2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模拟试题及答案详解

**(a)** A double circulatory system means that for every complete circuit of the body, blood passes through the heart twice (once through the pulmonary circulation to the lungs, and once through the systemic circulation to the rest of the body).
An advantage is that blood can be pumped to body organs and tissues at a much higher pressure than is possible in a single circulatory system, ensuring rapid delivery of oxygen and nutrients to tissues.

**(b)**
1. Biconcave disc shape: gives a large surface area to volume ratio, allowing oxygen to diffuse in and out of the cell very rapidly.
2. Lack of a nucleus: provides more internal space to pack in haemoglobin molecules, maximizing oxygen transport capacity.
(Other correct answers: highly flexible cell membrane to squeeze through narrow capillaries.)

**(c)(i)**
1. Arteries have thick, muscular, and elastic walls, whereas veins have thin walls with much less muscle and elastic tissue.
2. Arteries have a narrow lumen, whereas veins have a wider lumen. (Alternatively, veins contain valves while arteries do not.)

**(c)(ii)**
- Arteries carry blood directly from the heart under very high pressure, so they need thick walls to withstand this high pressure without bursting, and a narrow lumen to maintain the high pressure.
- Veins carry blood back to the heart under low pressure, so they do not need thick walls. The wider lumen reduces resistance to blood flow, helping blood return to the heart easily.

**(a)** [3 marks total]
- Blood passes through the heart twice for each complete circuit / has pulmonary and systemic circuits [1]
- Advantage: blood is delivered to body tissues at high pressure [1]
- Higher rate of oxygen / nutrient delivery / faster blood flow [1]

**(b)** [2 marks total]
- Any two from:
- Biconcave disc shape increases surface area to volume ratio [1]
- No nucleus provides more space for haemoglobin / oxygen [1]
- Flexible / can bend to fit through capillaries [1]
- Contains haemoglobin to bind oxygen [1]

**(c)(i)** [2 marks total]
- Any two structural differences:
- Artery has thicker muscular / elastic walls (or vein has thinner walls) [1]
- Artery has a narrower lumen (or vein has a wider lumen) [1]
- Vein has valves, artery does not [1]

**(c)(ii)** [3 marks total]
- Arteries carry blood under high pressure (from heart) [1]
- Thick walls prevent bursting / narrow lumen maintains pressure [1]
- Veins carry blood under low pressure / wide lumen reduces resistance / friction to flow [1]

**(a)** The balanced symbol equation is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light, chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

**(b)(i)**
- Relationship: As the distance of the light source increases, the rate of photosynthesis decreases.
- Explanation: Increasing distance decreases light intensity. Since light is required for photosynthesis, lower light intensity decreases the rate of reaction (light is a limiting factor).

**(b)(ii)**
- Why inaccurate: Bubbles can vary in size (volume), and some gas may dissolve in the water before forming bubbles, so counting them does not give an exact volume of oxygen.
- Improvement: Collect the oxygen gas produced in an inverted measuring cylinder / gas syringe to measure its actual volume over a set time period.

**(c)(i)**
- Chlorophyll absorbs light energy and transfers/converts this light energy into chemical energy for the synthesis of carbohydrates (glucose).

**(c)(ii)**
- Mineral ion: Magnesium (\(\text{Mg}^{2+}\)) ions.
- Appearance: The leaves turn yellow, particularly between the veins, a condition known as chlorosis.

