How this calculator works
Under Hardy-Weinberg assumptions, random union of gametes makes homozygote frequencies pi squared and distinct heterozygote frequencies twice pi times pj. The same rule works for two through five alleles. The recessive mode infers q from a modelled recessive genotype proportion of 1/N, then computes heterozygotes and AA. It is an educational population model and cannot determine a person's disease risk or confirm equilibrium from observations.
Formula
Sum(pi) = 1. f(AiAi) = pi2; f(AiAj) = 2pipj for i < j. Homozygosity = sum(pi2); heterozygosity = 1 - sum(pi2). For recessive proportion 1/N: q = sqrt(1/N).
Worked example
For p = q = 0.5, AA and BB each have frequency 0.25 and AB has frequency 0.5. A recessive genotype occurring in 1 in 100 gives q = 0.1, p = 0.9 and heterozygotes 0.18.
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Frequently asked questions
Does this test whether a real population is in equilibrium?
No. It gives model expectations. A goodness-of-fit test requires observed counts and appropriate statistical assumptions.
Should I enter percentages in the list?
Enter proportions, for example 0.7, 0.2, 0.1. The list must sum to one.
Does the multi-allele table imply dominance?
No. It reports unordered diploid genotype proportions. Dominance and phenotype mappings are separate biological information.