Level 1 — RecognitionDNA Structure & Replication

DNA Structure & Replication

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1 Examination: Recognition

Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each)

Select the single best answer.

Q1. In Griffith's experiment, the "transforming principle" that changed harmless bacteria into virulent ones was later identified (by Avery) as:

A) Protein
B) RNA
C) DNA
D) Lipid

Q2. The Hershey–Chase experiment used radioactive isotopes to track molecules. Which isotope labelled the DNA?

A) $^{35}S$
B) $^{32}P$
C) $^{14}C$
D) $^{3}H$

Q3. According to the Watson–Crick model, adenine pairs with thymine by:

A) 3 hydrogen bonds
B) 2 hydrogen bonds
C) 1 covalent bond
D) 2 ionic bonds

Q4. The two strands of DNA are described as antiparallel, meaning:

A) They run in the same $5' \to 3'$ direction
B) One runs $5' \to 3'$ , the other $3' \to 5'$
C) They are of unequal length
D) They never form base pairs

Q5. Chargaff's rule states that in double-stranded DNA:

A) $A = G$ and $T = C$
B) $A = T$ and $G = C$
C) $A + T = G + C$ always
D) $A = C$ and $T = G$

Q6. The Meselson–Stahl experiment demonstrated that DNA replication is:

A) Conservative
B) Dispersive
C) Semi-conservative
D) Random

Q7. DNA helicase functions to:

A) Join Okazaki fragments
B) Add nucleotides to the growing strand
C) Unwind and separate the double helix
D) Synthesize RNA primers

Q8. DNA polymerase adds new nucleotides only in the:

A) $3' \to 5'$ direction
B) $5' \to 3'$ direction
C) Both directions equally
D) $2' \to 5'$ direction

Q9. The lagging strand is synthesized:

A) Continuously toward the replication fork
B) Discontinuously in short fragments away from the fork
C) By telomerase only
D) Without any primer

Q10. Okazaki fragments are joined together by the enzyme:

A) Primase
B) Helicase
C) DNA ligase
D) Topoisomerase

Q11. Before DNA polymerase can begin, primase must synthesize a short:

A) DNA strand
B) RNA primer
C) Protein cap
D) Lipid bridge

Q12. Telomerase is an enzyme that:

A) Removes primers
B) Extends the repetitive ends (telomeres) of chromosomes
C) Proofreads the leading strand
D) Cuts DNA at restriction sites

Section B — Matching (1 mark each, 6 marks)

Q13. Match each enzyme/molecule (i–vi) to its function (P–U).

Enzyme/Molecule		Function
(i) Helicase		(P) Adds RNA primer
(ii) DNA polymerase		(Q) Seals nicks between fragments
(iii) Primase		(R) Unwinds the double helix
(iv) DNA ligase		(S) Extends chromosome ends
(v) Telomerase		(T) Adds & proofreads DNA nucleotides
(vi) Okazaki fragment		(U) Short DNA piece on lagging strand

Section C — True / False WITH Justification (2 marks each: 1 answer + 1 justification)

Q14. In DNA, the amount of purines equals the amount of pyrimidines. (True/False + justify)

Q15. DNA polymerase can start a new strand from scratch without a primer. (True/False + justify)

Q16. In the Meselson–Stahl experiment, after one round of replication in $^{14}N$ , all DNA was of intermediate (hybrid) density. (True/False + justify)

Q17. The leading strand requires many primers, one for each Okazaki fragment. (True/False + justify)

Q18. DNA proofreading by DNA polymerase reduces the error rate during replication. (True/False + justify)

Q19. Hershey and Chase found that the $^{35}S$ (protein) label entered the bacterial cells and directed viral reproduction. (True/False + justify)

Q20. If a DNA sample contains 30% adenine, it must contain 20% guanine. (True/False + justify)

END OF PAPER

Answer keyMark scheme & solutions

Section A (12 marks)

Q1. C) DNA — Avery, MacLeod & McCarty showed the transforming principle was destroyed by DNase but not by proteases/RNase, identifying DNA. (1)

Q2. B) $^{32}P$ — Phosphorus is found in DNA's phosphate backbone (not in protein), so $^{32}P$ tracks DNA. Sulfur ( $^{35}S$ ) tracks protein. (1)

Q3. B) 2 hydrogen bonds — A–T pairs share 2 H-bonds; G–C share 3. (1)

Q4. B) One runs $5' \to 3'$ , the other $3' \to 5'$ — Antiparallel orientation. (1)

Q5. B) $A = T$ and $G = C$ — Base-pairing complementarity. (1)

Q6. C) Semi-conservative — Each daughter molecule has one old and one new strand. (1)

Q7. C) Unwind and separate the double helix — Breaks H-bonds at the fork. (1)

Q8. B) $5' \to 3'$ direction — Nucleotides added to the free 3'-OH end. (1)

Q9. B) Discontinuously in short fragments away from the fork — Because synthesis must be $5'\to3'$ . (1)

Q10. C) DNA ligase — Seals the phosphodiester backbone between fragments. (1)

Q11. B) RNA primer — Provides a free 3'-OH for polymerase to extend. (1)

Q12. B) Extends the repetitive ends (telomeres) — Compensates for end-replication shortening. (1)

Section B (6 marks — 1 each)

Q13.

(i) Helicase → R (unwinds helix)
(ii) DNA polymerase → T (adds & proofreads DNA nucleotides)
(iii) Primase → P (adds RNA primer)
(iv) DNA ligase → Q (seals nicks)
(v) Telomerase → S (extends chromosome ends)
(vi) Okazaki fragment → U (short DNA piece on lagging strand)

Section C (14 marks — 1 answer + 1 justification each)

Q14. TRUE (1). Justification: By Chargaff, $A=T$ (both purine A + pyrimidine T counted) and $G=C$ ; purines $(A+G)$ = pyrimidines $(T+C)$ . (1)

Q15. FALSE (1). Justification: DNA polymerase can only extend an existing 3'-OH; it requires a primer laid down by primase. (1)

Q16. TRUE (1). Justification: After one replication in $^{14}N$ , every molecule has one heavy $^{15}N$ strand + one light $^{14}N$ strand = single intermediate band. (1)

Q17. FALSE (1). Justification: The leading strand is synthesized continuously and needs only one primer; it is the lagging strand that needs many primers. (1)

Q18. TRUE (1). Justification: The $3'\to5'$ exonuclease activity removes mismatched nucleotides, lowering the error rate. (1)

Q19. FALSE (1). Justification: The $^{32}P$ (DNA) label entered the cells; $^{35}S$ (protein coat) remained outside, proving DNA is the genetic material. (1)

Q20. TRUE (1). Justification: If $A=30\%$ then $T=30\%$ ; remaining $40\%$ splits equally: $G=C=20\%$ . (1)

[
  {"claim":"Q20: 30% A implies 20% G by Chargaff", "code":"A=30; T=A; rest=100-A-T; G=rest/2; C=rest/2; result=(G==20 and C==20)"},
  {"claim":"Q14: purines equal pyrimidines given A=T and G=C", "code":"A,G=Symbol('A'),Symbol('G'); T=A; C=G; purines=A+G; pyrimidines=T+C; result=simplify(purines-pyrimidines)==0"},
  {"claim":"Q3/Q5: A-T has 2 H-bonds, G-C has 3 H-bonds", "code":"AT=2; GC=3; result=(AT==2 and GC==3 and GC>AT)"},
  {"claim":"Q16: one strand heavy one strand light gives one hybrid band", "code":"heavy=1; light=1; bands_after_1_gen=1; result=(heavy==light==1 and bands_after_1_gen==1)"}
]