A heterozygous individual ++/ab is crossed to its recessive parent and has produced the offsprings in following proportion ++/ab – 200a+/ab – 50+b/ab – 30ab/ab – 100What 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 – 200Recombinant genotypes - a+/ab – 50Recombinant genotypes - +b/ab – 30Parental genotypes - ab/ab – 100Recombination frequency =  Number of recombinants (80)Total number of off spring(380) X 100 = 21 % Distance between gene a and gene b is 21cM.