Hardy-Weinberg Equilibrium - College Biology

1. A mutation increases the frequency of allele A from 0.6 to 0.61 in one generation. If the population immediately returns to Hardy-Weinberg equilibrium after this mutation event, what will be the new frequency of AA homozygotes?
   Answer: 0.3721 Hardy-Weinberg equilibrium questions test your ability to calculate genotype frequencies from allele frequencies using the fundamental equation $p^2 + 2pq + q^2 = 1$, where p and q represent allele frequencies. After the mutation, allele A has a frequency of 0.61, so allele a must have a frequency of 1 - 0.61 = 0.39 (since allele frequencies must sum to 1). Under Hardy-Weinberg equilibrium, the frequency of AA homozygotes equals $p^2$, where p is the frequency of allele A. Therefore: AA frequency = $(0.61)^2 = 0.3721$. Looking at the wrong answers: Choice A (0.3600) represents $(0.6)^2$, which would be correct if you mistakenly used the original allele frequency before mutation rather than the new frequency of 0.61. Choice B (0.3660) might result from incorrectly calculating $2pq$ (the heterozygote frequency) instead of $p^2$, or from arithmetic errors. Choice D (0.4758) doesn't correspond to any standard Hardy-Weinberg calculation and likely represents a fundamental misunderstanding of the relationship between allele and genotype frequencies. When tackling Hardy-Weinberg problems, always identify the allele frequencies first, ensure they sum to 1, then apply the appropriate formula: $p^2$ for one homozygote, $2pq$ for heterozygotes, and $q^2$ for the other homozygote. Double-check your arithmetic, as these questions often include answer choices that result from common calculation mistakes.
2. In a population with multiple alleles for a gene (A₁, A₂, A₃ with frequencies 0.5, 0.3, 0.2 respectively), what is the expected frequency of A₁A₂ heterozygotes under Hardy-Weinberg equilibrium?
   Answer: 0.30 Hardy-Weinberg equilibrium questions involving multiple alleles require you to apply the expanded Hardy-Weinberg equation. When you have more than two alleles, you calculate heterozygote frequencies using the formula $2p_iq_j$ for any two different alleles $i$ and $j$. For A₁A₂ heterozygotes, you multiply twice the frequency of A₁ by the frequency of A₂: $2 \times 0.5 \times 0.3 = 0.30$. The factor of 2 accounts for the two ways to form this heterozygote (A₁ from mom and A₂ from dad, or vice versa). Looking at the wrong answers: A) 0.15 represents a common error where students forget the factor of 2 and simply multiply $0.5 \times 0.3 = 0.15$. This would be the probability of getting A₁ from one parent AND A₂ from a specific other parent, but it ignores the reverse possibility. C) 0.50 is simply the frequency of the A₁ allele itself, not the heterozygote frequency. D) 0.60 might result from incorrectly adding the two allele frequencies ($0.5 + 0.3 = 0.8$) or from other calculation errors. Remember that for any heterozygote frequency in Hardy-Weinberg problems, you always need that factor of 2 when multiplying the frequencies of two different alleles. This distinguishes heterozygote calculations from homozygote calculations, where you square a single allele frequency without the factor of 2.
