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

a) The three main structures found in plant cells but absent in animal cells are the cellulose cell wall, chloroplasts, and a large permanent vacuole. b) The root hair cell is adapted by: 1. Having a long extension (hair) which greatly increases the surface area for rapid absorption of water (by osmosis) and mineral ions (by active transport). 2. Having no chloroplasts, which is an adaptation because it is located underground where there is no light, so chloroplasts would be an unnecessary use of cell resources. c) Convert image size from mm to \(\mu\text{m}\): \(48 \text{ mm} \times 1000 = 48000 \mu\text{m}\). Use the formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\). Substitute the values: \(\text{Actual size} = \frac{48000}{800} = 60 \mu\text{m}\).

(a) 1 mark for each correct structure up to 3 max: cellulose cell wall, chloroplast, large/permanent vacuole. [Max 3] (b) 1 mark for identifying a structure (e.g., long projection/extension), 1 mark for linking it to increasing surface area, 1 mark for stating this allows increased absorption of water/mineral ions. Alternatively, 1 mark for noting the lack of chloroplasts and 1 mark for explaining they are not needed underground. [Max 3] (c) 1 mark for converting mm to \(\mu\text{m}\) (showing 48,000). 1 mark for correct formula \(\text{Actual} = \text{Image} / \text{Magnification}\). 1 mark for correct substitution and calculation (60). 1 mark for correct unit (\(\mu\text{m}\)). [Max 4]

a) Asexual reproduction involves only one parent and produces genetically identical offspring (clones). Sexual reproduction involves two parents and produces genetically diverse offspring. b) Sperm adaptations: 1. Flagellum (tail) allows it to swim actively towards the egg. 2. Midpiece contains many mitochondria, which release energy from aerobic respiration to power the movement of the tail. 3. Acrosome in the head contains digestive enzymes to penetrate the outer protective jelly coat of the egg cell. 4. Haploid nucleus containing half the normal number of chromosomes to fuse with the egg nucleus. c) i) Fertilisation is defined as the fusion of the nuclei of a male gamete (sperm) and a female gamete (egg). ii) Fertilisation takes place in the oviduct (or fallopian tube). iii) The zygote undergoes mitotic cell division (mitosis) repeatedly to form a ball of cells called an embryo.

(a) 1 mark for comparing parents (asexual = 1, sexual = 2). 1 mark for comparing variation (asexual = no variation/clones, sexual = genetic variation). [Max 2] (b) 1 mark for structural feature, 1 mark for its explanation, up to 2 features: flagellum/tail allows swimming (1+1); mitochondria provide energy/ATP (1+1); acrosome has enzymes to digest egg outer layer (1+1); haploid nucleus carries half genetic material (1+1). [Max 4] (c) (i) Fusion (1) of the nuclei of male and female gametes (1). [2] (ii) Oviduct / fallopian tube (1). [1] (iii) Divides by mitosis to form a ball of cells / embryo (1). [1]

a) A dominant allele is an allele that is expressed if it is present in the genotype (it masks the expression of a recessive allele). b) Heterozygous means having two different alleles of a particular gene (e.g., Bb). c) The genetic diagram is structured as follows: - Parental Phenotypes: Black fur x Black fur - Parental Genotypes: Bb x Bb - Gametes: B and b from each parent - Offspring Genotypes: BB, Bb, Bb, bb - Offspring Phenotypes: BB (black fur), Bb (black fur), bb (brown fur). d) The expected phenotypic ratio is 3 black fur : 1 brown fur. e) The actual ratio may differ because: 1. Fertilisation is a random process where any sperm can fertilise any egg by chance. 2. The sample size (number of offspring in a single litter) is small, which easily allows for statistical deviation from the expected ratio.

