Tomato Testcross: Independent Assortment or Linkage?
In a tomato genetics lab, true-breeding plants with purple stems and round fruit are crossed with true-breeding plants with green stems and elongated fruit. All F1 plants have purple stems and round fruit. A student testcrosses an F1 plant with a green-stemmed, elongated-fruit plant. The testcross offspring are counted: 410 purple, round; 92 purple, elongated; 88 green, round; 410 green, elongated. Total = 1,000. A second student says the results are close enough to a 1:1:1:1 ratio and that the two genes assort independently. A third student notices that the two largest classes are the same combinations seen in the original parents. The class is asked to decide whether the two genes are linked and to estimate the distance between them. The testcross parent is known to be homozygous recessive for both traits and produces only one kind of gamete.
1. Evaluate the claim that the genes assort independently. Include the expected numbers under independent assortment and an explanation of what the observed numbers indicate.
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If the two genes assorted independently, the testcross should produce four phenotypic classes in a 1:1:1:1 ratio. With 1,000 offspring, the expected number in each class would be 250. The observed numbers are far from that: the two parental combinations, purple-round and green-elongated, total 820, while the two recombinant combinations total only 180. A difference this large is extremely unlikely to be due to chance.
The testcross parent contributes only recessive alleles, so each offspring's phenotype directly reflects the gamete produced by the F1 parent. The excess of parental combinations means the F1 parent produced mostly purple-round and green-elongated gametes. That is the signature of linkage: the two genes are on the same chromosome, and the parental allele combinations tend to stay together. The recombinant classes arise from crossing over between the two loci.
2. Calculate the recombination frequency and explain what it represents physically.
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Recombination frequency = (number of recombinant offspring / total offspring) x 100.
Recombinants are purple-elongated (92) and green-round (88):
(92 + 88) / 1,000 = 180 / 1,000 = 0.18 = 18%.
The recombination frequency estimates the proportion of gametes in which a crossover occurred between the two genes during meiosis. An 18% value means the genes are relatively close together on the same chromosome, about 18 map units (centimorgans) apart. The value is not a direct physical distance in nanometers; it is a relative measure based on crossover frequency. Genes farther apart have more opportunity for crossing over and therefore higher recombination frequencies.
3. Predict the testcross results if the F1 had been produced from a different cross in which purple was linked to elongated and green to round (the alleles in trans). Give approximate numbers for 1,000 offspring and explain.
Read the worked answer
If the alleles were in the trans arrangement, the parental gametes from the F1 would be purple-elongated and green-round, not purple-round and green-elongated. The two parental classes would again be the most common, but their identities would switch. With the same 18% recombination frequency, about 82% of gametes would be parental and 18% recombinant.
For 1,000 offspring, the approximate numbers would be:
- purple, elongated: about 410
- green, round: about 410
- purple, round: about 90
- green, elongated: about 90
The key point is that linkage phase determines which combinations are overrepresented. The recombination frequency stays about the same because it depends on the distance between the loci, not on which alleles are on which chromosome.