What does this test answer?

It tests whether the observed genotype counts in a sample match the Hardy-Weinberg expectation for a given allele frequency. A large χ² with a small p-value means the population is not in Hardy-Weinberg equilibrium.

Why is df = 1 for three genotype classes?

Degrees of freedom = (number of classes − 1) − (number of parameters estimated from the data). Three classes minus 1 minus 1 estimated parameter (the allele frequency p) gives df = 1.

What is the rejection rule?

If χ² exceeds the critical value at the chosen α level, reject H₀. For df = 1: χ²crit = 3.841 at α = 0.05 and 6.635 at α = 0.01.

Can I use this for any goodness-of-fit?

The maths is the same for any expected vs observed comparison, but this tool hardcodes the three Hardy-Weinberg categories. For a 9:3:3:1 dihybrid ratio you'd use a different df and different expected proportions.

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NSW · HSCBiology

Allele frequency chi-square test

Test whether your observed AA, Aa, aa counts match Hardy-Weinberg expectations for a given allele frequency p. Returns the per-cell contributions, total χ², and the verdict at α = 0.05 and α = 0.01.

Formula

χ² = Σ (O − E)² / E

where for Hardy-Weinberg:

E(AA) = N · p²
E(Aa) = N · 2pq
E(aa) = N · q², with q = 1 − p

Degrees of freedom: df = 3 − 1 − 1 = 1.

Inputs

Observed AACount of homozygous dominant individuals.Observed AaCount of heterozygous individuals.Observed aaCount of homozygous recessive individuals.Expected p (dominant allele frequency)Hardy-Weinberg expectation: p² (AA), 2pq (Aa), q² (aa).

Goodness-of-fit test versus Hardy-Weinberg expectation. df = 3 − 1 − 1 = 1 (one parameter estimated).

Result

Total N = 100

p = 0.600, q = 0.400

Expected counts

E(AA) = N·p² = 36.00

E(Aa) = N·2pq = 48.00

E(aa) = N·q² = 16.00

Cell contributions (O − E)² / E

AA: 1.000

Aa: 1.021

aa: 0.063

χ² = 2.083 (df = 1)

Critical values at df=1: 3.841 (α=0.05), 6.635 (α=0.01).

Fail to reject H₀ at α = 0.05 (consistent with Hardy-Weinberg).

Worked example

A sample of 100 wallabies has 30 AA, 55 Aa, 15 aa. The expected dominant allele frequency from prior work is p = 0.6, so q = 0.4.

Expected counts:

E(AA) = 100 · 0.36 = 36
E(Aa) = 100 · 2 · 0.6 · 0.4 = 48
E(aa) = 100 · 0.16 = 16

Cell contributions:

AA: (30 − 36)² / 36 = 1.000
Aa: (55 − 48)² / 48 = 1.021
aa: (15 − 16)² / 16 = 0.063

χ² = 2.084, df = 1. Since 2.084 < 3.841 we fail to reject H₀ at α = 0.05: the sample is consistent with Hardy-Weinberg.

How this calculator works

The calculator takes your three observed counts and an expected dominant-allele frequency p. It builds the Hardy-Weinberg expected counts E = N·(p², 2pq, q²), sums (O − E)²/E across the three classes, and compares the total to the χ² critical values at df = 1.

Common questions

What does this test answer?: It tests whether the observed genotype counts in a sample match the Hardy-Weinberg expectation for a given allele frequency. A large χ² with a small p-value means the population is not in Hardy-Weinberg equilibrium.
Why is df = 1 for three genotype classes?: Degrees of freedom = (number of classes − 1) − (number of parameters estimated from the data). Three classes minus 1 minus 1 estimated parameter (the allele frequency p) gives df = 1.
What is the rejection rule?: If χ² exceeds the critical value at the chosen α level, reject H₀. For df = 1: χ²crit = 3.841 at α = 0.05 and 6.635 at α = 0.01.
Can I use this for any goodness-of-fit?: The maths is the same for any expected vs observed comparison, but this tool hardcodes the three Hardy-Weinberg categories. For a 9:3:3:1 dihybrid ratio you'd use a different df and different expected proportions.