Question 1
Describe the disadvantages of asexual reproduction for plants in the wild.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic B15.3 — Sexual reproduction in plants (Part (a)(i)–(iv))
• Topic B15.1 — Asexual reproduction (Part (b))
▶️ Answer/Explanation
(a)(i) Anther
The anther is the part of the stamen that produces pollen grains.
A label line should be drawn to one of the anther structures at the tip of the stamens in Fig. 1.1.
The label must be named “anther” for full credit.
(a)(ii) X placed on the ovary
Fertilisation is the fusion of the male and female gametes.
This occurs inside the ovary, where the ovule is located.
The ovary is the rounded structure at the base of the flower in Fig. 1.1.
(a)(iii) Any two from: feathery; hangs outside of flower; large
Structure \( Y \) is the stigma of a wind-pollinated flower.
It is feathery and/or large, increasing the surface area to catch airborne pollen.
It hangs outside the flower, exposing it to air currents carrying pollen.
(a)(iv) Any two from: larger; stickier; rougher surface; heavier
Insect-pollinated pollen is adapted to stick to an insect’s body.
It is typically larger, stickier, and has a rougher surface than wind-pollinated pollen.
Wind-pollinated pollen is light and smooth so that it can be carried easily by air currents.
(b) Any three from: no/less genetic variation; poor adaptation to changing environment; risk of extinction from disease; disadvantageous traits inherited by all offspring
Asexual reproduction produces genetically identical offspring (clones).
This means there is little or no genetic variation within the population.
A lack of variation makes the population vulnerable to disease or environmental change, since all individuals share the same weaknesses.
Question 2
Calculate the average current in the lightning strike.
Explain why an observer sees the light before they hear the sound.
State the name of the particles that provide this negative charge.
Use the phrases to complete the sentences.
You may use each phrase once, more than once or not at all.
The wavelength of gamma radiation is ……………………………… the wavelength of visible light.
The frequency of X-rays is ……………………………… the frequency of gamma radiation. [2]
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic P1.6.1 — Energy (Part (a))
• Topic P4.2.2 — Electric current (Part (b)(i))
• Topic P3.4 — Sound (Part (b)(ii))
• Topic P4.1 — Simple phenomena of magnetism (Part (c)(i)–(ii))
• Topic P3.3 — Electromagnetic spectrum (Part (d))
• Topic P2.1.2 — Particle model (Part (e))
▶️ Answer/Explanation
(a) 11 700 J
Change in GPE is calculated using \( \Delta GPE = mgh \).
\( \Delta GPE = 225 \times 10 \times 5.2 = 11\,700 \, \text{J} \).
This energy is transferred to other forms (e.g. kinetic, sound) as the branch falls.
(b)(i) 30 000 A
Current is calculated using \( I = \dfrac{Q}{t} \).
\( I = \dfrac{6000}{0.20} = 30\,000 \, \text{A} \).
This extremely high current reflects the huge charge transfer typical of a lightning strike.
(b)(ii) Light travels faster than sound
Light travels at approximately \( 3 \times 10^8 \, \text{m/s} \), while sound travels at only about \( 340 \, \text{m/s} \) in air.
Both waves travel the same distance from the storm to the observer.
Over that distance, the much slower speed of sound means it arrives noticeably later than the light.
(c)(i) Electrons
Negative charge at the base of a cloud is provided by electrons.
These accumulate due to collisions between ice crystals and water droplets within the cloud.
(c)(ii) A region in which charged particles experience a force
An electric field is defined as a region of space around a charge in which another charged particle experiences a force.
The field lines in Fig. 2.2 point from the positive charge towards the negative charge.
(d) the same as; less than; less than
All electromagnetic waves, including visible light and X-rays, travel at the same speed in a vacuum, \( c = 3 \times 10^8 \, \text{m/s} \).
Gamma radiation has a shorter wavelength than visible light, since it lies further along the electromagnetic spectrum.
