2025 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模擬試題連答案詳解

(a) A reflex action is a rapid, automatic response to a stimulus that does not involve conscious thought.

(b) (i) The effector is the muscle (specifically, muscles in the arm/biceps that contract to pull the hand away).
(ii)
1. A receptor in the skin detects the thermal stimulus (heat) and generates an electrical impulse.
2. The sensory neurone transmits this impulse from the receptor into the central nervous system (spinal cord).
3. In the spinal cord, the impulse is passed across a synapse to a relay neurone.
4. The relay neurone passes the impulse across another synapse to a motor neurone.
5. The motor neurone transmits the impulse out of the spinal cord to the effector (muscle), which contracts.

(c) Three key differences:
- Transmission: Nervous uses electrical impulses through neurones, whereas endocrine uses chemical hormones transported in the blood plasma.
- Speed: Nervous transmission and response are extremely fast, whereas endocrine responses are typically much slower.
- Duration: Nervous responses are short-term/instantaneous, whereas endocrine responses are longer-lasting.
- Location: Nervous impulses target precise localized effectors, while endocrine hormones can target multiple organs widely across the body.

(a)
- Rapid / automatic / involuntary response; [1 mark]
- To a stimulus; [1 mark]

(b)(i)
- Muscle (accept biceps / arm muscle); [1 mark] (Reject: 'hand', 'skin', 'effector organ' without specifying muscle)

(b)(ii)
- Receptor generates impulse AND sensory neurone transmits impulse to CNS/spinal cord; [1 mark]
- Relay neurone (in spinal cord/grey matter) receives and transmits impulse; [1 mark]
- Motor neurone transmits impulse from CNS/spinal cord to effector; [1 mark]
- Mention of synapses (chemical transmission between neurones); [1 mark]

(c) Any three from: [3 marks maximum]
- Nervous: electrical / impulses VS Endocrine: chemical / hormones; [1]
- Nervous: via neurones VS Endocrine: via blood; [1]
- Nervous: very rapid VS Endocrine: slower speed of flow/action; [1]
- Nervous: short-lived effect VS Endocrine: long-lasting effect; [1]
- Nervous: localized / precise target VS Endocrine: widespread / target organs; [1]

(a) (i) Aerobic respiration chemical equation: \(C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\). (Energy can be written on the right side but is not required for the balanced chemical equation marks).
(ii) Anaerobic respiration in human muscle: \(\text{glucose} \rightarrow \text{lactic acid}\).

(b) Anaerobic respiration in yeast (fermentation) produces ethanol and carbon dioxide as products. In contrast, anaerobic respiration in human muscle produces only lactic acid. Yeast fermentation is also used to make bread rise due to carbon dioxide, whereas muscle anaerobic respiration does not produce gas.

(c) (i) Thin alveolar walls: The wall of each alveolus is only one cell thick. This means the diffusion pathway is extremely short (only two cells thick including the capillary wall), which increases the rate of diffusion of \(O_2\) and \(CO_2\).
(ii) Dense network of capillaries: The blood capillaries wrap around the alveoli. The continuous flow of blood removes oxygenated blood from the lungs and supplies blood high in carbon dioxide. This maintains a steep concentration gradient for both gases between the air inside the alveolus and the blood, ensuring rapid, continuous gas exchange.

(a)(i)
- Correct formulae for reactants (\(C_6H_{12}O_6\) and \(O_2\)) and products (\(CO_2\) and \(H_2O\)); [1 mark]
- Fully balanced: \(C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\); [1 mark]

(a)(ii)
- glucose \(\rightarrow\) lactic acid; [1 mark]

(b)
- Yeast produces ethanol and carbon dioxide (muscle produces lactic acid only / no carbon dioxide); [1 mark]
- Yeast releases slightly more energy per glucose molecule / reference to different enzymes involved; [1 mark]

(c)(i)
- Walls are one cell thick / extremely thin; [1 mark]
- Shortens the diffusion distance / path; [1 mark]

(c)(ii)
- Brings blood high in \(CO_2\) / low in \(O_2\) (or removes oxygenated blood); [1 mark]
- Maintains a steep concentration gradient (for both oxygen and carbon dioxide); [1 mark]
- Allows rapid / continuous diffusion of gases; [1 mark]

(a) (i) Genotype is defined as the genetic makeup of an organism in terms of the alleles present.
(ii) Homozygous means having two identical alleles of a particular gene (e.g., \(RR\) or \(rr\)).

