Question

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

Answer 1

25 mu

Step-by-step 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.