A tomato testcross: chi-squared, linkage and map distance
In tomato, fruit shape is controlled by a gene with alleles R (round, dominant) and r (elongated, recessive). Leaf type is controlled by a second gene with alleles C (cut leaf, dominant) and c (potato leaf, recessive). A plant heterozygous for both genes, RrCc, is crossed with a plant that has elongated fruit and potato leaves, rrcc. The cross produces 1000 offspring: 428 round, cut-leaf; 72 round, potato-leaf; 68 elongated, cut-leaf; 432 elongated, potato-leaf. A chi-squared test is carried out with the null hypothesis that the two genes assort independently. Critical values are 7.82 for 3 degrees of freedom at p = 0.05 and 11.34 at p = 0.01. The student writes that the result “proves the genes are on the same chromosome and are 14 map units apart”.
1. Calculate the chi-squared value, state the degrees of freedom and give the conclusion of the test.
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If the genes assort independently, a testcross should give four phenotypic classes in a 1:1:1:1 ratio. With 1000 offspring, the expected number in each class is 250. The chi-squared value is the sum of (observed − expected)² ÷ expected. For round, cut: (428 − 250)² ÷ 250 = 126.7. For round, potato: (72 − 250)² ÷ 250 = 126.7. For elongated, cut: (68 − 250)² ÷ 250 = 132.5. For elongated, potato: (432 − 250)² ÷ 250 = 132.5. The total is 518.4. There are four classes, so degrees of freedom = 4 − 1 = 3. The calculated value, 518.4, is far greater than the critical value of 11.34 at p = 0.01. The probability that the deviation from the expected ratio is due to chance is less than 1%. The null hypothesis is rejected: the observed results do not fit independent assortment. The two large classes are the parental types, and the two small classes are the recombinant types.
2. Explain why a testcross was used and what the recombinant classes show about the behaviour of the genes in meiosis.
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A testcross uses a homozygous recessive parent, rrcc, which can donate only r c gametes. Therefore the phenotypes of the offspring directly reveal the gamete genotypes produced by the heterozygous parent. If the genes were on different chromosomes and assorted independently, the heterozygous parent would produce R C, R c, r C and r c gametes in equal numbers, giving four equal offspring classes. The observed result is not equal: the R C and r c classes are much larger than the R c and r C classes. This indicates that the R and C alleles tend to stay together, and r and c tend to stay together, because the two genes are linked on the same chromosome. The R C and r c combinations are parental. The smaller classes, R c and r C, are recombinants produced by crossing over between the two loci during prophase I of meiosis. The recombinant frequency is (72 + 68) ÷ 1000 × 100 = 14%.
3. Evaluate the student's statement that the result “proves” linkage and shows a distance of 14 map units.
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The chi-squared test shows that the data are very unlikely to fit independent assortment, so linkage is strongly supported. But a statistical test does not prove physical linkage. The deviation could also be caused by differences in viability among the four phenotypes, misclassification of leaf type or fruit shape, or non-random fertilisation. A control cross or a viability test could check for these. The estimate of 14 map units comes from the recombinant frequency, 140 ÷ 1000 × 100 = 14%. For linked genes, 1% recombination is approximately 1 map unit, so 14% suggests about 14 cM. However, this is an estimate. Double crossovers between the two loci can produce parental combinations and would not be counted as recombinants in a two-point testcross, so the true map distance may be slightly greater than 14 cM. Sampling error also matters with 1000 offspring. The student should write that the data support linkage at a distance of approximately 14 cM, not that they prove it.