Genes cut (ct) and dusky (dy) are X-linked in the fruit fly Drosophila melanogas

Question

Genes cut (ct) and dusky (dy) are X-linked in the fruit fly Drosophila melanogaster. The normal wing-shape allele (ct+) is dominant to the cut-wing allele (ct-). The transparent-wing-color allele (dy+) is dominant to the dusky-wing allele (dy-). An F1 female fruit fly with long transparent wings (ct+dy+/ct-dy-) is mated to a male with cut and dusky wings (ct- dy-/Y). Assume that the two genes (ct and dy) segregate independently in the mother, that there is no recombination between the X and the Y chromosomes in the father and that 200 offspring are observed. Fill in the following table with the expected numbers of each genotype and phenotype among the male and female offspring of the cross.

Female Genotype

Expected Number

Male Genotype

Expected Number

Female Phenotype

Expected Number

Male Phenotype

Expected Number

I've been trying to figure it out with the class notes and articles on the internet for two days, but so far haven't had any luck. I think it could be figured out with a punnett square, but I'm not sure how to do one with two x-linked traits. I know the male is only supposed to give one gene or set of genes in this case I suppose since there are two. I think there is also usually a somewhat significant difference in the number of each genotype/phenotype based on gender. Other than that I'm completely lost. We're supposed to fill out the chart above. Any help would be greatly appreciated. Thanks!

Female Genotype

Expected Number

Male Genotype

Expected Number

Female Phenotype

Expected Number

Male Phenotype

Expected Number

Explanation / Answer

F1 female fruit fly with long transparent wings (ct+dy+/ct-dy-) is mated to a male with cut and dusky wings (ct- dy-/Y). X ct+dy+ X ct-dy-   x      X ct- dy- Y The progeny are X ct+dy+ X ct-dy- = Normal and transparent wings female. X ct+dy+ Y = Normal and transparent wings male X ct-dy- X ct-dy- = Cut and dusky winged female. X ct-dy- Y = Cut and dusky winged male. Female genotype = X ct+dy+ X ct-dy-     = expected number = 50, as there are 200 offsprings.                               =X ct-dy- X ct-dy         = expected number = 50, as there are 200 offsprings Male genotype = X ct+dy+ Y   = expected number = 50, as there are 200 offsprings.                           = X ct-dy- Y     = expected number = 50, as there are 200 offsprings. Female phenotype = Normal and transparent wings female and cut and dusky winged female are 100 in number. Male phenotype = Normal and transparent wings male and cut and dusky winged male are 100 in number. These are the expected numbers and they may vary with th eobserved phenotypes.

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