2026 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

(a) (i) Structure P is the petal, structure Q is the anther, and structure R is the stigma.
(ii) Structure S is the sepal; its primary function is to protect the unopened flower bud.
(b) Pollen grains from wind-pollinated flowers are typically smaller, lighter, and have a smooth surface compared to the larger, stickier, or spiked pollen of insect-pollinated flowers. They are also produced in much greater quantities.
(c) The essential environmental conditions required for seed germination are water/moisture, oxygen, and a suitable/favourable temperature (any two).

(a) (i) 1 mark for each correct structure [max 3]:
- P: petal [1];
- Q: anther (ALLOW stamen) [1];
- R: stigma (ALLOW carpel / pistil) [1].

(a) (ii) 1 mark for:
- protects the flower / internal parts when in bud / before opening [1];
(IGNORE attracts insects, produces food).

(b) 1 mark per valid structural difference [max 2]:
- wind-pollinated pollen grains are lighter / smaller / have lower mass (ALLOW insect-pollinated are heavier / larger) [1];
- wind-pollinated pollen grains are smooth / non-sticky (ALLOW insect-pollinated are sticky / rough / spiky / have hooks) [1];
- wind-pollinated pollen grains are produced in larger quantities / numbers [1].
(REJECT differences describing petals or stigmas).

(c) Any two from [max 2]:
- water / moisture [1];
- oxygen [1];
- suitable temperature / warm temperature / warmth (ALLOW optimum temperature) [1];
(REJECT sunlight / light, soil, nutrients, fertiliser).

(a) Heart rate can be monitored by feeling/counting the pulse (e.g., at the wrist/radial artery or neck/carotid artery), using a stethoscope to listen to heart sounds/valve closures, or via an electrocardiogram (ECG).

(b) (i)
- Initial resting heart rate of Person B (at \(t = 0\text{ min}\)) = \(72\text{ bpm}\)
- Heart rate at end of exercise (at \(t = 6\text{ min}\)) = \(162\text{ bpm}\)
- Increase in heart rate = \(162 - 72 = 90\text{ bpm}\)
- Percentage increase = \(\frac{\text{increase}}{\text{initial}} \times 100 = \frac{90}{72} \times 100 = 125\%\)

(ii)
- Exercise stopped at \(t = 6\text{ min}\).
- Person A reached their baseline resting rate of \(58\text{ bpm}\) at \(t = 12\text{ min}\).
- Time taken after exercise stopped = \(12 - 6 = 6\text{ minutes}\).

(c) Any two differences:
1. Person A has a lower resting heart rate (\(58\text{ bpm}\)) compared to Person B (\(72\text{ bpm}\)).
2. Person A reaches a lower maximum heart rate during exercise (\(125\text{ bpm}\)) than Person B (\(162\text{ bpm}\)) / Person B's heart rate increases more steeply.
3. Person A returns to their resting heart rate within \(6\text{ minutes}\) of recovery, whereas Person B does not return to their resting rate within the \(6\text{ minutes}\) (remains at \(76\text{ bpm}\)).

(d)
- During exercise, muscle cells contract more frequently / work harder.
- Muscle cells require more energy, so the rate of aerobic respiration increases.
- The heart beats faster to pump more blood to the muscles.
- This delivers oxygen and glucose at a faster rate (and removes carbon dioxide and heat more rapidly).

(a) Any 1 from:
- (feeling / counting) pulse (at wrist / neck / radial / carotid artery) ;
- listening to heart / valve sounds with a stethoscope ;
- ECG / electrocardiogram ;
[1 mark]
[IGNORE: blood pressure monitor]

(b)(i)
- Correct values identified from table: \(72\) and \(162\) (or difference = \(90\)) [1] ;
- Correct calculation method: \(\frac{162 - 72}{72} \times 100\) OR \(\frac{90}{72} \times 100\) [1] ;
- Correct answer: \(125\) (\%)[1] ;
[3 marks]

(b)(ii)
- \(6\) (minutes) ;
[1 mark]
[ALLOW: \(12 - 6 = 6\)]

