Question 1
State two other main components of blood.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B15.5 — Sexually transmitted infections (Part (a))
• Topic B9.4 — Blood (Part (b))
• Topic B15.4 — Sexual reproduction in humans (Part (c))
▶️ Answer/Explanation
(a)(i) 31%
Reading from Fig. 1.1: infections in 2000 \( = 3.2 \) million, in 2010 \( = 2.2 \) million.
Decrease \( = 3.2 – 2.2 = 1.0 \) million.
Percentage decrease \( = \dfrac{1.0}{3.2} \times 100 = 31\% \).
(a)(ii) Any three from:
Discovery of routes of transmission.
Increased use of barrier contraception / abstinence.
Screening of blood transfusions.
Reduced drug use / use of clean needles.
Monitoring / testing / screening for HIV.
Increased education / awareness.
(b)(i) Phagocytosis; antibody production.
White blood cells defend the body by engulfing pathogens (phagocytosis).
They also produce antibodies that neutralise pathogens or toxins.
(b)(ii) Red blood cells; plasma (also accept platelets).
Blood is made up of red blood cells, white blood cells, platelets, and plasma.
Red blood cells carry oxygen, and plasma is the liquid that transports cells and dissolved substances.
(c) Placenta
The placenta separates the mother’s and fetus’s blood supplies.
It allows exchange of nutrients, oxygen, and waste while acting as a barrier to many toxins and pathogens.
Question 2
Draw a dot-and-cross diagram to show the ions formed when elements B and D react together.
Include the charges on the ions.
Describe the lattice structure of ionic compounds.
Table 2.1 shows the relative charge and relative mass of a proton.
Complete the table to show the relative charge and relative mass of a neutron.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C2.2 — Atomic structure and the Periodic Table (Parts (a)(i)–(v))
• Topic C2.4 — Ions and ionic bonds (Parts (a)(vi)–(vii))
• Topic C2.2 — Atomic structure and the Periodic Table (Part (b))
▶️ Answer/Explanation
(a)(i) Element C has 3 electrons in its outer shell.
Group number corresponds to the number of electrons in the outermost shell.
C has 3 outer-shell electrons, so it belongs to Group III.
(a)(ii) E
An ion with charge \(-3\) forms by gaining 3 electrons to complete its outer shell.
Element E has 5 electrons in its outer shell, so it gains 3 more to reach a full outer shell of 8.
(a)(iii) 2.8.5
Element E has 2 electrons in the first shell, 8 in the second shell, and 5 in the third (outer) shell.
(a)(iv) D
A proton number of 9 corresponds to fluorine, which has the electronic structure 2.7.
Element D matches this structure in the diagram.
(a)(v) Element A has a full outer shell.
Atoms with a full (complete) outer shell of electrons are stable and do not readily react.
This is why element A, a noble gas, is unreactive.
(a)(vi)
Element B loses 2 electrons to form a \(2+\) ion.
Element D gains 1 electron (per atom, two D atoms needed) to form a \(1-\) ion, giving the ionic compound BD₂.
(a)(vii) Regular, alternating arrangement of positive and negative ions.
Ionic lattices consist of oppositely charged ions arranged in a repeating, ordered 3D pattern.
Strong electrostatic forces of attraction hold the ions together throughout the lattice.
(b)
A neutron has no electrical charge, so its relative charge is 0.
A neutron has approximately the same mass as a proton, so its relative mass is 1.
Question 3
Calculate the total kinetic energy of the man, his luggage, and the small boat when their speed reaches \(4.0 \, \text{m/s}\).
The man exerts a downwards force \(F\) which keeps the boat in equilibrium as shown in Fig. 3.3.
The wheels of the trolley act as a pivot.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.2 — Motion (Part (a)(i), (ii))
• Topic P1.6.1 — Energy (Part (a)(iii))
• Topic P1.5.2 — Turning effect of forces (Part (b))
▶️ Answer/Explanation
(a)(i) Constant speed from 0–50 s, covering 200 m; then stationary from 50–100 s.
The graph is a straight line from 0 to 50 s, showing constant speed as distance increases steadily from 0 to 200 m.
From 50 s to 100 s the line is horizontal at 200 m, showing the boat has stopped moving.
(a)(ii) 4.0 m/s
Speed \( = \dfrac{\text{distance}}{\text{time}} \).
