Hardy-Weinberg Equilibrium

Punnett squares tell you what offspring to expect from two parents. Hardy-Weinberg equilibrium does the same trick for a whole population: it predicts genotype frequencies from allele frequencies, as long as the population behaves "ideally." It is the baseline you compare real data against.

The Equations And What Each Symbol Means

For a gene with two alleles, let $p$ be the frequency of one allele and $q$ the frequency of the other. The allele frequencies must account for everything:

The expected genotype frequencies then follow from squaring that sum:

Here $p^2$ is the expected frequency of one homozygote, $2pq$ is the expected frequency of the heterozygote, and $q^2$ is the expected frequency of the other homozygote.

Why Squaring The Allele Frequencies Works

The genotype equation is not a separate rule you memorize. It is just $(p + q)^2$ written out. If mating is random, the chance of pulling allele $A$ for the first copy is $p$ and again $p$ for the second copy, so an $AA$ individual appears with probability $p \times p = p^2$. The heterozygote can form two ways (allele from either parent), which is where the factor of $2$ in $2pq$ comes from. Expanding $(p+q)^2 = p^2 + 2pq + q^2$ reproduces all three terms at once. Seeing it as a squared sum is what keeps allele frequency and genotype frequency from blurring together.

Worked Example: From Allele Frequency To Genotype Frequency

Suppose a gene has two alleles, $A$ and $a$, with allele frequency $p = 0.7$ for $A$ and $q = 0.3$ for $a$.

First confirm the allele frequencies sum to one:

Then compute the expected genotype frequencies:

These add to $1$, as they must:

So under Hardy-Weinberg assumptions you expect about 49% $AA$, 42% $Aa$, and 9% $aa$.

Now Try It With Different Numbers

Take $p = 0.8$ and $q = 0.2$ and work the same three steps: square $p$, double the product $pq$, and square $q$. You should land on $0.64$, $0.32$, and $0.04$, which sum to $1$. Then ask the harder question: if real genotype counts came out far from those values, which assumption would you suspect first?

Where Calculations Go Wrong

Confusing allele frequency with genotype frequency

$p$ and $q$ describe alleles, not the fraction of individuals. The genotype frequencies are $p^2$, $2pq$, and $q^2$. Mixing these up is the single most common error.

Treating the equation as proof of equilibrium

The identity $p^2 + 2pq + q^2 = 1$ holds algebraically for any two-allele setup. A real population is only in equilibrium if the assumptions are reasonable and observed frequencies match the expected ones closely enough.

Forgetting the model is conditional

The classic model assumes random mating, no selection, no mutation introducing alleles, no migration, and a population large enough that drift is negligible. If any of these fail, the prediction may not hold, and a mismatch is a clue to investigate rather than a finished explanation.

When You Use This

You will meet Hardy-Weinberg in population genetics, evolution, and introductory biology, often when estimating carrier frequencies or checking whether genotype counts match a simple expectation before asking which evolutionary force is at work.

If you want another case to practice, try solving a similar population-genetics problem step by step with GPAI Solver.