In a population of $1000$ individuals, the frequency of a recessive allele $a$ is $q=0.4$. Assuming Hardy-Weinberg equilibrium, what is the expected number of heterozygous individuals in the next generation?

Solution

1. Given the frequency of recessive allele $q=0.4$, the frequency of dominant allele $$\begin{array}{rl}p& =1-0.4\\ & =0.6\end{array}$$.
2. The frequency of heterozygotes in Hardy-Weinberg equilibrium is $2pq$.
3. Calculating $$\begin{array}{rl}2pq& =2\times 0.6\times 0.4\\ & =0.48\end{array}$$.
4. The number of heterozygous individuals is $0.48\times 1000=480$.
5. Hence the answer is (B).