Hardy-Weinberg and allele frequencies: WACE Year 12 Biology

Syllabus dot point

“Use the Hardy-Weinberg principle to calculate allele and genotype frequencies and test for change”

7 min readUpdated 2 Jun 2026

Quick answer

The Hardy-Weinberg principle gives the expected allele and genotype frequencies in a population that is not evolving. Allele frequencies follow p plus q equals 1, and genotype frequencies follow p squared plus 2pq plus q squared equals 1. If a real population's frequencies differ from these expected values, one of the equilibrium conditions has been broken and the population is evolving.

What this dot point is asking

SCSA wants you to apply the two Hardy-Weinberg equations to calculate frequencies, state the conditions for equilibrium, and interpret a deviation from equilibrium as evidence of evolution. A strong answer defines each symbol and shows the working.

The principle and its purpose

The Hardy-Weinberg principle provides a null model: it predicts what allele and genotype frequencies should be if no evolutionary forces are acting. It is useful precisely as a baseline, because real populations rarely meet all the conditions, so comparing observed frequencies with the predicted ones reveals whether and how a population is changing.

The two equations

For a gene with two alleles:

Let p be the frequency of the dominant allele and q be the frequency of the recessive allele. Then p plus q equals 1.
The genotype frequencies are given by p squared plus 2pq plus q squared equals 1, where p squared is the frequency of homozygous dominant individuals, 2pq is the frequency of heterozygotes, and q squared is the frequency of homozygous recessive individuals.

Hardy-Weinberg equations

Allele frequencies: p plus q equals 1.
Genotype frequencies: p squared plus 2pq plus q squared equals 1.
Here p squared is homozygous dominant, 2pq is heterozygous, q squared is homozygous recessive.

The conditions for equilibrium

A population stays at Hardy-Weinberg equilibrium only if five conditions hold:

no mutation,
no gene flow (no migration in or out),
no natural selection,
random mating,
a very large population (no genetic drift).

If any condition is broken, allele frequencies can change and the population evolves.

The principle is a null model

Hardy-Weinberg does not describe what real populations do; it predicts what they would do if nothing were changing them. Its value is as a baseline: any difference between the predicted and observed frequencies points to an evolutionary force at work.

Worked calculation

Finding allele frequencies from phenotype

Solve for the allele and genotype frequencies

In a population, 9 percent of individuals show a recessive condition, so q squared equals 0.09. Taking the square root, q equals 0.3. Since p plus q equals 1, p equals 0.7. The frequency of heterozygous carriers is 2pq, which is 2 times 0.7 times 0.3, equal to 0.42, or 42 percent. The frequency of homozygous dominant individuals is p squared, which is 0.49, or 49 percent. So in this population almost half the people are unaffected carriers, a result you could not see by phenotype alone.

Interpreting deviation as evolution

The power of the principle is in the comparison. If the observed genotype frequencies match p squared, 2pq and q squared, the population is at equilibrium and is not evolving for that gene. If the observed frequencies differ significantly, then one of the five conditions has been broken, for example selection against the recessive phenotype or gene flow from another population, and the population is evolving.

A practical caution

The recessive allele frequency must be found from the homozygous recessive group (q squared), because that is the only genotype you can identify by phenotype alone under simple dominance. You cannot tell homozygous dominant from heterozygous individuals just by looking, which is exactly why the equation is needed to estimate carrier numbers.

Key terms

Allele frequency: the proportion of one allele among all copies of a gene in a population. Genotype frequency: the proportion of individuals with a particular genotype. Equilibrium: the state in which allele frequencies stay constant. Null model: the expected result if no evolution is occurring.

Why this matters for continuity

Hardy-Weinberg turns evolution into something measurable. By calculating expected frequencies and comparing them with reality, biologists can detect selection, migration or drift in a real population, estimate the hidden number of carriers of a genetic condition, and quantify how fast a gene pool is changing. It connects the abstract idea of evolution to numbers that can be tested.

Common traps

Finding p from the dominant phenotype: You cannot, because dominant phenotype includes both p squared and 2pq. Always start from q squared, the recessive phenotype.
Forgetting to square root: q squared gives the recessive phenotype frequency; q is its square root.
Listing fewer than five conditions: Equilibrium needs no mutation, no gene flow, no selection, random mating and a large population.
Saying the population is evolving when it matches: Matching the equations means no evolution for that gene; deviation indicates evolution.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style6 marks

In a population, 16% of individuals show a recessive condition controlled by a single gene with two alleles. Assuming Hardy-Weinberg equilibrium, calculate the frequency of the recessive allele, the frequency of heterozygous carriers, and the frequency of homozygous dominant individuals. Show your working.

Show worked answer →

A 6 mark calculation answer must show each step.

Recessive allele: The recessive phenotype frequency is $q^2 = 0.16$ , so $q = \sqrt{0.16} = 0.4$ .
Dominant allele: Since $p + q = 1$ , $p = 1 - 0.4 = 0.6$ .
Heterozygous carriers: Carrier frequency is $2pq = 2 \times 0.6 \times 0.4 = 0.48$ , i.e. $48\%$ .
Homozygous dominant: $p^2 = 0.6^2 = 0.36$ , i.e. $36\%$ .
Check: $0.36 + 0.48 + 0.16 = 1.00$ , as required.

Markers reward starting from $q^2$ , the square root step, correct use of $p + q = 1$ and correct $2pq$ and $p^2$ values.

WACE 2023-style5 marks

State the conditions required for a population to remain at Hardy-Weinberg equilibrium, and explain how a difference between observed and expected genotype frequencies can be used as evidence that a population is evolving.

Show worked answer →

A 5 mark answer needs the five conditions plus the interpretation.

Conditions. A population stays at equilibrium only if: there is no mutation; no gene flow (no migration); no natural selection; mating is random; and the population is very large (so genetic drift is negligible).

Using deviation. The equations $p + q = 1$ and $p^2 + 2pq + q^2 = 1$ predict the genotype frequencies expected if no evolutionary force acts (a null model). If the observed genotype frequencies match these predictions, the population is not evolving for that gene. If they differ significantly, at least one condition has been broken (for example selection against the recessive phenotype or gene flow), so the population is evolving.

Markers reward all five conditions and the explicit logic that a significant deviation from the predicted frequencies indicates an evolutionary force.

Practise this

Exam-style questions 2 on this page
Flashcards Biology deck

Sources & how we know this

Biology ATAR course: syllabus and support materials — SCSA
Biology: past ATAR course examinations — SCSA