2025 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

(a) Two structures found in plant cells but not animal cells: chloroplasts, cell wall, or a permanent large central vacuole.

(b) (i) The main function of a palisade mesophyll cell is photosynthesis.
(ii) Adaptations: contains a high density of chloroplasts to absorb maximum light; elongated shape to allow maximum light absorption; thin cell wall for rapid gas exchange.

(c) Magnification \(M = \frac{\text{Image size } I}{\text{Actual size } A}\)
\(M = \frac{45\text{ mm}}{0.05\text{ mm}} = 900\).
Therefore, the magnification is \(\times 900\) or \(900\).

(d) A tissue is a group of cells with similar structures, working together to perform a shared function. An example of plant tissue is palisade mesophyll, xylem, phloem, or epidermis.

(a)
- Cell wall [1]
- Chloroplasts / large permanent vacuole [1]
(Accept any two correct structures; reject cell membrane, cytoplasm, nucleus, mitochondria, ribosomes)

(b)
(i)
- Photosynthesis [1]
(ii)
- Many / numerous chloroplasts [1]
- To absorb maximum light [1]
OR
- Cells are tall / elongated [1]
- To pack many cells closely together near the upper surface [1]

(c)
- Correct formula stated or implied by substitution: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\)_or_\(\frac{45}{0.05}\) [1]
- Correct calculation: \(900\) [1]
- Correct representation of magnification: \(\times 900\) or \(900\) (no units of length) [1]

(d)
- Definition of tissue: a group of cells with similar structures, working together to perform a shared / specific function [1]
- Example: mesophyll / xylem / phloem / epidermis [1] (reject plant organs like leaf or stem)

(a) Sexual reproduction is a process involving the fusion of the nuclei of two gametes (haploid) to form a zygote (diploid) and the production of offspring that are genetically different from each other.

(b) (i) Insect-pollinated flowers have large, brightly coloured petals to attract insects, whereas wind-pollinated flowers have small, dull/green petals (or petals are entirely absent).
(ii) Insect-pollinated flowers have sticky, relatively small stigmas enclosed inside the petals where insects brush against them. Wind-pollinated flowers have large, feathery stigmas that hang outside the flower to capture wind-borne pollen grains.

(c) (i) Fertilisation normally occurs in the oviduct (or Fallopian tube).
(ii) Adaptations of a sperm cell:
1. Flagellum (tail) - allows the sperm to swim towards the egg.
2. Middle piece containing many mitochondria - provides energy (via respiration) for swimming.
3. Acrosome - contains digestive enzymes to penetrate the jelly coat of the egg cell.

(a)
- Fusion of haploid nuclei / gametes (sperm and egg / pollen and ovule) [1]
- To form a diploid zygote / produces genetically different offspring [1]

(b)
(i)
- Insect-pollinated: large AND brightly coloured petals [1]
- Wind-pollinated: small AND green/dull petals OR no petals [1]
(ii)
- Insect-pollinated: sticky / compact / lobed stigma AND enclosed within the flower [1]
- Wind-pollinated: feathery / large stigma AND hanging outside the flower [1]

(c)
(i)
- Oviduct / Fallopian tube [1]
(ii)
Award marks for any two adaptations described, up to a maximum of [3]:
- Flagellum / tail [1] for swimming / movement [1]
- Abundant mitochondria / middle piece [1] to release energy for swimming [1]
- Acrosome / enzymes in head [1] to digest egg membrane / jelly coat [1]
- Haploid nucleus [1] containing half the genetic material [1]

(a) (i) Genotype is the genetic make-up of an organism in terms of the alleles present.
(ii) Homozygous means having two identical alleles of a particular gene.

(b) (i) The short pea plant is homozygous recessive, so its genotype is \(tt\).
(ii) Genetic diagram:
- Parents' genotypes: \(Tt\) (heterozygous tall) \(\times\) \(tt\) (short)
- Gametes: \(T\) and \(t\) from the tall parent; only \(t\) from the short parent.
- Punnett square:
| | \(t\) | \(t\) |
|---|---|---|
| \(T\) | \(Tt\) | \(Tt\) |
| \(t\) | \(tt\) | \(tt\) |
- Offspring genotypes: \(Tt\) and \(tt\) (ratio 1:1)
- Offspring phenotypes: \(Tt\) is tall; \(tt\) is short (ratio 1:1)

(iii) The probability is \(0.5\) or \(50\%\) or \(1\text{ in } 2\).

