In a diploid population, allele $A$ has frequency $p=0.4$ and allele $a$ has frequency $q=0.6$. Suppose the fitnesses are $w_{AA}=1$, $w_{Aa}=1$, and $w_{aa}=1-s$ with $s=0.3$ (selection against $aa$). What is the allele frequency of $a$ after one generation of selection?

Step-by-step solution

1. Compute initial genotype frequencies using Hardy–Weinberg: $\text{AA}=p^2=0.16$, $\text{Aa}=2pq=0.48$, $\text{aa}=q^2=0.36$.
2. Apply selection: ${\text{AA}}^{\prime }=0.16\times 1=0.16$, ${\text{Aa}}^{\prime }=0.48\times 1=0.48$, ${\text{aa}}^{\prime }=0.36\times (1-0.3)=0.36\times 0.7=0.252$.
3. Sum to find total fitness: $\bar{w}=0.16+0.48+0.252=0.892$.
4. Normalize genotype frequencies: ${\text{AA}}^{\prime \prime }=0.16/0.892=0.1795$, ${\text{Aa}}^{\prime \prime }=0.48/0.892=0.5384$, ${\text{aa}}^{\prime \prime }=0.252/0.892=0.2825$.
5. Compute new allele frequency of $a$: $q^{\prime }=({\text{Aa}}^{\prime \prime }/2)+{\text{aa}}^{\prime \prime }=0.5384/2+0.2825=0.2692+0.2825=0.5517$.
6. Round to three decimal places: $0.552$.

Concept tested: Allele frequency change due to selection

Published 19 September 2026 · How this solution was prepared and checked