(b) (i) Yellow fruit is a recessive phenotype, so its genotype must be homozygous recessive: \(rr\).
(ii)
- Heterozygous parent genotype: \(Rr\), producing gametes: \(R\) and \(r\).
- Yellow parent genotype: \(rr\), producing gametes: \(r\).
- Crossing them yields:
- \(50\%\) heterozygous \(Rr\) (red fruit)
- \(50\%\) homozygous recessive \(rr\) (yellow fruit)

(iii) The phenotypic ratio is 1 red-fruited plant : 1 yellow-fruited plant (or 1:1, or 50% red and 50% yellow).

(c) This is called a test cross. A breeder uses a test cross to determine the genotype of an organism showing the dominant phenotype (red fruit, which could be \(RR\) or \(Rr\)). By crossing it with a homozygous recessive individual (yellow fruit, \(rr\)), the offspring phenotypes reveal the parent's genotype:
- If any offspring are yellow (\(rr\)), the parent must be heterozygous (\(Rr\)).
- If all offspring are red (\(Rr\)), the parent is homozygous dominant (\(RR\)).

(a)(i)
- Genetic makeup of an organism / alleles present; [1 mark]

(a)(ii)
- Having two identical alleles of a gene; [1 mark]

(b)(i)
- \(rr\); [1 mark]

(b)(ii)
- Correct parental genotypes identified: \(Rr\) and \(rr\); [1 mark]
- Correct gametes shown: (\(R\) and \(r\)) and (\(r\) and \(r\), or just \(r\)); [1 mark]
- Correct offspring genotypes derived: \(Rr\) and \(rr\); [1 mark]

(b)(iii)
- 1 red : 1 yellow (or 1:1, or 50% red and 50% yellow); [1 mark]

(c)
- Name: Test cross / back cross; [1 mark]
- Purpose: To determine whether the red-fruited parent is homozygous dominant (\(RR\)) or heterozygous (\(Rr\)); [1 mark]
- Explanation: If any yellow offspring (\(rr\)) are produced, the unknown parent must be heterozygous / if all are red, it is homozygous dominant; [1 mark]

(a) The principal source of energy input is sunlight / solar radiation.

(b) Energy transfer between trophic levels is inefficient (only about 10% of energy is passed on). Energy is lost at each level due to:
- Organisms respiring (energy lost as heat).
- Energy lost in excretory products (urea) and egestion (undigested waste/faeces).
- Not all parts of an organism being eaten.
Because of these continuous losses, the amount of energy available decreases dramatically at each level, leaving insufficient energy to support a sixth trophic level.

(c) Eutrophication sequence:
1. Nitrates from fertilisers leach into the lake water, acting as nutrients.
2. This causes rapid, excessive growth of algae on the water surface (algal bloom).
3. The thick layer of algae blocks sunlight from penetrating deeper into the lake.
4. Submerged aquatic plants can no longer photosynthesise, so they die.
5. Decomposing bacteria feed on the dead organic matter and multiply rapidly.
6. The bacteria respire aerobically, consuming vast amounts of dissolved oxygen from the water.
7. The water becomes severely depleted of oxygen (anoxic), causing fish to die due to lack of oxygen for respiration.

(a)
- Sunlight / light energy (from the Sun); [1 mark]

(b)
- Energy is lost at each trophic level / transfer is inefficient (only ~10% transferred); [1 mark]
- Energy lost via respiration / heat / movement / excretion / egestion / uneaten parts; [1 mark]
- Insufficient energy remains at the top of the chain to support another level; [1 mark]

(c)
- Nitrates cause rapid growth of algae / algal bloom; [1 mark]
- Algae block sunlight from reaching plants below; [1 mark]
- Under-water plants cannot photosynthesise and die; [1 mark]
- Bacteria / decomposers feed on dead plants and multiply; [1 mark]
- Bacteria respire aerobically / use up dissolved oxygen; [1 mark]
- Fish suffocate / die due to lack of oxygen; [1 mark]

### Part (a)
(i) The gas produced at the anode (+) is chlorine.
* Test: Place damp blue litmus paper in the gas.
* Observation: The paper turns red and then bleaches white.
(ii) The reaction at the cathode involves the reduction of hydrogen ions:
\(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\)

### Part (b)
* Anode (+): Pure silver.
* Electrolyte: Aqueous silver nitrate (\(\text{AgNO}_3\)) or another soluble silver salt.
* Observation at cathode (-): The steel spoon becomes coated in a shiny grey/silver solid layer.

