Hardy-Weinberg equilibrium test: chi-square, exact test, inbreeding coefficient F and de Finetti diagram

Test whether one sample's genotype counts fit Hardy-Weinberg proportions. Gives allele frequencies, observed and expected heterozygosity, the inbreeding coefficient F, chi-square with optional continuity correction, and the exact test (two-sided and one-sided), with a de Finetti diagram. Also handles three or more alleles by chi-square.

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Genotypes
p and q
HWE test
Diagram

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Genotype counts

AA AB BB Allele names

Continuity correction (Yates) for the chi-square test

Hardy-Weinberg equilibrium predicts genotype shares p², 2pq and q² from the allele frequencies. The chi-square test is an approximation that needs expected counts of about 5 or more; Yates's correction makes it more cautious in small samples. The exact test (two alleles) is correct at any sample size and is the one to report. F = 1 − Ho/He: above 0 means too few heterozygotes, below 0 too many. The example counts are made up.

Result

Observed and expected

de Finetti diagram

Each corner is a genotype; a sample is a point whose distances from the three sides are its genotype shares. The curve is every population in Hardy-Weinberg equilibrium. Points below the curve have too few heterozygotes; above it, too many.

Exact test: every possible heterozygote count

With these allele counts fixed, the chance of each possible number of heterozygotes under Hardy-Weinberg. The exact p-value adds up every count no more likely than the one observed.

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What it does

In a large population with random mating and no selection, migration or mutation at a gene, genotype shares settle at p², 2pq and q², where p and q are the two allele frequencies. This is Hardy-Weinberg equilibrium (HWE). Testing it is a classic teaching exercise. It is also a routine quality check for genotyping: a SNP far out of HWE in healthy samples often means a genotyping error.

This page tests one sample at one gene (locus). It gives the allele frequencies, the expected counts, the chi-square test (with Yates’s continuity correction if you want it), the exact test of Wigginton and others (two-sided, plus one-sided for a heterozygote deficit or excess), the inbreeding coefficient F = 1 − Ho/He, and a de Finetti diagram.

How to use it

Two alleles: type the counts of the three genotypes, such as AA, AB and BB. You can rename the alleles; the labels follow.
Three or more: one row per genotype, written with a slash (A/B), then its count. Genotypes you leave out count as zero. Paste from a spreadsheet or open a CSV or Excel file. This mode uses chi-square with k(k − 1)/2 degrees of freedom; the exact test and the diagram are for two alleles.
Continuity correction: subtracts ½ from each |O − E| before squaring. This makes the chi-square test more cautious in small samples. The exact test does not need it.

Reading the de Finetti diagram

The triangle’s corners are the three genotypes. Each sample is one point, placed so that its distances to the sides are its genotype shares. The curve is the Hardy-Weinberg parabola: every possible population in equilibrium. Points below the curve have too few heterozygotes, as from inbreeding or mixing populations; points above it have too many. The second chart shows the exact test itself: the probability of each possible heterozygote count, given the allele counts.

Worked example

The made-up sample has 12 AA, 70 AB and 18 BB, so p(A) = 0.47 and Hardy-Weinberg expects 22.09, 49.82 and 28.09. There are 70 heterozygotes where 49.82 are expected: Ho = 0.7 against He = 0.4982, so F = −0.4051, a heterozygote excess. Chi-square is 16.41 (p = 5.11 × 10⁻⁵), or 15.21 with the continuity correction. The exact test gives p = 0.000109 two-sided. The one-sided test for too many heterozygotes gives p = 5.67 × 10⁻⁵, and the one for too few gives p = 1.00. The point sits well above the curve in the diagram.

In Three or more, the made-up three-allele sample gives chi-square 4.254 with 3 degrees of freedom, p = 0.235, and F = 0.1179.

This tool or the allele frequency calculator?

Use this Hardy-Weinberg tester for one sample at one locus: learning HWE, checking a SNP’s genotype calls, one-sided tests, the continuity correction, more than two alleles, and the de Finetti picture. Use the allele frequency calculator when you have several populations and want to compare them with F-statistics (FIS, FST, FIT, by Weir and Cockerham or Nei).

FAQ

Why only 1 degree of freedom for 3 genotypes? One is lost to the fixed total and one to the allele frequency, which is estimated from the same data.

Which p-value should I report? The exact test: it is valid at every sample size. Use the one-sided versions only if you decided the direction before looking.

Is my data uploaded? No. Everything runs in your browser.

Where the numbers come from

Hardy-Weinberg principle (G. H. Hardy 1908; W. Weinberg 1908). Exact test: J. E. Wigginton, D. J. Cutler and G. R. Abecasis (2005), Am. J. Hum. Genet. 76: 887-893. Continuity correction: F. Yates (1934). de Finetti diagram: B. de Finetti (1926). Formulas only; no tables are copied.
Checked in the tests: two-sided and one-sided exact p-values against a full enumeration with exact fractions in Python; Yates-corrected chi-square against NumPy and scipy.stats.chi2; three-allele chi-square, He and F against an independent NumPy calculation.
The example counts are made up.

Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.

Last updated 2026-11-09.