An individual heterozygous for recessive mutations at three

An individual heterozygous for recessive mutations at three genes (AaBbCc) was test-crossed and the progeny phenotypes are listed below. Construct a linkage map and calculate the interference.

Solution

ab+ 216

++c 178

+bc 60

a++ 57

+b+ 40

a+c 33

+++ 10

abc 6

The individuals with highest number would be parental and lowest would be recombinant.

Interference (I) can be calculated using Coefficient of coincidence (C)

I = 1- C

where C = Observed number of double cross over / Expected number of double cross over.