Level 2 — RecallProkaryotic vs Eukaryotic Cells

Prokaryotic vs Eukaryotic Cells

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Chapter 2.2 — Prokaryotic vs Eukaryotic Cells

Level: 2 — Recall (definitions, standard textbook problems, short derivations) Time Limit: 30 minutes Total Marks: 40

Instructions: Answer all questions. Show working for calculation questions. Use diagrams where helpful.

Question 1. State three structural differences between a prokaryotic cell and a eukaryotic cell. (3 marks)

Question 2. Define each of the following bacterial structures and give one function of each: (a) Nucleoid (2 marks) (b) Plasmid (2 marks) (c) Capsule (2 marks)

Question 3. The bacterial cell wall provides shape and protection. (a) Name the polymer that makes up the bacterial cell wall. (1 mark) (b) State two functions of bacterial flagella. (2 marks)

Question 4. Copy and complete the table below comparing plant and animal cells by writing "Present" or "Absent". (4 marks)

Structure	Plant Cell	Animal Cell
Cell wall
Chloroplast
Large central vacuole
Centrioles

Question 5. The endosymbiotic theory explains the origin of certain organelles. (a) State which two organelles are explained by this theory. (2 marks) (b) Give three pieces of evidence that support the endosymbiotic theory. (3 marks)

Question 6. Define the term surface-area-to-volume ratio (SA:V). (2 marks)

Question 7. A cell is modelled as a cube with side length $2\ \mu m$ . (a) Calculate its surface area. (2 marks) (b) Calculate its volume. (2 marks) (c) Calculate its surface-area-to-volume ratio. (2 marks)

Question 8. Consider two cube-shaped cells, one of side $1\ \mu m$ and one of side $4\ \mu m$ . (a) Calculate the SA:V ratio of each. (4 marks) (b) State which cell exchanges materials with its surroundings more efficiently, and why. (2 marks)

Question 9. Explain, in terms of surface-area-to-volume ratio, why cells remain microscopic. (3 marks)

Question 10. Give one reason each why a larger organism does not simply have larger cells, but instead has more cells. (1 mark)

End of Paper

Answer keyMark scheme & solutions

Chapter 2.2 — Prokaryotic vs Eukaryotic Cells

Question 1. (3 marks — 1 each, max 3) Any three of:

Prokaryotes have no membrane-bound nucleus (DNA free in cytoplasm); eukaryotes have a true nucleus. ✓
Prokaryotes lack membrane-bound organelles (no mitochondria, ER, Golgi); eukaryotes have them. ✓
Prokaryotic cells are smaller (~1–5 µm) vs eukaryotic (~10–100 µm). ✓
Prokaryotic DNA is usually circular; eukaryotic DNA is linear and associated with histones. ✓
Prokaryotic ribosomes are 70S; eukaryotic 80S. ✓ Why: These are the defining organisational contrasts of the two cell types.

Question 2. (6 marks) (a) Nucleoid — the region of the cytoplasm containing the bacterial (circular) DNA, not membrane-bound. ✓ Function: carries the genetic information / controls cell activities. ✓ (b) Plasmid — a small circular piece of extra-chromosomal DNA that replicates independently. ✓ Function: carries extra genes e.g. antibiotic resistance. ✓ (c) Capsule — an outer slime/mucous layer surrounding the cell wall. ✓ Function: protection / prevents desiccation / aids adhesion / evades host immune system. ✓

Question 3. (3 marks) (a) Peptidoglycan (murein). ✓ (1) (b) Any two: enables movement/locomotion; propels cell toward stimulus (chemotaxis); movement away from harmful conditions. ✓✓ (2)

Question 4. (4 marks — ½ each cell, rounded; award 1 mark per fully correct row)

Structure	Plant Cell	Animal Cell
Cell wall	Present ✓	Absent
Chloroplast	Present ✓	Absent
Large central vacuole	Present ✓	Absent
Centrioles	Absent	Present ✓

Why: Plants have cellulose walls, chloroplasts and a large vacuole for support/photosynthesis; centrioles are typically found in animal cells.

Question 5. (5 marks) (a) Mitochondria ✓ and chloroplasts ✓. (2) (b) Any three: they have their own circular DNA ✓; they have their own 70S (bacterial-type) ribosomes ✓; they are surrounded by a double membrane ✓; they divide by binary fission independently of the cell ✓. (max 3) Why: These features resemble free-living prokaryotes, supporting engulfment origin.

Question 6. (2 marks) The ratio of a cell's total surface area to its volume ✓, indicating how much exchange surface is available per unit of internal volume ✓.

Question 7. (6 marks) Cube side $a = 2\ \mu m$ . (a) Surface area $= 6a^2 = 6(2)^2 = 24\ \mu m^2$ . ✓ (formula ✓, answer ✓) (b) Volume $= a^3 = 2^3 = 8\ \mu m^3$ . ✓✓ (c) SA:V $= \dfrac{24}{8} = 3\ (:1)$ , i.e. $3\ \mu m^{-1}$ . ✓✓

Question 8. (6 marks) (a) Side $1\ \mu m$ : SA $=6(1)^2=6$ , V $=1^3=1$ , SA:V $=6:1$ . ✓✓ Side $4\ \mu m$ : SA $=6(4)^2=96$ , V $=4^3=64$ , SA:V $=96/64=1.5:1$ . ✓✓ (b) The $1\ \mu m$ cell exchanges more efficiently ✓ because it has a higher SA:V ratio, giving more surface per unit volume for diffusion. ✓

Question 9. (3 marks)

As a cell grows, volume increases faster (cubed) than surface area (squared), so SA:V falls. ✓
A low SA:V means the surface cannot supply/remove enough materials by diffusion for the internal volume. ✓
Therefore cells stay small to maintain a high SA:V for efficient exchange of nutrients, gases and wastes. ✓

Question 10. (1 mark) Smaller cells keep a high SA:V ratio for efficient exchange / diffusion distances stay short — so growth is by adding more cells, not enlarging them. ✓ (any valid reason)

[
  {"claim":"Q7a cube side 2: surface area = 24", "code":"a=2\nresult = (6*a**2 == 24)"},
  {"claim":"Q7b/c cube side 2: volume=8 and SA:V=3", "code":"a=2\nsa=6*a**2\nv=a**3\nresult = (v==8) and (Rational(sa,v)==3)"},
  {"claim":"Q8a cube side 1 gives SA:V 6 and side 4 gives 1.5", "code":"r1=Rational(6*1**2,1**3)\nr4=Rational(6*4**2,4**3)\nresult = (r1==6) and (r4==Rational(3,2))"},
  {"claim":"Larger cube has smaller SA:V than smaller cube", "code":"r1=Rational(6*1**2,1**3)\nr4=Rational(6*4**2,4**3)\nresult = (r1 > r4)"}
]