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Explain with the help of… (Class 12 Biology 2025, Q33B)

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BiologyCBSE Class 12 · Board exam 2025, set 57/1/1 · 5 marks · Long answer

CBSE Class 12 Biology 2025: question 33(B)

Set in the CBSE Class 12 board exam of 2025, paper 57/1/1, section E, question 33(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(B) · 5 marks · Long answerMolecular Basis of Inheritance

Explain with the help of well-labelled diagrams how lac operon operates in E. coli :
(i) In presence of an inducer.
(ii) In absence of an inducer.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 57/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

List the genes of the lac operon in order and note which one makes the repressor.

Hint 2

Ask whether the repressor can sit on the operator when lactose is present.

Hint 3

Remember that RNA polymerase can transcribe z, y, a only when the operator is free.

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. Structure of the lac operon: It has a regulator gene i (with its own promoter), a promoter (p), an operator (o) and three structural genes z, y, a. These code for β-galactosidase, permease and transacetylase respectively. The i gene constitutively produces a repressor protein.

    ½ mark

  2. (i) Presence of inducer — step 1: Lactose, or its derivative allolactose, acts as the inducer. Permease lets it enter the cell, where it binds to the repressor and inactivates it.

    ½ mark

  3. (i) Presence of inducer — step 2: The inactivated repressor cannot bind to the operator, so the operator is free.

    ½ mark

  4. (i) Presence of inducer — step 3: RNA polymerase binds the promoter and transcribes z, y, a into a polycistronic mRNA. This is translated into β-galactosidase, permease and transacetylase. β-galactosidase hydrolyses lactose into glucose and galactose. The operon is switched ON (inducible operon).

    ½ mark

  5. Diagram (i), operon ON: Draw DNA in order i – p – o – z – y – a.

    • Show repressor mRNA from the i gene giving the repressor protein.
    • Show the repressor bound to the inducer (lactose), forming an inactive complex that is not attached to the operator.
    • Show RNA polymerase moving from p through z, y, a, producing lac mRNA.
    • Show that mRNA translated into β-galactosidase, permease and transacetylase.

    Label every part.

    1 mark

  6. (ii) Absence of inducer — step 1: The repressor produced by the i gene stays active and binds to the operator region.

    ½ mark

  7. (ii) Absence of inducer — step 2: The bound repressor blocks RNA polymerase, so z, y, a are not transcribed and no enzymes are made. The operon is switched OFF (negative regulation).

    ½ mark

  8. Diagram (ii), operon OFF: Draw the same DNA, i – p – o – z – y – a.

    • Show the repressor protein from the i gene attached to the operator.
    • Show RNA polymerase on the promoter but blocked, with a cross mark meaning no transcription.
    • Show no lac mRNA and no enzymes formed.

    Label every part.

    1 mark

Answer: (i) With an inducer: lactose/allolactose binds and inactivates the repressor, the operator is free, RNA polymerase transcribes z, y, a, and β-galactosidase, permease and transacetylase are made (operon ON). (ii) Without an inducer: the active repressor binds the operator, RNA polymerase is blocked, and there is no transcription (operon OFF).

Where the marks are

  • Components of the lac operon: i, p, o, z, y, a and the repressor (½ mark)
  • Lactose/allolactose acts as the inducer and binds the repressor (½ mark)
  • Inactive repressor cannot bind the operator (½ mark)
  • RNA polymerase transcribes z, y, a, producing the three enzymes (½ mark)
  • Labelled diagram in the presence of inducer (1 mark)
  • In the absence of inducer, the active repressor binds the operator (½ mark)
  • RNA polymerase is blocked, so there is no transcription (½ mark)
  • Labelled diagram in the absence of inducer (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Molecular Basis of Inheritance, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 57/1/1: every question, by section.

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.