What is the maximum number of amino acids that could result from the following mRNA sequence?...

What is the maximum number of amino acids that could result from the following mRNA sequence?

5′ AUGAGACCGUCG 3′

A. 0

B. 4

C. 7

D. 10

Expert Solution

The maximum number of amino acids that could result from the sequence of mRNA 5' AUGAGACCGUCG 3' is

4 (Option B)

Codons are the genetic codes,represented by sequence of three nitrogenous babses in nucleic acids. Each codon ( thee nitogenous bases) codes for one aminoacid. After transcription the codons in mRNA is responsible for the sequence of aminoacids in the proteins during traslation. AUG is always the start codon which codes for methionine. Protein traslation continues according to the codon sequence in the mRNA from 5' to 3' till a stop codons is encountered . Stop codons include UAG, UGA, UAA which donot have a corresponding amino acid. This terminates the protein translation. In the given mRNA sequence 5' AUGAGACCGUGC 3', grouping the nitrogenous bases in threes starting from 5' end , we obtain four codons - AUG, AGA, CCG and UGC. These codons code for four amino acids- Methionine, Arginine, Proline and Serine respectively.

The option A (0) is wrong because the sequence starts with AUG a start codon so traslation will progress. The other options , C (7) is wrong beause for 7 amino acids, 7 x 3 = 21 bases are required. We are given only 12 bases. Option D) 10 would require 10 x 3 = 30 nitrogenous bases, which is not the case in the given sequence. (we are multiplying by three beacuse for each amino acid there has to be one codon and each codon comprises of three nitrogenous bases). Hence the answer is option B , 4.

gladiator answered 2 years ago

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