(c) Any 2 from:
- Person A has a lower resting heart rate (than Person B) / Person B has a higher resting heart rate ;
- Person A has a lower maximum/peak heart rate / smaller overall increase (than Person B) ;
- Person A's heart rate rises less steeply / at a lower rate during exercise ;
- Person A recovers faster / returns to resting heart rate by 12 min whereas Person B does not ;
[2 marks]
[Note: Comparisons must be explicit to gain credit]

(d) Any 3 from:
- (skeletal) muscles contract more / work harder ;
- increased rate of (cellular / aerobic) respiration ;
- need more oxygen AND more glucose (delivered to muscle cells) ;
- blood pumped faster / increased blood flow (to transport substances) ;
- increased removal of carbon dioxide / waste products / heat ;
[Max 3 marks]

(a) (i) The primary input of energy entering almost all natural ecosystems is sunlight from the Sun, which is trapped by photosynthetic producers.

(ii) Organisms that consume only photosynthetic producers are classified as primary consumers or herbivores.

(b) (i) Plant and algal cells store excess glucose in the form of starch, an insoluble polysaccharide.

(ii) The food test for starch requires adding a few drops of iodine solution (or iodine dissolved in potassium iodide). A positive result is indicated by a colour change from yellow-brown / orange to blue-black.

(c) As energy passes along food chains, roughly 90% is lost at each trophic level. Energy is lost as heat released during cellular respiration, utilized for metabolic processes and movement, and lost in unconsumed parts or excreted waste products (faeces and urine).

(d) The invasive predatory fish will feed on the sticklebacks, causing the stickleback population to decrease. Since sticklebacks prey on water snails, a decline in sticklebacks reduces predation pressure on the water snails, causing the snail population to increase.

(a)(i) Sun / sunlight / light energy from the Sun [1]
IGNORE: light alone / heat
REJECT: Sun and rain

(a)(ii) herbivore / primary consumer [1]

(b)(i) starch [1]
REJECT: glycogen / cellulose

(b)(ii) add iodine (solution) / aqueous iodine [1];
(colour changes to) blue-black / blue / black [1]
ALLOW: colour change from brown / orange / yellow-brown to blue-black
REJECT: purple / brick-red

(c) any two from:
- energy is lost at each trophic level / between trophic levels [1];
- energy lost as heat / thermal energy (from cellular respiration) [1];
- energy lost in excretion / egestion / faeces / urine / uneaten material [1];
- energy used for life processes / movement [1];

(d) (snail population) increases / rises [1];
(invasive fish eat sticklebacks so) fewer sticklebacks to eat / prey on snails OR predation (pressure) on snails decreases [1]

(a)(i) The balanced chemical equation with state symbols is:
\[2\text{Na(s)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{NaCl(s)}\]
- Correct formulas and balancing: 1 mark
- Correct state symbols: 1 mark

(a)(ii) Sodium is a metal in Group I, so each sodium atom loses 1 electron to achieve a stable outer shell, forming a sodium ion (\(\text{Na}^+\)). Chlorine is a non-metal in Group VII, so each chlorine atom gains 1 electron to complete its outer shell, forming a chloride ion (\(\text{Cl}^-\)).

(b)(i) In a molecule of \(\text{HCl}\):
- There is 1 shared pair of electrons (one single covalent bond) between the hydrogen atom and the chlorine atom.
- The chlorine atom has 6 non-bonding outer-shell electrons (3 lone pairs), giving it a full outer shell of 8 electrons, while hydrogen has 2 outer electrons.

(b)(ii)
- Sodium chloride has a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions (\(\text{Na}^+\) and \(\text{Cl}^-\)).
- Hydrogen chloride consists of simple molecules with weak intermolecular forces between the molecules.
- Significantly more energy is required to overcome the strong ionic bonds in sodium chloride than the weak intermolecular forces in hydrogen chloride.

