S5 Biology Practice Paper: Cell Cycle, Enzymes, Transport, Immunity And Biotechnology

S5 Biology Practice Paper: Cell Cycle, Enzymes, Transport, Immunity and Biotechnology

2026-11-12

S5 Biology – Cells, Transport and Biotechnology

S5 Biology practice paper on the cell cycle and mitosis, enzymes, transport, immunity and vaccines, plant transport and biotechnology, with full model answers and marks.

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Paper details　S5 Biology practice paper with model answers on the cell cycle, enzymes, transport, immunity and biotechnology.

Part 1: Cell cycle and mitosis (20 marks)

Four questions. Show working and give suitable units.

#	Question	Answer and marking notes
1	(a) A somatic cell has 24 chromosomes. Find the number of chromatids at metaphase of mitosis, and state the chromosome number and DNA content (in C, where a G1 cell has 2C) of each daughter cell after division.	Answer: at metaphase each chromosome has two sister chromatids, so chromatids = 24 x 2 = 48. Each daughter cell receives 24 chromosomes and 2C DNA. (2 marks chromatids, 2 marks chromosomes, 1 mark DNA)
2	(b) State the correct order of the four stages of mitosis, name the stage in which sister chromatids separate, and describe the role of the spindle fibres.	Answer: prophase, metaphase, anaphase, telophase. Sister chromatids separate in anaphase. Spindle fibres are microtubules that shorten during anaphase and pull the chromatids to opposite poles, so each daughter cell receives a complete set of chromosomes. (2 marks order, 1 mark stage, 2 marks spindle role)
3	(c) A tissue has a cell cycle of 9 h in G1, 6 h in S, 4 h in G2 and 1 h in M. Find the total length of the cycle and the percentage of the cycle spent in S and in M.	Answer: total = 9 + 6 + 4 + 1 = 20 h. S phase = 6 / 20 x 100 = 30.0%; M phase = 1 / 20 x 100 = 5.0%. (2 marks total, 2 marks S, 1 mark M)
4	(d) In a root tip squash, 30 out of 600 cells are in mitosis. Calculate the mitotic index and explain what this value indicates about the tissue.	Answer: mitotic index = 30 / 600 = 5.0% (0.05). It is the proportion of cells in mitosis; 5.0% is a high value, consistent with active division in the root tip meristem. (2 marks calculation, 2 marks meaning, 1 mark tissue link)

Part 2: Enzymes and metabolism (16 marks)

Four questions. Use correct enzyme kinetics terminology in your explanations.

#	Question	Answer and marking notes
5	(a) Explain how an enzyme lowers the activation energy of a reaction and why the active site is specific to its substrate.	Answer: the active site binds the substrate to form an enzyme-substrate complex, holding substrate molecules close together and in the correct orientation, so less activation energy is needed. The active site shape and charge are complementary to one substrate only, giving specificity. (2 marks activation energy, 1 mark complex, 1 mark specificity)
6	(b) An enzyme has an initial rate of 12 units per minute at 25 C and 30 units per minute at 35 C. Find how many times faster the reaction is at 35 C.	Answer: ratio = 30 / 12 = 2.5. Higher temperature gives molecules more kinetic energy, so substrate molecules collide with active sites more often. (2 marks calculation, 2 marks explanation)
7	(c) The same enzyme has a rate of 3 units per minute at 60 C. Find this rate as a percentage of the rate at the optimum temperature (35 C) and explain the effect of high temperature on the enzyme.	Answer: percentage = 3 / 30 x 100 = 10%. High temperature breaks the hydrogen and ionic bonds holding the tertiary structure, the active site is distorted and the enzyme is irreversibly denatured, so the substrate can no longer bind. (2 marks calculation, 2 marks denaturation)
8	(d) Distinguish competitive from non-competitive inhibitors and state which type heavy metal ions usually belong to.	Answer: a competitive inhibitor resembles the substrate and occupies the active site, so raising substrate concentration reverses it. A non-competitive inhibitor binds elsewhere and changes the enzyme shape, so more substrate does not help. Heavy metal ions such as mercury bind sulfhydryl groups and are non-competitive. (1.5 marks each mechanism, 1 mark heavy metals)

Part 3: Transport of substances (18 marks)

Four questions. Show steps in osmosis and surface area calculations.

#	Question	Answer and marking notes
9	(a) Human red blood cells are placed in distilled water and in 3% sodium chloride solution (plasma is about 0.9%). Describe the changes and explain them in terms of water potential.	Answer: distilled water has a higher water potential than the cytoplasm, so water enters by osmosis and the cell swells until the membrane bursts (haemolysis). The 3% solution has a lower water potential, so water leaves and the cell shrinks and becomes crenated. In 0.9% solution there is no net change because the water potentials are equal. (2 marks each case, 1 mark water potential)
10	(b) Two potato strips are weighed: strip A is 4.0 g and becomes 4.6 g after 30 minutes in 0.2 mol/dm3 sucrose; strip B is 5.0 g and becomes 4.4 g in 0.6 mol/dm3 sucrose. Calculate the percentage change in mass of each and state whether water entered or left the cells.	Answer: strip A = (4.6 – 4.0) / 4.0 x 100 = +15.0%, mass increased, so cells gained water. Strip B = (4.4 – 5.0) / 5.0 x 100 = -12.0%, mass decreased, so cells lost water. (2.5 marks each strip)
11	(c) Compare diffusion, osmosis and active transport in terms of energy requirement, need for membrane proteins and direction of net movement.	Answer: diffusion and osmosis are passive and need no metabolic energy, moving down a concentration or water potential gradient. Active transport needs ATP from respiration and carrier proteins, and can move substances against a gradient. Osmosis refers specifically to water crossing a partially permeable membrane. (1 mark each comparison, 1 mark osmosis definition)
12	(d) Two cubes have sides of 2 cm and 4 cm. Calculate the surface area to volume ratio of each and explain which loses heat more easily through its surface.	Answer: the 2 cm cube has area 6 x 2^2 = 24 cm2 and volume 8 cm3, ratio 3.0 per cm. The 4 cm cube has area 96 cm2 and volume 64 cm3, ratio 1.5 per cm. The smaller cube has the larger ratio, so more surface per unit volume and faster heat loss. (1.5 marks each calculation, 1 mark conclusion)

Part 4: Immunity and vaccines (16 marks)

Four questions. Use correct immunological terms.

