Hardy-Weinberg p² + 2pq + q²
Enter the dominant allele frequency p and get the three genotype frequencies at Hardy-Weinberg equilibrium: AA as p², Aa as 2pq and aa as q², with q = 1 − p.
AA / Aa / aa
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Hardy-Weinberg equilibrium
The Hardy-Weinberg principle says that, in an idealised population, allele and genotype frequencies hold steady from one generation to the next: p² + 2pq + q² = 1 with p + q = 1. Here p² is the share of homozygous dominants (AA), 2pq the heterozygotes (Aa) and q² the homozygous recessives (aa). For the population to stay put and not evolve, five things all have to hold at once: infinite size, no mutation, no selection, no migration, and random mating. You estimate the allele frequency with p = (2·N_AA + N_Aa) / (2·N). To see it in action, suppose p = 0.6 in our Example; then q = 0.4, which works out to AA = 0.36, Aa = 0.48 and aa = 0.16.
Applications
It is the bedrock of population genetics. Conservation biologists use it in species conservation to spot inbreeding and dwindling diversity. It lets you estimate how common recessive diseases are; cystic fibrosis, for instance, affects ~1/2,500 Caucasians, so q ≈ 1/50 and the carrier rate 2pq ≈ 4%. And in forensics it underlies the probability calculations behind DNA profiles.
FAQ
Does any real population fit Hardy-Weinberg? Hardly any do. The model is meant as an idealised yardstick, and any departure from it points to evolution at work, whether through selection, drift or migration.
How is the carrier frequency estimated? Say the disease shows up in 1/10,000 people (q² = 0.0001). Then q = 0.01 and 2pq ≈ 0.02, which means roughly 2% of the population carries the allele as a heterozygote.
What is the chi-square test used for? It tells you whether the genotype frequencies you actually observe stray significantly from what Hardy-Weinberg predicts. When they do, you know one of the five conditions has broken down.
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