A Tomato Testcross That Refuses to Fit 1:1:1:1
A school genetics project uses two true-breeding tomato lines: one tall with round, red fruit, the other dwarf with oblong, yellow fruit. The F1 from the cross is uniformly tall with round red fruit. To locate the two genes, the students testcross the F1 to the dwarf oblong yellow parent and score 1,000 progeny: 428 tall red, 79 tall yellow, 71 dwarf red, 422 dwarf oblong yellow. The teacher calls this 'linkage', but one student argues that the shortage of the two recombinant classes could simply mean those seeds germinated poorly in the school nursery. Another points out that the total tall:dwarf ratio is 507:493 and the red:yellow ratio is 499:501. The students must decide how much confidence to place in the linkage conclusion, estimate the distance between the loci if linkage is real, and justify why they used a testcross rather than allowing the F1 to self-pollinate.
1. Calculate the expected counts under independent assortment, test the fit, and decide whether the data support linkage. Use the single-locus ratios in your argument about differential viability.
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Under independent assortment the testcross should give 1:1:1:1, i.e. 250 in each of the four classes. The chi-square is Σ(O − E)²/E = (428 − 250)²/250 + (79 − 250)²/250 + (71 − 250)²/250 + (422 − 250)²/250 = 126.7 + 117.0 + 117.0 + 126.7 = 487.4 with 3 degrees of freedom. The 5% critical value is 7.81, so the deviation is far too large to be chance.
The viability objection can be tested from the marginal ratios. Tall:dwarf = 507:493 against 500:500 gives χ² = 49/500 + 49/500 = 0.196 (df = 1), not significant. Red:yellow = 499:501 gives χ² ≈ 0.004, also not significant. So there is no evidence that either single locus distorts survival. More importantly, poor germination would scatter losses across all four classes and make the test less powerful; it would not create the complementary pattern seen here - a large excess of the two parental combinations and a matching deficit of the two recombinant combinations. That reciprocal excess/deficit is the signature of linkage in coupling phase. The conclusion of linkage is therefore well supported, although the students should state that the chi-square test assumes independent viability and that the parental classes are tall red and dwarf oblong yellow.
2. Estimate the recombination frequency and map distance. Explain why this is an estimate rather than an exact distance, and what would change it.
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The recombinant classes are tall yellow (79) and dwarf red (71), so recombination frequency = (79 + 71)/1000 = 150/1000 = 0.15, or 15%. For a testcross this is converted directly to a map distance of about 15 centimorgans (cM). The loci are in coupling (cis) phase in the F1 because the parental combinations are the tall red and dwarf yellow classes.
The value is an estimate for three reasons. First, scoring only recognises an odd number of crossovers between the two loci; double crossovers restore the parental combination and are counted as non-recombinants, so 15% underestimates the true crossover frequency. The discrepancy is small at short distances but grows with distance, which is why recombination frequency is not a strictly additive map distance. Second, sampling error is appreciable: the standard error of the proportion is √(0.15 × 0.85/1000) ≈ 1.1%, giving an approximate 95% confidence interval of 13–17%. Third, misclassification or viability differences would bias the estimate. A larger progeny set, scoring with molecular markers rather than phenotypes, or a three-point cross would refine the distance and reveal double crossovers.
3. Why is a testcross the right design here, and what cross would you do to test whether the two loci are actually linked rather than affected by a third factor?
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A testcross to the double recessive parent is the correct design because the tester contributes only recessive alleles. The phenotypes of the progeny therefore reveal the gametic frequencies of the F1 directly, with no masking by dominance. If the F1 were allowed to self-pollinate, the F2 phenotypes would depend on dominance at both loci and parental and recombinant gametes could not be read off without progeny testing. The testcross also gives a simple expected 1:1:1:1 under independent assortment, which makes the statistical test straightforward.
To check that the deviation is linkage and not a third factor, several controls are useful. A reciprocal testcross, using F1 as male and as female, tests for parent-of-origin or gametic viability effects; true linkage should give similar recombination frequencies in both directions. Selling the F1 and scoring F2 against the 9:3:3:1 expectation with linkage provides an independent estimate. The single-locus ratios should also be checked in every family, since a third factor such as gamete lethality often distorts one locus on its own. Finally, a direct gamete assay - pollen staining or molecular markers on gametes - would separate gametic viability from recombination itself.