(a) An allele that is expressed if present / always expressed (1). [1] (b) Having two different alleles of a gene (1). [1] (c) Parental genotypes: Bb and Bb (1); gametes identified: B and b for both (1); offspring genotypes: BB, Bb, Bb, bb (1); offspring phenotypes correctly matched (BB/Bb = black, bb = brown) (1); layout structured as a logical cross/diagram (1). [Max 5] (d) 3 : 1 / 3 black to 1 brown (1). [1] (e) Fertilisation is a random process / chance fusion of gametes (1); small sample size / small number of offspring in a litter (1). [Max 2]

(a) The Sun is the ultimate source of energy entering almost all ecosystems via photosynthesis.

(b) The correct food chain starts with the producer: phytoplankton \(\rightarrow\) zooplankton \(\rightarrow\) small fish \(\rightarrow\) leopard seal. The arrows must point in the direction of energy flow (from the organism being eaten to the organism eating it).

(c) A producer is defined as an organism that makes its own organic nutrients (such as glucose), usually using energy from sunlight (by photosynthesis).

(d) Energy transfer between trophic levels is highly inefficient, with only about 10% of energy passed on to the next level. Energy is lost due to processes such as respiration (as heat), movement, excretion, egestion (feces), and because not all parts of an organism are eaten. At higher levels (above 4 or 5), there is not enough energy left to support a viable population.

(e) If the population of small fish decreases, the leopard seals will have less food. This leads to starvation, reduced reproduction, or increased migration, ultimately causing a decrease in the leopard seal population.

(a) [1 mark]
- Sun / sunlight (accept light / solar energy; reject heat)

(b) [2 marks]
- Correct sequence of organisms: phytoplankton \(\rightarrow\) zooplankton \(\rightarrow\) small fish \(\rightarrow\) leopard seal [1]
- Arrows pointing in the correct direction (from prey to predator) [1]

(c) [2 marks]
- Organism that makes its own organic nutrients / food [1]
- (Using energy from) sunlight / by photosynthesis [1]

(d) [3 marks] (Any three from:)
- Energy is lost between trophic levels / only approx. 10% of energy is transferred [1]
- Energy lost through respiration / as heat [1]
- Energy lost through movement / metabolic processes [1]
- Energy lost in excretion / egestion / feces / uneaten parts [1]
- Insufficient energy remains at higher trophic levels to support another level [1]

(e) [2 marks]
- Population (of leopard seals) decreases [1]
- Less food / energy source available (leading to starvation / death) [1]

(a) The Sun is the ultimate source of energy entering almost all ecosystems via photosynthesis.

(b) The correct food chain starts with the producer: phytoplankton \(\rightarrow\) zooplankton \(\rightarrow\) small fish \(\rightarrow\) leopard seal. The arrows must point in the direction of energy flow (from the organism being eaten to the organism eating it).

(c) A producer is defined as an organism that makes its own organic nutrients (such as glucose), usually using energy from sunlight (by photosynthesis).

(d) Energy transfer between trophic levels is highly inefficient, with only about 10% of energy passed on to the next level. Energy is lost due to processes such as respiration (as heat), movement, excretion, egestion (feces), and because not all parts of an organism are eaten. At higher levels (above 4 or 5), there is not enough energy left to support a viable population.

(e) If the population of small fish decreases, the leopard seals will have less food. This leads to starvation, reduced reproduction, or increased migration, ultimately causing a decrease in the leopard seal population.

(a) [1 mark]
- Sun / sunlight (accept light / solar energy; reject heat)

(b) [2 marks]
- Correct sequence of organisms: phytoplankton \(\rightarrow\) zooplankton \(\rightarrow\) small fish \(\rightarrow\) leopard seal [1]
- Arrows pointing in the correct direction (from prey to predator) [1]

(c) [2 marks]
- Organism that makes its own organic nutrients / food [1]
- (Using energy from) sunlight / by photosynthesis [1]

(d) [3 marks] (Any three from:)
- Energy is lost between trophic levels / only approx. 10% of energy is transferred [1]
- Energy lost through respiration / as heat [1]
- Energy lost through movement / metabolic processes [1]
- Energy lost in excretion / egestion / feces / uneaten parts [1]
- Insufficient energy remains at higher trophic levels to support another level [1]

(e) [2 marks]
- Population (of leopard seals) decreases [1]
- Less food / energy source available (leading to starvation / death) [1]

(a) A covalent bond is formed by a shared pair of electrons between two non-metal atoms, resulting in electrostatic attraction between the shared electrons and the positive nuclei.

