Hardy-Weinberg Equilibrium Calculator

How p and q Are Calculated

1. Count alleles. 500 plants carry 1,000 alleles. Y alleles: 2 × 245 + 210 = 700; y alleles: 2 × 45 + 210 = 300.
2. Divide: p = 700 ÷ 1,000 = 0.7 and q = 300 ÷ 1,000 = 0.3. They add to 1.
3. Predict genotypes: p² = 0.49 (YY), 2pq = 0.42 (Yy), q² = 0.09 (yy), which add to 1.
4. Expected counts: multiply by 500 to get 245, 210 and 45. Here they match the observed counts exactly.

OpenStax’s follow-up question uses p = 0.8 and q = 0.2 in 500 plants: 0.64 × 500 = 320 VV, 0.32 × 500 = 160 Vv and 0.04 × 500 = 20 vv, so 480 violet and 20 white flowers. Choose “dominant allele frequency p” to reproduce it.

Testing for Equilibrium with Chi-Square

Compare observed and expected genotype counts with χ² = Σ (observed − expected)² ÷ expected. There are three genotype classes, minus 1 for the total and minus 1 because p was estimated from the same data, which leaves 1 degree of freedom. The 5% critical value is 3.84.

The Cornell Statistical Consulting Unit’s handout works the 725 / 123 / 7 example to χ² ≈ 0.49 and says the population is consistent with equilibrium. Its printed p-value, “about 0.104,” does not match: a χ² of 0.49 on 1 degree of freedom has an upper-tail p-value of about 0.48. The conclusion is the same either way. With an expected count near 5 in the rare genotype class, an exact test is the safer choice, as the handout also notes.

From a Recessive Trait to Carrier Frequency

When only the recessive phenotype can be seen, its frequency is q². Take the square root for q, then 2pq gives the share of carriers. These rows are illustrative frequencies, not data for a particular disease:

Carriers far outnumber affected people for rare recessive traits: at 1 in 2,500, about 1 person in 26 carries the allele.

Common Mistakes

• Taking q from the recessive phenotype directly. The recessive frequency is q², so q is its square root: 0.09 means q = 0.3, not 0.09.
• Counting individuals instead of alleles. Each diploid individual carries two alleles, so divide by 2n.
• Using 2 degrees of freedom. Estimating p from the data costs one, leaving 1.
• Reading “not significant” as proof of equilibrium. A small sample may simply lack the power to detect a departure.
• Applying it where the assumptions fail. OpenStax lists them: a very large population, random mating, and no mutation, migration or selection.

What This Calculator Does Not Cover

It handles one gene with two alleles in a diploid population. Multiple alleles, X-linked genes, exact tests for small samples and inbreeding coefficients need other methods.

Frequently Asked Questions

How to Use This Calculator

1. Choose what you know — genotype counts, the dominant allele frequency p, or the recessive phenotype frequency q².
2. Enter the numbers — counts for each genotype, or p or q² as a decimal or 1/N.
3. Set a population size — for expected counts when starting from a frequency.
4. Read the results — p, q, p², 2pq, q², expected counts and, from counts, the chi-square test.