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.

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Does anyone know if there is a list of sex linked diseases we are meant to remember or are we to always assume the disease is non sex linked unless the question specifies?

No need to memorize any disease, they will mention it in the text if needed.

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hey Ari, a little confused because i assumed since CF is autosomal recessive, and not sex linked the probability of child’s sex does not matter?

Question asks for the childs sex! Asks for the probability the child is a boy on top of the child not having cystic fibrosis

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