Two genes in Drosophila code for body color and eye color Th

Two genes in Drosophila code for body color and eye color. The wild type body trait is gray (b+) and the mutant version is black (b). The wild type eye color is red (p+) and the mutant version is purple. These genes are right next to each other on the same chromosome and for the sake of this question, let\'s assume they never cross over. If you test cross a heterozygote for both genes (p+p/b+b) with a homozygous mutant for both genes (pp/bb), what is the expected ratio of offspring?

I understand how I got 50% gray/ red eyes and 50% black and purple eyes BUT the question I have is:

For the same question above When you test cross a heterozygote for both genes with a homozygous mutant for both genes, you obtain the following progeny:

gray/red: 48

gray/purple: 11

black/red: 13

black/purple: 51

How many map units apart are the genes? Im really having trouble with map units, and coupling/repulsion. Also what would this be? I think it would be coupling but im not 100% sure.

Solution

The recombination frequency is directly proportional to distance between genes. Here the total recombinant progeny is 11+13= 24. So, the frequency of recombinants 24/123 x 100= 19.5
(The total population size is 48+11+13+51= 123). So, the genes are 19.5mu apart.