An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V1) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
Biology 部分
Answer all questions covering human reproduction, cell transport, plant nutrition, ecology, and circulation.
9 题目 · 81 分
题目 1 · Structured
9 分
Human reproduction involves fertilization followed by the development of a fetus.
(a) Define fertilization. [2]
(b) Describe how the structure of a human sperm cell is adapted to its function. [3]
(c) State two functions of the placenta during gestation. [2]
(d) Explain why it is important that the blood systems of the mother and the fetus are kept separate. [2]
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解题
(a) Fertilization is the fusion of the nuclei of male and female gametes (sperm and egg) to form a zygote.
(b) Adaptations of sperm: 1. Flagellum (tail) for swimming towards the egg. 2. High concentration of mitochondria in the middle piece to provide energy for movement. 3. Acrosome in the head containing digestive enzymes to penetrate the outer layer of the egg.
(c) Placenta functions: 1. Exchange of nutrients (glucose, amino acids) and oxygen from mother to fetus, and removal of waste products (urea, carbon dioxide). 2. Secretion of hormones (e.g. progesterone) to maintain pregnancy.
(d) Importance of blood separation: 1. Prevents high blood pressure of the maternal circulation from damaging delicate fetal capillaries. 2. Prevents maternal immune system (antibodies) from attacking the foreign proteins on fetal cells / avoids blood group incompatibility issues.
评分标准
(a) Fusion of nuclei [1]; of male and female gametes / haploid gametes / sperm and egg [1].
(b) Max [3] from: - Flagellum/tail [1] + for movement/swimming [1]; - Mitochondria [1] + to release/provide energy (via respiration) [1]; - Acrosome/enzymes in head [1] + to digest/penetrate egg membrane [1]; - Streamlined shape [1] + reduces resistance/helps swim faster [1].
(c) Any [2] from: - Exchange of oxygen/nutrients (to fetus) [1]; - Removal of carbon dioxide/urea (from fetus) [1]; - Secretion of progesterone/hormones (to maintain pregnancy) [1]; - Transfer of passive immunity/antibodies [1].
(d) Any [2] from: - Prevents damage to fetal blood vessels by high maternal blood pressure [1]; - Prevents immune attack/rejection of fetus by mother's white blood cells [1]; - Prevents mixing of incompatible blood groups [1].
题目 2 · Structured
9 分
Cracking is a reaction used in the petrochemical industry to break down large hydrocarbons into more useful smaller ones. Ethene is a common product of this process.
(a) Decane, \(\text{C}_{10}\text{H}_{22}\), is cracked to produce ethene, \(\text{C}_2\text{H}_4\), and octane, \(\text{C}_8\text{H}_{18}\).
(i) Write a balanced chemical equation for this cracking reaction. [2]
(ii) State the temperature and catalyst conditions required for catalytic cracking. [2]
(b) Ethene is an unsaturated hydrocarbon.
(i) State the colour change observed when ethene is bubbled through aqueous bromine. [2]
(ii) Draw the displayed structure of the product formed when ethene reacts with bromine, showing all atoms and covalent bonds. [3]
(a)(ii) High temperature (approx 450 - 800 °C) and a catalyst (such as alumina, silica, or zeolite).
(b)(i) Orange/brown/yellow to colourless (decolourises).
(b)(ii) The product is 1,2-dibromoethane. Displayed structure has two Carbon atoms single-bonded to each other, with two Hydrogens and one Bromine atom single-bonded to each Carbon: H H | | H-C - C-H | | Br Br
评分标准
(a)(i) Correct reactant and products: \(\text{C}_{10}\text{H}_{22}\) and \(\text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\) [1]; Correct balancing (already balanced) [1].
(a)(ii) High temperature / range 450 - 800 °C [1]; Catalyst (alumina/silica/zeolite/porous pot) [1].
(b)(i) Orange/brown/red-brown [1]; to colourless (reject 'clear') [1].
(b)(ii) C-C single bond [1]; four C-H single bonds and two C-Br single bonds correctly shown [1]; correct valence of all atoms (C with 4 bonds, H with 1, Br with 1) and all atoms labelled [1].
