Question

In Drosophila, males from a true-breeding stock with raspberry-colored eyes were mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sable-colored bodies. When F1 males and females were mated, the F2 generation was composed of 216 females with wild-type eyes and bodies, 223 females with wild-type eyes and sable bodies, 191 males with wild-type eyes and sable bodies, 188 males with raspberry eyes and wild-type bodies, 23 males with wild-type eyes and bodies, and 27 males with raspberry eyes and sable bodies. Explain these results by diagramming the crosses, and calculate any relevant map distances.

Answer 1

This results are explained by the fact that these two genes are in the X chromosome and both mutations are recessive.

There is a estimated map distance of 5.76 cM.

Step-by-step explanation:

First step is to know what are the initial phenotypes, they say "males from a true-breeding stock" that means that they are homozigous for both genes. Also, they say that in F1 generation, all females had the same phenotype but all males had a different phenotype than females, so this suggest a sex-linked inheritance. In the F1 generation, females had a wild type phenotype, but we know these females had to carry the mutant copies of the two genes, so mutant alleles must be recessive.

Therefore, if r is the allele for raspberry-colored eyes and s for sable-colored bodies, the genotypes of parents should be:

- Males with raspberry-colored eyes (wild type for body color):

`X^(rS)Y^{}`

- Females with sable-colored bodies (wild type for eyes color):

`X^(Rs)X^(Rs)`

If they are crossed, we obtain the results on the Punnet Square.

If there is not recombination, we expect F2 as in Punnter Square, but in the results, there are two phenotypes that were not expected: wild type males (23) and males with raspberry-colored eyes with sable-colored bodies (27). These two should be recombinants. To calculate the distantance between the two genes, we use:

`Distance=(recombinants)/(Total )*100`

`Distance=(23+27)/(868)*100=(50)/(868)*100=5.76 cM`

Therefore, distance between s and r is 5.76 centimorgans.