2025 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

The cell membrane is a partially permeable barrier that controls the entry and exit of substances in both plant and animal cells. The cell wall provides structural support, the cytoplasm is where chemical reactions occur, and the vacuole stores cell sap.

Award 1 mark for identifying the cell membrane as the structure controlling the entry and exit of substances.

An enzyme works best at its optimum pH, where the rate of reaction is highest. The rate of reaction is highest when the time taken to complete the reaction is shortest. In this investigation, the reaction was completed fastest (in 2 minutes) at pH 7.

Award 1 mark for selecting the pH with the shortest reaction time (pH 7).

The proton number (atomic number) is 11, so the atom has 11 protons and, being neutral, 11 electrons. The nucleon number (mass number) is 23, so the number of neutrons is the nucleon number minus the proton number: \(23 - 11 = 12\) neutrons.

Award 1 mark for the correct number of protons (11), neutrons (12), and electrons (11).

Using larger marble chips of the same total mass decreases the surface area exposed to the acid. A smaller surface area reduces the frequency of collisions between reactant particles, thereby decreasing the rate of the reaction.

Award 1 mark for identifying that larger chip size (lower surface area) decreases the rate of reaction.

Alkanes are saturated hydrocarbons because they contain only single covalent carbon-carbon bonds. Alkenes contain a double carbon-carbon bond (unsaturated) and are the ones that decolourise aqueous bromine, with the general formula \(C_nH_{2n}\).

Award 1 mark for identifying the correct definition of saturated hydrocarbons (alkanes).

The change in gravitational potential energy is calculated using the formula:
\(\Delta E_p = mgh\)
\(\Delta E_p = 2.0\text{ kg} \times 9.8\text{ m/s}^2 \times 3.0\text{ m} = 58.8\text{ J}\)
This rounds to \(59\text{ J}\).

Award 1 mark for the correct calculation of gravitational potential energy change.

In the electromagnetic spectrum, ordered from longest to shortest wavelength: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Therefore, radio waves have the largest wavelength among the options.

Award 1 mark for identifying radio waves as the region with the largest wavelength.

For resistors in series, the total resistance is the sum of individual resistances:
\(R_{\text{total}} = 4.0\ \Omega + 8.0\ \Omega = 12\ \Omega\)
Using Ohm's law, \(I = \frac{V}{R}\):
\(I = \frac{12\text{ V}}{12\ \Omega} = 1.0\text{ A}\).

Award 1 mark for calculating the total resistance and applying Ohm's law to find the correct current.

A typical plant cell contains a cell wall, chloroplasts, and a large permanent vacuole, which are absent in typical animal cells. Both cell types share a cell membrane, cytoplasm, and a nucleus.

1 mark for the correct answer C.

Evaporation is the change of state from liquid to gas that occurs at temperatures below the boiling point. Condensation is gas to liquid, freezing is liquid to solid, and melting is solid to liquid.

1 mark for the correct answer B.

Average speed is calculated by dividing the total distance by the total time taken: \(\text{average speed} = \frac{\text{distance}}{\text{time}} = \frac{1500\text{ m}}{60\text{ s}} = 25\text{ m/s}\).

1 mark for the correct answer B.

Enzymes are biological catalysts made of proteins. They speed up reactions without being consumed or permanently changed, and they are highly specific to their substrates.

1 mark for the correct answer A.

Group VII of the Periodic Table contains the halogens (such as fluorine, chlorine, bromine, and iodine). Group I contains the alkali metals, Group VIII contains the noble gases, and the middle block contains the transition metals.

1 mark for the correct answer C.

Convection involves the movement of the particles themselves within a fluid (liquids and gases) to transfer thermal energy. It does not occur in solids as their particles are in fixed positions, nor in a vacuum because there are no particles.

1 mark for the correct answer B.

The raw materials required for photosynthesis are carbon dioxide and water. In the presence of sunlight and chlorophyll, they react to produce glucose and oxygen.

1 mark for the correct answer B.

Unsaturated hydrocarbons contain at least one double carbon-carbon bond (alkenes). Ethene is an alkene and is therefore unsaturated. Methane, ethane, and propane are saturated alkanes.

1 mark for the correct answer C.

Chloroplasts are the site of photosynthesis and require light to function. Since palisade mesophyll cells are located in leaves exposed to sunlight, they contain chloroplasts. Root hair cells are underground and receive no light, so they do not develop chloroplasts.

Award 1 mark for the correct option (C).

The seedling grows larger, which shows 'growth' (1). It bends towards a light source in response to a stimulus, which shows 'sensitivity' (3). Since it is a young seedling, it does not undergo 'reproduction' (2).

1 mark for the correct option B (1 and 3 only).