**(a)** [2 marks total]
- Correct reactants and products (\(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2\)) [1]
- Correct balancing (\(6\text{CO}_2\), \(6\text{H}_2\text{O}\), \(6\text{O}_2\)) [1]

**(b)(i)** [2 marks total]
- As distance increases, rate of photosynthesis decreases / light intensity decreases [1]
- Light is a reactant / energy source / limiting factor [1]

**(b)(ii)** [2 marks total]
- Inaccuracy: bubbles are different sizes / gas dissolves in water [1]
- Improvement: collect gas in a gas syringe / measuring cylinder to measure volume [1]

**(c)(i)** [2 marks total]
- Absorbs light energy [1]
- Converts / transfers it to chemical energy (for glucose synthesis) [1]

**(c)(ii)** [2 marks total]
- Magnesium (ions) [1]
- Yellowing of leaves / chlorosis [1]

**(a)(i)**
In ethene (\(\text{C}_2\text{H}_4\)), there is a double covalent bond between the two carbon atoms (sharing 4 electrons), and single covalent bonds between each carbon atom and two hydrogen atoms (sharing 2 electrons each).
Diagram layout:
- A central C=C region with 2 crosses and 2 dots representing the double bond.
- Four C-H regions, each containing 1 dot and 1 cross.
- No other valence electrons are left unshared.

**(a)(ii)**
- Test: Add aqueous bromine (bromine water).
- Observation with Alkane (ethane): The solution remains orange/brown (no reaction in the dark).
- Observation with Alkene (ethene): The orange/brown solution is rapidly decolourised (becomes colourless).

**(b)(i)**
- Temperature: \(300\ ^\circ\text{C}\) (accept \(250\text{--}350\ ^\circ\text{C}\))
- Catalyst: Phosphoric acid (\(\text{H}_3\text{PO}_4\))

**(b)(ii)**
- It is an addition reaction (specifically, hydration).

**(c)(i)**
- Addition polymerisation is a reaction where many small, unsaturated monomer molecules (containing double bonds) join together to form a long-chain polymer molecule as the single product, without the formation of any by-products.

**(c)(ii)**
- The repeating unit of poly(ethene) is drawn by converting the double bond of ethene into a single bond, with open-ended single bonds extending outwards from each carbon atom, enclosed in square brackets with an 'n' subscript:
`-[CH2-CH2]-`

**(a)(i)** [2 marks total]
- Four shared electrons (2 pairs) shown between the two C atoms [1]
- One pair of shared electrons shown between each C and its two H atoms (total 4 C-H bonds) [1]

**(a)(ii)** [3 marks total]
- Add bromine water / aqueous bromine [1]
- Alkene: decolourises / turns colourless [1]
- Alkane: remains orange / yellow / brown / no change [1]

**(b)(i)** [2 marks total]
- \(300\ ^\circ\text{C}\) (accept \(250\text{--}350\ ^\circ\text{C}\)) [1]
- Phosphoric acid catalyst (accept \(\text{H}_3\text{PO}_4\)) [1]

**(b)(ii)** [1 mark total]
- Addition / hydration [1]

**(c)(i)** [1 mark total]
- Small molecules / monomers join to form a large molecule / polymer as the only product [1]

**(c)(ii)** [1 mark total]
- Correctly drawn repeating unit showing single C-C bond with open bonds extending beyond brackets [1]

**(a)(i)**
Using Ohm's Law:
\(I = \frac{V}{R} = \frac{24\text{ V}}{12\ \Omega} = 2.0\text{ A}\)

**(a)(ii)**
Using the power formula:
\(P = V \times I = 24\text{ V} \times 2.0\text{ A} = 48\text{ W}\)
(Or \(P = I^2 R = 2.0^2 \times 12 = 48\text{ W}\))

**(b)**
- The resistance of the wire is halved (becomes \(6.0\ \Omega\)).
- Explanation: Resistance is inversely proportional to the cross-sectional area of a wire (\(R \propto \frac{1}{A}\)). Doubling the area doubles the number of conduction pathways, which halves the resistance.

**(c)(i)**
- The diagram must show a battery connected in series with an ammeter.
- The current then splits into three parallel branches, each containing one of the \(6.0\ \Omega\) resistors.
- The branches recombine to return to the battery.
- A voltmeter must be connected in parallel across the combined resistor network (or across any of the parallel branches).

