Two linked genes, (A) and (B), are separated by 18 m.u. A man with genotype Aa Bb marries a woman who is aa bb. The man's father was AA BB. What is the probability that their first two children will both be ab/ab?
A) 0.168
B) 0.0081
C) 0.032
D) 0.062
E) 0.13

Question

Quizplus · Accepted Answer

The correct answer is A.The probability of the first child being ab/ab is 0.084. The probability of the second child being ab/ab is also 0.084. The probability of both children being ab/ab is 0.084 * 0.084 = 0.007056.

Two Linked Genes, (A) and (B), Are Separated by 18