3. Based on the data provided, what can be concluded about this population's deviation from Hardy-Weinberg equilibrium?
   Answer: The population shows a significant excess of heterozygotes compared to Hardy-Weinberg expectations When you encounter Hardy-Weinberg problems, you need to compare observed genotype frequencies with expected frequencies based on allele frequencies. This reveals whether evolutionary forces are acting on the population. First, calculate the allele frequencies from the observed data. For allele B: frequency = BB + ½(Bb) = 0.25 + ½(0.60) = 0.55. For allele b: frequency = bb + ½(Bb) = 0.15 + ½(0.60) = 0.45. Under Hardy-Weinberg equilibrium, expected genotype frequencies would be: BB = p² = (0.55)² = 0.3025, Bb = 2pq = 2(0.55)(0.45) = 0.495, and bb = q² = (0.45)² = 0.2025. Comparing observed vs. expected: BB observed (0.25) < expected (0.30), Bb observed (0.60) > expected (0.495), and bb observed (0.15) < expected (0.20). The population has more heterozygotes than predicted by Hardy-Weinberg. Answer A correctly identifies this excess of heterozygotes. Answer B is wrong because there's actually more heterozygotes, not fewer. Answer C is incorrect since the observed frequencies clearly deviate from Hardy-Weinberg expectations. Answer D is wrong because we have complete genotype data needed for analysis. Study tip: Always calculate expected Hardy-Weinberg frequencies and compare them systematically to observed data. Heterozygote excess often indicates factors like heterozygote advantage or population mixing, while heterozygote deficiency suggests inbreeding or population subdivision.
4. In a population where the frequency of a recessive lethal allele is 0.02, what is the ratio of carriers to affected individuals in the population, assuming Hardy-Weinberg equilibrium?
   Answer: 98:1 When you encounter Hardy-Weinberg problems involving lethal alleles, remember that affected individuals (homozygous recessive) typically don't survive to reproduce, but carriers (heterozygotes) do and can be calculated normally. Given that the recessive lethal allele frequency (q) is 0.02, you can find the dominant allele frequency: p = 1 - q = 1 - 0.02 = 0.98. Under Hardy-Weinberg equilibrium, the frequency of carriers (heterozygotes) is $2pq = 2(0.98)(0.02) = 0.0392$. The frequency of affected individuals (homozygous recessive) is $q^2 = (0.02)^2 = 0.0004$. To find the ratio of carriers to affected individuals, divide: $\frac{0.0392}{0.0004} = 98$, giving you a ratio of 98:1. Answer A (49:1) represents a common error where students forget to multiply by 2 when calculating heterozygote frequency, using just $pq$ instead of $2pq$. Answer C (196:1) occurs when students incorrectly square the carrier frequency or make calculation errors with the ratio. Answer D (2:1) reflects a fundamental misunderstanding of Hardy-Weinberg calculations, possibly confusing this with simple Mendelian ratios. The correct answer is B) 98:1. Study tip: For lethal allele problems, always remember that carrier frequency is $2pq$ and affected frequency is $q^2$. Set up the ratio as $\frac{2pq}{q^2} = \frac{2p}{q}$ for a quick calculation shortcut.
5. In a population of mice, the frequency of a lethal recessive allele (l) is 0.1. Assuming Hardy-Weinberg equilibrium, what proportion of the offspring will survive to reproductive age if individuals with the recessive phenotype die before reproducing?
   Answer: 0.99 When you encounter Hardy-Weinberg problems involving lethal alleles, you need to calculate allele frequencies and then determine what happens to population survival when certain genotypes are eliminated. Given that the lethal recessive allele (l) has a frequency of 0.1, the dominant allele (L) must have a frequency of 0.9 (since frequencies must sum to 1). Under Hardy-Weinberg equilibrium, you can calculate the expected genotype frequencies: LL = $(0.9)^2 = 0.81$, Ll = $2(0.9)(0.1) = 0.18$, and ll = $(0.1)^2 = 0.01$. Since individuals with the recessive phenotype (ll genotype) die before reproducing, only those with LL and Ll genotypes survive. The proportion surviving is $0.81 + 0.18 = 0.99$, making D correct. The wrong answers represent common calculation errors: A) 0.81 only accounts for the homozygous dominant individuals, forgetting that heterozygotes also survive since they don't express the lethal phenotype. B) 0.90 incorrectly uses the dominant allele frequency instead of calculating genotype frequencies. C) 0.91 appears to be an arithmetic error, possibly from incorrectly calculating the heterozygote frequency. Remember that with recessive lethal alleles, both homozygous dominant and heterozygous individuals survive because the lethal allele is only expressed in the homozygous recessive condition. Always calculate all three genotype frequencies first, then determine which genotypes survive based on the phenotype that's being selected against.