### Part (c)
In solid sodium chloride, the ions (\(\text{Na}^+\) and \(\text{Cl}^-\)) are held in fixed positions in a giant ionic lattice and cannot move. In molten sodium chloride, the lattice is broken, allowing the ions to move freely and carry the electric current.

### Part (a)
* (i) [3 marks]:
* Name: chlorine [1]
* Test: damp blue litmus paper [1]
* Observation: bleaches / turns white [1]
* (ii) [2 marks]:
* \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\)
* Species correct [1]
* Correctly balanced [1]

### Part (b)
* [3 marks]:
* Anode: silver / \(\text{Ag}\) [1]
* Electrolyte: aqueous silver nitrate / soluble silver salt [1]
* Observation: silver/grey solid deposit on spoon [1]

### Part (c)
* [2 marks]:
* Solid: ions are fixed / held in a lattice and cannot move [1]
* Molten: ions are free to move [1]

### Part (a)
(i) Short-wavelength solar radiation passes through the atmosphere and warms the Earth. The Earth then re-emits longer-wavelength infrared radiation. Greenhouse gases absorb this infrared radiation, trapping heat and warming the atmosphere.
(ii) Major man-made sources of methane include livestock farming (cattle), decomposition of organic waste in landfills, or flooding of rice fields.

### Part (b)
(i) Under the high temperature and pressure in car engines, nitrogen and oxygen from the intake air react together to form nitrogen oxides.
(ii) The reaction in catalytic converters is:
\(2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2\)

### Part (c)
(i) Filtration: to remove insoluble solids and particles.
(ii) Chlorination: to kill harmful microorganisms/bacteria.

### Part (a)
* (i) [3 marks]:
* Solar/short-wavelength radiation warms Earth, which re-emits longer-wavelength/infrared (IR) radiation [1]
* Greenhouse gases absorb / trap this infrared radiation [1]
* Preventing heat escaping into space / warming the atmosphere [1]
* (ii) [1 mark]:
* Livestock farming / landfills / rice paddies / gas pipeline leaks [1]

### Part (b)
* (i) [2 marks]:
* High temperature / pressure in engine [1]
* Nitrogen and oxygen (from air) react together [1]
* (ii) [2 marks]:
* \(2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2\)
* Correct species [1]
* Correctly balanced [1]

### Part (c)
* (i) [1 mark]:
* To remove insoluble solids / sand [1]
* (ii) [1 mark]:
* To kill bacteria / microbes [1]

### Part (a)
(i) Excess copper(II) carbonate is used to make sure that all the sulfuric acid is fully neutralised and none remains in the solution.
(ii) Filtration is used to remove the unreacted insoluble copper(II) carbonate.
(iii) To obtain dry, pure crystals from the filtrate:
1. Heat the filtrate to evaporate some of the water until a saturated solution is formed (crystallisation point).
2. Allow the hot saturated solution to cool and crystallise slowly.
3. Filter the crystals, wash them with a small amount of cold distilled water, and dry them using filter paper.
(iv) The reaction equation with state symbols is:
\(\text{CuCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{CuSO}_4(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)

### Part (b)
(i) The ionic equation for neutralisation is:
\(\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\)

### Part (a)
* (i) [1 mark]:
* To ensure all the sulfuric acid has reacted / neutralised [1]
* (ii) [1 mark]:
* Filtration / filtering [1]
* (iii) [3 marks]:
* Heat/evaporate filtrate until crystallization point / saturated [1]
* Cool to allow crystals to form [1]
* Filter (to separate crystals), wash with small amount of cold distilled water, and dry using filter paper [1]
* (iv) [3 marks]:
* \(\text{CuCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{CuSO}_4(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
* Correct species [1]
* Balanced [1]
* All correct state symbols [1]

### Part (b)
* (i) [2 marks]:
* \(\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\)
* Correct species [1]
* Correct state symbols [1]

### Part (a)
(i) Catalytic cracking requires high temperature (around \(450-800\text{ }^\circ\text{C}\)) and a catalyst (such as alumina, silica, or zeolites).
(ii) One molecule of hexadecane produces one molecule of octane and four molecules of ethene:
\(\text{C}_{16}\text{H}_{34} \rightarrow \text{C}_8\text{H}_{18} + 4\text{C}_2\text{H}_4\)
Name of alkene: ethene.