**(c)(ii)**
For resistors in parallel:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\)
\(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} + \frac{1}{6.0} = \frac{3}{6.0} = 0.5\ \Omega^{-1}\)
\(R_p = \frac{6.0}{3} = 2.0\ \Omega\)

**(a)(i)** [1 mark total]
- \(2.0\text{ A}\) (accept with correct units) [1]

**(a)(ii)** [2 marks total]
- Correct formula used, e.g., \(P = V I\) or \(P = I^2 R\) [1]
- \(48\text{ W}\) (correct value with unit) [1]

**(b)** [2 marks total]
- Resistance decreases / is halved / becomes \(6.0\ \Omega\) [1]
- Resistance is inversely proportional to cross-sectional area [1]

**(c)(i)** [3 marks total]
- Three resistors connected correctly in parallel [1]
- Ammeter connected in series to measure total current [1]
- Voltmeter connected in parallel across the resistors [1]

**(c)(ii)** [2 marks total]
- Evidence of formula: \(\frac{1}{R} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\) [1]
- \(2.0\ \Omega\) (correct calculation with unit) [1]

**(a)**
According to the lock and key hypothesis:
1. The enzyme has a specifically shaped region called the **active site**.
2. The **substrate** molecule has a complementary shape that fits precisely into the active site, like a key into a lock.
3. They bind to form an enzyme-substrate complex, where the reaction occurs, releasing products.

**(b)(i)**
- Between \(10\ ^\circ\text{C}\) and \(40\ ^\circ\text{C}\), as temperature increases, the kinetic energy of the enzyme and substrate molecules increases.
- This causes them to move faster, leading to more frequent collisions per second, resulting in a higher rate of successful collisions and thus a faster rate of reaction.

**(b)(ii)**
- Above the optimum temperature (which is around \(40\ ^\circ\text{C}\)), high temperatures cause the weak bonds holding the enzyme's protein structure together to break.
- This alters the specific shape of the active site, meaning the substrate molecule can no longer fit into it.
- The enzyme is denatured. At \(60\ ^\circ\text{C}\), all amylase molecules are completely denatured, so the rate drops to zero.

**(c)**
- Each enzyme has an optimum pH at which it works most efficiently.
- Moving away from the optimum pH (making it too acidic or too alkaline) disrupts the ionic bonds in the enzyme, changing the shape of its active site (denaturing it) and decreasing its activity.

**(a)** [3 marks total]
- Active site has a specific / complementary shape to substrate [1]
- Substrate binds to active site [1]
- To form enzyme-substrate complex / reaction occurs [1]

**(b)(i)** [2 marks total]
- Molecules gain kinetic energy / move faster [1]
- More frequent (successful) collisions between enzyme and substrate [1]

**(b)(ii)** [3 marks total]
- Above optimum temperature, enzyme shape is altered / bonds break [1]
- Active site changes shape so substrate no longer fits [1]
- Enzyme is denatured / completely denatured at \(60\ ^\circ\text{C}\) [1]

**(c)** [2 marks total]
- Each enzyme has an optimum pH / works best at specific pH [1]
- Extreme pH changes shape of active site / denatures the enzyme [1]

**(a)(i)**
\(M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100\)

**(a)(ii)**
Using the formula:
\(\text{Moles} = \frac{\text{mass}}{M_r} = \frac{5.0\text{ g}}{100} = 0.050\text{ mol}\)

**(a)(iii)**
From the stoichiometry of the balanced equation:
\(1\text{ mol}\) of \(\text{CaCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\).
Therefore, \(0.050\text{ mol}\) of \(\text{CaCO}_3\) produces \(0.050\text{ mol}\) of \(\text{CO}_2\).

