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

S4 Biology Practice Paper: Cell Structure, Transport and Enzymes (12 questions, 100 marks, 75 minutes)

2026-11-09

HKDSE Biology: Cell Structure, Transport and Enzymes

12 questions, 100 marks, 75 minutes on cell structure, microscopy, osmosis, water potential, active transport and enzyme activity, with full 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 30 minutes on Section A, then work through osmosis and transport in Section B, and finish with enzyme activity in Section C.

Section A　Cell Structure and the Microscope

Section A: Show the formula and working for each calculation, with correct units.

#	Question	Answer and marking notes
1	Data 1: a student uses a microscope with a ×10 eyepiece and a ×40 objective to observe onion epidermal cells. (a) Calculate the total magnification. (b) An image of a cell in the field of view has a diameter of 20 mm; find the actual diameter of the cell in µm. (c) Another cell has an actual diameter of 60 µm; find the diameter of its image in mm at the same magnification. (d) State one advantage of staining the onion epidermis before observation.	(a) Total magnification = eyepiece magnification × objective magnification = 10 × 40 = 400 times (2 marks). (b) Actual diameter = image diameter ÷ magnification = 20 mm ÷ 400 = 0.05 mm = 50 µm (3 marks). (c) Image diameter = 60 µm × 400 = 24000 µm = 24 mm (3 marks). (d) Staining increases the contrast of structures such as the cell wall and nucleus, so the cell boundary and internal structures are easier to identify (1 mark).
2	(a) A cube-shaped cell has sides of 20 µm. Calculate its surface area, its volume and the surface area to volume ratio in µm⁻¹. (b) Repeat the three calculations for a cube-shaped cell with sides of 10 µm. (c) Explain why a cell with a smaller volume exchanges substances more efficiently by diffusion. (d) State one limitation of using the surface area to volume ratio alone to judge diffusion efficiency.	(a) For sides of 20 µm: surface area = 6 × 20² = 2400 µm², volume = 20³ = 8000 µm³, ratio = 2400 ÷ 8000 = 0.30 µm⁻¹ (3 marks). (b) For sides of 10 µm: surface area = 600 µm², volume = 1000 µm³, ratio = 600 ÷ 1000 = 0.60 µm⁻¹ (2 marks). (c) A cell with a smaller volume has a larger surface area to volume ratio, so there is more membrane area per unit volume for exchange and the diffusion distance is shorter, giving more efficient diffusion (3 marks). (d) The ratio ignores folds in the cell surface, the number of transport proteins in the membrane and differences in metabolic rate (1 mark).
3	(a) State one main function each of the mitochondrion, the chloroplast and the Golgi apparatus. (b) Heart muscle cells contain far more mitochondria than cheek epithelial cells; explain why. (c) State the role of ribosomes on the rough endoplasmic reticulum. (d) Describe two functions of lysosomes in a cell.	(a) Mitochondria carry out aerobic respiration and release energy for the cell; chloroplasts carry out photosynthesis and convert light energy into chemical energy; the Golgi apparatus modifies, packages and secretes proteins (3 marks). (b) Heart muscle cells contract continuously and need a large supply of energy, so they contain more mitochondria to produce enough ATP (2 marks). (c) Ribosomes are the site of protein synthesis, joining amino acids into polypeptide chains (1 mark). (d) Lysosomes contain hydrolytic enzymes that digest worn-out organelles or material taken into the cell, and they also take part in programmed cell death (2 marks).
4	(a) Describe the fluid mosaic model of the cell membrane, including the arrangement of the phospholipid bilayer and proteins. (b) Explain why lipid-soluble molecules cross the membrane more easily than charged ions. (c) State one function of cholesterol in animal cell membranes. (d) Explain what is meant by the membrane being selectively permeable.	(a) Phospholipids form a bilayer with hydrophilic heads facing outwards and hydrophobic tails facing inwards; protein molecules are embedded in or attached to the bilayer and can move sideways, so the model is described as a fluid mosaic (3 marks). (b) The interior of the bilayer is hydrophobic, so lipid-soluble molecules dissolve in it and pass through directly; charged ions are hydrated and insoluble in lipid, so they need channel or carrier proteins (2 marks). (c) Cholesterol sits between phospholipid molecules and regulates membrane fluidity, keeping the membrane stable at different temperatures (2 marks). (d) The membrane allows some substances to pass through while restricting or preventing others, and this property is called selective permeability (2 marks).