The enzyme works best at the pH where the reaction rate is fastest, which corresponds to the shortest time taken for digestion. This occurs at pH 7, where the digestion takes only 2 minutes.

1 mark for the correct option C (pH 7).

The proton number is 15, so the atom has 15 protons and 15 electrons. The nucleon number is 31, so the number of neutrons is \(31 - 15 = 16\).

1 mark for the correct option B (15 protons, 16 neutrons, 15 electrons).

The rate of bubble production indicates reactivity. W is most reactive (very rapid), followed by Z (moderate), then Y (slow), and finally X is least reactive (no reaction).

1 mark for the correct option B.

Since the hand is to the side and not touching the beaker, conduction (requiring contact) and convection (which primarily moves hot fluids upwards) are not the main mechanisms. The primary mechanism of sideways thermal transfer through the air gap without contact is radiation.

1 mark for the correct option D (radiation).

The total resistance in a series circuit is the sum of individual resistances: \(R = 4.0\ \Omega + 8.0\ \Omega = 12.0\ \Omega\).
Using Ohm's law, \(I = \frac{V}{R} = \frac{6.0\text{ V}}{12.0\ \Omega} = 0.50\text{ A}\).

1 mark for the correct option A (0.50 A).

Microwaves are used for both satellite communication (due to their ability to pass through the Earth's atmosphere) and in microwave ovens for cooking food.

1 mark for the correct option B (microwaves).

Decreasing the temperature decreases the kinetic energy of the particles, leading to fewer collisions per second and a lower proportion of successful collisions, thus decreasing the rate of reaction.

1 mark for the correct option B.

Enzymes are defined as proteins that function as biological catalysts. This means they speed up biochemical reactions without being consumed in the process.

1 mark for the correct option (B).

Sensitivity is defined as the ability to detect or sense changes in the internal or external environment and to make appropriate responses.

1 mark for the correct option (D).

For a neutral atom, the number of protons equals the atomic number (11), which also equals the number of electrons (11). The number of neutrons is the nucleon number minus the proton number: 23 - 11 = 12.

1 mark for the correct option (A).

The reaction of a metal carbonate with an acid always produces a salt, carbon dioxide, and water. Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water.

1 mark for the correct option (B).

Alkenes contain a carbon-carbon double bond (C=C), making them unsaturated hydrocarbons. Alkanes contain only single covalent bonds (C-C) and are saturated. Unsaturation allows alkenes to rapidly decolourise aqueous bromine water.

1 mark for the correct option (C).

The increase in gravitational potential energy is calculated using G.P.E. = m * g * h. Substituting the values: G.P.E. = 5.0 kg * 9.8 m/s² * 3.0 m = 147 J.

1 mark for the correct option (C).

An ammeter must be connected in series with the component to measure the current passing through it. A voltmeter must be connected in parallel across the component to measure the potential difference across it.

1 mark for the correct option (D).

All electromagnetic waves (radio, microwave, infrared, visible, ultraviolet, X-ray, gamma) are transverse waves that travel at the same constant speed of approximately 3.0 * 10^8 m/s in a vacuum.

1 mark for the correct option (B).

An exothermic reaction releases thermal energy to the surroundings because the energy released when new bonds are formed in the products is greater than the energy absorbed to break the bonds in the reactants.

1 mark for the correct explanation that bond-making releases more energy than is absorbed during bond-breaking.

Bromine ($X_2$) is more reactive than iodine ($Y_2$). When added to potassium iodide ($\text{KI}$), bromine displaces iodine to form potassium bromide and iodine ($Y_2$). Dissolved iodine is brown in aqueous solution, causing the observed colour change.

1 mark for identifying the correct halogen pair where $X_2$ is more reactive than $Y_2$, producing a brown iodine solution.

Hydrocarbon $P$ ($\text{C}_3\text{H}_6$) is propene, which is an alkene. Alkenes react with steam in the presence of a catalyst to form alcohols (hydration). Hydrocarbon $Q$ ($\text{C}_3\text{H}_8$) is propane, which is an alkane. Alkanes do not undergo addition reactions.

1 mark for identifying the correct statement regarding propene's reaction with steam.

In the electromagnetic spectrum, ultraviolet radiation has a higher frequency and therefore a shorter wavelength than visible light. All electromagnetic waves travel at the same speed in a vacuum ($3.0 \times 10^8\text{ m/s}$).

1 mark for the correct comparison of ultraviolet and visible light properties.

By conservation of energy, the loss in gravitational potential energy equals the gain in kinetic energy:
$$\text{KE} = mgh = 0.50\text{ kg} \times 9.8\text{ m/s}^2 \times 20\text{ m} = 98\text{ J}$$
To find the speed $v$:
$$\text{KE} = \frac{1}{2}mv^2 \implies 98 = \frac{1}{2}(0.50)v^2 \implies v^2 = 392 \implies v \approx 20\text{ m/s}$$
Alternatively, $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 20} \approx 20\text{ m/s}$.

