2026 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

Structure P is the bronchus (plural: bronchi), which branches off the trachea to carry air into each lung. Structure Q is the alveolus (plural: alveoli), the microscopic air sac where diffusion of oxygen and carbon dioxide takes place between air and blood.

P: bronchus / bronchi ; [1]
Q: alveolus / alveoli / air sac ; [1]
NOT: bronchiole for P; NOT: lung / capillary for Q.

Structure X is the left ventricle, which has a thick muscular wall to pump oxygenated blood at high pressure throughout the systemic circulation. Structure Y is the pulmonary artery, which transports deoxygenated blood from the right ventricle to the lungs.

X: left ventricle ; [1]
Y: pulmonary artery ; [1]
NOT: ventricle alone for X; NOT: pulmonary vein / aorta for Y.

Part J is the stigma, the sticky terminal portion of the female reproductive organ (carpel) that receives pollen during pollination. Part K is the anther, the part of the male reproductive organ (stamen) where pollen grains are produced and released.

J: stigma ; [1]
K: anther ; [1]
NOT: style / carpel for J; NOT: filament / stamen for K.

(a) \(\text{Rate} = \frac{\text{Volume}}{\text{Time}} = \frac{0.4\text{ cm}^3}{5\text{ min}} = 0.08\text{ cm}^3\text{ per min}\).

(b) As the distance of the lamp increases (from 10 cm to 50 cm), the rate of oxygen production decreases from 0.90 cm\(^3\)/min to 0.08 cm\(^3\)/min (or decreases by 0.82 cm\(^3\)/min).

(a) 0.08 (cm\(^3\) per min); [1]

(b) as distance increases, the rate of oxygen production decreases / inverse relationship; [1]
correct comparative data quote with units (e.g. rate is 0.90 cm\(^3\)/min at 10 cm and drops to 0.08 cm\(^3\)/min at 50 cm / decreases by 0.82 cm\(^3\)/min / rate is halved from 20 cm to 20 cm); [1]
(ALLOW ECF from (a))

(a) The optimum temperature is 40 \(^\circ\text{C}\) because the rate of reaction is at its highest (50 arbitrary units).

(b) Above the optimum temperature, excessive kinetic energy disrupts the enzyme structure, causing the enzyme to denature. The active site changes shape so that the substrate (starch) can no longer bind to form enzyme-substrate complexes, reducing and eventually stopping the reaction.

(a) 40 (\(^\circ\text{C}\)); [1]

(b) enzyme denatures / active site changes shape; [1]
substrate / starch no longer fits into the active site / enzyme-substrate complexes can no longer form; [1]
(REJECT 'enzyme is killed' / 'enzyme dies')

(a) In the alveoli, oxygen diffuses down its concentration gradient from the alveolar air into the blood capillaries to be transported to body cells for aerobic respiration. Carbon dioxide, produced as a waste product of cellular respiration in body tissues, is transported by the blood to the lungs, where it diffuses from the capillary blood into the alveoli and is exhaled.

(b) Feature 1: The alveolar walls are extremely thin (one cell thick), providing a very short diffusion pathway for gases.
Feature 2: There are millions of alveoli, providing a very large total surface area to maximize the rate of gas diffusion between air and blood (or a moist lining that allows gases to dissolve, or an extensive network of surrounding blood capillaries maintaining a steep concentration gradient).

(a) 2 marks total:
• Oxygen diffuses from alveoli into blood / used by cells for (aerobic) respiration; [1]
• Carbon dioxide is produced by respiration and diffuses from blood into alveoli (to be exhaled); [1]

(b) 2 marks total (scaled to 1.75 marks equivalent):
Award 1 mark per pair of feature and linked explanation (max 2 pairs):
• (Feature:) thin wall / one cell thick / short distance; (Explanation:) gives a short diffusion pathway / fast diffusion; [1]
• (Feature:) large surface area / large surface area to volume ratio; (Explanation:) allows more gas molecules to diffuse per unit time; [1]
• (Feature:) moist surface / film of moisture; (Explanation:) allows oxygen to dissolve before diffusing; [1]
• (Feature:) dense capillary network / good blood supply; (Explanation:) maintains a steep concentration gradient; [1]
IGNORE: references to breathing movements alone.

(a) An artery has a significantly thicker wall containing more smooth muscle and elastic fibres compared to a vein, and it has a narrower central lumen. Veins have thinner muscular walls, a wider lumen, and possess internal semi-lunar valves which are absent in arteries.

