Hardy–Weinberg Calculator
Derive allele and genotype frequencies from a recessive phenotype fraction under ideal equilibrium, with optional expected counts.
A QUICK WALKTHROUGH
How to use this tool
- Enter the values and choose their units.
- Enter recessive phenotype fraction q² from 0 to 1, and optionally a population size N: integer 1–1,000,000,000. This two-allele diploid model assumes that the recessive phenotype identifies aa exactly, random mating, a sufficiently large population, and no selection, mutation or migration. q=√(q²), p=1−q; AA=p², Aa=2pq, aa=q². Expected counts equal frequency×N and may be fractional; they are not observed counts. This does not test equilibrium or provide medical or genetic diagnosis.
- Review the calculated result.
Formula and assumptions
Enter recessive phenotype fraction q² from 0 to 1, and optionally a population size N: integer 1–1,000,000,000. This two-allele diploid model assumes that the recessive phenotype identifies aa exactly, random mating, a sufficiently large population, and no selection, mutation or migration. q=√(q²), p=1−q; AA=p², Aa=2pq, aa=q². Expected counts equal frequency×N and may be fractional; they are not observed counts. This does not test equilibrium or provide medical or genetic diagnosis.
Inputs stay in this browser; nothing is sent.
Inputs stay in this browser; nothing is sent.
GOOD TO KNOW
Common questions
How is the result calculated?
Enter recessive phenotype fraction q² from 0 to 1, and optionally a population size N: integer 1–1,000,000,000. This two-allele diploid model assumes that the recessive phenotype identifies aa exactly, random mating, a sufficiently large population, and no selection, mutation or migration. q=√(q²), p=1−q; AA=p², Aa=2pq, aa=q². Expected counts equal frequency×N and may be fractional; they are not observed counts. This does not test equilibrium or provide medical or genetic diagnosis.