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QUESTION 33 What amino acid sequence would be found in the polypeptide produced from this mRNA? Follow the normal rules for g
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Answer #1

The given mRNA sequence is...

5' - A G C A A U A U G G C G A U U C G C U G A G U A C C U - 3'

According to the normal rules of gene expression, the the transcription will begin from the start codon and will end in the stop codon. Means, the first amino acid incorporated to the polypeptide chain will be carried by the tRNA which has the anticodon to the start codon. And there's no tRNA with the complementary anticodon to stop codon. So, there the elongation of polypeptide chain will stop and the translation will ultimately terminate. AUG, GUG, UUG are the start codons, among which the AUG is most abundant which codes for methionine in eukaryotes and formylated methionine in prokaryotes. UAA, UAG and UGA are the stop codons.

Following the above rules of gene expression, the translated polypeptide chain from the above alluded mRNA will have the sequence as following...

Met (strat) - Ala - Ile - Arg - (stop)

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