1 mark for calculating the correct kinetic energy ($98\text{ J}$) and speed ($20\text{ m/s}$).

First, calculate the equivalent resistance of the parallel combination ($R_p$):
$$\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} \implies R_p = 3.0\ \Omega$$
Next, find the total resistance of the circuit ($R_{\text{total}}$) by adding the series resistor:
$$R_{\text{total}} = R_p + 3.0\ \Omega = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega$$
Finally, use Ohm's law to find the current ($I$):
$$I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}$$
Thus, the current drawn is $2.0\text{ A}$.

1 mark for calculating the correct current ($2.0\text{ A}$) through parallel and series combination.

Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration (against a concentration gradient) using energy from respiration. Osmosis is a passive process that does not require energy from respiration.

1 mark for identifying active transport and its direction from lower to higher concentration.

Efficient gas exchange surfaces require a large surface area to maximise the space for diffusion, a thin barrier (one cell thick) to minimise the diffusion distance, and a good blood supply (along with ventilation) to maintain a steep concentration gradient.

1 mark for identifying the correct combination of surface area, barrier thickness, and blood supply.

The overall energy change (\( \Delta H \)) of a reaction is calculated as:
\( \Delta H = \text{Energy of products} - \text{Energy of reactants} \)
\( \Delta H = 110\text{ kJ/mol} - 250\text{ kJ/mol} = -140\text{ kJ/mol} \)

Since \( \Delta H \) is negative, energy is released to the surroundings, meaning the reaction is exothermic.

1 mark for the correct calculation of \( \Delta H = -140\text{ kJ/mol} \) and identifying the reaction as exothermic (Option B).

Water leaves the xylem vessels and enters the mesophyll cells of the leaf. It then evaporates from the surface of these mesophyll cells into the air spaces, forming water vapour, which finally diffuses out of the leaf through the stomata into the atmosphere.

1 mark for correctly identifying the sequence as mesophyll cells \(\rightarrow\) air spaces \(\rightarrow\) stomata.

In Group VII (the halogens), reactivity decreases as you go down the group. Based on their physical states at room temperature:
- \( X \) (gas) must be at the top of the group (most reactive).
- \( Z \) (liquid) is in the middle of the group (intermediate reactivity).
- \( Y \) (solid) is near the bottom of the group (least reactive).

Therefore, the order of reactivity from most to least reactive is \( X \rightarrow Z \rightarrow Y \).

1 mark for the correct order of decreasing reactivity down the group: \( X \rightarrow Z \rightarrow Y \).

1. Find the equivalent resistance (\( R_p \)) of the parallel combination:
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0} \)
\( R_p = 3.0\ \Omega \)

2. Find the total resistance (\( R_{\text{total}} \)) of the circuit by adding the series resistor:
\( R_{\text{total}} = R_p + R_3 = 3.0\ \Omega + 2.0\ \Omega = 5.0\ \Omega \)

3. Calculate the total current (\( I \)) supplied by the battery using Ohm's Law:
\( I = \frac{V}{R_{\text{total}}} = \frac{10.0\text{ V}}{5.0\ \Omega} = 2.0\text{ A} \)

1 mark for the correct calculations leading to \( 2.0\text{ A} \).

1. Calculate acceleration (\( a \)) using Newton's second law:
\( a = \frac{F}{m} = \frac{12\text{ N}}{4.0\text{ kg}} = 3.0\text{ m/s}^2 \)

2. Calculate the final velocity (\( v \)) after \( 3.0\text{ s} \) starting from rest (\( u = 0 \)):
\( v = u + at = 0 + (3.0\text{ m/s}^2 \times 3.0\text{ s}) = 9.0\text{ m/s} \)

3. Calculate kinetic energy (\( E_k \)):
\( E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 4.0\text{ kg} \times (9.0\text{ m/s})^2 = 2.0 \times 81 = 162\text{ J} \)

1 mark for calculating the acceleration, final velocity and final kinetic energy correctly to get \( 162\text{ J} \).

A is incorrect because alkenes decolourise aqueous bromine via an addition reaction, not substitution.
B is incorrect because alkanes are saturated hydrocarbons containing only single covalent bonds.
C is correct because alkenes are unsaturated hydrocarbons containing double carbon-carbon bonds, which can open up to link together in addition polymerisation.
D is incorrect because alkenes (not alkanes) react with steam to form alcohols.

1 mark for identifying that alkenes can undergo addition polymerisation to form polymers.

