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Given below is a stretch of… (Class 12 Biology 2026, Q32b)

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BiologyCBSE Class 12 · Board exam 2026, set 57/1/1 · 5 marks · Long answer

CBSE Class 12 Biology 2026: question 32(b)

Set in the CBSE Class 12 board exam of 2026, paper 57/1/1, 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 answerMolecular Basis of Inheritance

Given below is a stretch of DNA showing the coding strand of structural gene of transcription unit.
5' - ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA 3'

  1. (i)

    Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity.

    2 marks
  2. (ii)

    If GUA of transcribed mRNA is an intron, then depict the sequence involved in formation of mRNA / mature / processed hnRNA strand :
    (1) In a bacterium (1)
    (2) In humans (1)

    2 marks
  3. (iii)

    How many amino acids the resulting polypeptide will have after the process of translation in humans ?

    1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 57/1/1.

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 looks like the coding strand.

Hint 2

Prokaryotes have no introns. Eukaryotic hnRNA undergoes splicing, capping and tailing.

Hint 3

After removing every GUA, count the codons that remain; one codon gives one amino acid.

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) Template strand. It is complementary to the coding strand and has the opposite polarity:

    3' – TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT – 5'

    1 mark

  2. (i) mRNA. The mRNA is copied from the template strand, so it has the same sequence and polarity as the coding strand, with U in place of T:

    5' – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA – 3'

    1 mark

  3. (ii)(1) In a bacterium. Prokaryotic genes are not split, so there is no splicing. The transcript itself works as mRNA:

    5' – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA – 3'

    1 mark

  4. (ii)(2) In humans. The hnRNA is processed in three ways:

    • Splicing: all three GUA introns are removed and the exons are joined.
    • Capping: methyl guanosine triphosphate (m7Gppp) is added at the 5' end.
    • Tailing: a poly-A tail is added at the 3' end.

    Mature mRNA: 5' – m7Gppp – AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA – poly A tail – 3'

    1 mark

  5. (iii) The mature human mRNA has 10 codons after the start codon AUG: AUG, ACC, UUU, UCU, GUG, CCC, CUU, CAG, GCA, UUA. Each codon codes for one amino acid, so the polypeptide will have 10 amino acids.

    1 mark

Answer: (i) Template 3'-TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT-5'; mRNA 5'-AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA-3'. (ii)(1) Bacterium: same mRNA, unchanged; (2) Humans: 5'-m7Gppp-AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA-poly A-3'. (iii) 10 amino acids

Where the marks are

  • Writes the template strand with correct bases and 3'→5' polarity (1 mark)
  • Writes the mRNA with U for T and 5'→3' polarity (1 mark)
  • States that the bacterial mRNA is unchanged because there is no splicing (1 mark)
  • Gives the human mature mRNA with the GUA introns removed, 5' cap and poly-A tail (1 mark)
  • States 10 amino acids (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 paper2026 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.