Zoology - Evolution Question with Solution | TestHub

In a stable population, for a gene with two alleles, 'A' (dominant) and 'a' (recessive), the frequency of ' ${\mathrm{A}}^{\prime }=\mathrm{p}$ the frequency of ' $a$ ' $=q$, According Hardy-Weinberg equation: $${\mathrm{p}}^2+2pq+q^2=1$$ where ${\mathrm{p}}^2=$ Frequency of $\mathrm{A}$ (homozygous dominant) individuals ${\mathrm{q}}^2=$ Frequency of aa (homozygous recessive) individuals $2\mathrm{pq}=$ Frequency of $\mathrm{Aa}$ (heterozygous) individuals. $$\begin{array}{rl}& \therefore 2\mathrm{pq}(\text{ frequency of heterozygote })=\\ & 2\times 0.6\times 0.4=0.48\end{array}$$