Using the graph, at \(t = 20\,\text{s}\), distance \( = 80\,\text{m}\), so speed \( = \dfrac{80}{20} = 4.0 \, \text{m/s}\).
(a)(iii) 800 J
Kinetic energy \( KE = \dfrac{1}{2}mv^2 \).
\( KE = \dfrac{1}{2} \times 100 \times (4.0)^2 = \dfrac{1}{2} \times 100 \times 16 = 800 \, \text{J} \).
(b) 240 N
By the principle of moments, clockwise moment = anticlockwise moment about the pivot.
\( 600 \, \text{N} \times 40 \, \text{cm} = F \times 100 \, \text{cm} \), so \( F = \dfrac{600 \times 40}{100} = 240 \, \text{N} \).
Question 4
Four paper discs, each soaked with a different antibiotic, are placed on an agar plate containing the bacteria.
The shaded areas show where the bacteria grow.
The clear areas show where no bacteria grow. Fig. 4.1 shows the results.
Complete the sentences to describe how strains with antibiotic resistance develop.
Different strains of bacteria will show ………………….. in their ability to resist antibiotics. When antibiotics are used some of the bacteria will survive and some will ……………………… .
Those that survive will pass on their ……………………… to the next generation.
The next generation will also show ……………………… to antibiotics.
Eventually the whole population will have this feature.
The box on the left contains the term evolution.
The boxes on the right show some sentence endings.
Draw two lines from the box on the left to make two correct sentences.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B17.2 — Selection (Part (a))
• Topic B17.2 — Selection (Part (b))
• Topic B17.1 — Variation (Part (c))
▶️ Answer/Explanation
(a)(i) B
The most effective antibiotic produces the largest clear zone where no bacteria grow.
Antibiotic B has the largest clear area around its disc, so it inhibits this bacterial strain the most.
(a)(ii) variation; die; alleles; resistance
Bacteria show natural variation in their ability to resist antibiotics.
When antibiotics are applied, non-resistant bacteria die while resistant ones survive.
Surviving bacteria pass their resistance alleles to offspring, so the next generation also shows resistance.
(b) Evolution → increases suitability to the environment; Evolution → is a change of adaptive features.
Evolution describes gradual changes in a population’s adaptive features across many generations.
It does not occur in a single generation, is not restricted to bacteria, and does not change individual organisms during their own lifetime.
(c)(i) A mutation is a change in the gene or chromosome.
Mutations are random changes to the DNA sequence.
Such changes can create new alleles, including ones that confer antibiotic resistance.
(c)(ii) Ionising radiation
Ionising radiation (e.g. X-rays, gamma rays, UV) can damage DNA directly.
This damage increases the rate at which mutations occur.
Question 5
The scientist also analyses four known food colourings, A, B, C, and D.
Fig. 5.1 shows the chromatogram produced. The result for food colouring D is not shown.
Calculate the distance travelled by food colouring D.
Table 5.1 shows the melting point of each substance.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C12.3 — Chromatography (Parts (a), (b), (c))
• Topic C12.4 — Separation and purification (Part (d))
• Topic C3.3 — The mole and the Avogadro constant (Part (e))
▶️ Answer/Explanation
(a) 0.90
From the chromatogram, distance travelled by B \( = 4.5\,\text{cm}\), distance travelled by solvent \( = 5.0\,\text{cm}\).
\( R_f = \dfrac{4.5}{5.0} = 0.90 \).
(b) 2.8 cm
Rearranging the formula: distance travelled by D \( = R_f \times \) distance travelled by solvent.
Distance \( = 0.56 \times 5.0 = 2.8\,\text{cm} \).
(c) A
Food colouring X shows spots matching B and C at the same heights as their reference spots.
A’s spot does not appear anywhere in X’s chromatogram, so A is not present in X.
(d) V and Y are pure.
Pure substances melt at one specific, fixed temperature rather than over a range.
V (98 °C) and Y (82 °C) each have a single melting point, while W and Z melt over a range, showing they are impure.
(e) 0.10 mol/dm³
Moles \( = \dfrac{\text{mass}}{M_r} = \dfrac{4.8}{192} = 0.025 \, \text{mol} \).
Volume \( = 250\,\text{cm}^3 = 0.25\,\text{dm}^3 \).
Concentration \( = \dfrac{0.025}{0.25} = 0.10 \, \text{mol/dm}^3 \).
