VCE Biology — Unit 4 AoS 2
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 2, 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.
Key Points
- Hardy-Weinberg models allele and genotype frequencies in a population NOT undergoing evolution: p + q = 1 (alleles) and p² + 2pq + q² = 1 (genotypes).
- Five assumptions: no mutation, no migration, no natural selection, random mating, infinitely large population. Violation of ANY assumption = evolution.
- If you know q² (homozygous recessive frequency), take √ → q; then p = 1 − q; then 2pq gives the carrier frequency. This is the standard VCAA calculation.
- Real populations rarely meet all assumptions — Hardy-Weinberg is a NULL MODEL, not a description of reality. Departures tell you what evolutionary force is acting.
- Bottleneck effect (population crash) and founder effect (small colonising group) both reduce genetic diversity via drift.
- VCAA exam skill: given genotype counts, calculate observed frequencies, expected Hardy-Weinberg frequencies, and identify which assumption is most likely violated.
Common Mistakes to Avoid
- Applying Hardy-Weinberg to populations that don't meet the five assumptions (small, non-random mating, selection) and expecting correct answers.
- Confusing p² (homozygous dominant) with q² (homozygous recessive) — q² is the one you're usually given because recessive phenotypes are visible.
- Forgetting that 2pq is the HETEROZYGOTE frequency, not a carrier "rate" — they're different if the question asks about disease probability.
- Using the wrong equation: p² + 2pq + q² = 1 for GENOTYPES; p + q = 1 for ALLELES.
- Mixing up "allele frequency" and "genotype frequency" — allele frequency is about the pool of gametes; genotype is about individuals.
Exam Strategy
VCAA Unit 4 AOS 1 Hardy-Weinberg questions are calculation-heavy. Start by extracting the given data: is it a phenotype count, genotype count, or allele count? Convert to q² first, then q, then p, then use 2pq or p² as required. Always state your working explicitly. For "which assumption is violated?" questions, link the scenario to a specific evolutionary force (e.g. "inbreeding increases homozygote frequencies — violates random mating").
Revision Tip
Hardy-Weinberg is all practice — drill Revizi calculation flashcards until you can move from phenotype frequency to carrier frequency in under 90 seconds.
Related Concepts
Last updated: March 2026