A geneticist crosses tall pea plants with short pea plants. All the F₁ plants are tall. The F₁ plants are then allowed to self-fertilize, and the F₂ plants are classified by height: 62 tall and 26 short. From these results, the geneticist concludes that shortness in peas is due to a recessive allele (s) and that tallness is due to a dominant allele (S). On this hypothesis, 2/3 of the tall F₂ plants should be heterozygous Ss. To test this prediction, the geneticist uses pollen from each of the 62 tall plants to fertilize the ovules of emasculated flowers on short pea plants. The next year, three seeds from each of the 62 crosses are sown in the garden and the resulting plants are grown to maturity. If none of the three plants from a cross is short, the male parent is classified as having been homozygous SS; if at least one of the three plants from a cross is short, the male parent is classified as having been heterozygous Ss. Using this system of progeny testing, the geneticist concludes that 29 of the 62 tall F₂ plants were homozygous SS and that 33 of these plants were heterozygous Ss.
(a) Using the chi-square procedure, evaluate these results for goodness of fit to the prediction that 2/3 of the tall F₂ plants should be heterozygous.
(b) Informed by what you read in A Milestone in Genetics: Mendel’s 1866 Paper, which you can find in the Student Companion Site, explain why the geneticist’s procedure for classifying tall F₂ plants by genotype is not definitive.
(c) Adjust for the uncertainty in the geneticist’s classification procedure and calculate the expected frequencies of homozygotes and heterozygotes among the tall F₂ plants.
(d) Evaluate the predictions obtained in (c) using the chisquare procedure.