Question 1
- the number of people with the genotype \(aa\)
- the number of people with the sex chromosomes \(XY\)
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B16.3 — Monohybrid inheritance (Part (a))
• Topic B11.1 — Gas exchange in humans (Part (b))
• Topic B11.1 — Gas exchange in humans (Part (c))
▶️ Answer/Explanation
(a)(i) 2 people with genotype \(aa\); 4 people with sex chromosomes \(XY\)
Genotype \(aa\) corresponds to individuals shown as affected (filled symbols) in the pedigree.
There are 2 such individuals: the affected son and the affected daughter of P and Q.
\(XY\) chromosomes are carried by all males in the diagram, of which there are 4.
(a)(ii) Homozygous recessive
A genotype with two identical alleles of a gene is described as homozygous.
Since both alleles here are the recessive \(a\) allele, the genotype \(aa\) is homozygous recessive.
(a)(iii) 
Both parents P and Q must be heterozygous (\(Aa\)), since they already have an affected child.
Cross: \(Aa \times Aa \rightarrow AA : Aa : Aa : aa\).
Only 1 in 4 offspring is \(aa\) (affected), giving a percentage likelihood of \(\frac{1}{4} \times 100 = 25\%\).
(b)(i) Goblet cells
Goblet cells are specialised cells lining the airways that secrete mucus.
This mucus normally traps dust and pathogens before they reach the lungs.
(b)(ii) Thick mucus prevents cilia from clearing pathogens
The mucus traps pathogens and bacteria that enter the airways.
Because the mucus is abnormally thick and sticky, the cilia are unable to remove it effectively.
Pathogens therefore build up in the airways, leading to frequent infections.
(c) (Tobacco) smoking
Smoking introduces carcinogenic chemicals into the lungs over long periods of exposure.
These chemicals damage lung cells and can cause uncontrolled cell division, leading to lung cancer.
This is recognised as the major cause of lung cancer.
Question 2
These pollutants cause problems.
Draw one line from each pollutant to the problem it causes.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C11.1 — Formulas and terminology (Part (a))
• Topic C11.4 — Alkanes (Part (b))
• Topic C11.3 — Fuels (Part (c))
• Topic C10.2 — Air quality and climate (Part (d))
▶️ Answer/Explanation
(a) A compound that contains only carbon and hydrogen atoms
A hydrocarbon is defined strictly by its elemental composition.
No other elements, such as oxygen or nitrogen, may be present in the molecule.
(b)(i) Single covalent
Alkanes contain only single bonds between all carbon and hydrogen atoms.
This is why alkanes are described as saturated hydrocarbons.
(b)(ii) \(C_3H_8\)
A saturated hydrocarbon contains only single C–C bonds, following the general formula \(C_nH_{2n+2}\).
\(C_3H_8\) fits this formula, while \(C_2H_2\), \(C_2H_4\) and \(C_4H_8\) all contain double or triple bonds.
(c)(i) Alkene
Cracking breaks larger saturated alkanes into smaller alkanes and alkenes.
Alkenes contain a carbon–carbon double bond and are unsaturated.
(c)(ii) High temperature; catalyst
Cracking requires a high temperature to break the strong C–C bonds in large alkane molecules.
A catalyst is used to speed up the reaction and allow it to occur at a lower temperature.
(c)(iii) \(C_{24}H_{50} \rightarrow C_{10}H_{22} + 2C_6H_{12} + C_2H_4\)
Balancing carbon atoms: \(24 = 10 + 2(6) + 2\), which checks out.
Balancing hydrogen atoms: \(50 = 22 + 2(12) + 4\), which also checks out.
The equation balances with a coefficient of 2 for \(C_6H_{12}\) and the missing product being \(C_2H_4\).
(d) 
Carbon monoxide is toxic because it binds to haemoglobin, reducing oxygen transport in the blood.
Sulfur dioxide dissolves in atmospheric water to form acidic solutions, causing acid rain.
Question 3
Barium-141 decays by emitting a beta-particle.
Blade A has a surface area of \(0.12 \, \text{m}^2\).
Show that the force acting on blade A is \(2.2 \times 10^{6} \, \text{N}\).
Calculate the distance \(d\), from the centre of blade A to the pivot of the turbine.
