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BiologyCBSE Class 12 · Sample paper, session 2023-24 · 4 marks · Case-based

CBSE Class 12 Biology sample paper 2023-24: question 29

From the official CBSE sample question paper for 2023-24, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 4 marks · Case-basedCompetency-basedBiotechnology: Principles and Processes

The structure below shows pUC18 which is similar to pBR322 in its function. However, they differ in some of their restriction sites and number of ori. The ori number for pBR322 is approximately 20.

  1. a)

    How are puc18 and pBR322 used in biotechnological studies?

    0 marks
  2. OR

    a)

    What will be the impact if ori in the above structure gets damaged?

    0 marks
  3. b)

    The lac z gene has many recognition sites. Study the segment of DNA given below and answer the questions
    5’... ATC GTA AAG CTT CAT…3’
    3’... TAG CAT TTC GAA GTA…5’
    i) Applying your knowledge of palindrome sequences identify and mark the possible region where the restriction enzyme X will act.
    ii) Restriction enzyme Y was used to extract gene of interest from a plant. This gene needs to be inserted in the given DNA segment which has been treated with restriction enzyme X. Will there be a successful recombination? Explain with a reason.

    0 marks
  4. c)

    Which one of the two (pUC18 and pBR322) would you prefer for biotechnological studies? Justify.

    0 marks

Figure in the paper: Circular map of plasmid pUC18 showing AmpR gene, lacZ gene and ori (label: ori more than or equal to 100). See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

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

Hint 1

Look for a six-base sequence that reads the same 5'→3' on both strands.

Hint 2

For ligation, the sticky ends must be complementary. When does that happen?

Hint 3

Compare the ori copy numbers of the two plasmids.

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) pUC18 and pBR322 are used as cloning vectors. The gene of interest from a desired organism is inserted into them and carried into a host cell, where it is replicated or expressed.

    a) (OR) If ori is damaged, the plasmid cannot replicate in the host, so no copies of the linked DNA (gene of interest) will be made.

    1 mark
  2. b)(i) The palindrome, which reads the same 5'→3' on both strands, is the HindIII site:
    5'… ATC GTA /AAG CTT/ CAT …3'
    3'… TAG CAT /TTC GAA/ GTA …5'
    Enzyme X acts at 5'-AAGCTT-3' / 3'-TTCGAA-5'.

    1 mark
  3. b)(ii) No, recombination will not be successful. Enzyme Y produces sticky ends different from those made by enzyme X, so the ends are not complementary and cannot be joined by ligase. The same restriction enzyme must cut both the vector and the source DNA.

    1 mark
  4. c) pUC18 is preferred because its copy number is higher (ori ≥ 100 copies, against about 20 for pBR322). It therefore gives many more copies of the recombinant DNA.

    1 mark

Answer: a) Used as cloning vectors (OR: damaged ori means no replication, so no copies). b)(i) AAGCTT/TTCGAA; (ii) No, different enzymes give non-complementary sticky ends. c) pUC18, due to its higher copy number.

Where the marks are

  • Used as vectors to carry gene of interest / ori damage means no replication (1 mark)
  • Correct palindrome marked on both strands (1 mark)
  • No, the same restriction enzyme is needed for complementary sticky ends (0.5 + 0.5 marks)
  • pUC18 with justification of higher copy number (0.5 + 0.5 marks)

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.