- Amplitude is defined as the maximum displacement of a particle from its undisturbed position, which is directly given as \( 4.0\text{ cm} \).
- The distance from a crest to the next consecutive trough corresponds to half of a full wavelength (\( \frac{\lambda}{2} \)). Therefore, the full wavelength is:
\( \lambda = 2 \times 10.0\text{ cm} = 20.0\text{ cm} \).

1 mark for identifying the correct amplitude of \( 4.0\text{ cm} \) and wavelength of \( 20.0\text{ cm} \).

- Cell wall is present in both plant cells (made of cellulose) and bacterial cells (made of peptidoglycan), but is entirely absent in animal cells.
- Cell membrane is present in all three types of cells.
- Chloroplast is present in some plant cells, but absent in both bacterial and animal cells.
- Nucleus is present in plant and animal cells (eukaryotes) but absent in bacterial cells (prokaryotes).

1 mark for identifying the cell wall as the shared structure.

During aerobic respiration in cells, glucose reacts with oxygen to produce carbon dioxide and water, releasing energy. Carbon dioxide is carried by the blood to the lungs, where it diffuses into the alveoli and is exhaled. Thus, expired air contains a significantly higher concentration of carbon dioxide (about 4%) compared to inspired air (about 0.04%).

1 mark for correct option A.

Ethanol is manufactured by the hydration of ethene (catalytic addition of steam to ethene) under the following conditions: phosphoric acid catalyst, high temperature (approx. 300 degrees Celsius), and high pressure (approx. 60 atm). Option C describes the fermentation of glucose, and option D describes the Haber process conditions.

1 mark for correct option A.

Matte black surfaces are both excellent absorbers and excellent emitters of infrared (thermal) radiation. Shiny silver surfaces are poor emitters of thermal radiation. Therefore, Can X (matte black) will radiate thermal energy to the surroundings much faster than Can Y (shiny silver), causing the water in Can X to cool down more quickly.

1 mark for correct option A.

First, find the equivalent resistance of the two parallel resistors: Rp = (12 * 12) / (12 + 12) = 6.0 ohms. Next, find the total resistance of the circuit by adding the series resistor: R_total = Rp + 12 ohms = 6.0 ohms + 12 ohms = 18.0 ohms. Using Ohm's law, calculate the total current leaving the battery: I = V / R_total = 9.0 V / 18.0 ohms = 0.50 A. Since the third resistor is connected in series with the battery, the total current flows through it. Therefore, the current in the third resistor is 0.50 A.

1 mark for correct option A.

Increasing the concentration of a solution means there are more reactant particles in a given volume. This reduces the average distance between particles, resulting in more frequent collisions per unit time (increased collision frequency), thereby increasing the rate of reaction. Option B describes the effect of temperature, not concentration.

1 mark for correct option A.

Energy is lost at each stage of a food chain because organisms use energy for life processes such as movement, active transport, and maintaining body temperature. This energy is ultimately lost to the environment as heat during respiration. Additionally, some energy is lost in undigested food (feces) and excreted waste, or is left behind in uneaten parts of organisms.

1 mark for correct option A.

As you go down Group VII: (1) The elements get darker in colour (chlorine is a pale green gas, bromine is a red-brown liquid, iodine is a grey-black solid). (2) Their reactivity decreases because the outer shell is further from the nucleus, making it harder to attract and gain an incoming electron. (3) Their melting and boiling points increase.

1 mark for correct option A.

Redshift is the increase in the observed wavelength (shift towards the red end of the spectrum) of electromagnetic radiation emitted from receding stars and galaxies. This shows that distant galaxies are moving away from us, with more distant galaxies moving faster, which indicates the expansion of space and provides strong evidence for the Big Bang theory.

1 mark for correct option A.

The distance travelled can be calculated by finding the area under the speed-time graph:
1. Area of acceleration phase (triangle): \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0\text{ s} \times 20\text{ m/s} = 50\text{ m}\).
2. Area of constant speed phase (rectangle): \(\text{base} \times \text{height} = 10\text{ s} \times 20\text{ m/s} = 200\text{ m}\).
3. Area of deceleration phase (triangle): \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0\text{ s} \times 20\text{ m/s} = 30\text{ m}\).

Total distance = \(50\text{ m} + 200\text{ m} + 30\text{ m} = 280\text{ m}\).

[1] Mark is awarded for the correct option B.

In a cracking reaction, the total number of carbon and hydrogen atoms in the reactants must equal the total number of carbon and hydrogen atoms in the products.

Total Carbon atoms in products = \(3 + 3 + (2 \times 2) = 10\).
Total Hydrogen atoms in products = \(8 + 6 + (2 \times 4) = 22\).

Therefore, the molecular formula of the reactant alkane \(Y\) is \(\text{C}_{10}\text{H}_{22}\).

[1] Mark is awarded for the correct option C.

