Reading a coding sequence: For the RNA 5′-GAU GGU UGA UGU-3′, how many amino acids are translated before termination (standard code, single reading frame)?
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AOne
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BTwo
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CThree
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DFour
Answer
Correct Answer: Two
Explanation
Introduction / Context:Decoding an mRNA requires grouping nucleotides into codons and applying the standard genetic code, including recognition of stop codons that terminate translation.
Given Data / Assumptions:
- Sequence is 5′-GAU GGU UGA UGU-3′.
- Translation starts at the first codon shown and proceeds in-frame.
- Standard genetic code applies; no reinitiation after a stop.
Concept / Approach:Translate codons left to right: GAU, GGU, UGA, UGU. GAU codes Aspartate; GGU codes Glycine; UGA is a stop signal; translation halts upon encountering UGA, so codons after the stop are not translated.
Step-by-Step Solution:
Group into codons: GAU | GGU | UGA | UGU.Apply code: GAU → Asp; GGU → Gly.Encounter UGA → stop; terminate translation.Count amino acids produced before stop: 2 (Asp, Gly).Verification / Alternative check:Check a codon table: UGA is one of the three termination codons (UAA, UAG, UGA). No further elongation occurs after a stop in a single continuous ORF.
Why Other Options Are Wrong:
- One: ignores the first two sense codons.
- Three or Four: incorrectly translate past the stop codon.
Common Pitfalls:Reading out of frame, forgetting that translation stops at the first in-frame stop, or assuming reinitiation within the same short stretch.
Final Answer:Two