In a village where the proportion of individuals who are sus

In a village where the proportion of individuals who are susceptible to malaria (genotype HbA/HbA) is 0.53, and the population is assumed to be at Hardy-Weinberg equilibrium, what proportion of the population should be heterozygous HbA/HbS?

Solution

Given, p² = 0.53

so p = 0.728

q = 1-p = 1 - 0.728 = 0.272

Therefore proportion of heterozygous HbA/HbS = 2 x 0.728 x 0.272 = 0.396