Evolution · Lesson 7 of 10
Hardy–Weinberg Equilibrium and Population Calculations
“From individual genotypes to a population gene pool; Why the Hardy–Weinberg formula has three terms; A worked calculation from a recessive phenotype; Conditions and interpretation; Using population counts as evidence”
• Define gene pool, allele frequency and genotype frequency. • Derive the expected p², 2pq and q² proportions from random gamete union. • Solve Hardy–Weinberg problems with explicit assumptions. • Distinguish deviations of genotype proportions from changes in allele frequency.
From individual genotypes to a population gene pool
A diploid individual carries two copies of an autosomal gene locus. Suppose that the locus has two alleles, A and a. Let p denote the fraction of all allele copies that are A and q the fraction that are a. Since these are the only two alleles at that locus, p + q = 1. The gene pool refers to the alleles present across members of the breeding population, and allele frequencies summarise their relative occurrence.
A genotype frequency is different: it is the proportion of individuals who are AA, Aa or aa. If a population contains 100 AA individuals, 60 Aa individuals and 40 aa individuals, it contains 200 people but 400 allele copies. There are 2×100 + 60 = 260 A copies and 60 + 2×40 = 140 a copies. Therefore p = 260/400 = 0.65 and q = 140/400 = 0.35. These allele frequencies sum to one even though the observed genotype frequencies need not equal Hardy–Weinberg predictions.
Why the Hardy–Weinberg formula has three terms
Imagine a very large population in which the frequency of A in eggs and sperm is p and the frequency of a is q, and their union is random with respect to this locus. An AA zygote requires A from each gamete, so its probability is p×p = p². An aa zygote requires a from both, so its probability is q×q = q². An Aa zygote can arise as A from egg and a from sperm or the reverse, so its probability is pq + qp = 2pq.
The three alternatives exhaust the possible genotypes, yielding p² + 2pq + q² = (p + q)² = 1. This is a probability identity for the expected genotype frequencies under the model's assumptions. It is not a general statement that every natural population has exactly those proportions. Deviations can be informative, but sampling error and the suitability of the model must be evaluated.
A worked calculation from a recessive phenotype
Suppose 9 percent of individuals in a large, randomly mating population display a recessive phenotype caused by aa, with complete recessivity, no selection and two alleles at an autosomal locus. Under equilibrium, q² = 0.09. Taking the square root gives q = 0.30, so p = 1 − 0.30 = 0.70. The predicted AA frequency is p² = 0.49, the heterozygote Aa frequency is 2pq = 2×0.70×0.30 = 0.42, and the aa frequency is 0.09.
The values add to 0.49 + 0.42 + 0.09 = 1.00. In a sample of 1,000 individuals, the model predicts about 490 AA, 420 Aa and 90 aa. The conclusion is conditional: a recessive phenotype directly indicates q² only when the genotype–phenotype assumptions apply and the population is sufficiently close to equilibrium. The heterozygote frequency is not 0.30; 0.30 is the frequency of the a allele among all allele copies.
| Genotype | Random union event | Expected frequency (p=0.70, q=0.30) |
|---|---|---|
| AA | A × A | p² = 0.49 |
| Aa | A × a, or a × A | 2pq = 0.42 |
| aa | a × a | q² = 0.09 |
Conditions and interpretation
The idealised equilibrium assumes a large breeding population, random mating with respect to the locus, no natural selection at that locus, no mutation introducing or converting alleles, and no migration that changes its allele pool. Under these conditions the allele frequencies remain stable between generations, and the predicted autosomal genotype proportions appear after random union of gametes. Real populations can approximate the model while violating some assumptions weakly.
Nonrandom mating can change the proportions of homozygotes and heterozygotes without necessarily changing p or q. Genetic drift, selection, mutation and gene flow can change allele frequencies. Recombination usually alters multilocus associations but does not by itself change the frequency of one allele at a particular locus. Therefore an observed departure from p²:2pq:q² does not by itself prove that the allele frequency has evolved, and failure to detect a departure does not prove that no evolutionary forces are acting.
Using population counts as evidence
To calculate frequencies from genotype counts, first count individuals and convert them to allele copies. For N diploid individuals, the total number of copies at an autosomal locus is 2N. Thus p = (2NAA + NAa)/(2N), with NAA and NAa denoting genotype counts, and q = (2Naa + NAa)/(2N). Expected equilibrium proportions are then computed from p and q and compared with observed proportions.
For the 100 AA, 60 Aa and 40 aa individuals above, observed genotype fractions are 0.50, 0.30 and 0.20. Equilibrium expectations from p = 0.65 and q = 0.35 are 0.4225, 0.4550 and 0.1225. The differences suggest a departure from the simple genotype model, but further data are needed to identify why. Nonrandom mating, substructure, selection, sampling or genotyping error can produce deviations.
Quick Quiz
Quiz
If p = 0.7 and q = 0.3, what is the expected heterozygote frequency?
If q² = 0.16 under the model, what is p?
What does p represent in this model?
What can nonrandom mating do even without changing allele frequencies?
A population contains 30 AA, 40 Aa and 30 aa individuals. What is p?
Practice Problems
- Derive the Hardy–Weinberg genotype equation rather than stating it.
- If 4 percent of a population has an autosomal recessive phenotype, estimate the carrier frequency under equilibrium.
- From 100 AA, 60 Aa and 40 aa individuals, calculate allele and expected genotype frequencies.
- Explain why departure from the expected genotype proportions is not by itself proof of changing allele frequencies.
- If p=0.8, what number of aa individuals is expected among 500 under equilibrium?
Under random union of A and a gametes at frequencies p and q, AA occurs with probability p²; aa occurs with q²; and heterozygotes arise in two orders, A then a or a then A, each with probability pq. Thus P(Aa) = 2pq. All genotypes together have probability p² + 2pq + q² = (p+q)² = 1.
Key Takeaways
• Allele frequency and genotype frequency measure different properties of a population. • For two alleles, p + q = 1. • Random gamete union gives expected frequencies p², 2pq and q². • Heterozygotes have two possible parental gamete arrangements. • The equation requires assumptions including random mating and a sufficiently large population. • Nonrandom mating can alter genotype proportions without necessarily changing allele frequency. • Drift, selection, mutation and gene flow can alter allele frequencies.