题目 3 · Calculation
9 分
A remote-controlled car of mass 0.50 kg is tested on a straight, level track. It accelerates uniformly from rest to a speed of 6.0 m/s in a time of 4.0 s.
(a) (i) Calculate the acceleration of the car. [2]
(ii) Calculate the resultant force acting on the car during this acceleration. [2]
(b) The car then encounters a ramp. It travels up the ramp at a constant speed of 3.0 m/s until it reaches a platform at a vertical height of 1.2 m above the track. Take the acceleration of free fall \(g = 9.8\ \text{m/s}^2\).
(i) Calculate the increase in the gravitational potential energy (GPE) of the car. [2]
(ii) State the change in kinetic energy (KE) of the car as it climbs the ramp and explain your answer. [3]
(b)(ii) The change in kinetic energy is zero (or constant KE). The speed of the car is constant (3.0 m/s), and since mass is constant, \(\text{KE} = \frac{1}{2}mv^2\) remains unchanged.
评分标准
(a)(i) \(a = \Delta v / t\) or \(6.0 / 4.0\) [1]; \(1.5\ \text{m/s}^2\) [1].
(b)(i) \(\Delta \text{GPE} = mgh\) or \(0.50 \times 9.8 \times 1.2\) [1]; \(5.9\ \text{J}\) or \(5.88\ \text{J}\) [1].
(b)(ii) Change in KE is zero / constant KE [1]; speed is constant (at 3.0 m/s) [1]; since KE depends on speed/velocity (\(E_k = \frac{1}{2}mv^2\)), KE remains unchanged [1].
题目 4 · Structured
9 分
The mammalian circulatory system is described as a double circulation.
(a) Explain what is meant by a double circulation. [2]
(b) Describe and explain how the structure of the left ventricle of the heart differs from that of the right ventricle. [3]
(c) Coronary heart disease (CHD) occurs when the coronary arteries become narrowed.
(i) Explain how this narrowing can lead to a heart attack. [2]
(ii) State two dietary lifestyle factors that increase the risk of developing CHD. [2]
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解题
(a) Double circulation means that for every complete circuit of the body, blood passes through the heart twice (once through the pulmonary system to the lungs, and once through the systemic system to the rest of the body).
(b) The left ventricle has a much thicker muscular wall than the right ventricle. This is because the left ventricle must pump blood under high pressure to the entire body (systemic circulation), whereas the right ventricle only pumps blood under lower pressure to the lungs (pulmonary circulation).
(c)(i) Narrowing of the coronary arteries reduces blood flow to the heart muscle. This decreases the supply of oxygen and glucose to the heart muscle cells, preventing aerobic respiration. The cells are forced to respire anaerobically, producing lactic acid, which damages or kills the muscle tissue, causing a myocardial infarction (heart attack).
(c)(ii) High intake of saturated fats and high intake of salt/sugar.
评分标准
(a) Blood passes through the heart twice [1]; for one complete circuit / to make one trip around the body [1].
(b) Left ventricle wall is thicker / more muscular [1]; to generate higher pressure [1]; to pump blood a further distance / to the whole body (rather than just to the lungs) [1].
(c)(i) Less blood flow / oxygen / glucose delivered to heart muscle [1]; heart muscle cells cannot respire aerobically / respire anaerobically / die [1].
(c)(ii) High saturated fat/cholesterol diet [1]; high salt intake / high calorie diet leading to obesity [1].
题目 5 · Structured
9 分
Methane, \(\text{CH}_4\), is the main constituent of natural gas. Its combustion is a highly exothermic process.
(a) (i) State what is meant by an exothermic reaction in terms of energy transfer and temperature change of the surroundings. [2]
(ii) Explain, in terms of bond breaking and bond forming, why the combustion of methane is exothermic. [3]
(b) Draw a labeled energy level diagram for an exothermic reaction. On your diagram, show and clearly label: - the reactants and products - the activation energy, \(E_a\) - the overall energy change, \(\Delta H\) [4]
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解题
(a)(i) An exothermic reaction is one that transfers thermal energy to the surroundings, causing the temperature of the surroundings to increase.
(a)(ii) Bond breaking requires energy (endothermic) and bond forming releases energy (exothermic). In an exothermic reaction, more energy is released when new bonds (C=O and H-O in \(\text{CO}_2\) and \(\text{H}_2\text{O}\)) are formed than is absorbed to break the existing bonds (C-H and O=O in \(\text{CH}_4\) and \(\text{O}_2\)).
