AP Biology: Unit 7, Natural Selection: Mechanisms & Evidence (Drill 27)

Question 1. Based on the Generation 1 data in the table, which of the following correctly identifies the frequency of the recessive allele (q) in this population?

• A) 0.09
• B) 0.30 ✓
• C) 0.42
• D) 0.70

Explanation: Correct: (B) 0.30. The frequency of the homozygous recessive genotype (aa) in Generation 1 is 0.09. Under Hardy-Weinberg equilibrium, freq(aa) = q2. Therefore q = the square root of 0.09 = 0.30. The recessive allele (a) is present at a frequency of 0.30 in this population. Incorrect: (A) 0.09 is the genotype frequency q2, not the allele frequency q. The square root must be taken to move from genotype frequency to allele frequency. Incorrect: (C) 0.42 is the frequency of the heterozygous genotype Aa, which equals 2pq. This is a genotype frequency, not an allele frequency. Incorrect: (D) 0.70 is p, the dominant allele frequency (p = 1 minus q = 1 minus 0.30 = 0.70). The question asks for q, not p.

Question 2. Using the Generation 1 data, which of the following is closest to the number of recessive alleles (a) present specifically in heterozygous individuals (Aa) in this population?

• A) 84
• B) 120
• C) 168 ✓
• D) 240

Explanation: Correct: (C) 168. Each heterozygous individual (Aa) carries exactly one copy of the recessive allele (a). The table shows 168 Aa individuals in Generation 1. Therefore recessive alleles in Aa individuals = 168 times 1 = 168. This differs from the total recessive alleles in the population, which also includes the two 'a' alleles carried by each of the 36 aa individuals (72 additional), giving 240 total. Incorrect: (A) 84 results from dividing 168 by 2. Each Aa individual carries one 'a' allele, not half an allele. Dividing by 2 has no biological basis here. Incorrect: (B) 120 may arise from multiplying q (0.30) by N (400), which conflates allele frequency with a different count. This does not represent recessive alleles in heterozygotes specifically. Incorrect: (D) 240 is the total number of recessive alleles in the entire population: 168 from Aa individuals plus 72 from aa individuals. The question asks only about recessive alleles in heterozygous individuals.

Question 3. A student calculates the allele frequencies from Generation 3 observed data without assuming equilibrium, finding q = 0.32 and p = 0.68. The expected genotype counts under Hardy-Weinberg for these allele frequencies are: AA = 185, Aa = 174, aa = 41. Which of the following conclusions is best supported by comparing these expected counts to the observed Generation 3 counts in the table?

• A) The population is in Hardy-Weinberg equilibrium in Generation 3 because the dominant allele frequency did not change from Generation 1.
• B) The observed excess of homozygotes and deficit of heterozygotes relative to expectations is consistent with a violation of Hardy-Weinberg assumptions such as nonrandom mating or population subdivision. ✓
• C) The population cannot be evaluated for Hardy-Weinberg equilibrium without first knowing the cause of the allele frequency change.
• D) The Generation 3 observed counts are close enough to the expected counts to conclude the population remains in Hardy-Weinberg equilibrium.

Explanation: Correct: (B) Comparing observed to expected in Generation 3: AA observed (196) exceeds expected (185) by 11; Aa observed (152) falls below expected (174) by 22; aa observed (52) exceeds expected (41) by 11. Both homozygous classes are in excess while heterozygotes are deficient. This pattern, excess homozygotes paired with a heterozygote deficit, is consistent with violations of Hardy-Weinberg assumptions such as nonrandom mating (inbreeding or assortative mating) or population subdivision, either of which reduces heterozygote frequency below expectations. Incorrect: (A) Stability of p alone is insufficient to conclude Hardy-Weinberg equilibrium. Equilibrium requires all five assumptions to hold and is evaluated by comparing observed genotype frequencies to expected frequencies. Incorrect: (C) Hardy-Weinberg equilibrium is evaluated by comparing observed to expected genotype frequencies derived from observed allele frequencies. Knowing the cause of allele frequency change is a separate question that does not prevent the equilibrium test. Incorrect: (D) The deviations are not negligible: the Aa count is 22 below expectation (a deficit of approximately 13%) and the aa count is 11 above expectation (an excess of approximately 27%). These are meaningful deviations supporting the conclusion that the population is not in equilibrium.

