S4 Biology Practice Paper: Cells, Enzymes, Transport And Cell Division (100 Marks)

S4 Biology Practice Paper: Cells, Enzymes, Transport and Cell Division (100 marks)

2026-11-12

Biology – Cells, Enzymes and Transport

S4 Biology: 6 questions, 26 parts, 100 marks, 60 minutes. Magnification, enzyme rate, osmosis, active transport, mitotic index and error. Full answers.

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Question 1 – Magnification and Cell Size (18 marks)

A student observes onion epidermal cells with a compound microscope. The eyepiece is 10x and the objective is 40x. The image of one cell is 8 mm long.

#	Question	Answer and marking notes
1	(a) Find the total magnification of the microscope.	Answer: total magnification = eyepiece x objective = 10 x 40 = 400 times. Marking: formula 2, value with unit 2.
2	(b) Find the actual length of the cell in micrometres.	Answer: first convert the image length: 8 mm = 8,000 micrometres. Actual length = image length / magnification = 8,000 / 400 = 20 micrometres. Marking: unit conversion 2, division by magnification 2.
3	(c) Assume the cell is a sphere of radius 10 micrometres. Find its surface area to volume ratio (per micrometre). What is the ratio if the radius increases to 20 micrometres? Explain the change.	Answer: surface area = 4 pi r squared = 4 pi x 100, about 1,257 square micrometres; volume = 4/3 pi r cubed, about 4,189 cubic micrometres; ratio = 1,257 / 4,189, about 0.30 per micrometre. At radius 20 micrometres, surface area rises with r squared but volume with r cubed, so the ratio becomes 3/r = 3 / 20 = 0.15 per micrometre, that is halved. Marking: surface area 1, volume 1, each ratio 1, explanation 1.
4	(d) If a cell 20 micrometres long is magnified 1,500 times, find the image length in millimetres.	Answer: image length = 20 x 1,500 = 30,000 micrometres; 30,000 micrometres = 30 mm. Marking: multiplication 2, conversion to mm 1.
5	(e) Give one reason for starting observation with the low power objective.	Answer: the low power objective gives a wider field of view, so the specimen is easier to locate on the slide, and the objective is less likely to hit the slide or crush the specimen. Marking: wider field or easier location 1, avoids damage 1.

Question 2 – Enzyme Activity and Rate (17 marks)

A student uses amylase to break down starch and measures the volume of reducing sugar produced in 4 minutes at different temperatures: 30 degrees C gives 12 cm3, 40 degrees C gives 18 cm3 and 60 degrees C gives 0 cm3. The pH and enzyme concentration are the same in all tubes.

#	Question	Answer and marking notes
6	(a) Find the mean rate of reaction at 30 degrees C and at 40 degrees C in cm3 per minute.	Answer: rate = volume / time. At 30 degrees C: 12 / 4 = 3.0 cm3 per minute; at 40 degrees C: 18 / 4 = 4.5 cm3 per minute. Marking: formula 1, each value 1.
7	(b) Find the temperature coefficient Q10 between 30 degrees C and 40 degrees C.	Answer: Q10 = rate at the higher temperature / rate at the lower temperature = 4.5 / 3.0 = 1.5. Marking: correct pairing of rates 2, value 2.
8	(c) Explain why no product is formed at 60 degrees C.	Answer: an enzyme is a protein. High temperature breaks bonds within the molecule and changes its three dimensional shape, including the shape of the active site, so the substrate can no longer bind. This is denaturation and it is irreversible, so the reaction stops. Marking: denaturation of protein 2, active site change stops binding 2.
9	(d) At 40 degrees C, how many minutes are needed to produce 27 cm3 of reducing sugar?	Answer: time = volume / rate = 27 / 4.5 = 6 minutes, assuming the rate stays constant over that period. Marking: formula 2, value with unit 1.
10	(e) Apart from temperature, state two variables that must be kept constant and explain why.	Answer: (1) pH, because each enzyme has an optimum pH and a shift alters the charges at the active site and lowers the rate; (2) substrate or enzyme concentration, because both directly affect the rate, so a change would make the result impossible to attribute to temperature. Marking: each variable 1, reason for control 1.

Question 3 – Diffusion and Osmosis (17 marks)

Four groups of potato strips each weighing 2.50 g are kept for 30 minutes in sucrose solutions of 0.0, 0.2, 0.4 and 0.6 mol per dm3. Their masses afterwards are 2.80 g, 2.60 g, 2.40 g and 2.20 g.

