CBSE Class 12 Biology sample paper 2025-26: question 31(A)
From the official CBSE sample question paper for 2025-26, section E, question 31(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.
Q31(A) · 5 marks · Long answerMolecular Basis of Inheritance
A. Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below.
B. How is transcription a more complex process in eukaryotic cells? Explain the additional processes that a precursor mRNA has to undergo in these organisms.
(For visually impaired students: A. Why are both the strands not copied during transcription? B. How is transcription a more complex process in eukaryotic cells? Explain the additional processes that a precursor mRNA has to undergo in these organisms.)
Figure in the paper: Template strand with bases A T G C A T G C A T A C written in a row, with arrow at left end of the strand. See it in the official paper.
Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.
Hints
Open one hint at a time and try again before the next.
Hint 1
The template strand has 3′→5′ polarity; write the complementary coding strand 5′→3′.
Hint 2
The promoter is placed relative to the coding strand – upstream or downstream?
Hint 3
For eukaryotes, think about how many RNA polymerases there are and what is done to hnRNA.
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.
- 1 mark
A. The template strand reads 3′→5′. The coding strand is complementary to it and runs 5′→3′:
Template: 3′ – A T G C A T G C A T A C – 5′
Coding: 5′ – T A C G T A C G T A T G – 3′The mRNA formed has the coding-strand sequence with U for T: 5′-UACGUACGUAUG-3′.
- 1 mark
Transcription unit:
5′ —[Promoter]— TACGTACGTATG —[Terminator]— 3′ (coding strand)
3′ —[Promoter]— ATGCATGCATAC —[Terminator]— 5′ (template strand)The promoter lies upstream, at the 5′ end of the coding strand, and is where RNA polymerase binds. The terminator lies downstream, at the 3′ end of the coding strand, and ends transcription.
- 1 mark
B. Transcription in eukaryotes uses three RNA polymerases in the nucleus:
- RNA polymerase I transcribes rRNAs (28S, 18S, 5.8S).
- RNA polymerase II transcribes hnRNA, the precursor of mRNA.
- RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs.
- 1 mark
Splicing: The primary transcript (hnRNA) contains both exons (coding) and introns (non-coding). The introns are removed and the exons are joined in a defined order.
- 1 mark
Capping and tailing: In capping, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′ end of hnRNA. In tailing, about 200–300 adenylate residues are added at the 3′ end without a template. The fully processed hnRNA, now mRNA, is transported out of the nucleus for translation.
Answer: A. Coding strand 5′-TACGTACGTATG-3′ with promoter at its 5′ (upstream) end and terminator at its 3′ (downstream) end. B. Three RNA polymerases; hnRNA undergoes splicing, 5′ capping and 3′ poly-A tailing.
Where the marks are
- Correct complementary coding strand with polarity (1 mark)
- Promoter upstream (5′ of coding strand) and terminator downstream (1 mark)
- Three RNA polymerases and their products (1 mark)
- Splicing – removal of introns, joining of exons (1 mark)
- Capping (methyl guanosine triphosphate at 5′) and tailing (poly-A at 3′) (1 mark)
Revise the chapter
- Chapter: Molecular Basis of Inheritance (notes, formulas, where marks are lost and every past question)
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