Since \( c = f\lambda \) is constant, a shorter wavelength for gamma rays corresponds to a higher frequency, so the frequency of X-rays is less than that of gamma radiation.
(e) Volume increases; molecules move faster and further apart
As temperature increases, air molecules gain more kinetic energy and move faster.
This causes the molecules to spread further apart on average.
As a result, the volume of the air increases (or expands) at constant pressure.
Question 3
Describe how the rate of reaction changes during the experiment.
Explain your answer using ideas about collisions between particles.
This time they use warm dilute hydrochloric acid instead of cold dilute hydrochloric acid. The reaction is much faster.
Explain why the reaction is much faster by using ideas about collisions between particles.
Acid rain reacts very slowly with marble buildings.
Suggest why the reaction is so slow.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic C6.2 — Rate of reaction (Part (a), (b)(i)–(ii))
• Topic C10.1 — Water (Part (c)(i))
• Topic C10.2 — Air quality and climate (Part (c)(ii))
• Topic C7.3 — Preparation of salts (Part (d))
▶️ Answer/Explanation
(a) 100 \( \text{cm}^3 \)
From Table 3.1, the volume of gas remains constant at \( 100 \, \text{cm}^3 \) from 35 s onwards.
This constant reading shows that the reaction has stopped producing gas.
(b)(i) Rate decreases as the reaction proceeds
As the reaction proceeds, the concentration of hydrochloric acid decreases, so there are fewer acid particles in the same volume.
This means the particles are less crowded, so there are fewer collisions per second between reacting particles.
Since rate depends on the frequency of collisions, the rate of reaction decreases over time.
(b)(ii) Warm acid increases particle energy and collision frequency
In warm acid, particles move faster and have more kinetic energy than in cold acid.
This results in more frequent collisions between particles.
A greater proportion of these collisions have energy greater than or equal to the activation energy, giving more successful collisions per second.
(c)(i) Acid rain is very dilute / marble has a small surface area
Acid rain is a very dilute acid, meaning it contains a low concentration of \( \text{H}^+ \) ions.
Marble buildings also present a relatively small surface area compared to their volume.
Both factors reduce the frequency of collisions between acid particles and the marble surface, slowing the reaction.
(c)(ii) Combustion of fossil fuels
Sulfur dioxide is released mainly from the combustion of fossil fuels that contain sulfur compounds, such as coal and some petroleum products.
This sulfur dioxide dissolves in atmospheric water to form acid rain.
(d) \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \)
The equation must balance for calcium, chlorine, hydrogen, carbon, and oxygen atoms.
Two moles of \( \text{HCl} \) are needed to supply the two chloride ions in \( \text{CaCl}_2 \) and balance the hydrogens across water.
The carbon dioxide gas produced is what is collected in the gas syringe in Fig. 3.1.
Question 4
Breathing rate increases between rest and exercise by ……………………………… breaths per minute.
An increase in the breathing rate is caused by an increase in carbon dioxide concentration in the ……………………………… .
During exercise the working ……………………………… require more energy for contraction.
Oxygen is required for ……………………………… ……………………………… to release the energy required.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic B12 — Respiration (Part (a))
• Topic B11 — Gas exchange in humans (Part (b)(i)–(ii))
• Topic B14 — Drugs (Part (c)(i)–(iii))
▶️ Answer/Explanation
(a) 31; blood; muscles; aerobic respiration
Breathing rate increases from 46 to 77 breaths per minute, a rise of \( 77 – 46 = 31 \) breaths per minute.
This increase is triggered by a rise in carbon dioxide concentration detected in the blood.
During exercise, working muscles require more energy for contraction, and oxygen is needed for aerobic respiration to release this energy.
(b)(i) Any two from: shorter cilia less effective at moving mucus; increased risk of infection/breathing difficulties/coughing; less efficient gas exchange
Shorter cilia in smokers are less effective at sweeping mucus (and trapped particles/bacteria) out of the airways.
