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AP Biology Help: Hardy Weinberg Equilibrium

Review real example questions for Hardy Weinberg Equilibrium in AP Biology.

Question 1 / 10

0 of 10 answered

Gene flow breaks HWE. 100 members with p=0.6p=0.6 merge with 400 with p=0.4p=0.4. New pp:

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Question 1

Gene flow breaks HWE. 100 members with p=0.6p=0.6 merge with 400 with p=0.4p=0.4. New pp:

  1. 0.50
  2. 0.40
  3. 0.44 (correct answer)
  4. 0.60

Explanation: Combine the allele contributions: 100 members with p=0.6 give 60, and 400 with p=0.4 give 160. Total is 220 out of 500, so new p = 220/500 = 0.44. The tempting wrong answer is 0.50, but that just averages 0.6 and 0.4 equally. The larger group has more influence, so you must weight by group size.

Question 2

A locus shows significantly more heterozygotes than HWE predicts. Safest conclusion:

  1. Selection favors heterozygotes
  2. Allele frequencies thus change
  3. Mating is random at this locus
  4. HWE is rejected at this locus (correct answer)

Explanation: Significantly more heterozygotes than HWE predicts means the genotype frequencies do not match Hardy-Weinberg expectations, so the null hypothesis of HWE is rejected. The tempting wrong answer is that selection favors heterozygotes, since that can cause excess heterozygotes, but it is only one possible explanation and not the safest conclusion. Other forces could also produce the deviation.

Question 3

In 500 organisms, 200 are AA, 200 Aa, 100 aa. HWE-predicted counts:

  1. 200 AA, 200 Aa, 100 aa
  2. 80 AA, 240 Aa, 180 aa
  3. 180 AA, 240 Aa, 80 aa (correct answer)
  4. 320 AA, 160 Aa, 20 aa

Explanation: First find allele frequencies: A = (2 x 200 + 200) / 1000 = 0.6, so a = 0.4. Under HWE, expected AA = 0.36 x 500 = 180, Aa = 2 x 0.6 x 0.4 x 500 = 240, aa = 0.16 x 500 = 80. The tempting wrong answer is 200/200/100, but that is just the observed starting counts, not the HWE-predicted equilibrium based on allele frequencies.

Question 4

A HWE population with p=q=0.5p=q=0.5 starts complete selfing. Next generation:

  1. 25% Aa; pp remains 0.5 (correct answer)
  2. 50% Aa; pp remains 0.5
  3. 12.5% Aa; pp remains 0.5
  4. 25% Aa; pp falls to 0.25

Explanation: In HWE with p=q=0.5, heterozygotes are 2pq = 50% of the population. Under complete selfing, each Aa individual produces 1/4 AA, 1/2 Aa, and 1/4 aa, so next-generation Aa is 50% x 1/2 = 25%. Selfing changes genotype frequencies but not allele frequencies, so p stays 0.5. The tempting 50% Aa answer confuses the initial heterozygote frequency with the offspring after selfing.

Question 5

For a recessive trait in HWE with q=0.1q=0.1, chance two phenotypically normal parents have an affected child:

  1. ~0.032
  2. ~0.0083 (correct answer)
  3. ~0.045
  4. ~0.0025

Explanation: For a recessive trait, affected is q^2 = 0.01. A phenotypically normal person is a carrier with probability 2pq/(1−q21-q^2) = 0.18/0.99 = 0.1818. Both parents must be carriers and then pass on the recessive allele: (0.1818)^2 x 1/4 = 0.0083. The tempting ~0.032 is the chance both parents are carriers, not the chance their child is affected; you still need the 1/4 chance each passes the recessive allele.

Question 6

A bird population has a beak-shape locus with alleles A and a. In a given year, allele frequencies are p(A)=0.30p(A)=0.30 and q(a)=0.70q(a)=0.70. The population is large, and mating is random with respect to beak shape. A severe drought occurs, and individuals with genotype aa have substantially lower survival to reproduction than individuals with AA or Aa. No migration is detected during the drought year. Assume allele frequencies are measured among the breeders that produce the next generation.

Which change in allele frequency is most likely after one generation?

  1. The frequency of allele a will decrease because aa individuals contribute fewer alleles to the next generation. (correct answer)
  2. The frequency of allele a will increase because allele a is initially more common than allele A.
  3. The frequency of allele A will decrease because selection acts only on homozygous dominant genotypes.
  4. Allele frequencies will remain constant because random mating prevents selection from changing pp and qq.
  5. Allele frequencies will oscillate each generation because p+qp+q is less than 1 in drought conditions.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. The frequency of allele a will decrease because aa individuals have lower survival, contributing fewer a alleles to the next generation and shifting q from 0.70 toward a lower value while p increases from 0.30. This selection against the recessive homozygote alters allele frequencies under non-equilibrium conditions, despite random mating and large population size. No migration reinforces that selection is the driving force for the change in breeders' allele frequencies. A tempting distractor is choice B, which predicts allele a will increase because it is initially more common, stemming from the misconception that majority alleles always rise without considering selection's directional effect. To predict allele frequency changes, calculate relative fitness contributions of genotypes and track allele inputs to the next generation.

