VCE Biology — Unit 4 AoS 1
Hardy-Weinberg Equilibrium — Flashcards & Quiz
The Hardy-Weinberg equilibrium describes conditions under which allele frequencies remain constant across generations: no mutation, random mating, no selection, large population size, and no migration. In VCE Biology Unit 4 AoS 1, you use the equations p + q = 1 and p² + 2pq + q² = 1 to calculate genotype and allele frequencies. Violations of Hardy-Weinberg assumptions indicate evolution is occurring. Exam questions test calculations, interpretation of data, and explaining which assumption is violated in a given scenario.
Sample Flashcards
Q1: State the Hardy-Weinberg equilibrium and its conditions.
Hardy-Weinberg predicts that allele and genotype frequencies remain constant in a population across generations IF: 1) No natural selection. 2) No mutations. 3) No gene flow (migration). 4) Random mating. 5) Large population size (no genetic drift). Equations: p + q = 1 (allele frequencies); p² + 2pq + q² = 1 (genotype frequencies), where p = dominant allele frequency, q = recessive allele frequency.
Q2: How can Hardy-Weinberg be used to detect if evolution is occurring?
Hardy-Weinberg provides a null model — expected genotype frequencies IF no evolution is occurring. By comparing observed genotype frequencies in a population to Hardy-Weinberg predictions, deviations indicate that evolutionary forces are acting. Significant deviation from expected p² + 2pq + q² frequencies suggests one or more conditions are violated: natural selection, mutation, gene flow, non-random mating or small population size.
Sample Quiz Questions
Q1: Hardy-Weinberg equilibrium requires that the population is very small for allele frequencies to remain stable.
Answer: FALSE
Hardy-Weinberg requires a LARGE population to minimise the effects of genetic drift. Small populations experience random fluctuations in allele frequencies (genetic drift), violating the equilibrium.
Q2: In Hardy-Weinberg equilibrium, q² represents the frequency of the homozygous recessive genotype.
Answer: TRUE
In the Hardy-Weinberg equation p² + 2pq + q² = 1: p² = homozygous dominant frequency, 2pq = heterozygous frequency, q² = homozygous recessive frequency. q is the recessive allele frequency.
Last updated: March 2026 · 2 flashcards · 2 quiz questions