This increases the risk of infection, coughing, or breathing difficulties.
Reduced mucus clearance can also make gas exchange less efficient overall.
(b)(ii) Any two from: trachea; bronchi/bronchus; bronchioles
The trachea, bronchi, and bronchioles are all lined with ciliated epithelial cells.
These cilia move mucus (containing trapped dust and bacteria) up and away from the lungs.
(c)(i) 
Fig. 4.1 shows that as cigarette consumption rose, lung cancer deaths also rose, indicating a strong correlation.
There is a noticeable time delay between the rise in cigarettes smoked and the corresponding rise in lung cancer deaths.
The other statements are not fully supported, since the decrease in smoking and cancer deaths do not occur in the same year, the maxima are not at the same value, and correlation does not prove cigarettes are the only causal factor.
(c)(ii) Tar
Tar is the component of tobacco smoke known to be carcinogenic (cancer-causing).
(c)(iii) COPD / coronary heart disease
Smoking tobacco is linked to diseases such as chronic obstructive pulmonary disease (COPD) and coronary heart disease (CHD), in addition to various cancers.
Question 5
An atom that gains electrons to get a full outer shell and become stable.
An atom that shares electrons to get a full outer shell and become stable
An atom that loses electrons to get a full outer shell and become stable.
lattice oxygen sodium
molecular polymer strong
negative positive weak
An example of an atom gaining 1 electron to complete its outer shell is …………………………… .
During the formation of ionic bonds there is a ……………………………… attraction between ions because of their ……………………………… electrical charges. The ions form a regular arrangement of alternating ions called a ……………………………… structure.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic C2.7 — Metallic bonding (Part (a))
• Topic C2.4 — Ions and ionic bonds (Part (b))
• Topic C2.5 — Simple molecules and covalent bonds (Part (c)(i)–(iii))
▶️ Answer/Explanation
(a) An atom that loses electrons to get a full outer shell and become stable 
A metal atom becomes stable by losing its outer-shell electrons, forming a positive ion.
This is different from non-metal atoms, which usually gain or share electrons to achieve a stable outer shell.
(b) negative; chlorine; strong; opposite; lattice
Gaining electrons gives an atom an overall negative charge, forming a negative ion.
Chlorine is a common example of an atom that gains one electron to complete its outer shell.
Ionic bonds arise from a strong electrostatic attraction between oppositely charged ions, which pack together into a regular repeating pattern called a lattice structure.
(c)(i) Covalent
In a water molecule, the oxygen and hydrogen atoms share pairs of electrons.
This sharing of electrons is characteristic of covalent bonding.
(c)(ii) 
Each nitrogen atom needs three more electrons to complete its outer shell.
This is achieved by sharing three pairs of electrons between the two nitrogen atoms, forming a triple covalent bond (\( \text{N} \equiv \text{N} \)).
(c)(iii) Weak intermolecular forces require little energy to break
Water and nitrogen are simple molecular substances held together internally by strong covalent bonds.
However, the forces of attraction between separate molecules (intermolecular forces) are weak.
Only a small amount of energy is needed to overcome these weak intermolecular forces, resulting in low melting points.
Question 6
Explain why the average acceleration of shape A is not \( 10 \, \text{m/s}^2 \). Use ideas about forces in your explanation.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic P1.2 — Motion (Part (a)(i)–(iii))
• Topic P1.4 — Density (Part (b)(i)–(ii))
▶️ Answer/Explanation
(a)(i) 2.2 m/s
Average speed is calculated using \( \text{speed} = \dfrac{\text{distance}}{\text{time}} \).
\( \text{speed} = \dfrac{1.5}{0.68} = 2.2 \, \text{m/s} \).
(a)(ii) 8.5 \( \text{m/s}^2 \)
Average acceleration is calculated using \( a = \dfrac{\Delta v}{t} \), where \( \Delta v \) is the change in speed from rest.
