Linked Genes in a Tomato Test Cross
A tomato breeder crosses a plant heterozygous for two genes with a dwarf plant with yellow fruit. Red fruit (R) is dominant to yellow (r), and tall stem (T) is dominant to dwarf (t). The heterozygous parent was produced by crossing a true-breeding red tall plant with a true-breeding yellow dwarf plant. The test-cross offspring are: red tall 83; red dwarf 17; yellow tall 19; yellow dwarf 81. The breeder expected a 1:1:1:1 ratio and is surprised that red tall and yellow dwarf are so common. She wonders whether the two genes are on the same chromosome and whether she can use the offspring to estimate the distance between them. She plans to self-pollinate the heterozygous parent to produce a new variety.
1. Explain whether the two genes are linked. Use the offspring numbers to support your answer.
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The offspring are not close to a 1:1:1:1 ratio. If the genes were on different chromosomes and assorted independently, the four classes should be roughly equal, about 50 each out of 200. Instead, the parental combinations red tall (83) and yellow dwarf (81) make up 164 of 200, while the recombinant combinations red dwarf (17) and yellow tall (19) make up only 36. This excess of parental types shows that the two genes are linked on the same chromosome. The heterozygous parent inherited R with T from the red tall parent and r with t from the yellow dwarf parent, so RT and rt are the common gametes. Crossing over between the genes produces the less common Rt and rT gametes.
2. Calculate the recombination frequency and interpret it in terms of chromosome map distance.
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Recombinant offspring are red dwarf plus yellow tall: 17 + 19 = 36. Total offspring = 83 + 17 + 19 + 81 = 200. Recombination frequency = 36/200 × 100 = 18%. By convention, 1% recombination is approximately 1 map unit or centimorgan, so the genes are about 18 map units apart. This is an estimate: the value assumes the test cross reveals gamete frequencies directly, and double crossovers would make the true map distance slightly larger. The result is still strong evidence for linkage, because independent assortment would give about 50% recombinants.
3. The breeder self-pollinates the heterozygous parent. Predict why the offspring will include some red dwarf and yellow tall plants, and why the proportion may differ from a test cross.
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Self-pollination means the heterozygous plant provides both male and female gametes. Crossing over still produces some Rt and rT gametes, so some offspring can receive a recombinant gamete and become red dwarf or yellow tall. However, unlike a test cross, the other parent is not homozygous recessive; it can also contribute dominant alleles. Dominant alleles can mask recessive ones, and many different gamete combinations are possible, so the phenotypes do not directly count the gametes. A test cross is therefore better for estimating recombination frequency: the homozygous recessive parent contributes only r and t alleles, so each offspring phenotype shows the genotype of the gamete from the heterozygous parent. Selfing is useful for producing a new variety but less useful for mapping the genes.