### Part (b)
(i)
* Reagent: Aqueous bromine / bromine water.
* With propene: The orange/brown bromine water is decolourised (turns colourless).
* With propane: The orange/brown bromine water remains unchanged (remains orange/brown).
(ii) Poly(propene) consists of single C-C bonds in the backbone, with alternating carbon atoms having a methyl (\(-\text{CH}_3\)) group and a hydrogen atom, and the others having two hydrogen atoms. For two repeating units, the structure is written as:
\(-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\)
showing all C-H and C-C bonds and open-ended continuation bonds at both ends.

### Part (a)
* (i) [2 marks]:
* High temperature [1]
* Catalyst (silica / alumina / zeolite) [1]
* (ii) [3 marks]:
* \(\text{C}_{16}\text{H}_{34} \rightarrow \text{C}_8\text{H}_{18} + 4\text{C}_2\text{H}_4\) [2]
* Award [1] for correct species (\(\text{C}_2\text{H}_4\) identified as product).
* Award [1] for correct balancing (coefficient of 4).
* Name of alkene: ethene [1]

### Part (b)
* (i) [3 marks]:
* Reagent: aqueous bromine / bromine water [1]
* Propene: turns colourless / decolourises [1] (reject: turns clear)
* Propane: remains orange / yellow / brown / no change [1]
* (ii) [2 marks]:
* Carbon-carbon single bonds in a four-carbon chain backbone with open-ended continuation bonds at both ends [1]
* Alternating hydrogen and methyl (\(-\text{CH}_3\)) groups on the 1st and 3rd carbons, and two hydrogens on the 2nd and 4th carbons [1]

(a) (i) Acceleration \(a = \frac{v - u}{t} = \frac{6.0 - 0}{4.0} = 1.5\text{ m/s}^2\).

(ii) The total distance is the area under the velocity-time graph:
Distance during acceleration: \(s_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 6.0\text{ m/s} = 12.0\text{ m}\).
Distance during constant velocity: \(s_2 = \text{time} \times \text{velocity} = 6.0\text{ s} \times 6.0\text{ m/s} = 36.0\text{ m}\).
Total distance = \(12.0 + 36.0 = 48.0\text{ m}\).

(b) (i) Kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 2.0\text{ kg} \times (6.0\text{ m/s})^2 = 36.0\text{ J}\).

(ii) Using the work-energy theorem, Work Done = Change in Kinetic Energy.
\(W = F \times d = \Delta E_k\)
\(F \times 4.5\text{ m} = 36.0\text{ J}\)
\(F = \frac{36.0}{4.5} = 8.0\text{ N}\).

Part (a)(i) [2 marks]:
- 1.5 (1)
- m/s^2 (1)

Part (a)(ii) [3 marks]:
- Attempt to find area under graph or use equations of motion (e.g. 12 m or 36 m) (1)
- Summing both parts of motion (1)
- Correct answer: 48.0 m (1)

Part (b)(i) [2 marks]:
- Recall of formula E_k = 1/2 m v^2 (1)
- Correct calculation: 36.0 J (1)

Part (b)(ii) [3 marks]:
- Equating work done to loss in kinetic energy (or using equations of motion v^2 = u^2 + 2as to find deceleration a = 4.0 m/s^2 then F = ma) (1)
- Substitution of values F * 4.5 = 36 (1)
- Correct final answer: 8.0 N (1)

(a) (i) Since the components are in parallel, the potential difference across the \(30\ \Omega\) resistor is the full supply voltage of \(12\text{ V}\).
\(I = \frac{V}{R} = \frac{12}{30} = 0.40\text{ A}\).

(ii) In a parallel circuit, total current is the sum of currents in each branch:
\(I_X = I_{\text{total}} - I_R = 0.80 - 0.40 = 0.40\text{ A}\).

(iii) Resistance of component X:
\(R_X = \frac{V}{I_X} = \frac{12}{0.40} = 30\ \Omega\).

