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BiologyCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Biology sample paper 2025-26: question 32(B)

From the official CBSE sample question paper for 2025-26, section E, question 32(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(B) · 5 marks · Long answerBiotechnology: Principles and Processes

BamH1 is a restriction enzyme which recognizes the sequence- 5' - GGATCC - 3'. The restriction activity of this enzyme is between G and G.
A. Construct the palindrome for the above sequence.
B. Draw a labeled diagram to show the formation of recombinant DNA (rDNA) using BamH1.
C. PBR322 is a plasmid that has a restriction site for this enzyme at the tetracycline resistant gene. If BamH1 were to be used, how will it impact the response of the transformant with rDNA to antibiotics- ampicillin and tetracycline. Justify.

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

Write the complementary strand and read both strands from 5' to 3'.

Hint 2

Cutting between G and G on both strands leaves single-stranded overhangs. Which enzyme joins the matching ends?

Hint 3

Ask which antibiotic resistance gene is disrupted by the insert and which one remains intact.

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. A. Palindrome: A palindromic sequence reads the same on both strands when each is read in the same direction (5'→3'). The complementary strand of GGATCC, read 5'→3', is also GGATCC.

    5' – G G A T C C – 3'
    3' – C C T A G G – 5'

    1 mark
  2. B. Diagram (described in words): formation of rDNA using BamH1.
    1. Draw the vector DNA (plasmid) and the foreign DNA, each carrying the site 5'-GGATCC-3' / 3'-CCTAGG-5'. Label: vector DNA, foreign DNA, BamH1 recognition site.
    2. Show BamH1 cutting each strand between G and G, giving fragments with single-stranded overhangs 5'-GATC-3'. Label: BamH1 (restriction endonuclease) and sticky ends:

    • 5'–G GATCC–3'
    • 3'–CCTAG G–5'

    3. Show the sticky ends of the foreign DNA fragment pairing by complementary bases with the sticky ends of the cut vector. Label: complementary base pairing.
    4. Show DNA ligase sealing the sugar-phosphate backbone. Label: DNA ligase, recombinant DNA (rDNA).

    2 marks
  3. C. Ampicillin: BamH1 cuts pBR322 inside the tetracycline resistance gene (), so the ampicillin resistance gene () stays intact. The transformant carrying rDNA therefore grows on a medium containing ampicillin.

    1 mark
  4. C. Tetracycline: The foreign DNA inserted into breaks up its coding sequence (insertional inactivation), so tetracycline resistance is lost. The recombinant transformant cannot grow (dies) on a medium containing tetracycline. Non-recombinants grow on both media, which lets recombinants be selected.

    1 mark

Answer: A. 5'-GGATCC-3' / 3'-CCTAGG-5'. B. Labelled diagram: BamH1 cuts vector and foreign DNA leaving GATC sticky ends, which pair and are sealed by DNA ligase to form rDNA. C. The transformant survives on ampicillin ( intact) but dies on tetracycline ( inactivated by the insert).

Where the marks are

  • Writes the palindrome with the complementary strand 3'-CCTAGG-5' and correct polarity (1 mark)
  • Diagram shows BamH1 cutting vector and foreign DNA, sticky ends and their joining (1 mark)
  • Labels: BamH1, sticky ends, DNA ligase, recombinant DNA (1 mark)
  • Transformant resistant to ampicillin because is intact (1 mark)
  • Transformant sensitive to tetracycline because of insertional inactivation of (1 mark)

Revise the chapter

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.