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About This Drill
AP Biology — Unit 7 — Natural Selection: Mechanisms & Evidence — Drill 27 is a practice drill. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice Hardy-Weinberg equilibrium calculations and evolutionary reasoning with this AP Biology drill. You will calculate allele frequencies from genotype data, compare observed versus expected genotype distributions, and evaluate whether population data are consistent with Hardy-Weinberg assumptions. Includes quantitative practice with the equations p + q = 1 and p2 + 2pq + q2 = 1, plus higher-order questions on genetic drift, natural selection, and habitat fragmentation.
Passage
A research team studied the inheritance of a recessive condition in a prairie wildflower, Silphium integrifolium. Individuals homozygous recessive (aa) survive to reproductive maturity at a rate 60% lower than individuals carrying at least one dominant allele (AA or Aa), representing a substantial fitness disadvantage. In a continuous meadow population of 400 adult plants, the team recorded genotype counts and frequencies over three consecutive generations. The team verified that the population satisfied Hardy-Weinberg assumptions at Generation 1: random mating, no migration, no mutation, and a population large enough to minimize drift.
Observed genotype data:
| Generation 1 — N=400: AA count=196, Aa count=168, aa count=36 | Freq AA=0.49, Freq Aa=0.42, Freq aa=0.09 |
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| Generation 2 — N=400: AA count=194, Aa count=170, aa count=36 | Freq AA=0.49, Freq Aa=0.43, Freq aa=0.09 |
| Generation 3 — N=400: AA count=196, Aa count=152, aa count=52 | Freq AA=0.49, Freq Aa=0.38, Freq aa=0.13 |
Hardy-Weinberg Reference: p + q = 1 and p2 + 2pq + q2 = 1
where p = freq(A), q = freq(a), p2 = freq(AA), 2pq = freq(Aa), q2 = freq(aa)
Allele frequency from observed data (no equilibrium assumption required): q = freq(aa) + 0.5 x freq(Aa)
Questions in This Drill
- 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?
- 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 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?
- 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?
- 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?