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Old 03-06-2011
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DNA NBME 1 block 4 qn 17

The nucleotide sequence encoding four amino acids near the amino terminus of a 10- kd protein is 5- ACT GAT TGC GTT -3. This sequence is mutated by insertion of a single nucleotide (A) in the second codon and becomes 5- ACT GAA TTG CGT -3. Which of the following consequences of this mutation is most likely?

A ) Altered cell-cycle checkpoint

B ) Decreased rate of transcription

C ) Protein truncation

D ) Single amino acid substitution
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Old 03-06-2011
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I'm thinking either A or D.
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Old 03-07-2011
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Quote:
Originally Posted by rahul View Post
I'm thinking either A or D.
Yes, me too, on D, its a singe point mutation but of the same (corresponding) nucleotide pair.
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Old 03-07-2011
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It cannot be D, because we are adding a new nucleotide (A) leading to change in the downstream sequence, a Frame shift mutation.
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Old 03-07-2011
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Originally Posted by doc_study View Post
It cannot be D, because we are adding a new nucleotide (A) leading to change in the downstream sequence, a Frame shift mutation.
Yes, you are right, I just noticed it was a frame shift mutation
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Old 03-07-2011
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Default C.

I would choose C - a frameshift mutation is likely to result in a premature stop codon.

I can see why A would be tempting, but I think that a mutation means that the proofreading of the replication machinery has already failed, and I don't know if a single nucleotide's difference would affect the G2 checkpoint.
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Old 03-07-2011
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C.... Frame Shift mutation, addition or deletion of a single or two nucleotide at a time can alter the sequence down stream. Eventually during translation a stop codon is encountered prematurely that will lead to truncated protein which will be dysfunctional and eliminated by protein degradation machinery. One example of deletion of a nucleotide is Cystic Fibrosis.
Altered cell cycle check point can be= mutation in Rb gene or P53 gene - i am guessing ; so not the choice here.
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Old 03-07-2011
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The ans key says that choice C) Protein truncation is the correct one.

What I know about Protein truncation is that a premature termination of translation and whatever the proteins made are dysfunctional.

In the qn above, taking about an insertion of a single nucleotide into the functional codon. I know the fact that an insertion or deletion of nucleotide bases in a functional mRNA (originally from DNA transcription) will result in frame shift mutation, that may lead to longer (trinucleotide repeats) or shorter (stop codon) protein production.

In this case we are dealing with an insertion of a single nucleotide.

Before insertion 5- ACT GAT TGC GTT -3

After insertion 5- ACT GAA TTG CGT -3

if we translate this codon into a simple mRNA we'll get ...

5' - ACU GAU UGC GUU - 3' as we see the new codons, there is no stop codons, which means the translation will not stop, hence no truncation.

So my conclusion is choice C) can not be the correct ans.

what do you guys think????????
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Old 03-07-2011
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Quote:
Originally Posted by aktorque View Post
The ans key says that choice C) Protein truncation is the correct one. What I know about Protein truncation is that a premature termination of translation and whatever the proteins made are dysfunctional.
In the qn above, taking about an insertion of a single nucleotide into the functional codon. I know the fact that an insertion or deletion of nucleotide bases in a functional mRNA (originally from DNA transcription) will result in frame shift mutation, that may lead to longer ... or shorter (stop codon) protein production.

In this case we are dealing with an insertion of a single nucleotide.
Before insertion 5- ACT GAT TGC GTT -3
After insertion 5- ACT GAATTG CGT -3
if we translate this codon into a simple mRNA we'll get ...
5' - ACU GAU UGC GUU - 3' as we see the new codons, there is no stop codons, which means the translation will not stop, hence no truncation.
So my conclusion is choice C) can not be the correct ans.
what do you guys think????????
Well, it's not just one nucleotide - it's a frameshift near the N-terminus of a the sequence encoding a 10-kD protein. The polypeptide will be a little less than 100 AA's long, so the coding region of the mRNA is going to be a little less than 300 nucleotides long - the 12 bp part given in the question would only produce 4 peptides, so would be a little over 0.4 kD; it's not the whole thing. So even though we haven't seen a stop codon yet, the probability is still great that one will arise.

We learn as biological canon that frameshift mutations usually result in truncated and nonfunctioning proteins - even counting the fact that some won't ever be translated (inhibitory UTR's), frameshifts are only within one exon (the spliceosome doesn't count in threes), and many will misfold and remain in the ER (UPR). I think that sometimes we know so much stuff that we out-think ourselves and make easy questions hard. In the case of this question, I think they are just testing the detail of our knowledge of the basics, and the best thing is just to follow the basic maxim: Frameshift mutation usually results in early termination.

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Old 03-08-2011
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Thank you very much Mondoshawan. I think I was little exaggerated my thinking toward the qn. I didn't realize that they were just testing the simple fact. I knew that frame shift mutation is simply a premature termination and result in protein truncation. Straight forward qn...
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