Hardy–Weinberg equilibrium and chi-square calculator

How to use (3 steps)

Choose the input mode: genotype counts (AA, Aa, aa) or allele frequencies p, q.
Enter counts or p (and optional sample size N). Default values use a simple teaching example.
Tap Compute to see p, q, expected genotype frequencies, chi-square (Genotype mode), and the steps. In Allele mode only expected frequencies are shown. Copy URL shares the exact setup.

Interpretation: a non-significant χ² means the sample does not show a detectable deviation from Hardy–Weinberg equilibrium at the chosen α.

Inputs

Input mode

Genotype counts Allele frequencies

Use observed genotype counts to compute p, q, and χ².

Assume p (and optional N) to see HWE expected frequencies only.

Allele 1 label

Allele 2 label

Genotype counts

Count of homozygous for allele 1 (e.g. AA) individuals

Count of heterozygous (e.g. Aa) individuals

Count of homozygous for allele 2 (e.g. aa) individuals

Significance level α

Auto-calculate

Results

Chi-square test

df = 1 because p is estimated from the sample.

Visualizations

Observed vs expected counts

Hardy–Weinberg genotype frequencies at this p

How it is calculated

Formulas used: p = (2·AA + Aa)/(2N); q = 1 − p; expected counts: N·p², N·2pq, N·q²; χ² = Σ (O−E)²/E with df = 1.

FAQ

What is Hardy–Weinberg equilibrium?

It is an idealized model where allele and genotype frequencies stay constant under assumptions like random mating and no selection, mutation, or migration. This calculator uses that assumption to compute expected frequencies.

Does a non-significant chi-square prove equilibrium?

No. It only means the sample does not show a statistically detectable deviation at the chosen α. It does not prove perfect equilibrium.

What if expected counts are small?

When expected counts are small, the chi-square approximation weakens. Consider exact tests such as Fisher's exact test for very small counts.

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