Questions
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?
detailed solution
Correct option is B
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.Similar Questions
The tendency of genes to be inherited together as a result of their location on same chromosome is
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