AP Biology Quiz: Hardy Weinberg Equilibrium

What this quiz covers

This quiz focuses on Hardy Weinberg Equilibrium, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

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 $p$ and $q$.
5. Allele frequencies will oscillate each generation because $p+q$ 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 2

1. 0.09
2. 0.42
3. 0.49
4. 0.51 (correct answer)
5. 0.91

Explanation: This question tests Hardy-Weinberg equilibrium analysis by calculating the frequency of individuals carrying at least one dominant allele. With p(M) = 0.3 and q(m) = 0.7, we need MM + Mm frequencies. Under Hardy-Weinberg: MM = p² = 0.09 and Mm = 2pq = 0.42, so MM + Mm = 0.09 + 0.42 = 0.51. Alternatively, we can calculate as 1 - q² = 1 - 0.49 = 0.51, since q² represents individuals with no M alleles. Choice C (0.49) represents mm homozygotes who lack the M allele entirely. To find carriers of at least one dominant allele, calculate 1 - q² or sum the frequencies of both genotypes containing that allele.

Question 3

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 4

1. It equals $2pq=0.18$.
2. It equals $q^2=0.01$. (correct answer)
3. It equals $p^2=0.81$.
4. It equals $pq=0.09$.
5. It equals $1-q^2=0.99$.

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 5

1. The population is in equilibrium because $p=q=0.50$.
2. The population is in equilibrium because homozygotes are equally frequent.
3. The population is not in equilibrium because observed heterozygotes are fewer than $2pq$. (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 6

1. $p(M)=0.32$
2. $q(m)=0.20$ (correct answer)
3. $p(M)=0.40$
4. $q(m)=0.04$
5. $p(M)=0.64$

Explanation: This question tests calculating allele frequencies from genotype frequencies under Hardy-Weinberg equilibrium. Given genotype frequencies MM: 0.64, Mm: 0.32, mm: 0.04, we can verify these follow Hardy-Weinberg proportions. Since mm = q² = 0.04, we get q(m) = √0.04 = 0.20 (option B). We can verify: p(M) = 1 - 0.20 = 0.80, so MM = p² = 0.64 ✓ and Mm = 2pq = 2(0.80)(0.20) = 0.32 ✓. Option C (0.40) might result from incorrectly taking the square root of MM frequency (√0.64 = 0.80) and subtracting from 1. Always use the homozygous recessive frequency (q²) to find q directly by taking its square root.

Question 7

1. $0.16$
2. $0.04$
3. $0.64$
4. $0.32$ (correct answer)
5. $0.80$

Explanation: This question applies Hardy-Weinberg equilibrium to calculate heterozygote frequency from allele frequencies. Given p(B) = 0.20 and q(b) = 0.80, the frequency of heterozygotes (Bb) under Hardy-Weinberg equilibrium is 2pq = 2(0.20)(0.80) = 0.32. This makes option D correct. Option B (0.04) represents p² (frequency of BB homozygotes), not heterozygotes—a common error where students forget the factor of 2 in the heterozygote formula. Remember the complete Hardy-Weinberg equation: p² + 2pq + q² = 1, where 2pq specifically represents heterozygote frequency.

Question 8

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.

Question 9

1. No mutation, because patchy plant distributions increase mutation rates at the R locus.
2. No gene flow, because clustered patches guarantee alleles move between patches each generation.
3. Random mating, because limited pollen movement can increase mating among nearby relatives. (correct answer)
4. Large population size, because $p=0.80$ indicates a small population.
5. No selection, because heterozygote deficits always indicate selection against heterozygotes.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. Random mating is most directly violated because limited pollen movement between clustered patches increases mating among nearby relatives, leading to a deficit of heterozygotes compared to the expected 2pq = 2(0.80)(0.20) = 0.32. This nonrandom mating disrupts genotype frequency expectations without necessarily changing allele frequencies p = 0.80 and q = 0.20. The patchy distribution promotes inbreeding, violating the random mating assumption despite large population size and no selection. A tempting distractor is choice E, which claims no selection is violated because heterozygote deficits indicate selection against them, based on the misconception that deviations always imply selection, but inbreeding can cause similar patterns without fitness differences. When heterozygote deficits occur, evaluate mating patterns or population structure before assuming selection as a transferable strategy.