Section B　Transport Across the Cell Membrane

Section B: Answer explanation questions using water potential, concentration gradient or membrane proteins, in full sentences.

#	Question	Answer and marking notes
5	Data 2: three similar potato strips are placed in distilled water, 0.2 mol dm⁻³ sucrose solution and 0.6 mol dm⁻³ sucrose solution for 30 minutes. The first strip changes from 2.0 g to 2.3 g and the second changes from 2.5 g to 2.3 g. (a) Calculate the percentage change in mass of the first strip. (b) Calculate the percentage change in mass of the second strip. (c) State which strip was most likely placed in distilled water and explain your answer. (d) Explain why the surface of the strips is blotted dry before the experiment.	(a) Percentage change = (2.3 − 2.0) ÷ 2.0 × 100% = +15% (2 marks). (b) Percentage change = (2.3 − 2.5) ÷ 2.5 × 100% = −8% (2 marks). (c) The first strip was most likely in distilled water because distilled water has a higher water potential than the cell sap, so water enters the cells by osmosis and the strip gains mass; the second strip loses mass, showing that water leaves the cells (3 marks). (d) Blotting removes surface water so that the balance reading is not affected and the initial and final masses are compared under the same conditions (2 marks).
6	(a) Explain what happens to human red blood cells when placed in distilled water and in 0.9% sodium chloride solution. (b) Define water potential and state its value for pure water. (c) Explain why a solution with a higher solute concentration has a lower water potential. (d) A plant cell placed in concentrated sucrose solution shows plasmolysis; explain this in terms of water potential.	(a) In distilled water the water potential is higher than that of the cytoplasm, so water enters the red blood cells by osmosis, the cells swell and haemolysis occurs; in 0.9% sodium chloride solution the water potentials are similar, so the cells keep their shape (2 marks). (b) Water potential is the free energy of water per unit volume, a measure of the tendency of water to move; pure water has a water potential of zero (2 marks). (c) Solute particles occupy space and bind water molecules, reducing the number of free water molecules, so the tendency of water to leave the solution is lower and its water potential is lower (2 marks). (d) The sucrose solution has a lower water potential than the cell sap, so water leaves the vacuole and cytoplasm, the vacuole shrinks and the cell membrane pulls away from the cell wall, producing plasmolysis (2 marks).
7	(a) Compare simple diffusion, facilitated diffusion and active transport in three respects: whether membrane proteins are needed, whether metabolic energy is used, and whether movement against the concentration gradient is possible. (b) Give one example of active transport in the human body. (c) Explain why a low temperature slows down both facilitated diffusion and active transport.	(a) Simple diffusion needs no membrane protein, uses no metabolic energy and moves substances only down the concentration gradient; facilitated diffusion needs channel or carrier proteins, uses no metabolic energy and moves substances only down the gradient; active transport needs carrier proteins, uses metabolic energy and can move substances against the gradient (3 marks). (b) Absorption of glucose by intestinal epithelial cells, or reabsorption of glucose in the kidney tubules (2 marks). (c) A low temperature reduces the kinetic energy of molecules, slowing diffusion and the shape changes of carrier proteins, and it also lowers the respiration rate and so the ATP supply, slowing both facilitated diffusion and active transport (3 marks).
8	Data 3: a root hair is modelled as a cylinder 0.5 mm long and 0.02 mm in diameter; take π as 3.14. (a) Calculate the lateral surface area of one root hair in mm². (b) Calculate the total lateral surface area of 1 000 root hairs. (c) Compare this with a smooth root surface of area 1 mm² and state one advantage for water uptake. (d) State the process by which root hairs mainly absorb water.	(a) Lateral surface area = π × diameter × length = 3.14 × 0.02 × 0.5 = 0.0314 mm², about 0.031 mm² (2 marks). (b) Total area of 1 000 root hairs = 1 000 × 0.0314 = 31.4 mm² (2 marks). (c) The total area is about 31 times that of a smooth surface of 1 mm², greatly increasing the absorbing surface so that more water and mineral salts are absorbed per unit time (3 marks). (d) Water enters root hairs mainly by osmosis (1 mark).

