OxfordAQA International AS and A-level Biology: Allele frequencies, drift and Hardy-Weinberg (9610) – Revision Notes

These notes condense sections 3.3.7.1 and 3.3.7.2 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, for International A-level exams May/June 2018 onwards. Unit 3 is listed under the International A-level assessments, not the International AS. For full explanations use the study guide. Then try the practice questions, check your coverage on the printable checklist, see the course hub and find gaps with the free diagnostics.

Definitions to learn word for word

Term	Definition
Population	Organisms of one species that could interbreed and that live in the same place at the same time
Gene pool	Every allele of a gene present in one population
Allele frequency	Share of the gene copies in the population that are a chosen allele
Genetic bottleneck	A severe reduction in population size, caused by an environmental event or human activity, which reduces the genetic variety of the gene pool
Genetic drift	A change in the frequency of alleles in a population due to chance
Hardy-Weinberg principle	A mathematical model predicting constant allele frequencies across generations, provided certain conditions apply

Selection for high-yielding breeds

• Breeders choose parents with the wanted phenotype (for example high milk yield or grain yield) and breed only from them.
• Alleles for the wanted phenotype become more frequent. Other alleles become rarer or are lost.
• The gene pool becomes smaller and less varied, especially if few parents are used.
• Advantage: more reliable high yield.
• Disadvantage: little variation for selection to act on if conditions change, for example a new disease. More individuals may be susceptible at once. Breeding between relatives makes harmful homozygous recessive genotypes more likely.

Bottleneck versus drift

	Genetic bottleneck	Genetic drift
What it is	Event that sharply cuts population size	Chance change in allele frequency from generation to generation
Cause	Environmental event or human activity	Random sampling of alleles into gametes and zygotes, and random survival to breed
Effect on gene pool	Reduces genetic variety; allele frequencies in survivors may differ from the original	Changes frequencies by chance; larger effect when an allele has few copies
Linked to advantage?	No, survival is largely by chance	No

Link them in an answer: a bottleneck leaves a small population, and drift acts strongly in a small population, so rare alleles are easily lost and the recovered population is genetically uniform.

Hardy-Weinberg: conditions and results

Conditions (learn all five):

1. Large population.
2. Random mating.
3. No mutation.
4. No migration (no gene flow).
5. No selection.

If they hold, allele frequencies stay constant from generation to generation.

Quantity	Symbol or expression
Frequency of dominant allele	p
Frequency of recessive allele	q
Two alleles together	p + q = 1
All genotypes together	p² + 2pq + q² = 1
Frequency of AA type	p²
Frequency of Aa type	2pq
Frequency of aa type	q²
Recessive phenotype (complete dominance)	q²
Dominant phenotype (complete dominance)	p² + 2pq, or 1 − q²

Method in steps

1. Find the genotype you can identify directly. With complete dominance, this is the recessive phenotype, so its frequency is q².
2. Frequency = number ÷ total.
3. Take q as the square root of q². Then p = 1 − q.
4. Work out p², 2pq and q², then check that they add to 1.
5. To find how many individuals, multiply by the total in the population. Round only at the end.

Worked reminders

Reminder 1: from q. If q = 0.08, then p = 1 − 0.08 = 0.92. Then p² = 0.8464, 2pq = 2 × 0.92 × 0.08 = 0.1472 and q² = 0.0064. Check: 0.8464 + 0.1472 + 0.0064 = 1.

Reminder 2: from the recessive phenotype. In a population of 2500 insects, 49 are wing-less, a recessive trait. q² = 49 ÷ 2500 = 0.0196, so q = 0.14 and p = 0.86. Carriers (2pq) = 2 × 0.86 × 0.14 = 0.2408, which is 0.2408 × 2500 = 602 insects. Homozygous dominant = 0.86² × 2500 = 1849. Check: 1849 + 602 + 49 = 2500.

Reminder 3: from genotype counts. When you can count all three genotypes (codominance, or a genotype test), find allele frequency directly: allele copies ÷ 2N. You do not need the square root. Then compare the observed genotype numbers with p², 2pq and q² multiplied by N to see whether the population fits the model.

