In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn + and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct + is crossed to a sn+ct male. The F1 flies are testcrossed. What is the map distance between sn and ct? The F2 males are distributed as follows

sn ct

13

sn ct+

36

sn + ct

39

sn + ct+

12

25 mu

Explanation:

When two genes are linked, they show the tendency to get inherited together. Thus, the resulting progeny has more parental combinations than the recombinant ones. Here, snct and sn+ct + are in lowest numbers hence they are the recombinants.

Map distance between two genes = recombination percentage = (Number of recombinants / Total progeny) * 100

= [ (13 + 12) / (13 + 12 + 36 + 39) ] * 100

= (25 / 100) * 100

= 0.25 * 100 = 25%

Since there is 25% recombination, map distance between sn and ct genes is 25 mu or 25 Cm.