(b) As voltage increases, current increases, which causes the filament temperature to rise. The metal lattice ions in the filament vibrate with greater amplitude and frequency. This increases the rate of collisions between the conduction electrons and the vibrating metal ions, offering greater resistance to electron flow.

(c) If the current exceeds the rating of the fuse, the wire inside the fuse gets hot and melts. This breaks the circuit, stopping any current from flowing and preventing damage or fire.

Part (a)(i) [2 marks]:
- Recall of I = V/R (1)
- Correct calculation: 0.40 A (1)

Part (a)(ii) [2 marks]:
- Subtracting current: 0.80 - 0.40 (1)
- Correct answer: 0.40 A (1)

Part (a)(iii) [2 marks]:
- Recall of R = V/I (1)
- Correct calculation: 30 ohms (1)

Part (b) [2 marks]:
- Higher temperature increases vibrations of metal ions/atoms in the lattice (1)
- This increases collisions with free/conduction electrons (1)

Part (c) [2 marks]:
- High current causes the fuse wire to melt (1)
- This breaks the circuit / stops the current (1)

(a) (i) The balanced nuclear equation is:
\(^{14}_{\
6}\text{C} \rightarrow\ ^{14}_{\
7}\text{N} + \ ^{0}_{-1}\text{e}\)

(ii) A beta particle is a high-speed/energetic electron.

(b) (i) Number of half-lives elapsed: \(\frac{17,100}{5,700} = 3\).
After 1 half-life: \(4.0 \times 10^{12}\) atoms.
After 2 half-lives: \(2.0 \times 10^{12}\) atoms.
After 3 half-lives: \(1.0 \times 10^{12}\) atoms.

(ii) Background radiation is low-level ionizing radiation that is constantly present in the environment around us. Two sources of background radiation include radon gas (from rocks/soil), cosmic rays (from space), medical procedures (X-rays), or nuclear fallout.

Part (a)(i) [3 marks]:
- Correct reactants and products shown (14/6 C, 14/7 N, 0/-1 e) (1)
- Top numbers balance: 14 = 14 + 0 (1)
- Bottom numbers balance: 6 = 7 - 1 (1)

Part (a)(ii) [1 mark]:
- High-speed / energetic electron (1)

Part (b)(i) [3 marks]:
- Determines 3 half-lives (1)
- Halving the original quantity three times (or (1/2)^3) (1)
- Correct answer: 1.0 x 10^12 (1)

Part (b)(ii) [3 marks]:
- Meaning of background radiation: constant ionizing radiation in the environment (1)
- Any two sources from: radon gas/rocks/soil, cosmic rays, medical equipment, food, nuclear waste (2 marks - 1 for each source)

(a) (i) Using \(v = f \lambda\):
\(\lambda = \frac{v}{f} = \frac{0.20}{5.0} = 0.040\text{ m}\) (or \(4.0\text{ cm}\)).

(ii) The frequency remains constant / does not change.

(iii) Using \(v = f \lambda\) with \(f = 5.0\text{ Hz}\) and \(\lambda = 0.030\text{ m}\):
\(v = 5.0 \times 0.030 = 0.15\text{ m/s}\).

(iv) The wavefronts bend towards the normal (or change direction to become more perpendicular to the boundary) because the wave speed is slower in shallow water.

(b) (i) Infrared radiation (or IR).

(ii) Television remote controls, thermal imaging, cooking (grills/toasters), optical fibre communications.

(iii) \(3.0 \times 10^8\text{ m/s}\).

Part (a)(i) [2 marks]:
- Recall of v = f * lambda (1)
- Correct calculation: 0.040 m (or 4.0 cm) (1)

Part (a)(ii) [1 mark]:
- Constant / unchanged (1)

Part (a)(iii) [2 marks]:
- Correct use of v = f * lambda with frequency constant (1)
- Correct calculation: 0.15 m/s (1)

Part (a)(iv) [2 marks]:
- Direction bends towards the normal / angle of refraction is smaller than angle of incidence (1)
- Explanation: because the wave travels slower in shallow water (1)

Part (b)(i) [1 mark]:
- Infrared / IR (1)

Part (b)(ii) [1 mark]:
- Valid use (e.g. remote controls, thermal imaging, short-range communication, heating) (1)

Part (b)(iii) [1 mark]:
- 3.0 x 10^8 m/s (accept 3 x 10^8 m/s) (1)