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Two linked genes, A and B, are separated by 18 cM. A man with genotype AB/ab marries a woman who is ab/ab. The man’s father was AB/AB.a. What is the probability that their first child will be Ab/ab?b. What is the probability that their first two children will both be ab/ab?

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Answer:

a) 9%.  

b) 16.8%.

Explanation:

a).

We are provided with the information that Two linked genes, A and B, are separated by 18 cM (centiMorgan). i.e the recombinant frequency is 18%

Also , the man's genotype is AB/ab... This only result to one explanation, that The man will definitely produce 18% of recombinant gametes which entails

9% Ab & 9% aB

i.e 0.09 Ab & 0.09 aB

On the other-hand, The mother ab/ab have tendency to produce just one single type of gamete which is ab

∴

The probability that their first child will be Ab/ab will be

Pr ( Ab/ab) = (0.09) x (1)

= 0.09

= 9%.  

b).

If the father produces 18% of recombinant gametes which entails

9% Ab & 9% aB , this typically implies that the number of the non-recombinant gametes will be;

100%-18% = 82%  ( non-recombinant gametes)

i.e genotype AB/ab = 82%

AB =41%; ab = 41%

AB = 0.41 ; ab = 0.41

Now, the probability that their first two children will both be ab/ab:

Using Multiplication Rule to calculate the probability that their first two children (ab/ab); we have:

 (0.41)(1) ×(0.41)(1)

= 0.1681

= 16.8%.

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