S4 Biology Practice Paper: Cells, Enzymes And Transport Of Substances (12 Questions, 100 Marks, 75 Minutes, With Answers)

S4 Biology Practice Paper: Cells, Enzymes and Transport of Substances (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-09

HKDSE Biology: Cells, Enzymes and Transport of Substances

S4 Biology practice paper on cell structure, surface area to volume ratio, microscopy, enzyme activity, diffusion, osmosis, active transport and the mitotic index, with full answers.

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Section A: Data and Calculation Questions — Cells, Microscopy and Enzymes (32 marks)

Study the information in each question and answer as instructed.

#	Question	Answer and marking notes
1	Data: three cube-shaped cell models have sides of 1 cm, 2 cm and 3 cm, with surface area to volume ratios of about 6:1, 3:1 and 2:1. (a) State the surface area to volume ratio of the 1 cm and of the 3 cm model. (2 marks) (b) Explain why a small cell volume favours the movement of substances into and out of a cell. (4 marks)	(a) 6:1 for the 1 cm model and 2:1 for the 3 cm model. (2 marks) (b) A large surface area to volume ratio gives more membrane area per unit volume, and the diffusion path is short, so nutrients and wastes reach or leave the centre of the cell quickly and exchange is efficient. (4 marks)
2	A student used a compound microscope with a 10 times eyepiece and a 40 times objective. The image of a cell measured 40 mm across while the actual cell was 0.05 mm across. (a) Calculate the magnification and state two correct steps to take before using the high power objective. (4 marks) (b) Explain why specimens are stained before observation and name one common stain. (4 marks)	(a) Magnification is 40 mm divided by 0.05 mm, which is 800 times; focus with the low power objective first and then rotate to high power; after switching, focus only with the fine adjustment and adjust light with the iris and lamp. (4 marks) (b) Staining gives contrast to colourless transparent structures so that the nucleus and cell wall are easier to identify; methylene blue or iodine solution is commonly used. (4 marks)
3	In an experiment, 1% starch solution was mixed with amylase at 37 degrees C and the volume of product was recorded every 10 seconds; 12 cubic cm of product formed in 60 seconds, after which the rate fell clearly. (a) Calculate the average rate over the first 60 seconds and explain one reason for the later fall in rate. (4 marks) (b) Explain why the reaction nearly stops when the temperature is raised to 70 degrees C. (4 marks)	(a) Average rate is 12 cubic cm divided by 60 s, which is 0.2 cubic cm per second; the substrate is gradually used up, so its concentration falls and collisions between enzyme and substrate become less frequent, lowering the rate. (4 marks) (b) High temperature disrupts the three-dimensional structure of the enzyme, the active site changes shape and the substrate no longer fits; the enzyme is denatured and activity is not fully restored on cooling. (4 marks)
4	Three potato strips of the same size were placed in distilled water, 0.2 mol per dm3 sucrose solution and 0.5 mol per dm3 sucrose solution for 30 minutes. The strip in distilled water rose in mass from 2.4 g to 2.7 g. The water potential of the cell sap is -150 kPa. (a) Calculate the percentage change in mass of that strip. (3 marks) (b) Explain why its mass increased in distilled water. (4 marks) (c) Predict and explain what happens to the strip in 0.5 mol per dm3 sucrose solution. (3 marks)	(a) Percentage change is (2.7 minus 2.4) divided by 2.4 times 100%, which is plus 12.5%. (3 marks) (b) The water potential of distilled water is higher than that of the cell sap; water moves by osmosis through the partially permeable membrane into the cell, the vacuole swells, the cell becomes turgid and its mass rises. (4 marks) (c) The sucrose solution has a lower water potential than the cell sap, so water leaves the cell, turgor is lost, the cell becomes flaccid with plasmolysis and the mass falls. (3 marks)

Section B: Structured Questions — Transport of Substances and Cell Division (36 marks)

Answer in points, applying biological concepts.

