About this O-Level Science practice test

Combined Science means switching between biological processes, chemical properties, and mechanics within the same paper, often the same page. These drills mix all three the way the real exam does, with a focus on the precise scientific vocabulary Cambridge examiners actually look for in full-mark answers.

O-Level Science Practice Test sample questions

These starter questions help you launch a science mock test quickly. Swap them with your own worksheet, notebook, or textbook questions any time.

1. 1. Which cell structure controls the movement of substances into and out of a cell, and what is it made of?

2. 2. A student places a fresh potato chip into a concentrated sugar solution for 30 minutes. Predict what happens to the mass of the chip and explain why.

3. 3. Write the balanced word equation for photosynthesis, and state two factors (other than light intensity) that can limit its rate.

4. 4. Explain why anaerobic respiration in muscle cells releases far less energy per glucose molecule than aerobic respiration.

5. 5. Describe how the structure of a villus in the small intestine is adapted for the absorption of digested food, and name the blood vessel that carries absorbed glucose away from the intestine.

6. 6. Name the two main waste products removed from the blood by the kidneys, and state where each is produced in the body.

7. 7. Explain how the body responds when blood glucose concentration rises after a meal, including the role of a named hormone and its target organs.

8. 8. State two differences between sexual and asexual reproduction in terms of the genetic variation produced in offspring.

9. 9. In pea plants, tall (T) is dominant over dwarf (t). A heterozygous tall plant (Tt) is crossed with a dwarf plant (tt). Using a genetic diagram, work out the expected ratio of tall to dwarf offspring.

10. 10. In the food chain grass -> grasshopper -> frog -> snake, explain why there is usually much less biomass at the snake trophic level than at the grasshopper trophic level.

11. 11. An atom of chlorine has atomic number 17 and mass number 35. State the number of protons, neutrons and electrons in this atom.

12. 12. Explain, in terms of structure and bonding, why magnesium oxide has a very high melting point.

13. 13. Write a balanced symbol equation, including state symbols, for the reaction between zinc metal and dilute sulfuric acid.

14. 14. Describe how you would prepare a pure, dry sample of copper sulfate crystals starting from copper oxide and dilute sulfuric acid.

15. 15. Using collision theory, explain why powdered calcium carbonate reacts faster with hydrochloric acid than the same mass of marble chips.

16. 16. During the electrolysis of molten lead(II) bromide using inert electrodes, name the products formed at each electrode and explain why electrolysis only works when the compound is molten (or dissolved), not solid.

17. 17. During the electrolysis of concentrated aqueous sodium chloride using inert electrodes, name the gas produced at the cathode and the gas produced at the anode.

18. 18. Name the homologous series to which ethene belongs, and state the type of bond between the two carbon atoms that allows it to take part in addition reactions.

19. 19. Describe a simple chemical test that distinguishes an alkene from an alkane, including the observation for each.

20. 20. Explain why the reactivity of Group I metals (the alkali metals) increases going down the group.

21. 21. A cyclist accelerates uniformly from rest to 12 m/s in 6 seconds. Calculate the acceleration and the distance travelled during this time.

22. 22. A box of weight 50 N rests on a horizontal table. A horizontal force of 15 N is applied to the box, but it does not move. State the size and direction of the frictional force acting on the box, and explain why the box stays stationary.

23. 23. A solid block exerts a force of 40 N on a surface through a base area of 0.5 m^2. It is then turned onto its side so the contact area becomes 0.2 m^2. Calculate the pressure exerted in each position and explain the difference.

24. 24. A 2 kg object is lifted 5 m above the ground (take g = 10 N/kg). Calculate its gain in gravitational potential energy, and state the energy transfer that occurs as it falls freely back to the ground.

25. 25. Explain, in terms of wave behaviour, why an echo is heard when a sound wave reflects off a distant wall, and state one factor that would increase the time delay between the original sound and its echo.