**(a)(iv)**
Using the gas volume formula:
\(\text{Volume} = \text{moles} \times 24\text{ dm}^3\)
\(\text{Volume} = 0.050\text{ mol} \times 24\text{ dm}^3 = 1.2\text{ dm}^3\)

**(b)(i)**
From the balanced equation, the mole ratio of \(\text{CaCO}_3\) to \(\text{HCl}\) is \(1 : 2\).
- Moles of \(\text{CaCO}_3\) needed = \(\frac{\text{moles of HCl}}{2} = \frac{0.10\text{ mol}}{2} = 0.050\text{ mol}\).
- Mass of \(\text{CaCO}_3\) = \(\text{moles} \times M_r = 0.050\text{ mol} \times 100 = 5.0\text{ g}\).

**(b)(ii)**
- Test: Bubble the gas into limewater (aqueous calcium hydroxide).
- Observation: The limewater turns cloudy/milky.

**(a)(i)** [1 mark total]
- \(100\) [1]

**(a)(ii)** [2 marks total]
- Formula: \(\text{moles} = \frac{\text{mass}}{\text{formula mass}}\) [1]
- \(0.050\text{ mol}\) (accept \(0.05\)) [1]

**(a)(iii)** [1 mark total]
- \(0.050\text{ mol}\) (error carried forward from (a)(ii)) [1]

**(a)(iv)** [2 marks total]
- Formula: \(\text{volume} = \text{moles} \times 24\) [1]
- \(1.2\text{ dm}^3\) (correct calculation with unit) [1]

**(b)(i)** [3 marks total]
- Mole ratio calculation: \(0.10\text{ mol HCl} : 0.050\text{ mol CaCO}_3\) [1]
- Calculation: \(0.050 \times 100\) [1]
- \(5.0\text{ g}\) (correct value with unit) [1]

**(b)(ii)** [1 mark total]
- Bubble through limewater AND turns milky / cloudy [1]

**(a)(i)**
The critical angle is the angle of incidence in an optically denser medium (such as water) which produces an angle of refraction of \(90\ ^\circ\) in the optically less dense medium (such as air).

**(a)(ii)**
Using the refractive index formula for critical angle:
\(\sin(c) = \frac{1}{n}\)
\(\sin(c) = \frac{1}{1.33} \approx 0.7519\)
\(c = \sin^{-1}(0.7519) \approx 48.8\ ^\circ\) (or \(49\ ^\circ\))

**(b)(i)**
- The ray of light is refracted as it exits into the air, bending away from the normal.

**(b)(ii)**
- Total internal reflection occurs; the light is completely reflected back into the water.

**(c)(i)**
- Any one common property, such as: they all travel at the same speed in a vacuum (\(3.0 \times 10^8\text{ m/s}\)), they can all travel through a vacuum, or they are all transverse waves.

**(c)(ii)**
Using the wave equation:
\(v = f \lambda \implies \lambda = \frac{v}{f}\)
\(\lambda = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^6\text{ Hz}} = 200\text{ m}\)

**(a)(i)** [2 marks total]
- Angle of incidence in denser medium [1]
- Angle of refraction is \(90\ ^\circ\) (along the boundary) [1]

**(a)(ii)** [3 marks total]
- Formula: \(\sin(c) = \frac{1}{n}\) [1]
- Substitution: \(\sin(c) = \frac{1}{1.33}\) [1]
- \(48.8\ ^\circ\) (accept \(49\ ^\circ\)) [1]

**(b)(i)** [1 mark total]
- Refracted / passes out into air (bending away from normal) [1]

**(b)(ii)** [1 mark total]
- Total internal reflection (reflected back into water) [1]

**(c)(i)** [1 mark total]
- Any valid property, e.g., travel at \(3 \times 10^8\text{ m/s}\) in vacuum / can travel through vacuum / are transverse [1]

**(c)(ii)** [2 marks total]
- Formula: \(\lambda = \frac{v}{f}\) [1]
- \(200\text{ m}\) (correct calculation with unit) [1]

