PTC Survey and a Premature Hardy–Weinberg Claim
In a 120-student ninth-grade survey, 78 students report tasting PTC paper and 42 report no taste. A student calculates q² = 42/120 = 0.35, q = 0.59, p = 0.41, then writes that the class is not in Hardy–Weinberg equilibrium because p is not 0.50 and that selection must be favoring the nontasting allele. The survey was run after lunch; students with colds were told to guess; 19 students are siblings from eight families; and two students said they tasted only the paper, not the strip. You have one 50-minute period, a chi-square table, and no access to parent genotypes. The student wants to present the claim at a school science fair. The class has not yet studied assumptions of Hardy–Weinberg.
1. Evaluate whether the data can support the student's claim. Address the Hardy–Weinberg assumptions and the survey design.
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The claim is not supported. Hardy–Weinberg equilibrium describes an idealized population with no selection, no mutation, no migration, random mating, and infinite size. A single class is not a population, and 19 siblings from eight families mean the sampled individuals are not independent. Students with colds who guessed may be misclassified, and the two students who tasted only the paper create ambiguity about the phenotype. More fundamentally, the calculation is circular: the student used the nontaster frequency to estimate q and then declared disequilibrium because p was not 0.50. There is no independent expectation that p should equal 0.50, and a difference from 0.50 is not evidence of selection. Selection would require comparing allele or phenotype frequencies across generations or between populations with known differential survival or reproduction. At most, the survey estimates the nontaster phenotype frequency in this group, and even that estimate is weakened by nonindependent sampling and self-report error.
2. Calculate expected phenotype counts under Hardy–Weinberg using p = 0.41 and q = 0.59, and explain what a chi-square test would and would not show with these data.
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With p = 0.41 and q = 0.59, the expected genotype frequencies are TT = p² = 0.1681, Tt = 2pq = 0.4838, and tt = q² = 0.3481. For 120 students, expected counts are about 20.2 TT, 58.1 Tt, and 41.8 tt. Because tasters include TT and Tt, the expected taster count is 20.2 + 58.1 = 78.3, and expected nontasters are 41.8. The observed counts are 78 tasters and 42 nontasters, so a phenotype-level chi-square is essentially zero and would not reject Hardy–Weinberg. But this test is largely circular because q² was estimated from the same nontaster count. It also tests only the phenotype fit, not whether selection, nonrandom mating, or migration is occurring. With only two phenotype classes and one degree of freedom, the test cannot distinguish the underlying genotype frequencies, and the violated sampling assumptions make the p-value unreliable. The appropriate conclusion is that the data are consistent with the estimate used to generate them, not that the class is evolving.
3. Propose a stronger study design that could actually test for selection at the PTC locus.
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Sample unrelated individuals from a defined population, record phenotypes with a standardized blind protocol, and avoid testing immediately after meals or during illness. Include a control strip and repeat ambiguous cases. Estimate allele frequencies from phenotype data only if the genetic model is justified, and report confidence intervals. To test selection, compare allele or phenotype frequencies across generations, between populations with different diets or environments, or before and after a documented selective event. If the goal is only to describe the school, use a random sample and report the sampling method. A science-fair claim would be stronger as an estimate of phenotype frequency with limitations, not as a claim that selection favors the nontasting allele.