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2 answers:
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Answer:
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:
=
=
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= 0.3