Hardy–Weinberg Equilibrium: Allele Frequencies, p²+2pq+q², and the χ² Test

Hardy–Weinberg says that in a population that isn’t evolving, genotype frequencies follow p² + 2pq + q² = 1 from the allele frequencies p and q. Get p and q from the genotype counts (p = (2·AA + Aa) / 2N), work out the expected counts, and a chi-square test tells you whether the population is at equilibrium. Do all of it — allele frequencies, expected counts and the χ² verdict — with the Hardy–Weinberg calculator; here’s the reasoning.

The idea in one picture

Step 1 — allele frequencies from genotype counts

Count the three genotypes and turn them into allele frequencies. Each homozygote carries two copies of its allele, each heterozygote one of each:

p = (2·AA + Aa) / (2N) and q = 1 − p

where N is the total number of individuals. For a classic textbook sample of 298 AA, 489 Aa and 213 aa (N = 1000): p = (2×298 + 489) / 2000 = 1085 / 2000 = 0.5425, so q = 0.4575.

Step 2 — expected genotype frequencies

If the population is at equilibrium, the genotype frequencies come straight from p and q:

• p² — homozygous dominant (AA)
• 2pq — heterozygous (Aa)
• q² — homozygous recessive (aa)

These sum to 1 because (p + q)² = p² + 2pq + q² = 1. Multiply each by N to get expected counts. For our example: p²·N ≈ 294, 2pq·N ≈ 496, q²·N ≈ 209.

Step 3 — test with chi-square

Now compare what you observed to what’s expected under Hardy–Weinberg with a chi-square goodness-of-fit test:

χ² = Σ (observed − expected)² / expected

with 1 degree of freedom (three genotype classes, minus one for the total, minus one for the estimated allele frequency). Our example gives a small χ² (≈ 0.22) and a p-value around 0.6 — well above 0.05 — so the population is consistent with Hardy–Weinberg equilibrium. A p-value at or below 0.05 would signal a significant departure. The Hardy–Weinberg calculator runs this test and gives the plain-English verdict.

What a departure means

Hardy–Weinberg is a null model: it describes a population where nothing is changing the allele frequencies. So when observed genotypes depart significantly from the expectation, it means one of the model’s assumptions is being violated:

• Natural selection — some genotypes survive or reproduce better.
• Non-random mating — for example inbreeding, which raises homozygotes.
• Migration (gene flow) — alleles entering or leaving the population.
• Mutation — new alleles appearing.
• Genetic drift — random change in a small population.

The test doesn’t tell you which — it tells you that the simple no-evolution model doesn’t fit, which is the starting point for asking why.

Quick summary

Hardy–Weinberg predicts genotype frequencies from allele frequencies: p + q = 1 and p² + 2pq + q² = 1. Get p and q from counts with p = (2·AA + Aa)/2N, compute expected counts, and test observed-versus-expected with a chi-square goodness-of-fit at 1 degree of freedom — p > 0.05 means the population is consistent with equilibrium. Run the frequencies, expected counts and the χ² verdict with the Hardy–Weinberg calculator, and see where a single cross comes from with the Punnett square calculator.

Sources: Hardy (1908) and Weinberg (1908), the Hardy–Weinberg principle; standard population-genetics texts. NHGRI — genetics glossary. Educational information.