Question 10

1. The population is not in equilibrium because the allele frequencies are not $p=0.50$ and $q=0.50$.
2. The population is in Hardy-Weinberg equilibrium because observed and expected genotype frequencies match closely. (correct answer)
3. The population is not in equilibrium because the heterozygote frequency equals $2pq$ only by coincidence.
4. The population is not in equilibrium because the SS genotype must be less common than the Ss genotype.
5. The population is in equilibrium only if mutation is increasing the s allele frequency each generation.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. The population is in Hardy-Weinberg equilibrium because the observed genotype frequencies of 0.49 SS, 0.42 Ss, and 0.09 ss closely match the expected frequencies of p² = (0.70)² = 0.49, 2pq = 2(0.70)(0.30) = 0.42, and q² = (0.30)² = 0.09. These allele frequencies p = 0.70 and q = 0.30 remain consistent with no deviations, and the absence of migration supports equilibrium conditions. The matching frequencies indicate no violations of assumptions like random mating or no selection. A tempting distractor is choice A, which states the population is not in equilibrium because allele frequencies are not p = q = 0.50, arising from the misconception that equilibrium requires equal allele frequencies, but Hardy-Weinberg holds for any p and q as long as conditions are met. To assess equilibrium, compute expected genotype frequencies from observed allele frequencies and compare them to observed counts for discrepancies.

Question 11

1. The population is not in equilibrium because heterozygotes are more common than homozygotes.
2. The population is in Hardy-Weinberg equilibrium because observed genotype frequencies match $p^2:2pq:q^2$. (correct answer)
3. The population is not in equilibrium because allele frequencies must be $p=0.36$ and $q=0.16$.
4. The population is not in equilibrium because genetic drift must be occurring in any wild population.
5. The population is in equilibrium only if $R$ is dominant to $r$.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in a population. The observed genotype frequencies are RR=0.36, Rr=0.48, and rr=0.16, yielding allele frequencies p(R)=0.60 and q(r)=0.40 through the calculation p = freq(RR) + 0.5×freq(Rr). The expected genotype frequencies under HWE are p²=0.36 for RR, 2pq=0.48 for Rr, and q²=0.16 for rr, which precisely match the observed values. Given that the population meets all HWE conditions such as random mating, no selection, no mutation, and isolation, it is in equilibrium. A tempting distractor is C, which incorrectly sets p=0.36 and q=0.16 by confusing genotype frequencies with allele frequencies, a common misconception in frequency calculations. To evaluate Hardy-Weinberg equilibrium, always derive allele frequencies from observed genotypes and compare them to expected p²:2pq:q² ratios.

Question 12

1. Random mating
2. No mutation (correct answer)
3. No migration
4. Large population size
5. No natural selection

Explanation: This question assesses the skill of identifying violations of Hardy-Weinberg equilibrium conditions. The genotype frequencies EE=0.25, Ee=0.50, and ee=0.25 match HWE for p=0.50 and q=0.50, but a lab assay detects mutation from E to e alleles each generation. This violates the no-mutation condition, preventing long-term equilibrium as new e alleles alter frequencies over time. Other conditions like random mating and isolation are met, isolating mutation as the cause. A tempting distractor is A, random mating, but mating is random, a misconception from assuming frequency match implies all conditions are satisfied. To maintain HWE, verify no mutations occur, as even low rates can accumulate and disrupt equilibrium over generations.

