Question

The pedigree above shows a cross between an affected female and unaffected male for Huntington's disease. Huntington's disease is autosomal dominant. In this case, what is the chance that the two parents at the top have a child with Huntington's disease?
Responses
A 0%0%
B 5%5%
C 25%25%
D 50%50%
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The pedigree above shows a cross between an affected female and unaffected male for Huntington's disease. Huntington's disease is autosomal dominant. In this case, what is the chance that the two parents at the top have a child with Huntington's disease?
Responses
A 0%0%
B 5%5%
C 25%25%
D 50%50%
Skip to navigation

Answer 1

In the case of Huntington's disease, which is an autosomal dominant condition, an affected individual (female) will have at least one copy of the dominant allele (let's denote it as \(H\)). The unaffected male will have two copies of the recessive allele (let's denote it as \(hh\)).

The genotypes for the parents would be:
- Affected female: \(Hh\) (since she is affected, she must have at least one dominant allele)
- Unaffected male: \(hh\)

To determine the possible genotypes of the children, we can set up a simple Punnett square:

| | h | h |
|------|------|------|
| **H** | Hh | Hh |
| **h** | hh | hh |

From the Punnett square, we can see the following outcomes for their potential children:
- 50% chance of being \(Hh\) (affected by Huntington's disease)
- 50% chance of being \(hh\) (unaffected)

Thus, the chance that the two parents will have a child with Huntington's disease is **50%**.

The answer is: **D 50%**.