(a)(i)
- \(2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}\) [1]
- Correct state symbols: \(\text{(s)}\) for \(\text{Na}\), \(\text{(g)}\) for \(\text{Cl}_2\), and \(\text{(s)}\) for \(\text{NaCl}\) [1]
(ALLOW state symbol mark only if formulas are correct or balanced with correct chemical species)

(a)(ii)
- Sodium (atom) loses one electron / electron transferred from sodium [1]
- Chlorine (atom) gains one electron / electron transferred to chlorine [1]
(ALLOW 'Na forms \(\text{Na}^+\)' and 'Cl forms \(\text{Cl}^-\)' if electron loss/gain is explicitly stated)

(b)(i)
- One pair of shared electrons between H and Cl [1]
- 6 unshared / non-bonding electrons on Cl (and no extra electrons on H) [1]
(ALLOW a correct dot-and-cross description/diagram representation)

(b)(ii)
- (Sodium chloride has) strong electrostatic attractions between oppositely charged ions / strong ionic bonds [1]
- (Hydrogen chloride has) weak intermolecular forces / weak forces between molecules [1] (REJECT 'weak covalent bonds')
- More energy needed to separate ions / break ionic bonds in sodium chloride than to overcome weak intermolecular forces in hydrogen chloride [1]

(a) The general word equation for an acid reacting with a metal carbonate is:
$$\text{acid} + \text{metal carbonate} \rightarrow \text{salt} + \text{water} + \text{carbon dioxide}$$
Therefore, hydrochloric acid reacts with magnesium carbonate to produce magnesium chloride, carbon dioxide, and water.

(b) The gas produced is carbon dioxide (\(\text{CO}_2\)). The test for carbon dioxide is bubbling the gas through limewater (aqueous calcium hydroxide). A positive result is the limewater turning milky or cloudy (forming a white precipitate of calcium carbonate).

(c) According to collision theory, increasing the temperature increases the average kinetic energy of the reacting particles (they move faster). This leads to two effects:
1. Particles collide more frequently (higher collision frequency).
2. A greater proportion of colliding particles have energy greater than or equal to the activation energy, so a higher percentage of collisions are successful/effective.

(a) magnesium chloride, carbon dioxide, water ;
[2 marks for all three correct; 1 mark for any two correct]
ALLOW correct chemical formulas: \(\text{MgCl}_2\), \(\text{CO}_2\), \(\text{H}_2\text{O}\)

(b) test: bubble through / add to limewater / aqueous calcium hydroxide ;
result: (turns) milky / cloudy / white precipitate ;
REJECT 'clear' or 'extinguishes a lighted splint' for the observation/test credit.

(c) Any three from:
- particles gain kinetic energy / move faster ;
- increased frequency of collisions / collisions occur more frequently / more collisions per second / per unit time ;
- a greater fraction / proportion of particles have energy greater than or equal to the activation energy (\(E_a\)) ;
- greater frequency of successful collisions / more successful collisions per second / per unit time ;
[Total: 7]

(a)
(i) Fractional distillation separates substances based on differences in their boiling points (or boiling temperature ranges).

(ii) The fraction removed at the bottom of the column is bitumen (or residue). Its primary commercial use is for surfacing roads (asphalt) or roofing.

(b)
(i) Balancing the carbon and hydrogen atoms:
$$\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4$$
Decane contains 10 carbons and 22 hydrogens. Octane accounts for 8 carbons and 18 hydrogens, leaving \(10 - 8 = 2\) carbon atoms and \(22 - 18 = 4\) hydrogen atoms, forming **\(\text{C}_2\text{H}_4\)** (ethene).

(ii) Industrial cracking requires:
1. High temperature / heat (typically \(450\text{--}750\,^\circ\text{C}\)).
2. A catalyst (such as aluminium oxide / alumina / silica / zeolite / porous pot).

(c)
(i) Incomplete combustion occurs when there is insufficient oxygen, producing carbon monoxide (\(\text{CO}\)).

(ii) Oxides of nitrogen dissolve in water droplets in clouds to form acid rain. Acid rain lowers the pH of freshwater lakes (killing fish and aquatic life), damages vegetation/trees, or accelerates the corrosion of metal and limestone structures.

