Which of the following statements regarding the below mentioned mRNA sequence is/are TRUE?
5-UGAUGAGCCUUAACCGGGAACGAAUUUAAG-3
It contains nine codons in the reading frame
It contains ten codons in the reading frame
It codes for eight amino acids
It codes for nine amino acids
ANSWER
(A) and (C). The reading frame starts at the first AUG (nucleotides 3–5) and ends at the in-frame stop codon UAA. That is 9 codons (AUG AGC CUU AAC CGG GAA CGA AUU UAA), and they code for 8 amino acids: Met-Ser-Leu-Asn-Arg-Glu-Arg-Ile.
Official answer: A, CSource: Official answer key, GATE 2024 BT (IISc Bengaluru)
Number the 30 nucleotides of 5'-UGAUGAGCCUUAACCGGGAACGAAUUUAAG-3' from the 5' end. Translation does not start at nucleotide 1. The reading frame is set by the start codon AUG, and the ribosome begins at the first AUG it finds scanning from the 5' end.
Look for AUG from the 5' end. Nucleotides 1–3 are UGA and 2–4 are GAU. Nucleotides 3–5 are AUG (U-G-A-U-G-A…), so the first AUG is at 3–5 and it defines the reading frame.
Read triplets from nucleotide 3: AUG | AGC | CUU | AAC | CGG | GAA | CGA | AUU | UAA, which leaves one G (nucleotide 30) unread.
Decode each codon with the standard genetic code: AUG = Met (start), AGC = Ser, CUU = Leu, AAC = Asn, CGG = Arg, GAA = Glu, CGA = Arg, AUU = Ile, UAA = stop (ochre).
The reading frame runs from the start codon AUG to the in-frame stop codon UAA, which gives codons. Option (A) 'nine codons in the reading frame' is TRUE. Option (B) 'ten codons' is FALSE, because ten codons come only from cutting the 30 nucleotides into triplets from nucleotide 1. That frame (UGA UGA GCC …) begins with a stop codon and has no start codon, so it is not a reading frame.
A stop codon has no tRNA and specifies no amino acid; release factors read it. Only the 8 sense codons before UAA add amino acids, giving the peptide Met-Ser-Leu-Asn-Arg-Glu-Arg-Ile. Option (C) 'eight amino acids' is TRUE and option (D) 'nine amino acids' is FALSE.
The correct options are (A) and (C).
KEY IDEA
A reading frame (open reading frame) is defined by the start codon AUG, not by the first nucleotide of the sequence. Find the first AUG from the 5' end, read triplets from there up to the first in-frame stop codon, and count the stop codon as a codon but not as an amino acid.
WATCH OUT FOR
The usual error is chopping the sequence into triplets from nucleotide 1 (UGA UGA GCC …), which gives 10 codons and suggests option (B). That frame starts with a stop codon and has no AUG, so it is not a reading frame. The second trap is counting the stop codon UAA as an amino acid, which gives 9 amino acids instead of 8.
From the GATE 2024 BT paper. Every question of this paper · Download the full paper · All 21 Gene expression questions, with notes