Question 13

1. The population deviates from equilibrium because $p=0.64$ and $q=0.04$.
2. The population is in equilibrium because observed genotypes equal $p^2:2pq:q^2$ for $p=0.80$. (correct answer)
3. The population deviates from equilibrium because $HH$ must equal $2pq$.
4. The population is in equilibrium only if $H$ is recessive to $h$.
5. The population deviates from equilibrium because heterozygotes should be the most common genotype.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in a population. From counts HH=640, Hh=320, hh=40 in 1,000 rabbits, frequencies are 0.64, 0.32, 0.04, yielding p(H)=0.80 and q(h)=0.20. Expected HWE frequencies are p²=0.64, 2pq=0.32, and q²=0.04, matching observed values exactly. Conditions like large size, isolation, no selection, and no mutation support equilibrium. A tempting distractor is A, claiming deviation with p=0.64 and q=0.04, a misconception from equating homozygote frequencies directly to allele frequencies. To confirm HWE, compute allele frequencies from all genotypes and verify if they generate the observed ratios.

Question 14

1. Mutation introducing many new $b$ alleles in one generation
2. Gene flow from the river population into the lake population (correct answer)
3. Random mating increasing heterozygote frequency above $2pq$
4. Genetic drift due to a very large effective population size
5. Stabilizing selection maintaining allele frequencies at $p=0.60$

Explanation: This question assesses the skill of analyzing evolutionary forces affecting Hardy-Weinberg equilibrium. Allele frequencies shifted from p(B)=0.60 and q(b)=0.40 to p=0.52 and q=0.48 after the storm connected the lake to the river, indicating an influx of b alleles. Gene flow from the river population, likely with higher q, caused this change, as other conditions like random mating and no selection remained unchanged. No other force, such as mutation or drift, fits the rapid, directional shift post-storm. A tempting distractor is D, genetic drift due to large population size, but large size reduces drift's effect, misconceptions arise from ignoring population size in drift. To identify forces altering allele frequencies, consider environmental changes like habitat connections that enable gene flow.

Question 15

1. Gene flow increased $D$ frequency because survivors immigrated from another island.
2. Mutation increased $D$ frequency because storms increase mutation rates at specific loci.
3. Genetic drift changed allele frequencies due to a population bottleneck. (correct answer)
4. Nonrandom mating changed allele frequencies by increasing homozygote production.
5. Directional selection increased $D$ frequency because hurricanes favor dominant alleles.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium disruptions by evolutionary forces. Allele frequencies changed from p(D)=0.50 and q(d)=0.50 to p=0.67 and q=0.33 after a hurricane reduced the population to 12 individuals, indicating a bottleneck effect. Genetic drift in small populations randomly alters frequencies, fitting the sudden change without migration, mutation, or selection. The small survivor count amplifies drift's impact, causing the observed shift. A tempting distractor is A, gene flow, but no immigration is mentioned, a misconception from assuming disasters always involve migration. To detect genetic drift, look for population size reductions like bottlenecks that amplify random frequency changes.

Question 16

1. $0.49$
2. $0.21$
3. $0.42$ (correct answer)
4. $0.30$
5. $0.09$

Explanation: This question assesses the skill of calculating expected frequencies under Hardy-Weinberg equilibrium. With allele frequencies p(C)=0.70 and q(c)=0.30, the expected heterozygote frequency is 2pq = 2×0.70×0.30 = 0.42. This holds because the population meets HWE conditions like isolation, random mating, no selection, and no mutation, predicting stable frequencies. The calculation directly uses the HWE equation for heterozygotes in the next generation. A tempting distractor is A, 0.49, which is p² instead, a misconception from confusing homozygote and heterozygote terms. For predicting genotype frequencies, apply the HWE formulas p², 2pq, and q² after confirming assumptions are met.

