S4 Biology Paper: Cells, Enzymes And Metabolism (100 Marks)

S4 Biology Paper: Cells, Enzymes and Metabolism (100 marks)

2026-11-11

S4 Biology – cells, enzymes and metabolism

S4 Biology paper: 6 questions, 22 parts, 100 marks, 75 minutes on enzyme kinetics, osmosis, photosynthesis, cell division and energy flow, with full marking notes.

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Paper details　S4 Biology practice paper covering enzyme kinetics, osmosis and water potential, photosynthesis, cell division, energy flow and microscope measurement. 6 questions, 22 parts, 100 marks, 75 minutes, with answers and marking notes.

Section A: Enzymes and Reaction Rates (18 marks)

An enzyme has Vmax = 120 umol min-1 and Km = 4.0 mM. Show your working in every part.

#	Question	Answer and marking notes
1	Calculate the reaction rate when the substrate concentration is 4.0 mM using the Michaelis-Menten equation.	Answer: rate = 120 x 4.0 / (4.0 + 4.0) = 60 umol min-1, exactly half of Vmax.
2	What is the reaction rate when the substrate concentration rises to 36 mM?	Answer: rate = 120 x 36 / 40 = 108 umol min-1, which is ninety per cent of Vmax.
3	When the substrate concentration rises from 4.0 mM to 12 mM, by what factor does the rate increase, and what percentage of Vmax is reached?	Answer: at 12 mM the rate is 90 umol min-1, a factor of 90 / 60 = 1.5; that is 90 / 120 x 100 = 75.0 per cent of Vmax.
4	State the values and units of Vmax and Km, and explain which property of the enzyme Km reflects.	Answer: Vmax = 120 umol min-1 and Km = 4.0 mM. Km is the substrate concentration giving half of Vmax, so it reflects the affinity of the enzyme for its substrate: a lower Km means higher affinity.
5	On a Lineweaver-Burk plot the slope equals Km divided by Vmax. Calculate the slope and give its unit.	Answer: slope = 4.0 / 120 = 0.033 min umol-1, that is 1/30, with units of minutes per umol.
6	A competitive inhibitor raises Km to 12 mM while Vmax stays unchanged. Calculate the new slope and describe how the plot changes.	Answer: new slope = 12 / 120 = 0.10 min umol-1. The line becomes steeper while the vertical intercept 1/Vmax is unchanged, so both lines meet the same point on the y-axis.

Section B: Osmosis and Water Potential (16 marks)

A piece of plant tissue of mass 2.40 g is placed in 0.35 M sucrose for 30 minutes and then weighs 2.16 g. Temperature is 298 K and the gas constant is 8.31 J mol-1 K-1.

#	Question	Answer and marking notes
7	Calculate the percentage change in mass of the tissue.	Answer: change = (2.16 – 2.40) / 2.40 x 100 = -10.0 per cent, a net loss of one tenth of the mass.
8	Calculate the water potential of 0.35 M sucrose at 298 K in MPa.	Answer: water potential = -(0.35 x 1000 x 8.31 x 298) / 10^6 = -0.867 MPa, since 0.35 mol dm-3 is 350 mol m-3.
9	The same tissue shows no change in mass in 0.15 M sucrose. Work out the water potential of that solution and hence estimate the water potential of the cell sap.	Answer: water potential = -(150 x 8.31 x 298) / 10^6 = -0.371 MPa. No change in mass means the two solutions are isotonic, so the cell sap is about -0.371 MPa.
10	An animal cell bursts in distilled water but a plant cell does not. Explain using water potential and cell structure.	Answer: distilled water has a water potential of zero, higher than the cell sap, so water enters. An animal cell has only a plasma membrane and no wall to restrain it, so it swells until it bursts; in a plant cell the vacuole swells and the wall generates pressure potential that opposes further entry.

Section C: Photosynthesis and Respiration (18 marks)

A leaf of area 0.025 m2 absorbs 1.50 mmol of carbon dioxide in 30 minutes in the light and releases 0.30 mmol in 30 minutes in the dark.