\( a = \dfrac{5.2}{0.61} = 8.5 \, \text{m/s}^2 \).
(a)(iii) Air resistance opposes the fall
Shape A experiences air resistance as it falls through the air.
This air resistance acts upwards, in the opposite direction to the shape’s weight.
Because the resultant force is less than the weight alone, the acceleration is less than \( 10 \, \text{m/s}^2 \).
(b)(i) Displacement method
The shape is placed into the measuring cylinder, fully submerging it in the water.
The volume of water displaced is found from the difference between the new water level and the original level (\( 200 – 125 = 75 \, \text{cm}^3 \)).
(b)(ii) 1.8 \( \text{g/cm}^3 \)
Density is calculated using \( \text{density} = \dfrac{\text{mass}}{\text{volume}} \).
\( \text{density} = \dfrac{135}{75} = 1.8 \, \text{g/cm}^3 \).
Question 7
in the boxes to identify all the examples of voluntary actions.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic B13.1 — Coordination and response (Part (a)(i)–(iii), (b))
• Topic B13.2 — Hormones (Part (c))
• Topic B13.3 — Homeostasis (Part (d))
▶️ Answer/Explanation
(a)(i) receptor: retina; effector: circular/radial muscles in the iris
The receptor detecting the change in light intensity is the retina.
The effector responsible for changing the pupil size is the circular or radial muscle in the iris.
(a)(ii) Adrenaline
Fig. 7.1a shows a wider pupil (dilation), which can be caused by the hormone adrenaline.
Adrenaline prepares the body for a “fight or flight” response, including pupil dilation.
(a)(iii) It is automatic; it is rapid
A reflex action is automatic, meaning it requires no conscious thought or decision-making.
It is also rapid, allowing the body to respond very quickly to a stimulus, such as bright light.
(b) Eating and talking
Voluntary actions are consciously controlled, such as eating and talking.
Heart beating, sneezing, and sweating are all involuntary/automatic actions, not under conscious control.
(c) Hormonal effects are longer lasting; hormonal responses are slower
Hormonal control acts more slowly than nervous control, since hormones travel via the bloodstream rather than as fast electrical impulses.
However, the effects of hormones tend to be longer lasting compared to the fast, short-lived responses of the nervous system.
(d) Insulin / glucagon
Insulin and glucagon are both hormones released by the pancreas that regulate blood glucose concentration.
Insulin lowers blood glucose, while glucagon raises it, working together to maintain homeostasis.
Question 8
Explain why farmers add nitrogen-containing fertilisers to crops.
Determine the formula of ammonium sulfate.
The reactants are potassium hydroxide, \( \text{KOH} \), and sulfuric acid, \( \text{H}_2\text{SO}_4 \).
[\( A_r \): H, 1; K, 39; O, 16; S, 32]
It is made by the Haber process from the reaction of nitrogen with hydrogen.
Explain why a temperature of 450°C is used in an ammonia factory.
Use ideas about the position of the equilibrium and the rate of reaction.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic B6.1 — Photosynthesis (Part (a))
• Topic C3.1 — Formulas (Part (b))
• Topic C3.3 — The mole and the Avogadro constant (Part (c))
• Topic C6.2 — Rate of reaction (Part (d)(i)–(ii))
▶️ Answer/Explanation
(a) Increased crop yield / better-quality crops
Nitrogen is needed by plants for the synthesis of amino acids and proteins.
Adding nitrogen-containing fertilisers replaces nutrients in the soil, leading to increased growth rate and crop yield.
(b) \( (\text{NH}_4)_2\text{SO}_4 \)
The \( \text{SO}_4^{2-} \) ion has a charge of \(-2\), so two \( \text{NH}_4^+ \) ions are needed to balance the charge.
This gives the formula \( (\text{NH}_4)_2\text{SO}_4 \).
