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Q.Assume that the given mRNA (start site is not depicted) is theoretically translated in two reading frames.

(a) Translation starting from the first nucleotide (Reading frame 1)
(b) Translation starting from the second nucleotide (Reading frame 2) Frame 1 5′ – CUCGCUUGCCGAUCAAGGGUUA – 3′ Frame 2 5′ – GUGGCACUCAGUCCUUAAUGGCG – 3′ Answer the following question : How many amino acids will be specified in case
(a) and case
(b) on translation ? Justify your answer.
CBSECBSE Class XII Board 2025Subjective· 2mImportance★★★★★
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Read as triplet codons from the stated starting point, both reading frames give 7 amino acids: Frame 1 (from the 1st nucleotide) has 7 complete codons and no stop codon, and Frame 2 (from the 2nd nucleotide, as instructed) also has 7 complete codons with no stop codon. The genetic code being a triplet code is the key justification.

The genetic code is read as a triplet code: the ribosome groups the mRNA into non-overlapping codons of three nucleotides, beginning at a defined point, and adds one amino acid per sense codon until it meets a stop codon (UAA, UAG or UGA) or the readable sequence ends. The critical instruction here is where each frame begins.

Case (a): Reading frame 1 — start at the first nucleotide

Sequence: 5′–CUCGCUUGCCGAUCAAGGGUUA–3′

Grouping from the first base:

CUC | GCU | UGC | CGA | UCA | AGG | GUU | A

That is seven complete triplets with a single leftover base (A) that cannot form a codon. None of the seven triplets is a stop codon (in particular UCA codes for serine, it is not a stop codon). So all seven are translated → 7 amino acids.

Case (b): Reading frame 2 — start at the SECOND nucleotide

Sequence: 5′–GUGGCACUCAGUCCUUAAUGGCG–3′

The question specifies that translation begins at the second nucleotide, so we skip the first G and start at the U:

UGG | CAC | UCA | GUC | CUU | AAU | GGC | (G left over) …

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