From a diploid population of 120 individuals, genotype counts are: 30 $AA$, 70 $Aa$, and 20 $aa$. (i) Calculate the allele frequencies of $A$ and $a$. (ii) Using a chi–square test with $ch{i}_{0.05,1}^2=3.841$, determine whether the population is in Hardy–Weinberg equilibrium. Provide the chi-square value and the conclusion.

Step-by-step solution

1. (i) Count total alleles: $2\times 120=240$. Count $A$ alleles: $2\times 30+70=130$. Thus $p=\frac{130}{240}=0.5417$ and $q=1-p=0.4583$.
2. (ii) Expected genotype frequencies: $p^2=0.2935$, $2pq=0.4965$, $q^2=0.2100$.
3. Convert to expected counts for 120 individuals: $35.22$ $AA$, $59.58$ $Aa$, $25.20$ $aa$.
4. Compute chi-square: ${\chi }^2=\frac{(30-35.22{)}^2}{35.22}+\frac{(70-59.58{)}^2}{59.58}+\frac{(20-25.20{)}^2}{25.20}=0.774+1.823+1.073=3.67$.
5. Compare with critical value 3.841: $3.67<3.841$, so we fail to reject null hypothesis. The population is in Hardy–Weinberg equilibrium.

Concept tested: Hardy–Weinberg equilibrium test

Published 1 October 2026 · How this solution was prepared and checked