(b) The energy level diagram should show: 1. Reactants at a higher energy level than products. 2. A curve rising from reactants to a peak (representing transition state) and falling to products. 3. Activation energy \(E_a\) labeled with an arrow pointing from the reactant level up to the peak. 4. Enthalpy change \(\Delta H\) labeled with an arrow pointing downward from the reactant level to the product level.
评分标准
(a)(i) Transfers thermal energy to surroundings [1]; temperature of surroundings increases [1].
(a)(ii) Bond breaking absorbs energy AND bond forming releases energy [1]; more energy is released during bond forming than is taken in/absorbed during bond breaking [1]; reference to specific reactants/products (methane/oxygen, carbon dioxide/water) [1].
(b) Reactants line higher than products line [1]; a curved path going up to a peak and then down to products [1]; activation energy, \(E_a\), shown as an arrow from reactant level to peak of curve [1]; overall energy change, \(\Delta H\), shown as downward arrow from reactant level to product level [1].
题目 6 · Calculation
9 分
An electrical circuit contains a 12.0 V battery of negligible internal resistance connected to a parallel combination of two resistors. The resistance values are \(R_1 = 4.0\ \Omega\) and \(R_2 = 6.0\ \Omega\).
(a) Calculate:
(i) the combined resistance of the parallel combination of \(R_1\) and \(R_2\). [2]
(ii) the total current flowing from the battery. [2]
(iii) the current flowing through resistor \(R_2\). [2]
(b) Calculate the total electrical power supplied by the battery to the circuit. State the unit of your answer. [3]
(a)(ii) \(I = V / R\) or \(12.0 / 2.4\) [1]; \(5.0\ \text{A}\) [1].
(a)(iii) \(I = V / R_2\) or \(12.0 / 6.0\) [1]; \(2.0\ \text{A}\) [1].
(b) \(P = VI\) or \(12.0 \times 5.0\) or \(I^2 R_p\) or \(V^2 / R_p\) [1]; \(60\) (or \(60.0\)) [1]; \(\text{W}\) or Watts [1].
题目 7 · Structured
9 分
Substances move into and out of plant and animal cells by different transport mechanisms.
(a) Contrast diffusion and active transport. Your answer should refer to concentration gradients and energy requirements. [4]
(b) Plant cells are immersed in different solutions to study osmosis.
(i) Explain, in terms of water potential, why a plant cell placed in distilled water becomes turgid. [3]
(ii) State what would happen to a human red blood cell if it were placed in distilled water. Explain why this response differs from that of a plant cell. [2]
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解题
(a) 1. Diffusion is passive and does not require energy from respiration; active transport requires energy in the form of ATP. 2. Diffusion occurs down a concentration gradient (from higher to lower concentration); active transport moves substances against a concentration gradient (from lower to higher concentration).
(b)(i) Distilled water has a higher water potential than the cytoplasm of the plant cell. Water moves into the cell down a water potential gradient by osmosis across the partially permeable cell membrane. As water enters, pressure (turgor pressure) builds up inside the cell, pressing the cell membrane against the rigid cell wall, making the cell turgid.
(b)(ii) The red blood cell would swell and burst (lyse). This occurs because animal cells lack a cell wall to withstand the internal osmotic pressure.
评分标准
(a) Diffusion does not require energy / is passive [1] AND active transport requires energy (from respiration) [1]; Diffusion is down concentration gradient [1] AND active transport is against concentration gradient [1].
(b)(i) Distilled water has higher water potential than cell cytoplasm / cell has lower water potential [1]; water enters cell by osmosis [1]; cell wall prevents cell from bursting / turgor pressure increases [1].
(b)(ii) Red blood cell swells and bursts / lyses [1]; because animal cells do not have a cell wall (to withstand turgor pressure) [1].
题目 8 · Structured
9 分
Iron is extracted from the ore hematite, which contains iron(III) oxide, \(\text{Fe}_2\text{O}_3\), in a blast furnace.
(a) Carbon monoxide, \(\text{CO}\), acts as a reducing agent in the furnace.
