Question

Messenger RNAs (mRNAs) translate to generate polypeptides and the translation terminates at the UGA, UAG or UAA stop codons. The AGA codon acts as additional stop codon in the mitochondria. The longest possible polypeptide(s) that can be hypothetically translated in the cytosol and/or mitochondria, from the given full length mRNA sequence (36 nucleotides) is/are

5'-AACACCAUGACCCAUGUGGCGAGACGGUAGUUAAAA-3'

• A. 7 amino acids long in the cytosol
• B. 5 amino acids long in the mitochondria
• C. 6 amino acids long in the mitochondria
• D. 8 amino acids long in the cytosol

Hint:

Find every AUG, read forward in that frame to the first stop codon under each compartment's rules, then keep only the longest result per compartment.

Answer 1

The Correct Option is A, C

Step 1: Understanding the Question:
We have one 36-base mRNA and must find, for each possible starting AUG, how many amino acids get made before a stop codon is hit, separately for the cytosol (stops: UGA, UAG, UAA) and the mitochondria (same three stops, plus AGA). We then keep only the longest result in each compartment, since the question asks for the longest possible polypeptide(s).

Step 2: Key Formula or Approach:
List the full sequence with position numbers, find every AUG, and for each one write out the codon table for that reading frame until the first stop codon (checking the cytosol rule first, then the mitochondria rule).

Step 3: Detailed Explanation:
Sequence: $5'\text{-AACACCAUGACCCAUGUGGCGAGACGGUAGUUAAAA-3}'$, positions 1 to 36.
Scanning for AUG, position 7 (A-U-G at 7,8,9) and position 14 (A-U-G at 14,15,16) are the only two start codons present.
Frame from position 7: AUG-ACC-CAU-GUG-GCG-AGA-CGG-UAG. Codon 6 in this frame is AGA (positions 22-24).
$\quad$ Cytosol: AGA is not a stop there, so translation runs to codon 8, UAG, giving $7$ amino acids (Met-Thr-His-Val-Ala-Arg-Arg).
$\quad$ Mitochondria: AGA is a stop here, so translation halts right after codon 5 (Ala), giving only $5$ amino acids.
Frame from position 14: AUG-UGG-CGA-GAC-GGU-AGU-UAA. Codon 3 in this frame is CGA (positions 20-22), which is not AGA (the base order is different), so this frame contains no mitochondrial-specific stop at all.
$\quad$ Both cytosol and mitochondria: translation runs until codon 7, UAA, giving $6$ amino acids (Met-Trp-Arg-Asp-Gly-Ser), the same result in either compartment.

Step 4: Final Answer:
Collecting the largest value per compartment: cytosol's best is $7$ amino acids (from the position-7 start), and mitochondria's best is $6$ amino acids (from the position-14 start, since the position-7 start is cut down to only $5$ amino acids by the extra AGA stop there). So the longest possible polypeptides are $7$ amino acids in the cytosol and $6$ amino acids in the mitochondria, which are options (A) and (C), i.e. answer $1,3$.