(b) In a water molecule, there are two single covalent bonds. This means there are two shared pairs of electrons (one pair between the oxygen atom and each of the two hydrogen atoms). The central oxygen atom has four non-bonding outer-shell electrons (two lone pairs) remaining, completing its octet.

(c) (i) Energy must be absorbed (endothermic process) to break the existing chemical bonds in the reactants (\(\text{H}-\text{H}\) and \(\text{O}=\text{O}\) bonds). Energy is released (exothermic process) when new bonds (\(\text{O}-\text{H}\) bonds) are formed in the products. Since the energy released during bond making is greater than the energy absorbed during bond breaking, the overall reaction is exothermic.

(ii) In the energy level diagram:
- The reactants (\(2\text{H}_2 + \text{O}_2\)) are placed on a higher horizontal energy level than the products (\(2\text{H}_2\text{O}\)).
- The activation energy (\(E_a\)) is represented by an upward-pointing arrow from the reactants level to the peak of the energy barrier curve.
- The enthalpy change (\(\Delta H\)) is represented by a downward-pointing arrow from the reactants level to the products level, indicating a negative value.

Part (a) [2 marks]
- Shared pair of electrons (1)
- Between two non-metal atoms / electrostatic attraction between nuclei and shared electrons (1)

Part (b) [2 marks]
- Two shared pairs of electrons (one per H-O bond) (1)
- Four non-bonding outer-shell electrons on the oxygen atom (1)

Part (c)(i) [3 marks]
- Bond breaking is endothermic / takes in energy AND bond making is exothermic / releases energy (1)
- More energy is released when making bonds than is absorbed when breaking bonds (1)
- Conclusion that overall energy is released to surroundings (1)

Part (c)(ii) [3 marks]
- Reactants shown at higher energy level than products (1)
- Activation energy shown as energy barrier from reactants level to peak (1)
- Enthalpy change (\(\Delta H\)) shown as downward arrow from reactants to products (1)

(a) Method: Place equal volumes and concentrations of dilute hydrochloric acid into three separate test-tubes. Add equal surface areas/masses of copper, magnesium, and zinc to each test-tube and compare the rate of bubbling (effervescence).
Observations:
- Magnesium: Reacts rapidly/vigorously, with rapid effervescence and the metal dissolving quickly.
- Zinc: Reacts moderately, with a steady stream of gas bubbles produced.
- Copper: No reaction, no bubbles of gas are produced.
This shows reactivity order: Magnesium > Zinc > Copper.

(b) (i) Equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)

(ii) Carbon or carbon monoxide cannot be used because aluminium is more reactive than carbon. Therefore, carbon is not a strong enough reducing agent to displace/reduce aluminium from its oxide.

(iii) Aluminium is extracted by electrolysis of a molten mixture of aluminium oxide (alumina) dissolved in molten cryolite. The main raw material ore is bauxite.

Part (a) [4 marks]
- Experimental method: Add dilute hydrochloric acid to each metal under controlled conditions (equal acid concentration/volume or equal metal size) (1)
- Magnesium observation: Rapid/vigorous bubbling (1)
- Zinc observation: Moderate/slow bubbling (1)
- Copper observation: No bubbles/no reaction (1)

Part (b)(i) [2 marks]
- Correct formulae for reactants and products: \(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\) (1)
- Correct balancing: \(1:3 \rightarrow 2:3\) (1)

Part (b)(ii) [2 marks]
- State that it cannot be used because aluminium is more reactive than carbon (1)
- Explain that carbon cannot reduce/displace aluminium from its oxide (1)

Part (b)(iii) [2 marks]
- Extraction method: Electrolysis (1)
- Ore raw material: Bauxite (1)

(a) (i) Two changes that would increase the rate:
1. Increase the temperature.
2. Decrease the particle size / increase the surface area of the calcium carbonate solid (by grinding the chips into a powder).
(Note: adding a catalyst is also acceptable).

