AP Biology: Hardy?Weinberg Equilibrium (p² + 2pq + q² = 1, Conditions, Applications)

Hardy?Weinberg Equilibrium (p² + 2pq + q² = 1, Conditions, Applications)

Concept Summary

• Hardy-Weinberg Equilibrium (HWE): A mathematical model predicting allele and genotype frequencies in a non-evolving population, serving as a null hypothesis for evolutionary studies.
• Allele frequency (p, q): Proportion of a specific allele (e.g., A or a) in a population’s gene pool; p + q = 1 for two alleles.
• Genotype frequency (p², 2pq, q²): Proportion of homozygous dominant (p²), heterozygous (2pq), and homozygous recessive (q²) individuals in a population; p² + 2pq + q² = 1.
• Conditions for HWE: Five requirements (no mutation, no migration, no selection, large population, random mating) that must be met for allele frequencies to remain constant.
• Applications: Used to detect evolution (deviations from HWE), estimate carrier frequencies (e.g., 2pq for recessive disorders), and model genetic drift.

Core Questions

WHAT (definitional)

Q: What is the Hardy-Weinberg equation? A: A formula (p² + 2pq + q² = 1) that predicts genotype frequencies in a population where allele frequencies are stable. Trap/Clarification: The equation describes expected frequencies; real populations rarely meet all conditions.

Q: What does p represent in HWE? A: The frequency of the dominant allele (A) in the gene pool. Trap/Clarification: p is the allele frequency, not the frequency of dominant phenotypes (which includes p² + 2pq).

WHY (causal/explanatory)

Q: Why is HWE considered a "null model"? A: It provides a baseline to test for evolution: deviations from HWE suggest evolutionary forces (e.g., selection, drift) are acting. Trap/Clarification: HWE itself does not cause evolution; it describes a hypothetical non-evolving population.

Q: Why is random mating required for HWE? A: Non-random mating (e.g., inbreeding) alters genotype frequencies (e.g., increases homozygosity) without changing allele frequencies. Trap/Clarification: Random mating affects genotype frequencies, not allele frequencies (which remain p and q).

HOW (process/application)

Q: How do you calculate allele frequencies from genotype counts? A: For a gene with alleles A and a: - p = (2 × AA + Aa) / (2 × total individuals) - q = (2 × aa + Aa) / (2 × total individuals) Trap/Clarification: Count alleles, not individuals (e.g., each AA individual contributes 2 A alleles).

Q: How is HWE used to estimate carrier frequency for a recessive disorder? A: If q² (frequency of affected individuals) is known, 2pq (carrier frequency)-2q (since p-1 for rare disorders). Trap/Clarification: This shortcut only works if q is very small (e.g., q < 0.05); otherwise, use 2pq with p = 1 – q.

CAN (conditions/possibilities)

Q: Can a population be in HWE if it has a small population size? A: No; genetic drift (random allele frequency changes) violates HWE in small populations. Trap/Clarification: Even large populations can deviate from HWE if other conditions (e.g., selection) are violated.

Q: Under what conditions does p² + 2pq + q² = 1 hold true? A: Only if all five HWE conditions are met: no mutation, no migration, no selection, large population, and random mating. Trap/Clarification: The equation is always mathematically true for p and q, but genotype frequencies only match expectations if conditions are met.

Quick Facts & Traps

• Fact: HWE applies to a single locus with two alleles; it cannot predict frequencies for multiple genes or polygenic traits.
• Trap: Assuming p² = frequency of dominant phenotypes-Reality: Dominant phenotypes include p² + 2pq (heterozygotes).
• Fact: Allele frequencies (p, q) remain constant in HWE, but genotype frequencies (p², 2pq, q²) may change if mating is non-random.
• Trap: Confusing allele frequency (q) with genotype frequency (q²)-Reality: q is the proportion of a alleles; q² is the proportion of aa individuals.
• Fact: Selection against a recessive allele (e.g., aa individuals die) slows but does not stop its elimination because heterozygotes (Aa) retain the allele.
• Trap: Ignoring the 5% rule for chi-square tests-Reality: If p or q < 0.05, use 2q to estimate 2pq (carriers) for rare disorders.

Rapid-Fire True/False

• Statement: If a population is in HWE, it is not evolving. Answer: TRUE Why the common mistake happens: Students confuse HWE (a model) with real populations, which often violate its conditions.

• Statement: The frequency of the recessive phenotype (q²) can be used to directly calculate p. Answer: TRUE (p = 1 – ?q²) Why the common mistake happens: Students forget to take the square root of q² to find q before calculating p.

• Statement: A population with 100% heterozygotes (2pq = 1) can be in HWE. Answer: FALSE Why the common mistake happens: Students overlook that p² and q² must also exist (even if rare) for HWE to hold.