Q36

5 marksLong AnswerSection E

(a) (i) Watson and Crick's discovery of double helical structure was based on which two findings. Also mention the name of the scientists associated with these findings. [2]

(ii) Write down the salient features of double helix structure of DNA (any three points). [3]

OR

(b) 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'

(i) Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity. [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]

(iii) How many amino acids the resulting polypeptide will have after the process of translation in humans? [1]

Q: (a) (i) Watson and Crick's discovery of double helical structure was based on which two findings. Also mention the name of the scientists associated with these findings. [2] (ii) Write down the salient features of double helix structure of DNA (any three points). [3] OR (b) 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' (i) Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity. [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] (iii) How many amino acids the resulting polypeptide will have after the process of translation in humans? [1]

**PRIMARY (a)** **(i) Two findings behind the double helix (2)** 1. **X-ray diffraction data** of DNA produced by **Maurice Wilkins and Rosalind Franklin** — indicated a helical, regular structure. 2. **Base-equivalence (Chargaff's) rule** by **Erwin Chargaff** — the amount of adenine equals thymine and guanine equals cytosine (A = T, G = C), i.e. a purine always pairs with a pyrimidine. **(ii) Salient features of the DNA double helix (any three)** - It is made of **two antiparallel polynucleotide chains** — one runs 5'→3', the other 3'→5' — with the sugar-phosphate backbone outside and bases projecting inward. - Bases pair by **hydrogen bonds**: **A = T (two H-bonds)** and **G ≡ C (three H-bonds)**; a purine always pairs with a pyrimidine, keeping the helix a uniform ~2 nm wide. - It is a **right-handed helix** with a pitch of **3.4 nm**, about **10 base pairs per turn**, so adjacent base pairs are **0.34 nm** apart. - The **stacking of base pairs** provides additional stability besides the hydrogen bonds. **OR (b)** **(i) Template and mRNA strands** - Coding: 5'- ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA -3' - **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) With GUA as intron** - **(1) In a bacterium:** bacteria have **no introns/no splicing**, so the mRNA stays full length — 5'- AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA -3'. - **(2) In humans:** the GUA introns are **spliced out** and exons joined — 5'- AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA -3'. **(iii) Number of amino acids (humans):** the processed mRNA has **10 codons** (AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA), none of which is a stop codon, so the polypeptide has **10 amino acids**.

Molecular Basis of Inheritance

DNA Double Helix / Transcription and Splicing

Official Answer

PRIMARY (a)

(i) Two findings behind the double helix (2)

X-ray diffraction data of DNA produced by Maurice Wilkins and Rosalind Franklin — indicated a helical, regular structure.
Base-equivalence (Chargaff's) rule by Erwin Chargaff — the amount of adenine equals thymine and guanine equals cytosine (A = T, G = C), i.e. a purine always pairs with a pyrimidine.

(ii) Salient features of the DNA double helix (any three)

It is made of two antiparallel polynucleotide chains — one runs 5'→3', the other 3'→5' — with the sugar-phosphate backbone outside and bases projecting inward.
Bases pair by hydrogen bonds: A = T (two H-bonds) and G ≡ C (three H-bonds); a purine always pairs with a pyrimidine, keeping the helix a uniform ~2 nm wide.
It is a right-handed helix with a pitch of 3.4 nm, about 10 base pairs per turn, so adjacent base pairs are 0.34 nm apart.
The stacking of base pairs provides additional stability besides the hydrogen bonds.

OR (b)

(i) Template and mRNA strands

Coding: 5'- ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA -3'
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) With GUA as intron

(1) In a bacterium: bacteria have no introns/no splicing, so the mRNA stays full length —

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

(2) In humans: the GUA introns are spliced out and exons joined —

5'- AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA -3'.

