A heterozygous individual ++/ab is crossed to its recessive parent and has produced the offsprings in following proportion.

++/ab – 200
a+/ab – 50
+b/ab – 30
ab/ab – 100

What is the expected distance between the two gene loci?

When a heterozygous individual ++/ab is crossed to its recessive parent, the offspring with the following genotypes are formed.

Parental genotypes - ++/ab – 200
Recombinant genotypes - a+/ab – 50
Recombinant genotypes - +b/ab – 30
Parental genotypes - ab/ab – 100

Recombination frequency =  $\frac{\mathrm{Number} \mathrm{of} \mathrm{recombinants} \left(80\right)}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{off} \mathrm{spring}\left(380\right)} \mathrm{X} 100 = 21 \%$ 

Distance between gene a and gene b is 21cM.