A 30‑year‑old woman undergoes predictive genetic testing for Huntington’s disease (HD) because her father was affected. She is currently asymptomatic. Her test reveals 38 CAG repeats in the HD gene.

Based on the lecture material, which statement best reflects this result?

Question

A 30‑year‑old woman undergoes predictive genetic testing for Huntington’s disease (HD) because her father was affected. She is currently asymptomatic. Her test reveals 38 CAG repeats in the HD gene.

Based on the lecture material, which statement best reflects this result?

BlackTom AI · Accepted Answer

An HTT allele with 38 CAG repeats falls into the 36–39 repeat “reduced (incomplete) penetrance” category, so clinical outcome is uncertain and offspring remain at risk.

1) Relevant knowledge points and concepts
- Huntington disease (HD) is caused by expanded CAG repeats in the HTT gene.
- Repeat categories: normal ~9–35 (no disease), reduced-penetrance 36–39 (may or may not develop symptoms), full-penetrance ≥40 (very likely to develop HD).
- HD is autosomal dominant: each child of a carrier has a 50% chance of inheriting the allele.
- CAG repeats are unstable in transmission; expansions (especially via paternal transmission) can change repeat length between generations.

2) Step-by-step reasoning
- The patient’s test: 38 CAG repeats.
- Map 38 into the established categories: 36–39 = reduced penetrance.
- Reduced penetrance implies some carriers develop symptoms (often later onset or variable severity) while others remain asymptomatic.

3) Calculations / intermediate steps
- Inheritance probability: autosomal dominant → each child has 1/2 (50%) probability of inheriting the 38-repeat allele.
- No simple numeric lifetime-risk percentage is universally assigned for 36–39 repeats; clinical expression is variable and age-dependent, so the result is probabilistic, not absolute.

4) Why the correct answer is correct
- Option C correctly states that 38 repeats lie in the 36–39 reduced-penetrance range, so the woman may or may not develop HD and her children are at risk (50% transmission probability); repeat instability can alter offspring risk/severity.

5) Why other options are incorrect
- A is wrong: ≥36 does not uniformly mean full penetrance—36–39 is reduced penetrance; only ≥40 is considered full penetrance.
- B is wrong: 38 is outside the normal 9–35 range.
- D is wrong: reduced-penetrance alleles are transmitted; they do confer risk to offspring.

6) Conclusion
The correct interpretation is option C: she has a reduced-penetrance allele (36–39 repeats), so she may or may not develop symptoms, but her children are at risk. This matches the provided answer.

A 30‑year‑old woman undergoes predictive genetic testing for Huntington’s disease (HD) because her father was affected. She is currently asymptomatic. Her test reveals 38 CAG repeats in the HD gene. Based on the lecture material, which statement best reflects this result?单项选择题

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