10 free AS Level Biology Cells & Molecules questions written in the style of Cambridge Biology papers: 3 multiple-choice and 7 structured, each with its answer and mark scheme. 3 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Cell structure, Biological molecules, Enzymes, Cell membranes & transport, The mitotic cell cycle, Nucleic acids & protein synthesis. Work through the questions below, then generate a full Cells & Molecules paper with a fresh set every time.
Generate a full Cells & Molecules paper →An eyepiece graticule is calibrated against a stage micrometer: 100 eyepiece divisions equal 10 stage-micrometer divisions, and each stage division is 10 µm. A row of 3 identical cells measures 15 eyepiece divisions.
What is the length of one cell?
Answer: B
A cube of agar with sides of 4 cm is cut from a block.
What is its surface area to volume ratio?
Answer: C
A student tests a solution with Benedict’s solution and it stays blue. After boiling with dilute acid and neutralising, the Benedict’s test gives a brick-red precipitate.
What does the solution contain?
Answer: B
An enzyme has a maximum rate of reaction (Vmax) of 80 µmol min−1 and a Michaelis–Menten constant (Km) of 2 mmol dm−3.
Explain what is meant by Vmax and by Km.
State the rate at a substrate concentration of 2 mmol dm−3.
Explain what a low Km value indicates about the enzyme.
Describe and explain the effect of a competitive inhibitor on Vmax and Km.
Describe how a non-competitive inhibitor differs in its effect.
Enzymes can be immobilised in alginate beads. Suggest two advantages of using immobilised enzymes in industry.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | Vmax: maximum rate when all active sites occupied; Km: substrate concentration at half Vmax Vmax: the maximum rate, when all active sites are occupied (saturated) B1 Km: the substrate concentration at which the rate is half Vmax B1 | 2 |
| (b) | 40 µmol min−1 40 µmol min−1 B1 | 1 |
| (c) | high affinity for substrate; reaches half Vmax at low concentration high affinity for its substrate B1 the enzyme–substrate complex forms readily even at low substrate concentration B1 | 2 |
| (d) | Km increases, Vmax unchanged; inhibitor similar shape, binds active site; overcome by high substrate Km increases B1 Vmax unchanged B1 the inhibitor has a similar shape to the substrate and competes for the active site B1 at high substrate concentration the substrate out-competes the inhibitor so Vmax is still reached B1 | 4 |
| (e) | binds elsewhere (allosteric), changes active site shape; Vmax decreases, Km unchanged binds at a site other than the active site, changing its shape B1 Vmax is reduced (and cannot be overcome by more substrate) while Km is unchanged B1 | 2 |
| (f) | reused; product not contaminated; more stable to temperature/pH Any 2 of: enzyme can be reused / easily recovered; the product is not contaminated with enzyme; more stable over a wider range of temperature and pH B2 | 2 |
Starch, glycogen and cellulose are polysaccharides of glucose.
Compare the structures of amylose and cellulose.
Explain how glycogen is suited to energy storage in animals.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | α vs β glucose; helix vs straight chains; H-bonds between chains amylose: α-glucose, 1,4 glycosidic bonds, helix B1 cellulose: β-glucose, alternate molecules inverted B1 straight chains linked by hydrogen bonds into microfibrils B1 | 3 |
| (b) | branched, compact, insoluble, rapid hydrolysis highly branched so many ends for rapid hydrolysis B1 compact and insoluble so no osmotic effect B1 | 2 |
Mitosis is part of the cell cycle.
Describe the events of metaphase and anaphase.
Explain the role of telomeres.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | chromosomes at equator; centromeres split, chromatids pulled apart metaphase: chromosomes line up at the equator attached to spindle fibres B1 anaphase: centromeres divide B1 sister chromatids pulled to opposite poles B1 | 3 |
| (b) | prevent loss of genes during replication telomeres are repeated sequences at chromosome ends B1 they prevent loss of genes when DNA is replicated B1 | 2 |
This question is about biological molecules.
Describe how two α-glucose molecules join to form maltose.
Explain how the structure of glycogen makes it suited for energy storage in animal cells.
Explain why collagen is suited to its role in tendons.