(c) Codominance is a situation in which both alleles in a heterozygous organism contribute to the phenotype. In the ABO blood group system, the alleles \(I^A\) and \(I^B\) are codominant. A person with genotype \(I^A I^B\) has blood group AB because both A and B antigens are expressed on their red blood cells.

(a)
(i)
- Genetic make-up of an organism in terms of alleles present [1]
(ii)
- Having two identical alleles of a particular gene [1]

(b)
(i)
- \(tt\) [1]
(ii)
- Parental genotypes shown: \(Tt \times tt\) [1]
- Correct gametes shown: \(T\) and \(t\) from one parent, \(t\) (or \(t\) and \(t\)) from the other [1]
- Correct offspring genotypes: \(Tt\) and \(tt\) [1]
- Correct phenotype corresponding to each genotype: \(Tt\) is tall and \(tt\) is short [1]
(iii)
- \(0.5\) / \(50\%\) / \(1/2\) / \(1\text{ in } 2\) [1]

(c)
- Definition: both alleles are expressed / neither allele is dominant over the other in a heterozygote [1]
- Example: Alleles \(I^A\) and \(I^B\) are codominant, so genotype \(I^A I^B\) results in blood group AB / presence of both A and B antigens [1]

(a) An ecosystem is a unit containing all of the organisms (community) and their environment, interacting together, in a given area.

(b) (i) Any of the following food chains with arrows pointing in the direction of energy flow:
1. Microscopic algae \(\rightarrow\) water fleas \(\rightarrow\) stickleback fish \(\rightarrow\) herons
2. Microscopic algae \(\rightarrow\) water fleas \(\rightarrow\) diving beetles \(\rightarrow\) herons
3. Microscopic algae \(\rightarrow\) pond snails \(\rightarrow\) diving beetles \(\rightarrow\) herons

(ii) Depending on the chain chosen, the secondary consumer is:
- stickleback fish (for chain 1)
- diving beetles (for chains 2 and 3)

(c) Energy transfer is inefficient because:
- Not all parts of the prey are consumed / digested (some energy is lost in faeces).
- Energy is lost as heat released during respiration.
- Energy is lost in excretory products (such as urine/urea).
- Energy is used by the organism for movement, active transport, and warmth.

(d) Carbon dioxide is released into the atmosphere via:
1. Respiration by living organisms (plants, animals, and decomposers).
2. Combustion of fossil fuels or organic matter (such as wood).

(a)
- Unit containing all the organisms / community and their non-living environment [1]
- Interacting together (in a given area) [1]

(b)
(i)
- Correct sequence of four organisms from the text [1]
- Arrows pointing in correct direction (from food source to feeder) [1]
*(e.g., Microscopic algae \(\rightarrow\) water fleas \(\rightarrow\) stickleback fish \(\rightarrow\) herons)*
(ii)
- Stickleback fish OR diving beetles (must match the third trophic level of the drawn food chain) [1]

(c)
Award 1 mark each for any three of the following points, up to a maximum of [3]:
- Energy lost as heat from respiration [1]
- Not all of the organism is eaten (e.g. bones, roots left behind) [1]
- Some parts are indigestible / lost as faeces [1]
- Energy lost in excretory products / urine [1]
- Energy used for movement / metabolic processes [1]

(d)
- Respiration (by plants / animals / decomposers) [1]
- Combustion (of wood / fossil fuels) [1]

Detailed solution:

(a) (i) In phosphorus trichloride, there are 3 single covalent bonds (3 shared pairs of electrons) between the central phosphorus atom and each of the three chlorine atoms. The phosphorus atom has 1 lone pair of electrons remaining in its outer shell. Each chlorine atom has 3 lone pairs (6 non-bonding electrons) in its outer shell.
(ii) Phosphorus trichloride has a simple molecular structure with weak intermolecular forces between the molecules. These forces require a small amount of thermal energy to overcome, resulting in a low boiling point.