(i) Write a balanced chemical equation for the reduction of iron(III) oxide by carbon monoxide. [2]
(ii) Define redox in terms of oxygen transfer, and identify which substance is oxidized and which is reduced in this reaction. [4]
(b) Aluminium is more reactive than iron. Explain why aluminium cannot be extracted from its oxide, alumina (\(\text{Al}_2\text{O}_3\)), using carbon in a blast furnace. [3]
(a)(ii) Oxidation is the gain of oxygen and reduction is the loss of oxygen. \(\text{Fe}_2\text{O}_3\) is reduced because it loses oxygen to become Fe. \(\text{CO}\) is oxidized because it gains oxygen to become \(\text{CO}_2\).
(b) Aluminium is more reactive than carbon. This means aluminium has a higher affinity for oxygen than carbon does, so carbon cannot reduce aluminium oxide. Aluminium must be extracted using electrolysis, which requires a large amount of electricity.
评分标准
(a)(i) Correct formulae of all reactants and products (\(\text{Fe}_2\text{O}_3, \text{CO}, \text{Fe}, \text{CO}_2\)) [1]; Correct balancing [1].
(a)(ii) Oxidation is gain of oxygen AND reduction is loss of oxygen [1]; \(\text{Fe}_2\text{O}_3\) is reduced [1]; \(\text{CO}\) is oxidized [1]; linked to loss and gain of oxygen respectively [1].
(b) Aluminium is more reactive than carbon / carbon is less reactive than aluminium [1]; carbon cannot displace/remove oxygen from aluminium oxide [1]; aluminium has a stronger affinity for oxygen [1].
题目 9 · structured
9 分
A toy rocket of mass \(0.25\text{ kg}\) is launched vertically upwards from rest.
(a) The rocket accelerates vertically upwards with a constant acceleration of \(4.0\text{ m/s}^2\) for the first \(3.0\text{ s}\).
(i) Calculate the speed of the rocket at \(t = 3.0\text{ s}\). Show your working. speed = ..................................................... \(\text{m/s}\) [2]
(ii) Calculate the kinetic energy of the rocket at \(t = 3.0\text{ s}\). Show your working. kinetic energy = ..................................................... \(\text{J}\) [2]
(iii) Calculate the height of the rocket above the launch pad at \(t = 3.0\text{ s}\). Show your working. height = ..................................................... \(\text{m}\) [2]
(b) After \(3.0\text{ s}\), the rocket engine turns off. The rocket continues to move upwards to its maximum height. Ignore air resistance. Take \(g = 9.8\text{ m/s}^2\).
(i) State the energy transfer that takes place as the rocket rises after the engine has turned off. [1]
(ii) Calculate the maximum height reached by the rocket above the launch pad. Show your working. maximum height = ..................................................... \(\text{m}\) [2]
(b) (i) Kinetic energy is transferred to gravitational potential energy.
(ii) Using conservation of energy for the flight after the engine cuts off: \(m g \Delta h = E_k\) \(0.25 \times 9.8 \times \Delta h = 18\) \(2.45 \Delta h = 18 \implies \Delta h \approx 7.35\text{ m}\) Total height above launch pad = \(18 + 7.35 = 25.35\text{ m}\) (accepts \(25.3\text{ m}\) or \(25.4\text{ m}\))
评分标准
(a) (i) - 1 mark for correct substitution into formula: \(4.0 \times 3.0\) - 1 mark for correct final value: \(12\text{ m/s}\)
(ii) - 1 mark for correct substitution into kinetic energy formula: \(0.5 \times 0.25 \times 12^2\) (allow ecf from (a)(i)) - 1 mark for correct final value: \(18\text{ J}\)
(iii) - 1 mark for correct substitution into distance formula: \(0.5 \times 4.0 \times 3.0^2\) - 1 mark for correct final value: \(18\text{ m}\)
(b) (i) - 1 mark for kinetic (energy) to gravitational potential (energy)
(ii) - 1 mark for calculating additional height: \(\Delta h \approx 7.3\text{ m}\) or \(7.4\text{ m}\) (allow ecf) - 1 mark for correct final total height: \(25.3\text{ m}\) or \(25.4\text{ m}\) (allow ecf)