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.

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.

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.

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.

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.

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.

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.

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.

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.

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.