26. 26. Two 4 ohm resistors are connected in series with a 12 V battery. Calculate the total resistance of the circuit and the current flowing through it.

27. 27. Explain why connecting two identical lamps in parallel across a battery makes each lamp shine as brightly as if it were alone, whereas connecting them in series makes both lamps dimmer.

28. 28. Describe how you would use a plotting compass to determine the shape of the magnetic field around a bar magnet.

29. 29. A radioactive source emits beta particles. Describe what a beta particle is, and explain why it is more penetrating than an alpha particle but less penetrating than gamma rays.

30. 30. A radioactive isotope has an initial activity of 800 counts per minute. After 24 days, the activity has fallen to 100 counts per minute. Given that the half-life of the isotope is 8 days, show working to verify that this data is consistent.

Answer key & quick explanations

Short answers for the sample questions above. Use this to self-check before generating a fresh AI-built mock test.

1. 1. Cell membrane function

   The cell membrane (a partially permeable membrane made mainly of a phospholipid bilayer with embedded proteins).

   The cell membrane surrounds every cell and regulates which substances enter or leave, allowing diffusion and osmosis of small molecules while proteins embedded in it control active transport and cell signalling. Without this selective control, cells could not maintain the internal conditions needed for their chemical reactions.

2. 2. Potato chip in sugar solution

   The mass of the chip decreases, because it loses water.

   The sugar solution is more concentrated than the fluid inside the potato cells, so it has a lower water potential outside than inside. Water therefore moves out of the cells, across the partially permeable cell membranes, by osmosis, down the water potential gradient, causing the tissue to lose water and mass and become flaccid.

3. 3. Photosynthesis equation and limiting factors

   Carbon dioxide + water -(light energy, chlorophyll)-> glucose + oxygen. Limiting factors besides light intensity: carbon dioxide concentration and temperature.

   Photosynthesis converts carbon dioxide and water into glucose and oxygen using light energy trapped by chlorophyll. Its rate is controlled by whichever factor is in shortest supply at the time — low CO2 concentration limits the raw material available, while temperature affects the rate of the enzyme-controlled reactions involved, so either can become the limiting factor even when light is plentiful.

4. 4. Anaerobic vs aerobic energy release

   Anaerobic respiration only partially breaks glucose down into lactic acid without using oxygen, so far less energy is released per glucose molecule than in aerobic respiration.

   Aerobic respiration fully oxidises glucose into carbon dioxide and water, releasing all the energy stored in its bonds. Anaerobic respiration in muscles stops at lactic acid because no oxygen is available to complete the breakdown, so most of the chemical energy in glucose remains locked in the lactic acid rather than being released.

5. 5. Villus adaptations and glucose transport

   Villi have a thin (one-cell-thick) surface, a large surface area from folding and microvilli, and a rich capillary network; the hepatic portal vein carries absorbed glucose away to the liver.

   These features maximise the rate of diffusion and active transport of digested food into the blood: a short diffusion path, huge surface area, and a constant blood supply that maintains a steep concentration gradient by carrying absorbed nutrients away quickly. Glucose specifically enters the capillaries of the villus and is transported via the hepatic portal vein to the liver before entering general circulation.

6. 6. Kidney excretory waste products

   Urea (produced in the liver by deamination of excess amino acids) and excess water and dissolved ions/salts (from diet and metabolism).

   The liver breaks down surplus amino acids that cannot be stored, removing the nitrogen-containing amino group and converting it into urea, which is then filtered out of the blood by the kidneys. The kidneys also regulate water and ion balance by excreting whatever excess is not needed, keeping the blood's composition stable.

7. 7. Regulation of rising blood glucose

   The pancreas detects the rise and secretes insulin, which causes liver and muscle cells to take up glucose from the blood and convert it into glycogen, lowering blood glucose back to normal.

   This is an example of negative feedback: a change (rising glucose) triggers a response (insulin release) that reverses the change. Insulin binds to receptors on liver and muscle cells, increasing their uptake of glucose and stimulating the enzyme-controlled conversion of glucose to stored glycogen (glycogenesis).

