Question

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?

(1 point)
Responses

25%
25%

100%
100%

0%
0%

50%
50%

Answer 1

To determine the probability that the daughters of a man (XBY) who is unaffected and a woman who is a carrier (XBXb) for cystic fibrosis will be carriers of the disease, we can look at the potential genetic combinations from their union.

The woman's genotype is XBXb (where XB represents the normal allele and Xb represents the cystic fibrosis allele). The man's genotype is XBY.

When they have children, the possible combinations of X and Y chromosomes can be analyzed as follows:

- Daughters receive one X chromosome from each parent.
- The possible combinations for daughters would be:
- XBXB (normal)
- XBXb (carrier)

So, the combinations are:
1. From the mother (XBXb), she can pass on either XB or Xb.
2. From the father (XBY), he always passes on his X chromosome (X is relevant for a daughter since daughters receive X from both parents).

Thus, the potential genotypes for the daughters are:
- XBXB (normal)
- XBXb (carrier)

The breakdown is as follows:
- 50% chance for XBXB (normal)
- 50% chance for XBXb (carrier)

Therefore, the probability that a daughter will be a carrier (XBXb) is **50%**.

The correct response is:
**50%**