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Two linked genes, A and B, are separated by 20 cM. A man with genotype AB/ab marries a woman who is ab/ab. What is the probability that their first child will be Ab/ab?

Respuesta :

Answer:

10%

Explanation:

Given:

man  AB/ab

woman ab/ab

gene distance  20 cM

Genetic distances within a given linkage group  are measured in centiMorgans.

Children will be AB/ab, Ab/ab, aB/ab and ab/ab

Here , as given gene distance is 20 cM, so the recombination frequency is 20%.

So the frequency of recombinant progeny is 20 %.

The the probability that their first child will be  Ab/ab is  10%.

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