IMAT 2017 Q35 [Cystic Fibrosis]

A mother and a father, neither of whom has cystic fibrosis, conceive a child who has the condition.

What is the likelihood that the same parents will have another child who is a boy without cystic fibrosis?

A. 1 in 4
B. 3 in 4
C. 1 in 2
D. 3 in 8
E. 1 in 8

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The correct answer is D.

Before we start answering this question, we have to know how to write proper formal genetics :

  • A is a gene and can have 2 alleles (versions) : Ax and Ay
  • Because humans are diploid, they will have 2 alleles of the same gene which is represented by (Ax/Ay)
  • Depending on the alleles a subject has, it will either be [healthy] or [sick]
  • The letter F represents the generation we are at, in other words :
    • F0 for the parents
    • F1 for their offspring

In this situation, we will hypothesize that there is only one gene A responsible for cystic fibrosis. It can be found in 2 different versions :

  • A+ which is the healthy and dominant allele
  • A- which is the sick and recessive allele

Because both parents are healthy but ended up giving birth to a sick child, it is safe to assume that the mother and the father both have one healthy allele and a sick allele.

Therefore, we can write the following :

  • F0 : :female_sign: (A+/A-) [healthy] x :male_sign: (A+/A-) [healthy] → F1
F1 A+ A-
A+ (A+/A+) (A+/A+)
A- (A+/A-) (A-/A-)

How do I read that table ? The first row represents the mother’s genes, and the first column the father’s. During sexual reproduction, each parent gives half of their genetic pool to their offspring : each child is a combination of both of its parents’ genes.

  • F1 therefore has :
    • A 1 out of 4 probability of being sick
    • A 3 out of 4 probability of being healthy

But the question doesn’t end here : we want to know the probability of having a boy AND a healthy child. The probability of having a boy is the following :

F1 X X
X (X/X) (X/X)
Y (X/Y) (X/Y)

Therefore, there is a 1 out of 2 probability for the child to be either a boy or a girl, and like we’ve said before, a 3 out of 4 probability of being healthy. We can then determine that the probability of the child to be both a boy and healthy is : (1/2) * (3/4) = 3/8.