#	Question	Answer and marking notes
11	Calculate the net rate of photosynthesis in the light in mmol m-2 h-1.	Answer: net rate = 1.50 / 0.025 / 0.5 = 120 mmol m-2 h-1, since 30 minutes is 0.5 hour.
12	Calculate the respiration rate in the dark in mmol m-2 h-1.	Answer: respiration rate = 0.30 / 0.025 / 0.5 = 24 mmol m-2 h-1.
13	Work out the gross rate of photosynthesis in the light and explain why the respiration rate is added to the net rate.	Answer: gross rate = 120 + 24 = 144 mmol m-2 h-1. In the light the leaf respires as well as photosynthesises, so the carbon dioxide taken in is only the net difference between the two processes.
14	Assuming all the carbon is converted to glucose, with 6 mol of carbon dioxide giving 1 mol of glucose of molar mass 180.16 g mol-1, calculate the glucose yield per hour per square metre.	Answer: glucose = 144 / 6 = 24 mmol m-2 h-1; mass = 24 x 180.16 / 1000 = 4.32 g m-2 h-1.

Section D: Cell Division and Genetic Material (18 marks)

A cell has a cycle of 20 hours, of which the synthesis phase lasts 6 hours and the mitotic phase 1.5 hours.

#	Question	Answer and marking notes
15	Calculate the percentage of the cell cycle occupied by the synthesis phase and by the mitotic phase.	Answer: synthesis phase = 6 / 20 x 100 = 30.0 per cent; mitotic phase = 1.5 / 20 x 100 = 7.5 per cent.
16	An organism has 8 chromosomes in a body cell. State the number of possible chromosome combinations in one gamete and the number of genotype combinations formed when two gametes fuse.	Answer: combinations in one gamete = 2^4 = 16; when two gametes fuse the combinations are 16 x 16 = 256.
17	Human body cells have 46 chromosomes. Ignoring crossing over, how many chromosome combinations can one sperm carry?	Answer: the number of combinations is 2^23 = 8 388 608.
18	Compare the DNA content and chromosome number of the daughter cells produced by mitosis and by meiosis, and give one example of the role of each process.	Answer: mitosis produces two daughter cells of 2C from a 2C cell after replication, with the same chromosome number as the parent, and serves growth and repair; meiosis gives four 1C gametes after two divisions with the chromosome number halved, generating genetic variation.

Section E: Energy Flow in Ecosystems (14 marks)

Producers in an ecosystem fix 48 000 kJ per square metre per year, and only one tenth of the energy at each trophic level passes to the next.

#	Question	Answer and marking notes
19	Calculate the energy available per square metre per year to the primary, secondary and tertiary consumers, and state the transfer efficiency at each level.	Answer: primary consumers = 48 000 x 0.1 = 4 800 kJ; secondary = 480 kJ; tertiary = 48 kJ, with an efficiency of 10 per cent at each step.
20	A primary consumer ingests 6 000 kJ, assimilates 4 800 kJ and respires 3 840 kJ. Calculate its net production and the percentage of assimilated energy that becomes net production.	Answer: net production = 4 800 – 3 840 = 960 kJ; percentage = 960 / 4 800 x 100 = 20.0 per cent, the other eighty per cent being lost as heat from respiration.

Section F: Microscopy and Experimental Design (16 marks)

A microscope has a 10x eyepiece. With the 10x objective the field of view measures 1.8 mm in diameter; the same specimen is then viewed with the 40x objective.

#	Question	Answer and marking notes
21	Calculate the total magnification with the 40x objective and the new field diameter in millimetres and micrometres.	Answer: total magnification = 10 x 40 = 400; field diameter = 1.8 x 10 / 40 = 0.45 mm, that is 450 um. Magnification rises fourfold and the field diameter falls to one quarter.
22	At 400x the image of a cell measures 90 mm long. Calculate the actual length of the cell and state two possible sources of error in this measurement.	Answer: actual length = 90 / 400 = 0.225 mm = 225 um. Errors arise because the cell boundary is blurred so the end points are uncertain, and because the eyepiece graticule may not be calibrated, while a cell that is not lying flat can make the length appear greater.

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