Explain why the velocity of blade A changes.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.6.3 — Energy resources (Part (a))
• Topic P5.2.3 — Radioactive decay (Part (b)(i))
• Topic P5.2.4 — Half-life (Part (b)(ii))
• Topic P1.7 — Pressure (Part (c)(i))
• Topic P1.5.2 — Turning effect of forces (Part (c)(ii))
• Topic P1.2 — Motion (Part (c)(iii))
▶️ Answer/Explanation
(a) Does not release greenhouse gases / does not contribute to global warming
Nuclear fission does not involve burning fossil fuels.
This means no carbon dioxide is released, so it does not contribute to global warming or climate change.
(b)(i) \(_{56}^{141}\textrm{Ba}\rightarrow \, _{57}^{141}\textrm{La}+\,_{-1}^{0}\beta\)
Mass number is conserved: \(141 = 141 + 0\).
Charge (proton number) is conserved: \(56 = 57 + (-1)\).
This gives lanthanum-141 and a beta particle of mass number 0 and charge \(-1\).
(b)(ii) 72 minutes
The mass falls from 160 g to 10 g, a reduction by a factor of 16, i.e. \(2^4\).
This corresponds to 4 half-lives.
Time \(= 4 \times 18 = 72\) minutes.
(c)(i) Force \(= 1.8 \times 10^7 \times 0.12 = 2.2 \times 10^6 \, \text{N}\)
Force is calculated using \(F = P \times A\).
Substituting the given pressure and area confirms the stated value of \(2.2 \times 10^6 \, \text{N}\).
(c)(ii) \(d = 0.61\) m (or 0.63 m)
Moment of a force is given by \(\text{moment} = F \times d\), so \(d = \dfrac{\text{moment}}{F}\).
\(d = \dfrac{1.35 \times 10^6}{2.2 \times 10^6} \approx 0.61 \, \text{m}\) (using the rounded force, \(0.63\,\text{m}\) is also accepted).
(c)(iii) The direction of blade A changes
Velocity is a vector quantity, having both magnitude (speed) and direction.
As the blade moves in a circle, its direction continuously changes even though its speed stays constant, so its velocity changes.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B12.1 — Respiration (Parts (a), (b) — aerobic vs anaerobic respiration, yeast)
• Topic B1.1 — Characteristics of living organisms (Part (c))
▶️ Answer/Explanation
(a)(i) Sugar C
Table 4.1 shows sugar C produced the largest volume of gas, 82 cm³, in 2 hours.
This is higher than sugars A, B, D and E.
(a)(ii) 0.075 cm³/min
2 hours must be converted to minutes: \(2 \times 60 = 120\) minutes.
Rate \( = \dfrac{9}{120} = 0.075 \, \text{cm}^3/\text{min}\).
(a)(iii) Carbon dioxide
Anaerobic respiration in yeast produces ethanol and carbon dioxide.
The gas collected and measured in this investigation is carbon dioxide.
(a)(iv) Bread making (or any named practical use)
In bread making, carbon dioxide produced by yeast causes dough to rise.
Other accepted uses include brewing alcoholic drinks.
(b) Any three differences (aerobic vs anaerobic respiration in humans)
Aerobic respiration requires oxygen, while anaerobic respiration does not.
Aerobic respiration releases more energy per glucose molecule than anaerobic respiration.
Aerobic respiration produces carbon dioxide and water, whereas anaerobic respiration in humans produces lactic acid instead.
(c) Any one other characteristic of living organisms
Accepted characteristics include movement, sensitivity, growth, reproduction, excretion, and nutrition.
Any one of these, besides respiration, is a valid answer.
Question 5
Aqueous sodium hydroxide is defined as a base because it is ……………… ……………… .
Copper chloride, \(CuCl_2\), water and carbon dioxide are made.
Construct the balanced symbol equation for this reaction.
Copper sulfate, \(CuSO_4\), is made.
Use your answers to determine the limiting reactant in this reaction.
Show your working.
[\(A_r\) : H, 1; Cu, 64; O, 16; S, 32]
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C7.1 — The characteristic properties of acids and bases (Parts (a), (b))
• Topic C3.1 — Formulas (Part (c)(i))
• Topic C12.5 — Qualitative analysis (Part (c)(ii))
• Topic C3.3 — The mole and the Avogadro constant (Part (d))
▶️ Answer/Explanation
(a) 1
A strong acid has a very low pH, close to 0–1.
Of the given values, 1 represents the strongest acid, since pH increases with decreasing acidity.
(b) Proton donor; proton acceptor
An acid is defined as a substance that donates a proton (\(H^+\)) in a reaction.
A base is defined as a substance that accepts a proton (\(H^+\)) in a reaction.