Must-know distinctions

• Allele frequency, genotype frequency, phenotype frequency. Allele frequency counts gene copies. Genotype frequency counts individuals with a given genotype. Phenotype frequency counts individuals showing a characteristic, which for a dominant trait includes heterozygotes.
• p and q, or p² and q². The first pair are allele frequencies. The second pair are homozygote frequencies.
• Hardy-Weinberg as a prediction versus as a description. The equation predicts what stays constant if the conditions hold. A population that departs from the prediction is being affected by selection, drift, migration, mutation or non-random mating.
• Drift versus selection. Selection depends on the effect of an allele on survival or reproduction. Drift is chance and does not depend on any advantage.
• Natural versus artificial selection. In artificial selection, people choose the parents. In natural selection, the environment determines which individuals survive and reproduce.

Quick self-test

1. State what is meant by the gene pool of a population.
2. A sample of 50 plants has 30 AA, 15 Aa and 5 aa. Calculate the frequency of allele A.
3. In a population in Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype is 0.0324. Calculate the frequency of heterozygotes.
4. In a population of 6400, 64 individuals show a recessive phenotype. Assuming Hardy-Weinberg conditions, calculate how many are heterozygous.
5. In a population, 75% of individuals show the dominant phenotype. Calculate q.
6. A small island population receives a few individuals from the mainland each year. Name two Hardy-Weinberg conditions that this population breaks.
7. Explain why genetic drift has a larger effect when an allele has few copies.
8. Explain why a population that recovers in number after a bottleneck can still be at risk from a new disease.
9. An allele has frequency 0.125 in a population of 12 individuals. Calculate how many copies of the allele are present.
10. Give one advantage and one disadvantage of selecting only the highest-yielding animals as parents.

Answers

1. The complete set of alleles for a gene in a single population.
2. Copies of A = (2 × 30) + 15 = 75. Total copies = 2 × 50 = 100. Frequency of A = 0.75.
3. q² = 0.0324, so q = 0.18 and p = 0.82. 2pq = 2 × 0.82 × 0.18 = 0.2952.
4. q² = 64 ÷ 6400 = 0.01, so q = 0.1 and p = 0.9. 2pq = 0.18, and 0.18 × 6400 = 1152.
5. The dominant phenotype is p² + 2pq = 0.75, so q² = 1 − 0.75 = 0.25 and q = 0.5.
6. It breaks the conditions of a large population and no migration (gene flow).
7. With few copies, each chance event (an individual failing to breed, or one allele not being passed on) is a large fraction of all the copies, so the frequency can change a lot or the allele can be lost. With many copies, the chance effects average out.
8. The few survivors carried only part of the original gene pool, so the new population has low genetic variety. If few individuals have an allele giving resistance, most may be susceptible, and the variety lost is not restored quickly.
9. Total copies of the gene = 2 × 12 = 24. Copies of the allele = 0.125 × 24 = 3.
10. Advantage: the yield of offspring is high and more consistent. Disadvantage: reduced genetic variety, so more individuals may be susceptible to a new disease or a change in conditions; breeding between relatives raises the chance of harmful recessive genotypes.

Where marks are usually lost

• Treating the recessive phenotype frequency as if it were q, when it is q².
• Counting individuals rather than alleles when working out an allele frequency.
• Forgetting the factor 2 in 2pq, or leaving out the heterozygotes when giving the frequency of a dominant phenotype.
• Writing “the frequency” without saying which allele or genotype it refers to.
• Leaving out units or the final multiplication by population size when the question asks for a number of individuals.
• Answering “explain why drift matters” with “natural selection” ideas. Drift is chance, not advantage.
• Naming a Hardy-Weinberg condition without saying how breaking it changes allele frequencies, for example “migration brings in new alleles or changes their proportions”.
• Treating a bottleneck as selection against weak individuals rather than a random reduction in numbers and variety.
• Rounding p and q early, so that the final number of individuals is slightly out.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, published by OxfordAQA. Sections 3.3.7.1 and 3.3.7.2.