A Testcross That Does Not Behave: Linkage in Drosophila
A student in a genetics teaching lab sets up a testcross: 40 heterozygous female Drosophila (wild-type body colour, normal wings) are crossed to ebony-bodied, vestigial-winged males. The two loci are on the same chromosome, but the student has not been told whether they are linked. After three weeks the class pools its data - 1,000 progeny in total - and the instructor asks for an analysis before the next practical:
- wild-type body, normal wings - 425
- ebony body, normal wings - 78
- wild-type body, vestigial wings - 77
- ebony body, vestigial wings - 420
The student's partner argues that because all four classes appear, the genes must be assorting independently and the deviation from 1:1:1:1 is just sampling noise from the small number of parents. The instructor also notes that the reciprocal cross (heterozygous males × double-recessive females) was set up at the same time but has not yet been scored.
1. Test the partner's claim formally. State the expected numbers, the test statistic, the degrees of freedom, and your conclusion.
Read the worked answer
Hypothesis (H₀): the two loci assort independently, so the testcross yields four phenotypic classes in a 1:1:1:1 ratio. With n = 1,000, each expected count is 250.
Chi-square:
| Class | Observed | Expected | (O−E)²/E |
|---|---|---|---|
| wild, normal | 425 | 250 | 122.5 |
| ebony, normal | 78 | 250 | 118.3 |
| wild, vestigial | 77 | 250 | 119.7 |
| ebony, vestigial | 420 | 250 | 115.6 |
χ² = 476.1 with df = 4 − 1 = 3. The 5% critical value for df = 3 is 7.81, so H₀ is rejected overwhelmingly.
Interpretation, not just arithmetic: the departure is not random scatter - it is directional. The two classes that reproduce the parental combinations (wild-type and double mutant) together make up 845/1000, while the recombinant classes make up only 155/1000. That excess of parental types is the signature of linkage in coupling (cis) configuration, i.e. the wild-type alleles sit on one homologue and the mutant alleles on the other. The partner's argument confuses 'all four classes present' with 'all four classes equally likely'; independent assortment is a quantitative prediction, not a list of possible outcomes.
2. Estimate the distance between the two loci. Explain carefully what that number means and in which direction it is likely to be biased.
Read the worked answer
Estimate: recombination frequency (RF) = recombinants ÷ total = (78 + 77)/1000 = 0.155, i.e. 15.5 map units (centimorgans).
What it means: among meioses in these females, about 15.5% of the gametes scored carry a chromosome on which the two markers have been recombined relative to the parental arrangement. Because 1 cM is defined as 1% recombination, the naive map distance is 15.5 cM.
Direction of bias: this is an underestimate of the true distance. Only an odd number of crossovers between the two loci produces a detectable recombinant class; a double crossover (or any even number) restores the parental combination and is scored as non-recombinant. The effect is negligible for closely linked loci but grows with distance, which is why RF saturates below 50% and never exceeds it even for genes far apart on the same chromosome. Correction functions (Haldane, Kosambi) map 15.5% to roughly 16–17 cM.
Practical check: 15.5% is comfortably below 50%, so the loci are linked but not tightly; the data cannot tell you which side of either locus anything else lies without a third marker.
3. Predict the outcome of the reciprocal cross (heterozygous males × double-recessive females) and explain the consequence for the student's plan to confirm the map distance with that cross.
Read the worked answer
Prediction: essentially two classes only - wild-type and double mutant, in roughly equal numbers - with the single-mutant recombinant classes near zero.
Reasoning: in Drosophila, crossing over is absent in males (no synaptonemal complex/chiasma formation in spermatocytes). The heterozygous male therefore transmits only the two parental combinations that were assembled in his mother, regardless of the physical distance between the loci. This is not a failure of linkage analysis; it is a property of the organism.
Consequence for the plan: the reciprocal cross cannot be used to confirm 15.5 cM, and scoring it would produce a spurious RF of ~0%. If the student pooled the two reciprocal crosses, the map distance would be diluted toward half its true value and the recombination estimate would be meaningless. The correct design is to keep the crosses separate, use only the female-heterozygote testcross for recombination, and use the male testcross (if used at all) as a check on the parental configurations.
This is a real judgement the exam expects: an apparent absence of recombinants in one sex is evidence about meiosis in that sex, not evidence that the genes are inseparable.