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 0.31, and the population is assumed to be at Hardy-Weinberg equilibrium, what proportion of the population should be heterozygous HbA/HbS?

Solution

Answer :

susceptible to malaria (genotype HbA/HbA) is 0.31

q2=0.31

q= sq root of 0.31=0.55

p+q=1

p=1-q=1-0.55=0.45

Non susceptible to malaria (genotype HbS/HbS)= 0.45

heterozygous HbA/HbS

Hardy weinberg equation

p^2 +2pq+q^2=1

2pq= (HbS/HbS) = 2 x 0.45 x 0.55=0.49

Proportion of the population which is heterozygous HbA/HbS=49%