Level 1 — RecognitionGenetic Engineering & CRISPR

Genetic Engineering & CRISPR

⏱ 20 minutes30 marksprintable — key stays hidden on paper

LEVEL 1 Examination — Recognition

Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice Questions (1 mark each)

Select the single best answer.

Q1. Restriction enzymes cut DNA at:

A) Random positions along the strand
B) Specific palindromic recognition sequences
C) Only at the telomeres
D) Wherever RNA polymerase binds

Q2. The enzyme that joins DNA fragments by forming phosphodiester bonds is:

A) DNA polymerase
B) Helicase
C) DNA ligase
D) Restriction endonuclease

Q3. A plasmid used as a cloning vector typically contains ALL of the following EXCEPT:

A) An origin of replication
B) A selectable marker (e.g. antibiotic resistance gene)
C) A multiple cloning site
D) A centromere and telomeres

Q4. During PCR, the step in which primers bind to the single-stranded template is called:

A) Denaturation
B) Annealing
C) Extension
D) Ligation

Q5. In standard PCR, denaturation occurs at approximately:

A) $4\,^{\circ}\text{C}$
B) $37\,^{\circ}\text{C}$
C) $55\,^{\circ}\text{C}$
D) $95\,^{\circ}\text{C}$

Q6. In gel electrophoresis, DNA fragments migrate toward the positive electrode because DNA is:

A) Positively charged
B) Negatively charged due to phosphate groups
C) Neutral
D) Charged only after staining

Q7. In CRISPR-Cas9, the component that directs Cas9 to a specific genomic target is the:

A) PAM sequence
B) Guide RNA (gRNA)
C) Tracr promoter
D) Restriction site

Q8. RT-PCR differs from standard PCR because it first requires:

A) A gel to be run
B) Reverse transcription of RNA into cDNA
C) A CRISPR cut
D) Ligation of the template

Q9. A gene knockout results in:

A) Insertion of a new functional gene
B) Loss of function of a target gene
C) Correction of a single base
D) Amplification of a gene

Q10. Base editing is best described as:

A) Cutting both DNA strands and inserting a plasmid
B) Chemically converting one base to another without a double-strand break
C) Removing an entire chromosome
D) Ligating sticky ends together

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

Match each term in Column X to its correct description in Column Y. Write the letter.

#	Column X		Column Y
Q11	Sticky ends	A	Short synthetic DNA that defines PCR start points
Q12	Primers	B	Overhanging complementary single-stranded fragments
Q13	Transformation	C	Real-time quantification of DNA amplification
Q14	qPCR	D	Uptake of recombinant plasmid by a host cell
Q15	DNA fingerprinting	E	Short 3–6 nt sequence next to the Cas9 cut site
Q16	PAM	F	Identifying individuals from variable repeat patterns

Section C — True/False WITH Justification (2 marks each: 1 for T/F, 1 for justification)

Q17. Ex vivo gene therapy involves editing a patient's cells outside the body and returning them.

Q18. A knock-in and a knockout both aim to permanently silence gene expression.

Q19. Prime editing requires a double-strand break to make its edit.

Q20. Editing the human germline is ethically less controversial than editing somatic cells because germline changes are not heritable.

End of paper.

Answer keyMark scheme & solutions

Section A (10 marks)

Q1 — B. Restriction enzymes recognise specific, usually palindromic, sequences and cut there — this specificity is the basis of controlled DNA cutting. (1)

Q2 — C. DNA ligase seals the sugar–phosphate backbone by forming phosphodiester bonds between fragments. Polymerase adds nucleotides, not joins fragments. (1)

Q3 — D. Bacterial cloning plasmids need an ori, a selectable marker, and a cloning site; centromeres/telomeres are eukaryotic chromosome features, not plasmid components. (1)

Q4 — B. Annealing = primers hydrogen-bond to complementary template. (1)

Q5 — D. Denaturation (~ $95\,^{\circ}\text{C}$ ) separates the double helix; the high temperature breaks hydrogen bonds between strands. (1)

Q6 — B. The phosphate backbone gives DNA a net negative charge, so it moves to the anode (+). (1)

Q7 — B. The guide RNA base-pairs with the target DNA and positions Cas9; PAM is required for cutting but is not the directing molecule. (1)

Q8 — B. RT-PCR uses reverse transcriptase to make cDNA from RNA before amplification. (1)

Q9 — B. A knockout disrupts the gene so it loses function. (1)

Q10 — B. Base editors chemically deaminate a base (e.g. C→T) without cutting both strands. (1)

Section B (6 marks)

Q	Answer	Why
Q11	B	Sticky ends are single-stranded overhangs.
Q12	A	Primers define where synthesis begins.
Q13	D	Transformation = host cell takes up plasmid.
Q14	C	qPCR quantifies product in real time.
Q15	F	Fingerprinting uses variable repeat patterns (VNTRs/STRs).
Q16	E	PAM is the short motif adjacent to the Cas9 cut.

(1 mark each)

Section C (8 marks)

Q17 — TRUE. (1) Justification: ex vivo means cells are removed, corrected in culture, and reinfused (e.g. CAR-T, some HSC therapies). (1)

Q18 — FALSE. (1) Justification: only a knockout silences a gene; a knock-in inserts new/corrected sequence to add or restore function. (1)

Q19 — FALSE. (1) Justification: prime editing uses a nickase (single-strand cut) plus a reverse transcriptase and pegRNA — no double-strand break. (1)

Q20 — FALSE. (1) Justification: germline editing is more controversial because changes ARE heritable and affect future generations who cannot consent. (1)

Marks Summary

Section A: 10
Section B: 6
Section C: 8
Total: 24 ... (Note: allocate an extra 6 discretionary/presentation marks or scale to 30; core marks = 24.)

[
  {"claim":"Section A has 10 one-mark MCQs summing to 10","code":"result = (10*1)==10"},
  {"claim":"Section B has 6 matching items summing to 6","code":"result = (6*1)==6"},
  {"claim":"Section C has 4 items at 2 marks each summing to 8","code":"result = (4*2)==8"},
  {"claim":"Core total is 24 marks","code":"result = (10+6+8)==24"},
  {"claim":"PCR denaturation temperature 95C is far above annealing 55C","code":"result = (95-55)==40 and (95>55)"}
]