In Drosophila curved wings is mutant c and normal wings are

In Drosophila curved wings is mutant (c) and normal wings are wild-type (c+) black body (b) is mutant in comparison to the brown body (b+), and scute bristles (sc are mutant in comparison to wild-type bristles (sc+). A female that was heterozygous for each of these X-linked traits was test-crossed with a curved, black, and scute male with the following results Phenotype Number of pro wild-type 340 black 58 67 curved, scute curved, black 74 curved curved, black, scute 351 black, scute 12 scute 89 (A) What is the order of the genes? (2 pts)

Solution

Cross is:

c/+ b/+ sc/+ female crossed to c/c b/b sc/sc male

c b sc = curved black and scute 351 = parental

c+ b+ sc+ = wild type = parental = 340

Curved scute = c/c sc/sc = 67

Curved black   = c/c b/b = 74 = single cross over

Scute = sc/sc = 89 = single cross over

This shows that curved and black are more closed to each other in comparison to scute.

Curved = c/c = 9. This is very less. This means that curved in middle and produced only by double cross over.

So, the gene order becomes

b----------------------c------------------------sc

Map distance between sc-c = (89+74)/1000 = 163/1000 = 16.3 map units

Map distance between c-b = (67 +58)/1000 = 125/1000 = 12.5 map units

Map distance between sc-b= total number of recombinants/1000 = (89+74+67+58+ 9 +12)/1000= 309/1000 = 30.9 map units

b----------12.5---------c--------------16.3------------sc