Question

The genes for sepia eye color, short bristles, and dark body coloration are on the same chromosome of Drosophila melanogaster. Each gene has two alleles: wild type, which is dominant, or mutant, which is recessive.

            se+ is dominant and causes red eyes; se is recessive and causes sepia eyes

            sb+ is dominant and causes long bristles; sb is recessive and causes short bristles

            b+ is dominant and causes gray body coloration; b is recessive and causes dark body coloration

The sepia gene and short bristle gene are separated by 25 map units; the short bristle gene and the dark bodygene are separated by 12 map units. The sepia gene and dark body gene are separated by (25 + 12)map units. (please add 25 and 12 together to get the distance; Canvas won't do the addition itself).

If you crossed a fly, which was homozygous for the wild type allele of each gene with a homozygous recessive fly for each gene, the F₁ generation would entirely consist of flies that had the wild type traits, but were heterozygous for each gene.

If you were to cross these heterozygous F₁ flies to recessive homozygous flies, how many flies would you predict to have the double cross-over phenotype? (You will show all of your work and phenotypic classes in the next problem. Just show the NUMBER of double cross-over flies predicted by the map unit data here. Assume there are 1000 progeny. Round to a whole number)

Answer 1

First we draw the genetic map of the three genes. The genetic map is shown in the below picture.

se 12 m.u 25 m.u

Now, from the genetic map we know that 25% gametes produced by the heterozygous F1 flies at se & sb loci will be recombinants & 12% gametes produced by the heterozygous F1 flies at sb & b loci will be recombinants (As 1 m.u = 1% recombination).

Now, we can find the probability of the double cross-over gametes between se & b by multiplying the probability of recombination between se & sb & the probability of recombination between sb & b (applying multiplication rule of probability).

So, probability of double cross-over = Probability of recombination between se & sb x Probability of recombination between sb & b = 0.25 x 0.12 = 0.03

So, expected number of double cross-over flies = 1000 x 0.03 (As given progeny number = 1000) = 30