The Hardy–Weinberg Principle
Where does p² + 2pq + q² = 1 come from, and what does it tell you about a population? Step through the algebra and work real allele-frequency calculations.
A Punnett square predicts one cross. The Hardy–Weinberg principlescales the same idea up to a whole population — and makes a bold claim: left undisturbed, the mix of alleles never changes. It is the “do nothing” baseline against which biologists measure evolution.
Why the frequencies stay put
Picture every copy of a gene in a population tipped into one enormous pot. Some fraction of them are the dominant allele — call that fraction p — and the rest are the recessive allele, q. Hardy and Weinberg showed that if nothing interferes, those proportions are simply reshuffled into the next generation unchanged: parents pass on the very mix they inherited.
Because nothing drives the frequencies up or down, the population sits at equilibrium— generation after generation, p and q hold steady. That makes the principle a null hypothesis: if real data drift away from what it predicts, some evolutionary force (selection, migration or mutation) must be at work.
Hardy–Weinberg is the genetic version of “an object stays still unless a force pushes it”.
See it: build the equation
The whole principle turns on one move. If alleles pair up at random when gametes meet, the chance of any genotype is just the product of its two allele frequencies. Squaring p + q = 1 does that bookkeeping for all three genotypes at once.
Step through the derivation below and watch each term appear.
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A step-by-step derivation. Starting from the allele-frequency equation p + q = 1, squaring both sides models combining two alleles at random and expands to p² + 2pq + q² = 1 — the genotype frequencies of homozygous dominant (p²), heterozygous (2pq) and homozygous recessive (q²).
The two equations
Everything you need is two linked equations. The first counts alleles; the second counts genotypes.
Alleles: p + q = 1. Every allele at this locus is either the dominant or the recessive version, so the two fractions must add up to 1 (that is, to 100% of all the alleles).
Genotypes: p² + 2pq + q² = 1, where p² is the frequency of homozygous dominant (AA), 2pq of heterozygous carriers (Aa), and q² of homozygous recessive (aa) individuals.
Worked example
A recessive genetic condition affects 1 in 100 people in a population at equilibrium. What fraction are unaffected carriers?
Only affected people (aa) have a known genotype, so start there. One in a hundred show the recessive phenotype, so q² = 0.01. Square-root it: q = √0.01 = 0.1. Then p = 1 − q = 0.9.
Carriers are the heterozygotes (Aa), whose frequency is 2pq = 2 × 0.9 × 0.1 = 0.18, or 18%. Notice the twist students love: the 18% of hidden carriers vastly outnumber the 1% who are actually affected.
Common mistakes
Practice
In a population at equilibrium, the homozygous recessive genotype has a frequency of q² = 0.04. Find q, p and the frequency of heterozygous carriers.
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q = √0.04 = 0.2, so p = 1 − 0.2 = 0.8. Carriers are 2pq = 2 × 0.8 × 0.2 = 0.32 — that is 32% of the population.
Where next?
Hardy–Weinberg tracks the same alleles a Punnett square handles, but across an entire population rather than a single cross — so if the algebra here feels abstract, working through a cross or two first makes the p and q click into place.
Frequently asked questions
What are the two Hardy–Weinberg equations?+
How do you find allele frequencies from a phenotype?+
What conditions are needed for Hardy–Weinberg equilibrium?+
Why is 2pq used for heterozygotes?+
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