What is allele frequency?

Allele frequency is the proportion of all chromosomes in the population carrying a particular allele, denoted p for A and q for a in a two-allele system.

What assumptions underlie Hardy–Weinberg equilibrium?

Random mating, infinite/large population, no selection, no mutation, no migration, and no genetic drift.

Can I test for HWE from allele counts only?

Not exactly. Allele counts can give p and q, but without genotype counts you cannot perform a valid HWE goodness-of-fit test.

Why might my data deviate from HWE?

Population structure, assortative mating, selection, inbreeding, genotyping error, or small sample size.

Minor allele frequency (MAF) is the frequency of the less common allele: min(p, q).

What confidence interval is reported for p?

A Wilson score interval for k successes out of n=2N allele trials at 95% confidence.

Is this tool suitable for clinical diagnostics?

No. It is intended for educational and research use. Clinical decisions require validated pipelines and expert oversight.

Allele Frequency Calculator

This professional-grade allele frequency calculator helps students, educators, and researchers compute allele frequencies (p and q), expected genotype frequencies under Hardy–Weinberg equilibrium (HWE), heterozygosity, minor allele frequency (MAF), a chi-square HWE test (when genotype counts are provided), and 95% confidence intervals — all with WCAG-compliant, mobile-first UX.

Interactive Calculator

Choose input mode

Genotype counts

Allele counts

Known allele frequency p

AA count *

Aa count *

aa count *

Results

Sample size (N) —

Allele frequency p (A) —

Allele frequency q (a) —

Minor allele frequency (MAF) —

Expected genotypes (HWE): AA, Aa, aa —

Observed genotypes: AA, Aa, aa —

Heterozygosity (2pq) —

95% CI for p (Wilson) —

HWE chi-square (df=1), p-value —

Interpretation Provide genotype counts to test HWE.

Note: HWE test is computed only when observed genotype counts are provided and N > 0 with valid expected counts.

Data Source and Methodology

Authoritative sources:

G.H. Hardy (1908), Mendelian proportions in a mixed population, Science 28(706):49–50. Link
W. Weinberg (1908), Über den Nachweis der Vererbung beim Menschen, Jahreshefte des Vereins für vaterländische Naturkunde in Württemberg.
J.F. Crow and M. Kimura (1970), An Introduction to Population Genetics Theory. Harper & Row.
Wigginton, Cutler, Abecasis (2005), A note on exact tests of Hardy–Weinberg equilibrium, AJHG 76:887–893. Link
Nature Education: Hardy–Weinberg Equilibrium. Link

All calculations are strictly based on the formulas and data provided by these sources.

The Formula Explained

If AA, Aa, aa are genotype counts and N = AA + Aa + aa:

Allele frequencies: $$p = \\frac{2\\,AA + Aa}{2N}, \\quad q = 1 - p$$

Hardy–Weinberg expected genotype frequencies: $$P(AA) = p^2,\\quad P(Aa) = 2pq,\\quad P(aa) = q^2$$

Expected counts: $$E[AA] = N p^2,\\quad E[Aa] = N (2pq),\\quad E[aa] = N q^2$$

Heterozygosity: $$H = 2pq$$

Goodness-of-fit statistic (chi-square): $$\\chi^2 = \\sum_{g\\in\\{AA,Aa,aa\\}} \\frac{(O_g - E_g)^2}{E_g} \\quad \\text{with 1 degree of freedom}$$

For df = 1, the p-value is: $$p\\text{-value} = \\operatorname{erfc}\\!\\left(\\sqrt{\\chi^2/2}\\right)$$

Wilson 95% CI for p using k A-alleles in n = 2N trials: $$\\hat{p}_W = \\frac{\\hat{p} + z^2/(2n)}{1 + z^2/n},\\quad \\text{margin} = \\frac{z}{1+z^2/n}\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n} + \\frac{z^2}{4n^2}}$$

Glossary of Variables

AA, Aa, aa: observed genotype counts.
N: sample size in individuals (N = AA + Aa + aa; for allele counts, N = (A + a)/2).
p: frequency of allele A; q: frequency of allele a (q = 1 − p).
MAF: minor allele frequency, min(p, q).
Expected counts: E[AA] = N p^2, E[Aa] = N 2pq, E[aa] = N q^2 (assuming HWE).
Heterozygosity (H): 2pq, the expected fraction of heterozygotes under HWE.
Chi-square and p-value: measure agreement between observed and expected genotypes under HWE (df = 1).
95% CI for p: Wilson interval treating 2N chromosomes as trials and number of A alleles as successes.

How It Works: A Step-by-Step Example

Suppose you observed AA = 36, Aa = 48, aa = 16 (N = 100). Then:

Compute p: p = (2×36 + 48) / (2×100) = 0.60; q = 0.40.
Expected genotype frequencies: AA = p^2 = 0.36, Aa = 2pq = 0.48, aa = q^2 = 0.16.
Expected counts: E[AA] = 36, E[Aa] = 48, E[aa] = 16.
HWE test: χ² = Σ (O−E)²/E = 0; p-value = 1 → no evidence against HWE.
MAF = min(0.60, 0.40) = 0.40; Heterozygosity = 2pq = 0.48. The 95% CI for p is computed using the Wilson method.

Frequently Asked Questions (FAQ)

What input should I use: genotype counts, allele counts, or p?

Use genotype counts to compute everything including an HWE test. If you only have allele counts, you can compute p and expected genotypes but not a valid HWE test. If you already know p and N, you can generate expected counts and heterozygosity.

When is the chi-square HWE test appropriate?

It is appropriate for bi-allelic loci with sufficiently large expected counts (common rule of thumb: all expected ≥ 5). For small samples or rare alleles, consider exact tests (e.g., Wigginton et al., 2005).

How is the 95% CI for p calculated?

We use a Wilson score interval treating the 2N chromosomes as binomial trials with k = 2×AA + Aa A-alleles.

Can I change allele labels?

This tool uses A and a for clarity. You can map them to your locus alleles (e.g., reference vs alternate).

Why do I get “even sum required” with allele counts?

In diploid organisms, the total number of alleles is 2×N. Therefore, A + a must be even to correspond to an integer number of individuals.

Is deviation from HWE always biologically meaningful?

No. Deviations can arise from technical issues (e.g., genotyping error) or sampling noise, not only from evolutionary forces. Always consider context and replicate data quality checks.

Tool developed by Ugo Candido. Content verified by CalcDomain Editorial Board.
Last reviewed for accuracy on: September 15, 2025.