8. 8. Sexual vs asexual reproduction and variation

   Sexual reproduction combines genetic material from two parents through fertilisation, producing genetically varied offspring; asexual reproduction involves only one parent and no fusion of gametes, producing genetically identical offspring (clones).

   In sexual reproduction, each parent contributes a gamete with a unique mix of alleles (from meiosis), so offspring inherit a new combination of genes from both parents, increasing variation. Asexual reproduction uses mitosis from a single parent with no mixing of genetic material, so offspring are genetically identical to that parent unless a mutation occurs.

9. 9. Tt x tt genetic cross ratio

   1 tall : 1 dwarf (50% Tt, 50% tt).

   The heterozygous parent (Tt) produces gametes carrying either T or t; the dwarf parent (tt) can only produce gametes carrying t. Combining these in a Punnett square gives offspring genotypes Tt, Tt, tt, tt — two tall (Tt) and two dwarf (tt) out of four, a 1:1 ratio, because a test cross like this directly reveals the gametes produced by the heterozygous parent.

10. 10. Biomass loss along a food chain

    Biomass decreases at each trophic level because energy and material are lost through respiration, movement, heat loss, and undigested/egested waste, so less biomass is available to build the bodies of organisms further along the chain.

    Only a fraction of the biomass an organism eats is actually converted into its own new tissue; the rest is used for life processes like respiration and movement (lost mostly as heat) or passed out as faeces. Because this loss happens at every step, the snake — being three steps removed from the grass — has access to only a small fraction of the original biomass produced by the grass.

11. 11. Chlorine-35 subatomic particles

    17 protons, 18 neutrons, 17 electrons.

    The atomic number (17) always equals the number of protons, and in a neutral atom the number of electrons equals the number of protons. The number of neutrons is found by subtracting the atomic number from the mass number: 35 − 17 = 18.

12. 12. High melting point of magnesium oxide

    Magnesium oxide is a giant ionic lattice held together by very strong electrostatic forces of attraction between Mg2+ and O2- ions in all directions, so a large amount of energy is needed to break these bonds and melt it.

    Ionic compounds like MgO form regular giant lattices where each ion is strongly attracted to several oppositely charged neighbours. Because magnesium and oxide ions carry two charges each (rather than one, as in NaCl), the electrostatic attraction is even stronger, which is why MgO has an unusually high melting point compared to many other ionic solids.

13. 13. Zinc and dilute sulfuric acid equation

    Zn(s) + H2SO4(aq) -> ZnSO4(aq) + H2(g)

    Zinc is more reactive than hydrogen, so it displaces hydrogen from the acid, forming the soluble salt zinc sulfate and releasing hydrogen gas. The equation is already balanced with one zinc, one sulfate group and two hydrogen atoms on each side, and the state symbols show the acid and salt are aqueous while hydrogen escapes as a gas.

14. 14. Preparing copper sulfate crystals

    Add excess copper oxide to warm dilute sulfuric acid until no more dissolves, filter to remove the unreacted excess oxide, gently evaporate the filtrate to the point of crystallisation, then leave it to cool so crystals form, and filter and dry them.

    Using excess insoluble base ensures all the acid is used up, so no acid remains as an impurity in the final product — this is why the excess solid, not the acid, is filtered off. Evaporating to saturation and then cooling allows copper sulfate crystals to grow slowly out of solution, which are then separated by filtration and dried without heating strongly (to avoid losing their water of crystallisation).

15. 15. Powdered vs lump calcium carbonate rate

    Powdered calcium carbonate has a much greater surface area exposed to the acid than the same mass of marble chips, so acid particles collide with carbonate particles more frequently, increasing the rate of reaction.

    Collision theory states that a reaction's rate depends on the frequency and energy of collisions between reacting particles. Breaking the marble into powder exposes far more carbonate particles at the surface to the acid at any given moment, so more successful collisions occur per second even though the total mass of reactant is unchanged.

