Hardy-Weinberg Equilibrium - AP Biology

1. In a large population of snails, allele frequencies are $p=0.8$ and $q=0.2$. Individuals preferentially mate with the same phenotype. No selection, mutation, or migration occurs. Which change is most likely after several generations?
   Answer: Genotype frequencies will deviate from $p^2:2pq:q^2$, with fewer heterozygotes than expected. This question tests Hardy-Weinberg equilibrium analysis under assortative mating conditions. When individuals preferentially mate with similar phenotypes, this violates the random mating assumption of Hardy-Weinberg equilibrium. Assortative mating doesn't change allele frequencies but redistributes genotypes, increasing homozygotes (AA and aa) and decreasing heterozygotes (Aa) below the expected 2pq. With p = 0.8 and q = 0.2, the expected heterozygote frequency is 2pq = 0.32, but assortative mating will reduce this. Choice D incorrectly claims heterozygotes will increase, misunderstanding that like-with-like mating produces more homozygous offspring. To predict effects of nonrandom mating, remember that positive assortative mating increases homozygosity without changing allele frequencies.
2. A bird population has a beak-shape locus with alleles A and a. In a given year, allele frequencies are $p(A)=0.30$ and $q(a)=0.70$. 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?
   Answer: The frequency of allele a will decrease because aa individuals contribute fewer alleles to the next generation. 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.
3. In a population of flowering plants, petal pigment is influenced by alleles R and r at one locus. Researchers estimate allele frequencies as $p(R)=0.80$ and $q(r)=0.20$. The population is large, and no differences in survival or fecundity among genotypes are detected. However, pollen-transfer observations show that pollinators move primarily between nearby plants, and plants occur in clustered patches separated by open ground. Genotype frequencies in the combined population show a deficit of heterozygotes compared with $2pq$. Which Hardy-Weinberg condition is most directly violated?
   Answer: Random mating, because limited pollen movement can increase mating among nearby relatives. 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.
4. In a large population of beetles, a locus with alleles C and c is studied. At the start of the season, allele frequencies are $p(C)=0.70$ and $q(c)=0.30$. Adults mate randomly, and no migration is detected. During gamete formation, a rare mutation converts C alleles to c alleles at a measurable rate, while the reverse mutation is negligible. No differences in survival or reproduction among genotypes are detected. Researchers predict allele frequencies among zygotes in the next generation. Which prediction is most consistent with these conditions?
   Answer: The frequency of allele c will increase because mutation introduces additional c alleles each generation. This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. The frequency of allele c will increase because mutation converts C to c alleles during gamete formation, raising q from 0.30 while decreasing p from 0.70 in the next generation's zygotes. This directional mutation violates the no-mutation assumption, altering allele frequencies despite random mating and no selection. Negligible reverse mutation ensures a net gain for c. A tempting distractor is choice C, which predicts constant frequencies because random mating prevents mutation effects, based on the misconception that mating alone stabilizes alleles, but mutation directly changes them. To predict under violations, model the specific force's impact on allele frequencies while holding other assumptions constant.
5. In a large population of frogs, allele frequencies at a locus are $p(F)=0.80$ and $q(f)=0.20$. The population is isolated and mates randomly. A parasite reduces reproductive success of $ff$ individuals relative to $FF$ and $Ff$, while no mutation is detected. Which outcome is most consistent with this violation of Hardy-Weinberg conditions over multiple generations?
   Answer: The frequency of allele $f$ is expected to decrease due to selection against $ff$. This question assesses the skill of predicting outcomes from violations of Hardy-Weinberg equilibrium. With selection against ff homozygotes reducing their reproductive success, the frequency of allele f (q=0.20) is expected to decrease over generations. Since ff has lower fitness, f alleles are removed more often, shifting frequencies despite random mating and isolation. This directional selection violates HWE, leading to declining q. A tempting distractor is D, suggesting f increases because ff is recessive, a misconception ignoring that selection acts on phenotypes regardless of dominance. When selection is present, model allele frequency changes based on relative fitness of genotypes.
6. In a population of mice, coat pattern is determined by alleles $E$ and $e$. A sample shows genotype frequencies: $EE=0.25$, $Ee=0.50$, and $ee=0.25$. The population is large, mates randomly, and is isolated from other populations. However, a lab assay detects that new $e$ alleles arise from $E$ alleles at a measurable rate each generation. Which Hardy-Weinberg condition is violated, best explaining why equilibrium will not be maintained over time?
   Answer: No mutation 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.
7. In a large population of beetles, color is controlled by two alleles: $B$ and $b$. A random sample of 1,000 beetles shows genotype counts: 360 $BB$, 480 $Bb$, and 160 $bb$. Assume mating is random and there is no migration. Which conclusion is best supported about Hardy-Weinberg equilibrium in this population?