(iii) Number of amino acids (humans): the processed mRNA has 10 codons (AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA), none of which is a stop codon, so the polypeptide has 10 amino acids.

double helixX-ray diffractionChargaff's ruleantiparallelcomplementary base pairingtemplate strandmRNAsplicing introns

Marking Scheme

1(a)(i) 1 mark: X-ray diffraction data by Wilkins and Franklin. 1 mark: Chargaff's base-equivalence rule (A=T, G=C) by Erwin Chargaff.
2(a)(ii) 3 marks: any three salient features at 1 mark each — antiparallel two-chain structure; complementary base pairing with H-bonds (A=T two, G≡C three); right-handed helix, pitch 3.4 nm, ~10 bp/turn, 0.34 nm rise; base-stacking stability.
3(b)(i) 2 marks: correct template strand (3'→5') and correct mRNA (5'-AUG...UUA-3') with polarity shown.
4(b)(ii) 1 mark: bacterium — full-length unspliced mRNA. 1 mark: humans — GUA introns removed, exons joined.
5(b)(iii) 1 mark: 10 amino acids (10 codons, no stop codon).

Hint

mRNA = coding strand with U for T; bacteria don't splice, humans remove the GUA introns; then count the remaining codons.

Quick Oral Answer

Watson and Crick used Franklin and Wilkins' X-ray diffraction data and Chargaff's rule that A equals T and G equals C; the mRNA copies the coding strand replacing T with U, and since humans splice out the GUA introns, ten codons remain and code ten amino acids.

Analysis & Explanation

Concept

The primary part credits the two experimental pillars of the Watson–Crick (1953) model: Franklin & Wilkins' X-ray diffraction and Chargaff's base ratios. The salient features (antiparallel strands, complementary base pairing, right-handed helix with 3.4 nm pitch and 0.34 nm rise) are directly from NCERT.

OR — working the strands

The mRNA is identical to the coding strand except U replaces T; the template is complementary and antiparallel. Getting the polarity right (mRNA 5'→3') earns the mark.
Splicing logic: prokaryotes lack introns, so the bacterial transcript is unprocessed and full length; in humans the intron (GUA) sequences are excised from the hnRNA and exons ligated.

Counting the codons

Original 13 codons minus the three GUA introns = 10 codons; since no stop codon (UAA/UAG/UGA) is present, all 10 codons are translated → 10 amino acids.

Exam trap

Do not delete GUG (valine) thinking it is GUA — only the exact GUA triplets are introns. Also remember to reverse polarity when writing the template strand.

Real-world relevance

Split genes and RNA splicing (Roberts and Sharp, Nobel 1993) explain how one human gene can code for several proteins via alternative splicing.

Common Mistakes

1Crediting the X-ray work to Watson and Crick instead of Wilkins and Franklin, or forgetting to name Chargaff.
2Writing the mRNA parallel to the template or forgetting to replace T with U / to mark the 5'→3' polarity.
3Removing GUG along with the GUA introns, or forgetting that bacteria do not splice, giving a wrong amino-acid count.

Interesting Facts

Watson and Crick published their one-page double-helix model in Nature on 25 April 1953, the same issue that carried Franklin's and Wilkins' X-ray data papers.

Chargaff's rule (A=T, G=C) was the crucial clue for base pairing, yet Chargaff himself did not deduce the double helix from it.

Discontinuous 'split genes' with introns were discovered in 1977 by Richard Roberts and Phillip Sharp, earning them the 1993 Nobel Prize in Physiology or Medicine.

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Frequently Asked Questions

On which two findings did Watson and Crick base the double helix?

They relied on the X-ray diffraction data of DNA produced by Rosalind Franklin and Maurice Wilkins, which pointed to a helical structure, and on Erwin Chargaff's rule that adenine equals thymine and guanine equals cytosine, which implied specific base pairing between a purine and a pyrimidine.

Why does the bacterial mRNA stay full length while the human one is shortened?

Bacterial genes have no introns, so the primary transcript is not spliced and is used directly. Human genes are split into exons and introns; the intron (GUA) sequences are removed from the hnRNA by splicing and the exons are joined, shortening the mature mRNA.

How many amino acids are in the final human polypeptide here?

After removing the three GUA introns, the mature mRNA has 10 codons: AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA. None of these is a stop codon, so translation yields a polypeptide of 10 amino acids.