Describe a test to show that a solution contains a non-reducing sugar.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | condensation; glycosidic bond between C1 and C4; water removed condensation reaction B1 forming a 1,4-glycosidic bond B1 a molecule of water is removed B1 | 3 |
| (b) | branched, many ends for rapid hydrolysis; compact; insoluble, no osmotic effect highly branched with 1,6 links so many ends for rapid hydrolysis by enzymes B1 compact / coiled so much stored in small space B1 insoluble so does not affect water potential B1 | 3 |
| (c) | three chains in triple helix; H bonds between chains; cross-links between molecules; staggered fibrils high tensile strength Any 3 of: three polypeptides wound into a triple helix held by hydrogen bonds; covalent cross-links between molecules form fibrils; ends staggered so no weak points; high tensile strength B3 | 3 |
| (d) | Benedict’s negative; boil with dilute HCl, neutralise with NaHCO₃; Benedict’s again, red-brown precipitate test with Benedict’s first and get a negative result B1 boil with dilute hydrochloric acid then neutralise with sodium hydrogencarbonate B1 heat with Benedict’s again: orange-red precipitate B1 | 3 |
A student calibrates an eyepiece graticule using a stage micrometer with divisions 10 µm apart. 25 eyepiece units line up with 10 stage divisions.
Explain why the eyepiece graticule must be calibrated at each magnification.
Calculate the actual length represented by one eyepiece unit.
A cell measures 12 eyepiece units. Use your answer to (b) to calculate its actual length.
State three structural differences between a prokaryotic cell and a eukaryotic cell.
Explain why an electron microscope has a much higher resolution than a light microscope, and give one disadvantage of electron microscopy.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | graticule fixed in eyepiece; apparent size of each division on specimen changes with objective the graticule scale stays the same size while the image is magnified differently by each objective B1 so the actual length represented by one eyepiece unit changes B1 | 2 |
| (b) | 4.00 µm 10 × 10 µm = 100 µm M1 ÷ 25 = 4.00 µm A1 | 2 |
| (c) | 48.0 µm 12 × 4.00 = 48.0 µm B1 | 1 |
| (d) | no nucleus (circular DNA); no membrane-bound organelles; 70S ribosomes; peptidoglycan wall; plasmids Any 3 of: prokaryotes: no nucleus (DNA free in cytoplasm, circular); no membrane-bound organelles (e.g. mitochondria); smaller (70S) ribosomes; cell wall of peptidoglycan; plasmids B3 | 3 |
| (e) | electrons have much shorter wavelength; resolution limited by wavelength; specimens dead / vacuum / artefacts electrons have a much shorter wavelength than light B1 resolution is limited to about half the wavelength, so finer detail is distinguished B1 disadvantage: specimens must be dead / in a vacuum / artefacts / no colour B1 | 3 |
Electron micrographs show the ultrastructure of eukaryotic cells.
Explain why an electron microscope has a higher resolution than a light microscope.
Describe the roles of the rough endoplasmic reticulum and the Golgi body in producing a secreted protein.
State two structural differences between prokaryotic and eukaryotic cells.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | shorter wavelength of electrons electrons have a much shorter wavelength than light B1 so points closer together can be distinguished B1 | 2 |
| (b) | RER synthesises; vesicles; Golgi modifies and packages ribosomes on RER synthesise protein which enters the RER B1 transported in vesicles to the Golgi B1 Golgi modifies (e.g. adds carbohydrate) and packages it into secretory vesicles B1 | 3 |
| (c) | no nucleus; 70S ribosomes; no membrane-bound organelles; circular DNA Any 2 of: prokaryotes: no nucleus; no membrane-bound organelles; smaller (70S) ribosomes; circular DNA B2 | 2 |
Protein synthesis involves transcription and translation.
Describe the process of transcription.
Explain the role of tRNA in translation.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | RNA polymerase; complementary base pairing; mRNA leaves nucleus DNA unwinds and hydrogen bonds break B1 one strand is the template B1 RNA polymerase joins free RNA nucleotides by complementary base pairing (A–U) B1 mRNA leaves via nuclear pore B1 | 4 |
| (b) | anticodon pairs with codon; carries amino acid tRNA carries a specific amino acid B1 its anticodon pairs with the complementary codon on mRNA B1 amino acids joined by peptide bonds B1 | 3 |
In the 9700 syllabus, Cells & Molecules covers Cell structure, Biological molecules, Enzymes, Cell membranes & transport, The mitotic cell cycle, Nucleic acids & protein synthesis.
This page has 10 Cells & Molecules questions with answers. For more, the free generator on mentorriyaz.in builds a new Cells & Molecules paper with a full mark scheme each time you click.
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in