7.5 Hardy–Weinberg Equilibrium

Syllabus
2025
Topic
7.5
Level

Learning objectives

Using Hardy–Weinberg as a Null Model

Hardy–Weinberg equilibrium predicts allele and genotype frequencies in a non-evolving population. Its conditions are an idealized null hypothesis: a reference against which observed population frequencies can be compared.

Equilibrium condition If the condition is violated
Large population Genetic drift can change allele frequencies
No migration Gene flow can add or remove alleles
No new mutations Mutation can introduce new variation
Random mating Genotype proportions can depart from random-mating expectations
No natural selection Differential reproductive success can change allele frequencies

p+q=1$p$ = frequency of allele 1; $q$ = frequency of allele 2.

p^2+2pq+q^2=1$p^2$ = expected frequency of allele-1 homozygotes; $2pq$ = expected heterozygote frequency; $q^2$ = expected frequency of allele-2 homozygotes.

Worked example for a hypothetical equilibrium population: if p = 0.70, then q = 1 − 0.70 = 0.30. Expected genotype frequencies are p² = (0.70)² = 0.49, 2pq = 2(0.70)(0.30) = 0.42, and q² = (0.30)² = 0.09. Check: 0.49 + 0.42 + 0.09 = 1.00. These are unitless proportions, equivalent to 49%, 42%, and 9%.

The equations produce expected frequencies only when Hardy–Weinberg conditions apply. A difference between observed and expected genotype frequencies does not identify the cause by itself; it shows that the equilibrium model or its assumptions should be investigated.