#	Question	Answer and marking notes
13	(a) Define antigen and antibody and name the cell type that produces antibodies.	Answer: an antigen is a foreign substance, usually a protein or polysaccharide on a pathogen surface, that triggers a specific immune response. An antibody is an immunoglobulin secreted by plasma cells and binds to one specific antigen. Antibodies are produced by plasma cells derived from B lymphocytes. (1.5 marks each definition, 1 mark cell type)
14	(b) After a first infection the antibody titre is 1:100; after a second infection with the same pathogen it rises to 1:1600. Find the fold increase and explain why the second response is faster and stronger.	Answer: fold increase = 1600 / 100 = 16. Memory B cells remain after the first response; on re-exposure they divide and differentiate rapidly into plasma cells, giving a larger and faster antibody response, the secondary response. (2 marks calculation, 1 mark memory cells, 1 mark features)
15	(c) A community has 480 000 people and needs 75% immunity for herd immunity. Calculate the number of people who must be vaccinated.	Answer: number vaccinated = 480 000 x 0.75 = 360000. Herd immunity reduces the proportion of susceptible hosts so the pathogen cannot spread easily, protecting those who cannot be vaccinated. (2 marks calculation, 2 marks mechanism)
16	(d) Explain why antibiotics cannot treat influenza and why the influenza vaccine is updated each year.	Answer: influenza is caused by a virus, which has no cell wall, ribosomes or metabolic pathways targeted by antibiotics, so antibiotics work only against bacteria. The surface antigens of the virus mutate frequently (antigenic drift), so antibodies from an old vaccine no longer recognise the new strain and the vaccine must be updated each year. (2 marks antibiotics, 2 marks antigenic drift)

Part 5: Transport in plants (16 marks)

Four questions. Show steps and units in the potometer calculation.

#	Question	Answer and marking notes
17	(a) Explain how transpiration pull and root pressure together move water up the xylem.	Answer: transpiration from the stomata lowers the water potential of mesophyll cells, water is drawn from the xylem and the resulting tension (transpiration pull) pulls the continuous water column upwards. Root pressure arises when roots actively secrete mineral ions into the xylem, lowering its water potential so water enters and pushes water up, mainly at night or in humid air. (2 marks each)
18	(b) A potometer shows the air bubble moving 12 mm in 30 minutes; the capillary has a cross sectional area of 1.0 mm2. Find the volume of water taken up in 30 minutes and per hour, in cm3.	Answer: volume = 12 mm x 1.0 mm2 = 12 mm3. In one hour the bubble moves 24 mm, giving 24 mm3 = 0.024 cm3 per hour. (2 marks each)
19	(c) Explain the pressure flow hypothesis of translocation and why sugars accumulate in the bark above a girdled ring.	Answer: at the source, sucrose is actively loaded into sieve tubes, lowering water potential so water enters from the xylem and raises the hydrostatic pressure. At the sink, sucrose is unloaded, water leaves and pressure falls, so sap flows from source to sink. Girdling removes the phloem and breaks the path from leaves to roots, so sugars accumulate above the ring. (2 marks each)
20	(d) List three environmental factors affecting the rate of transpiration and briefly explain each mechanism.	Answer: low humidity increases the water potential gradient between the leaf and the air, speeding transpiration. Wind removes the humid layer at the leaf surface and maintains the vapour gradient. Light opens the stomata and raises leaf temperature. (1 mark each factor, 1 mark each mechanism, maximum 4 marks)

Part 6: Biotechnology (14 marks)

Four questions. Give technical steps in the correct order.

#	Question	Answer and marking notes
21	(a) Name the three steps of PCR and state what happens in each, then calculate the number of copies produced from one DNA molecule after 30 cycles (give the power of 2 and the value).	Answer: denaturation near 95 C separates the two strands; annealing lets primers bind to complementary sequences; extension by a heat-stable DNA polymerase synthesises new strands. Copies = 2^30 = 1,073,741,824 (about 1.07 x 10^9). (1 mark each step, 1 mark calculation)
22	(b) In gel electrophoresis, which fragment moves further, 200 bp or 2000 bp? Explain.	Answer: the 200 bp fragment moves further. The gel is a mesh and DNA is negatively charged, so it migrates towards the positive electrode; smaller fragments meet less resistance and travel further, allowing size estimation. (1 mark conclusion, 1 mark charge, 2 marks size and resistance)
23	(c) Outline the three main steps in producing human insulin using genetically modified bacteria.	Answer: cut out the human insulin gene and the plasmid with the same restriction enzyme; join the gene into the plasmid using DNA ligase to form a recombinant plasmid; introduce the plasmid into bacteria, select transformed cells, culture them and extract and purify the insulin produced. (1 mark each step)
24	(d) State two ecological or ethical concerns about genetically modified food.	Answer: ecologically, herbicide resistance genes may spread by pollen to wild relatives, creating superweeds, or harm non-target insects. Ethically, consumers worry about long term safety and informed choice, and some object to patenting genetic resources. (1.5 marks each)

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