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 yields the following progeny: Phenotypes Number Blue shell, long antenna 82 Green shell, short antenna 78 Blue shell, short antenna 37 Green shell, long antenna 43 Total 240 Assuming that the genes are linked, what is the map distance between them in m.u.

Answer 1

The map distance = 33.3 cM

Step-by-step 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 = {
`(Number of recombinants)/(Total offspring)`}*100

= (80/240)*100 = 33.33 %

Thus, the map distance = 33.3 cM