Question 6
Fig. 6.2 shows the circuit used in the electric fence.
State the effect of doubling the cable length on the resistance of the fence.
The zebra has black and white stripes.
The infrared camera shows that the black stripes are a different temperature to the white parts of the zebra.
Describe and explain the difference in temperature recorded.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.2.4 — Resistance (Part (a)(i), (a)(iii))
• Topic P4.3.2 — Series and parallel circuits (Part (a)(ii))
• Topic P3.4 — Sound (Part (b)(i))
• Topic P1.2 — Motion (Part (b)(ii))
• Topic P2.3.3 — Radiation (Part (c))
▶️ Answer/Explanation
(a)(i) 25 000 Ω
Current \( = 80\,\text{mA} = 0.08\,\text{A} \).
Using \( R = \dfrac{V}{I} \): \( R = \dfrac{2000}{0.08} = 25000\,\Omega \).
(a)(ii) 50 000 Ω
For two identical resistors in parallel: \( \dfrac{1}{R_{total}} = \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{2}{R} \).
So \( R = 2 \times R_{total} = 2 \times 25000 = 50000\,\Omega \).
(a)(iii) The resistance doubles.
Resistance of a wire is directly proportional to its length.
Doubling the length (with the same thickness/material) doubles the resistance.
(b)(i) Sound is transmitted through compressions and rarefactions.
Vibrating sources cause air molecules to oscillate back and forth.
This creates regions of compression and rarefaction that travel through the air as a longitudinal wave.
(b)(ii) 120 m
Distance \( = \text{speed} \times \text{time} \).
Distance \( = 16 \times 7.5 = 120\,\text{m} \).
(c) Black stripes show a higher temperature than white stripes.
Black surfaces absorb more infrared radiation than white surfaces, which reflect more of it.
Because black absorbs (and also emits) more radiation, the infrared camera records a higher temperature for the black stripes than the white ones.
Question 7
Circle the two words in their definition that are not correct.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B19.1 — Habitat destruction (Parts (a), (b))
• Topic B18.2 — Food chains and food webs (Part (c))
▶️ Answer/Explanation
(a) The incorrect words are “offspring” and “time”.
An ecosystem includes all organisms (not just their offspring) together with their non-living environment.
It is an ongoing interacting system, not something restricted to “a given time”.
(b)(i) Country C
Reading the bar chart, country C has the tallest bar, at about 2.0 kg per person per day.
This is higher than any other country shown.
(b)(ii) 0.75 kg
Reading directly from the bar chart, the bar for country E reaches approximately 0.75 kg.
(c)(i) Surface producers block light from reaching underwater producers.
A dense layer of surface producers (e.g. algal bloom) covers the water surface.
This prevents light from penetrating to underwater producers, so they cannot photosynthesise and eventually die.
(c)(ii) Decomposition of dead producers uses up oxygen, killing aquatic animals.
Dead underwater producers are broken down by decomposers (bacteria).
These decomposers respire aerobically, using up dissolved oxygen in the water.
The resulting drop in oxygen levels means aquatic animals cannot respire and die.
Question 8
Suggest the pH of the dilute sulfuric acid.
State what the student observes.
Sodium sulfate, \(Na_2SO_4\), and water are made.
Construct the balanced symbol equation for this reaction.
Define an acid in terms of proton transfer.
Calculate the concentration of the solution in g/dm³.
[\(A_r\): H, 1; O, 16; S, 32]
State two other conditions chosen for stage 2.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C7.1 — Acids and bases (Parts (a) and (b))
• Topic C3.3 — Moles / concentration (Part (c))
• Topic C6.2 — Rate of reaction / Contact process (Part (d))
▶️ Answer/Explanation
(a)(i) Any pH below 7.
Sulfuric acid is a strong acid.
Universal Indicator shows a low pH value (well below 7) for strong acids.
(a)(ii) The red litmus paper turns blue.
Sodium hydroxide is an alkali.
Red litmus paper turns blue in the presence of an alkaline solution.
(a)(iii) \( H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \)
This is a neutralisation reaction between an acid and a base.
Balancing requires 2 moles of NaOH to react with 1 mole of \(H_2SO_4\), producing 1 mole of salt and 2 moles of water.
(b) An acid is a proton donor.
In the Brønsted–Lowry theory, acids donate \(H^+\) ions during a reaction.