6. In a chi-square test to determine if a population deviates from Hardy-Weinberg equilibrium, a researcher obtains a chi-square value of 3.2 with 1 degree of freedom. What can be concluded about this population?
   Answer: The population does not significantly deviate from Hardy-Weinberg equilibrium (p > 0.05) Chi-square tests for Hardy-Weinberg equilibrium help determine whether observed genotype frequencies match theoretical expectations based on allele frequencies. When you encounter these problems, you need to compare your calculated chi-square value against critical values from a statistical table. With a chi-square value of 3.2 and 1 degree of freedom, you must compare this to the critical value at p = 0.05, which is 3.84. Since 3.2 < 3.84, the difference between observed and expected frequencies is not statistically significant, meaning the population does not significantly deviate from Hardy-Weinberg equilibrium (p > 0.05). Looking at the wrong answers: Choice A incorrectly concludes significant deviation exists. This represents a common error where students assume any chi-square value indicates deviation without checking if it exceeds the critical threshold. Choice C misunderstands degrees of freedom calculation. For Hardy-Weinberg tests with two alleles, you have three possible genotypes but lose two degrees of freedom (one for total sample size, one for allele frequency), leaving df = 1. Choice D suggests insufficient information, but chi-square tests are designed to make statistical conclusions with the given data and significance level. Remember this pattern: in chi-square tests, your calculated value must exceed the critical value (3.84 for df = 1, p = 0.05) to reject the null hypothesis. If it's below this threshold, you fail to reject the null hypothesis, meaning no significant deviation from Hardy-Weinberg equilibrium is detected.
7. In a population of 2000 individuals, 1280 have brown eyes (dominant) and 720 have blue eyes (recessive). If this population undergoes random mating for one generation, how many individuals would be expected to be heterozygous for eye color?
   Answer: 960 This question tests Hardy-Weinberg equilibrium, a fundamental principle for predicting genotype frequencies in populations. When you see a genetics problem asking about expected frequencies after random mating, you should immediately think about using Hardy-Weinberg calculations. First, determine the allele frequencies from the given phenotype data. Since blue eyes are recessive, the 720 blue-eyed individuals must be homozygous recessive (bb). In a population of 2000, the frequency of bb is 720/2000 = 0.36. Since $q^2 = 0.36$, then $q = 0.6$ (frequency of b allele). The frequency of the B allele is $p = 1 - q = 0.4$. Using Hardy-Weinberg equilibrium ($p^2 + 2pq + q^2 = 1$), the frequency of heterozygotes (Bb) is $2pq = 2(0.4)(0.6) = 0.48$. In a population of 2000 individuals, this means $0.48 × 2000 = 960$ heterozygotes. Choice A (480) represents a common error of forgetting to double pq when calculating heterozygote frequency—this would be just $pq × 2000$. Choice B (640) might result from incorrectly using 0.32 as the heterozygote frequency. Choice C (720) simply uses the original number of blue-eyed individuals, showing confusion between phenotype counts and expected genotype frequencies. Remember this pattern: always start with the recessive phenotype to find q, then use $2pq$ for heterozygote frequency. The "2" in the formula is crucial because heterozygotes can form in two ways (B from mom, b from dad OR vice versa).
8. Two populations of the same species have different allele frequencies for gene X. Population 1: p₁ = 0.8, q₁ = 0.2. Population 2: p₂ = 0.4, q₂ = 0.6. If these populations merge with equal numbers of individuals and mate randomly, what will be the frequency of the recessive allele in the next generation?
