The recessive phenotype's frequency is q², never q. Take the square root before anything else.
Hardy-Weinberg is the 'nothing is happening' baseline for one gene in a population. When real counts miss that baseline, some force is at work, and the shape of the miss hints at which one.
Key terms
- Allele frequency — the share of all copies of a gene that are one allele. With two alleles, p + q = 1.
- Genotype frequency — the share of individuals with each genotype. The baseline predicts p² AA, 2pq Aa, and q² aa.
- Hardy-Weinberg equilibrium — allele frequencies holding steady from one generation to the next. A population in equilibrium is not evolving at that gene.
- Carrier — a heterozygote (Aa). It holds one recessive allele but shows the dominant trait.
- Inbreeding — mating with relatives. It makes homozygotes more common and heterozygotes rarer. On its own, it leaves allele frequencies unchanged.
How to run the numbers
- Given genotype counts, count alleles. Each AA holds two A's and each Aa holds one. p = A copies ÷ (2 × individuals), and q = 1 − p.
- Given only phenotypes, start with the recessive one. Only aa individuals show it, so its frequency is q². Take the square root to get q.
- Predict genotypes: p² AA, 2pq Aa, q² aa, which add up to 1. Multiply each by the population size for expected counts.
- Compare observed with expected. If the gap is too big to be chance (a chi-square test decides), at least one condition is broken.
- The conditions: a large population, random mating, no gene flow, no mutation, and no selection. Break any one, and the prediction can fail.
Why it matters
Hardy-Weinberg is the baseline for the rest of this unit. Selection, drift, gene flow, and mutation are each a way to break a condition. Exam questions hand you counts, ask you to compute frequencies, then ask what broke. It also explains why harmful recessive alleles linger. When q is small, 2pq dwarfs q², so almost every copy sits in a healthy carrier, where selection can't reach it.
Exam traps
| Trap | Actually |
|---|
| 4% of people show the recessive trait, so q = 0.04. | That 4% is q², since only aa individuals show it. The square root gives q = 0.2. |
| The dominant allele is the common one, or it will become common. | Dominance decides which allele shows, not how many copies exist. Under Hardy-Weinberg conditions, a rare dominant allele stays rare. |
| Equilibrium means the two alleles are equally common. | Any p and q can be at equilibrium. What matters is that they hold steady and genotypes match p², 2pq, q². |
| Too few heterozygotes means selection is favoring one allele. | A symmetric pattern, with both homozygotes inflated, is the classic sign of inbreeding. It regroups alleles into homozygotes without changing their frequencies. |
Worked example
A large wildflower population is at Hardy-Weinberg equilibrium. Red (R) is dominant to white (r), and 9% of the plants are white. A gardener pulls up every white-flowered plant before it can shed pollen or set seed. The remaining plants mate at random. About what percent of the next generation will be white?
- White plants are rr, so q² = 0.09. That gives q = 0.3 and p = 0.7.
- Picture 100 plants: 49 RR, 42 Rr, 9 rr. Pull the 9 white ones. 91 plants remain, carrying 182 alleles.
- Only the 42 Rr plants carry r, one copy each. New q = 42 ÷ 182 ≈ 0.23.
- Random mating again: white = q² ≈ 0.23 × 0.23 ≈ 0.05.
About 5% white: down from 9%, but nowhere near zero.
Pulling white plants removes only the r alleles sitting in rr plants. The 42 red Rr plants still pass r to half their gametes, and two carriers can produce a white seedling. The near miss is 0%, which assumes the allele lived only in the plants you could see. Selection against a recessive trait slows down as the allele gets rarer, because more of its copies hide in carriers.