Question

if 36% of a population is homozygous recessive, what percentage of the population is expected to be heterozygous? 24% 36% 48% 64%

Explanation:

Step1: Calculate the frequency of recessive allele ($q$)

According to Hardy - Weinberg equilibrium, if the frequency of homozygous recessive ($q^{2}$) is $0.36$, then $q=\sqrt{0.36}=0.6$.

Step2: Calculate the frequency of dominant allele ($p$)

Since $p + q=1$, then $p = 1 - q=1 - 0.6 = 0.4$.

Step3: Calculate the frequency of heterozygous ($2pq$)

Substitute $p = 0.4$ and $q = 0.6$ into $2pq$, we get $2\times0.4\times0.6=0.48 = 48\%$.

Answer:

C. 48%