Tomato Wilt, Fruit Size, and a Spoiled Bench
A greenhouse worker is studying two tomato varieties. Variety A is resistant to a fungal wilt but produces small fruit; variety B is susceptible to wilt but produces large fruit. A cross between pure-breeding A and pure-breeding B yields F1 plants that are all resistant and all large-fruited. When F1 plants are self-pollinated, 240 offspring are scored: 135 resistant large, 45 resistant small, 45 susceptible large, 15 susceptible small. However, one bench of 20 seedlings was accidentally kept in cool, damp conditions and 8 of those died from wilt before scoring. The worker wants to know whether resistance and fruit size are inherited independently and whether the 240 plants support a 9:3:3:1 ratio. A colleague claims that because F1 all looked resistant and large, the two traits must be linked. The worker also plans to use a test cross to identify a resistant large F1 plant’s genotype.
1. Analyze the F2 data. State the likely inheritance pattern, assign symbols, and test whether the observed numbers fit a 9:3:3:1 ratio. Show your expected counts and a simple comparison.
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The data fit two independently assorting genes with complete dominance. Let R = resistant and r = susceptible; L = large fruit and l = small fruit. Pure A is RRll (resistant, small), pure B is rrLL (susceptible, large), and F₁ is RrLl (resistant, large). Selfing F₁ gives the expected 9 R_L_ : 3 R_ll : 3 rrL_ : 1 rrll. For 240 offspring, expected counts are 9/16×240 = 135; 3/16×240 = 45; 45; and 1/16×240 = 15. The observed counts are exactly 135, 45, 45, 15, so the data fit 9:3:3:1 perfectly. This supports independent assortment, not linkage. However, a perfect fit to expected numbers is unusual in real data and should be checked for scoring bias; also the 20 excluded seedlings mean the effective sample is not a clean random sample of all F₂ offspring.
2. The colleague says F1 uniformity proves linkage. Evaluate this claim. What would the F1 phenotypes be if the genes were completely linked in coupling (resistance with small fruit)?
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The colleague’s claim is not valid. F₁ uniformity shows only that the resistant and large-fruit traits are dominant and that the parents were pure-breeding; it does not show whether the genes are linked. If the genes were completely linked in coupling, with resistance inherited together with small fruit (R–l) and susceptibility with large fruit (r–L), the pure parents would be RRll and rrLL, and F₁ would still be RrLl and uniformly resistant large. Thus the F₁ phenotype is the same under independent assortment and under complete linkage. Linkage would change the F₂ or test-cross ratios by reducing the recombinant classes (resistant large and susceptible small) and increasing the parental classes. The observed F₂ data, with 135 and 15 in the recombinant classes, instead match independent assortment. A test cross is the cleaner next step.
3. Explain why the 20 seedlings on the cool damp bench should not be included in the F2 ratio, and describe a test cross to determine whether a resistant large F2 plant is homozygous or heterozygous for both genes. Include expected offspring ratios for one possible genotype.
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The 20 cool, damp seedlings should be excluded from the main ratio because the environment was not the same for all plants. Cool, damp conditions favour the fungal wilt, so susceptible plants may die before scoring; this is differential mortality, not a fair test of genetic ratios. Including them would underestimate susceptible classes and bias the result. To test a resistant large F₂ plant, cross it with a susceptible small plant of genotype rrll. If the unknown is RrLl, the offspring should be approximately 1 resistant large : 1 resistant small : 1 susceptible large : 1 susceptible small. If it is RRLL, all offspring should be resistant large. If it is RRLl, expect 1 resistant large : 1 resistant small. If it is RrLL, expect 1 resistant large : 1 susceptible large. The test cross reveals the unknown’s genotype because the rrll parent contributes only recessive alleles.