Authoritative Content and Methodology
Data Source and Methodology
Primary sources:
- G. H. Hardy (1908). “Mendelian Proportions in a Mixed Population.” Science, 28(706):49–50. doi: 10.1126/science.28.706.49
- W. Weinberg (1908). “Über den Nachweis der Vererbung beim Menschen.” Jahreshefte des Vereins für vaterländische Naturkunde in Württemberg 64:368–382. (English translations available)
- A. W. F. Edwards (2008). “G. H. Hardy (1908) and Hardy–Weinberg Equilibrium.” Genetics, 179(3):1143–1150. doi link
All expectations are computed from p and q under random mating. The hypothesis test is Pearson’s chi-square goodness-of-fit with 1 degree of freedom when p is estimated from the sample.
Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.
The Formulas Explained
Allele frequencies from genotype counts:
$$ p = \frac{2 \cdot n_{AA} + n_{Aa}}{2N}, \quad q = 1 - p, \quad N = n_{AA} + n_{Aa} + n_{aa} $$
Expected genotype frequencies under HWE:
$$ P(AA) = p^2, \quad P(Aa) = 2pq, \quad P(aa) = q^2 $$
Expected counts (given N):
$$ E(AA) = N p^2,\quad E(Aa) = N \cdot 2pq,\quad E(aa) = N q^2 $$
Pearson chi-square test statistic:
$$ \chi^2 = \sum_{g \in \{AA,Aa,aa\}} \frac{(O_g - E_g)^2}{E_g}, \quad \text{df} = 1 \text{ when } p \text{ is estimated} $$
Glossary of Variables
- AA, Aa, aa (Observed counts): Number of individuals with each genotype.
- N: Total sample size, N = AA + Aa + aa.
- p: Frequency of allele A; q: frequency of allele a; q = 1 − p.
- Expected frequencies: p² (AA), 2pq (Aa), q² (aa).
- Expected counts: N times the expected frequencies.
- χ², p-value: Chi-square statistic and its p-value for testing HWE.
- α: Significance level used to accept/reject the null hypothesis of HWE.
Come Funziona: Un Esempio Passo-Passo
Inputs: AA = 48, Aa = 32, aa = 20 (N = 100), α = 5%.
- Compute p and q: p = (2·48 + 32)/(2·100) = 128/200 = 0.64; q = 1 − 0.64 = 0.36.
- Expected frequencies: p² = 0.4096; 2pq = 0.4608; q² = 0.1296.
- Expected counts: E(AA) = 40.96; E(Aa) = 46.08; E(aa) = 12.96.
- Chi-square: χ² = (48−40.96)²/40.96 + (32−46.08)²/46.08 + (20−12.96)²/12.96 ≈ 7.50.
- With df = 1, p ≈ 0.006. Since p < 0.05, reject H0 (the sample deviates from HWE).
Frequently Asked Questions (FAQ)
What assumptions underlie HWE?
Random mating, infinitely large population (no drift), no selection, no mutation, and no migration.
Is χ² suitable for small expected counts?
When any expected count is below ~5, the chi-square approximation may be unreliable. Consider an exact test.
Can I supply only p and N to run the test?
No. You need observed genotype counts to compare against expectations.
Do you apply Yates’ continuity correction?
No. This calculator reports the uncorrected Pearson χ² (df = 1), commonly used in HWE screening.
Why don’t p and q change in allele-frequency mode?
In that mode, p is provided by you and q is computed as 1 − p; there are no observed counts to re-estimate p.
How many decimals are shown?
Frequencies to 4–6 decimals and counts to 2 decimals for readability.
Can this handle multiple alleles?
No. This tool focuses on a bi-allelic locus. Multiallelic HWE requires different formulas and degrees of freedom.