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%.
