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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?

Respuesta :

Answer:

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

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