Answer:
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
B Potential energy! Hope it helps :)
B:-Leukemia is the answer.