Question 17

1. The population is in Hardy-Weinberg equilibrium because observed genotypes match $p^2:2pq:q^2$. (correct answer)
2. The population is not in equilibrium because allele frequencies must be $p=q=0.50$.
3. The population is not in equilibrium because heterozygotes should be the most common genotype.
4. The population is not in equilibrium because genotype frequencies cannot be calculated from genotype counts.
5. The population is in equilibrium only if the rr genotype is absent from the population.

Explanation: This question tests Hardy-Weinberg equilibrium analysis by checking if observed genotype frequencies match expected frequencies. From the counts (490 RR, 420 Rr, 90 rr), we calculate allele frequencies: p(R) = (2×490 + 420)/(2×1000) = 1400/2000 = 0.70 and q(r) = 0.30. Under Hardy-Weinberg, expected frequencies are p² = 0.49 (RR), 2pq = 0.42 (Rr), and q² = 0.09 (rr), which match the observed frequencies of 0.49, 0.42, and 0.09. Option B incorrectly assumes equilibrium requires equal allele frequencies, confusing a special case with the general principle. Remember: Hardy-Weinberg equilibrium means observed genotype frequencies match p²:2pq:q² predictions, regardless of specific allele frequency values.

Question 18

1. Frequency of Tt is $0.24$.
2. Frequency of TT is $0.36$. (correct answer)
3. Frequency of tt is $0.60$.
4. Frequency of Tt is $0.40$.
5. Frequency of TT is $0.60$.

Explanation: This question requires applying Hardy-Weinberg equilibrium to predict genotype frequencies from given allele frequencies. With p(T) = 0.60 and q(t) = 0.40, we calculate expected genotype frequencies using the Hardy-Weinberg equation. For TT homozygotes, the frequency is p² = (0.60)² = 0.36, making option B correct. For Tt heterozygotes, it's 2pq = 2(0.60)(0.40) = 0.48 (not 0.24 or 0.40), and for tt homozygotes, it's q² = (0.40)² = 0.16 (not 0.60). Option C incorrectly uses the allele frequency (0.60) as a genotype frequency, a common misconception. Always square allele frequencies for homozygotes and use 2pq for heterozygotes.

Question 19

1. 0.01
2. 0.09
3. 0.18 (correct answer)
4. 0.81
5. 0.90

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. Under Hardy-Weinberg equilibrium, the expected frequency of heterozygotes (Gg) is 2pq = 2(0.10)(0.90) = 0.18, derived from allele frequencies p(G) = 0.10 and q(g) = 0.90. This calculation assumes random mating, leading to genotype frequencies of p² = 0.01 GG, 2pq = 0.18 Gg, and q² = 0.81 gg. Conditions like large population, no migration, and negligible mutation support the equilibrium prediction. A tempting distractor is choice B (0.09), which might be confused with q² = 0.81 or miscalculated as pq = 0.09, arising from the misconception of omitting the factor of 2 in heterozygote frequency. Always use the full Hardy-Weinberg equations p² + 2pq + q² = 1 to compute expected frequencies and verify assumptions hold.

Question 20

1. No natural selection
2. Random mating (correct answer)
3. No mutation
4. No migration (gene flow)
5. Very large population size

Explanation: This question requires Hardy-Weinberg equilibrium analysis to identify which condition is violated. First, calculate allele frequencies: p(A) = (2×200 + 50)/(2×500) = 0.45 and q(a) = (2×250 + 50)/(2×500) = 0.55. Expected genotype frequencies are: AA = p² = 0.2025, Aa = 2pq = 0.495, aa = q² = 0.3025. Observed frequencies are: AA = 200/500 = 0.40, Aa = 50/500 = 0.10, aa = 250/500 = 0.50. The heterozygote frequency (0.10) is drastically lower than expected (0.495), indicating nonrandom mating where individuals preferentially mate with similar genotypes. Choice A (no selection) is incorrect because selection would change allele frequencies over generations, not just genotype frequencies. To identify violated Hardy-Weinberg conditions, check if heterozygote frequency deviates significantly from 2pq.