Answer: Option B
:
B
According to Hardy-Weinberg equation,
$p^2+2pq+q^2=1andp+q=1$
Assuming that ‘p’ is the frequency of occurrence of the dominant allele ‘A’ and ‘q’ is the frequency of occurrence of the recessive allele ‘a’, we could say that the frequency of occurrence of the genotype aa=36%
$\Rightarrow q\times q=36$
$\Rightarrow q^2=0.36$
$\Rightarrow q=\sqrt{0.36}$
$\Rightarrow q=0.6$
Therefore the frequency of recessive allele a = 0.6
Applying the equation p+q = 1
$\Rightarrow p=1-q$
$\Rightarrow p=1-0.6=0.4$
Therefore, the frequency of dominant allele A = 0.4
The frequency of the homozygous genotype $\left(AA\right)=p^2$
$p^2=(0.4\times 0.4)=0.16$
The frequency of the heterozygous genotype (Aa)=2pq
$2pq=2\times 0.4\times 0.6=0.48$