(c)(i) \(2HCl + CuCO_3 \rightarrow CuCl_2 + H_2O + CO_2\)
All formulae must be correct: \(HCl\), \(CuCO_3\), \(CuCl_2\), \(H_2O\), \(CO_2\).
The equation is balanced by placing a coefficient of 2 in front of \(HCl\), since \(CuCl_2\) contains two chlorine atoms.
(c)(ii) Test: limewater; Positive result: milky/cloudy or white precipitate
Carbon dioxide gas is bubbled through limewater.
If carbon dioxide is present, the limewater turns milky or cloudy due to the formation of a white precipitate of calcium carbonate.
(d) Copper oxide is the limiting reactant
Relative formula mass: \(M_r(H_2SO_4) = 98\), \(M_r(CuO) = 80\).
Moles of \(H_2SO_4 = \dfrac{2.45}{98} = 0.025\, \text{mol}\); moles of \(CuO = \dfrac{1.60}{80} = 0.02\, \text{mol}\).
Since the reaction is 1:1 and there are fewer moles of copper oxide, copper oxide is the limiting reactant.
Question 6
Calculate the kinetic energy of the bee when it is moving at \(5.8 \, \text{m/s}\).
Suggest how this charge is produced.
Describe how Brownian motion provides evidence for the kinetic molecular model of matter.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.2 — Motion (Part (a)(i))
• Topic P1.6.1 — Energy (Part (a)(ii))
• Topic P3.3 — Electromagnetic spectrum (Part (b))
• Topic P4.2.1 — Electrical charge (Part (c))
• Topic P2.1.2 — Particle model (Part (d))
▶️ Answer/Explanation
(a)(i) 348 m (or 350 m)
Distance is calculated using \(d = v \times t\).
\(d = 5.8 \times 60 = 348 \, \text{m}\).
(a)(ii) \(3.4 \times 10^{-3} \, \text{J}\)
The mass must first be converted to kg: \(0.20\, \text{g} = 0.00020\, \text{kg}\).
Kinetic energy is calculated using \(KE = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 0.00020 \times 5.8^2\).
This gives \(KE \approx 0.0034 \, \text{J}\) or \(3.4 \times 10^{-3} \, \text{J}\).
(b) Similarity: both travel at the speed of light and are transverse waves; Difference: visible light has a lower frequency (longer wavelength) than ultraviolet light
Visible light and ultraviolet light are both parts of the electromagnetic spectrum.
They share the same fundamental wave properties, such as travelling at the speed of light in a vacuum.
They differ in frequency and wavelength, with ultraviolet light having a shorter wavelength and higher frequency.
(c) Friction with air causes electrons to move off the bee’s surface
As the bee flies, friction occurs between the bee and air particles.
This friction causes negative electrons to be transferred off the surface of the bee.
Losing electrons leaves the bee with an overall positive charge.
(d) Random motion of pollen grains caused by collisions with faster-moving, smaller water molecules
Pollen grains suspended in water are observed to move randomly and erratically.
This motion is caused by collisions with fast-moving, invisible water molecules.
Since the molecules must be moving quickly to cause this visible motion, Brownian motion supports the kinetic molecular model of matter.
Question 7
Amylase breaks down the substrate ………………………. into smaller molecules of ………………………. .
As temperature increases, the activity of amylase increases until it reaches its optimum temperature of ………………………………. °C.
As the temperature increases, the amylase particles gain …………………………………………. energy.
This results in more frequent successful collisions between amylase and its substrate.
At temperatures above 75°C, all of the amylase has become …………………………………………. .
This means the …………………………………………. of amylase has changed shape and is no longer complementary to the substrate.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B5.1 — Enzymes (Part (a))
• Topic B7.2 — Digestive system (Part (b))
• Topic B4.1 — Biological molecules (Part (c))
▶️ Answer/Explanation
(a) starch; simpler sugars; 25 °C; kinetic; denatured; active site
Amylase breaks down starch into simpler sugars.
Enzyme activity increases with temperature, up to an optimum of 25°C, as particles gain kinetic energy and collide more frequently.
Above 75°C, amylase becomes denatured, meaning its active site changes shape and no longer fits the substrate.
(b) Salivary glands; pancreas (or small intestine)
Amylase is secreted at two main sites along the alimentary canal.
Any two of: salivary glands, pancreas, or small intestine are accepted.
(c)(i) Carbon, hydrogen, oxygen, nitrogen
All proteins are made up of amino acids, which contain these four elements.
