Interpreting a Deviant Testcross Ratio in Tomato
Two true-breeding tomato lines were crossed: tall with smooth fruit × dwarf with hairy fruit. All F1 plants were tall with smooth fruit. A student testcrossed the F1 to a dwarf hairy plant and scored 100 progeny: tall smooth 41, dwarf hairy 39, tall hairy 9, dwarf smooth 11. About six seedlings died after transplanting and were never scored. One classmate says the result disproves Mendel's law of independent assortment; another says the experiment was simply done badly, because the four classes ought to be equal. The student wonders whether the dead seedlings could explain the shortage of the two rarer classes.
1. Test the result against the expectation of independent assortment, showing expected numbers, and state your conclusion.
Read the worked answer
For a testcross of a double heterozygote with independent assortment, the four phenotypic classes should appear in a 1:1:1:1 ratio - 25 plants of each class in a sample of 100.
Observed: 41, 39, 9, 11. Deviations are +16, +14, −16, −14. The two parental-type classes (tall smooth, dwarf hairy) are in large excess, and the two new combinations (tall hairy, dwarf smooth) are markedly deficient.
A chi-square test makes this quantitative: χ² = 16²/25 + 14²/25 + 16²/25 + 14²/25 = (256 + 196 + 256 + 196)/25 = 36.2 with 3 degrees of freedom, far above the 7.8 critical value at p = 0.05. The probability that this is chance deviation from independent assortment is very small.
The conclusion is linkage: the two gene pairs lie on the same chromosome and tend to be inherited together, so parental combinations are over-represented and recombinant combinations are under-represented. A rough excess of the parental classes is enough for the exam; the chi-square confirms that it is not sampling noise.
2. Before accepting linkage, rule out differential viability using the same data, and then calculate recombination frequency, map distance and the linkage configuration.
Read the worked answer
If one genotype died preferentially, that class would be short while the others stayed equal; linkage produces a paired excess of parental classes and a paired deficit of recombinants. Check each gene separately.
Height: tall = 41 + 9 = 50; dwarf = 39 + 11 = 50. Exact 1:1. Fruit surface: smooth = 41 + 11 = 52; hairy = 39 + 9 = 48. Close to 1:1.
Neither single-gene ratio is distorted, so plants of both phenotypes survived and were counted normally. The six lost seedlings would have been distributed across classes; a general loss cannot create a specific deficit of the two recombinant classes. Differential viability is therefore not the explanation, and linkage stands.
Recombination frequency = (9 + 11)/100 × 100 = 20%, i.e. 20 map units (centimorgans) between the two loci.
Configuration: the original parents were tall smooth and dwarf hairy, and these are the frequent classes, so the F1 carries both dominant alleles on one chromosome and both recessive alleles on the other - coupling, or cis, arrangement. Gametes: TS 40%, tf 40%, Tf 10%, tF 10%.
3. The student wants a true-breeding dwarf smooth line. Which progeny should be chosen and how should homozygosity be confirmed?
Read the worked answer
Dwarf smooth plants from the testcross are the 11 individuals of that class. Genotypically they are tF/tf: recessive for height (tt), and heterozygous for fruit surface because the testcross parent contributed f, while the dominant F came from the F1 parent.
Selfing one of these plants gives gametes tF and tf in equal proportion, so the progeny are tt FF and tt Ff, both dwarf smooth, and tt ff, dwarf hairy, in a 3:1 ratio. Selection on appearance alone cannot separate a homozygous tt FF plant from a heterozygous tt Ff plant, since both look dwarf smooth.
The practical procedure is to self each of the 11 plants separately, raise the progeny, and discard any family that produces a single hairy-fruited plant. A family that breeds true, giving only dwarf smooth offspring, identifies the parent as tt FF. Progeny testing is essential because the hairy allele can be hidden; and because only 11 plants are available, several may be heterozygous, so every one should be tested rather than pooling seed.