What does chi-square (χ²) tell me?

It measures how far observed genotype counts are from Hardy–Weinberg expected counts. Larger χ² indicates a larger deviation, but interpretation depends on course conventions and assumptions.

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Hardy–Weinberg Calculator

Calculate allele frequencies (p, q), expected genotype frequencies (p², 2pq, q²), and (optionally) a quick χ² check to compare observed vs expected genotype counts. Includes quick picks, sanity meter, and optional step-by-step.

Background

Under Hardy–Weinberg conditions (large population, random mating, no selection/migration/mutation/drift), allele frequencies p and q (where p + q = 1) predict genotype frequencies: AA = p², Aa = 2pq, aa = q². If you measured genotype counts, you can also compare observed vs expected with χ².

Enter values

What do you have?

Genotype counts (AA, Aa, aa)

Allele frequency (p or q)

Genotype frequencies / % (AA, Aa, aa)

Tip: If you have counts, we can compute p, q, expected counts, and optionally χ².

Observed genotype counts

We use: p = (2·AA + Aa) / (2N) and q = 1 − p.

Nice-to-have outputs

Compute χ² (observed vs expected)

χ² uses Σ (O − E)² / E. (We show df=1 as the common intro-bio convention.)

Options:

Round results

Show step-by-step

Result:

No results yet. Enter values and click Calculate.

How to use this calculator

Pick what you have: counts, allele frequency, or genotype frequencies.
Enter values (and N if you want expected counts).
Click Calculate. Turn on Step-by-step to see the exact math.
If using counts, keep χ² checked to compare observed vs expected.

Quick tip: if your allele frequencies don’t add to 1 exactly (rounding), this tool will normalize.

How this calculator works

Alleles → genotypes: AA=p², Aa=2pq, aa=q²
Counts → alleles: p=(2AA+Aa)/(2N), q=1−p
χ² check: χ²=Σ(O−E)²/E

Note: χ² is a quick check. In real bio/stats work, interpretation depends on assumptions, sample size, and how the test is taught in your course.

Formula & Equation Used

Hardy–Weinberg: p + q = 1

Expected genotype frequencies: p², 2pq, q²

Expected counts: E(AA)=Np², E(Aa)=N·2pq, E(aa)=Nq²

Allele freq from counts: p=(2AA+Aa)/(2N)

Chi-square: χ² = Σ (O − E)² / E

Example Problem & Step-by-Step Solution

Example 1 — genotype counts

Observed: AA=48, Aa=32, aa=20.

Total N = 48 + 32 + 20 = 100
p = (2·48 + 32)/(2·100)=0.64, q=0.36
Expected freqs: p²=0.4096, 2pq=0.4608, q²=0.1296

Example 2 — allele frequency given

Given allele frequency p = 0.70 in a population of N = 200.

Compute q = 1 − p = 1 − 0.70 = 0.30
Expected genotype frequencies: AA = p² = 0.49, Aa = 2pq = 0.42, aa = q² = 0.09
Expected genotype counts: AA = 200·0.49 = 98, Aa = 200·0.42 = 84, aa = 200·0.09 = 18

Example 3 — genotype frequencies (%)

Given genotype frequencies: AA = 40.96%, Aa = 46.08%, aa = 12.96% (sample size N = 250).

Convert percentages to proportions: AA = 0.4096, Aa = 0.4608, aa = 0.1296
Compute allele frequencies: p = AA + ½Aa = 0.4096 + 0.2304 = 0.64, q = 1 − p = 0.36
Expected genotype frequencies: p² = 0.4096, 2pq = 0.4608, q² = 0.1296
Expected counts: AA = 250·0.4096 = 102.4, Aa = 115.2, aa = 32.4

Frequently Asked Questions

Q: Do p and q have to add to 1?

Yes. If your inputs are slightly off due to rounding, we’ll normalize them.

Q: What does χ² tell me?

It measures how far observed counts are from Hardy–Weinberg expected counts. Bigger χ² means bigger deviation, but interpretation depends on your class conventions and assumptions.

Q: Can I compute expected counts without N?

You can always compute expected frequencies. Counts require a sample size N.