How this calculator works
At a diploid locus with two alleles, p and q sum to one. Each homozygote contributes two copies of its allele and each heterozygote contributes one. Inferring q from a recessive genotype frequency, or from heterozygote frequency, requires Hardy-Weinberg assumptions. Heterozygote frequency has two possible allele-frequency solutions; the branch selector makes that ambiguity explicit. These are population models, not individual clinical risk estimates.
Formula
p + q = 1. At equilibrium: f(AA) = p2, f(Aa) = 2pq, f(aa) = q2. From counts: q = (2naa + nAa)/(2N). From heterozygotes H: q = (1 ± sqrt(1 - 2H))/2.
Worked example
If q = 0.2 then p = 0.8, expected AA = 64%, Aa = 32% and aa = 4%. If recessive genotype frequency is 9%, q = 0.3 and the expected heterozygote frequency is 42%.
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Frequently asked questions
Does allele frequency equal genotype frequency?
No. Alleles are counted among gene copies; genotype frequency is counted among individuals.
Can heterozygotes exceed 50%?
Observed heterozygotes can, but a two-allele Hardy-Weinberg model cannot. This inverse mode rejects a larger value.
What assumptions are needed for expected genotype proportions?
Random mating at the locus and a large population without selection, migration or mutation effects are the usual idealized conditions.