(d)
Add aqueous bromine (or bromine water). An alkene reacts by addition and causes the bromine water to turn from orange/brown to colourless (decolourises), whereas an alkane remains orange/brown.

(a)(i)
* (different) boiling points / boiling ranges; [1]
(IGNORE melting points)

(a)(ii)
* Bitumen; [1]
* (surfacing) roads / asphalt / tarmac / roofing / waterproofing; [1]
(ALLOW residue / asphalt for name)

(b)(i)
* \(\text{C}_2\text{H}_4\) / \(\text{H}_2\text{C=CH}_2\); [1]

(b)(ii)
* high temperature / heat / \(400\text{--}800\,^\circ\text{C}\); [1]
* catalyst / named catalyst (e.g. alumina / aluminium oxide / silica / zeolite / broken pot); [1]
(IGNORE pressure / high pressure)

(c)(i)
* carbon monoxide / \(\text{CO}\); [1]
(REJECT carbon dioxide / soot)

(c)(ii)
* forms acid rain / reacts with water/rain to form nitric/acidic solution; [1]
* (causes) acidification of lakes / kills aquatic organisms / damages trees / leaches nutrients from soil / damages limestone (buildings) / corrodes metal; [1]
(IGNORE global warming / greenhouse effect)

(d)
* (add) aqueous bromine / bromine water AND (turns from orange/brown to) colourless / decolourises; [1]
(ALLOW bromine in organic solvent; REJECT turns clear unless colourless is stated)

(a) Speed of electromagnetic waves in a vacuum = \(3.0 \times 10^8\text{ m/s}\).

(b) Use the wave equation \(v = f \lambda\):
\[\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{1.2 \times 10^8\text{ Hz}} = 2.5\text{ m}\]

(c) Use the equation \(v = \frac{s}{t}\):
\[t = \frac{s}{v} = \frac{2.4 \times 10^{11}\text{ m}}{3.0 \times 10^8\text{ m/s}} = 800\text{ s}\]

(d)(i) Satellite communication / satellite television / mobile phone signals / Wi-Fi transmissions.
(d)(ii) Internal heating of body cells / body tissues.

(a) \(3.0 \times 10^8\text{ m/s}\) [1]
(ALLOW \(3 \times 10^8\text{ m/s}\) or \(300\,000\,000\text{ m/s}\); IGNORE missing units if numerical value is correct)

(b)
- Correct formula or substitution: \(\lambda = \frac{v}{f}\) OR \(\frac{3.0 \times 10^8}{1.2 \times 10^8}\) [1]
- Correct answer with unit: \(2.5\text{ m}\) [1]
(ALLOW ECF from incorrect speed in (a))

(c)
- Correct formula or substitution: \(t = \frac{s}{v}\) OR \(\frac{2.4 \times 10^{11}}{3.0 \times 10^8}\) [1]
- Correct answer: \(800\text{ s}\) [1]
(ALLOW ECF from speed in (a))

(d)(i) Any one from: [1]
- Satellite communications / satellite television
- Mobile phones / cell phones
- Wi-Fi / wireless networks / Bluetooth
- Radar
(IGNORE cooking / microwave ovens as question specifies communication)

(d)(ii) Internal heating of body tissues / body cells / burns [1]
(REJECT ionisation / causes cancer / mutates DNA / radiation poisoning)

(a)(i) Weight \( W = mg \) or \( \text{weight} = \text{mass} \times \text{gravitational field strength} \).

(a)(ii) \( W = 350\text{ kg} \times 9.8\text{ N/kg} = 3430\text{ N} \).

(b) Total contact area \( A = 2 \times 0.28\text{ m}^2 = 0.56\text{ m}^2 \).
Pressure \( p = \frac{F}{A} = \frac{3430\text{ N}}{0.56\text{ m}^2} = 6125\text{ Pa} \) (or \( \text{N/m}^2 \)).

(c)(i) \( \text{Resultant force } F = 840\text{ N} - 210\text{ N} = 630\text{ N} \).

(c)(ii) Using \( F = ma \):
\( a = \frac{F}{m} = \frac{630\text{ N}}{350\text{ kg}} = 1.8\text{ m/s}^2 \).