First, calculate the equivalent resistance of the two \(6.0\ \Omega\) parallel resistors:
\(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} \implies R_p = 3.0\ \Omega\).

Next, calculate the total resistance of the circuit (series plus parallel):
\(R_{\text{total}} = 2.0\ \Omega + 3.0\ \Omega = 5.0\ \Omega\).

Finally, calculate the total current using Ohm's law:
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{5.0\ \Omega} = 2.4\text{ A}\).

[1] Mark is awarded for the correct option C.

At high temperatures above the optimum (typically around \(40^\circ\text{C}\)), the heat energy breaks the bonds maintaining the three-dimensional shape of the enzyme molecule. This alters the shape of the active site permanently (denaturation), so the substrate molecule can no longer fit into it, causing the rate of reaction to drop rapidly.

[1] Mark is awarded for the correct option C.

In this reaction, iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) loses oxygen to become iron, meaning it is reduced. Carbon monoxide (\(\text{CO}\)) gains oxygen to become carbon dioxide, meaning it is oxidized. Therefore, \(\text{CO}\) acts as the reducing agent because it removes oxygen from \(\text{Fe}_2\text{O}_3\).

[1] Mark is awarded for the correct option B.

Use the wave equation:
\(v = f \lambda\)

Rearranging to find wavelength (\(\lambda\)):
\(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{1.0 \times 10^8\text{ Hz}} = 3.0\text{ m}\).

[1] Mark is awarded for the correct option B.

Energy at each trophic level is calculated as follows:
- Trophic level 1 (Producers): \(20\ 000\text{ kJ}\)
- Trophic level 2 (Primary consumers): \(20\ 000\text{ kJ} \times 10\% = 2000\text{ kJ}\)
- Trophic level 3 (Secondary consumers): \(2000\text{ kJ} \times 10\% = 200\text{ kJ}\).

[1] Mark is awarded for the correct option C.

Fertilisation is defined as the fusion of the nuclei of two haploid gametes (the male sperm cell and the female egg cell) to form a diploid zygote.

[1] Mark is awarded for the correct option B.

The orbital speed \(v\) is given by \(v = \frac{2\pi r}{T}\).
For Planet X: \(v_X = \frac{2\pi r}{T}\).
For Planet Y: \(v_Y = \frac{2\pi (2r)}{3T} = \frac{4\pi r}{3T}\).
Therefore, the ratio is:
\(\frac{v_X}{v_Y} = \frac{\frac{2\pi r}{T}}{\frac{4\pi r}{3T}} = 2 \times \frac{3}{4} = 1.5\).

Award 1 mark for the correct option B.

Alkenes undergo addition reactions with steam to produce alcohols. Therefore, propene reacts with steam to form propanol. This hydration reaction requires an acid catalyst, commonly phosphoric acid.

Award 1 mark for the correct option A.

A concentrated sugar solution has a lower water potential than the cell sap. Water moves out of the cells down a water potential gradient by osmosis, causing the vacuole and cytoplasm to shrink and pull the cell membrane away from the cell wall. This state is known as plasmolysis.

Award 1 mark for the correct option B.

Useful GPE gained by the load = \(mgh = 25\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 2940\text{ J}\).
Useful power output = \(\frac{\text{Useful GPE}}{t} = \frac{2940\text{ J}}{15\text{ s}} = 196\text{ W}\).
Since \(\text{Efficiency} = \frac{\text{Useful power output}}{\text{Power input}} \times 100\%\):
\(80\% = \frac{196}{\text{Power input}} \times 100\%\)
\(\text{Power input} = \frac{196}{0.80} = 245\text{ W}\).

Award 1 mark for the correct option C.

Isotopes are atoms of the same element that contain the same number of protons but different numbers of neutrons. Particles W and X both have 17 protons, meaning they belong to the same element (chlorine). They have different numbers of neutrons (18 and 20 respectively), making them isotopes of each other.

Award 1 mark for the correct option A.

The electromagnetic spectrum in order of increasing frequency is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Therefore, infrared and microwaves lie between radio waves and visible light. Since infrared is an option, it is correct.

Award 1 mark for the correct option A.

Increasing concentration increases the number of reacting particles per unit volume. This makes collisions more frequent, which increases the rate of reaction. The kinetic energy of the particles is affected by temperature, not concentration.

Award 1 mark for the correct option B.

First, calculate the resistance of the parallel combination (\(R_p\)):
\(R_p = \frac{4.0 \times 12.0}{4.0 + 12.0} = \frac{48.0}{16.0} = 3.0\ \Omega\).

Next, calculate the total resistance of the series circuit:
\(R_{\text{total}} = R_p + 5.0\ \Omega = 3.0\ \Omega + 5.0\ \Omega = 8.0\ \Omega\).

