How Allele Frequency is Calculated
At a locus with two alleles, A (dominant) and a (recessive), each individual carries exactly two copies. The frequency of allele A (denoted p) is the proportion of all allele copies in the sample that are A:
p = (2 x N_AA + N_Aa) / (2 x N_total)
q = 1 - p
where N_AA = count of AA individuals, N_Aa = count of Aa individuals, N_total = N_AA + N_Aa + N_aa
The denominator is 2 x N_total because each individual carries two alleles. For example, if you have 320 AA, 160 Aa, and 20 aa individuals (total = 500), the total number of alleles is 1,000. The A alleles are: (2 x 320) + 160 = 800, so p = 800 / 1,000 = 0.8. Then q = 1 - 0.8 = 0.2.
Hardy-Weinberg Expected Genotype Frequencies
Under Hardy-Weinberg equilibrium, the expected frequencies of the three genotypes are determined entirely by the allele frequencies:
Expected frequency of AA = p²
Expected frequency of Aa = 2pq
Expected frequency of aa = q²
Expected count of AA = p² x N
Expected count of Aa = 2pq x N
Expected count of aa = q² x N
In the worked example (320 AA, 160 Aa, 20 aa; N = 500; p = 0.8; q = 0.2): expected AA = 0.64 x 500 = 320; expected Aa = 2 x 0.8 x 0.2 x 500 = 160; expected aa = 0.04 x 500 = 20. The observed counts match the HWE expected counts exactly in this example, giving a chi-square of 0 and a non-significant HWE test.
Chi-Square Test for Hardy-Weinberg Equilibrium
The chi-square goodness-of-fit test compares observed genotype counts to HWE expected counts. The statistic is:
X² = sum of [(Observed - Expected)² / Expected]
X² = (O_AA - E_AA)² / E_AA + (O_Aa - E_Aa)² / E_Aa + (O_aa - E_aa)² / E_aa
The degrees of freedom for the HWE test is 1 (three genotype classes minus one allele frequency estimated from the data minus one constraint that frequencies sum to 1, giving 3 - 1 - 1 = 1). At a significance level of 0.05, the critical value is 3.841. If the chi-square exceeds this threshold, the population departs significantly from HWE at the chosen significance level.
The chi-square test is only reliable when expected counts are at least 5 for each genotype class. With small samples, Fisher's exact test or a permutation test is preferable.
What Causes Departure from Hardy-Weinberg Equilibrium?
| Cause | Effect on genotype frequencies |
|---|
| Inbreeding / non-random mating | Excess homozygotes (AA and aa), deficit of heterozygotes (Aa) |
| Positive assortative mating | Excess homozygotes |
| Outbreeding / hybrid vigour | Excess heterozygotes |
| Natural selection against a genotype | Deficit of the selected-against genotype |
| Recent migration or admixture | Excess heterozygotes (Wahlund effect reversed) |
| Wahlund effect (hidden subpopulations) | Excess homozygotes |
| Genotyping error | Often excess homozygotes due to allele dropout |
Worked Example
A researcher genotypes 500 individuals at a single nucleotide polymorphism (SNP) and finds: AA = 320, Aa = 160, aa = 20. To test whether this SNP locus is in Hardy-Weinberg equilibrium:
- Total alleles = 2 x 500 = 1,000
- A alleles = (2 x 320) + 160 = 800; so p = 800/1,000 = 0.8000
- q = 1 - 0.8 = 0.2000
- Expected AA = 0.8² x 500 = 320.0
- Expected Aa = 2 x 0.8 x 0.2 x 500 = 160.0
- Expected aa = 0.2² x 500 = 20.0
- Chi-square = (320-320)²/320 + (160-160)²/160 + (20-20)²/20 = 0.000
- Conclusion: chi-square (0.000) is below the critical value of 3.841, so the locus is in HWE (p-value = 1.000).
The default inputs in this calculator are set to match this worked example exactly.
Related Calculators
Sources and method: Hardy GH (1908) "Mendelian proportions in a mixed population." Science 28:49-50. Weinberg W (1908) "Uber den Nachweis der Vererbung beim Menschen." Jahreshefte des Vereins fur vaterlandische Naturkunde in Wurttemberg 64:368-382. Chi-square critical value from standard chi-square distribution table (df = 1). Guo SW, Thompson EA (1992) "Performing the exact test of Hardy-Weinberg proportion for multiple alleles." Biometrics 48:361-372.
This calculator is for educational and research planning purposes. The chi-square test for HWE requires expected counts of at least 5 per genotype class. For small samples, use Fisher's exact test. Always consult a statistician or geneticist for formal population genetics analyses.