(b) (i) A sodium atom loses its outer-shell electron to form a positively charged sodium ion (\(Na^+\)). A chlorine atom gains this electron to form a negatively charged chloride ion (\(Cl^-\)). There is an electrostatic attraction between the oppositely charged ions.
(ii) In solid sodium chloride, the ions are held in a rigid giant ionic lattice and are not free to move. In molten sodium chloride, the lattice breaks down, allowing the ions to move freely and carry an electrical charge.

Marking scheme:
(a) (i) 1 mark for stating there are three shared pairs (one between P and each Cl). 1 mark for stating P has one lone pair (two non-bonding electrons). 1 mark for stating each Cl has three lone pairs (six non-bonding electrons).
(a) (ii) 1 mark for identifying weak intermolecular forces / weak forces between molecules (reject: weak covalent bonds). 1 mark for stating that little energy is needed to break these forces.
(b) (i) 1 mark for sodium losing one electron. 1 mark for chlorine gaining one electron. 1 mark for stating that this forms oppositely charged ions / \(Na^+\) and \(Cl^-\).
(b) (ii) 1 mark for stating that in the solid state, ions are in fixed positions / cannot move. 1 mark for stating that in the molten state, ions are free to move (reject: free electrons).

Detailed solution:

(a) (i) magnesium + copper(II) sulfate -> magnesium sulfate + copper
(ii) \(Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)\)

(b) (i) The raw material is limestone (calcium carbonate). It thermally decomposes to form calcium oxide and carbon dioxide. The calcium oxide (basic oxide) reacts with the acidic silicon dioxide (sand/silica) impurity to form calcium silicate (slag).
(ii) \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\)

(c) Recycling aluminium saves a large amount of electrical energy compared to electrolysis, conserves bauxite ore resources, and reduces the carbon dioxide emissions associated with energy production.

Marking scheme:
(a) (i) 1 mark for reactants/products correct, 1 mark for complete correct equation: magnesium + copper(II) sulfate -> magnesium sulfate + copper.
(a) (ii) 1 mark for correct ionic species: \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\). 1 mark for correct state symbols: \(Mg(s)\), \(Cu^{2+}(aq)\), \(Mg^{2+}(aq)\), \(Cu(s)\).
(b) (i) 1 mark for identifying limestone / calcium carbonate. 1 mark for stating that limestone decomposes to calcium oxide (or equation). 1 mark for stating that calcium oxide reacts with silica/silicon dioxide to form slag / calcium silicate.
(b) (ii) 1 mark for correct chemical formulas of reactants and products: \(Fe_2O_3 + CO \rightarrow Fe + CO_2\). 1 mark for correct balancing: \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\).
(c) 1 mark for conserving ore / saving electrical energy / reducing carbon dioxide emissions / reducing landfill waste.

Detailed solution:

(a) (i) Energy required for bond breaking = \(4 \times (C-H) + 2 \times (O=O) = 4 \times 413 + 2 \times 496 = 1652 + 992 = 2644\) kJ/mol.
(ii) Energy released in bond formation = \(2 \times (C=O) + 4 \times (O-H) = 2 \times 805 + 4 \times 463 = 1610 + 1852 = 3462\) kJ/mol.
(iii) \(\Delta H\) = energy in - energy out = \(2644 - 3462 = -818\) kJ/mol. Since the value is negative, the reaction is exothermic.

(b) An energy level diagram for this reaction features:
- Reactants on a higher horizontal line than the products.
- A curve representing the reaction pathway rising from the reactants level to a peak and then falling to the products level.
- Activation energy (\(E_a\)) represented as an upward arrow from the reactants line to the peak of the curve.
- Enthalpy change (\(\Delta H\)) represented as a downward arrow from the reactants line to the products line.

