Question

1. You are doing lab work with a new species of beetle. You isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny: Blue shell, long antenna 92 Green shell, short antenna 88 Blue shell, short antenna 27 Green shell, long antenna 33 Total 240 Assuming that the genes are linked, what is the map distance in cM between them?

   		33 cM
   		47.9 cM
   		25 cM
   		27 cM
   		69 Cm
   		92 cM
   		50.4
   		52 cM
   		46.5 cM
   		88 cM

Answer 1

Recombination frequnacy = no of recombinant / total no of offspring

= 27+33/23+33+92+88

60/240

=0.25

converting into map distnace =0.25x 100

=25cM

Option C