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 fro straight bristles and intact wings, respectively. A female homozygous for sn and ct + is crossed to a sn + ct male. The F1 flies are interbred. The F2 males are distributed as follows:

sn ct 13

sn ct + 36

sn + ct 39

sn + ct + 12

What is the map distance bwteen sn and ct?

The map distance between sn and ct is = 25 m. u or map unit.

Explanation:

Given, sn ct + * sn + ct

In F1 generation the progenies were interbred.

In F2 generation,

sn ct = 13 (recombinant)

sn ct + = 36 (parental)

sn + ct = 39 (parental)

sn + ct + = 12 (recombinant)

Linkage Map distance = (no. of Recombinant progeny/total progeny)

* 100

Recombination frequency = (no. of Recombinant progeny/total progeny)

* 100

= (13 + 12) / 100 * 100 m. u

= 25/100 * 100

= 25 %

∴ The distance between sn and ct = 25 m. u or map unit

∴ Linkage happened between sn and ct as the map distance is much less than 50