Population Genetics

Population Genetics is topic 7.4 of AP Biology, inside Natural Selection. This page works through three real practice questions on it, with the full reasoning behind each credited answer.

Practise this topic freeThe whole unit
29 questionsStatistical Tests and Data Analysis83% with a figure
Worked examples

Three real population genetics questions

From the practice pool, not the mock papers — each with the reasoning that produces the answer.

83% of them come with a figure. Reading the graph or diagram correctly is most of the work here before any content knowledge applies.

Concept Explanation

Which condition is required for a population to remain in Hardy-Weinberg equilibrium?

  1. AMutation rates must increase every generation.
  2. BMating must occur only among close relatives.
  3. CNo natural selection acts on the trait being analyzed.correct
  4. DIndividuals with one genotype must always have higher fitness.
Why C is correct

Hardy-Weinberg equilibrium assumes no evolution by selection, mutation, migration, genetic drift, or nonrandom mating for the gene being analyzed.

The stem asks students to explain a specific mechanism, not just recall a unit term: "Which condition is required for a population to remain in Hardy-Weinberg equilibrium". The correct choice is supported by Hardy-Weinberg equilibrium assumes no evolution by selection, mutation, migration, genetic drift, or nonrandom mating for the gene being analyzed. The tempting alternatives such as Mutation rates must increase every generation or Mating must occur only among close relatives fail because they do not produce the same outcome described in the stem.

Representing and Describing Data · with figure

The genotype counts for a population are shown. What is the frequency of allele A? Use Figure 1 to answer the question.

  1. A48/100 = 0.48
  2. B[2(36)+48]/200 = 0.60correct
  3. C36/100 = 0.36
  4. D(36+48)/100 = 0.84
Why B is correct

There are 100 individuals and 200 total alleles. A alleles = 2(36) + 48 = 120. The frequency of A is 120/200 = 0.60.

The stem asks for a numerical or quantitative judgment: "The genotype counts for a population are shown. What is the frequency of allele A". The safe route is to identify the counted event, use the full denominator, and then interpret the number biologically. Here the worked reasoning is A alleles = 2(36) + 48 = 120, so choices based on 48/100 = 0.48 or 36/100 = 0.36 lose the required setup.

Statistical Tests and Data Analysis

In a population in Hardy-Weinberg equilibrium, 16% of individuals show a recessive phenotype. What proportion of the population is expected to be heterozygous?

  1. Apq = 0.24
  2. B2pq = 0.48correct
  3. Cq^2 = 0.16
  4. Dp^2 = 0.36
Why B is correct

The recessive phenotype frequency is q^2 = 0.16, so q = 0.4 and p = 0.6. The heterozygote frequency is 2pq = 2(0.6)(0.4) = 0.48.

The stem asks for a numerical or quantitative judgment: "In a population in Hardy-Weinberg equilibrium, 16% of individuals show a recessive phenotype. What proportion of...". The safe route is to identify the counted event, use the full denominator, and then interpret the number biologically. Here the worked reasoning is The recessive phenotype frequency is q^2 = 0.16, so q = 0.4 and p = 0.6, so choices based on pq = 0.24 or q^2 = 0.16 lose the required setup.

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Last reviewed 2026-08-28. Topic and unit names follow the College Board course framework. Question counts describe the PrepScore practice bank, not the exam.
Same unit
Natural Selection52 questionsPhylogeny42 questionsSpeciation26 questionsHardy-Weinberg Equilibrium23 questions

Work population genetics until the reasoning is automatic.

Real AP questions with a full explanation on every answer, and a mistake bank that only clears when you get it right.

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