Q.78 The length of a coding region in an mRNA is 897 bases. How many amino acids will be there in the polypeptide synthesized using this mRNA? (A) 297 (B) 298 (C) 299 (D) 897

Q.78 The length of a coding region in an mRNA is 897 bases. How many amino acids will be there in the
polypeptide synthesized using this mRNA?
(A) 297 (B) 298 (C) 299 (D) 897

 Genetic Code Fundamentals

The coding region (CDS) of mRNA contains multiples of 3 nucleotides called codons. Each codon specifies one amino acid during translation at the ribosome.

  • Start codon (AUG = Met) begins translation
  • Stop codons (UAA/UGA/UAG) end translation without coding amino acids
  • “Coding region” length excludes UTRs and stop codons

✅ Correct Answer: (C) 299

897 bases ÷ 3 bases/codon = 299 amino acids

Number of amino acids = 897 / 3 = 299

This assumes perfect triplet reading frame with no stops in the coding sequence definition.

❌ Why Not the Other Options?

Option Calculation Why Wrong
(C) 299 897 ÷ 3 = 299 Correct
(A) 297 891 ÷ 3 ❌ Subtracts 6 bases (2 stop codons) prematurely
(B) 298 894 ÷ 3 ❌ Subtracts 3 bases (1 stop codon) from CDS
(D) 897 897 ÷ 1 ❌ Ignores triplet genetic code (3 bases = 1 amino acid)

📊 Translation Diagram

mRNA: 5′-AUG-XXX-XXX-…-XXX-UAA-3′
Coding: ||||||||||| <– 897 bases = 299 × 3
Protein: Met-AAA-AAA-…-AAA <– 299 amino acidsKey: CDS excludes start counting? NO! AUG codes Methionine (counts as aa #1)
Stop codon NOT counted in CDS length

🎯 Key Exam Concepts

✅ Perfect Score Checklist:

  • 897 divisible by 3 = valid CDS (no remainder)
  • No remainder = perfect reading frame
  • Stop codon NOT counted in CDS length
  • AUG counts as amino acid #1 (Methionine)

🧠 Exam Mnemonic

“897 Triplets = 299 Proteins”
8+9+7=24 → 2+4=6 → 6×50=300 → 300-1=299

Or simply: Divide by 3, pick the answer!

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