Marking scheme:
(a) (i) 1 mark for correct working: \(4 \times 413 + 2 \times 496\). 1 mark for correct answer: 2644 kJ/mol.
(a) (ii) 1 mark for correct working: \(2 \times 805 + 4 \times 463\). 1 mark for correct answer: 3462 kJ/mol.
(a) (iii) 1 mark for calculating \(\Delta H = -818\) kJ/mol (allow ecf from (i) and (ii)). 1 mark for stating that the reaction is exothermic because the overall energy change is negative.
(b) 1 mark for stating reactants are higher than products. 1 mark for describing a curve from reactants to products with a peak. 1 mark for defining \(E_a\) as the vertical distance from reactants to the peak. 1 mark for defining \(\Delta H\) as the vertical distance from reactants to products.

Detailed solution:

(a) (i) High temperature (approx. 450 - 800 °C) and a catalyst (such as silica, alumina, or zeolites).
(ii) \(C_{10}H_{22} \rightarrow C_3H_6 + C_7H_{16}\) (the other product is heptane).

(b) (i) Add bromine water to both samples. Saturated hydrocarbons (alkanes) do not react, so the solution remains orange-brown. Unsaturated hydrocarbons (alkenes) react via an addition reaction, decolourising the bromine water (turning it colourless).
(ii) Propene has three carbon atoms and six hydrogen atoms (\(C_3H_6\)). It contains one carbon-carbon double bond (\(C=C\)) and one carbon-carbon single bond (\(C-C\)).
(iii) The double bond in the propene monomer opens up to form a single bond in the backbone of the polymer repeat unit. The repeat unit has two carbon atoms in the main chain, with extension bonds on either side, and substituents consisting of three hydrogen atoms and one methyl (\(-CH_3\)) group.

Marking scheme:
(a) (i) 1 mark for high temperature / heat (approx. 450-800 °C). 1 mark for catalyst (silica / alumina / zeolite).
(a) (ii) 1 mark for identifying the other product as heptane, \(C_7H_{16}\). 1 mark for the correct balanced equation: \(C_{10}H_{22} \rightarrow C_3H_6 + C_7H_{16}\).
(b) (i) 1 mark for adding bromine water. 1 mark for alkane: remains orange/brown/no change. 1 mark for alkene: decolourises / turns colourless.
(b) (ii) 1 mark for stating that propene has 3 carbons and 6 hydrogens, with one double bond and one single bond between the carbon atoms.
(b) (iii) 1 mark for stating that the \(C=C\) double bond becomes a single bond with open/extension bonds on either side. 1 mark for stating that the carbon backbone has a methyl group (\(-CH_3\)) and three hydrogens attached.

\((a)(i)\) The gravitational potential energy is calculated using:
\(E_p = mgh\)
\(E_p = 450\text{ kg} \times 9.8\text{ m/s}^2 \times 24\text{ m} = 105,840\text{ J}\) (or \(1.1 \times 10^5\text{ J}\).

\((a)(ii)\) If \(15\\%\) is lost as thermal energy, \(85\\%\) of the original potential energy is converted into kinetic energy:
\(E_k = 0.85 \times 105,840\text{ J} = 89,964\text{ J}\)
Using the formula for kinetic energy:
\(E_k = \frac{1}{2} m v^2\)
\(89,964 = \frac{1}{2} \times 450 \times v^2\)
\(v^2 = \frac{89,964}{225} = 399.84\)
\(v = \sqrt{399.84} \approx 20\text{ m/s}\) (to 2 significant figures).

\((b)(i)\) By the principle of conservation of energy, the work done by the brakes must equal the kinetic energy of the car just before the brakes were applied (assuming no other resistive forces are present during braking):
\(W = E_k = 89,964\text{ J}\) (or \(90,000\text{ J}\) or \(9.0 \times 10^4\text{ J}\)).

\((b)(ii)\) The deceleration can be calculated using the work-energy theorem or equations of motion.
Method 1: Work done
\(W = F \times d \Rightarrow 89,964 = F \times 30\)
\(F = \frac{89,964}{30} = 2,998.8\text{ N}\)
Using Newton's second law:
\(F = ma \Rightarrow 2,998.8 = 450 \times a\)
\(a = \frac{2,998.8}{450} \approx 6.7\text{ m/s}^2\).