(b) Arteries receive blood directly pumped from the ventricles under high and pulsating pressure. The thick muscular and elastic walls withstand this high pressure without bursting, and elastic recoil helps maintain pressure between heart contractions. In contrast, blood in veins travels under very low pressure; the wider lumen reduces resistance to blood flow, and one-way valves prevent the backflow of blood as it returns towards the heart.

(a) 2 marks total:
Any two comparative points from:
• Artery has thicker wall / vein has thinner wall; [1]
• Artery has more elastic / muscle tissue / vein has less elastic / muscle tissue; [1]
• Artery has a narrower lumen / vein has a wider lumen; [1]
• Veins have valves / arteries do not have valves; [1]

(b) 2 marks total (scaled to 1.75 marks equivalent):
• Artery explanation: high pressure from ventricle contraction requires thick / elastic walls to withstand pressure / prevent bursting / maintain blood pressure by recoil; [1]
• Vein explanation: low pressure means blood moves slowly / needs valves to prevent backflow OR large lumen to reduce resistance to flow; [1]
NOT: crediting valves in arteries.

(a) Insect-pollinated flowers produce larger, sticky or spiky pollen grains in relatively smaller quantities so they easily adhere to an insect's body. Wind-pollinated flowers produce very small, lightweight, smooth pollen grains in massive quantities so they can be easily carried by air currents.

(b) In wind-pollinated flowers, the stigmas hang loosely outside the floral parts (petals) and have a feathery, branched structure. This exposed position allows them to be directly in the path of passing air currents, and the large feathery surface area acts as a net to trap airborne pollen grains blown by the wind.

(a) 2 marks total:
Any two comparative points (insect-pollinated vs wind-pollinated):
• Insect pollen is larger / wind pollen is smaller / lighter; [1]
• Insect pollen is sticky / spiky / rough / wind pollen is smooth / non-sticky; [1]
• Insect pollen produced in smaller numbers / wind pollen produced in large quantities; [1]

(b) 2 marks total (scaled to 1.75 marks equivalent):
• (Position & appearance:) stigma hangs outside / exposed AND is feathery / branched / large surface area; [1]
• (Explanation:) exposed to wind currents / acts as a net to catch / trap drifting pollen grains; [1]

(a) The enzyme in the stomach is protease (pepsin). It requires strongly acidic conditions (approximately pH 1.5 to 2.0), which are provided by the hydrochloric acid secreted in gastric juice.

(b) Starch is chemically digested by the enzyme amylase, not protease. Enzymes exhibit specificity because the active site of an enzyme is complementary in shape only to its specific substrate. Amylase functions optimally in neutral to slightly alkaline conditions (as found in saliva and the duodenum). When amylase swallowed with food reaches the highly acidic environment of the stomach, its active site is denatured (changes shape permanently). Therefore, starch cannot bind to form enzyme-substrate complexes, preventing starch digestion in the stomach.

(a) 2 marks total:
• (Enzyme:) protease / pepsin; [1]
• (Condition:) acidic / low pH / pH 1 to 3 / presence of hydrochloric acid; [1]

(b) 2 marks total (scaled to 1.75 marks equivalent):
• (Specificity:) amylase (which breaks down starch) has an active site complementary only to starch / protease cannot bind starch; [1]
• (Denaturation / conditions:) the acidic pH / stomach acid denatures amylase / changes the shape of its active site so no enzyme-substrate complexes can form; [1]
IGNORE: references to mechanical churning.

(a) Moving down Group VII, the physical state changes from gas (fluorine and chlorine) to liquid (bromine) to solid (iodine).

(b)(i) Bromine is more reactive than iodine, so it displaces iodide ions to form aqueous iodine, turning the solution from colourless to brown/orange-brown.
(b)(ii) Iodine is displaced by bromine.

(c) A chlorine atom gains 1 electron to form a stable full outer shell chloride ion (\(\text{Cl}^-\)), so its electronic structure becomes 2,8,8.

(a) (changes from) gas to liquid to solid / boiling point increases down the group [1];

(b)(i) colourless to brown / orange-brown / yellow-brown [1]
(IGNORE red / purple unless organic solvent specified);

(b)(ii) iodine / \(\text{I}_2\) [1]
(REJECT iodide / \(\text{I}^-\));

(c) 2,8,8 [1]

(a) Down Group I, the outer electron is further from the nucleus and shielded by more electron shells, making it easier to lose; thus reactivity increases down the group.

(b)(i) Potassium reacts much more vigorously than lithium, generating enough heat to ignite the hydrogen gas produced, burning with a characteristic lilac flame.
(b)(ii) The reaction produces potassium hydroxide, which is a strongly alkaline solution (pH > 11), turning universal indicator purple (or blue).