Section C　Enzymes and Metabolic Reactions

Section C: Consider both the shape of the active site and the rate of reaction, and give numerical answers to three significant figures.

#	Question	Answer and marking notes
9	Data 4: in an experiment, catalase catalyses the breakdown of hydrogen peroxide and 18 cm³ of oxygen is collected in the first 3 minutes. (a) Calculate the average rate of gas production in cm³ min⁻¹. (b) Predict the rate if the amount of enzyme is doubled with substrate in excess, and explain. (c) State one factor that would lower the rate between the fourth and fifth minutes.	(a) Average rate = 18 cm³ ÷ 3 min = 6.0 cm³ min⁻¹ (3 marks). (b) With double the enzyme the rate would be about 12 cm³ min⁻¹, because there are more active sites and, with substrate in excess, more substrate molecules can be catalysed at the same time (3 marks). (c) The substrate concentration falls as the reaction proceeds, products accumulate and inhibit the enzyme, or the enzyme gradually denatures (any one, 2 marks).
10	The initial rate of an enzyme reaction is 12 cm³ min⁻¹ at 20 °C and the Q₁₀ of the enzyme is 2. (a) Calculate the predicted rate at 30 °C. (b) Calculate the predicted rate at 40 °C. (c) Explain why the rate is close to zero at 60 °C.	(a) Predicted rate at 30 °C = 12 × 2 = 24 cm³ min⁻¹ (3 marks). (b) Predicted rate at 40 °C = 24 × 2 = 48 cm³ min⁻¹ (2 marks). (c) 60 °C is above the optimum temperature, so hydrogen bonds and other bonds in the enzyme break, the active site changes shape and the enzyme is permanently denatured, so the substrate can no longer bind and the rate is close to zero (3 marks).
11	(a) Explain how a change in pH affects enzyme activity and state what is meant by the optimum pH. (b) Explain why heavy metal ions inhibit enzyme activity. (c) State one difference between a competitive inhibitor and a non-competitive inhibitor. (d) Explain why an enzyme usually catalyses only one substrate, referring to the shape of the active site.	(a) A change in pH alters the charges on amino acid side chains, breaking hydrogen and ionic bonds so that the active site changes shape and fewer enzyme-substrate complexes form; the optimum pH is the pH at which enzyme activity is highest (2 marks). (b) Heavy metal ions bind to sulfhydryl groups of the enzyme and disrupt its three-dimensional structure, so the active site is distorted and the enzyme loses activity (2 marks). (c) A competitive inhibitor competes with the substrate for the active site and its effect can be reduced by increasing the substrate concentration, while a non-competitive inhibitor binds away from the active site, changes its shape, and increasing the substrate concentration does not relieve the inhibition (2 marks). (d) The active site has a specific three-dimensional shape and charge distribution, so only a substrate with a complementary shape can bind, and each enzyme therefore usually catalyses only one substrate (2 marks).
12	(a) Describe an experiment to measure the effect of temperature on enzyme activity, including the variables to be controlled. (b) State two variables that must be kept constant. (c) Explain why the same batch of enzyme and the same substrate concentration are used. (d) State one source of error when the rate is measured by the volume of gas collected.	(a) Pre-incubate equal volumes of enzyme solution and substrate solution separately in water baths at different temperatures, then mix them and start timing; measure the amount of product or the volume of gas at fixed intervals and calculate the rate (3 marks). (b) Enzyme concentration, substrate concentration, pH and reaction time (any two, 2 marks). (c) To keep the enzyme activity and substrate concentration the same in every trial, so that any difference in rate is caused only by the independent variable, temperature (2 marks). (d) Some gas may dissolve in the liquid or leak out, so the volume collected is less than the true volume (1 mark).

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