An unaffected man (XBY) marries a woman who is a carrier for cystic fibrosis (XBXb). What is the probability that the daughters will be carriers of cystic fibrosis?
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To determine the probability that the daughters will be carriers of cystic fibrosis, we need to consider the genetic inheritance pattern of cystic fibrosis.
Cystic fibrosis is an autosomal recessive disorder, which means that both copies of the gene need to be mutated for the disease to be present. The unaffected man (XBY) only has one unaffected copy of the gene (Xb) and he will pass either a Y or Xb chromosome to his daughters. The carrier woman (XBXb) has one normal copy of the gene (XB) and one mutated copy (Xb), and she will pass either an XB or Xb chromosome to her daughters.
In order for the daughters to be carriers of cystic fibrosis, they would need to inherit the mutated Xb chromosome from their carrier mother since their unaffected father does not have the mutated gene. There are 2 possibilities for the genotype of the daughters: XbY and XbXb.
Let's calculate the probability of each possible outcome:
- Probability of receiving the Xb chromosome from the father (XbY): 1/2
- Probability of receiving the Xb chromosome from the mother (XbXb): 1/2
The probability of the daughters being carriers of cystic fibrosis is the sum of the probabilities of each possible outcome:
1/2 + 1/2 = 1
Therefore, the probability that the daughters will be carriers of cystic fibrosis is 1, or 100%.