Bases, by contrast, accept these protons.
(c) 19.6 g/dm³
Molar mass of \(H_2SO_4 = (2\times1) + 32 + (4\times16) = 98\,\text{g/mol}\).
Concentration (g/dm³) = concentration (mol/dm³) × molar mass \( = 0.2 \times 98 = 19.6\,\text{g/dm}^3\).
(d)(i) \( 2SO_2 + O_2 \rightleftharpoons 2SO_3 \)
Sulfur dioxide reacts with oxygen in a reversible reaction to form sulfur trioxide.
The equation is balanced with 2 moles of \(SO_2\) and 1 mole of \(O_2\) producing 2 moles of \(SO_3\).
(d)(ii) Temperature of about 450 °C; vanadium(V) oxide (\(V_2O_5\)) catalyst.
A moderate temperature of 450 °C balances reaction rate against equilibrium yield.
A vanadium(V) oxide catalyst speeds up the reaction without being consumed.
Question 9
- forces between molecules
- distances between molecules
- motions of molecules
Use the words increases, decreases or remains constant.
You can use each word once, more than once or not at all.
The transformer is made up of a primary coil and a secondary coil, wrapped around an iron core.
Calculate the number of turns on the secondary coil.
This is a very high potential difference.
Explain why a very high potential difference is used.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.6.3 — Energy resources (Part (a))
• Topic P2.1.2 — Particle model (Part (b)(i))
• Topic P2.1.3 — Pressure changes (Part (b)(ii))
• Topic P4.5.6 — Transformers and power transmission (Part (c))
▶️ Answer/Explanation
(a) Advantage: high energy density / abundant / easy to store and transport.
Disadvantage: releases \(CO_2\), contributing to global warming (also releases \(SO_2\), causing acid rain).
Coal is a reliable, energy-dense fuel that is relatively cheap and easy to obtain.
However, burning it releases greenhouse gases and pollutants that damage the environment.
(b)(i) Forces: stronger in liquid than gas.
Distances: molecules closer together in liquid than gas.
Motions: molecules move more freely (faster and more randomly) in gas than liquid.
In a liquid, intermolecular forces are stronger, keeping molecules close together with restricted movement.
In a gas, forces are much weaker, so molecules are far apart and move rapidly and randomly in all directions.
(b)(ii)
| temperature | volume | pressure |
|---|---|---|
| increases | remains constant | increases |
| decreases | remains constant | decreases |
| remains constant | increases | decreases |
| remains constant | decreases | increases |
At constant volume, pressure rises with temperature and falls as temperature falls (Gay-Lussac’s Law).
At constant temperature, increasing volume decreases pressure and decreasing volume increases pressure (Boyle’s Law).
(c)(i) 2400 turns
Using \( \dfrac{V_p}{V_s} = \dfrac{N_p}{N_s} \), rearrange to \( N_s = \dfrac{V_s \times N_p}{V_p} \).
\( N_s = \dfrac{400 \times 120}{20} = 2400 \) turns.
(c)(ii) A high potential difference reduces the current, which reduces power loss as heat.
For a given power \(P = IV\), a higher voltage means a lower current is needed.
Since power loss in cables is \(I^2R\), a lower current greatly reduces energy wasted as heat during transmission.
Question 10
that provides structural support for the leaf ……………………
that transports mineral ions ……………………
where most photosynthesis occurs ……………………
that controls gas exchange ……………………
State which of these regions in a plant acts as a source and which as a sink.
region of growth ……………………
region of production ……………………
region of storage ……………………
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B6.2 — Leaf structure (Part (a))
• Topic B8.4 — Translocation (Part (b))
• Topic B6.1 — Photosynthesis (Part (c)(i))
• Topic B11.1 — Gas exchange in humans (Part (c)(ii))
▶️ Answer/Explanation
(a)(i) Structural support: B; transports mineral ions: B; most photosynthesis: A; controls gas exchange: D.
The vascular bundle (B) contains xylem, which gives structural support and transports water and mineral ions.
The palisade mesophyll (A) contains the most chloroplasts, so most photosynthesis occurs there.
The guard cells (D) surrounding the stomata control gas exchange by opening and closing the pore.
(a)(ii) Circle drawn around the vascular bundle (structure B) in the diagram.
The vascular bundle is the cluster of xylem and phloem tissue visible in the centre of the leaf cross-section.
