Answer:
The map distance = 33.3 cM
Explanation:
Let us take,
BBLL as blue shell, long antenna .........Parent 1
bbll as green shell, short antenna .......... Parent 2
Then, cross between parent 1 and parent 2, which is
BBLL X bbll
The F1 generation will be heterzygous
BbLl as blue shell, long antenna ...... F1 offspring
By the test cross,
BbLl X bbll :
given,
BL/bl = 82 : Parental
bl/bl = 78 : Parental
Bl/bl = 37 : Recombinant
bL/bl = 43 : Recombinant
Total = 240
Recombination frequency = { [tex]\frac{Number of recombinants}{Total offspring}[/tex]}*100
= (80/240)*100 = 33.33 %
Thus, the map distance = 33.3 cM