An individual who is heterozygous for two linked genes (with alleles A, a and B, b) is crossed with an AB/ab individual, and among the progeny are: 14 AB/ab 36 Ab/ab 34 aB/ab 16 ab/ab What is the frequency of recombination

The correct answer is - 0.30

Explanation:

In case of the same alleles, there will be only one type of gametes as there is no repulsion takes place while in heterozygous 4 types of gametes will form due to repulsion.

Always less number shows recombinant. A higher number shows parental. Since the linkage is possible just when removing between the gens is under 50 So, the number of recombinants are:

16+14 = 30

then:

Recombination recurrence = number of recombinants/Total offspring

Recombination Frequency = 30/100 = 0.30

Thus, the correct answer is - 0.30

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