Hardy-Weinberg Equilibrium, Explained

After reading this you can turn a single allele frequency into expected genotype frequencies, work allele frequencies back out from genotype counts, and test whether a population departs from equilibrium.

Imagine a wildflower where purple petals (allele A) are dominant over white (allele a). You count 500 plants and find 320 purple homozygotes, 160 heterozygotes, and 20 white recessives. Is that pattern what random mating would produce, or is something bending it? The Hardy-Weinberg model gives you the baseline expectation. Deviations from that baseline are the interesting part.

What the model says

A population carries two alleles at one locus, A and a, with frequencies p and q. Because every allele copy is one or the other, the two frequencies must add to one.

Here p is the fraction of all allele copies that are A, and q is the fraction that are a. If p = 0.6 then q = 0.4, always.

Now suppose mating is random: any allele copy pairs with any other, like drawing two marbles from a bag. The chance of drawing A twice is p \times p. The chance of two a copies is q \times q. A heterozygote can form two ways (A then a, or a then A), so its probability is 2pq. Those three genotype frequencies sum to one.

p^2 is the expected frequency of AA, 2pq of Aa, and q^2 of aa. That single line is the whole model. The rest is arithmetic and interpretation.

When to use it, and when not to

Use Hardy-Weinberg when you have one locus with two alleles and you want a reference point: what should the genotype counts look like with no evolutionary forces acting? It is the null hypothesis of population genetics. You reach for it to estimate carrier frequency for a recessive disease, to sanity-check genotyping data, or to detect selection or non-random mating by measuring the gap between expected and observed.

The equilibrium rests on five assumptions: a large population, random mating, and no selection, mutation, or migration. Do not apply it blindly when any of these clearly fails. A tiny population drifts by chance. Strong inbreeding inflates homozygotes. A gene under heavy selection will not sit at equilibrium. In those cases the model still helps, but as a comparison, not a description.

Intuition for the square

The 2pq term surprises people. Why double? Think of a 2 by 2 grid of gamete combinations, the same idea behind a Punnett square. One axis is the allele from parent one, the other axis the allele from parent two. Each axis has A with probability p and a with probability q. The four cells hold probabilities p^2, pq, pq, and q^2. The two off-diagonal cells are both heterozygotes, so they combine into 2pq.

One consequence worth remembering: heterozygotes peak at p = 0.5, where 2pq = 0.5. When an allele is rare, say q = 0.01, most copies of it hide in heterozygotes. The recessive phenotype frequency is q^2 = 0.0001, but the carrier frequency 2pq = 0.0198 is about 200 times larger. This is why rare recessive alleles persist: selection rarely sees them.

Worked example: the demo data

Run the tool in counts mode with the sample data: 320 AA, 160 Aa, and 20 aa. This shows how to recover allele frequencies and then test the fit.

Notice the shortcut hidden in step 3. The frequency of A equals the frequency of AA plus half the frequency of Aa: p = 0.64 + 0.5(0.32) = 0.8. Half of each heterozygote's alleles are A, which is why you add half the heterozygote count.

Reading and interpreting the result

When observed and expected match, as in the demo, the population is consistent with equilibrium. The more useful case is a mismatch. Suppose you instead observed 360 AA, 80 Aa, and 60 aa in 500 plants. The allele count is unchanged: 2(360) + 80 = 800, so p = 0.8 still. Expected counts are still 320, 160, 20. But observed homozygotes are inflated and heterozygotes are scarce (80 versus 160).

An excess of homozygotes with a deficit of heterozygotes is the classic signature of inbreeding or population structure. A deficit of homozygotes can signal heterozygote advantage. To decide whether a gap is real or just sampling noise, run a chi-square test: sum (O - E)^2 / E across the three genotypes and compare to a chi-square distribution with one degree of freedom (three categories minus one, minus one estimated allele frequency).

Common mistakes

Three errors account for most wrong answers.

Treating q^2 as the carrier frequency

q^2 is the frequency of affected recessive homozygotes. Carriers are heterozygotes at frequency 2pq. For a disease with q^2 = 0.0004, so q = 0.02, carrier frequency is 2(0.98)(0.02) = 0.0392, nearly 1 in 25.

Forgetting the factor of two on alleles

Total allele copies are twice the number of individuals. Dividing an allele count by the number of people instead of by twice that number halves your frequency.

Assuming equilibrium proves no evolution

Matching Hardy-Weinberg for one generation does not mean forces are absent. It means they are too weak to detect in your sample, or they cancel out. Equilibrium is a null, not a certificate.

Related tools

Once you have genotype frequencies, a Punnett Square Calculator shows the offspring ratios from a specific cross, the microscopic mechanism behind the 2pq term. If you work upstream at the sequence level, the DNA & RNA Sequence Toolkit handles transcription, translation, and GC content. For population dynamics over time rather than genotype proportions, the Bacterial Growth Calculator models exponential change, and the Michaelis-Menten Kinetics Calculator covers enzyme reaction rates.

Related tools

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Genotype and phenotype ratios of a cross.

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Bacterial Growth Calculator

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DNA & RNA Sequence Toolkit

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Michaelis-Menten Kinetics Calculator

Enzyme reaction rate from Vmax, Km and [S].

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Balance any chemical equation to whole numbers.