Using Ohm's law, find the total current flowing from the battery:
\(I = \frac{V}{R_{\text{total}}} = \frac{12.0\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).

Since the \(5.0\ \Omega\) resistor is in series with the parallel block, the total current of \(1.5\text{ A}\) flows directly through it.

Award 1 mark for the correct option C.

(a) Structure A is the cell wall, which provides support. B is the large permanent vacuole, containing cell sap. C represents a chloroplast, where photosynthesis occurs.

(b) Root hair cells have a long hair-like extension which significantly increases the surface area for absorption of water and mineral ions. They also have a thin cell wall which decreases the distance over which water and minerals must travel.

(c) The reactants of photosynthesis are carbon dioxide and water, and the products are glucose and oxygen in the presence of light and chlorophyll.

(d) Magnesium ions are essential for the synthesis of chlorophyll, the green pigment in chloroplasts. Without sufficient magnesium, plants cannot produce chlorophyll, leading to chlorosis (yellowing of leaves, especially between the veins).

1 (a) 1 mark for each correct identification: A = cell wall, B = vacuole (or permanent vacuole), C = chloroplast. [3]
1 (b) 1 mark for mentioning the elongated shape/projection/large surface area; 1 mark for thin wall/large number of mitochondria/large vacuole. [2]
1 (c) 1 mark for 'water' (first gap); 1 mark for 'glucose' (or sugar/starch) (second gap). [2]
1 (d) 1 mark for stating it is needed to make chlorophyll; 1 mark for yellowing of leaves / chlorosis. [2]

(a) (i) Propane contains 3 carbon atoms and 8 hydrogen atoms, so its molecular formula is \(\text{C}_3\text{H}_8\).
(ii) Propane has only single C-C bonds and fits the general formula \(\text{C}_n\text{H}_{2n+2}\), identifying it as an alkane.
(iii) 'Hydrocarbon' means it contains only hydrogen and carbon atoms. 'Saturated' means all carbon-carbon bonds are single covalent bonds (no double or triple bonds).

(b) Complete combustion of a hydrocarbon yields carbon dioxide and water: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\).

(c) (i) Ethene has a double bond between the two carbon atoms: \(\text{H}_2\text{C}=\text{CH}_2\).
(ii) Bromine water test: Alkenes like ethene readily react with bromine due to their double bond, decolourising orange/brown bromine water. Alkanes like propane do not react under normal conditions, so the bromine water remains orange/brown.

2 (a) (i) 1 mark for \(\text{C}_3\text{H}_8\). [1]
2 (a) (ii) 1 mark for alkane(s). [1]
2 (a) (iii) 1 mark for containing only single bonds; 1 mark for containing hydrogen and carbon only. [2]
2 (b) 1 mark for \(3\text{CO}_2\); 1 mark for \(4\text{H}_2\text{O}\). [2]
2 (c) (i) 1 mark for drawing ethene structure correctly with a carbon-carbon double bond and four single carbon-hydrogen bonds. [1]
2 (c) (ii) 1 mark for specifying bromine water / aqueous bromine; 1 mark for stating ethene turns it colourless (decolourises) AND propane remains orange/brown/no change. [2]

(a) (i) For resistors in series, the total resistance \(R_{\text{total}} = R_1 + R_2 = 3.0\ \Omega + 5.0\ \Omega = 8.0\ \Omega\).
(ii) Using Ohm's Law, \(I = \frac{V}{R} = \frac{12.0\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).
(iii) The potential difference across the \(5.0\ \Omega\) resistor is \(V = I \times R = 1.5\text{ A} \times 5.0\ \Omega = 7.5\text{ V}\).

(b) (i) In parallel, the combined resistance is less than the smallest individual resistance, so the total resistance decreases. Since \(I = \frac{V}{R}\), the total current increases.
(ii) A fuse contains a thin wire with a low melting point. When the current exceeds the designated threshold, the heating effect of the current causes the wire to melt, breaking the circuit and stopping any further current flow.

3 (a) (i) 1 mark for \(8.0\ \Omega\). [1]
3 (a) (ii) 1 mark for using \(I = \frac{V}{R}\); 1 mark for obtaining \(1.5\text{ A}\). [2]
3 (a) (iii) 1 mark for formula/calculation \(1.5 \times 5.0\); 1 mark for \(7.5\text{ V}\). [2]
3 (b) (i) 1 mark for 'increases'. [1]
3 (b) (ii) 1 mark for current heating the fuse wire; 1 mark for the wire melting; 1 mark for breaking/opening the circuit. [3]

(a) (i) Wholemeal bread is rich in dietary fiber, unlike refined products, butter, or meat.
(ii) Vitamin C is essential for collagen synthesis, maintaining healthy skin, gums, and blood vessels. A deficiency leads to scurvy, characterized by bleeding gums and poor wound healing.

