S4 Biology Practice Paper: Cell Structure, Transport Across Membranes and Enzymes (100 marks)

2026-11-10

S4 Biology – Cells and transport

S4 Biology paper: 6 questions, 30 parts, 100 marks, 75 min. Covers cell structure, transport across membranes, enzymes, photosynthesis, respiration and transport in plants.

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Section A – Cell structure and microscopy

Show your working and units in calculations.

#	Question	Answer and marking notes
1	A cell appears 8 mm across in a micrograph taken at 400x magnification. Calculate the actual diameter of the cell in micrometres.	Total magnification = 10 x 40 = 400; actual diameter = 8 mm / 400 = 0.02 mm = 20 micrometres.
2	Name three structures found in plant cells but not in animal cells.	Cell wall (cellulose), chloroplast, and a large central vacuole.
3	Explain why an electron microscope has a higher resolution than a light microscope.	Electrons have a much shorter wavelength than visible light, and resolution is about half the wavelength used; a light microscope resolves about 200 nm while an electron microscope reaches about 0.2 nm.
4	Why is the cell membrane described as a fluid mosaic model?	The phospholipid bilayer is fluid, and proteins, glycoproteins and cholesterol are embedded in or attached to it and can move laterally.
5	Why is a root tip squashed into a thin layer when observing mitosis? State which stage has the most cells, and why.	Squashing separates the cells into a single layer so light can pass through without cells overlapping; interphase occupies most of the cell cycle, so most cells are in interphase.

Section B – Transport across membranes

Use the terms water potential and concentration gradient where appropriate.

#	Question	Answer and marking notes
6	For cube-shaped cells with sides of 20 micrometres and 40 micrometres, calculate the surface area to volume ratio in per micrometre and state which diffuses more efficiently.	For 20 micrometres: surface area 6 x 20^2 = 2400, volume 8000, ratio 0.30 per micrometre. For 40 micrometres: surface area 9600, volume 64000, ratio 0.15 per micrometre. The smaller cell has the larger ratio and diffuses more efficiently.
7	Potato strips are placed in 0.5 M sucrose solution and in distilled water. How do they differ in appearance after 20 minutes?	In 0.5 M sucrose the cells lose water, so the strip becomes soft and shorter; in distilled water the cells take up water, turgor rises and the strip becomes firm and longer.
8	Explain the process of plasmolysis.	In a concentrated solution the external water potential is lower, so water leaves the vacuole through the cytoplasm and membrane; the vacuole shrinks and the membrane pulls away from the cell wall.
9	State three differences between active transport and simple diffusion.	Active transport uses ATP, needs carrier proteins and can work against the concentration gradient; simple diffusion needs no energy, no carrier and moves down the gradient.
10	A dialysis bag contains either 0.3 M sucrose or 0.3 M glucose. Which sugar is detected outside the bag sooner? Explain.	Glucose appears sooner: it is a smaller molecule and passes through carrier proteins in the membrane, whereas the larger sucrose molecule crosses much more slowly.

Section C – Enzymes and metabolic rate

Quote values from the data given in the question.

#	Question	Answer and marking notes
11	Enzymes are biological catalysts. Describe how an enzyme speeds up a reaction.	The enzyme binds its substrate to form an enzyme-substrate complex, lowering the activation energy so that more molecules have enough energy to react at the same temperature.
12	Rate of oxygen production by catalase at various temperatures in cubic centimetres per minute: 10 C = 12, 20 C = 28, 30 C = 46, 40 C = 52, 50 C = 30, 60 C = 8. (a) State the optimum temperature. (b) Explain why the rate falls at 50 C and 60 C.	(a) About 40 C, where the rate peaks at 52 cubic centimetres per minute. (b) Above the optimum the bonds holding the tertiary structure break, the active site changes shape and the enzyme denatures; denaturation is more severe at 60 C, so the rate is lower still.
13	By what percentage does the rate increase when the temperature rises from 20 C to 30 C?	(46 – 28) / 28 x 100 = 64.3 per cent, an increase of about 64 per cent.
14	Explain how pH affects enzyme activity.	Each enzyme has an optimum pH; away from it the charges on the amino acids at the active site change, ionic and hydrogen bonds break, and the enzyme denatures so activity falls.
15	How do heavy metal ions such as Cu2+ inhibit enzymes, and how does this differ from a competitive inhibitor?	Heavy metal ions bind to sulfhydryl groups and irreversibly change the protein structure, which is non-competitive inhibition; raising the substrate concentration does not restore activity. A competitive inhibitor competes with the substrate for the active site, so raising the substrate concentration relieves the inhibition.

