A 1 and w 22 = 0.85 for A 2 A 2 . Using heterozygote advantage concept , determine frequency of the mutant sickle - cell beta - hemoglobin allele (q) in this population assuming that genotype frequencies are at balance ( p = 0; balancing selection). What is the mea n fitness of this population for locus A? Show your calculations and equations used. Explain why this balance is not the same as Hardy - Weinberg balance

Explanation / Answer

Relative fitnesses (w) of homozygous genotypes in beta - hemoglobin locus are,

w 11 = 0.75 for A1A1 and

w 22 = 0.85 for A2A2

The mean fitness of this population at locus A is given by the formula,

p = t / (s+t)

1 - s = 0.75 and

1 - t = 0.85

Therefore s = 0.25 and t = 0.15 respectively.

Substituting these values in the equation,

p = 0.15 / (0.25 + 0.15)

= 0.15 /0.4

= 0.375

The frequency of mutant sickle-cell beta hemoglobin allele is given by the formula,

2pq = 1

2 (0.375) (q) = 1

0.75 (q) = 1

q = 1.33

In Hardy-weinberg balance, the sum of p and q is always 1. Here, p + q is not equal to 1. So, this is not in Hardy-Weinberg's balance.

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