(c) A potassium atom loses its single outer shell electron to form a \(\text{K}^+\) ion, resulting in the electron arrangement 2,8,8.

(a) (reactivity) increases / becomes more reactive [1];

(b)(i) catches fire / burns with a flame / lilac flame / explodes [1]
(ALLOW produces a lilac flame;
IGNORE moves faster / effervesces faster);

(b)(ii) purple / blue / violet [1]
(REJECT red / orange / yellow / green);

(c) 2,8,8 [1]

(a) A carbon atom has 4 outer electrons and needs 4 more electrons to achieve a full shell (octet). Each of the 4 hydrogen atoms has 1 electron and needs 1 more to complete its shell (duplet). Therefore, the central carbon atom forms 4 single covalent bonds, sharing one pair of electrons (one dot, one cross) with each of the 4 hydrogen atoms. There are no lone pairs on the carbon atom or the hydrogen atoms.

(b) A single covalent bond consists of a shared pair of electrons held together by the electrostatic attraction between the negatively charged shared pair of electrons and the positively charged nuclei of the two bonded atoms.

(a)
• Four shared pairs of electrons shown between the central C atom and each of the four H atoms (one dot and one cross per shared pair) [1];
• Fully correct outer shell configuration with no extra unshared electrons on C or H [1];

(b)
• Electrostatic attraction between the shared pair of electrons and the (positively charged) nuclei (of the bonded atoms) [1].

[Total: 3 marks]

(a) Oxygen is in Group VI and has 6 valence electrons; each hydrogen has 1 valence electron. The oxygen atom shares one electron with each of the two hydrogen atoms to form two single covalent bonds (two shared pairs). Oxygen retains 4 non-bonding outer electrons (two lone pairs), giving a complete outer shell of 8 electrons around oxygen and 2 electrons around each hydrogen.

(b) Water is a simple molecular substance. The covalent bonds within the molecules are strong, but the intermolecular forces (forces between molecules) are weak. Only a small amount of thermal energy is needed to overcome these weak attractive forces, resulting in a low boiling point.

(a)
• One pair of shared electrons (one dot, one cross) between the central O atom and each of the two H atoms [1];
• Two lone pairs (4 unbonded electrons) on the outer shell of the O atom and no unshared electrons on the H atoms [1];

(b)
• Weak intermolecular forces / weak forces between molecules (require little thermal energy to overcome) [1].
(ALLOW 'weak attractions between molecules'; REJECT 'covalent bonds break' / 'bonds between H and O break' for mark 3)

[Total: 3 marks]

(a) Balancing the equation: Sulfur dioxide reacts with oxygen to form sulfur trioxide: \(2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{SO}_3\text{(g)}\).
(b) Sulfur dioxide dissolves in atmospheric water to form acid rain, which damages vegetation, acidifies lakes and aquatic ecosystems, or corrodes limestone structures.

(a) 2 (before \(\text{SO}_2\)) AND 2 (before \(\text{SO}_3\)) [1.5];
(b) Causes acid rain / acidification of aquatic ecosystems / leaching of soil nutrients / corrosion of stone/limestone buildings [1];
IGNORE: general global warming or ozone depletion (NOT credited).

(a) Carbon monoxide is produced when fuels burn in an incomplete combustion process due to a limited or insufficient oxygen supply.
(b) Balancing the reaction: \(2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2\). Both nitrogen and carbon balance with 2 on each side, yielding 4 oxygen atoms on both sides.

(a) Incomplete combustion / limited or insufficient supply of oxygen/air [1];
(b) Fully balanced equation with coefficients 2, 2, 2 [1.5];
(Award [1] if two of the three stoichiometric coefficients are correct).

(a) Thermal decomposition of calcium carbonate: \(\text{CaCO}_3\text{(s)} \rightarrow \text{CaO}\text{(s)} + \text{CO}_2\text{(g)}\).
(b) Excess carbon dioxide acts as a greenhouse gas that absorbs thermal energy (infrared radiation) emitted by the Earth, leading to the enhanced greenhouse effect / global warming.

(a) \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) [1];
Correct state symbols: \(\text{(s)}\) for \(\text{CaCO}_3\) and \(\text{CaO}\), \(\text{(g)}\) for \(\text{CO}_2\) [0.5];
(b) Enhanced greenhouse effect / global warming / climate change [1];
REJECT: ozone depletion.

(a) \(\text{CaO}\text{(s)} + \text{SO}_2\text{(g)} \rightarrow \text{CaSO}_3\text{(s)}\).
(b) Powdered solid has a much larger total surface area exposed to the gas. Therefore, there are more frequent collisions (more collisions per unit time/second) between reacting particles, which increases the reaction rate.

