2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

(a) The human circulatory system is called a double circulation because blood passes through the heart twice for every one complete circuit of the body: once through the pulmonary circuit to the lungs, and once through the systemic circuit to the rest of the body. (b) Arteries carry blood away from the heart under high pressure. Their thick walls containing smooth muscle and elastic fibres withstand this high pressure. The elastic fibres stretch and recoil to smooth out blood flow and maintain high blood pressure. The narrow lumen also helps maintain high blood pressure. (c) The blood vessel is the pulmonary vein. It carries blood under much lower pressure than the aorta (a major artery). Consequently, the pulmonary vein has a much thinner wall with less muscle and elastic tissue, and a wider lumen relative to its wall thickness than the aorta. It also contains valves, which are absent in the aorta (except at its base).

(a) 1 mark for stating blood travels through the heart twice; 1 mark for specifying this is for one complete circuit of the body. (b) Max 4 marks: 1 mark for thick walls; 1 mark for containing muscle/elastic fibres; 1 mark for withstanding/maintaining high blood pressure; 1 mark for narrow lumen; 1 mark for elastic fibres stretching/recoiling. (c) 1 mark for identifying pulmonary vein; Max 3 marks for structural differences (thinner wall, less muscle/elastic tissue, wider lumen, presence of valves).

(a) Carbon dioxide concentration is increased by the combustion of fossil fuels (for electricity, transport, etc.) and deforestation (reducing carbon dioxide uptake). Methane concentration is increased by agricultural activities (such as digestive processes in cattle farming and decomposition in flooded rice fields) and waste disposal in landfill sites. (b) The purification process involves: 1. Filtration through sand and gravel beds to trap and remove suspended insoluble solids and impurities. 2. Chlorination, which involves adding chlorine gas or compounds to kill harmful bacteria, pathogens, and other microorganisms, making the water safe for human consumption. (c) Fossil fuels containing carbon and hydrogen are hydrocarbons. Their complete combustion in excess oxygen produces carbon dioxide and water: hydrocarbon + oxygen -> carbon dioxide + water.

(a) 1 mark for carbon dioxide + 1 mark for its human source; 1 mark for methane + 1 mark for its human source. (b) 1 mark for filtration step; 1 mark for explaining filtration removes insoluble solids; 1 mark for chlorination step; 1 mark for explaining chlorination kills pathogens/bacteria/microbes. (c) 1 mark for reactants (hydrocarbon/fuel + oxygen); 1 mark for products (carbon dioxide + water).

(a)(i) For two parallel resistors: \( 1/R_p = 1/R_1 + 1/R_2 = 1/6.0 + 1/12.0 = 2/12.0 + 1/12.0 = 3/12.0 = 1/4.0 \). Thus, \( R_p = 4.0\ \Omega \). (ii) Using Ohm's law: \( I = V / R_p = 12\text{ V} / 4.0\ \Omega = 3.0\text{ A} \). (b)(i) The total resistance \( R_{total} \) of the new circuit is the sum of the series resistor and the parallel combination: \( R_{total} = R_3 + R_p = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega \). (ii) The new total current is \( I_{new} = V / R_{total} = 12\text{ V} / 8.0\ \Omega = 1.5\text{ A} \). Since \( R_3 \) is in series, the full current of \( 1.5\text{ A} \) passes through it. The power dissipated in \( R_3 \) is: \( P = I_{new}^2 \times R_3 = (1.5\text{ A})^2 \times 4.0\ \Omega = 2.25 \times 4.0 = 9.0\text{ W} \).

(a)(i) 1 mark for formula \( 1/R_p = 1/R_1 + 1/R_2 \) or correct substitution; 1 mark for final answer of 4.0 ohms. (ii) 1 mark for formula \( I = V / R \) or correct substitution; 1 mark for final answer of 3.0 A. (b)(i) 1 mark for adding the parallel resistance to \( R_3 \); 1 mark for final answer of 8.0 ohms. (ii) 1 mark for calculating the new total current \( 1.5\text{ A} \); 1 mark for using power formula \( P = I^2 R \); 1 mark for correct substitution; 1 mark for final answer of 9.0 W.

(a)(i) Acceleration: \( a = (v - u) / t = (4.0\text{ m/s} - 0) / 2.0\text{ s} = 2.0\text{ m/s}^2 \). (ii) Distance: \( s = \text{average speed} \times t = \frac{u+v}{2} \times t = \frac{0+4.0}{2} \times 2.0 = 2.0 \times 2.0 = 4.0\text{ m} \). (b)(i) Gravitational potential energy lost: \( \Delta \text{GPE} = mgh = 0.50\text{ kg} \times 10\text{ m/s}^2 \times 1.2\text{ m} = 6.0\text{ J} \). (ii) Kinetic energy at the bottom: \( \text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.50\text{ kg} \times (4.0\text{ m/s})^2 = 0.25 \times 16 = 4.0\text{ J} \). (iii) The kinetic energy (4.0 J) is less than the GPE lost (6.0 J) because work is done against resistive forces like friction and air resistance. This energy is dissipated to the ramp and surroundings as thermal energy (heat).