(b) (i) Haemoglobin is the oxygen-transporting protein containing iron.
(ii) Red blood cells have a biconcave disc shape to maximize the surface-area-to-volume ratio, facilitating faster diffusion of oxygen. They lack a nucleus to provide more internal volume for carrying haemoglobin.

(c) Arteries carry blood away from the heart. Veins have valves to ensure one-way flow back to the heart under low pressure. Capillaries have walls that are one cell thick, providing a very short diffusion pathway for nutrient and gas exchange.

4 (a) (i) 1 mark for B (wholemeal bread). [1]
4 (a) (ii) 1 mark for healthy skin/gums/tissue repair; 1 mark for scurvy. [2]
4 (b) (i) 1 mark for haemoglobin. [1]
4 (b) (ii) 1 mark each for any two adaptations: biconcave shape (large surface area), lack of nucleus, or flexibility. [2]
4 (c) 1 mark for 'away from' (first gap); 1 mark for 'valves' (second gap); 1 mark for 'one' (third gap). [3]

(a) (i) The reaction between a metal and an acid produces hydrogen gas.
(ii) The word equation is: \(\text{zinc} + \text{hydrochloric acid} \rightarrow \text{zinc chloride} + \text{hydrogen}\).

(b) Hydrogen is tested using a lighted splint. The positive result is a characteristic 'squeaky pop' sound as the hydrogen combusts rapidly with oxygen.

(c) (i) Powdered zinc has a larger total surface area than the same mass of zinc lumps. This increases the contact area between the reactants, resulting in a higher frequency of successful collisions per unit time, which increases the rate of reaction.
(ii) Higher temperature gives the reacting particles more kinetic energy. They move faster, causing more frequent collisions. Furthermore, a larger fraction of particles possess energy greater than the activation energy, making a higher percentage of collisions successful.

5 (a) (i) 1 mark for hydrogen. [1]
5 (a) (ii) 1 mark for reactants (zinc + hydrochloric acid); 1 mark for products (zinc chloride + hydrogen). [2]
5 (b) 1 mark for lighted/burning splint; 1 mark for (squeaky) pop. [2]
5 (c) (i) 1 mark for rate increases / faster reaction; 1 mark for explanation involving larger surface area / more frequent collisions. [2]
5 (c) (ii) 1 mark for rate increases; 1 mark for explanation involving greater kinetic energy of particles / more frequent collisions / more successful collisions. [2]

(a) (i) The angle of incidence is measured between the normal line (which is perpendicular to the mirror surface, \(90^\circ\)) and the incident ray. Therefore, the angle of incidence \(= 90^\circ - 40^\circ = 50^\circ\).
(ii) According to the law of reflection, the angle of reflection equals the angle of incidence, so it is also \(50^\circ\).

(b) (i) The region between ultraviolet and gamma rays is X-rays.
(ii) Common uses of infrared include remote controls, thermal imaging cameras, and radiant heaters.
(iii) Wavelength is inversely proportional to frequency. Since radio waves have the lowest frequency, they have the longest wavelength.

(c) (i) In longitudinal waves, the direction of vibration of the particles is parallel to the direction of energy transfer (propagation). In transverse waves, the direction of vibration is perpendicular to the direction of energy transfer.
(ii) Sound is a mechanical wave that propagates by transferring kinetic energy through collisions of physical particles (atoms or molecules). In a vacuum, there are no particles, so sound cannot travel. Light is an electromagnetic wave consisting of oscillating electric and magnetic fields, which can propagate through empty space.

6 (a) (i) 1 mark for \(50^\circ\). [1]
6 (a) (ii) 1 mark for \(50^\circ\). [1]
6 (b) (i) 1 mark for X-rays. [1]
6 (b) (ii) 1 mark for any correct use of infrared (e.g., remote controls, heating, thermal imaging). [1]
6 (b) (iii) 1 mark for radio waves. [1]
6 (c) (i) 1 mark for longitudinal vibrations parallel to wave direction; 1 mark for transverse vibrations perpendicular to wave direction. [2]
6 (c) (ii) 1 mark for stating sound needs a medium/particles to vibrate; 1 mark for stating light/electromagnetic waves do not need a medium. [2]

(a) In yeast, anaerobic respiration (fermentation) breaks down glucose without oxygen, yielding ethanol and carbon dioxide: \(\text{glucose} \rightarrow \text{ethanol} + \text{carbon dioxide}\).

(b) (i) Structure X is the trachea (windpipe) and Y represents the alveoli (air sacs).
(ii) Alveoli are adapted for gas exchange by having extremely thin walls (only one cell thick) to minimize the diffusion path. They are also present in millions, providing an enormous total surface area, and they are surrounded by an extensive capillary network to maintain a steep concentration gradient.

