Question

A phenotypically normal individual with a sibling that has cystic fibrosis marries a known carrier of the CF mutation. If they have three children, what is the probability that two of their children will be unaffected and one will have cystic fibrosis?

A. 0.116
B. 0.250
C. 0.347

Answer 1

To calculate the probability of having two unaffected children and one child with cystic fibrosis for a couple where one is a known CF carrier and the other is a phenotypically normal but possible carrier, we use the autosomal recessive inheritance pattern and consider possible combinations. The final calculation yields a probability of approximately 28.125%, which is not among the provided answer choices.

Step-by-step explanation:

The question involves calculating the probability of a phenotypically normal person who may be a carrier of the cystic fibrosis (CF) gene, having children with a partner who is a known carrier. Since cystic fibrosis is an autosomal recessive disorder, an individual needs to inherit two copies of the faulty CFTR allele to express the disease.

To find the probability that two children are unaffected and one has cystic fibrosis, we must first determine the genotypes of the parents. If the normal individual's sibling has CF, there is a 2/3 chance that this normal individual is a carrier (as we assume the individual does not have CF, and one out of the three possibilities—being homozygous for the normal allele—is not possible). Therefore, the cross we will consider is between a carrier (Ff) and a known carrier (Ff).

Using the Punnett square, we find there's a 25% chance of having a child with CF (ff) and a 75% chance of having a child who is either a carrier or unaffected (Ff or FF).

The probability of having two unaffected (non-CF) children and one affected child can be found by the following combination: (3! / (2! * 1!)) * (0.75²) * (0.25¹), considering all the possible sequences of birth outcomes. Calculating this, we get 3 * 0.5625 * 0.25 = 0.421875. Since there is a 2/3 chance the normal individual is a carrier, the final probability is 0.421875 * 2/3 which equals approximately 0.28125 or 28.125%.

This was not one of the answer choices provided, indicating a potential difference in the assumptions made regarding the normal individual's carrier status or a simple misunderstanding of the question context.