AcornPrep

AP Biology Unit 7: Natural Selection — Worked Examples

Hardy-Weinberg Equilibrium and Detecting Selection

Hard

In a population of 500 wildflowers, flower color is determined by a single gene with two alleles: R (red, dominant) and r (white, recessive). A botanist counts 80 white-flowered plants. If the population is in Hardy-Weinberg equilibrium, what is the expected number of heterozygous (Rr) plants?

  1. 160
  2. 240 ✓ Correct
  3. 200
  4. 320
Solution

Start with the recessive phenotype: q2=80/500=0.16q^2 = 80/500 = 0.16, so q=0.4q = 0.4. Since p+q=1p + q = 1, p=0.6p = 0.6. The heterozygous frequency is 2pq=2(0.6)(0.4)=0.482pq = 2(0.6)(0.4) = 0.48. Expected heterozygous plants: 0.48times500=∗∗240∗∗0.48 \\times 500 = **240**. Choice A (160) results from using q2times2times500q^2 \\times 2 \\times 500 instead of 2pqtimes5002pq \\times 500. Choice C (200) uses 2q2times5002q^2 \\times 500 or another common arithmetic error. Choice D (320) likely uses p2times500p^2 \\times 500 or incorrectly counts all red-flowered plants as heterozygous.

Interpreting a Cladogram from Molecular Data

Hard

Scientists sequenced a 20-nucleotide region of a homologous gene in four species. The aligned sequences and a proposed cladogram are shown below. Based on the data, which pair of species shares the most recent common ancestor?

  1. W and X, because they differ by only 1 nucleotide
  2. X and Y, because they share a derived substitution at position 12 not found in W ✓ Correct
  3. Y and Z, because they have the most nucleotide differences from the outgroup
  4. W and Z, because they represent the most divergent lineages
Solution

To determine the most recent common ancestor, we look for shared derived characters (synapomorphies), not overall similarity. Comparing the sequences: W differs from X at position 16 (G->A). X differs from Y at position 12 (G->A). Y differs from Z at position 4 (G->A). Species X and Y share a derived substitution at position 12 (A instead of G) that is not found in W, indicating they share a more recent common ancestor than either does with W. Choice A uses overall similarity (fewest differences) rather than shared derived characters — W and X are similar but their similarity may be ancestral. Choice C confuses distance from outgroup with relatedness. Choice D describes the most distant pair, not the closest.