#	Question	Answer and marking notes
5	(a) Compare diffusion and active transport, giving one similarity and one difference. (3 marks) (b) Explain why the rate at which root cells absorb mineral ions falls sharply after a respiratory inhibitor is added, while oxygen uptake continues. (6 marks)	(a) Similarity: both move substances from a higher to a lower concentration. Difference: active transport works against the gradient and needs carrier proteins and energy from ATP, while diffusion is passive and needs no energy. (3 marks) (b) Mineral ion concentration in the soil is lower than inside root cells, so uptake must be by active transport; this depends on ATP from respiration, and the inhibitor stops aerobic respiration and the ATP supply, so uptake falls sharply; oxygen can still enter by diffusion because diffusion needs no energy. (6 marks)
6	Villi in the small intestine and root hairs in plants are surfaces for exchange. (a) Describe three structural features of villi and root hairs that favour absorption. (4 marks) (b) State the role of microvilli on villus epithelial cells. (2 marks) (c) Explain why a multicellular organism cannot rely on diffusion through its body surface alone. (3 marks)	(a) Large surface area, with finger-like projections on villi and long extensions on root hairs; thin walls, a single layer of epithelial cells, so the diffusion distance is short; a dense blood capillary network in villi carries absorbed substances away and keeps a steep concentration gradient. (4 marks) (b) Microvilli further increase the apical surface area of epithelial cells and raise the rate of glucose and amino acid absorption. (2 marks) (c) As size increases the surface area to volume ratio falls, so diffusion through the body surface cannot supply inner cells, and respiratory and circulatory systems are needed. (3 marks)
7	A student examined a slide of onion root tip and counted 40 cells at random, of which 10 were in mitosis. (a) Calculate the mitotic index and state one point to note when choosing fields of view. (4 marks) (b) Explain two reasons why the meristem of a root tip is suitable for observing mitosis. (5 marks)	(a) Mitotic index is 10 divided by 40 times 100%, which is 25%; cells must be counted in several fields chosen at random, not only fields full of dividing cells, to reduce bias. (4 marks) (b) Cells in the meristem divide continuously, so a high proportion are in mitosis; the cells are small and their chromosomes stain and resolve clearly, and they are packed closely so many cells are seen in one field. (5 marks)
8	(a) Name three structures found in plant cells but not in animal cells and state the function of one of them. (4 marks) (b) Compare mesophyll cells with muscle cells in the structures that supply energy, and explain why the difference matters for function. (5 marks)	(a) Cell wall, chloroplast and large vacuole; the cell wall is made of cellulose and supports the cell, keeping its shape and preventing bursting when water enters. (4 marks) (b) Mesophyll cells contain many chloroplasts and use light energy to make organic matter by photosynthesis; muscle cells contain many mitochondria and need aerobic respiration to make large amounts of ATP for contraction. Structure matches function: the form of energy conversion differs, so the organelles differ in number. (5 marks)

Section C: Essay Questions — Synthesis and Evaluation (32 marks)

Answer each question with reference to biological evidence.

#	Question	Answer and marking notes
9	‘Diffusion alone can explain all movement of substances into and out of cells.’ Do you agree? (8 marks)	For: small molecules such as oxygen and carbon dioxide cross the phospholipid bilayer by simple diffusion, and water moves by osmosis. Against: glucose, amino acids and ions are charged or large and need carrier proteins in facilitated diffusion; uptake against a concentration gradient, as in root hairs absorbing mineral ions or intestinal cells absorbing glucose, needs active transport and ATP; large molecules such as proteins enter and leave by endocytosis and exocytosis. Diffusion explains only part of the movement, so a balanced conclusion with examples is required. (8 marks)
10	‘Enzyme activity is affected only by temperature.’ Do you agree? (8 marks)	Against: a change in pH alters the charges on amino acid side chains in the active site, and departure from the optimum pH reduces binding and can denature the enzyme; substrate and enzyme concentrations change the frequency of collisions and the rate; heavy metal ions act as inhibitors by binding to the enzyme; a lack of coenzymes or cofactors also stops the enzyme working. For: temperature matters greatly, since above the optimum the enzyme denatures and at low temperature molecular kinetic energy falls. Temperature is a major factor but not the only one. (8 marks)
11	‘A multicellular organism must develop a transport system in order to survive.’ Explain with one example from plants and one from animals. (8 marks)	Plants: in a tall tree the roots and leaves are far apart, and the xylem carries water and mineral ions from root to leaf by transpiration pull while the phloem moves products of photosynthesis from source to sink; by diffusion alone water would move far too slowly over such a distance to supply transpiration and photosynthesis. Animals: a large body has a small surface area to volume ratio, so gases and nutrients cannot reach deep tissues by diffusion; mammals use a heart, blood vessels and blood to carry oxygen, glucose and hormones and to remove carbon dioxide and urea. A transport system overcomes the limits of distance and of surface area to volume ratio, so it is essential for survival. (8 marks)
12	‘Progress in microscopy was the key to the establishment of the cell theory.’ Do you agree? (8 marks)	For: in 1665 Hooke used a compound microscope on cork and saw box-like compartments, coining the term cell; better resolution of light microscopes allowed Brown to observe the nucleus, and Schleiden and Schwann concluded that plants and animals are made of cells; electron microscopes reaching nanometre resolution revealed organelles and membrane structure and confirmed the cell as the basic unit of life. Against: staining techniques, sectioning and microtomy were were equally essential, since without staining the nucleus and organelles cannot be distinguished; contemporary biological ideas also shaped the theory. Microscopy was one key condition among several. (8 marks)

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