(c)(iii) Using \( a = \frac{\Delta v}{t} \) (with initial speed \( u = 0 \)):
\( v = a \times t = 1.8\text{ m/s}^2 \times 6.0\text{ s} = 10.8\text{ m/s} \).

(a)(i) \( W = mg \) / \( \text{weight} = \text{mass} \times \text{gravitational field strength} \) [1];

(a)(ii) \( 3430\text{ [N]} \) [1]
(ALLOW \( 3500\text{ [N]} \) if \( g = 10\text{ N/kg} \) is used);

(b) Total contact area \( = 2 \times 0.28 = 0.56\text{ [m}^2\text{]} \) OR \( p = \frac{F}{A} \) seen [1];
\( 6125 \) (or \( 6250 \) if \( 3500\text{ N} \) used from (a)(ii)) (ALLOW ECF from (a)(ii)) [1];
\( \text{Pa} \) OR \( \text{N/m}^2 \) (or \( \text{N}\cdot\text{m}^{-2} \)) [1];

(c)(i) \( 630\text{ [N]} \) [1];

(c)(ii) \( [a = ] \frac{F}{m} \) OR \( \frac{630}{350} \) (ALLOW ECF from (c)(i)) [1];
\( 1.8\text{ [m/s}^2\text{]} \) [1];

(c)(iii) \( [v = ] a \times t \) OR \( 1.8 \times 6.0 \) (ALLOW ECF from (c)(ii)) [1];
\( 10.8\text{ [m/s]} \) (ALLOW \( 11\text{ [m/s]} \)) [1]

(a) (i) Metals contain delocalised (free) electrons. When heated, these electrons gain kinetic energy and move through the metal lattice, transferring thermal energy rapidly via collisions with metal ions and other electrons.

(ii) Air in contact with the heater absorbs thermal energy and warms up. The warm air expands, causing its density to decrease. This less dense, warm air rises above the heater. Cooler, denser air sinks to take its place, creating a continuous circulation of air known as a convection current.

(b) (i) For two identical resistors in parallel:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{46} + \frac{1}{46} = \frac{2}{46} = \frac{1}{23}\)
\(R_p = 23\ \Omega\)

(ii) Using Ohm's law:
\(I = \frac{V}{R_p} = \frac{230}{23} = 10\text{ A}\)

(iii) First convert time to seconds:
\(t = 15 \times 60 = 900\text{ s}\)
Then calculate energy:
\(E = IVt = 10 \times 230 \times 900 = 2\ 070\ 000\text{ J} = 2.07 \times 10^6\text{ J}\)

(a)(i)
- (presence/movement of) delocalised / free electrons [1];
- transfer energy / kinetic energy by colliding with (lattice) ions / other electrons [1];
ALLOW: free electrons move faster throughout the metal [1]
REJECT: atoms / ions move from one end of the metal to the other [0]

(a)(ii)
- air expands AND density decreases / air becomes less dense (as it is heated) [1];
- warm / less dense air rises [1];
- cool(er) / dense(r) air sinks / falls (to take its place / creating a convection current) [1];
IGNORE: heat rises

(b)(i)
- correct formula / substitution: \(\frac{1}{R} = \frac{1}{46} + \frac{1}{46}\) OR \(R = \frac{46 \times 46}{46 + 46}\) OR \(\frac{46}{2}\) [1];
- \(23\) (\(\Omega\)) [1];

(b)(ii)
- \(10\) (\(\text{A}\)) [1];
ALLOW: ECF from (b)(i) \(I = \frac{230}{\text{ans from (b)(i)}}\)

(b)(iii)
- conversion of time: \(15 \times 60 = 900\text{ (s)}\) OR use of \(E = IVt\) / \(E = Pt\) with \(P = 2300\text{ W}\) [1];
- \(2\ 070\ 000\text{ (J)}\) / \(2.07 \times 10^6\text{ (J)}\) / \(2.1 \times 10^6\text{ (J)}\) [1];
ALLOW: ECF from (b)(ii) for calculated energy using \(E = I \times 230 \times 900\)