Section D – Photosynthesis and respiration

Equations must be balanced; show mole calculations.

#	Question	Answer and marking notes
16	Write the balanced equation for photosynthesis, including the energy required.	6CO2 + 6H2O -> C6H12O6 + 6O2, using light energy absorbed by chlorophyll.
17	A plant absorbs 4.4 g of carbon dioxide per hour. Calculate the mass of glucose produced per hour (C = 12, H = 1, O = 16).	Moles of CO2 = 4.4 / 44 = 0.10 mol; 6 mol CO2 give 1 mol glucose, so 0.10 / 6 = 0.0167 mol; Mr of glucose is 180, so mass = 0.0167 x 180 = 3.0 g.
18	One leaf is tested with iodine solution for starch after 6 hours in light; another is kept in the dark for 24 hours first. Predict and explain both results.	The illuminated leaf turns blue-black because photosynthesis makes glucose which is stored as starch; the dark leaf shows no colour change because without photosynthesis the earlier starch has been used in respiration or exported.
19	State three differences between respiration and burning.	Respiration is enzyme-controlled and occurs at body temperature, releasing energy in many small steps stored as ATP; burning needs no enzyme, requires a high temperature and releases energy in one step as heat.
20	Why does muscle produce lactate during vigorous exercise? State one benefit of the reaction.	When oxygen supply is limited, lactate fermentation breaks down glucose to lactate and regenerates NAD+ so that glycolysis can continue making a small amount of ATP.

Section E – Transport in plants

Answer comparison questions point by point.

#	Question	Answer and marking notes
21	A root hair is 0.5 mm long and 10 micrometres in diameter. Calculate the lateral surface area of one root hair in square millimetres, to three significant figures.	Radius = 0.005 mm; lateral area = 2 x pi x 0.005 x 0.5 = 0.0157 square millimetres.
22	Hence calculate the total surface area of 1000 root hairs and state one advantage of root hairs for water uptake.	0.0157 x 1000 = 15.7 square millimetres; root hairs greatly increase the absorbing surface, speeding up osmosis and mineral uptake.
23	Explain how transpiration pull raises water from the root to the leaves.	Transpiration at the stomata lowers the water potential of mesophyll cells and creates a pull; cohesion between water molecules by hydrogen bonding lets the column withstand tension, and adhesion to the xylem walls keeps it rising.
24	Compare the substances carried and the directions of transport in xylem and phloem.	Xylem carries water and mineral salts upward from the root in one direction; phloem carries sucrose and amino acids from source to sink in either direction.
25	A ring of bark including the phloem is removed from a tree trunk. What happens above and below the ring? Explain.	Above the ring the bark swells as sugars accumulate; below it the phloem is cut, sugars cannot reach the roots and the roots may eventually die.

Section F – Experimental design and data handling

Identify the independent, dependent and controlled variables.

#	Question	Answer and marking notes
26	In an experiment comparing the effect of pH on amylase activity, state three variables that must be kept constant.	Temperature, volume and concentration of the enzyme, and volume and concentration of the starch solution, using the same batch of enzyme and the same reaction time.
27	0.2 g of potato with 5 cubic centimetres of hydrogen peroxide gives 24.0 cubic centimetres of oxygen in 60 s; 0.4 g of potato gives 48.0 cubic centimetres in 30 s. Calculate both mean rates and explain the difference.	0.2 g: 24.0 / 60 = 0.40 cubic centimetres per second; 0.4 g: 48.0 / 30 = 1.60 cubic centimetres per second. Doubling the enzyme raises the rate fourfold because more active sites are available so more substrate reacts at once.
28	Why should each treatment be repeated three times and a mean calculated?	Repeats reduce random error and make the result more reliable; a mean represents the trend better than a single reading.
29	10 cubic centimetres of 1.0 M sucrose solution is diluted with water to 50 cubic centimetres. Calculate the new concentration.	Moles = 1.0 x 0.010 = 0.010 mol; concentration = 0.010 / 0.050 = 0.20 M.
30	State the two elements a conclusion supported by experimental data must contain.	The relationship or trend between the independent and dependent variables, supported by data values or ranges.

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