Hardy-Weinberg Equilibrium Calculator

What the Hardy-Weinberg equation says

For a gene with two alleles, call the frequency of the dominant allele p and the frequency of the recessive allele q. Every allele in the gene pool is one or the other, so p + q = 1. If gametes pair at random, the chance that a zygote receives two A alleles is p × p, the chance of two a alleles is q × q, and the chance of one of each is 2pq (A from the egg and a from the sperm, or the reverse). That gives the genotype frequencies p² + 2pq + q² = 1, and, in the absence of evolutionary forces, both allele and genotype frequencies stay the same from one generation to the next. Hardy-Weinberg equilibrium is the null model of evolution: the population that is not changing.

The five conditions

A population stays in equilibrium only while all of these hold: (1) a very large population, so allele frequencies do not drift by chance; (2) random mating with respect to the gene; (3) no mutation creating or destroying alleles; (4) no migration bringing alleles in or out; (5) no natural selection, so every genotype survives and reproduces equally well. Real populations violate at least one, which is exactly why the equation is useful. Measure the genotypes, compare them with p², 2pq, and q², and any significant gap points to a force worth investigating.

Worked example: how many carriers?

Cystic fibrosis is recessive and affects about 1 in 2,500 newborns of European ancestry. Only aa individuals are affected, so q² = 1/2,500 = 0.0004 and q = √0.0004 = 0.02. Then p = 1 − 0.02 = 0.98, and the carrier frequency is 2pq = 2 × 0.98 × 0.02 = 0.0392, roughly 1 person in 25. Notice the asymmetry: the disease is rare, but carriers are nearly a hundred times more common than affected people, because a recessive allele hides in heterozygotes. This is why recessive alleles are almost impossible to eliminate by selection against the affected phenotype alone.

Allele frequency versus genotype frequency

Heterozygotes are most common when p = q = 0.5, where 2pq reaches its maximum of 0.50. As either allele becomes rare, almost every copy of it sits inside a heterozygote: at q = 0.1, only 1% of the population is aa but 18% carries the allele.

Reading the chi-square verdict

In “observed counts” mode the calculator estimates p directly from the sample by counting alleles: each AA individual contributes two A alleles and each Aa contributes one, so p = (2 × AA + Aa) / 2n. It then predicts the counts expected under equilibrium (p²n, 2pqn, q²n) and computes χ² = Σ (O − E)² / E. There are three genotype classes, but one degree of freedom is lost to the total and another to estimating p from the same data, so df = 1 and the α = 0.05 critical value is 3.841. A χ² above 3.841 means the observed genotypes would arise by chance less than 5% of the time if the population were in equilibrium. A heterozygote deficit often signals inbreeding or population structure; a heterozygote excess can point to heterozygote advantage, as with the sickle-cell allele in malaria regions.