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
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 : (A+/A-) [healthy] x (A+/A-) [healthy] → F1
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 :
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.