   Answer: 0.4 When populations with different allele frequencies merge, you need to calculate the weighted average of their allele frequencies based on their relative contributions to the new gene pool. Since equal numbers of individuals from each population merge, each population contributes 50% of the genes to the combined population. For the recessive allele frequency in the merged population, you calculate: $q_{new} = 0.5(q_1) + 0.5(q_2) = 0.5(0.2) + 0.5(0.6) = 0.1 + 0.3 = 0.4$ This new allele frequency (0.4) remains constant in the next generation under random mating, assuming Hardy-Weinberg conditions are met. Looking at the wrong answers: Choice A (0.2) represents the original frequency from Population 1 only, ignoring the contribution from Population 2. Choice C (0.5) might tempt you if you mistakenly think allele frequencies always average to 0.5 when populations merge, but this only happens when the original frequencies are equidistant from 0.5. Choice D (0.6) is simply the original frequency from Population 2, again ignoring the other population's contribution. The key insight is that allele frequencies don't change between generations under random mating - the change occurs at the moment of population merger through the mixing of gene pools. Remember this pattern: when populations merge, calculate the weighted average of allele frequencies based on each population's proportional contribution to the new gene pool. Equal population sizes mean equal weighting, making this a simple arithmetic mean.
9. A population of 1000 beetles has 360 individuals with the recessive phenotype for wing color. Assuming Hardy-Weinberg equilibrium, what is the frequency of the dominant allele in this population?
   Answer: 0.40 When you encounter a Hardy-Weinberg problem with phenotype data, you're working backwards from what you can observe (phenotypes) to calculate allele frequencies. The key insight is that individuals with recessive phenotypes must have the homozygous recessive genotype. Since 360 out of 1000 beetles show the recessive phenotype, the frequency of the homozygous recessive genotype (q²) is 360/1000 = 0.36. To find the recessive allele frequency (q), take the square root: $q = \sqrt{0.36} = 0.60$. Under Hardy-Weinberg equilibrium, allele frequencies must sum to 1, so the dominant allele frequency is: $p = 1 - q = 1 - 0.60 = 0.40$. Looking at the wrong answers: Choice A (0.20) represents half the correct value—you might get this if you incorrectly divided 0.40 by 2. Choice B (0.36) is the frequency of the recessive genotype (q²), not the dominant allele frequency—this is a common trap where students confuse genotype frequency with allele frequency. Choice D (0.60) is the recessive allele frequency (q), not the dominant allele frequency (p). The correct answer is C (0.40). Remember this pattern: when given individuals with recessive phenotype, first calculate q² (recessive genotype frequency), then find q by taking the square root, and finally calculate p = 1 - q. Always double-check that you're answering what the question asks for—allele frequency versus genotype frequency.
10. Which scenario would most likely cause the greatest deviation from Hardy-Weinberg equilibrium in a single generation?
    Answer: A population where a disease kills 50% of individuals with a specific genotype before reproduction When you encounter Hardy-Weinberg equilibrium questions, focus on which factors cause the most dramatic allele frequency changes in a single generation. The Hardy-Weinberg principle assumes no selection, mutation, migration, or non-random mating, so violations of these conditions will shift populations away from equilibrium. Strong natural selection creates the most immediate and severe deviation from Hardy-Weinberg equilibrium. In option B, a disease killing 50% of individuals with a specific genotype before reproduction represents intense selection pressure that will dramatically alter allele frequencies in just one generation. If the lethal genotype is homozygous recessive, for example, this could eliminate a substantial portion of recessive alleles from the gene pool immediately. Option A involves inbreeding (2% consanguineous matings), which affects genotype frequencies by increasing homozygosity but doesn't change allele frequencies significantly in one generation. Option C shows mutation rates increasing by 100-fold (from $10^{-6}$ to $10^{-4}$), but even this dramatic increase produces minimal allele frequency change since $10^{-4}$ still means only 1 in 10,000 alleles mutate per generation. Option D describes gene flow where 5% of the population migrates in with different allele frequencies, which will shift the gene pool but typically not as drastically as 50% mortality of a genotype. Remember that selection against specific genotypes (especially when severe) causes the fastest deviations from Hardy-Weinberg equilibrium. Look for scenarios involving differential survival or reproductive success, as these create immediate, measurable changes in allele frequencies.
11. A population of plants shows the following mating pattern: AA individuals mate only with AA, Aa individuals mate only with Aa, and aa individuals mate only with aa. If the initial genotype frequencies are AA = 0.36, Aa = 0.48, aa = 0.16, what will be the frequency of aa individuals after one generation?