(a)(i) 1 mark for formula \( a = (v - u)/t \) or correct calculation; 1 mark for final answer of 2.0 m/s^2 (with unit). (ii) 1 mark for formula \( s = vt \) (average speed) or correct integration/area under graph; 1 mark for final answer of 4.0 m. (b)(i) 1 mark for formula \( mgh \) or correct substitution; 1 mark for final answer of 6.0 J. (ii) 1 mark for formula \( \frac{1}{2}mv^2 \) or correct substitution; 1 mark for final answer of 4.0 J. (iii) 1 mark for mentioning work done against friction/air resistance; 1 mark for stating that energy is converted to thermal energy (heat).

(a) The balanced chemical equation for photosynthesis is: \( 6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \). (b) An investigation can be designed as follows: 1. Place a piece of aquatic plant (e.g. Cabomba) in a beaker of water containing sodium hydrogencarbonate (as a carbon source). 2. Position a light source at a measured distance (e.g., 10 cm) from the plant. This distance is the independent variable, which determines light intensity. 3. Allow the plant to adjust, then count the number of oxygen gas bubbles produced per minute (or measure the volume of gas using a syringe). This is the dependent variable representing the photosynthetic rate. 4. Move the light source to different distances (e.g., 20 cm, 30 cm, 40 cm) and repeat the bubble counts. 5. Control variables: Use a water bath to maintain a constant temperature, and keep the concentration of sodium hydrogencarbonate constant. (c) As light intensity increases, the rate of photosynthesis rises until it levels off (plateaus). At this stage, light intensity is no longer the limiting factor. The rate is now restricted by another limiting factor, such as temperature, carbon dioxide availability, or the concentration of photosynthetic enzymes.

(a) 3 marks: 1 mark for correct reactants (CO2 and H2O); 1 mark for correct products (glucose and O2); 1 mark for correct balancing. (b) Max 5 marks: 1 mark for describing how to vary light intensity (changing light source distance); 1 mark for specifying the independent variable (light intensity / distance); 1 mark for specifying the dependent variable (bubble rate / volume of oxygen per unit time); 1 mark for specifying a controlled variable (constant temperature or CO2 concentration); 1 mark for repeating measurements to calculate average. (c) 1 mark for stating light intensity is no longer limiting; 1 mark for explaining that another factor (e.g. temperature, CO2) has become the limiting factor.

(a)(i) The cracking reaction of decane is: \( \text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4 \). The alkene produced is ethene. (ii) Catalytic cracking requires a high temperature (typically around \( 450^\circ\text{C} \) to \( 800^\circ\text{C} \)) and the presence of a catalyst (such as zeolite, silica, or alumina). (b) Add bromine water (aqueous bromine) to both samples. With octane (an alkane), the mixture remains orange/brown (no reaction in the absence of UV light). With ethene (an alkene), the bromine water is rapidly decolourised, turning from orange/brown to colourless because an addition reaction occurs. (c) The alkene is ethene, \( \text{H}_2\text{C}=\text{CH}_2 \). It undergoes addition polymerisation to form poly(ethene). Two repeat units must be shown with single bonds between carbon atoms and open bonds at each end: \( -\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2- \) (all carbons must be bonded to 2 hydrogens, and there must be extension lines at both ends of the chain).

(a)(i) 2 marks: 1 mark for correct formula of ethene (C2H4); 1 mark for fully correct, balanced equation. (ii) 1 mark for high temperature (accept a range between 450-800 degrees C); 1 mark for catalyst (accept zeolite/silica/alumina/clay/catalyst). (b) 1 mark for reagent (bromine water / aqueous bromine); 1 mark for octane observation (no change / remains orange-brown); 1 mark for ethene observation (decolourises / turns colourless). (c) 3 marks: 1 mark for correct carbon-carbon single bond backbone; 1 mark for correct hydrogen atoms attached (4 hydrogens per repeat unit, total 8); 1 mark for extension bonds shown at both ends.

