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* nbme 7 block 4 Q 34
  student27 - 12/08/10 22:40
  Hi everyone... please help in this. I have taken NBME 7. Can please anyone explain what is the answer of the following question.

A 40 year old man is at risk for an autosomal dominant neurodegenerative disorder caused by Trinucleotide repeat expansion. Molecular diagnostic analysis of his DNA shows a normal allele and an allele with expansion of codon CAG encoding glutamine within exon 2 of the responsible gene as shown. Risk estimates for various CAG repeat numbers are shown in the graph Based on this information. Which of the following is the best estimate of the likelihood that this man will develop this disorder?
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* Re:nbme 7 block 4 Q 34
  littlebuddha - 12/13/10 01:05
  I go for 100 % .

The question stem says that this man is at risk for developing an autosomal dominant disease associated with CAG repeat (could be huntington or could be another CAG repeat diseases such as spinocerebellar ataxia ).
A person is said to be at risk for an autosomal dominant disease if he/she has a positive family history (unless it is a case of new mutation). The person in the question got the repeat from the parent who had the disease with the 15 CAG repeat or less. so he will also develop it.
it is true that the penetrance depends on the number of the trinucleotide repeat. for example for Huntington, 6-34 repeat is normal, 35-39 repeat will have variable penetrance and 40 or more will have 100 % penetrance.

60% does not seem to be the answer because the 15 repeat corresponds to 70% in the graph that is not in the choices. so it must be disease that has 100% penetrance even with 15 repeats.

anyone who did the question online, please tell what is the correct answer...

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* Re:nbme 7 block 4 Q 34
  sarim - 12/13/10 01:41
  60% is the BEST correct answer.  
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* Re:nbme 7 block 4 Q 34
  einst74 - 12/13/10 21:01
  Huntington's Disease is an AD disease which means that an affected individual TYPICALLY inherits a defective gene from an AFFECTED parent. If the parent has a trinucleotide repeat count that is normal (40 (full penetrance).

Back to the question. As littlebuddha asserted above it can be any trinuc expansion disorder. But used Hunt. disease to illustrate some points. First, risk doesn't necessarily reflect that the parent has the disorder but the parent can be an unaffected carrier with trinuc expansions that will not result in full penetrance. Second, the graph is very important. It shows that individuals inheriting a gene with a trinuc expan. of 20 have a 100% risk of the disease (full penetrance), as well as those with CAG repeats of 9 or less have 0% risk. The curve represents reduced penetrance (where some individuals fail to exhibit the trait even though they carry the abnormal allele. As the number of CAG repeats increase there is an increase in the percentage of individuals at risk of the disease. This implies that penetrance is increasing up to point where it becomes full or complete (all individuals who have the abnormal allele will manifest (signs/symptoms) the disease. The answer that best fits the curve is 60%
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