Inheritance of Two Traits in a Fruit Fly Cross
In Drosophila, body colour (grey dominant, G; ebony recessive, g) and wing shape (normal dominant, N; vestigial recessive, n) are being studied. A heterozygous grey, normal-winged female is test-crossed with an ebony, vestigial male. The offspring number 1000 and fall into four classes: grey normal 410, grey vestigial 90, ebony normal 90, ebony vestigial 410. A student concludes the two genes assort independently because both traits show a 3:1 pattern within each phenotype. A second student argues the genes are linked. The experiment is at 25 °C and no crossing over is assumed to be affected by temperature.
1. Analyse the offspring ratios and decide which student is correct. Support your answer with a chi-square style comparison to the independent-assortment expectation.
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For independent assortment of two unlinked genes in a test cross, the expected ratio is 1:1:1:1, i.e. 250 of each class out of 1000.
Observed: grey normal 410, grey vestigial 90, ebony normal 90, ebony vestigial 410.
The two parental classes (grey normal and ebony vestigial) are strongly over-represented, and the recombinant classes are strongly under-represented. The deviation is enormous; a chi-square calculation would give a value far beyond any critical value at 3 degrees of freedom, so independent assortment is rejected.
The first student's reasoning is flawed: a 3:1 ratio within each trait separately does not imply independent assortment between traits. Each trait alone is segregating 1:1 in a test cross, and the joint distribution is what tests linkage.
Judgement: linkage in coupling (cis) configuration is indicated, because the parental combinations dominate.
2. Estimate the recombination frequency and the map distance between the two genes, and explain what the recombinant classes represent.
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Recombinants are the grey vestigial and ebony normal classes, which arise from crossovers between the two loci.
Recombinant count = 90 + 90 = 180 out of 1000.
Recombination frequency = 180/1000 = 0.18, or 18%. Map distance = 18 map units (centimorgans).
The parental classes (410 + 410 = 820) represent non-crossover gametes. Because the heterozygous parent was in coupling phase (GN/gn), the parental gametes are GN and gn, and recombinants are Gn and gN.
Judgement: recombination frequency is calculated from recombinant/total, and 1% recombination equals 1 cM for small distances. At 18% the estimate is still reasonably reliable, though double crossovers begin to cause slight underestimation.
3. Predict the offspring if the same heterozygous female were crossed with a heterozygous male, and state one assumption that could fail.
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The female produces gametes GN 41%, gn 41%, Gn 9%, gN 9% (parental vs recombinant). The male, if also GN/gn, produces the same frequencies, assuming no sex difference in recombination.
Expected offspring: parental × parental combinations dominate. GN/GN and GN/gn give grey normal; gn/gn gives ebony vestigial. Roughly, grey normal ≈ 0.41×0.41×1 + cross terms, ebony vestigial ≈ 0.41×0.41 ≈ 0.168, and recombinants are rare. The precise fractions can be obtained from a Punnett square using the four gamete frequencies.
Assumption that could fail: recombination frequency may differ between the sexes. In Drosophila, males show no crossing over, so a heterozygous male would produce only GN and gn gametes, drastically changing the outcome and producing no recombinant phenotypes from the male side.
Judgement: sex-specific recombination is a real biological constraint and must be checked before applying a single recombination frequency to both parents.