(c) 43.5 g
Relative formula mass of \( \text{KOH} = 56 \) and of \( \text{K}_2\text{SO}_4 = 174 \).
Moles of \( \text{KOH} = \dfrac{28}{56} = 0.5 \, \text{mol} \).
From the 2:1 mole ratio, moles of \( \text{K}_2\text{SO}_4 = 0.25 \, \text{mol} \), so mass \( = 0.25 \times 174 = 43.5 \, \text{g} \).
(d)(i) The percentage of ammonia increases
From the 600°C curve in Fig. 8.1, the percentage of ammonia produced increases as pressure increases.
This is because increased pressure shifts the equilibrium towards the side with fewer gas molecules (ammonia).
(d)(ii) Position of equilibrium: low temperature favours ammonia; Rate of reaction: reaction is too slow at low temperature
A lower temperature (such as 200°C) moves the position of equilibrium further to the right, favouring greater ammonia yield.
However, at low temperatures the rate of reaction is very slow, so a compromise temperature of 450°C is used to give a reasonably high yield in a reasonable time.
Question 9
It takes 336 kJ of energy to heat some water from room temperature to 100°C.
Calculate the time it will take for the kettle to heat the water from room temperature to 100°C.
The temperature of the water in each cup is measured every minute for 15 minutes.
Fig. 9.3 shows the results.
A shows the results for the …………………………….. cup.
Convex lenses can form real and virtual images.
Describe the difference between a real image and a virtual image.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic P4.2.5 — Electrical energy and electrical power (Part (a), (b))
• Topic P2.3.3 — Radiation (Part (c))
• Topic P3.2.2 — Refraction of light (Part (d))
▶️ Answer/Explanation
(a) Resistance decreases, current/p.d. increases, power output increases
As the thermistor’s temperature increases, its resistance decreases.
This causes the current through (and potential difference across) the heater to increase, since the thermistor and heater are in the same circuit.
Since power depends on current and voltage, the power output of the heater increases.
(b) 112 s
Time is calculated using \( t = \dfrac{E}{P} \).
Converting energy to joules: \( 336 \, \text{kJ} = 336\,000 \, \text{J} \).
\( t = \dfrac{336\,000}{3000} = 112 \, \text{s} \).
(c) A is the white cup
A white surface emits less infrared radiation (thermal energy) than a black surface.
This means the white cup cools down more slowly over the 15 minutes, matching curve A, which stays at a higher temperature than curve B.
(d) A real image can be formed on a screen
A real image is formed where light rays actually converge and can be captured on a screen.
A virtual image, in contrast, cannot be formed on a screen because the light rays only appear to diverge from a point.
Question 10
Table 10.1 shows information about some large food molecules.
Complete Table 10.1.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic B5 — Enzymes (Part (a)(i)–(ii), (b), (c))
• Topic B4 — Biological molecules (Part (d))
▶️ Answer/Explanation
(a)(i) pH 8
Fig. 10.1 shows enzyme activity is at its peak at pH 8, which is therefore the optimum pH for this enzyme.
(a)(ii) Enzyme is denatured; active site shape changes; substrate no longer fits
At pH 2, the enzyme has zero activity because it has been denatured by the low, acidic pH.
The acidic conditions cause the shape of the enzyme’s active site to change.
As a result, the substrate is no longer complementary to the active site, so no enzyme-substrate complex can form.
(b) Amylase
Amylase is the digestive enzyme that breaks down starch into simpler sugars such as maltose.
(c) pH 1–3
The protease active in the stomach (pepsin) has an optimum pH in the acidic range of pH 1–3, matching the acidic conditions of the stomach.
(d)
Oils/fats are made from fatty acids and glycerol, tested using the ethanol emulsion test.
Proteins are made from amino acids, tested using biuret solution (which turns purple/violet if protein is present).
Starch is made from glucose (many glucose units joined together), and is tested using iodine solution (which turns blue-black if starch is present).