(ii) Increasing the concentration increases the number of acid particles (hydrogen ions) per unit volume. This means the reacting particles are closer together, leading to a higher frequency of collisions (collisions occur more often). Consequently, there is a higher frequency of successful collisions per unit time, which increases the reaction rate.

(b) (i) Functional group: alkene / carbon-carbon double bond (\(\text{C}=\text{C}\)). Displayed structure: Each carbon atom is double-bonded to the other carbon atom (\(\text{C}=\text{C}\)), and each carbon is also single-bonded to two hydrogen atoms, making four \(\text{C}-\text{H}\) single bonds in total.

(ii) Balanced equation with state symbols:
\(\text{C}_2\text{H}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{OH}(\text{g})\)
(Note: ethanol is produced as a gas/vapor in this industrial process).

(iii) Essential conditions (any one of the following):
- Catalyst: Phosphoric acid (\(\text{H}_3\text{PO}_4\))
- Temperature: \(300\,^\circ\text{C}\) (accept \(250\text{--}350\,^\circ\text{C}\))
- Pressure: \(60\,\text{atm}\) (accept \(50\text{--}70\,\text{atm}\))

Part (a)(i) [2 marks]
- Any two from: Increase temperature / Increase surface area of \(\text{CaCO}_3\) (decrease particle size) / Add a catalyst (1 mark for each)

Part (a)(ii) [3 marks]
- More particles per unit volume (1)
- Higher frequency of collisions / collisions occur more frequently (1)
- Higher frequency of successful/effective collisions (collisions with energy \(\ge\) activation energy) (1)

Part (b)(i) [2 marks]
- Functional group: alkene or carbon-carbon double bond / \(\text{C}=\text{C}\) (1)
- Displayed structure description: Central carbon-carbon double bond and four single C-H bonds (1)

Part (b)(ii) [2 marks]
- Correct chemical formulae: \(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\) (1)
- Correct state symbols for all species: \((\text{g})\) for ethene, \((\text{g})\) for steam, and \((\text{g})\) (or \((\text{l})\)) for ethanol (1)

Part (b)(iii) [1 mark]
- Phosphoric acid catalyst OR temperature of \(300\,^\circ\text{C}\) OR pressure of \(60\,\text{atm}\) (1)

Part (a)
(i) Since metal \(X\) does not react with zinc sulfate, zinc is more reactive than \(X\). Since metal \(X\) reacts with copper(II) sulfate, \(X\) is more reactive than copper. Thus, the order of decreasing reactivity is: Zinc, \(X\), Copper.
(ii) Copper(II) sulfate solution is blue. As copper ions are displaced and removed from the solution, the blue intensity decreases, making the solution paler or completely colourless.

Part (b)
Zinc is more reactive than copper and displaces it: \(\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}\). Standard state symbols must be included: (s) for solid metals and (aq) for aqueous solutions.

Part (c)
(i) An exothermic reaction is defined as one that releases thermal (heat) energy to its surroundings.
(ii) Energy is required to break chemical bonds (endothermic process) and energy is released when new bonds form (exothermic process). In an exothermic reaction, the energy released during bond formation is greater than the energy absorbed during bond cleavage.

Part (d)
Zinc powder consists of smaller particles than zinc pieces, giving it a much larger surface area for the same mass. A larger surface area allows more reactant particles (zinc atoms and copper ions) to come into contact simultaneously, increasing the frequency of successful collisions per unit time, thereby increasing the rate of reaction.

