In a population of 150 individuals the observed genotype numbers are: 36 $AA$, 72 $Aa$, and 42 $aa$. (i) Calculate the allele frequencies of $A$ and $a$. (ii) Using a chi‑square test, determine whether the population is in Hardy–Weinberg equilibrium (use ${\chi }_{0.05,1}^2=3.841$).

Step-by-step solution

1. Total number of alleles = $2\times 150=300$.
2. Count $A$ alleles: $2\times 36+72=144$. Hence $p=144/300=0.48$.
3. Count $a$ alleles: $2\times 42+72=156$. Hence $q=156/300=0.52$.
4. Compute expected genotype frequencies under H‑W: $p^2=0.2304$, $2pq=0.4992$, $q^2=0.2704$.
5. Convert frequencies to expected numbers: $A{A}_e=0.2304\times 150=34.56$, $A{a}_e=0.4992\times 150=74.88$, $a{a}_e=0.2704\times 150=40.56$.
6. Calculate chi‑square components: ${\chi }_{AA}^2=\frac{(36-34.56{)}^2}{34.56}=0.060$, ${\chi }_{Aa}^2=\frac{(72-74.88{)}^2}{74.88}=0.111$, ${\chi }_{aa}^2=\frac{(42-40.56{)}^2}{40.56}=0.051$.
7. Sum the components: ${\chi }^2=0.060+0.111+0.051=0.222$.
8. Compare with the critical value $3.841$ (df=1). Since $0.222<3.841$, the null hypothesis of Hardy–Weinberg equilibrium is not rejected.
9. Therefore, the population is in Hardy–Weinberg equilibrium.

Concept tested: Hardy–Weinberg equilibrium test

Published 24 September 2026 · How this solution was prepared and checked