Question 11
limewater sodium hydroxide
A polymer is a ……………………………… ……………………………… molecule formed from small units called ……………………………… .
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic C11.5 — Alkenes (Part (a), (b))
• Topic C11.4 — Alkanes (Part (c))
• Topic C11.7 — Polymers (Part (d)(i)–(ii))
• Topic C5.1 — Exothermic and endothermic reactions (Part (e)(i)–(iii))
▶️ Answer/Explanation
(a) Compound C
Compound C contains a carbon-carbon double bond (\( \text{C}=\text{C} \)), which makes it unsaturated.
Compounds A, B, and D contain only single carbon-carbon or carbon-oxygen bonds.
(b) Bromine
Bromine (bromine water) is used to test for unsaturation.
It changes from orange/brown to colourless in the presence of a carbon-carbon double bond.
(c) Ethanol
Compound B has the structure \( \text{CH}_3\text{CH}_2\text{OH} \), which is ethanol, since it contains an \( -\text{OH} \) group attached to a two-carbon chain.
(d)(i) Long chain; monomers
A polymer is a very long chain molecule.
It is formed by joining together many small repeating units called monomers.
(d)(ii) Ethene
Poly(ethene) is made from the monomer ethene, which contains a carbon-carbon double bond.
This double bond opens up during addition polymerisation, allowing the monomers to join in a long chain.
(e)(i) Products have less energy than reactants
In Fig. 11.2, the energy level of the products is lower than that of the reactants.
This drop in energy level shows that energy has been released to the surroundings.
(e)(ii) Exothermic
A reaction that releases energy to the surroundings is described as exothermic.
(e)(iii) More energy released in bond making than absorbed in bond breaking
Breaking the bonds in the reactants (compound B and oxygen) requires an input of energy, so bond breaking is endothermic.
Forming the new bonds in the products (carbon dioxide and water) releases energy, so bond making is exothermic.
Since more energy is released during bond making than is absorbed during bond breaking, there is a net release of energy overall.
Question 12
State the types of radiation which would follow the paths labelled P and Q.
Use the correct nuclide notation to complete the decay equation for americium-241.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic P4.5.1 — Electromagnetic induction (Part (a)(i)–(ii))
• Topic P5.2.2 — The three types of nuclear emission (Part (b)(i)–(iii))
▶️ Answer/Explanation
(a)(i) The wire experiences a changing magnetic field, inducing an emf
As the wire moves between the poles of the magnet, it cuts through the magnetic field lines.
This changing magnetic field induces an emf in the wire, which drives a current around the circuit.
(a)(ii) 
Moving the wire faster increases the rate of change of magnetic field, so the ammeter reading increases.
Reversing the direction of motion (right to left) reverses the induced current, so the reading becomes negative.
Keeping the wire stationary means there is no changing field, so the reading becomes zero.
Using a wire of lower resistance allows a larger current to flow for the same induced emf, so the reading increases.
(b)(i) P: beta; Q: gamma
Beta particles are deflected by a magnetic field, but less than alpha particles and in the opposite direction (since beta particles are negatively charged), matching path P.
Gamma radiation has no charge, so it is not deflected by the magnetic field and travels in a straight line, matching path Q.
(b)(ii) \( ^{241}_{95}\text{Am} \rightarrow \, ^{237}_{93}\text{Np} + \, ^{4}_{2}\alpha \)
In alpha decay, the mass number decreases by 4 and the atomic number decreases by 2.
So \( 241 – 4 = 237 \) and \( 95 – 2 = 93 \), giving neptunium-237 as the daughter nuclide.
(b)(iii) Least penetrating; easily stopped by smoke
Alpha particles are the least penetrating and have a short range in air, so they are easily absorbed within the small space of the detector.
This means alpha particles are easily stopped or scattered by smoke particles, causing the detectable change in current needed to trigger the alarm.