(a) Allele: an alternative form of a gene. Homozygous: having two identical alleles of a particular gene (e.g., RR or rr). (b)(i) The heterozygous round-seeded parent has the genotype \( Rr \). The wrinkled-seeded parent is homozygous recessive and has the genotype \( rr \). (ii) Setting up a genetic cross (Punnett square): Parental gametes: Parent 1 (Rr) produces gametes \( R \) and \( r \). Parent 2 (rr) produces gametes \( r \) and \( r \). Cross: \( R \) from Parent 1 and \( r \) from Parent 2 -> \( Rr \) (Round phenotype), and \( r \) from Parent 1 and \( r \) from Parent 2 -> \( rr \) (Wrinkled phenotype). The offspring genotypes are 50% \( Rr \) and 50% \( rr \). Therefore, the expected phenotypic ratio is 1 round-seeded plant : 1 wrinkled-seeded plant (or 50% round, 50% wrinkled). (c) Fertilisation is a random process with chance determining which gametes fuse. In small sample sizes, random statistical variations have a larger impact, making it unlikely to perfectly match the theoretical 1:1 ratio. Large sample sizes are required to approximate theoretical genetic ratios.

(a) 1 mark for allele definition (alternative form of a gene); 1 mark for homozygous definition (having two identical alleles of a gene). (b)(i) 1 mark for Rr; 1 mark for rr. (ii) 1 mark for correct gametes shown in Punnett square; 2 marks for correct offspring genotypes (Rr and rr); 1 mark for correct phenotype ratio (1 round : 1 wrinkled or 50% round and 50% wrinkled). (c) 1 mark for stating fertilisation is a random process / due to chance; 1 mark for explaining that small sample sizes are susceptible to random variation.

(a)(i) During the electrolysis of molten lead(II) bromide: At the anode, a brown gas/vapour (bromine) is evolved. At the cathode, a grey liquid or bead of metal (lead) is formed. (ii) The ionic half-equations are: At the anode (oxidation): \( 2\text{Br}^-(\text{l}) \rightarrow \text{Br}_2(\text{g}) + 2\text{e}^- \). At the cathode (reduction): \( \text{Pb}^{2+}(\text{l}) + 2\text{e}^- \rightarrow \text{Pb}(\text{l}) \). (b)(i) Hydrogen gas is produced at the cathode. Sodium is not produced because sodium ions are highly stable, and sodium is more reactive than hydrogen. Therefore, hydrogen ions (from water) are preferentially discharged/reduced at the cathode. (ii) The gas produced at the anode is chlorine. The chemical test is to expose the gas to damp blue litmus paper. The litmus paper will turn red first (due to acidic properties) and then quickly bleach white (confirming chlorine).

(a)(i) 1 mark for brown gas/vapour at the anode; 1 mark for grey liquid/bead at the cathode. (ii) 2 marks for anode equation (1 for correct species, 1 for correct balancing and state symbols); 2 marks for cathode equation (1 for correct species, 1 for correct balancing and state symbols). (b)(i) 1 mark for identifying hydrogen gas; 1 mark for explaining that sodium is more reactive than hydrogen (or hydrogen ions are preferentially discharged). (ii) 1 mark for using damp blue litmus paper; 1 mark for the observation (bleaches white).

(a) An alternating current (AC) is passed through the primary coil. This creates a continuously changing magnetic field in the soft iron core. This changing magnetic field passes through the secondary coil, cutting through its turns. As a result of this changing magnetic flux, an alternating electromotive force (EMF) / voltage is induced across the secondary coil.

(b) (i) Using the transformer equation:
\(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
\(V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{3000}{200} = 240 \times 15 = 3600\text{ V}\)

(ii) For a 100% efficient transformer, power in equals power out:
\(P = V_p \times I_p = V_s \times I_s\)
\(240 \times I_p = 3600 \times 0.4\)
\(240 \times I_p = 1440\)
\(I_p = \frac{1440}{240} = 6\text{ A}\)

(c) Transmitting electricity at high voltage means the current is much lower for the same amount of power transmitted. Since electrical energy lost as heat in the transmission wires is given by \(P = I^2R\), reducing the current significantly decreases energy/power loss, making transmission far more efficient.

(a)
- AC in primary coil produces a changing magnetic field (1)
- core guides/concentrates the magnetic field to the secondary coil (1)
- changing magnetic field cuts the secondary coil (1)
- induces an electromotive force (EMF) or voltage in secondary coil (1)

(b) (i)
- Correct formula or substitution shown: \(240 \times \frac{3000}{200}\) (1)
- Correct final value with unit: \(3600\text{ V}\) (1)

(b) (ii)
- Correct formula or substitution shown: \(240 \times I_p = 3600 \times 0.4\) or \(I_p = 0.4 \times 15\) (1)
- Correct final value with unit: \(6\text{ A}\) (1)

(c)
- High voltage leads to lower current (1)
- Lower current reduces heat/energy loss in cables (or refers to \(I^2R\)) (1)

(a) (i) An allele is an alternative form of a gene.
(ii) Homozygous means having two identical alleles of a particular gene.

