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

According to Hardy-Weinberg equilibrium, the allelic frequencies and proportion of individuals in a population can be calculated by

p² + 2pq + q² = 1

The given information in the question the homozygous recessive individuals HbA/HbA 0.53 and population is in equilibrium

recessive allelic frequency for malaria q2=0.53

q=0.72

p+q=1

dominant allelic frequency for malaria

p=1-0.72

p=0.28

Based on the equation, heterozygous individuals in the population

2pq=2x0.72x0.28

      =0.4032

Percentage or proportion of the population with heterozygousHbA/HbS for the malarial allele is 0.40 x 100 = 40%