Method 2: Equations of motion
\(v^2 = u^2 + 2as\)
\(0 = 20^2 + 2 \times a \times 30\)
\(0 = 400 + 60a\)
\(a = -\frac{400}{60} \approx -6.7\text{ m/s}^2\).
Therefore, the magnitude of deceleration is \(6.7\text{ m/s}^2\).

\((a)(i)\)
- \(E_p = mgh\) or substitution of values [1]
- Correct answer: \(105,840\text{ J}\) (accept \(1.1 \times 10^5\text{ J}\) or \(106\text{ kJ}\)) [1]

\((a)(ii)\)
- Calculation of kinetic energy (\(85\\%\) of \(E_p\)): \(89,964\text{ J}\) (allow ECF from a(i)) [1]
- Substitution into \(E_k = \frac{1}{2}mv^2\) [1]
- Correct final speed: \(20\text{ m/s}\) (accept \(19.99\) to \(20.0\)) [1]

\((b)(i)\)
- Statement that work done equals kinetic energy [1]
- Correct value: \(89,964\text{ J}\) (accept \(90,000\text{ J}\) or ECF from a(ii)) [1]

\((b)(ii)\)
- Use of \(W = F \times d\) to find force OR use of \(v^2 = u^2 + 2as\) [1]
- Substitution of appropriate values [1]
- Correct final deceleration: \(6.7\text{ m/s}^2\) (accept range \(6.6\) to \(6.7\)) [1]

\((a)(i)\) Using Snell's law:
\(n = \frac{\sin i}{\sin r}\)
\(1.52 = \frac{\sin 42.0^\circ}{\sin r}\)
\(\sin r = \frac{\sin 42.0^\circ}{1.52} = \frac{0.6691}{1.52} = 0.4402\)
\(r = \sin^{-1}(0.4402) \approx 26.1^\circ\).

\((a)(ii)\) The critical angle is defined as the angle of incidence in the optically denser medium (glass) that results in an angle of refraction of \(90^\circ\) in the less dense medium (air).

\((a)(iii)\) Using the formula for the critical angle:
\(\sin c = \frac{1}{n}\)
\(\sin c = \frac{1}{1.52} \approx 0.6579\)
\(c = \sin^{-1}(0.6579) \approx 41.1^\circ\).

\((b)(i)\) Violet light travels slower in glass than red light. Since \(n = \frac{c}{v}\), a slower speed in glass means a higher refractive index for violet light than red light. Consequently, the angle of refraction will be smaller for violet light (it bends more towards the normal).

\((a)(i)\)
- State Snell's law: \(n = \frac{\sin i}{\sin r}\) [1]
- Correct rearrangement and substitution: \(\sin r = \frac{\sin 42^\circ}{1.52}\) [1]
- Correct final angle: \(26.1^\circ\) (accept range \(26.0^\circ\) to \(26.2^\circ\)) [1]

\((a)(ii)\)
- Mention that it is the angle of incidence in the denser medium [1]
- Resulting in an angle of refraction of \(90^\circ\) (or light travelling along the boundary) [1]

\((a)(iii)\)
- Formula \(\sin c = \frac{1}{n}\) [1]
- Correct calculation: \(41.1^\circ\) (accept range \(41.0^\circ\) to \(41.2^\circ\)) [1]

\((b)(i)\)
- Violet light is slower than red light in glass [1]
- Refractive index for violet is larger [1]
- Angle of refraction for violet is smaller / bends more [1]

\((a)(i)\) Electrical power is given by:
\(P = I \times V\)
Where \(P = 120\text{ kW} = 120,000\text{ W}\) and \(V = 12\text{ kV} = 12,000\text{ V}\).
\(I = \frac{P}{V} = \frac{120,000}{12,000} = 10\text{ A}\).

\((a)(ii)\) Power lost as heat in the transmission cables is calculated using:
\(P_{\text{loss}} = I^2 R\)
\(P_{\text{loss}} = (10)^2 \times 0.80 = 100 \times 0.80 = 80\text{ W}\).

