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Hardy-Weinberg Equilibrium Calculator

Analyze genotype distributions. Enter either allele frequencies or observed genotype counts to determine expected distributions and test for equilibrium using a Chi-Square test.

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Input Mode Selection

Select how you want to input your data.

Allele Frequencies

Dominant Allele Frequency (p)

Recessive Allele Frequency (q)

Equilibrium Distribution

Calculated p / q 0.600 / 0.400

f(AA) = p² 0.360 (36.0%)

f(Aa) = 2pq 0.480 (48.0%)

f(aa) = q² 0.160 (16.0%)

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Expert Tip

If the Chi-Square statistic exceeds the critical value of 3.841 (df = 1, α = 0.05), the population deviates significantly from equilibrium. This suggests evolutionary forces like selection, mutation, or non-random mating are at play.

Methodology & Equations

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Hardy-Weinberg Formulas

For a single gene locus with two alleles (dominant A and recessive a) at frequencies p and q respectively:

p + q = 1

p² + 2pq + q² = 1

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Equilibrium Assumptions

Hardy-Weinberg equilibrium relies on five major assumptions:

1. No mutation occurs.
2. Mating is completely random.
3. No natural selection occurs.
4. The population size is infinitely large.
5. No gene flow (migration) occurs.

How to Check for Hardy-Weinberg Equilibrium: Step-by-Step

Below is a step-by-step example using observed counts of 36 AA, 48 Aa, and 16 aa individuals:

Calculate Allele Frequencies (p & q)

Find total population: N = 36 + 48 + 16 = 100.
Allele A count = (2 × 36) + 48 = 120. Frequency p = 120 / 200 = 0.60.
Allele a frequency q = 1 - 0.60 = 0.40.

Compute Expected Genotype Counts

Expected AA = p² × N = 0.36 × 100 = 36.
Expected Aa = 2pq × N = 0.48 × 100 = 48.
Expected aa = q² × N = 0.16 × 100 = 16.

Apply the Chi-Square Test

Evaluate differences between observed and expected counts:
χ² = (36-36)²/36 + (48-48)²/48 + (16-16)²/16 = 0.
Since 0 ≤ 3.841, observed counts match expectations perfectly. The population is in equilibrium.