Question 7

In a small island population of 50 rabbits, allele F has frequency p=0.50p=0.50 and allele f has frequency q=0.50q=0.50. There is no selection, migration, or mutation, and mating is random. After several generations, allele frequencies differ among replicate island populations founded the same way. Which evolutionary force best explains the differences?

  1. Genetic drift due to small population size. (correct answer)
  2. Heterozygote advantage maintaining both alleles.
  3. Gene flow equalizing allele frequencies among populations.
  4. Mutation rapidly converting F alleles into f alleles.
  5. Nonrandom mating changing allele frequencies directly each generation.

Explanation: This question identifies genetic drift as the evolutionary force causing allele frequency differences in small populations. With only 50 rabbits per island and no selection, migration, or mutation, random sampling effects (genetic drift) cause allele frequencies to fluctuate randomly each generation. Different islands experience different random changes, leading to divergent allele frequencies over time despite identical starting conditions. Option E incorrectly claims nonrandom mating changes allele frequencies directly, but nonrandom mating only affects genotype frequencies within a generation, not allele frequencies across generations. Remember: in small populations, genetic drift causes random allele frequency changes that accumulate over generations, while large populations buffer against these random effects.

Question 8

A mammal population has allele D frequency p=0.90p=0.90 and allele d frequency q=0.10q=0.10. The population is in Hardy-Weinberg equilibrium. Which statement best describes the expected frequency of dd individuals?

  1. It equals 2pq=0.182pq=0.18.
  2. It equals q2=0.01q^2=0.01. (correct answer)
  3. It equals p2=0.81p^2=0.81.
  4. It equals pq=0.09pq=0.09.
  5. It equals 1−q2=0.991-q^2=0.99.

Explanation: This question tests calculating homozygous recessive frequency under Hardy-Weinberg equilibrium. With allele d frequency q = 0.10, the frequency of dd individuals equals q² = (0.10)² = 0.01, making option B correct. This represents 1% of the population being homozygous recessive. Option A (0.18) represents 2pq, the heterozygote frequency, not the dd frequency—a common error where students confuse different genotype categories. Always match the genotype to its Hardy-Weinberg formula: DD = p², Dd = 2pq, dd = q².

Question 9

In a population of 800 snails, genotype counts are 320 GG, 160 Gg, and 320 gg. The population is large, and there is no mutation, migration, or selection at this locus. Which conclusion is best supported about Hardy-Weinberg equilibrium in this generation?

  1. The population is in equilibrium because p=q=0.50p=q=0.50.
  2. The population is in equilibrium because homozygotes are equally frequent.
  3. The population is not in equilibrium because observed heterozygotes are fewer than 2pq2pq. (correct answer)
  4. The population is not in equilibrium because allele frequencies cannot be computed from counts.
  5. The population is in equilibrium because heterozygotes are exactly half the population.

Explanation: This question tests detecting Hardy-Weinberg equilibrium violations by comparing observed and expected heterozygote frequencies. From counts (320 GG, 160 Gg, 320 gg), allele frequencies are p(G) = (640 + 160)/1600 = 0.50 and q(g) = 0.50. Under Hardy-Weinberg, expected Gg frequency = 2pq = 2(0.50)(0.50) = 0.50, meaning 400 heterozygotes expected. With only 160 observed (0.20 frequency vs 0.50 expected), there's a significant heterozygote deficit, indicating the population is not in equilibrium (option C). Option E incorrectly focuses on heterozygotes being "half the population" without checking if this matches 2pq expectations. Remember: even with equal allele frequencies, Hardy-Weinberg predicts specific genotype ratios—always calculate and compare to observations.

Question 10

In a population of island lizards, a single locus with alleles T and t affects scale pattern. Before a hurricane, allele frequencies are p(T)=0.50p(T)=0.50 and q(t)=0.50q(t)=0.50. Immediately after the hurricane, only 40 adults remain, and allele frequencies among the survivors are p(T)=0.65p(T)=0.65 and q(t)=0.35q(t)=0.35. The hurricane did not differentially damage habitats based on scale pattern, and no consistent differences in survival among genotypes were observed; the change is attributed to which individuals happened to survive. In the next generation, mating among survivors is random.

Which evolutionary force best explains the allele-frequency change?

  1. Gene flow, because new alleles entered the population after the hurricane.
  2. Natural selection, because the hurricane favored the T allele over the t allele.
  3. Genetic drift, because a chance reduction in population size altered allele frequencies. (correct answer)
  4. Mutation, because the t allele was converted into the T allele during the hurricane.
  5. Nonrandom mating, because survivors must preferentially mate with similar scale patterns.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. Genetic drift best explains the allele-frequency change because the hurricane caused a random reduction in population size to 40 individuals, leading to a chance shift from p(T) = 0.50 to 0.65 and q(t) = 0.50 to 0.35 among survivors. This bottleneck effect alters allele frequencies due to sampling error in small populations, without directional forces like selection or gene flow. The lack of differential survival by genotype and random post-hurricane mating further supports drift as the primary force. A tempting distractor is choice B, which attributes the change to natural selection favoring the T allele, based on the misconception that any environmental event like a hurricane must impose selection, but the question states no genotype-based survival differences. When evaluating evolutionary forces, identify random changes in small populations as genetic drift and rule out selection without evidence of fitness differences.