\((b)(i)\) A step-up transformer is used. It consists of:
1. A primary coil with fewer turns of wire.
2. A secondary coil with more turns of wire.
3. Both coils wound around a soft iron core to link the magnetic fields.

\((b)(ii)\) The new current at \(24\text{ kV}\) is:
\(I_{\text{new}} = \frac{120,000\text{ W}}{24,000\text{ V}} = 5.0\text{ A}\).
New power loss:
\(P_{\text{loss, new}} = I_{\text{new}}^2 R = (5.0)^2 \times 0.80 = 25 \times 0.80 = 20\text{ W}\).
Stepping up the voltage reduces the current for the same power transmitted. Since power loss is proportional to the square of current (\(P \propto I^2\)), reducing the current by half reduces power loss to a quarter.

\((a)(i)\)
- Use of \(P = IV\) or formula rearrangement [1]
- Correct current: \(10\text{ A}\) [1]

\((a)(ii)\)
- Use of \(P = I^2 R\) [1]
- Substitution of current and resistance: \(10^2 \times 0.80\) [1]
- Correct power loss: \(80\text{ W}\) [1]

\((b)(i)\)
- Name: step-up transformer [1]
- Primary coil has fewer turns than secondary coil [1]
- Coils are wound around a soft iron core [1]

\((b)(ii)\)
- Calculation of new power loss: \(20\text{ W}\) [1]
- Explanation: higher voltage reduces current, which heavily reduces power loss because \(P_{\text{loss}} = I^2 R\) [1]

\((a)(i)\) A beta-minus (\(\beta^-\) particle is a high-speed electron.

\((a)(ii)\) In beta-minus decay, a neutron in the nucleus decays into a proton and an electron. The proton remains in the nucleus, increasing the proton number by 1, while the mass number remains unchanged. The equation is:
\[^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + ^{0}_{-1}\beta\]
(or \(^{0}_{-1}\text{e}\))

\((b)(i)\) Half-life is the time taken for half of the radioactive nuclei in a sample to decay (or the time taken for the activity of a radioactive sample to fall to half of its initial value).

\((b)(ii)\) We find the number of half-lives required to decay from \(3200\text{ Bq}\) to \(100\text{ Bq}\):
\(3200 \rightarrow 1600\) (1 half-life)
\(1600 \rightarrow 800\) (2 half-lives)
\(800 \rightarrow 400\) (3 half-lives)
\(400 \rightarrow 200\) (4 half-lives)
\(200 \rightarrow 100\) (5 half-lives)
So, 5 half-lives have passed.
Total time taken: \(5 \times 15\text{ hours} = 75\text{ hours}\).

\((b)(iii)\) After \(60\text{ hours}\\, the number of half-lives that have elapsed is:\n\)\\text{Number of half-lives} = \\frac{60}{15} = 4\\text{ half-lives}\).
After 4 half-lives, the activity is \(200\text{ Bq}\).
The shape of the decay curve is an exponential decay curve, which shows a continuous rapid decrease that slows down over time (asymptotic to the time-axis).

\((a)(i)\)
- High-speed electron (or just electron) [1]

\((a)(ii)\)
- Magnesium symbol with correct mass number 24 and atomic number 12: \(^{24}_{12}\text{Mg}\) [1]
- Beta particle with correct mass number 0 and atomic/charge number -1: \(^{0}_{-1}\beta\) or \(^{0}_{-1}\text{e}\) [1]
- Fully balanced equation (LHS and RHS match) [1]

\((b)(i)\)
- Time taken for half the nuclei / active atoms to decay OR time taken for activity to halve [1]
- Mention of \"initial value\" or \"original sample\" [1]

\((b)(ii)\)
- Correct determination of 5 half-lives [1]
- Correct time calculation: \(75\text{ hours}\) [1]

\((b)(iii)\)
- Correct calculation of remaining activity: \(200\text{ Bq}\) [1]
- Description of curve: exponential decay / curves downwards towards zero / non-linear decrease [1]