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A scientist is… (Class 12 Biology SQP 2026-27, Q32A)

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BiologyCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Biology sample paper 2026-27: question 32(A)

From the official CBSE sample question paper for 2026-27, section E, question 32(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(A) · 5 marks · Long answerCompetency-basedBiotechnology: Principles and Processes

A scientist is screening bacterial transformants engineered with DNA fragments of the β-globin gene. A radioactive DNA probe complementary to the normal β-globin sequence is used. After autoradiography:

  • Plates A and C show strong and week radioactive signals respectively.
  • Plate B shows no radioactivity.

Answer the following:

  1. I.

    Why does plate B fail to show radioactivity, while A and C do?

    2 marks
  2. II.

    How can this experiment help in distinguishing a normal individual, a carrier, and a patient of sickle cell anaemia?

    2 marks
  3. III.

    Name one other commonly used technique (apart from autoradiography) to differentiate recombinants from non-recombinants.

    1 mark

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

A probe gives a signal only where it finds its complementary sequence.

Hint 2

How many normal β-globin alleles does a normal person, a carrier and a patient have?

Hint 3

Think of the screening that uses colour changes from the β-galactosidase gene.

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. I. The radioactive probe is complementary to the normal β-globin sequence. In plate B, the DNA carries the mutated (sickle-cell) β-globin sequence, so the probe cannot hybridise and no radioactivity shows on autoradiography.

    1 mark

  2. Plates A and C contain the normal complementary sequence, so the probe hybridises and gives a signal. In A the signal is strong, with more normal sequence; in C it is weak, with less normal sequence.

    1 mark

  3. II. The strength of probe hybridisation tells the three genotypes apart:

    • Normal individual (HbAHbA): the probe binds fully, giving a strong radioactive signal.
    • Carrier (HbAHbS): both normal and mutant alleles are present, so hybridisation is partial and the signal is weaker or mixed.
    1 mark
    • Patient (HbSHbS): only the mutant sequence is present, so there is no hybridisation and no radioactivity.
    1 mark

  5. III. Insertional inactivation, for example blue–white screening using the lacZ (β-galactosidase) gene: recombinants give white colonies and non-recombinants give blue colonies.

    1 mark

Answer: I. Plate B has the mutant (sickle) sequence, so the probe does not hybridise; A and C have the normal sequence. II. Strong signal = normal; weak/partial = carrier; none = patient. III. Insertional inactivation (blue–white screening).

Where the marks are

  • Plate B carries the mutated sequence, so there is no probe hybridisation (1 mark)
  • A and C have the normal complementary sequence, so the probe binds and gives a signal (1 mark)
  • Normal = strong signal; carrier = partial/weak signal (1 mark)
  • Patient = no hybridisation and no signal (1 mark)
  • Insertional inactivation/blue–white selection (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Biotechnology: Principles and Processes, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: 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.