Question

You are doing lab work with a new species of beetle. You have 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 82Green shell, short antenna 78Blue shell, short antenna 37Green shell, long antenna 43Total 240Assuming that the genes are linked, what is the map distance between them in cM?25.0 cM33.3 cM49.5 cMThe genes are actually assorting independently.8.0 cM

Answer 1

33.3 cM

Step-by-step explanation:

Parent 1: blue shell, long antenna : BBLL

Parent 2: green shell, short antenna : bbll

BBLL X bbll :

F1 : BbLl ( blue shell, long antenna )

BbLl X bbll :

BL/bl = 82 : Parental

bl/bl = 78 : Parental

Bl/bl = 37 : Recombinant

bL/bl = 43 : Recombinant

Total = 240

Recombination frequency = (Number of recombinants/ Total offspring)*100

= (80/240) * 100 = 33.33 %

Map distance = recombination frequency

Hence, map distance = 33.3 cM