Question

In Drosophila sable body (sb) and forked bristles (fr) are due to recessive mutations and the genes are linked. These genes are 7 map units (7 m.u) apart. If a true breeding sable body, forked bristles females are crossed a true breeding wild type flies and the F1 wild type looking hybrid females are crossed to wild type looking F1 hybrid males, what fraction (you can use a fraction or percentage for your answer) of the progeny would be just sable bodied (not any other phenotype, sable body and wild type bristles).

Answer 1

Given that there are two recessive mutations, sb (sable body) and fr (forked bristles).

Homozygous sable body, forked bristles females are crossed a true breeding wild type flies. This can be shown as:

sbsb/frfr × sb+sb+/fr+fr+

F1 would have the genotype sb+sb/fr+fr.

The required cross for F2 would be:

sb+sb/fr+fr × sb+sb/fr+fr

In the progeny, we will get four different phenotype. The non parental types or recombinants would be sable (sbab/fr+_) and forked (sb+_/frfr).

Since the genes are at 7 mu distance, thus the total proportion of recombinants will be 7%.

Half of this would be sable, i.e.

= (1/2) × 7

= 3.5%