The recessive mutations b (black body color), st (scarlet eye color), and hk (hooked bristles) identify ....?

The recessive mutations b (black body color), st (scarlet eye color), and hk (hooked bristles) identify three autosomal genes in D. melanogaster. The following progeny were obtained from a testcross of females heterozygous for all three genes.





black, scarlet 243


black 241


black, hooked 15


black, hooked, scarlet 10


hooked 235


hooked, scarlet 226


scarlet 12


wild-type 18


What conclusions are possible concerning the linkage relationship of these three genes? What was the genotype of the female heterozygote? Calculate any appropriate map distances. If you could explain how you came to your answers, I would appreciate it. ThanksThe recessive mutations b (black body color), st (scarlet eye color), and hk (hooked bristles) identify ....?
So I've been pretty much working on this since the question was posted like two days ago, and I'm not sure if I've got it, but here goes...





If you were to assume that all three genes were on the same chromosome and tried to map them, you would get really weird recombinant frequencies (often over 50%). RFs can never be more than 50%, so something is wrong. If you look at the four phenotypes that occurred frequently (black-scarlet, black, hooked, and hooked-scarlet) you will notice that these phenotypes either have the recesive black gene with the WT hooked gene (b hk+), or the WT black gene with the recessive hooked gene (b+ hk). It is only the scarlet gene that changes. I therefore assumed that the black and hooked genes were linked, while the scarlet gene was on a separate chromosome and did not affect recombinant frequencies.





If you take the remaining four phenotypes, you can simply add up all the numbers to get the frequency of recombinants for b-hk (I think) because the scarlet gene doesn't matter. You therefore get a frequency of crossing over of 55/1000 or 5.5%. This equals a 5.5 map unit distance between the black and the hooked genes.





The female was therefore b hk+/b+ hk ; st/st+. I am not sure if it matters which parent the female got her recessive st gene from (i.e. did she get b hk+ st from one parent and b+ hk st+ from the other, or b hk+ st+ from one parent and b+ hk st from the other).