Hardy-Weinberg equilibrium describes what happens to allele and genotype frequencies in a population when no evolutionary forces are acting. It gives biologists a baseline for predicting how common different genotypes should be from one generation to the next. This matters because real populations can be compared to the Hardy-Weinberg expectation to detect evolution.
If the observed genotype frequencies differ from the prediction, something is changing the gene pool.
The model uses two allele frequencies, p and q, for a gene with two alleles. The allele frequencies add to 1, and the genotype frequencies are predicted by expanding (p + q)^2 into p² + 2pq + q² = 1. The five conditions are no mutation, random mating, no natural selection, very large population size, and no migration.
Scientists often use the frequency of a recessive phenotype to calculate q², then find q, p, and the expected genotype frequencies.
Key Facts
- Allele frequency equation: p + q = 1.
- Genotype frequency equation: p² + 2pq + q² = 1.
- p represents the frequency of one allele, often the dominant allele.
- q represents the frequency of the other allele, often the recessive allele.
- For a recessive phenotype, frequency of aa = q², so q = square root of q².
- Hardy-Weinberg equilibrium requires no mutation, random mating, no selection, a very large population, and no migration.
Vocabulary
- Allele frequency
- The proportion of all gene copies in a population that are a particular allele.
- Genotype frequency
- The proportion of individuals in a population that have a particular genotype.
- Gene pool
- The complete set of alleles present in a population.
- Hardy-Weinberg equilibrium
- A condition in which allele and genotype frequencies stay constant across generations because no evolutionary forces are acting.
- Recessive phenotype
- A trait that appears only when an individual has two recessive alleles for a gene.
Common Mistakes to Avoid
- Using q instead of q² for the recessive phenotype frequency is wrong because the recessive phenotype usually represents the homozygous recessive genotype aa.
- Forgetting that p + q = 1 is wrong because allele frequencies are proportions of the total allele pool and must add to 100 percent.
- Treating 2pq as the frequency of one allele is wrong because 2pq is the genotype frequency of heterozygotes, not an allele frequency.
- Assuming Hardy-Weinberg equilibrium proves no evolution forever is wrong because it is a model prediction that depends on specific conditions being met in that generation.
Practice Questions
- 1 In a population, 9 percent of individuals show a recessive phenotype. Assuming Hardy-Weinberg equilibrium, find q, p, and the expected frequencies of AA, Aa, and aa.
- 2 A population has allele frequencies p = 0.7 and q = 0.3. Calculate the expected genotype frequencies p², 2pq, and q², then state the expected percent of heterozygotes.
- 3 A population has many more heterozygotes than Hardy-Weinberg equilibrium predicts. Give one possible biological reason for this pattern and explain which Hardy-Weinberg condition may be violated.