Some proteins may also contain sulfur, but this is not present in every protein.
(c)(ii) Biuret test
The biuret test is the standard chemical test used to detect the presence of proteins.
A positive result is shown by a colour change from blue to purple/violet.
Question 8
Choose from A, B, C or D.
Calculate the volume occupied by 224 g of oxygen gas.
The molar gas volume at room temperature and pressure is 24 dm³.
[\(A_r\) : O, 16]
Complete the dot-and-cross diagram in Fig. 8.1 to show the covalent bonding in ammonia.
Only show the outer shell electrons.
Fig. 8.2 shows the structures of graphite and diamond.
Use ideas about structure and bonding.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C2.5 — Simple molecules and covalent bonds (Parts (a), (c))
• Topic C3.3 — The mole and the Avogadro constant (Part (b)(i))
• Topic C9.3 — Alloys and their properties (Part (b)(ii))
• Topic C2.6 — Giant covalent structures (Part (d))
▶️ Answer/Explanation
(a) D
Oxygen is a simple covalent molecule with weak intermolecular forces.
This gives it a very low melting and boiling point, and it does not conduct electricity, matching the properties of substance D.
(b)(i) 168 dm³
Relative molecular mass of \(O_2 = 32\).
Moles of \(O_2 = \dfrac{224}{32} = 7\, \text{mol}\).
Volume \(= 7 \times 24 = 168\, \text{dm}^3\).
(b)(ii) Stainless steel does not rust (or rusts more slowly), and is stronger/harder/tougher
Adding chromium to iron forms an alloy that resists corrosion.
This makes stainless steel more durable and resistant to rusting compared with pure iron.
(c) 
Each N–H bond is a shared pair of electrons, one from nitrogen and one from hydrogen.
Nitrogen forms three such covalent bonds with the three hydrogen atoms, sharing one outer-shell electron pair per bond.
(d) Graphite has delocalised electrons that can move and carry charge; diamond does not
In graphite, each carbon atom bonds to only 3 other carbon atoms, leaving one delocalised electron per atom free to move.
These delocalised electrons allow graphite to conduct electricity.
In diamond, each carbon atom bonds to 4 other carbon atoms, so there are no delocalised electrons and it cannot conduct.
Question 9
- peak X is positive and peak Y is negative
- peak Y has a larger magnitude than peak X
On Fig. 9.2, sketch the data which would be obtained if a coil containing 400 turns was used with the same magnet.
Place ticks in Table 9.1 against each statement that is correct for potential difference and for electromotive force (e.m.f.).
You may place one or two ticks in each row. The first row has been done for you.
In a solid, the arrangement of atoms is ……………………… .
The forces between atoms are ……………………… which allows the atoms to ……………………… but keeps them in a ……………………… position.
Describe how thermal energy is transferred in copper.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.5.1 — Electromagnetic induction (Part (a))
• Topic P4.2.3 — Voltage (electromotive force and potential difference) (Part (b))
• Topic P2.1.1 — States of matter (Part (c))
• Topic P2.3.1 — Conduction (Part (d))
▶️ Answer/Explanation
(a)(i) The S pole causes peak X; the N pole causes peak Y; the magnet’s speed increases as it falls
As the magnet falls, its south pole passes through the coil first, inducing the positive peak X.
Its north pole passes through next, inducing the negative peak Y.
Peak Y is larger because the magnet has accelerated under gravity and is moving faster by the time its north pole passes through, inducing a greater e.m.f.
(a)(ii) Same shape graph, crossing the x-axis at the same point, but with both peaks lower
Halving the number of turns halves the induced e.m.f. at every point, since e.m.f. is proportional to the number of turns.
The overall shape and timing of the graph remain unchanged since the magnet’s motion is unaffected.
(b)
E.m.f. relates to the energy supplied by the source, while potential difference relates to the energy transferred by a circuit component.
Both are equal to work done per unit charge and are measured in volts.
(c) Regular/ordered (a lattice); strong forces; vibrate; fixed position
In a solid, atoms are arranged in a regular, ordered lattice.
Strong forces of attraction hold the atoms close together, allowing them only to vibrate about fixed positions rather than move freely.
(d) Atoms vibrate; vibrations pass to neighbouring atoms; free electrons also transfer energy
Atoms in copper vibrate more vigorously when heated.
These vibrations are passed on from atom to atom through the lattice.
Copper’s free (delocalised) electrons also move and transfer thermal energy quickly through the metal.