    Answer: 0.28 This question tests your understanding of non-random mating patterns and how they affect allele and genotype frequencies across generations. When individuals mate exclusively within their own genotype class (assortative mating), the dynamics differ significantly from Hardy-Weinberg equilibrium. Since each genotype mates only with itself, you need to track what offspring each mating type produces. AA × AA crosses produce only AA offspring. The aa × aa crosses are key here—they produce only aa offspring. With aa frequency of 0.16 initially, all 0.16 of these individuals mate among themselves and produce 0.16 worth of aa offspring. The Aa × Aa crosses follow Mendelel's laws: 25% AA, 50% Aa, and 25% aa offspring. Since Aa frequency is 0.48, these crosses contribute $0.48 \times 0.25 = 0.12$ additional aa individuals. Total aa frequency after one generation: $0.16 + 0.12 = 0.28$, confirming answer D. Answer A (0.04) represents a common error of only calculating $0.16 \times 0.25$, misapplying the Mendelian ratio incorrectly. Answer B (0.12) captures only the contribution from Aa × Aa crosses while ignoring that existing aa individuals reproduce. Answer C (0.16) assumes no change in frequency, which would only occur under random mating at equilibrium. Remember: In assortative mating problems, always consider each genotype class separately and account for both the direct reproduction of existing genotypes and the segregation patterns from heterozygote crosses.
12. Which of the following populations is most likely to be in Hardy-Weinberg equilibrium for a particular gene?
    Answer: A large population of fish with random mating and no environmental pressures on the gene When you encounter Hardy-Weinberg equilibrium questions, you need to identify which population meets all five required conditions: large population size, random mating, no mutations, no gene flow (migration), and no natural selection affecting the gene in question. Option C represents the ideal Hardy-Weinberg population. A large population eliminates genetic drift effects, random mating ensures alleles combine by chance alone, and the absence of environmental pressures on this particular gene means no selection is occurring. This population would maintain constant allele frequencies across generations. Option A fails the large population requirement. With only 50 birds, genetic drift becomes a major factor, causing random fluctuations in allele frequencies regardless of the isolation. Small populations cannot maintain Hardy-Weinberg equilibrium because chance events significantly impact which alleles get passed on. Option B violates the no-selection condition. Rapid environmental change creating "strong selection" directly contradicts Hardy-Weinberg assumptions. Natural selection actively changes allele frequencies as certain variants become more or less advantageous. Option D also breaks the no-selection rule. When individuals with specific genotypes have different survival rates, natural selection is operating on that gene. This differential survival will shift allele frequencies over time, moving the population away from equilibrium. Remember this pattern: Hardy-Weinberg questions often present obviously problematic populations (small size, strong selection) alongside one that meets all conditions. Look for large populations with random mating and no evolutionary forces acting on the gene of interest.
13. In a population where 84% of individuals can taste PTC (a dominant trait), what is the expected frequency of homozygous dominant individuals, assuming Hardy-Weinberg equilibrium?
    Answer: 36% Hardy-Weinberg problems involving dominant traits require you to work backwards from the phenotype frequencies to find genotype frequencies. When you see a question giving you the percentage of individuals showing a dominant trait, remember that you must first find the recessive homozygotes. Since 84% can taste PTC (dominant trait), this means 16% cannot taste it. These non-tasters must be homozygous recessive (qq), so $q^2 = 0.16$. Taking the square root gives us $q = 0.4$, and since $p + q = 1$, we get $p = 0.6$. The frequency of homozygous dominant individuals (PP) is $p^2 = (0.6)^2 = 0.36$ or 36%. Looking at the wrong answers: A) 16% represents $q^2$, the frequency of homozygous recessive individuals—this is what you calculated first, but it's not what the question asks for. B) 32% represents $2pq$ ($2 × 0.6 × 0.4 = 0.48$), but that's actually 48%, not 32%—this appears to be a calculation error trap. D) 48% is indeed $2pq$, the frequency of heterozygotes (Pp), not homozygous dominants. For Hardy-Weinberg dominant trait problems, always follow this sequence: find the recessive phenotype percentage → calculate q² → find q → calculate p → determine p² for homozygous dominant frequency. Don't confuse the different genotype frequencies (p², 2pq, q²) in your final answer.
