Question

An individual who is heterozygous for two linked genes (with alleles A, a and B, b) is crossed with an a b/a b individual, and among the progeny are the following

14 AB/ab
36 Ab/ab
34 aB/ab
16 ab/ab
What is the frequency of recombination?
a) 0.60
b) 0.30
c) 0.40
d) 0.70

Answer 1

b) 0.30

Explanation:

An individual who is heterozygous for two linked genes (with alleles A, a and B, b)

i.e Aa & Bb × ab/ab

This cross produce the following offspring :

14 AB/ab

36 Ab/ab

34 aB/ab

16 ab/ab

Now, the purpose of these calculation is to determine how close together on a gene (gene mapping) these recombinants are.

First, we need to determine the linked gene for the parental chromosomal configurations which are (Aa & Bb × ab/ab) and the recombinant configurations in the offspring which are ( AB/ab & ab/ab). This allows for the calculation of the recombinant frequency; and it is given by:

=
`(number of recombinant frequency)/(total number of offspring)`

=
`(14+16)/(14+36+34+16)`

=
`(30)/(100)`

= 0.3