S4 Biology Practice Paper: Cell Structure, Enzymes And Transport (12 Questions, 100 Marks, 75 Minutes, With Answers)

S4 Biology Practice Paper: Cell Structure, Enzymes and Transport (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-09

HKDSE Biology: Cell Structure, Enzymes and Transport

12 questions, 100 marks, 75 minutes on cell structure, magnification calculations, enzyme specificity, diffusion, osmosis, active transport and surface area to volume ratio, with answers.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.

Give numerical answers to three significant figures unless stated otherwise, and state the correct units.

Spend about 35 minutes on Section A, then use the remaining time for the enzyme data in Section B and the osmosis and design questions in Section C.

Section A: Cell Structure and the Microscope

Organelle functions, plant and animal cell comparison, magnification and specimen measurement

#	Question	Answer and marking notes
1	Compare three structural differences between animal cells and plant cells, and state the importance of each difference to plant life.	(a) Plant cells have a cellulose cell wall while animal cells do not (2 marks). (b) Plant cells have a large central vacuole while animal cells generally do not (2 marks). (c) Plant cells contain chloroplasts while animal cells do not (2 marks). (d) The cell wall provides support, so plants keep their shape without a skeleton (2 marks).
2	Describe the fluid mosaic model of the cell membrane, and explain how the model accounts for the membrane’s selective permeability.	(a) The phospholipid bilayer forms the basic framework of the membrane (2 marks). (b) Protein molecules are embedded in the bilayer and can move laterally (2 marks). (c) Some proteins form channels or carriers that admit only particular substances (2 marks). (d) The hydrophobic tails of the phospholipids prevent polar molecules and charged ions from passing freely, so the membrane is selectively permeable (2 marks).
3	Describe how the structure of mitochondria and chloroplasts is adapted to their functions.	(a) The inner membrane of a mitochondrion is folded into cristae, increasing the surface area for the electron transport chain (2 marks). (b) The matrix contains many enzymes for the Krebs cycle (2 marks). (c) Thylakoids in a chloroplast are stacked into grana, increasing the surface area for absorbing light energy (2 marks). (d) The stroma contains the enzymes needed for the light-independent reactions (2 marks).
4	A chloroplast is viewed with a 10x eyepiece and a 40x objective: (a) find the total magnification; (b) a cell in the image measures 20 mm, find its actual length in micrometres; (c) state one precaution to take after switching from low power to high power.	(a) Total magnification = 10 x 40 = 400 times (3 marks). (b) Actual length = 20 mm divided by 400 = 0.05 mm, that is 50 micrometres (3 marks). (c) After switching to high power use only the fine adjustment, and avoid moving the stage so that the slide is not cracked (3 marks).

Section B: Enzymes and Transport

Enzyme specificity and affecting factors, diffusion and osmosis, active transport

#	Question	Answer and marking notes
5	Explain why enzymes are specific, and describe how excessively high temperature and a pH away from the optimum affect enzyme activity.	(a) The active site is complementary in shape to the substrate, so only that substrate can bind (2 marks). (b) The enzyme and substrate form an enzyme-substrate complex, lowering the activation energy (2 marks). (c) Excessively high temperature denatures the enzyme, changing the shape of the active site, so activity falls irreversibly (2 marks). (d) A pH away from the optimum alters the charges on amino acid side chains in the active site, so the substrate cannot bind (2 marks).
6	An experiment collects 30 cm³ of oxygen in 10 minutes: (a) find the average rate of production; (b) when the temperature rises from 25°C to 37°C the rate becomes 3 times as great, find the new rate; (c) find the total volume collected in 10 minutes at 37°C.	(a) Average rate = 30 divided by 10 = 3 cm³ per minute (3 marks). (b) Rate at 37°C = 3 x 3 = 9 cm³ per minute (3 marks). (c) Total volume in 10 minutes = 9 x 10 = 90 cm³ (2 marks).
7	A dye travels 6 mm through agar in 15 minutes: (a) find the rate of diffusion; (b) explain how a rise in temperature affects the rate of diffusion; (c) state whether diffusion requires ATP.	(a) Rate of diffusion = 6 divided by 15 = 0.4 mm per minute (3 marks). (b) A rise in temperature increases the kinetic energy of the molecules, so they move faster and diffusion is quicker (2 marks). (c) The steeper the concentration gradient, the faster the net movement (2 marks). (d) Diffusion is passive and does not require ATP (1 mark).
8	A potato strip of mass 2.0 g is placed in distilled water and after 30 minutes has a mass of 2.4 g: (a) find the increase in mass; (b) find the percentage change in mass; (c) explain why the strip becomes firm; (d) explain why the same strip becomes soft in 1.0 mol dm⁻³ sucrose solution.	(a) Increase in mass = 2.4 minus 2.0 = 0.4 g (2 marks). (b) Percentage change = 0.4 divided by 2.0 x 100% = 20% (3 marks). (c) The water potential of distilled water is higher than that of the cell sap, so water enters the cells by osmosis and the cells become turgid (2 marks). (d) The sucrose solution has a lower water potential, so water leaves the cells and they lose turgor pressure and become soft (2 marks).

Section C: Applications and Experimental Design

Surface area to volume ratio, exchange surface adaptations, investigating temperature effects on enzymes

#	Question	Answer and marking notes
9	Compare active transport and passive transport in three respects.	(a) Active transport can move substances against a concentration gradient, whereas passive transport moves them down the gradient (2 marks). (b) Active transport requires ATP, whereas passive transport does not (2 marks). (c) Active transport requires carrier proteins, whereas simple diffusion does not (2 marks). (d) The rate of active transport is reduced by respiratory inhibitors, whereas passive transport is not affected (2 marks).
10	For a cube of side 2 cm and a cube of side 1 cm: (a) find the surface area, volume and surface area to volume ratio of each; (b) explain the significance of the surface area to volume ratio for gas exchange.	(a) Cube of side 2 cm: surface area = 6 x 2 squared = 24 cm² (1 mark), volume = 8 cm³ (1 mark), ratio = 24 divided by 8 = 3 (1 mark). (b) Cube of side 1 cm: surface area = 6 cm² (1 mark), volume = 1 cm³ (1 mark), ratio = 6 (1 mark). (c) The larger the ratio, the more surface is available per unit volume for gas exchange, so small organisms can survive on diffusion alone (2 marks).
11	Explain how the villi of the small intestine and the alveoli are adapted for exchange of materials, giving two adaptations for each.	(a) The surface of a villus has microvilli, which increase the surface area for absorption (2 marks). (b) The wall of a villus is one cell thick, which shortens the diffusion distance (2 marks). (c) Each villus has a rich blood supply, which maintains the concentration gradient (2 marks). (d) The alveolar wall is thin and moist, which allows oxygen and carbon dioxide to diffuse readily (2 marks). (e) There are very many alveoli, giving a large total surface area (1 mark).
12	Design an experiment to investigate the effect of temperature on the activity of amylase, stating four essential points of the design.	(a) Use the same concentration of starch solution and amylase, and set up five water baths at 20, 30, 40, 50 and 60°C (2 marks). (b) Allow each tube to reach the set temperature before adding the enzyme (2 marks). (c) Sample at fixed intervals and use iodine solution to find the time taken for starch to disappear; the shorter the time, the higher the activity (3 marks). (d) Control the variables: enzyme concentration, substrate concentration, pH and total volume must be the same in every tube (2 marks).

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