   Answer: The population is in Hardy-Weinberg equilibrium because observed genotype frequencies match $p^2:2pq:q^2$. This question tests Hardy-Weinberg equilibrium analysis by comparing observed genotype frequencies to expected frequencies. From the data, we calculate allele frequencies: p(B) = (2×360 + 480)/(2×1000) = 0.6 and q(b) = (2×160 + 480)/(2×1000) = 0.4. Expected genotype frequencies under Hardy-Weinberg are: BB = p² = 0.36, Bb = 2pq = 0.48, bb = q² = 0.16. Observed frequencies are: BB = 360/1000 = 0.36, Bb = 480/1000 = 0.48, bb = 160/1000 = 0.16, which exactly match the expected values. Choice E incorrectly confuses q² (frequency of bb homozygotes) with the frequency of heterozygotes Bb. When checking Hardy-Weinberg equilibrium, always calculate expected frequencies using p², 2pq, and q², then compare to observed frequencies.
8. In a coastal fish population, a locus with alleles M and m is monitored. At time 1, allele frequencies are $p(M)=0.55$ and $q(m)=0.45$. Over the next year, a nearby river floods and connects the lagoon to an upstream population in which $p(M)=0.20$ and $q(m)=0.80$. After the connection, researchers measure the lagoon population and find $p(M)=0.42$ and $q(m)=0.58$. No evidence indicates differential survival among genotypes in the lagoon during this period. Which Hardy-Weinberg condition is most clearly violated in the lagoon population?
   Answer: No gene flow, because movement of individuals between populations changed allele frequencies. This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. No gene flow is most clearly violated because the flood introduced individuals from an upstream population with different allele frequencies, shifting the lagoon's p(M) from 0.55 to 0.42 and q(m) from 0.45 to 0.58. This immigration mixes alleles, directly altering frequencies without evidence of selection or other forces. The connection via flooding confirms gene flow as the mechanism disrupting equilibrium. A tempting distractor is choice D, which claims no selection is violated because allele-frequency changes always indicate selection, based on the misconception that any shift implies fitness differences, but gene flow can cause changes independently. When allele frequencies change abruptly, investigate migration or gene flow before assuming internal forces like selection.
9. A population of freshwater snails has a shell-band gene with alleles S and s. In a sample of 500 snails, the genotype counts are 245 SS, 210 Ss, and 45 ss. From these counts, allele frequencies are $p(S)=0.70$ and $q(s)=0.30$. If the population were in Hardy-Weinberg equilibrium, expected genotype frequencies would be $p^2=0.49$, $2pq=0.42$, and $q^2=0.09$. The observed genotype frequencies are 0.49 SS, 0.42 Ss, and 0.09 ss. No migration into or out of the pond is detected during the sampling interval. Which conclusion is best supported by the data?
   Answer: The population is in Hardy-Weinberg equilibrium because observed and expected genotype frequencies match closely. 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.
10. In a large population of insects, wing color is controlled by alleles G and g. Researchers estimate allele frequencies as $p(G)=0.10$ and $q(g)=0.90$. Assuming Hardy-Weinberg conditions are met, expected genotype frequencies are $p^2$ for GG, $2pq$ for Gg, and $q^2$ for gg. The population is well mixed, and no migration is detected. Mutation is negligible over the time scale considered. Under Hardy-Weinberg equilibrium, what is the expected frequency of heterozygotes (Gg)?
    Answer: 0.18 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.
11. For a recessive trait in HWE with $q=0.1$, chance two phenotypically normal parents have an affected child:
    Answer: ~0.0083 For a recessive trait, affected is $q^2$ = 0.01. A phenotypically normal person is a carrier with probability 2pq/($1-q^2$) = 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.
12. A HWE population with $p=q=0.5$ starts complete selfing. Next generation:
    Answer: 25% Aa; $p$ remains 0.5 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.
13. Gene flow breaks HWE. 100 members with $p=0.6$ merge with 400 with $p=0.4$. New $p$:
    Answer: 0.44 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.
14. A population of 1,200 frogs has allele frequencies $p=0.6$ and $q=0.4$. Observed genotype frequencies are $AA=0.36$, $Aa=0.48$, and $aa=0.16$. No migration or mutation is detected. Which conclusion is best supported?
    Answer: The population is in equilibrium because observed genotype frequencies match Hardy-Weinberg expectations. This question tests Hardy-Weinberg equilibrium analysis by comparing observed and expected genotype frequencies. With p = 0.6 and q = 0.4, expected frequencies are: AA = p² = 0.36, Aa = 2pq = 0.48, aa = q² = 0.16. The observed frequencies (AA = 0.36, Aa = 0.48, aa = 0.16) exactly match these expectations, confirming the population is in Hardy-Weinberg equilibrium. Choice A incorrectly claims unequal p and q values prevent equilibrium, but Hardy-Weinberg makes no requirement for equal allele frequencies. To verify Hardy-Weinberg equilibrium, calculate expected genotype frequencies from allele frequencies and compare to observed values—exact or very close matches indicate equilibrium.
