Population genetics and Hardy–Weinberg
A population’s gene pool is all the alleles of all its genes. Allele frequency is the proportion of one allele among all the alleles of that gene. The Hardy–Weinberg principle predicts allele and genotype frequencies when nothing is changing them; comparing real populations with it shows when evolution is happening.
Key facts
- For a gene with two alleles, p is the frequency of the dominant allele and q the frequency of the recessive allele: p + q = 1.
- The genotype frequencies are given by p² + 2pq + q² = 1: p² is homozygous dominant, 2pq heterozygous and q² homozygous recessive.
- Only the recessive phenotype shows its genotype, so q² is found first from the number of homozygous recessive individuals, then q, then p.
- The principle applies only if the population is large, mating is random, and there is no mutation, no migration in or out, and no selection.
- Natural selection can be stabilising (the average is favoured), directional (one extreme is favoured) or disruptive (both extremes are favoured).
- Genetic drift is a random change in allele frequencies; it has the biggest effect in small populations, for example after the founder effect or a genetic bottleneck.
- Speciation: populations that are isolated stop interbreeding and their gene pools change differently. Allopatric speciation follows geographical separation; sympatric speciation happens in the same place, for example when groups breed at different times.
Worked examples
Example 1
In a population, 1 in 100 people has a recessive condition. Find the frequency of carriers.
- The recessive phenotype frequency is q². q² = 1 ÷ 100 = 0.01
- Take the square root. q = √0.01 = 0.1
- Use p + q = 1. p = 1 − 0.1 = 0.9
- Carriers are heterozygous. 2pq = 2 × 0.9 × 0.1 = 0.18
Answer: 0.18
Example 2
A population has 360 AA, 480 Aa and 160 aa individuals. Find the frequency of allele A.
- There are 1000 individuals, so 2000 alleles. 360 + 480 + 160 = 1000
- Each AA has two A alleles and each Aa has one. 2 × 360 + 480 = 1200
- Divide by the total number of alleles. 1200 ÷ 2000 = 0.6
Answer: 0.6
Example 3
9% of a population show a recessive phenotype. What percentage are homozygous dominant?
- q² is 9%. q² = 0.09
- Take the square root. q = 0.3
- Find p. p = 1 − 0.3 = 0.7
- Homozygous dominant is p². p² = 0.7² = 0.49 = 49%
Answer: 49%