CBSE Class 12 Biology sample paper 2026-27: question 5
From the official CBSE sample question paper for 2026-27, section A, question 5, for 1 mark. Try it first; the hints and our step-by-step solution are below.
Q5 · 1 mark · MCQMolecular Basis of Inheritance
The initiation step during the process of transcription in bacteria is shown below. Identify A, B, C and D by selecting the option:
- (A)A-RNA polymerase, B-Promoter, C-Sigma factor, D-DNA helix
- (B)A-Promoter, B-Sigma factor, C-RNA polymerase, D-DNA helix
- (C)A-Promoter, B-RNA polymerase, C-DNA helix, D-Sigma factor
- (D)A-Promoter, B-RNA polymerase, C-Sigma factor, D-DNA helix
Figure in the paper: DNA double helix with 3' and 5' ends labelled on both strands; an oval structure (B) bound near the left end with a smaller circle marked sigma (C) on it; A points to the left DNA region where the strands are separated (yellow); D points to the DNA helix at the right. See it in the official paper.
Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.
Hints
Open one hint at a time and try again before the next.
Hint 1
The small circle marked σ is attached to the large enzyme.
Hint 2
Think about where the enzyme binds to start transcription.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
Transcription initiation in bacteria works as follows: RNA polymerase (B), with its sigma factor (C) attached, binds at the promoter (A), where the strands open. The rest of the molecule is the DNA helix (D).
- 1 mark
This gives A – Promoter, B – RNA polymerase, C – Sigma factor, D – DNA helix. Option (D).
Answer: (D)
Where the marks are
- Correctly labels the promoter, RNA polymerase, sigma factor and DNA helix (1 mark)
Revise the chapter
- Chapter: Molecular Basis of Inheritance (notes, formulas, where marks are lost and every past question)
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