Hardy–Weinberg Equilibrium
- State the five conditions required for Hardy–Weinberg equilibrium
- Use p + q = 1 and p² + 2pq + q² = 1 to calculate allele and genotype frequencies
- Interpret Hardy–Weinberg as a null model whose violation signals evolution
A population that is NOT evolving
The Hardy–Weinberg principle describes an imaginary population in which allele frequencies stay constant from generation to generation — a population that is not evolving. It is a null hypothesis: if the real allele or genotype frequencies drift away from what Hardy–Weinberg predicts, something (selection, drift, migration, mutation, or nonrandom mating) is driving evolution. The model is powerful precisely because it tells you what to expect when nothing is happening, so any deviation becomes a signal.
Five conditions that must all hold
Equilibrium requires five conditions simultaneously. (1) No natural selection — all genotypes survive and reproduce equally. (2) No mutation — alleles are not converted into other alleles. (3) No gene flow — no migration of individuals into or out of the population. (4) Random mating — mates pair without regard to genotype. (5) A very large population — so that random sampling (genetic drift) does not shift frequencies by chance. Real populations rarely meet all five, which is exactly why real populations evolve. The five conditions are the mirror image of the five mechanisms of evolution.
The trick: start from the recessive phenotype
Homozygous dominant (p²) and heterozygous (2pq) individuals often look identical, so you usually cannot count them directly. But the homozygous recessive phenotype reveals its genotype (q²) exactly, because it is the only genotype that shows the recessive trait. So problems almost always begin the same way: take the fraction of the population showing the recessive phenotype, set it equal to q², and take the square root to get q. From q, everything else follows: p = 1 − q, then p², 2pq, and q².
In a field of 1000 snapdragon-type plants, flower color follows one gene with a dominant red allele and a recessive white allele. Exactly 160 plants have white flowers (the recessive phenotype). Assuming Hardy–Weinberg equilibrium, find the allele frequencies p and q, and the number of plants of each genotype.
- 1.White is the homozygous recessive phenotype, so its frequency equals q². q² = 160 ÷ 1000 = 0.16.
- 2.Take the square root to find the recessive allele frequency: q = √0.16 = 0.4.
- 3.Use p + q = 1 to find the dominant allele frequency: p = 1 − 0.4 = 0.6.
- 4.Homozygous dominant frequency: p² = (0.6)² = 0.36, so 0.36 × 1000 = 360 plants.
- 5.Heterozygous frequency: 2pq = 2 × 0.6 × 0.4 = 0.48, so 0.48 × 1000 = 480 plants.
- 6.Homozygous recessive frequency: q² = 0.16, so 0.16 × 1000 = 160 plants (matching the count given).
- 7.Check that the genotype counts sum to the total: 360 + 480 + 160 = 1000. ✓
In a large, randomly mating population at Hardy–Weinberg equilibrium, 9% of individuals show a recessive genetic condition (homozygous recessive). What is the frequency of the recessive allele, q?
Continuing that population (q = 0.3, so p = 0.7), what fraction of individuals are heterozygous carriers?
The single most common Hardy–Weinberg error is confusing an allele frequency with a genotype frequency. The percentage of individuals showing the recessive trait is q² (a genotype frequency); you must take its square root to get q (the allele frequency). And never forget the factor of 2 in the heterozygote term 2pq.
A biologist samples a population over several generations and finds the frequency of allele A steadily rising while allele a falls. What is the most valid conclusion?
On the AP exam, always begin Hardy–Weinberg problems from the recessive phenotype: set it equal to q², square-root to get q, then p = 1 − q. Show every step. Graders award points for the correct setup (q² = recessive frequency) even if arithmetic slips later.
Answer the 3 checkpoints as you read.
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