(a) \(\text{CaO} + \text{SO}_2 \rightarrow \text{CaSO}_3\) [1];
(b) Larger / increased surface area (of solid) [0.5];
More collisions per unit time / higher frequency of collisions (between particles) [1];
(Total 2.5 marks)

(a) First find the potential difference across the \(8.0\ \Omega\) resistor:
\(V_{\text{resistor}} = 12\text{ V} - 4.8\text{ V} = 7.2\text{ V}\)

Then use Ohm's law to find the current:
\(I = \frac{V}{R} = \frac{7.2}{8.0} = 0.90\text{ A}\)

(b) Calculate the power dissipated by the heating element using \(P = IV\):
\(P = 0.90\text{ A} \times 4.8\text{ V} = 4.32\text{ W}\) (or \(4.3\text{ W}\))

(a) \(V_{\text{resistor}} = 7.2\text{ V}\) OR \(I = \frac{12 - 4.8}{8.0}\) ;
\(I = 0.90\text{ A}\) ;

(b) \(P = IV\) OR \(0.90 \times 4.8\) ;
\(4.32\text{ W}\) / \(4.3\text{ W}\) [Allow ecf from (a)] ;

(a) For the lamp, \(P = IV\):
\(I_{\text{lamp}} = \frac{P}{V} = \frac{1.8}{6.0} = 0.30\text{ A}\)

(b) For the resistor in parallel, the potential difference across it is also \(6.0\text{ V}\):
\(I_{\text{resistor}} = \frac{V}{R} = \frac{6.0}{30} = 0.20\text{ A}\)

Total current is the sum of the branch currents:
\(I_{\text{total}} = I_{\text{lamp}} + I_{\text{resistor}} = 0.30 + 0.20 = 0.50\text{ A}\)

(a) \(I = \frac{P}{V}\) OR \(\frac{1.8}{6.0}\) ;
\(0.30\text{ A}\) ;

(b) Current in resistor \(I = \frac{6.0}{30} = 0.20\text{ A}\) ;
Total current \(= 0.50\text{ A}\) [Allow ecf from (a)] ;

(a) Convert time to seconds: \(t = 3.0 \times 60 = 180\text{ s}\)
Use the formula \(E = IVt\):
\(E = 2.5\text{ A} \times 24\text{ V} \times 180\text{ s} = 10\ 800\text{ J}\) (or \(10.8\text{ kJ}\) / \(1.1 \times 10^4\text{ J}\))

(b) The \(3\text{ A}\) fuse is the most suitable.
A \(2\text{ A}\) fuse would blow under normal operating current (\(2.5\text{ A}\)), whereas the \(3\text{ A}\) fuse allows normal operation while providing better protection than a \(5\text{ A}\) fuse.

(a) Conversion of \(3.0\text{ min}\) to \(180\text{ s}\) OR substitution into \(E = IVt\) (e.g. \(2.5 \times 24 \times 180\)) ;
\(10\ 800\text{ J}\) / \(10.8\text{ kJ}\) ;

(b) \(3\text{ A}\) (fuse) ;
(Reason:) Current of \(2.5\text{ A}\) would blow a \(2\text{ A}\) fuse / \(3\text{ A}\) is just above normal working current ;

(a) Use the wave equation:
\( v = f \lambda \)
Rearranging gives:
\( \lambda = \frac{v}{f} = \frac{0.45\text{ m/s}}{15\text{ Hz}} = 0.030\text{ m} \) (or \( 3.0\text{ cm} \))

(b) Waves transfer energy without transferring matter.

(a) Correct rearrangement or substitution: \( \lambda = \frac{v}{f} \) OR \( 0.45 = 15 \times \lambda \) [1]; Correct numerical value with matching unit: \( 0.030\text{ m} \) or \( 3.0\text{ cm} \) [1]
(b) Energy [1]

(a) The particles of gas are in continuous random motion and collide with the internal walls of the cylinder. Each collision exerts a force on the wall area, resulting in gas pressure (\( p = \frac{F}{A} \)).

(b) When heated, the temperature increases, so the particles gain kinetic energy and move faster. Consequently, the particles collide with the walls more frequently and with greater force/momentum change, leading to a greater total force per unit area and therefore higher pressure.

