Math July 13, 2026 · 8 Min Read

Allele Frequency Calculator – Guide & Formulas

Calculate allele frequencies (p and q) and expected genotype frequencies using Hardy-Weinberg equilibrium. Includes chi-square test for HWE and genotype frequency analysis.

Calculate allele frequencies and genotype proportions using Hardy-Weinberg equilibrium with chi-square testing for population genetics analysis.

Key Takeaway

Use the free Allele Frequency Calculator to calculate allele frequencies (p and q) and expected genotype frequencies using hardy-weinberg equilibrium. includes chi-square test for hwe and genotype frequency analysis. Get instant results with step-by-step explanations.

How to Use the Allele Frequency Calculator

  1. Enter the observed counts for each genotype: AA (homozygous dominant), Aa (heterozygous), and aa (homozygous recessive).
  2. Alternatively, enter total allele counts for p (A) and q (a) directly.
  3. Review the calculated allele frequencies p and q.
  4. Check the expected genotype frequencies under HWE: p², 2pq, and q².
  5. Review the chi-square statistic and p-value for HWE test.
  6. Interpret whether the population is in Hardy-Weinberg equilibrium.

The Formula

Hardy-Weinberg Equilibrium: p² + 2pq + q² = 1, where p + q = 1. p = frequency of A allele, q = frequency of a allele. Expected genotype counts = p² x N, 2pq x N, q² x N, where N is total individuals. Chi-square = Σ[(Observed - Expected)² / Expected] with 1 degree of freedom.

Variable Definitions

  • p: Frequency of the dominant (A) allele in the population
  • q: Frequency of the recessive (a) allele in the population
  • p²: Expected frequency of homozygous dominant (AA) genotype
  • 2pq: Expected frequency of heterozygous (Aa) genotype
  • q²: Expected frequency of homozygous recessive (aa) genotype
  • N: Total number of individuals in the population
  • χ²: Chi-square statistic measuring deviation from HWE

Testing a Population for Hardy-Weinberg Equilibrium

A population of 200 individuals: AA = 68, Aa = 104, aa = 28. Test whether this population is in HWE.

  1. Step 1: Calculate total alleles. Total A alleles = 2(68) + 104 = 240. Total a alleles = 2(28) + 104 = 160. Total alleles = 400.
  2. Step 2: Calculate allele frequencies. p = 240/400 = 0.60. q = 160/400 = 0.40. Verify: p + q = 1.0.
  3. Step 3: Calculate expected genotype frequencies. AA: p² = 0.36, Aa: 2pq = 0.48, aa: q² = 0.16.
  4. Step 4: Calculate expected counts. Expected AA = 0.36 x 200 = 72. Expected Aa = 0.48 x 200 = 96. Expected aa = 0.16 x 200 = 32.
  5. Step 5: Calculate chi-square. χ² = (68-72)²/72 + (104-96)²/96 + (28-32)²/32 = 0.222 + 0.667 + 0.500 = 1.389.
  6. Step 6: Compare to critical value. With 1 df, critical value at α = 0.05 is 3.841. Since 1.389 < 3.841, we fail to reject HWE — the population is in equilibrium.

Frequently Asked Questions

What is Hardy-Weinberg equilibrium?

Hardy-Weinberg equilibrium (HWE) is a principle stating that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of evolutionary influences such as mutation, selection, genetic drift, migration, and non-random mating.

How do I calculate allele frequency from genotype counts?

Count the number of each allele: A alleles = 2(AA count) + Aa count. a alleles = 2(aa count) + Aa count. Divide each by total alleles (2 x total individuals) to get p and q.

What does the chi-square test tell me about HWE?

The chi-square test compares observed genotype counts to those expected under HWE. A non-significant p-value (p > 0.05) means you fail to reject HWE — the population is consistent with equilibrium. A significant result indicates deviation from HWE.

What does it mean if a population is not in HWE?

Deviation from HWE suggests one or more assumptions are violated: selection, mutation, migration, non-random mating (e.g., inbreeding), or genetic drift. It can also indicate genotyping errors or population substructure.

Can I use this calculator with more than two alleles?

This calculator uses the two-allele (one locus) model. For multi-allele systems, use chi-square goodness-of-fit with n-1 degrees of freedom where n is the number of genotypes.

What is the significance of p² + 2pq + q² = 1?

This equation states that all genotype frequencies must sum to 1 (100% of the population). The binomial expansion represents the three possible genotype classes in a diploid organism with two alleles at one locus.

How many individuals do I need for a reliable HWE test?

At minimum, 50 individuals for detecting large deviations, but 100-200+ is recommended. With small samples, expected counts may be below 5 in some cells, reducing chi-square reliability. Use exact tests (Fisher) for small samples.

What if p or q is close to 0 or 1?

Extreme allele frequencies (near fixation) reduce the power of the chi-square test. Small expected counts in some genotype classes make the asymptotic chi-square approximation unreliable. Consider exact tests or increasing sample size.

How does inbreeding affect HWE?

Inbreeding increases homozygous genotypes (AA and aa) and decreases heterozygotes (Aa) compared to HWE expectations. This creates a deficit of heterozygotes, detectable as a significant chi-square result. The inbreeding coefficient F measures this deviation.

Can I enter raw allele counts instead of genotypes?

Yes, if you know the total count of A and a alleles in the population, enter them directly. The calculator will compute p and q and generate the expected genotype frequencies under HWE.