16. 16. Electrolysis of molten lead(II) bromide

    Lead is deposited at the cathode and bromine gas is formed at the anode. Electrolysis only works when ions are free to move, which happens when the compound is molten (or in solution) but not when it is a rigid solid lattice.

    At the negative cathode, Pb2+ ions gain electrons (reduction) to form lead metal, while at the positive anode Br- ions lose electrons (oxidation) to form bromine. In solid lead bromide the ions are locked in fixed positions in the lattice and cannot move to carry charge, but melting frees them to migrate to the electrodes, allowing conduction and electrolysis to occur.

17. 17. Electrolysis of concentrated brine

    Hydrogen gas is produced at the cathode and chlorine gas is produced at the anode (sodium hydroxide remains dissolved in the solution).

    At the cathode, hydrogen ions from water are discharged in preference to the more reactive sodium ions, which stay in solution. At the anode, because the chloride ion concentration is high, chloride ions are discharged in preference to hydroxide ions, releasing chlorine gas — this is the basis of the industrial chlor-alkali process.

18. 18. Ethene homologous series and bonding

    Ethene belongs to the alkenes; the two carbon atoms are joined by a carbon-carbon double covalent bond (C=C).

    Alkenes are unsaturated hydrocarbons because they contain at least one C=C double bond, unlike the fully saturated alkanes. This double bond can open up, allowing one of the pairs of shared electrons to form new bonds with other atoms, which is why alkenes readily undergo addition reactions while alkanes generally do not.

19. 19. Test to distinguish alkene from alkane

    Shake the hydrocarbon with orange bromine water: an alkene quickly decolourises it to colourless, while an alkane causes no colour change (it stays orange) in the absence of light.

    The bromine molecule adds across the C=C double bond in an alkene in an addition reaction, using up the bromine and removing its orange colour. Alkanes have no double bond to react with in this way, so the bromine water is not decolourised under normal conditions, making this a reliable distinguishing test.

20. 20. Reactivity trend down Group I

    Reactivity increases down Group I because atomic radius increases, so the single outer electron is further from the nucleus and more shielded by inner electron shells, making it easier to lose.

    Each element down the group has an extra electron shell, increasing the distance between the outer electron and the positively charged nucleus while also adding more shielding from inner shells. Both effects weaken the nucleus's pull on the outer electron, so it is lost more easily, which is why sodium reacts more vigorously with water than lithium, and potassium more vigorously than sodium.

21. 21. Cyclist acceleration and distance

    Acceleration = 2 m/s^2; distance travelled = 36 m.

    Acceleration is calculated as change in velocity divided by time: (12 − 0) / 6 = 2 m/s^2. Distance can then be found using the average velocity method, s = ((u+v)/2) x t = ((0+12)/2) x 6 = 36 m, which matches using s = ut + ½at^2.

22. 22. Static friction on a stationary box

    The frictional force is 15 N, acting horizontally in the opposite direction to the applied force.

    Since the box remains stationary, the resultant force on it must be zero, meaning all forces are balanced. Static friction adjusts itself up to a maximum value to exactly oppose an applied force, so as long as 15 N does not exceed this maximum, friction will match it exactly and the box stays still.

23. 23. Pressure with changing contact area

    80 Pa in the first position (40 N / 0.5 m^2) and 200 Pa in the second position (40 N / 0.2 m^2); pressure increases when the block is turned onto its side.

    Pressure is calculated as force divided by the area over which it acts, so for the same force, a smaller contact area results in a larger pressure. Turning the block onto its side reduces the area in contact with the surface without changing its weight, so the same force is now concentrated over a smaller area, increasing the pressure exerted.

24. 24. Gravitational potential energy and falling

    Gain in GPE = 100 J (using GPE = mgh = 2 x 10 x 5). As the object falls freely, this GPE is transferred into kinetic energy.

