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AP Biology Unit 7: Natural Selection — Worked Examples

Example 1Hard

Hardy-Weinberg Equilibrium and Detecting Selection

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. A160
  2. B240✓ Correct
  3. C200
  4. D320
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.

Example 2Hard

Interpreting a Cladogram from Molecular Data

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