(c) (i) Tar is a sticky, carcinogenic substance that paralyses the cilia in the airways, preventing the clearance of dust and microbes, and significantly increases the risk of lung cancer.
(ii) Carbon monoxide is a toxic gas that binds preferentially and irreversibly to haemoglobin in red blood cells, forming carboxyhaemoglobin. This reduces the amount of oxygen that can be transported, forcing the heart to work harder.

7 (a) 1 mark for glucose (reactant); 1 mark for ethanol + carbon dioxide (products). [2]
7 (b) (i) 1 mark for X = trachea; 1 mark for Y = alveoli (air sacs). [2]
7 (b) (ii) 1 mark each for any two adaptations: thin walls (one cell thick), large surface area, moist surface, or good blood supply/capillaries. [2]
7 (c) (i) 1 mark for stating tar causes cancer / clogs cilia / increases mucus production. [1]
7 (c) (ii) 1 mark for stating it binds to haemoglobin; 1 mark for explaining it reduces the oxygen-carrying capacity of the blood. [2]

(a) (i) Weight \(W = m \times g = 4.0\text{ kg} \times 9.8\text{ m/s}^2 = 39.2\text{ N}\) (rounded to \(39\text{ N}\)).
(ii) Work done \(W = F \times d = \text{weight} \times \text{height} = 39.2\text{ N} \times 3.0\text{ m} = 117.6\text{ J}\) (or \(120\text{ J}\) using \(39.2\text{ N}\)).

(b) Useful power output \(P = \frac{\text{Work Done}}{\text{time}} = \frac{117.6\text{ J}}{6.0\text{ s}} = 19.6\text{ W}\) (or \(20\text{ W}\)).

(c) Efficiency \(= \frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100\% = \frac{19.6\text{ W}}{30\text{ W}} \times 100\% = 65.3\%\) (or using \(20\text{ W}\) gives \(\frac{20}{30} \times 100\% = 66.7\%\)).

8 (a) (i) 1 mark for formula \(W = mg\); 1 mark for \(39.2\text{ N}\) (or \(39\text{ N}\)). [2]
8 (a) (ii) 1 mark for formula \(\text{Work} = F \times d\) (or \(mgh\)); 1 mark for \(117.6\text{ J}\) (or \(120\text{ J}\)). [2]
8 (b) 1 mark for formula \(P = \frac{W}{t}\); 1 mark for \(19.6\text{ W}\) (or \(20\text{ W}\)). [2]
8 (c) 1 mark for formula \(\text{Efficiency} = \frac{\text{Power Out}}{\text{Power In}} \times 100\); 1 mark for correct substitution; 1 mark for \(65\%\) or \(67\%\) (depending on rounding used). [3]

(a)(i)
The speed is constant at \(8.0\text{ m/s}\), so the motion is described as constant speed (or uniform speed / uniform velocity).

(a)(ii)
Using the formula for acceleration:
\(a = \frac{\Delta v}{t}\)
\(a = \frac{8.0\text{ m/s} - 0\text{ m/s}}{5.0\text{ s}} = 1.6\text{ m/s}^2\)

(a)(iii)
The distance travelled at a constant speed is given by the area under that section of the speed-time graph:
\(\text{time interval} = 15.0\text{ s} - 5.0\text{ s} = 10.0\text{ s}\)
\(\text{distance} = \text{speed} \times \text{time} = 8.0\text{ m/s} \times 10.0\text{ s} = 80\text{ m}\)

(b)
Work done is calculated using the formula:
\(W = F \times d\)
where \(F = 120\text{ N}\) and \(d = 80\text{ m}\) (from part a(iii)).
\(W = 120\text{ N} \times 80\text{ m} = 9600\text{ J}\) (or \(9.6\text{ kJ}\))

(a)(i)
- constant speed / steady speed / uniform speed / uniform velocity [1]
(Reject: terminal velocity / constant acceleration)

(a)(ii)
- formula or correct substitution: \(8.0 / 5.0\) [1]
- correct value: \(1.6\) [1]
- correct unit: \(\text{m/s}^2\) [1]

(a)(iii)
- evidence of using \(\text{time} = 10\text{ s}\) or showing \(8 \times 10\) [1]
- correct value: \(80\text{ m}\) [1]
*(Allow ecf if \(8 \times 15 = 120\text{ m}\) is used, award 1 mark)*

(b)
- correct formula: \(W = F \times d\) [1]
- correct substitution of \(120 \times \text{distance from a(iii)}\) [1]
- correct value and unit: \(9600\text{ J}\) or \(9.6\text{ kJ}\) [1]
(Allow ecf from (a)(iii)