In drosophila, a very prolific phenotypically normal female was crossed to a phenotypically normal male. all of the female offspring were normal. the 1,000 male offspring had some combination of abnormal traits, including purple (instead of the normal red) eyes, black (instead of the normal gray) bodies, and short (instead of the normal long) bristles. written using the symbols p for purple, b for black, s for short and for the respective normal alleles, the male offspring were: 321 pb 326 s 102sp b 106+ ++ 69++b 64sp+ 7+p+ 5 s+b what is the map distance between the s and b loci?

22 mu

Explanation:

Since maximum number of flies are observed with +pb and s++ phenotype, they are the parental combinations.

Minimum number of flies are observed with +p+ and s+b phenotype hence they are the result of double crossover.

Gene order would be +bp and s++ since it is the only case which would lead to production of above mentioned double crossover. Hence gene b is in the middle of genes s and p.

Single cross over between genes s and b will give progeny +++ and sbp.

Map distance between s and b loci = recombination frequency =

(number of recombinants/ total progeny)*100

= [(single cross over between s and b + double crossover)/total progeny]*100

= [(102+106+7+5/1000]*100

=(220/1000)*100

=0.22*100

=22 mu

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