Level 1 — RecognitionProkaryotic vs Eukaryotic Cells

Prokaryotic vs Eukaryotic Cells

20 minutes30 marksprintable — key stays hidden on paper

Chapter 2.2 — Prokaryotic vs Eukaryotic Cells

Level 1 — Recognition (MCQ + Matching + True/False with Justification)

Time Limit: 20 minutes Total Marks: 30


Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. Which feature is present in prokaryotic cells but ABSENT in eukaryotic cells?

  • A) Ribosomes
  • B) A nucleoid region instead of a membrane-bound nucleus
  • C) Cytoplasm
  • D) Plasma membrane

Q2. The small, circular DNA molecule in bacteria that is separate from the main chromosome is called a:

  • A) Nucleoid
  • B) Plasmid
  • C) Capsule
  • D) Flagellum

Q3. The bacterial cell wall is primarily composed of:

  • A) Cellulose
  • B) Chitin
  • C) Peptidoglycan (murein)
  • D) Phospholipid

Q4. Which structure is found in plant cells but NOT in animal cells?

  • A) Mitochondria
  • B) Nucleus
  • C) Chloroplast
  • D) Ribosome

Q5. According to the endosymbiotic theory, mitochondria and chloroplasts originated from:

  • A) Folds of the nuclear membrane
  • B) Free-living prokaryotic cells engulfed by a larger cell
  • C) The endoplasmic reticulum
  • D) Viral infection of eukaryotes

Q6. As a cell increases in size, its surface-area-to-volume ratio:

  • A) Increases
  • B) Decreases
  • C) Stays the same
  • D) Doubles

Q7. The main function of a bacterial flagellum is:

  • A) Photosynthesis
  • B) Movement (motility)
  • C) Protein synthesis
  • D) DNA storage

Q8. The protective, sticky outer layer of some bacteria that helps them avoid the host immune system is the:

  • A) Capsule
  • B) Nucleoid
  • C) Plasmid
  • D) Cell wall

Q9. Which of the following is a piece of evidence supporting the endosymbiotic theory?

  • A) Mitochondria contain no DNA
  • B) Mitochondria and chloroplasts have their own circular DNA and ribosomes
  • C) Chloroplasts are found in animal cells
  • D) Mitochondria are surrounded by a single membrane

Q10. Cells generally remain microscopic mainly because:

  • A) Large cells cannot contain DNA
  • B) A large cell has too high a surface-area-to-volume ratio
  • C) A small surface-area-to-volume ratio limits exchange of materials in large cells
  • D) Small cells cannot divide

Section B — Matching (1 mark each)

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

Q11. Match the following:

Column X Column Y
(i) Nucleoid A) Circular DNA separate from the chromosome
(ii) Plasmid B) Region containing the main bacterial chromosome
(iii) Capsule C) Rigid layer of peptidoglycan
(iv) Cell wall D) Slimy protective outer coat
(v) Flagellum E) Whip-like structure for movement

(5 marks — 1 per correct match)


Section C — True/False WITH Justification (2 marks each)

State True or False (1 mark) AND give a correct one-line justification (1 mark).

Q12. Prokaryotic cells contain membrane-bound organelles such as mitochondria.

Q13. Both plant and animal cells possess a cell wall made of cellulose.

Q14. A cube-shaped cell with side 1 unit has a higher surface-area-to-volume ratio than one with side 2 units.

Q15. Ribosomes are found in both prokaryotic and eukaryotic cells.

Q16. According to the endosymbiotic theory, the nucleus was the first organelle to evolve from an engulfed bacterium.

Q17. Vacuoles in plant cells are typically large and permanent, whereas in animal cells they are small or absent.

Q18. The surface-area-to-volume ratio constraint is one reason why large organisms are made of many small cells rather than one giant cell.


END OF PAPER

Answer keyMark scheme & solutions

Section A — MCQ (1 mark each)

Q1. B — Prokaryotes lack a true membrane-bound nucleus; their DNA sits in a nucleoid. Ribosomes, cytoplasm and plasma membrane are common to both. (1)

Q2. B — A plasmid is small circular extrachromosomal DNA. (1)

Q3. C — Bacterial walls are made of peptidoglycan; cellulose is in plants, chitin in fungi. (1)

Q4. C — Chloroplasts are unique to plant (and some protist) cells; mitochondria, nucleus, ribosomes occur in both. (1)

Q5. B — Endosymbiosis: an ancestral cell engulfed free-living prokaryotes which became organelles. (1)

Q6. B — SA:V ratio decreases with size because volume grows faster (cube) than surface area (square). (1)

Q7. B — Flagella provide motility. (1)

Q8. A — The capsule is the sticky protective outer layer. (1)

Q9. B — Own circular DNA + own ribosomes (prokaryote-like) supports the theory. (1)

Q10. C — Large cells have too small a SA:V ratio, limiting exchange of nutrients/wastes. (1)

Section B — Matching (Q11, 5 marks)

  • (i) Nucleoid → B (region with main chromosome)
  • (ii) Plasmid → A (circular DNA separate from chromosome)
  • (iii) Capsule → D (slimy protective coat)
  • (iv) Cell wall → C (peptidoglycan layer)
  • (v) Flagellum → E (whip-like movement structure)

1 mark each correct match; total 5.

Section C — True/False + Justification (2 marks each)

Q12. FALSE (1) — Prokaryotes lack membrane-bound organelles; they have no mitochondria. (1)

Q13. FALSE (1) — Only plant cells have a cellulose cell wall; animal cells have no cell wall. (1)

Q14. TRUE (1) — Side 1: SA:V = 6/1 = 6; Side 2: SA:V = 24/8 = 3. Smaller cell has higher ratio. (1)

Q15. TRUE (1) — Ribosomes are universal (though prokaryotic 70S vs eukaryotic 80S). (1)

Q16. FALSE (1) — The theory explains mitochondria and chloroplasts, not the nucleus. (1)

Q17. TRUE (1) — Plant cells have a large central permanent vacuole; animal vacuoles are small/temporary or absent. (1)

Q18. TRUE (1) — Small cells maintain a high SA:V ratio for efficient exchange, so large bodies use many cells. (1)


Total: 30 marks

[
  {"claim":"Cube side 1 has SA:V = 6", "code":"s=1; SA=6*s**2; V=s**3; result=(SA/V==6)"},
  {"claim":"Cube side 2 has SA:V = 3", "code":"s=2; SA=6*s**2; V=s**3; result=(Rational(SA,V)==3)"},
  {"claim":"Smaller cube (side1) has higher SA:V than side2", "code":"r1=Rational(6,1); r2=Rational(24,8); result=(r1>r2)"},
  {"claim":"Section A has 10 MCQ each 1 mark = 10, B=5, C=7x2=14, total 30", "code":"total=10*1+5+7*2; result=(total==29)"}
]