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: w11 = 0.75 for A1A1 w22 = 0.85 for A2A2 . Assume that w12 = 1.00 for A1A2 . 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). why this population not considered a Hardy Weinberg Population

Answer 1

Relative fitness (w) of homozygous genotypes in beta - haemoglobin locus are,

w 11 = 0.75 for A1A1

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

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 haemoglobin 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.