Population genetics & Hardy–Weinberg

Counting alleles in a population

To study evolution, you need a way to measure a population's genes. All the alleles of all the individuals in a population make up its gene pool. For a gene with two alleles, A and a, we write the allele frequencies as p (the fraction of alleles that are A) and q (the fraction that are a). Every allele is one or the other, so p + q = 1.

Take 500 pea plants: 245 are YY, 210 are Yy and 45 are yy. Each plant has two alleles, so there are 1,000 alleles in total. The Y alleles are 2 × 245 from the YY plants plus 210 from the Yy plants: 700. So p = 700 / 1,000 = 0.7 and q = 300 / 1,000 = 0.3.

The genotype frequencies are just the fractions of individuals with each genotype: here 0.49 YY, 0.42 Yy and 0.09 yy. Allele frequencies and genotype frequencies are different things, and keeping them apart is half the battle in this topic.

Why frequencies stay put

Imagine every individual releases gametes into one big pool, and gametes pair up at random. A fraction p of the gametes carry A and q carry a. The chance that an egg and a sperm are both A is p × p, so the frequency of AA offspring is p². The chance of two a's is q². A heterozygote can form two ways (A egg with a sperm, or a egg with A sperm), so its frequency is 2pq.

These three add up to everything: p² + 2pq + q² = 1. That's just (p + q)² multiplied out, the same logic as a Punnett square, but for a whole population instead of two parents.

Now count the alleles in the new generation. The frequency of A is p² + ½(2pq) = p² + pq = p(p + q) = p. It's exactly what it was before. Mendelian inheritance on its own simply reshuffles alleles; it doesn't favour any of them. A population where this holds is in Hardy–Weinberg equilibrium.

The five conditions, and what breaks them

Hardy–Weinberg equilibrium is a null model: a description of what happens when nothing is going on. It holds only if five conditions are all true. Each one that fails is a way for a population to evolve:

• No natural selection. If some genotypes survive or reproduce better, their alleles increase.
• No mutation. Mutations turn one allele into another, slowly changing p and q.
• No gene flow. Gene flow is the movement of alleles between populations, when individuals (or pollen) migrate in or out.
• A very large population. In a small population, chance alone changes allele frequencies from one generation to the next. This is genetic drift. A bottleneck, when a disaster leaves only a few survivors, and a founder effect, when a few individuals start a new population, are both drift in action.
• Random mating. If individuals choose mates like themselves, or mate with relatives, genotype frequencies shift away from p², 2pq and q².

No real population meets all five conditions perfectly. The value of the model is that it tells you what "no evolution" looks like, so a departure from it is a clue that something, whether selection, drift, migration, mutation or non-random mating, is at work.

From phenotypes to allele frequencies

Often you can't see genotypes. With a dominant and a recessive allele, AA and Aa look the same. But you can always spot the recessive homozygotes, and if the population is in equilibrium, their frequency is q².

1. 1Find the fraction of individuals with the recessive phenotype. This is q², not q.
2. 2Take the square root to get q.
3. 3Calculate p = 1 − q.
4. 4Calculate the carriers: 2pq. The dominant homozygotes are p².

For example, if 1 in 10,000 people has a recessive condition, q² = 0.0001, so q = 0.01 and p = 0.99. Carriers are 2 × 0.99 × 0.01 ≈ 0.02, or about 1 person in 50. Notice how many more carriers there are than affected people. Most copies of a rare recessive allele hide in heterozygotes, where selection can't see them.

Testing for equilibrium with chi-square

When you can count all three genotypes (for example, when alleles are codominant, or you have DNA data), you can test whether a population fits Hardy–Weinberg expectations. Real counts never match exactly, so you need to know whether the difference is bigger than chance would give. The chi-square (χ²) test does this.

1. 1Calculate p and q from the observed genotype counts.
2. 2Calculate expected counts: p² × N, 2pq × N and q² × N, where N is the number of individuals.
3. 3For each genotype, work out (observed − expected)² ÷ expected, and add the three values together to get χ².
4. 4Use 1 degree of freedom: 3 genotype classes, minus 1 because the total is fixed, minus 1 more because you estimated p from the same data.
5. 5Compare with the critical value of 3.84 (for p = 0.05). If χ² is below 3.84, the data are consistent with equilibrium. If it's above, reject equilibrium and look for the cause.

A real example: when the ecological geneticist E. B. Ford scored 1,612 scarlet tiger moths for a gene that affects their wing pattern, the counts of the three genotypes gave χ² ≈ 0.83, well below 3.84. That population looked like it was in equilibrium for that gene.

X-linked genes

For a gene on the X chromosome, males have only one copy. So the fraction of males with a recessive X-linked trait is simply q, while the fraction of females is q². If q = 0.08 for an X-linked colour-blindness allele, about 8% of males but only about 0.64% of females would be affected. That's why X-linked recessive traits are so much more common in males.

Key terms

Gene pool

All the alleles of all the individuals in a population.

Allele frequency (p, q)

The fraction of all copies of a gene in a population that are one particular allele.

Hardy–Weinberg equilibrium

The state in which allele and genotype frequencies stay constant from one generation to the next: p² + 2pq + q² = 1.

Genetic drift

Random change in allele frequencies from one generation to the next, strongest in small populations.

Founder effect / bottleneck

Drift caused by a few individuals starting a new population / by a sharp drop in population size.

Gene flow

Movement of alleles between populations through migration of individuals or gametes.

Non-random mating

Choosing mates by genotype or phenotype, or mating with relatives, which shifts genotype frequencies.

Chi-square test

A statistical test of whether observed counts differ from expected counts by more than chance would explain.

Check yourself

Try answering in your head before you open each answer.

• 1.About 1 in 2,500 babies in a population is born with a recessive condition. Assuming Hardy–Weinberg equilibrium, what fraction of the population are carriers?Show answerHide

  q² = 1/2,500 = 0.0004, so q = 0.02 and p = 0.98. Carriers: 2pq = 2 × 0.98 × 0.02 ≈ 0.039, or about 1 in 25 people. Don't use 0.0004 as q; that's the frequency of affected people.

• 2.Ten birds are blown onto an empty island and start a new population. A generation later, an allele that was at 0.1 on the mainland is at 0.3 on the island, and at first it has no effect on survival. Which Hardy–Weinberg condition was violated?Show answerHide

  The population was very small, so the change is best explained by genetic drift, specifically the founder effect. The ten founders happened to carry more copies of that allele than a typical mainland sample. Selection isn't needed to explain it, since the allele has no effect on survival.

• 3.A population of plants switches from cross-pollination to self-pollination. Predict what happens to p, q and the frequency of heterozygotes over the next few generations.Show answerHide

  p and q stay the same, because selfing doesn't add or remove alleles on its own. But heterozygotes become rarer each generation (a selfed Aa plant gives only half Aa offspring) and both homozygotes become more common. Genotype frequencies move away from Hardy–Weinberg expectations even though allele frequencies don't change.