Chemistry Question

[Pals_1010]

A simple protein has 88 amino acids in its primary structure. How many bases are required in the mRNA to produce this sequence? Explain your answer [NOV98/IV/3]

 

[HybridStasis]

I think the answer is 264 bases.
Since, 88 amino acids x 3 bases for each codon = 264 bases.

1 amino acid = 1 anti-codon (from tRNA which have 3 bases) + 1 codon (from mRNA which have 3 complementary bases to tRNA)

 

[Pals_1010]

Should we include the start and stop codon as well? That makes it 270 then?

 

[HybridStasis]

My earlier answer is incorrect, It should be 267.

The start codon codes for an amino acid which is methionine. When the polypeptide chain reaches it stop codon, It simply stops and does not code for an amino acid. The question asks for the amount of bases needed to produce the sequence, So the answer should be 88 x 3 = 264 (to clarify that there must be 264 anti-codons) and add it by 3 (to clarify that there must be a stop codon in the mRNA), so the final answer is 267.

It shouldn't be 270 since start codon codes for an amino acid.

 

[Pals_1010]

My earlier answer is incorrect, It should be 267.

The start codon codes for an amino acid which is methionine. When the polypeptide chain reaches it stop codon, It simply stops and does not code for an amino acid. The question asks for the amount of bases needed to produce the sequence, So the answer should be 88 x 3 = 264 (to clarify that there must be 264 anti-codons) and add it by 3 (to clarify that there must be a stop codon in the mRNA), so the final answer is 267.

It shouldn't be 270 since start codon codes for an amino acid.

Thanks a lot That was reallly helpful and clear ^^