Hardy–Weinberg
Go from phenotype to allele frequency to expected genotypes without mixing them up.
What to do when the question looks unfamiliar
- Step 1Label what you are given: phenotype, genotype, allele, or count.
- Step 2Use p + q = 1 for allele frequencies, and p² + 2pq + q² = 1 for expected genotypes.
- Step 3Multiply a frequency by sample size only when you need an expected count.
Know when to take a square root
For a single two-allele locus with complete dominance and Hardy–Weinberg proportions, the recessive phenotype frequency equals q². If 9% show it, q = √0.09 = 0.30 and p = 0.70. Expected heterozygotes are 2pq = 0.42.
The dominant phenotype combines p² and 2pq. It is not p² alone. If genotypes are directly counted, count two copies from each homozygote and one from each heterozygote; this does not require assuming equilibrium.
The model is a baseline, not a diagnosis
The standard equilibrium model assumes random mating with respect to the locus, a large population, and negligible selection, mutation, and migration. The expected genotype proportions are a comparison for observed data.
A mismatch can have several causes, including structure, selection, nonrandom mating, sampling, or measurement errors. Changing allele frequencies across generations documents evolution at that locus; a single sample’s genotype mismatch does not by itself show change over time.