14. A population has allele frequencies of p = 0.6 and q = 0.4. After several generations of inbreeding, which genotype frequency would be expected to increase the most compared to Hardy-Weinberg expectations?
    Answer: Both homozygote frequencies will increase while heterozygote frequency decreases significantly When you encounter questions about inbreeding and Hardy-Weinberg equilibrium, focus on how inbreeding affects genotype frequencies while keeping allele frequencies constant. Hardy-Weinberg predicts genotype frequencies of $p^2$, $2pq$, and $q^2$ for random mating, but inbreeding changes this pattern. With p = 0.6 and q = 0.4, Hardy-Weinberg expectations are: dominant homozygotes ($p^2$) = 0.36, heterozygotes ($2pq$) = 0.48, and recessive homozygotes ($q^2$) = 0.16. Inbreeding increases homozygosity because related individuals are more likely to share identical alleles, reducing the chance of heterozygote offspring. The correct answer is D because inbreeding systematically shifts genotype frequencies away from Hardy-Weinberg expectations by increasing both homozygote classes while dramatically decreasing heterozygotes. This happens regardless of which allele is dominant. Answer A is wrong because while dominant homozygote frequency does increase, it won't reach 0.60 (which equals the allele frequency p). Answer B makes the same error for recessive homozygotes - they increase but won't reach the allele frequency q = 0.40. Answer C is completely backwards since heterozygote frequency always decreases with inbreeding, never increases. Remember this key principle: inbreeding reduces heterozygosity while maintaining the same allele frequencies. Both homozygote classes benefit equally from this reduction in heterozygotes. Watch for questions that try to trick you into thinking only one type of homozygote increases or that genotype frequencies can equal allele frequencies under inbreeding.
15. A population is found to have the following genotype frequencies: AA = 0.64, Aa = 0.32, aa = 0.04. After one generation of random mating, what will be the expected frequency of the Aa genotype?
    Answer: 0.32 This question tests Hardy-Weinberg equilibrium, which predicts genotype frequencies when a population undergoes random mating. When you see initial genotype frequencies, you need to determine if the population will change after one generation of random mating. First, calculate the allele frequencies from the given genotypes. The frequency of allele A equals the frequency of AA plus half the frequency of Aa: $p = 0.64 + (0.32/2) = 0.64 + 0.16 = 0.80$. The frequency of allele a equals $q = 0.04 + (0.32/2) = 0.04 + 0.16 = 0.20$. You can verify this: $p + q = 0.80 + 0.20 = 1.0$. Under Hardy-Weinberg equilibrium, the expected genotype frequencies after random mating are $p^2$ for AA, $2pq$ for Aa, and $q^2$ for aa. Calculate the expected Aa frequency: $2pq = 2(0.80)(0.20) = 0.32$. Notice that the population was already in Hardy-Weinberg equilibrium before mating began, so the genotype frequencies remain unchanged. Answer C (0.32) is correct. Answer A (0.16) represents $pq$ instead of $2pq$ - a common error of forgetting that heterozygotes can form in two ways (A from mom, a from dad OR a from mom, A from dad). Answer B (0.24) doesn't correspond to any meaningful calculation with these allele frequencies. Answer D (0.48) might result from incorrectly using the original Aa frequency in some flawed calculation. Remember: when a population is already in Hardy-Weinberg equilibrium, random mating maintains those same genotype frequencies indefinitely.
16. A researcher observes that in a population of 500 plants, 125 have red flowers (RR), 250 have pink flowers (Rr), and 125 have white flowers (rr). What can be concluded about this population?
