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: Write out the mRNA and mark it in triplets.
The given 36-nucleotide mRNA is:
5'-AACACCAUGACCCAUGUGGCGAGACGGUAGUUAAAA-3'
Translation stops at UGA, UAG or UAA in the cytosol, and additionally at AGA in the mitochondria (so in mitochondria, UGA, UAG, UAA and AGA all act as stop codons). We must scan the sequence for every AUG start codon, since translation can begin at any AUG, then read forward in that same frame until a stop codon of the compartment being considered is reached.

Step 2: Locate every AUG start codon in the sequence.
Numbering the bases 1 to 36, there are exactly two AUG triplets: one at position 7-9, and another at position 14-16. These lie in two different reading frames, because \(14-7=7\), which is not a multiple of 3.

Step 3: Read the frame starting at position 7 (first AUG).
Grouping in triplets from position 7: AUG(7-9), ACC(10-12), CAU(13-15), GUG(16-18), GCG(19-21), AGA(22-24), CGG(25-27), UAG(28-30).
In the cytosol, AGA (22-24) is a normal sense codon (Arg), so the ribosome reads straight through it and only stops at UAG (28-30). The amino acids made are Met, Thr, His, Val, Ala, Arg, Arg, which is 7 amino acids before the stop. This matches option (A), 7 amino acids long in the cytosol.
In the mitochondria, AGA (22-24) is also a stop codon. Reading the same frame, the ribosome only gets through AUG, ACC, CAU, GUG, GCG (5 codons = 5 amino acids: Met, Thr, His, Val, Ala) before it hits AGA and stops, giving a 5 amino acid peptide from this start site in the mitochondria.

Step 4: Read the frame starting at position 14 (second AUG).
Grouping in triplets from position 14: AUG(14-16), UGG(17-19), CGA(20-22), GAC(23-25), GGU(26-28), AGU(29-31), UAA(32-34).
Codon 20-22 is CGA, not AGA (the letters are in a different order), so this frame never produces the special mitochondrial stop codon AGA at all. Reading in both the cytosol and the mitochondria therefore proceeds the same way here, stopping only at UAA (32-34), a universal stop codon. The amino acids made are Met, Trp, Arg, Asp, Gly, Ser, which is 6 amino acids before the stop. This matches option (C), 6 amino acids long in the mitochondria.

Step 5: Decide which candidates count as the "longest possible" polypeptide(s).
The question asks for the longest possible polypeptide(s), so for each compartment we take the maximum length achievable over all valid start sites, not every length that is technically possible.
In the cytosol: start 7 gives 7 amino acids; start 14 gives 6 amino acids. The longest in the cytosol is 7 amino acids, matching option (A).
In the mitochondria: start 7 gives only 5 amino acids (cut short by the AGA stop); start 14 gives 6 amino acids (no AGA stop in this frame). Since 6 is longer than 5, the longest possible peptide in the mitochondria is 6 amino acids, matching option (C), not the shorter 5 amino acid product of option (B).
Option (D), 8 amino acids in the cytosol, does not occur in either reading frame, so it is ruled out.

Final Answer:
The longest possible polypeptides are 7 amino acids in the cytosol (option A) and 6 amino acids in the mitochondria (option C).
\[ \boxed{\text{Options (A) and (C)}} \]