**(a)(i)**
An allele is an alternative or different version/form of a gene.

**(a)(ii)**
Heterozygous means having two different alleles of a particular gene (e.g., \(Tt\)).

**(b)(i)**
- **Parental phenotypes**: Tall plant x Dwarf plant
- **Parental genotypes**: \(Tt\) x \(tt\)
- **Gametes**: \(T\) and \(t\) from the tall parent; only \(t\) from the dwarf parent.
- **Punnett square / Cross results**:
- \(T\) combined with \(t\) \(\rightarrow Tt\) (Tall)
- \(t\) combined with \(t\) \(\rightarrow tt\) (Dwarf)
- **Offspring Genotypes**: \(Tt\) and \(tt\)
- **Offspring Phenotypes**: Tall and Dwarf

**(b)(ii)**
- The expected phenotypic ratio is \(1 : 1\) (or \(50\%\) tall to \(50\%\) dwarf).

**(c)(i)**
Codominance refers to a genetic scenario where neither allele is dominant over the other, so both alleles are fully expressed in the phenotype of a heterozygous individual (producing a combined or distinct phenotype).

**(c)(ii)**
- Example: Blood group \(\text{AB}\) in the ABO blood group system (where alleles \(I^A\) and \(I^B\) are codominant).

**(a)(i)** [1 mark total]
- Alternative form of a gene [1]

**(a)(ii)** [1 mark total]
- Having two different alleles of a gene [1]

**(b)(i)** [4 marks total]
- Correct parental genotypes (\(Tt\) and \(tt\)) [1]
- Correct gametes shown (\(T\) and \(t\) from one; \(t\) from the other) [1]
- Correct offspring genotypes (\(Tt\) and \(tt\)) [1]
- Genotypes correctly matched to phenotypes (\(Tt\) is tall, \(tt\) is dwarf) [1]

**(b)(ii)** [1 mark total]
- \(1 : 1\) or \(50\% : 50\%\) [1]

**(c)(i)** [2 marks total]
- Neither allele is dominant / recessive [1]
- Both alleles are expressed in the phenotype (of a heterozygote) [1]

**(c)(ii)** [1 mark total]
- ABO blood groups (accept Group AB) / sickle cell anaemia [1]

(a) The chemical equation 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) Palisade mesophyll cells are highly adapted because: 1. They are packed with chloroplasts to absorb maximum light. 2. They are closely packed together in an upright position. 3. They are located at the top of the leaf to receive the most sunlight. (c) (i) The gas produced is oxygen. (ii) Factors to control are temperature and carbon dioxide concentration. (iii) The rate of photosynthesis levels off because another factor, such as carbon dioxide concentration or temperature, has become the limiting factor.

(a) 1 mark for correct reactant and product chemical formulas; 1 mark for correct balancing. (b) 1 mark for mentioning numerous chloroplasts; 1 mark for vertical elongation/tight packing; 1 mark for location at the upper leaf surface. (c) (i) 1 mark for oxygen. (ii) 1 mark for temperature; 1 mark for carbon dioxide concentration. (iii) 1 mark for stating that light is no longer limiting; 1 mark for explaining that another factor (e.g., carbon dioxide concentration or temperature) is now the limiting factor.

(a) (i) Saturated hydrocarbons contain only C-C single bonds, while unsaturated hydrocarbons contain at least one C=C double bond. (ii) Test reagent: Bromine water (or aqueous bromine). Observation: The bromine water changes from orange/brown to colorless. (b) (i) Catalytic cracking is carried out because fractional distillation produces more large-chain hydrocarbons than are demanded by the market, and fewer short-chain hydrocarbons. Cracking converts the surplus larger, less useful molecules into smaller, higher-demand alkanes (for fuel) and alkenes (for chemical synthesis). (ii) Ethene has a double covalent bond between the two carbon atoms. The diagram shows: 1. Four shared electrons between the carbon atoms. 2. Two shared electrons between each carbon atom and each of the four hydrogen atoms. (c) (i) Catalyst: phosphoric acid (acid catalyst); Temperature: \(300^{\circ}\text{C}\). (ii) \(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\)

