Two-Gene Cross in a Fruit Fly Line
A genetics class maintains a fruit fly line with two linked autosomal genes: body color (G = gray dominant, g = ebony recessive) and wing shape (W = normal dominant, w = vestigial recessive). A heterozygous female (GgWw) is testcrossed to a homozygous recessive male (ggww). The 1,000 offspring are: gray-normal 415, gray-vestigial 85, ebony-normal 80, ebony-vestigial 420. A student expects a 1:1:1:1 ratio and concludes the genes are unlinked but that 'something went wrong' with the data. The instructor notes the two loci are 165 map units apart on the same chromosome.
1. Explain why the observed numbers deviate from 1:1:1:1, and identify which classes are parental and which are recombinant.
Read the worked answer
The deviation is expected because the two loci are linked on the same chromosome; 1:1:1:1 requires independent assortment. The parental (non-crossover) classes are the most frequent: gray-normal (415) and ebony-vestigial (420), totaling 835. The recombinant classes are gray-vestigial (85) and ebony-normal (80), totaling 165. Recombinant frequency = 165/1000 = 16.5%, which corresponds to a map distance of about 16.5 cM (centimorgans), not 165 cM. The student's 'something went wrong' conclusion is incorrect; the data are internally consistent with linkage.
2. The instructor says the loci are 165 map units apart. Explain why this figure is inconsistent with the offspring data and what the maximum observable recombination frequency is.
Read the worked answer
A map distance of 165 cM would require a recombinant frequency above 50%, which is impossible to observe directly. For two loci on the same chromosome, recombination frequency tops out near 50% because multiple crossovers (double, triple, etc.) between distant loci restore parental combinations; at large distances the loci behave as if unlinked. The observed 16.5% is the recombination frequency, giving ~16.5 cM. A 165 cM distance could only be an additive map distance across many intervening markers, not a single two-point cross. So the instructor's figure likely reflects summed intervals, and the data correctly show a relatively close linkage.
3. Predict the offspring proportions if the same heterozygous female were crossed to a heterozygous male (GgWw x GgWw), assuming the phase is coupling (GW/gw).
Read the worked answer
Because the female is in coupling phase (GW on one homolog, gw on the other) and recombination is 16.5%, her gametes are: GW 41.75%, gw 41.75%, Gw 8.25%, gW 8.25%. The male also produces these if he is coupling-phase heterozygous. Combining gametes gives mostly G-W- offspring (the double-dominant and parental combinations dominate), a small fraction of double recessives, and few recombinant phenotypes. Specifically, ggww requires gw from both parents: 0.4175 x 0.4175 ≈ 17.4%. The recombinant phenotypes (gray-vestigial and ebony-normal) are each roughly 2 x (0.4175 x 0.0825) plus (0.0825 x 0.0825) ≈ 7.6%. The key reasoning is that linkage in coupling phase inflates parental phenotypes and suppresses recombinant ones relative to 9:3:3:1.