(a) (i) Zinc, \(X\), copper (accept chemical symbols: \(\text{Zn}\), \(X\), \(\text{Cu}\)) [1 mark]
(ii) Blue colour of solution fades / becomes lighter / decolourises [1 mark] (Reject: solution turns brown - as this is the colour of the solid product, not the solution)

(b) \(\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}\)
- 1 mark for correct reactant and product formulas: \(\text{Zn}\), \(\text{CuSO}_4\), \(\text{ZnSO}_4\), \(\text{Cu}\)
- 1 mark for all correct state symbols: \(\text{(s)}\), \(\text{(aq)}\), \(\text{(aq)}\), \(\text{(s)}\)

(c) (i) Releases heat / thermal energy to surroundings (accept temperature of surroundings increases) [1 mark]
(ii) Bond breaking is endothermic and bond making is exothermic (or energy is taken in to break bonds and released when bonds are made) [1 mark]
- More energy is released when forming bonds than is taken in to break bonds [1 mark]

(d)
- Zinc powder has a larger surface area (than zinc pieces) [1 mark]
- More frequent collisions / more collisions per second / higher rate of collision (between reactant particles / zinc and copper ions) [1 mark]
- Resulting in an increased rate of reaction [1 mark]

(a) \(GPE = mgh = 1.2 \times 10 \times 0.80 = 9.6\text{ J}\).

(b)(i) \(KE = \frac{1}{2}mv^2 = 0.5 \times 1.2 \times (3.5)^2 = 0.6 \times 12.25 = 7.35\text{ J}\).

(b)(ii) As the car rolls down, gravitational potential energy is transferred to kinetic energy, and also to thermal energy / internal energy of the surroundings (and ramp) due to friction. Work done against friction is the energy loss: \(9.6\text{ J} - 7.35\text{ J} = 2.25\text{ J}\).

(c) Work done by braking force = \(KE\) at bottom = \(7.35\text{ J}\).
Since \(W = F \times d\), we have \(7.35 = F \times 1.4\).
Therefore, \(F = \frac{7.35}{1.4} = 5.25\text{ N}\).

(a)
- Formula: \(GPE = mgh\) or substitute numbers: \(1.2 \times 10 \times 0.80\) (1)
- Correct calculation: \(9.6\text{ J}\) (1)

(b)(i)
- Formula: \(KE = \frac{1}{2}mv^2\) or substitute numbers: \(0.5 \times 1.2 \times (3.5)^2\) (1)
- Correct calculation: \(7.35\text{ J}\) (1)

(b)(ii)
- Explanation: \(GPE\) transferred to \(KE\) and thermal energy (by work done against friction) (1)
- subtraction calculation: \(9.6 - 7.35\) (1)
- Correct answer: \(2.25\text{ J}\) (1)

(c)
- Equating work done by braking force to initial kinetic energy: \(W = 7.35\text{ J}\) (1)
- Rearranging formula: \(F = \frac{W}{d}\) or \(F = \frac{7.35}{1.4}\) (1)
- Correct calculation: \(5.25\text{ N}\) (1)

(a) High transmission voltage reduces the transmission current for a given power level, since \(P = IV\). Thermal energy/power loss in the cables is given by \(P_{\text{loss}} = I^2R\). Therefore, lowering the current significantly reduces energy losses in the transmission wires, making the system much more efficient.

(b)(i) The primary coil has more turns. In a step-down transformer, the primary voltage is higher than the secondary voltage, requiring \(N_p > N_s\) because the voltage ratio equals the turns ratio.

(b)(ii) Using \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\):
\(\frac{33\,000}{240} = \frac{N_p}{400}\)
\(N_p = 400 \times \frac{33\,000}{240} = 55\,000\text{ turns}\).

(c) Using \(P = IV\):
\(48\,000 = I \times 11\,000\)
\(I = \frac{48\,000}{11\,000} \approx 4.36\text{ A}\) (or \(4.4\text{ A}\)).