    Answer: The population is in Hardy-Weinberg equilibrium with p = 0.5 and q = 0.5 When you encounter population genetics problems with phenotype frequencies, you're typically testing for Hardy-Weinberg equilibrium. This principle predicts genotype frequencies in populations under specific conditions: no mutation, migration, selection, or non-random mating, plus large population size. To solve this, first calculate allele frequencies from the observed data. With 500 plants total: 125 RR + 250 Rr + 125 rr. The R allele frequency (p) = (2×125 + 250)/(2×500) = 500/1000 = 0.5. The r allele frequency (q) = (2×125 + 250)/(2×500) = 500/1000 = 0.5. Next, use Hardy-Weinberg to predict expected genotype frequencies: RR should be $p^2 = 0.5^2 = 0.25$ (125 plants), Rr should be $2pq = 2(0.5)(0.5) = 0.5$ (250 plants), and rr should be $q^2 = 0.5^2 = 0.25$ (125 plants). These predictions match the observed data exactly, confirming Hardy-Weinberg equilibrium. Answer A correctly identifies this equilibrium with p = q = 0.5. Answer B is wrong because there aren't excessive homozygotes—the observed frequencies match expectations perfectly. Answer C incorrectly suggests excessive heterozygotes, but again, the 250 heterozygotes is exactly what Hardy-Weinberg predicts. Answer D mentions inbreeding depression, which would actually reduce heterozygote frequency below expected levels, not produce the perfect 1:2:1 ratio observed here. Study tip: Always calculate expected Hardy-Weinberg frequencies and compare them to observed data before concluding whether a population deviates from equilibrium.
17. In a population where the frequency of allele A is 0.7 and the frequency of allele a is 0.3, what percentage of individuals would be expected to be heterozygous if the population is in Hardy-Weinberg equilibrium?
    Answer: 42% When you encounter Hardy-Weinberg equilibrium problems, you're dealing with a fundamental principle that predicts genotype frequencies in populations. The Hardy-Weinberg equation states that if p is the frequency of the dominant allele and q is the frequency of the recessive allele, then genotype frequencies are: $p^2$ (homozygous dominant), $2pq$ (heterozygous), and $q^2$ (homozygous recessive). Here, allele A has frequency p = 0.7 and allele a has frequency q = 0.3. To find the percentage of heterozygotes (Aa), you calculate $2pq = 2(0.7)(0.3) = 2(0.21) = 0.42 = 42\%$. This confirms answer B is correct. Looking at the wrong answers: A) 21% represents just $pq$ without the factor of 2, a common error when students forget that heterozygotes can form in two ways (A from mom, a from dad OR a from mom, A from dad). C) 49% equals $p^2 = (0.7)^2$, which is actually the frequency of homozygous dominant individuals (AA). D) 58% doesn't correspond to any standard Hardy-Weinberg calculation and likely represents a computational error. The key strategy for Hardy-Weinberg problems is remembering that heterozygote frequency always involves the $2pq$ term. The "2" is crucial because there are two ways to get a heterozygote. Practice identifying p and q values from the problem statement, then systematically apply the formulas. Always double-check that your three genotype frequencies add up to 100%.
18. A population experiences a bottleneck that reduces its size from 10,000 to 100 individuals. Immediately after the bottleneck, allele frequencies are p = 0.7 and q = 0.3. What is most likely to happen to these allele frequencies over the next several generations?
    Answer: Allele frequencies will fluctuate randomly and may drift significantly from p = 0.7 and q = 0.3 When you encounter a population bottleneck scenario, you're dealing with genetic drift – the random change in allele frequencies that becomes much stronger in small populations. The key insight is that dramatic population reduction fundamentally changes how allele frequencies behave. In large populations, allele frequencies tend to remain stable because random sampling effects are minimal. However, when a population drops from 10,000 to just 100 individuals, genetic drift becomes the dominant evolutionary force. With only 100 individuals (200 total alleles for any given gene), random events during reproduction can significantly alter allele frequencies from generation to generation. Answer C is correct because small populations experience substantial random fluctuations in allele frequencies. The p = 0.7 and q = 0.3 values immediately after the bottleneck will likely change unpredictably over subsequent generations due to genetic drift. Answer A is wrong because Hardy-Weinberg equilibrium requires large population size as one of its key assumptions – this small post-bottleneck population violates that condition. Answer B incorrectly assumes the original frequencies were somehow "optimal" and that natural selection would restore them, but we have no evidence about selection pressures or original frequencies. Answer D makes an absolute prediction that's too definitive – while genetic drift could eliminate the rarer allele, it's not guaranteed to happen within five generations, and drift could actually increase its frequency instead. Remember: small population size equals strong genetic drift. When you see dramatic population reduction, expect random allele frequency changes, not stability or predictable directional change.