Question 10
State two advantages of asexual reproduction.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B19.1 — Habitat destruction (Parts (a)(i), (a)(ii))
• Topic B18.1 — Energy flow (Parts (b), (c))
• Topic B15.1 — Asexual reproduction (Part (d))
▶️ Answer/Explanation
(a)(i) Fewer trees means less photosynthesis, so less carbon dioxide is removed from the atmosphere
Deforestation reduces the number of trees available for photosynthesis.
Since fewer trees are absorbing carbon dioxide, more carbon dioxide remains in the atmosphere.
(a)(ii) Any three effects, such as: extinction, migration, or destruction of habitats and food sources
Deforestation destroys habitats, shelter, and breeding grounds for animal species.
It also removes food sources, disrupting food chains and food webs in the area.
This can lead to species migrating elsewhere or, in severe cases, becoming extinct.
(b) A unit containing all of the organisms and their environment, interacting together in a given area
An ecosystem includes both the living organisms (biotic factors) and the non-living environment (abiotic factors).
These components interact together within a defined area.
(c) The Sun
The Sun is the principal source of energy that enters an ecosystem.
This energy is captured by producers through photosynthesis and passed along food chains.
(d) Any two advantages, such as speed and preservation of favourable traits
Asexual reproduction is faster and requires less energy than sexual reproduction.
It also preserves favourable genetic traits exactly, and allows reproduction even if a plant is isolated from others.
Question 11
Copper is formed at the cathode.
Construct the ionic half-equation for the formation of copper.
The student weighs the electrodes before the experiment to find their mass.
After the electrolysis, the student washes and dries the electrodes and then weighs the electrodes again to find their mass.
Table 11.1 shows the results.
Suggest the change in mass of the cathode in grams.
Write your answer in Table 11.1.
Explain why the anode loses mass.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C4.1 — Electrolysis (Parts (a), (b), (c))
▶️ Answer/Explanation
(a)(i) Bromine (\(Br_2\))
During electrolysis of molten lead(II) bromide, negative bromide ions move to the anode.
They lose electrons (are oxidised) to form bromine gas.
(a)(ii) Molten lead(II) bromide contains ions that are free to move
When molten, the ionic lattice breaks down and the \(Pb^{2+}\) and \(Br^-\) ions become mobile.
These free-moving ions can carry charge through the liquid, allowing it to conduct electricity.
(b) \(Cu^{2+} + 2e^- \rightarrow Cu\)
Copper(II) ions in solution gain two electrons at the cathode.
This reduction reaction deposits solid copper onto the cathode.
(c)(i) +0.62 g
With copper electrodes, the mass lost from the anode equals the mass gained at the cathode.
Since the anode lost 0.62 g, the cathode should gain 0.62 g, so the change in mass of the cathode is +0.62 g.
(c)(ii) Copper atoms at the anode dissolve and become copper ions that move into solution
At the anode, copper atoms lose electrons and are oxidised to form copper ions.
These copper ions then move into the surrounding solution, causing the anode to lose mass.
Question 12
Give your answer to 3 significant figures.
Red light has a wavelength of \(7.0 \times 10^{-7} \, \text{m}\).
Describe how Fig. 12.2 shows that red light travels faster through glass than violet light.
Use Fig. 12.3 to calculate the mass of the glass block.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P3.2.2 — Refraction of light (Parts (a), (b)(i))
• Topic P3.2.4 — Dispersion of light (Part (b)(ii))
• Topic P1.4 — Density (Part (c))
▶️ Answer/Explanation
(a) 1.55
Refractive index is calculated using \(n = \dfrac{\sin i}{\sin r}\).
\(n = \dfrac{\sin 53°}{\sin 31°} \approx 1.55\).
(b)(i) \(4.8 \times 10^{-7} \, \text{m}\)
Using the refractive index of 1.55 found in part (a), this value is read off the graph in Fig. 12.2.
This corresponds to a wavelength of approximately \(4.8 \times 10^{-7} \, \text{m}\).
(b)(ii) Refractive index is inversely proportional to speed
Fig. 12.2 shows that red light (longer wavelength) has a lower refractive index than violet light.
Since refractive index is inversely related to the speed of light in the medium, a lower refractive index means red light travels faster through the glass than violet light.
(c) 403 g
Volume of the glass block \(= 12.0 \times 2.0 \times 6.0 = 144 \, \text{cm}^3\).
Mass is calculated using \(\text{mass} = \text{density} \times \text{volume} = 2.80 \times 144\).
This gives a mass of approximately \(403 \, \text{g}\).