BIOB 160 Principles of Living Systems · Unit 3: Cell division and inheritance · on Exam 3
3.5 Odds and ends: genes in populations
Study notes for Odds and ends: genes in populations in BIOB 160 Principles of Living Systems at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in StudyCat.
Big ideas
- A population’s gene pool is all its alleles. Microevolution is a change in allele frequencies over generations.
- The Hardy–Weinberg equation describes a population that is not evolving: allele frequencies p + q = 1 and genotype frequencies p² + 2pq + q² = 1 (AA, Aa, aa).
- Hardy–Weinberg equilibrium assumes no mutations, random mating, no natural selection, a very large population (no genetic drift) and no gene flow. Real populations violate these, which is how evolution happens.
- To solve problems, start from the homozygous recessive frequency: q² = fraction showing the recessive trait, so q = √(q²), p = 1 − q, and carriers = 2pq.
- Genetic variation comes from mutation and sexual reproduction; natural selection, genetic drift (founder and bottleneck effects) and gene flow change allele frequencies.
Key formulas
Allele frequencies
$$p + q = 1$$
Genotype frequencies
$$p^2 + 2pq + q^2 = 1$$
Worked example
In a population in Hardy–Weinberg equilibrium, 16% of people show a recessive trait. What fraction are heterozygous carriers?
Show the solution
q² = 0.16, so q = 0.4 and p = 0.6. Carriers = 2pq = 2 × 0.6 × 0.4 = 0.48 (48%).
Common mistakes
- Setting q equal to the fraction with the recessive trait. That fraction is q²; take the square root.
- Forgetting the 2 in 2pq.
- Assuming the dominant allele must be the more common one. Frequencies depend on the population, not on dominance.
Exam tipReading: Campbell ch. 23 (pp. 487–491), listed as “odds and ends”. Always find q first from the recessive phenotype.
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