A test-cross ratio that is not 1:1:1:1: linkage, or bad germination?
A breeder test-crosses an F1 plant to the double recessive to study two simply inherited traits: leaf margin (serrate vs entire) and stem colour (purple vs green). The 1,000 test-cross progeny scored were: serrate-purple 428, serrate-green 72, entire-purple 68, entire-green 432. Seed set was uniform across the cross and all progeny were scored at the same growth stage. A colleague argues that the departure from a 1:1:1:1 ratio is an artefact of poor germination and wants to repeat the experiment with 5,000 plants before any conclusion is drawn. The same F1 was also selfed, and in the F2 the two parental combinations were again much more frequent than the recombinant ones. The breeder needs at least 20 serrate-green plants for a yield trial next season and can accommodate about 1,000 progeny per cross.
1. Test the data against independent assortment and estimate the recombination frequency.
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Check the single loci first. Serrate = 428 + 72 = 500, entire = 68 + 432 = 500. Purple = 428 + 68 = 496, green = 72 + 432 = 504. Both fit 1:1, so there is no gametic or zygotic distortion to explain the joint ratio.
Under independent assortment each class should be 250. Expected χ² = Σ(O − E)²/E:
(178²/250) + (178²/250) + (182²/250) + (182²/250) = 126.7 + 126.7 + 132.5 + 132.5 ≈ 518.5 with 3 df, against a 5% critical value near 7.8. Independent assortment is rejected decisively.
Recombination frequency = (72 + 68)/1,000 = 0.14, i.e. 14 cM. The two parental classes (428, 432) are almost equal and the two recombinant classes (72, 68) are almost equal - the signature of linkage in coupling phase without differential viability.
2. Is the colleague's demand for a 5,000-plant repeat justified? What would a repeat actually add?
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The objection is answerable with the data already in hand. Germination failure acts on single loci too: if one class were selectively lost, the marginal ratios would drift away from 1:1, and they do not (500 : 500 and 496 : 504). The joint deviation is also far too large and too structured to be sampling noise (χ² ≈ 519).
Precision: SE of the recombination fraction = √(0.14 × 0.86/1000) = 0.011, so the estimate is 14 ± 2.2 cM at 95% confidence - ample for concluding linkage.
Judgement: a 5,000-plant repeat is not needed to detect linkage. It would be worth doing, or a reciprocal cross, for two narrower purposes: to check that 14 cM is stable, and to test for segregation distortion or viability selection, which would show as unequal parental class sizes. A 5,000-plant repeat reduces the SE to about 0.5 cM, which only matters if a mapping population is being built.
3. How many progeny should be raised to be reasonably sure of getting at least 20 serrate-green plants?
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Observed frequency of serrate-green = 72/1000 = 0.072. Simple expectation gives 20/0.072 ≈ 278 plants - but that leaves only a 50% chance of actually reaching 20.
Use a one-sided 95% margin with the normal approximation to the binomial: 0.072n − 1.645√(0.072 × 0.928n) ≥ 20.
- n = 400: mean = 28.8, SD = √(26.7) = 5.2, and 28.8 − 1.645(5.2) = 20.3 - satisfied.
- n = 380: mean = 27.4, SD = 5.04, and 27.4 − 8.3 = 19.1 - fails.
So about 400 progeny are needed.
Judgement: 278 is the 'expected value' answer a breeder should not rely on, because field germination and transplanting losses act on top of the Mendelian ratio. Planting 400–450 gives a real margin without wasting a season on a trial that falls one plant short.