In a human population at Hardy-Weinberg equilibrium, the recessive gene is homozygous in 64% of the population. What will be the frequency of heterozygotes in the population?

The correct option is B 32%
The frequency of recessive gene is 64% that is 0.64, and it is represented as $q^2$
q= $\sqrt{q^2}$ = $\sqrt{0.64}$= 0.8
Since p + q = 1
p + 0.8 = 1
p = 1 - 0.8 = 0.2

Using the Hardy-Weinberg equation $p^2+2\mathrm{pq}+q^2=1$,
heterozygotes are represented 2pq
Therefore, 2pq= $2\times 0.8\times 0.2$
2pq = 0.32 or 32%
So, 32% of the individuals in the population are heterozygous for the given gene.