    Gravitational potential energy depends on mass, gravitational field strength and height, so multiplying these values gives the energy gained when the object is lifted. As it falls without air resistance, this stored energy converts into kinetic energy, and by conservation of energy the object's kinetic energy just before landing equals the GPE it had at the top.

25. 25. Echo formation and time delay

    An echo is heard because the sound wave reflects off the wall and travels back to the listener as a distinct, delayed repeat of the original sound; increasing the distance between the source and the wall increases the time delay.

    Sound waves obey the law of reflection when they meet a hard, flat surface like a wall, bouncing back rather than being absorbed. Because sound travels at a fixed speed, a longer path length to the wall and back takes more time, so the further away the wall is, the longer the delay before the echo is heard.

26. 26. Series resistors and current

    Total resistance = 8 ohms; current = 1.5 A.

    Resistors in series simply add together, so 4 ohms + 4 ohms = 8 ohms total resistance. Applying Ohm's law, current = voltage / resistance = 12 V / 8 ohms = 1.5 A, and this same current flows through both resistors since there is only one path for charge to take.

27. 27. Parallel vs series lamp brightness

    In parallel, each lamp has the full battery voltage across it, so each shines as brightly as if it were alone; in series, the same voltage is shared between the lamps, so each gets less voltage and power, making them dimmer.

    In a parallel circuit, each branch is connected directly across the battery terminals, so the voltage across each lamp equals the total supply voltage regardless of the other lamp. In a series circuit there is only one loop, so the total voltage is divided between the components in the loop, meaning each lamp receives less voltage (and therefore less power) than it would if it were the only component in the circuit.

28. 28. Plotting compass field mapping

    Place the compass at points near the magnet, mark the direction the needle points, move the compass along that direction and repeat to build a continuous line from the north pole to the south pole; repeating from different starting points reveals the full field pattern.

    A plotting compass aligns itself with the magnetic field at its location because its needle is itself a small magnet, so tracking the direction it points at successive positions traces out a magnetic field line. Field lines are conventionally drawn from north to south outside the magnet, and are more closely spaced near the poles where the field is strongest.

29. 29. Beta particle nature and penetration

    A beta particle is a fast-moving electron released from the nucleus when a neutron changes into a proton; it penetrates further than an alpha particle (which is larger and more strongly charged) but less far than gamma rays (high-energy electromagnetic radiation with no mass or charge).

    Penetrating power depends on how strongly a radiation type interacts with the atoms of the material it passes through: alpha particles are heavy and highly charged so they collide often and are stopped quickly (by paper or a few cm of air), beta particles are much smaller and less charged so they penetrate further (stopped by a few mm of aluminium), and gamma rays, having no mass or charge, interact only weakly and need thick lead or concrete to absorb them.

30. 30. Half-life consistency check

    24 days / 8-day half-life = 3 half-lives; activity falls 800 -> 400 -> 200 -> 100 counts per minute, which matches the given final activity of 100 cpm, so the data is consistent.

    Each half-life is the time taken for the activity of a radioactive sample to fall to half its previous value, so the number of half-lives elapsed is found by dividing the total time by the half-life. Halving the activity that many times in succession (three times here) predicts the activity at 24 days, and since this calculated value matches the measured value of 100 cpm, the data confirms an 8-day half-life.

Curriculum Mapping & Learning Guide

Use this breakdown to identify which skills each question tests and guide post-test review.

O-Level Science units covered

1. Chapter 1: Biology: cells, osmosis/diffusion, photosynthesis, respiration, digestion, transport
2. Chapter 2: Biology: excretion, homeostasis, reproduction, genetics, ecosystems
3. Chapter 3: Chemistry: atomic structure, bonding, equations, acids/bases/salts, rates of reaction
4. Chapter 4: Chemistry: electrolysis, organic chemistry basics, periodic table trends
5. Chapter 5: Physics: motion, forces, pressure, energy, waves, light and sound
6. Chapter 6: Physics: electricity, magnetism, radioactivity, half-life

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