(b)
- Parents (F1) Genotypes: \(Bb\) and \(Bb\)
- Gametes: \(B\) and \(b\) from parent 1, and \(B\) and \(b\) from parent 2.
- F2 Genotypes: \(BB\), \(Bb\), \(Bb\), \(bb\)
- F2 Phenotypes: \(BB\) and \(Bb\) = Black fur; \(bb\) = Brown fur
- Phenotypic Ratio: 3 Black : 1 Brown (or 75% Black : 25% Brown)

(c) To perform a test cross, the breeder must cross the unknown black-furred mouse (genotype \(B\_\)) with a homozygous recessive brown-furred mouse (genotype \(bb\)).
- Outcome 1: If the unknown black mouse is homozygous (\(BB\)), all offspring will have black fur (genotype \(Bb\)).
- Outcome 2: If the unknown black mouse is heterozygous (\(Bb\)), approximately half (or some) of the offspring will have brown fur (genotype \(bb\)).

(a) (i) alternative form / version of a gene (1)
(a) (ii) having two identical/same alleles (of a gene) (1)

(b)
- Parental genotypes correct: \(Bb \times Bb\) (1)
- Gametes correct: \(B\) and \(b\) (from both) (1)
- F2 offspring genotypes correct: \(BB\), \(Bb\), \(bb\) (1)
- Offspring phenotypes and ratio correct: 3 black : 1 brown (or equivalent percentages) (1)

(c)
- Cross the unknown black mouse with a homozygous recessive / brown mouse (\(bb\)) (1)
- Identify that brown offspring will be produced if the parent is heterozygous (1)
- Identify that only black offspring will be produced if the parent is homozygous (1)
- Credit detail: reference to genotypes/phenotypes in outcomes (1)

(a) (i) Product: Oxygen (or \(O_2\)). Observation: Bubbles of a colorless gas.
(ii) Product: Copper (or \(Cu\)). Observation: A pink/brown/reddish-brown solid deposit on the electrode.

(b) (i) At the cathode: \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\)
(ii) At the anode: \(4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-\) (or \(2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-\))

(c) If copper electrodes are used, the anode is no longer inert. Instead of oxygen gas forming, copper atoms from the anode are oxidized and enter the solution as ions. Thus, the anode dissolves (decreases in mass / becomes smaller), and no bubbles/gas are observed. The reaction at the cathode remains the deposition of copper metal, so it still increases in mass.

(a) (i)
- Product: oxygen / \(O_2\) (1)
- Observation: bubbles / effervescence / gas given off (1)

(a) (ii)
- Product: copper / \(Cu\) (1)
- Observation: pink / brown / red-brown solid layer (1)

(b) (i)
- \(Cu^{2+} + 2e^- \rightarrow Cu\) (1)
- Correct state symbols: \(Cu^{2+}(aq)\) and \(Cu(s)\) (1)

(b) (ii)
- \(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\) (1 for correct species, 1 for balancing) (2)

(c)
- At the anode, no gas/bubbles are produced (1)
- The copper anode dissolves / decreases in mass (1)

(a) The balanced chemical equation for photosynthesis is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) Adapting features include:
1. Many chloroplasts (containing chlorophyll) to absorb maximum light.
2. Column-shaped / vertically elongated structure so cells can be tightly packed together near the upper surface to absorb more light.
3. Large central vacuole which pushes the chloroplasts to the edges of the cell, decreasing the diffusion path for carbon dioxide.
4. Thin cell walls to facilitate rapid diffusion of carbon dioxide and water.

(c) (i) The rate can be measured by counting the number of gas (oxygen) bubbles produced per minute, or by measuring the volume of oxygen gas collected in a gas syringe over a set time period.

(ii) As light intensity increases, the rate of photosynthesis initially increases linearly. This is because at low light intensities, light is the limiting factor.
Eventually, the rate of photosynthesis reaches a maximum and plateaus (levels off). Beyond this point, further increases in light intensity do not increase the rate. This is because another factor (such as carbon dioxide concentration or temperature) has become the limiting factor.

(a)
- Correct formulas for reactants and products (1)
- Correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\) (1)

(b)
- Three features and explanations (1 mark per pair, max 3):
- feature: numerous/many chloroplasts AND explanation: to trap/absorb maximum light
- feature: closely packed / column-shaped AND explanation: to maximize absorption of light entering leaf
- feature: large central vacuole AND explanation: pushes chloroplasts to cell boundary for better gas exchange/light
- feature: thin cell walls AND explanation: allows fast diffusion of CO2 / gases

(c) (i)
- count bubbles of gas per unit time / measure volume of gas produced (1)

(c) (ii)
- Describe graph shape: rises at first then plateaus / levels off (1)
- At low light intensities, light is the limiting factor (increasing light increases rate) (1)
- At the plateau, light is no longer limiting / rate is constant (1)
- State that another factor is limiting, such as carbon dioxide concentration or temperature (1)