(a)
- Stating that higher voltage means lower current (for same power transmission) (1)
- Referencing formula \(P = I^2R\) (thermal power loss depends on current squared) (1)
- Concluding that lower current reduces energy loss / increases efficiency (1)

(b)(i)
- Correctly stating primary coil has more turns than secondary coil (1)
- Explanation referencing the turns ratio being proportional to voltage ratio: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) (1)

(b)(ii)
- Substitute values in formula: \(\frac{33\,000}{240} = \frac{N_p}{400}\) (1)
- Correct calculation: \(55\,000\) (1)

(c)
- Conversion of units: \(48\text{ kW} = 48\,000\text{ W}\) (1)
- Rearranging \(P = IV\) to \(I = \frac{P}{V}\) (1)
- Correct calculation: \(4.36\text{ A}\) or \(4.4\text{ A}\) (1)

(a)(i) The speed of light decreases (it slows down).

(a)(ii) Using Snell's Law:
\(n = \frac{\sin(i)}{\sin(r)}\)
\(n = \frac{\sin(42^\circ)}{\sin(26^\circ)} = \frac{0.6691}{0.4384} \approx 1.53\).

(b)(i) Using the wave equation \(v = f\lambda\):
\(3.0 \times 10^8 = 4.6 \times 10^{14} \times \lambda\)
\[\lambda = \frac{3.0 \times 10^8}{4.6 \times 10^{14}} \approx 6.52 \times 10^{-7}\text{ m}\ (\text{or } 6.5 \times 10^{-7}\text{ m})\].

(b)(ii) Electromagnetic radiation with wavelengths slightly longer than red light is infrared radiation.
Common uses include: remote controls, optical fiber communications, thermal imaging, intruder alarms, or radiant heaters.

(a)(i)
- Decreases / slows down (1)

(a)(ii)
- Formula: \(n = \frac{\sin(i)}{\sin(r)}\)(1)
- Correct substitution: \(\frac{\sin(42^\circ)}{\sin(26^\circ)}\)(1)
- Correct answer: \(1.53\) (accept range 1.52 to 1.54) (1)

(b)(i)
- Formula: \(v = f\lambda\) or \(\lambda = \frac{v}{f}\)(1)
- Correct substitution: \(\frac{3.0 \times 10^8}{4.6 \times 10^{14}}\)(1)
- Correct answer: \(6.52 \times 10^{-7}\text{ m}\) (accept \(6.5 \times 10^{-7}\text{ m}\)) (1)

(b)(ii)
- Infrared (1)
- Any one valid use, e.g. remote controls / thermal imaging / infrared heater (2)

(a) Using the formula \(W = mg\): \(W = 75\text{ kg} \times 9.8\text{ m/s}^2 = 735\text{ N}\).
(b) Using the formula \(\Delta E_p = mgh\): \(\Delta E_p = 75\text{ kg} \times 9.8\text{ m/s}^2 \times 4.0\text{ m} = 2940\text{ J}\).
(c)(i) Using the formula \(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\): \(\text{Efficiency} = \frac{2940\text{ J}}{4500\text{ J}} \times 100 = 65.33\%\), which rounds to \(65.3\%\) (or \(65\%\)).
(c)(ii) Using the formula \(W = Fd\): \(W = 350\text{ N} \times 12\text{ m} = 4200\text{ J}\).
(d) The work done by the tension force is greater because some work must be done against friction between the crate and the slope. This work is dissipated as thermal energy. The energy transferred to the thermal store is \(4200\text{ J} - 2940\text{ J} = 1260\text{ J}\).

(a) \(735\text{ N}\) [1] (Accept \(750\text{ N}\) if \(g = 10\text{ m/s}^2\) is used).
(b) Formula \(mgh\) or correct substitution [1]; \(2940\text{ J}\) [1] (Accept \(3000\text{ J}\) if \(g = 10\text{ m/s}^2\) is used).
(c)(i) Correct efficiency formula or ratio substitution [1]; \(65.3\%\) or \(65\%\) (or \(0.653\) / \(0.65\)) [1] (Allow error carried forward from (b)).
(c)(ii) Formula \(W = Fd\) or substitution [1]; \(4200\text{ J}\) [1].
(d) Explanation mentioning work done against friction / resistive forces [1]; Subtraction of GPE from work done by tension (\(4200 - 2940\)) [1]; \(1260\text{ J}\) [1] (Allow error carried forward from (b) and (c)(ii)).