(b)(i) amino acids and sucrose.
Phloem transports the soluble organic products of photosynthesis (sucrose) and amino acids made from them.
Glucose, starch, glycogen, and fatty acids are not the substances transported by phloem.
(b)(ii) Region of growth: sink; region of production: source; region of storage: sink.
A source is where sugars are made, e.g. photosynthesising leaves — the region of production.
A sink is where sugars are used or stored, e.g. growing regions and storage organs.
(c)(i) CO₂
Carbon dioxide is the raw material combined with water in photosynthesis to produce glucose and oxygen.
(c)(ii) Any two from: large surface area; thin (surface); good blood supply; good ventilation with air.
A large surface area and thin walls maximise the rate of diffusion of gases.
A good blood supply maintains a steep concentration gradient, and good ventilation keeps oxygen/CO₂ levels favourable.
Question 11
The alkenes are all hydrocarbons.
State two other features that the alkenes in the homologous series have in common.
Fig. 11.2 shows the equation for the reaction.
Complete the equation, by drawing the structure of the compound formed.
Fig. 11.3 shows the structure of propene.
Complete the structure of poly(propene) in Fig. 11.4. Include all the atoms and bonds of the repeating unit.
Polyesters are a group of polymers made in a condensation polymerisation reaction.
Describe the differences between addition polymerisation and condensation polymerisation.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C11.4 — Alkanes (Part (a))
• Topic C11.5 — Alkenes (Parts (b) and (c))
• Topic C11.7 — Polymers (Parts (d) and (e))
▶️ Answer/Explanation
(a) 
Ethane is a saturated hydrocarbon with formula \(C_2H_6\).
Both carbon atoms are joined by a single covalent bond, each carbon also bonding to three hydrogen atoms.
(b) Contain at least one C=C double bond; have the general formula \(C_nH_{2n}\).
Alkenes are unsaturated hydrocarbons due to the presence of a carbon–carbon double bond.
This double bond means they have fewer hydrogens per carbon than the corresponding alkane.
(c) 
Ethene’s C=C double bond opens up during addition with steam.
A hydrogen atom bonds to one carbon and an –OH group bonds to the other, forming ethanol.
(d) 
The C=C double bond in propene breaks and opens up to form single bonds linking monomers together.
Each repeat unit retains a methyl (\(CH_3\)) side branch from the original propene monomer.
(e) Addition polymerisation uses one type of unsaturated monomer and produces only a polymer (no by-product); condensation polymerisation uses two different monomers and produces a polymer plus a small molecule (e.g. water).
In addition polymerisation, alkene monomers join via their double bonds with no atoms lost.
In condensation polymerisation, monomers with two functional groups react and lose a small molecule such as water at each link.
Question 12
Suggest why the count rate recorded by the laptop increases.
State what is meant by a transverse wave.
Calculate the wavelength of the visible light wave.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P5.2.2 — The three types of nuclear emission / penetrating power (Part (a))
• Topic P3.2.1 — Reflection of light (Part (b)(i))
• Topic P3.1 — Properties of waves (Part (b)(ii), (iii))
▶️ Answer/Explanation
(a)(i) α-particles cannot penetrate the thick lead shield; the low count is background radiation.
Alpha particles have very low penetrating power and are stopped by the thick lead shield.
The small remaining count comes from background radiation, such as cosmic rays and naturally occurring radioactive materials.
(a)(ii) Gamma rays are far more penetrating than alpha particles.
Unlike alpha particles, gamma rays can pass through thick lead relatively easily.
This allows more radiation to reach the detector, increasing the count rate.
(b)(i) 
Light from the radioactive source travels to the mirror and reflects according to the law of reflection.
The reflected ray travels from the mirror into the teacher’s eye, allowing the source to be seen.
(b)(ii) A transverse wave is one where the oscillations are perpendicular to the direction of energy transfer.
In a transverse wave, the vibrations occur at right angles to the direction the wave travels.
Visible light is transverse because its oscillating electric and magnetic fields are perpendicular to its direction of propagation.
(b)(iii) \(6.0 \times 10^{-7}\,\text{m}\)
Using \( \lambda = \dfrac{v}{f} \):
\( \lambda = \dfrac{3.0 \times 10^8}{5.0 \times 10^{14}} = 6.0 \times 10^{-7}\,\text{m} \).