First slide

Inheritance of Two Genes

Question

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?

Moderate

A

47 cM
B

21 cM
C

2.1 cM
D

80 cM

Solution

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 = $\frac{Number of recombinants (80)}{Total number of off spring (380)} X 100 = 21 %$
Distance between gene a and gene b is 21cM.

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