15. In a small island population of birds, a locus has alleles $D$ and $d$. In generation 1, allele frequencies are $p(D)=0.50$ and $q(d)=0.50$. The birds mate randomly, and no migration, mutation, or selection is detected. A hurricane then reduces the population to 12 surviving individuals, and in generation 2 the allele frequencies are $p(D)=0.67$ and $q(d)=0.33$. Which conclusion is best supported about the cause of the allele-frequency change?
    Answer: Genetic drift changed allele frequencies due to a population bottleneck. 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.
16. A bird population has allele B at frequency $p=0.20$ and allele b at frequency $q=0.80$. The population meets Hardy-Weinberg conditions. Which value is expected for the frequency of heterozygotes (Bb)?
    Answer: $0.32$ 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.
17. In a population of 2,000 insects, genotype counts at a locus are 1,280 CC, 640 Cc, and 80 cc. Conditions for Hardy-Weinberg equilibrium are met. Which allele frequency is expected for c in this population?
    Answer: $q=0.20$ This question requires calculating allele frequency from genotype counts assuming Hardy-Weinberg equilibrium. From the counts (1280 CC, 640 Cc, 80 cc), we calculate: frequency of c allele = (640 + 2×80)/(2×2000) = 800/4000 = 0.20. This makes q = 0.20 (option A) correct. We can verify: with q = 0.20, expected cc frequency = q² = 0.04, which matches observed (80/2000 = 0.04). Option E (0.04) incorrectly uses the genotype frequency of cc as the allele frequency, confusing q² with q. Always count alleles, not genotypes: each heterozygote contributes one copy of each allele, while homozygotes contribute two copies of the same allele.
18. A population of fish has alleles $B$ and $b$ at a locus. In year 1, a random sample shows allele frequencies $p(B)=0.60$ and $q(b)=0.40$. The population is large, mating is random, and no selection is detected. In year 2, after a storm connects the lake to a nearby river, the allele frequencies change to $p(B)=0.52$ and $q(b)=0.48$. No other conditions are known to change. Which evolutionary force is best supported as causing the allele-frequency change?
    Answer: Gene flow from the river population into the lake population 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.
19. A large rabbit population has allele frequencies $p(F)=0.5$ and $q(f)=0.5$ for a gene that affects how efficiently the kidneys conserve water. During a severe drought, researchers found that $FF$ and $Ff$ rabbits survived and produced offspring at similar rates, and both groups survived and produced more offspring than $ff$ rabbits. The population remained large, mating remained random, no rabbits entered or left the population, and no new mutations at this gene were detected. In the next generation, $p=0.6$. Which evolutionary mechanism is most directly supported by these data?
    Answer: Natural selection Choice B is correct. The data show heritable variation at a gene, differential survival and reproduction among genotypes during the drought ($FF$ and $Ff$ outperformed $ff$), and a change in allele frequency from $p=0.5$ to $p=0.6$ in the next generation. Differential reproductive success of genotypes that leads to a change in allele frequency is natural selection. The size of the change is consistent with selection: starting from Hardy–Weinberg genotype frequencies (0.25, 0.50, 0.25), if $ff$ rabbits leave about one-third as many offspring as the other genotypes, $p$ becomes $0.50/0.833 \approx 0.6$. Choice A is incorrect because the population remained large, so random sampling effects are expected to be small, and the change matches the measured fitness differences rather than chance. Choice C is incorrect because mating remained random. Choice D is incorrect because no new mutations at this gene were detected, and mutation alone changes allele frequencies far too slowly to shift $p$ by 0.1 in one generation. Choice E is incorrect because balanced polymorphism maintains both alleles, for example through a heterozygote advantage; here $FF$ and $Ff$ performed similarly and $ff$ was selected against, which is directional selection favoring $F$. Strategy: rule out the other Hardy–Weinberg violations using the conditions given, then check whether measured differences in survival and reproduction account for the allele-frequency change.
20. A population of rabbits has a fur-density locus with alleles F and f. In generation 0, allele frequencies are $p(F)=0.50$ and $q(f)=0.50$. The population is large, with no migration and no detectable selection on fur density. However, field observations show that individuals with similar fur density are more likely to mate with each other than expected by chance. In generation 1, genotype frequencies show fewer heterozygotes than $2pq$ and more homozygotes than $p^2$ and $q^2$, while allele frequencies remain approximately 0.50 and 0.50. Which outcome is most consistent with these observations?
    Answer: Nonrandom mating can change genotype frequencies without necessarily changing allele frequencies in one generation. This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. Nonrandom mating can change genotype frequencies without necessarily changing allele frequencies in one generation, as assortative mating reduces heterozygotes below 2pq = 0.50 while keeping p(F) ≈ 0.50 and q(f) ≈ 0.50, increasing homozygotes above p² and q². This violation of random mating alters genotype distributions but preserves allele ratios due to no selection or migration. The large population size minimizes drift, highlighting mating patterns as the cause. A tempting distractor is choice A, which states allele frequencies must change when heterozygote frequency decreases, based on the misconception that genotype shifts always affect alleles immediately, but nonrandom mating impacts genotypes first. To distinguish effects, compare allele frequencies across generations and check for mating biases when genotypes deviate without allele changes.

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