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).
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
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