480 Tomato Plants and a Doubtful Ratio
A school agriculture club grows tomatoes. They cross a pure tall plant with round fruit (TTRR) with a pure dwarf plant with elongated fruit (ttrr). All the F1 plants are tall with round fruit. The F1 plants are self-pollinated, and the club counts 480 F2 plants: 268 tall round, 92 tall elongated, 88 dwarf round, 32 dwarf elongated. One member says the result is wrong because it is not exactly 9:3:3:1; another says dwarfness must have been lost in the F1 because no dwarf plant appeared there, and that the few dwarf plants in F2 are accidents of the season. The club also wants to keep seed from one F2 tall round plant and needs to know whether that plant is pure for both traits before using it as a parent. No DNA test is available; only further crosses are possible.
1. Test the first member's claim that the ratio is wrong. Calculate expected numbers and comment on the deviation.
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For a dihybrid cross between pure parents, F2 is expected in the ratio 9:3:3:1. With 480 plants: 9/16 × 480 = 270 tall round; 3/16 × 480 = 90 tall elongated; 90 dwarf round; 1/16 × 480 = 30 dwarf elongated.
Observed: 268, 92, 88, 32. Deviations: -2, +2, -2, +2 - all within a couple of plants of expectation.
A chi-square check: X² = (2²/270) + (2²/90) + (2²/90) + (2²/30) = 0.015 + 0.044 + 0.044 + 0.133 ≈ 0.24, with 3 degrees of freedom. The 5% critical value for 3 d.f. is 7.8, so the observed counts do not depart significantly from 9:3:3:1; such small deviations are expected from random fertilisation and from the fact that only 480 plants were counted.
Conclusion: the member is confusing an expected ratio with an exact count. The data fit a 9:3:3:1 result, which is evidence that the two gene pairs assort independently - the genes concerned are on different chromosomes or far apart on the same one.
2. Explain to the second member where dwarfness "went" in the F1 and why it reappeared in the F2.
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Dwarfness never disappeared. Each F1 plant received T from the tall parent and t from the dwarf parent, so its genotype is Tt. T is dominant, so the plant is tall, but the t allele is still present in every cell and is copied into half of the gametes the plant makes.
When F1 plants are self-pollinated, gametes carrying T and t combine at random: TT, Tt and tt appear in roughly 1:2:1. Only tt plants are dwarf, so about one-quarter of the F2 should be dwarf. Here 88 + 32 = 120 dwarf plants out of 480 - exactly one-quarter, which matches the prediction. In the same way, the fruit-shape alleles R and r segregate, so dwarfs appear with both round and elongated fruit.
The reappearance of a trait that was hidden for a generation is strong evidence that inheritance is by particles (genes) that stay intact and are not blended or diluted. The occurrence of dwarf plants in the F2 is not an accident of the season; it is the expected outcome of segregation.
3. Design a crossing plan to find out whether the chosen F2 tall round plant is pure for both traits, and state what result would indicate purity.
Read the worked answer
Use a test cross: cross the unknown tall round plant with a dwarf elongated plant (ttrr), which makes only gametes carrying t and r. Whatever alleles the unknown plant contributes will therefore be visible directly in the offspring, because the tester contributes no dominant allele to mask them. Grow at least 50 offspring, since small numbers fluctuate.
Possible outcomes:
- All offspring tall round → the parent made only TR gametes → genotype TTRR, pure for both traits.
- About half tall round and half dwarf round → parent made TR and tR gametes → TtRR, heterozygous for height only.
- About half tall round and half tall elongated → parent made TR and Tr gametes → TTRr, heterozygous for fruit shape only.
- Four types in roughly equal numbers - tall round, tall elongated, dwarf round, dwarf elongated → parent made TR, Tr, tR and tr gametes → TtRr, heterozygous for both.
If any recessive offspring appear, the parent is not pure and its seed must not be used as a pure breeding line. The logic holds for any organism: the tester contributes only recessive alleles, so the offspring phenotypes reveal the unknown parent's gametes.