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Given below is a… (Class 12 Biology SQP 2023-24, Q32a)

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BiologyCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Biology sample paper 2023-24: question 32(a)

From the official CBSE sample question paper for 2023-24, 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 answerMolecular Basis of Inheritance

Given below is a stretch of DNA showing the coding strand of a structural gene of a transcription unit?
5’--ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TAA—3’
a) Write the corresponding template strand and the mRNA strand that will be transcribed, along with its polarity.
b) If GUA of the transcribed mRNA is an intron, depict the sequence involved in the formation of mRNA /the mature processed hnRNA strand.
i. In a bacterium
ii. In humans
c) Upon translation, how many amino acids will the resulting polypeptide have? Justify.

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

The template strand is complementary and antiparallel to the coding strand; the mRNA matches the coding strand with U for T.

Hint 2

Do prokaryotic genes contain introns? What three processing steps happen in eukaryotic hnRNA?

Hint 3

Count the codons in the mature mRNA and remember what UAA does.

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) Template strand (complementary and antiparallel to the coding strand):
    3'– TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT ATT –5'

    1 mark

  2. mRNA (same as the coding strand, with U in place of T):
    5'– AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA –3'

    1 mark

  3. b)(i) In a bacterium: prokaryotic genes are not split (no introns), so no splicing occurs. The mRNA is used as transcribed:
    5'– AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA –3'

    1 mark

  4. b)(ii) In humans: the hnRNA is processed in three ways:

    • Splicing: all three GUA introns are removed and the exons are joined.
    • Capping: a methyl guanosine cap is added at the 5' end.
    • Tailing: a poly-A tail is added at the 3' end.

    Mature mRNA: 5'– m⁷GpppAUG ACC UUU UCU GUG CCC CUU CAG GCA UAA – poly(A) tail –3'

    1 mark

  5. c) The mature human mRNA has 10 codons. UAA is a stop codon and codes for no amino acid, so the polypeptide has 9 amino acids. (For the unspliced bacterial mRNA, 12 codons come before UAA, giving 12 amino acids.)

    1 mark

Answer: a) Template 3'-TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT ATT-5'; mRNA 5'-AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA-3'. b) Bacterium: unchanged; humans: 5'-m⁷GpppAUG ACC UUU UCU GUG CCC CUU CAG GCA UAA-polyA-3'. c) 9 amino acids (UAA is a stop codon).

Where the marks are

  • Correct template strand with polarity (1 mark)
  • Correct mRNA with polarity (1 mark)
  • Bacterial mRNA unchanged, as no splicing occurs (1 mark)
  • Human mature mRNA with GUA removed, 5' cap and poly-A tail (1 mark)
  • 9 amino acids because UAA is a stop codon (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 paperSample paper 2023-24: 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.