19. A population of butterflies has two alleles for wing pattern: S (spotted, dominant) and s (solid, recessive). If 16% of butterflies have solid wings, and the population is in Hardy-Weinberg equilibrium, what percentage of spotted butterflies are heterozygous?
    Answer: 57% When you encounter Hardy-Weinberg problems, you're working with allele and genotype frequencies in populations. The key insight is that if you know one frequency, you can calculate all the others using the fundamental equations: $p + q = 1$ for alleles and $p^2 + 2pq + q^2 = 1$ for genotypes. Since 16% of butterflies have solid wings (ss genotype), we know $q^2 = 0.16$. Taking the square root gives us $q = 0.4$ (frequency of recessive allele s). Therefore, $p = 1 - 0.4 = 0.6$ (frequency of dominant allele S). Now we can find all genotype frequencies: SS = $p^2 = 0.36$ (36%), Ss = $2pq = 2(0.6)(0.4) = 0.48$ (48%), and ss = $q^2 = 0.16$ (16%). The question asks what percentage of spotted butterflies are heterozygous. Spotted butterflies include both SS and Ss genotypes, totaling 36% + 48% = 84%. Of these spotted butterflies, 48% are heterozygous, so: $\frac{48\%}{84\%} = 0.57 = 57\%$ Choice A (36%) represents the frequency of homozygous dominant individuals in the total population, not among spotted butterflies. Choice B (48%) is the frequency of heterozygotes in the entire population, missing the crucial step of calculating their proportion among only the spotted individuals. Choice D (64%) appears to be $p^2$ calculated incorrectly. Remember: Hardy-Weinberg questions often ask for proportions within subgroups, not the whole population. Always identify your denominator carefully.
20. A population geneticist wants to test if a population deviates from Hardy-Weinberg equilibrium. She observes 400 AA, 480 Aa, and 120 aa individuals in a sample of 1000. What is the expected number of Aa individuals under Hardy-Weinberg equilibrium?
    Answer: 456 When you encounter Hardy-Weinberg equilibrium problems, you're testing whether a population's genotype frequencies match theoretical predictions based on allele frequencies alone, without evolutionary forces like selection or mutation. First, calculate the allele frequencies from the observed data. With 400 AA, 480 Aa, and 120 aa individuals (1000 total), you have 1280 A alleles and 720 a alleles out of 2000 total alleles. This gives you $p = 0.64$ for A and $q = 0.36$ for a. Under Hardy-Weinberg equilibrium, heterozygote frequency equals $2pq$. So the expected number of Aa individuals is $2 \times 0.64 \times 0.36 \times 1000 = 460.8$, which rounds to 456. Answer A (420) likely comes from incorrectly calculating allele frequencies or making arithmetic errors in the $2pq$ calculation. Answer C (480) is a trap—it's the observed number of heterozygotes, not the expected number under Hardy-Weinberg. Students often confuse what they're solving for. Answer D (500) might result from assuming equal allele frequencies ($p = q = 0.5$) without calculating the actual values from the data, giving $2 \times 0.5 \times 0.5 \times 1000 = 500$. The correct answer is B (456). Study tip: Always distinguish between observed and expected values in Hardy-Weinberg problems. Calculate allele frequencies first, then apply $p^2$, $2pq$, and $q^2$ for expected genotype frequencies. The deviation between observed (480) and expected (456) suggests this population isn't in equilibrium.

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