The fruit fly Drosophila melanogaster has a haploid chromosome number of n = 4,

Question

The fruit fly Drosophila melanogaster has a haploid chromosome number of n = 4, and 2n = 8. Chromosome IV in this species is a tiny autosome. Flies trisomy for chromosome IV are fertile and have no apparent defects. the eyeless (ey) and gawky (gw) loci are tightly linked on chromosome IV. Loss-of-function ey^-and g^w^- alleles are recessive to ey and gw, respectively. Flies homozygous for ey lack eyes and flies homozygous for gw have disrupted circadian rhythms. Consider a male fly trisomy for chromosome IV, with each of the three chromosome copies bearing different allele combinations for these two loci: for the progeny of this cross to be wild-type, they need to receive both the (ey^- gw^+) and (ey^+ gw^-) chromosomes from the trisomy male. This combination is present in 1/6 of the trisomy male's sperm. If this trisomy male is crossed with a (ey^- ey^-, gw^- gw^-) female, what proportion of the progeny will be phenotypically eyeless? Express your answer as a fraction (example: 3/5). This question will be shown after you complete previous question(s).

Explanation / Answer

For this question, cross the above sperm genotype probabilities with a female egg which will have the genotype 'ey'.
The resulting offspring will have the genotypes and the proportions of:
1/3 +eyey
1/3 +ey
1/6 ++ey
1/6 eyey
Basically you just added an 'ey' to each sperm genotype. The end result is 5/6 are wild-type, and 1/6 are eyeless. This assumes there is no crossing over and that the +eyey genotype makes a wild-type fly.

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