Question 4. The aa genotype has a substantial fitness disadvantage: individuals survive to reproductive maturity at a rate 60% lower than individuals with at least one dominant allele. Yet the data show that the frequency of aa increased from Generation 2 to Generation 3, and the recessive allele frequency rose from q = 0.30 to q = 0.32. Which of the following best explains why this observed pattern is inconsistent with natural selection acting against the aa genotype as the primary evolutionary force between Generations 2 and 3?

• A) Natural selection cannot reduce the frequency of a recessive allele when that allele is sheltered in heterozygous carriers, so aa frequency would be unaffected by selection.
• B) The Hardy-Weinberg model predicts that allele frequencies remain stable each generation when selection is operating, so no change is expected under selection.
• C) A population of 400 plants is too small for natural selection to produce detectable changes in allele frequency within a single generation.
• D) Natural selection against aa would be expected to decrease the frequency of the recessive allele over successive generations, not increase it as is observed in Generation 3. ✓

Explanation: Correct: (D) If selection consistently acts against aa individuals, those individuals reproduce at a substantially lower rate and contribute fewer offspring. Over successive generations this directional pressure reduces the frequency of the recessive allele q and therefore the aa genotype frequency. The data show the opposite: q increased from 0.30 to 0.32 and aa frequency rose from 0.09 to 0.13. An increase in a disadvantaged genotype's frequency is inconsistent with selection against it as the primary force, suggesting another mechanism such as genetic drift may be responsible for the Generation 3 shift. Incorrect: (A) This is a common misconception. Natural selection does reduce recessive allele frequencies over time even when the allele is sheltered in heterozygotes; it acts more slowly when the allele is rare, but it still exerts consistent downward pressure on q. The statement overgeneralizes. Incorrect: (B) This reverses Hardy-Weinberg logic. H-W predicts allele frequencies remain stable in the absence of evolutionary forces. Natural selection is precisely one of the forces that causes deviation from H-W equilibrium; it does not stabilize allele frequencies. Incorrect: (C) A population of 400 is large enough that strong selection could produce detectable change over time, though the exact magnitude depends on selection strength, dominance relationships, and population structure. A 60% fitness disadvantage is substantial and would generally be expected to exert downward pressure on q across generations.

Question 5. The research team later discovered that the meadow had been fragmented by a new road between Generations 2 and 3, dividing the original population into two smaller subpopulations that could no longer interbreed. Which of the following best predicts the most likely long-term evolutionary consequence of this fragmentation for allele frequencies within each subpopulation?

• A) The allele frequencies in each subpopulation will diverge from each other over time due to independent genetic drift events in each isolated group. ✓
• B) Natural selection will increase the frequency of the recessive allele in both subpopulations because isolated populations tend to accumulate deleterious alleles.
• C) The recessive allele will be eliminated from both subpopulations within a few generations because the aa genotype has substantially lower fitness.
• D) The allele frequencies in both subpopulations will remain identical over time because both groups began with the same ancestral gene pool.

Explanation: Correct: (A) When a population is fragmented into smaller isolated subpopulations, genetic drift becomes a more powerful evolutionary force because random chance events affect a proportionally larger fraction of a smaller population. Without gene flow between them, each subpopulation will experience independent random fluctuations in allele frequency. Over successive generations these independent drift events will cause the subpopulations to diverge genetically from each other, even if they began with identical allele frequencies. Loss of gene flow is the key condition allowing this divergence to accumulate. Incorrect: (B) Natural selection does not systematically increase deleterious allele frequencies. Isolation reduces gene flow but does not reverse selection direction. Small isolated populations can accumulate slightly deleterious alleles through drift, but this is a drift effect, not selection. Incorrect: (C) Rapid elimination of a recessive allele by selection alone is extremely unlikely because selection cannot efficiently act on recessive alleles sheltered in heterozygous carriers. Even strong selection takes many generations to substantially reduce q, and in small populations drift often overrides selection. Incorrect: (D) A shared ancestral gene pool does not cause isolated populations to maintain identical allele frequencies. Once gene flow ceases, independent drift and potentially different local selection pressures will cause divergence. Shared ancestry is a starting point, not an ongoing stabilizing force.