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How Do You Use the Hardy-Weinberg Equation?

Learn how to use the Hardy-Weinberg equations to calculate allele and genotype frequencies, with a worked AQA A Level Biology example and a common exam mistake.

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Use the Hardy-Weinberg equations to estimate allele and genotype frequencies in a population. If you know the proportion of individuals showing a recessive phenotype, start with q2q^2q2, then work back to qqq and ppp.

What the letters mean

For a gene with two alleles, ppp is the frequency of the dominant allele and qqq is the frequency of the recessive allele. An allele frequency is the proportion of copies of that gene that are a particular allele. Because these are the only two alleles:

p+q=1 p+q=1 p+q=1

The expected genotype frequencies are:

p2+2pq+q2=1 p^2+2pq+q^2=1 p2+2pq+q2=1

Here, p2p^2p2 is the proportion homozygous dominant, 2pq2pq2pq is the proportion heterozygous, and q2q^2q2 is the proportion homozygous recessive. Homozygous means having two copies of the same allele; heterozygous means having one of each.

A recessive phenotype identifies the homozygous recessive genotype. A dominant phenotype does not identify a single genotype: an individual showing it could be homozygous dominant or heterozygous. That is why the recessive group is often the easiest starting point.

Worked example

Suppose 16%16\%16% of a population shows a recessive trait. Assuming the trait is controlled by one gene with two alleles and complete dominance, those individuals are homozygous recessive. Convert the percentage to a decimal first: 16%=0.1616\%=0.1616%=0.16. Frequencies are proportions between 000 and 111, so do not substitute 161616 into the equation. Therefore:

q2=0.16,q=0.16=0.4 q^2=0.16,\qquad q=\sqrt{0.16}=0.4 q2=0.16,q=0.16​=0.4

The dominant allele frequency is p=1−0.4=0.6p=1-0.4=0.6p=1−0.4=0.6. The expected heterozygote frequency is:

2pq=2(0.6)(0.4)=0.48 2pq=2(0.6)(0.4)=0.48 2pq=2(0.6)(0.4)=0.48

So 48%48\%48% of the population is expected to be heterozygous. As a check, the expected homozygous dominant frequency is p2=0.36p^2=0.36p2=0.36, and 0.36+0.48+0.16=10.36+0.48+0.16=10.36+0.48+0.16=1.

Exam mistake to avoid: the proportion showing the recessive trait is q2q^2q2, not qqq. Take its square root before finding ppp. These are Hardy-Weinberg predictions: they assume conditions such as random mating, a large population, and no selection, mutation or migration affecting allele frequencies. Random mating means mates are not chosen by genotype. If these conditions do not hold, observed genotype frequencies may differ from the predictions.

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