Question

Fragile x syndrome is an X-linked Dominant trait. A woman who is Heterozygous for this trait marries a normal man.

What are the chances their child has the disorder?

Answer 1

Fragile X syndrome is an X-linked dominant disorder, which means that the gene responsible for the condition is located on the X chromosome. A woman who is heterozygous for this trait has one normal X chromosome (X^N) and one X chromosome with the mutation (X^F), while a normal man has one normal X chromosome (X^N) and one Y chromosome (Y).

Here are the possible combinations of gametes (sperm and eggs) that can result from this union:

- The mother (heterozygous): X^N (normal) or X^F (fragile X).
- The father (normal): X^N (normal) or Y (normal).

Using a Punnett square, we can figure out the genotypes of their potential children:

1. **Mother's X^N and Father's X^N:** Child will have X^N X^N (normal female).
2. **Mother's X^N and Father's Y:** Child will have X^N Y (normal male).
3. **Mother's X^F and Father's X^N:** Child will have X^F X^N (normal female but carrier).
4. **Mother's X^F and Father's Y:** Child will have X^F Y (affected male).

The potential outcomes for each child are:

- **X^N X^N:** Normal female (no disorder)
- **X^N Y:** Normal male (no disorder)
- **X^F X^N:** Normal female, carrier (no disorder)
- **X^F Y:** Affected male (has disorder)

From these combinations, the children can be classified as follows:

- 50% chance of a normal female (X^N X^N or X^F X^N)
- 25% chance of an affected male (X^F Y)
- 25% chance of a normal male (X^N Y)

Therefore, the chance that their child has Fragile X syndrome is **25%**.