#	Question	Answer and marking notes
11	(a) Find the percentage change in mass for the 0.0 and 0.6 mol per dm3 groups, and state what the sign means.	Answer: percentage change = (final mass – initial mass) / initial mass x 100. For 0.0: (2.80 – 2.50) / 2.50 x 100 = +12.0 per cent; for 0.6: (2.20 – 2.50) / 2.50 x 100 = -12.0 per cent. A positive sign means net uptake of water and a gain in mass; a negative sign means net loss of water. Marking: formula 1, each value 1.5, meaning of sign 1.
12	(b) Estimate the isotonic sucrose concentration for the potato cells and explain your method.	Answer: the 0.2 group gained 4.0 per cent and the 0.4 group lost 4.0 per cent, so the sign of the change flips between them and the isotonic concentration lies in between. By linear interpolation the midpoint is 0.30 mol per dm3, where net water movement is zero and the mass does not change. Marking: between 0.2 and 0.4, 2; midpoint with zero net water movement, 2.
13	(c) Use water potential and plasmolysis to explain why the strips in 0.6 mol per dm3 became soft.	Answer: the 0.6 mol per dm3 solution has a lower water potential than the cell sap, so water leaves the cell across the partially permeable membrane. The cell shrinks and the membrane pulls away from the cell wall, which is plasmolysis, and the loss of turgor pressure makes the tissue soft. Marking: lower external water potential 2, plasmolysis and loss of turgor 2.
14	(d) If the strips from the 0.0 group were then placed in 0.6 mol per dm3 solution for 30 minutes, predict the change in mass and explain.	Answer: the mass would fall, roughly by 12 per cent. The cells were fully hydrated, so their sap was diluted and still had a higher water potential than the external solution; the gradient is now reversed and water leaves the cells until they approach isotonic conditions. Marking: prediction of a fall 2, explanation by reversed water potential gradient 2.

Question 4 – Active Transport and Respiration (16 marks)

Intestinal epithelial cells are kept in a glucose medium and the glucose absorbed by each 100 mg of cells in 10 minutes is measured in micromoles: control 6.0; with cyanide 1.8; at 10 degrees C 2.4; control with doubled substrate concentration 9.6.

#	Question	Answer and marking notes
15	(a) Find the uptake rate of the control in micromoles per minute per 100 mg.	Answer: rate = uptake / time = 6.0 / 10 = 0.60 micromoles per minute per 100 mg. Marking: formula 2, value with unit 2.
16	(b) Find the uptake rate with cyanide as a percentage of the control and state the conclusion.	Answer: the cyanide rate = 1.8 / 10 = 0.18 micromoles per minute; as a percentage of the control, 0.18 / 0.60 x 100 = 30 per cent. Conclusion: cyanide blocks respiration and ATP production, so the sharp fall shows the process needs energy. Marking: calculation 2, conclusion 2.
17	(c) Explain why the uptake rate at 10 degrees C is lower.	Answer: low temperature reduces the activity of respiratory enzymes, so less ATP is produced, and the movement of molecules and the shape changes of carrier proteins are slower. Active transport needs ATP, so the rate falls to 40 per cent of the control. Marking: less ATP 2, slower enzyme and carrier activity 2.
18	(d) Doubling the substrate concentration raises the rate only 1.6 times. Explain.	Answer: active transport is limited by the number of carrier proteins in the membrane. Once most carriers are occupied the system is close to saturation, so a further rise in substrate concentration cannot raise the rate; carrier number is now the limiting factor. Marking: limited number of carriers 2, saturation as limiting factor 2.

Question 5 – Mitosis and the Cell Cycle (16 marks)

A slide of onion root tip is examined and 250 cells are counted. Twenty of them are in mitosis: 8 in prophase, 5 in metaphase, 4 in anaphase and 3 in telophase.

#	Question	Answer and marking notes
19	(a) Find the mitotic index as a percentage.	Answer: mitotic index = cells in mitosis / total cells x 100 = 20 / 250 x 100 = 8.0 per cent. Marking: formula 2, value 2.
20	(b) Find the number and percentage of cells in interphase.	Answer: cells in interphase = 250 – 20 = 230; percentage = 230 / 250 x 100 = 92.0 per cent. Marking: number 2, percentage 2.
21	(c) Which stage lasts longest, and how does the data support this?	Answer: interphase. The number of cells seen in a stage is proportional to how long that stage lasts; interphase accounts for 92.0 per cent, far above any mitotic stage (prophase, the highest, is only 8 / 250 x 100 = 3.2 per cent), so interphase occupies most of the cell cycle. Marking: interphase 2, argument from counts proportional to time 2.
22	(d) A cell in G1 contains 2C of DNA. State the DNA content after S phase and in each daughter cell after mitosis, and give the ratio of the two.	Answer: S phase replicates the DNA, raising the content to 4C; after mitosis the chromosomes are shared equally between two daughter cells, each returning to 2C. The ratio is 4C : 2C = 2 : 1. Marking: 4C and 2C 1 each, ratio 2.

Question 6 – Microscopic Measurement and Error (16 marks)

A student measures cells with an eyepiece graticule: 40 divisions equal 0.10 mm, and one cell covers 12 divisions. The published length of this cell type is 32 micrometres.

#	Question	Answer and marking notes
23	(a) Find the actual length represented by one graticule division in micrometres.	Answer: 0.10 mm = 100 micrometres; each division = 100 / 40 = 2.5 micrometres. Marking: conversion of mm to micrometres 2, division by the number of divisions 2.
24	(b) Find the measured length of the cell in micrometres.	Answer: measured length = 12 x 2.5 = 30 micrometres. Marking: multiplication 2, value with unit 2.
25	(c) Find the percentage error of the measurement relative to the published value.	Answer: percentage error = \|measured – published\| / published x 100 = \|30 – 32\| / 32 x 100 = 6.25 per cent. Marking: formula 2, value 2.
26	(d) Suggest two ways to improve the accuracy of the measurement.	Answer: (1) measure many cells and take the mean, which reduces random error; (2) use a higher power objective and recalibrate the graticule for finer resolution; locating the specimen at low power before switching up also reduces the risk of measuring the wrong cell. Marking: each valid method with its effect 2.

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