Hardy-Weinberg Calculator
p, q, p², 2pq and q² from genotype counts or the recessive phenotype frequency, with a chi-square test for equilibrium.
p² + 2pq + q² = 1 and p + q = 1. If 4% of a population shows a recessive trait, q² = 0.04 so q = 0.2, p = 0.8 and carriers are 2pq = 0.32, about 1 in 3.
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How to use it
- Choose what you know: genotype counts, the share of the population with the recessive trait, or p.
- Enter the numbers.
- Read p, q, carrier frequency and, for counts, the chi-square test result.
What you enter
- Genotype counts, q², or p
What you get
- p and q
- p², 2pq, q²
- Carrier frequency
- Expected counts
- Chi-square and p-value
Formula
p + q = 1 | p² + 2pq + q² = 1 | p = (2AA + Aa) / 2N | χ² = Σ (O − E)² / E, df = 1
Worked example: genotype counts
Input: AA 360, Aa 480, aa 160
Result: p = 0.6, q = 0.4; χ² = 0, in equilibrium
Observed counts match expected exactly, so there is no evidence of selection or non-random mating.
Questions
How do you solve a Hardy-Weinberg problem?
Start from what you know. If you know the share showing the recessive phenotype, that is q²; take its square root for q, then p = 1 − q, and genotype frequencies are p², 2pq and q².
How do you calculate allele frequency from genotype counts?
p = (2 × AA + Aa) ÷ (2 × total). Each individual carries two alleles.
How do I test for Hardy-Weinberg equilibrium?
Compare observed genotype counts with expected (p²N, 2pqN, q²N) using chi-square with 1 degree of freedom. p < 0.05 suggests the population is not in equilibrium.
What are the Hardy-Weinberg assumptions?
Random mating, a large population, and no mutation, migration or selection.
Limitations
- Two alleles at one locus only.
- Chi-square is unreliable when an expected count is below 5.
References
- Griffiths AJF et al. (2020) Introduction to Genetic Analysis, 12th ed. W.H. Freeman, ch. 18.
Related reading
More Genetics
All free tools · By Dr. Zubair Khalid, DVM, MS, PhD.