A Test-Cross That Refused to Give 1:1:1:1
In a brinjal (Solanum melongena) improvement programme at a Tamil Nadu agricultural college, two genes are being followed: P, purple fruit (dominant), p green fruit; S, spiny calyx (dominant), s spineless calyx. True-breeding purple spiny plants are crossed with true-breeding green spineless plants, and the F1 (PpSs) is test-crossed to a green spineless parent. Among 400 test-cross progeny the students count 178 purple spiny, 22 purple spineless, 26 green spiny and 174 green spineless. A second F1, this time made from purple spineless × green spiny, is also test-crossed: of 400 progeny, 24 are purple spiny, 172 purple spineless, 176 green spiny and 28 green spineless. A student writes in the record that the first result is "a bad 1:1:1:1 caused by experimental error" and asks for the experiment to be repeated. The teacher instead demands a statistical test and an interpretation.
1. Test the first test-cross result against independent assortment using chi-square, state the degrees of freedom and the conclusion, and give the recombination frequency.
Read the worked answer
Hypothesis (H0): the two genes assort independently, so the test-cross gives 1:1:1:1, i.e. 100 of each class out of 400.
| Class | Observed (O) | Expected (E) | O − E | (O − E)²/E |
|---|---|---|---|---|
| Purple spiny | 178 | 100 | +78 | 60.84 |
| Purple spineless | 22 | 100 | −78 | 60.84 |
| Green spiny | 26 | 100 | −74 | 54.76 |
| Green spineless | 174 | 100 | +74 | 54.76 |
χ² = 60.84 + 60.84 + 54.76 + 54.76 = 231.2.
Degrees of freedom = number of classes − 1 = 3. At the 5% level the critical value for df = 3 is 7.815. Since 231.2 ≫ 7.815, H0 is rejected decisively; this is not experimental error, because random error would scatter around 100 in both directions, and here two classes are grossly over-represented and two grossly under-represented in a symmetrical pattern.
Interpretation: the parental combinations (purple spiny and green spineless) are in excess and the new combinations are in deficit - the two genes are linked on the same chromosome, in the coupling (cis) arrangement in this F1.
Recombination frequency = (22 + 26)/400 × 100 = 12%, i.e. the genes are about 12 map units apart.
2. Interpret the second test-cross and explain why it gives almost the same map distance but a different set of abundant classes.
Read the worked answer
In the second cross the abundant classes are purple spineless (172) and green spiny (176), and the scarce classes are purple spiny (24) and green spineless (28). The recombinant total is 24 + 28 = 52, so the recombination frequency is 52/400 × 100 = 13%.
This is the same pair of genes on the same chromosome, but the two dominant alleles entered the cross from different parents, so in this F1 the alleles lie in the repulsion (trans) arrangement - P is on one homologue with s, and p is on the other with S. The parental (non-cross-over) gametes are therefore Ps and pS, and the rare cross-over gametes are PS and ps. That is exactly why the abundant classes swap over compared with the first cross.
The 12% and 13% estimates are statistically the same value; with n = 400 the standard error of a recombination estimate near 12% is roughly 1.6%, so a one-unit difference is well within sampling error. Recombination frequency is a property of the distance between the loci, not of which alleles are coupled, so both crosses give about 12–13 cM.
The practical point: if a student judges these data by the identity of the common classes rather than by their proportion, the two crosses look contradictory. They are not - they are the classic reciprocal demonstration that linkage phase changes the gamete frequencies but not the map distance.
3. The breeder wants a true-breeding purple spineless line. Predict the F2 phenotypic proportions from selfing the first F1, and explain why selecting purple spineless plants in F2 is not enough.
Read the worked answer
With 12% recombination the F1 gametes are PS 0.44, Ps 0.06, pS 0.06, ps 0.44.
- green spineless (pp ss) = 0.44² = 0.1936
- purple spineless (P– ss) = 0.0036 (PP ss) + 2(0.44)(0.06) = 0.0528 (Pp ss) = 0.0564
- green spiny (pp S–) = 0.0564 by symmetry
- purple spiny (P– S–) = 1 − (0.1936 + 0.0564 + 0.0564) = 0.6936
So the F2 approximates 69.4 : 5.6 : 5.6 : 19.4, nothing like 9:3:3:1. Tight linkage keeps the parental combinations together and makes the desired recombinant class rare.
Why selection alone fails: of the purple spineless plants, only the PP ss individuals breed true. Their frequency is 0.0036, so the proportion of homozygotes within the selected class is 0.0036/0.0564 ≈ 6.4% - the other ~94% are Pp ss and will segregate on selfing.
The breeder must therefore (i) grow a very large F2 to capture enough recombinants, (ii) self each selected purple spineless plant separately, and (iii) progeny-test the families, keeping only those that breed true for both traits. Anther culture to produce doubled haploids, or marker-assisted selection with markers flanking P and S, would shorten the programme considerably.