(a) (i) 1 mark for saturated having only single C-C bonds and unsaturated having at least one C=C double bond. (ii) 1 mark for bromine water / aqueous bromine; 1 mark for orange/brown to colorless. (b) (i) 1 mark for breaking down large molecules to smaller ones; 1 mark for explaining that demand for smaller fractions is higher than their supply from distillation. (ii) 1 mark for showing four shared electrons in the C=C double bond; 1 mark for showing two shared electrons in each of the four C-H single bonds, with no other electrons on the outer shells. (c) (i) 1 mark for phosphoric acid; 1 mark for \(300^{\circ}\text{C}\) (accept \(250-350^{\circ}\text{C}\)). (ii) 1 mark for the correct balanced equation.

(a) (i) Combined resistance \(R\) is given by: \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0} = \frac{1}{3.0}\). Thus, \(R = 3.0\ \Omega\). (ii) Total current \(I\) is calculated using Ohm's Law: \(I = \frac{V}{R} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A}\). (iii) In a parallel circuit, the voltage across each resistor is equal to the battery voltage. Therefore, \(V_2 = 6.0\text{ V}\). Power \(P = \frac{V^2}{R_2} = \frac{6.0^2}{12.0} = \frac{36.0}{12.0} = 3.0\text{ W}\). (b) (i) Ways to increase e.m.f.: 1. Spin the coil faster. 2. Use a stronger magnet. 3. Use more turns of wire on the coil. 4. Use a coil with a larger area. (ii) As the coil rotates, the wire sides move up and down through the magnetic field. Each side cuts the field lines in one direction, then in the opposite direction during the next half-cycle. This periodically reverses the direction of the induced electromotive force, producing an alternating current.

(a) (i) 1 mark for showing correct formula \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\); 1 mark for \(3.0\ \Omega\). (ii) 1 mark for \(I = \frac{V}{R}\) substitution; 1 mark for 2.0 A. (iii) 1 mark for identifying voltage is 6.0 V (or finding current through \(R_2\) as 0.5 A); 1 mark for 3.0 W. (b) (i) 1 mark for each correct method (up to 2). (ii) 1 mark for noting that rotation causes wire sides to cut magnetic field lines in opposite directions every half turn; 1 mark for explaining that this reverses the direction of the induced e.m.f. / current.

(a) In a double circulatory system, blood travels through the heart twice for every complete journey around the body: once to the lungs (pulmonary circuit) and once to the rest of the body (systemic circuit). (b) Differences between arteries and veins: 1. Arteries have a thicker outer wall of elastic and muscle fibers than veins, allowing them to withstand and maintain high pressure. 2. Veins have a larger lumen (inner space) than arteries because blood flows under low pressure and less resistance is needed. 3. Veins have valves to ensure blood flows toward the heart against gravity, whereas arteries do not have valves (except at the base of the aorta and pulmonary artery). (c) (i) Valves ensure that blood flows in only one direction and prevents backflow. (ii) The valve is the bicuspid (or mitral) valve. (d) The left ventricle has a much thicker muscular wall because it must generate enough pressure to pump blood to the entire body. The right ventricle only pumps blood to the lungs, which are nearby, requiring much less pressure.

(a) 1 mark for blood passing through the heart twice; 1 mark for per complete circuit of the body. (b) 1 mark for artery having thicker muscular/elastic walls; 1 mark for relating this to high blood pressure. 1 mark for vein having a wider lumen or valves; 1 mark for relating this to lower blood pressure or preventing backflow. (c) (i) 1 mark for preventing backflow / maintaining one-way flow. (ii) 1 mark for bicuspid / mitral valve. (d) 1 mark for left ventricle pumping blood to the rest of the body / systemic circulation; 1 mark for right ventricle only pumping to the lungs / pulmonary circulation (shorter distance / lower pressure).