(a) Particles collide with the walls of the container / cylinder [1]; Exerting a force (over an area) [1]
(b) Particles move faster / have more kinetic energy AND collide with walls more often / more frequently (or with greater force) [1]

(a) Using the wave equation:
\( \lambda = \frac{v}{f} = \frac{1500\text{ m/s}}{50\,000\text{ Hz}} = 0.030\text{ m} \)

(b) The total distance travelled by the pulse to the seabed and back is:
\( d_{\text{total}} = v \times t = 1500\text{ m/s} \times 0.16\text{ s} = 240\text{ m} \)
Depth is half of the total distance:
\( \text{Depth} = \frac{240\text{ m}}{2} = 120\text{ m} \)
Alternatively:
\( t_{\text{one-way}} = \frac{0.16\text{ s}}{2} = 0.08\text{ s} \)
\( \text{Depth} = 1500\text{ m/s} \times 0.08\text{ s} = 120\text{ m} \)

(a) Correct calculation: \( \lambda = \frac{1500}{50\,000} = 0.030\text{ m} \) (or \( 3.0\text{ cm} \)) [1]
(b) Correct calculation of total distance (\( 240\text{ m} \)) OR half-time (\( 0.08\text{ s} \)) [1]; Final depth = \( 120\text{ m} \) [1]

(a) Acceleration is given by:
\[a = \frac{\Delta v}{t} = \frac{5.0 - 0}{2.5} = 2.0\text{ m/s}^2\]

(b) Kinetic energy is given by:
\[E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.60\text{ kg} \times (5.0\text{ m/s})^2 = 0.30 \times 25 = 7.5\text{ J}\]

(c) By conservation of energy, maximum gravitational potential energy gained equals initial kinetic energy:
\[\Delta E_p = mg\Delta h = E_k\]
\[\Delta h = \frac{E_k}{mg} = \frac{7.5}{0.60 \times 9.8} = \frac{7.5}{5.88} \approx 1.28\text{ m} \text{ (or } 1.3\text{ m to 2 s.f.)}\]

(a) \(a = \frac{\Delta v}{t}\) OR \(\frac{5.0}{2.5}\) [1]; \(2.0\text{ m/s}^2\) [1]
(b) \(E_k = \frac{1}{2}mv^2\) OR \(\frac{1}{2} \times 0.60 \times 5.0^2\) AND \(7.5\text{ J}\) [1]
(c) Use of \(\Delta E_p = mg\Delta h\) with \(E_p = 7.5\text{ J}\) (or ECF from (b)) leading to \(1.3\text{ m}\) or \(1.28\text{ m}\) [1]

(a) (i) Distance is given by:
\[s = v \times t = 0.75\text{ m/s} \times 6.0\text{ s} = 4.5\text{ m}\]

(ii) Gravitational potential energy gain:
\[\Delta E_p = mg\Delta h = 80\text{ kg} \times 9.8\text{ N/kg} \times 4.5\text{ m} = 3528\text{ J} \approx 3500\text{ J}\]

(b) Efficiency calculation:
\[\text{Efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%\]
\[\text{Efficiency} = \frac{3528}{4800} \times 100\% = 73.5\% \text{ (or } 74\% \text{ to 2 s.f.)}\]

(a)(i) \(s = v \times t = 0.75 \times 6.0 = 4.5\text{ m}\) [1]
(a)(ii) \(\Delta E_p = mg\Delta h = 80 \times 9.8 \times 4.5\) [1]; \(3528\text{ J}\) or \(3500\text{ J}\) [1]
(b) \(\frac{3528}{4800} \times 100\) (allow ECF from (a)(ii)) = \(73.5\%\) or \(74\%\) (or \(0.735\) / \(0.74\)) [1]

(a) Gravitational potential energy decrease:
\[\Delta E_p = mg\Delta h = 0.50\text{ kg} \times 9.8\text{ N/kg} \times 1.2\text{ m} = 5.88\text{ J} \approx 5.9\text{ J}\]

(b) (i) Kinetic energy at the bottom:
\[E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.50\text{ kg} \times (4.2\text{ m/s})^2 = 0.25 \times 17.64 = 4.41\text{ J} \approx 4.4\text{ J}\]

(ii) Energy dissipated to surroundings:
\[E_{\text{dissipated}} = \Delta E_p - E_k = 5.88\text{ J} - 4.41\text{ J} = 1.47\text{ J} \approx 1.5\text{ J}\]

(a) \(\Delta E_p = mg\Delta h = 0.50 \times 9.8 \times 1.2\) [1]; \(5.88\text{ J}\) or \(5.9\text{ J}\) [1]
(b)(i) \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.50 \times 4.2^2 = 4.41\text{ J}\) or \(4.4\text{ J}\) [1]
(b)(ii) \(\Delta E_p - E_k\) evaluated correctly: \(5.88 - 4.41 = 1.47\text{ J}\) or \(1.5\text{ J}\) (allow ECF from (a) and (b)(i)) [1]