The 1,600-Seed Count: Linkage, or a Scoring Problem?
In a higher secondary school in Nagaon, 28 second-year students pool their counts from a genetics exercise on a dihybrid cross in tomato (seed shape and cotyledon colour). Both genes are on the same chromosome, and the heterozygous parent had the dominant alleles in coupling. The class expected a 9:3:3:1 ratio in 1,600 F2 seeds. The pooled counts are: round-yellow 1,150; round-green 100; wrinkled-yellow 120; wrinkled-green 230. A student, Priya, says the seed lot must be wrong because the ratio is nowhere near 9:3:3:1. Another, Rahul, says the double-recessive class is hard to score because shrivelled seeds are easily missed, so the count there is probably low. The teacher has one 45-minute period to settle the question, and the practical record must be completed before the external examiner visits next week. Eight students did the scoring, without a written scoring convention.
1. Test the pooled data against the 9:3:3:1 expectation. Compute the chi-square statistic, state the degrees of freedom and the 5% critical value (7.82), and state your conclusion.
Read the worked answer
Expected numbers in 1,600 seeds: 900, 300, 300, 100.
chi-square = (1150-900)^2/900 + (100-300)^2/300 + (120-300)^2/300 + (230-100)^2/100 = 62500/900 + 40000/300 + 32400/300 + 16900/100 = 69.4 + 133.3 + 108.0 + 169.0 = 479.7
Degrees of freedom = 4 - 1 = 3. The 5% critical value at 3 df is 7.82. Since 479.7 is far greater than 7.82, we reject the hypothesis that the F2 phenotypes fit 9:3:3:1. The deviations are not random noise; something systematic is present, and the next job is to decide what.
Notice the sign pattern before rushing to an explanation: round-yellow is in excess by 250, wrinkled-green is in excess by 130, while both single-dominant classes are deficient (by 200 and 180). A mere failure to fit tells you that a hypothesis is wrong; it does not tell you which alternative to adopt. That judgement needs a second, independent check.
2. Rahul blames under-scoring of the shrivelled class. Test this explanation, and describe the additional check that should be run before the class writes 'linkage' in the record.
Read the worked answer
Rahul's explanation fails on its own numbers. Wrinkled-green is the class he says is missed, yet it is observed 230 against an expectation of 100, an excess of 130, not a deficit. Under-counting can never produce an excess; it can only push a class down. A second weakness: careless scoring usually inflates some classes and deflates others roughly at random, but here both recombinant classes (round-green, wrinkled-yellow) are low by similar amounts and both parental classes are high, which is the signature of linkage in coupling.
However, the data cannot be accepted as they stand, because linkage is not the only systematic effect. Check each gene separately. Round to wrinkled is 1250:350 against an expected 1200:400 (chi-square 8.3, above 3.84 at 1 df); yellow to green is 1270:330 (chi-square 16.3). Both single-gene ratios are themselves distorted, and a single-gene distortion cannot be caused by linkage between the two genes. That points to classification error, most likely in distinguishing green from yellow cotyledons in immature seeds.
The additional check: re-score a fresh sample of at least 400 seeds, blind (two students score independently, a written rule for 'wrinkled' and for green, seed coats removed), and see whether each single-gene ratio returns to 3:1. Only then is an estimate of recombination worth recording.
3. Assuming a corrected re-scoring confirms the pattern, calculate the recombination fraction from the pooled data and state two reasons for reporting it with a caveat.
Read the worked answer
Recombinant phenotypes are the two single-dominant classes: 100 + 120 = 220 out of 1,600.
Recombination fraction = 220/1,600 = 0.1375, that is 13.75%, or about 13.8 map units if the genes are treated as linked in coupling.
Two caveats. First, this is an F2 from a selfed heterozygote, not a testcross, so the recombinant class is not read directly; the estimate rests on the assumption that all four classes are recognised with equal accuracy, which is precisely what the marginal ratios bring into doubt. Second, the figure is pooled across eight scorers over several sessions. Pooling hides scorer-specific bias, and errors of classification tend to depress the apparent recombinant classes and so inflate the impression of linkage. A defensible record states the number as provisional, gives the scoring convention used, and reports the single-gene ratios alongside, so that the examiner can see the data were checked rather than merely tabulated.