Hardy-Weinberg equilibrium
A model for a population that is not evolving. Allele frequencies stay constant if there is no selection, mutation, drift or gene flow, and mating is random. If p and q are the two allele frequencies, p + q = 1 and genotype frequencies are p^2 + 2pq + q^2 = 1.
Last updated
Analogy
A control experiment: what would happen if nothing changed?
Example
1 in 2,500 have disease.q^2 = 0.0004, so q = 0.02.p = 0.98.2pq = 0.039.About 1 in 25 carry it.
In practice
It lets us estimate how many carriers of a recessive disease are in a population.
Practice
Flashcards
- State the Hardy-Weinberg equations.p + q = 1 and p^2 + 2pq + q^2 = 1.
Interview questions
- List the conditions for Hardy-Weinberg equilibrium and explain why real populations rarely meet them.
- No mutation
- No selection
- No gene flow
- Large population, no drift
- Random mating
Questions people ask
Why is Hardy-Weinberg useful if populations never meet it?
It gives a baseline. If real allele frequencies differ from the predictions, something is acting on the population, such as selection, drift or non-random mating. It is also used to estimate carrier numbers for recessive diseases from the frequency of affected individuals.
How do I use the Hardy-Weinberg equation?
Start from what you can count, usually the frequency of the recessive phenotype, which equals q squared. Take the square root to get q, then calculate p as 1 minus q. Heterozygotes are 2pq, and the dominant homozygotes are p squared.