📚 GCSE Edexcel Science: Typical Example Questions Explained | GCSE Edexcel 科学：典型例题详解
This article walks through common GCSE Edexcel Science example questions across Biology, Chemistry and Physics. For each worked example, you will find the question, step-by-step solution and key exam tips. Every explanation is given in both English and Chinese to support bilingual learners, ensuring you understand the core concepts and can apply them confidently in your exams.
本文梳理了 GCSE Edexcel 科学考试中生物、化学和物理的典型例题。每道题都配有题目、分步解答和关键应试提示，且所有讲解均提供中英双语对照，帮助双语学习者透彻理解核心概念，在考试中自信应对。
1. Osmosis and Potato Cylinders | 渗透作用与土豆条实验
A student placed potato cylinders of equal initial mass into different concentrations of sucrose solution. After 30 minutes, the cylinders were reweighed. The table shows the results.
一名学生将初始质量相等的土豆条放入不同浓度的蔗糖溶液中。30分钟后重新称量，结果如下表。
| Sucrose concentration (mol/dm³) | Initial mass (g) | Final mass (g) |
|---|---|---|
| 0.0 | 2.00 | 2.20 |
| 0.2 | 2.00 | 2.08 |
| 0.4 | 2.00 | 1.95 |
| 0.6 | 2.00 | 1.80 |
| 0.8 | 2.00 | 1.70 |
(a) Calculate the percentage change in mass for the cylinder in 0.4 mol/dm³ sucrose solution. Give your answer to 2 significant figures.
(a) 计算在 0.4 mol/dm³ 蔗糖溶液中土豆条的质量变化百分比，结果保留两位有效数字。
Step 1: Write the formula. Percentage change = (final mass – initial mass) / initial mass × 100.
步骤一：写出公式。质量变化百分比 = (最终质量 – 初始质量) ÷ 初始质量 × 100。
Step 2: Substitute the values: (1.95 – 2.00) / 2.00 × 100 = –0.05 / 2.00 × 100 = –2.5%. The negative sign shows a decrease in mass.
步骤二：代入数值：(1.95 – 2.00) ÷ 2.00 × 100 = –0.05 ÷ 2.00 × 100 = –2.5%。负号表示质量减少。
(b) Explain which solutions caused water to enter the potato cells and why.
(b) 解释哪些溶液使水进入土豆细胞并说明原因。
The 0.0 and 0.2 mol/dm³ solutions led to an increase in mass, showing water entered the cells. This is because the water potential of the solution was higher (less negative) than that of the potato cytoplasm; water moved by osmosis down the water potential gradient into the cells.
0.0 和 0.2 mol/dm³溶液使质量增加，说明水进入了细胞。这是因为溶液的 water potential (水势) 高于土豆细胞质，水通过渗透作用顺着水势梯度进入细胞。
Exam tip: Always include the ‘–’ sign when mass decreases and mention osmosis and water potential gradient for full marks.
应试提示：质量减少时务必标上负号，并提到渗透作用和水势梯度才能拿满分。
2. Enzyme Activity and Rate of Reaction | 酶活性与反应速率
A student investigated the effect of pH on amylase activity. Five test tubes containing starch and amylase were buffered at different pH values. The time taken for the starch to be completely digested (iodine test no longer turns blue-black) was recorded. The rate of reaction was calculated as 1/time.
一名学生研究了 pH 对淀粉酶活性的影响。五支含淀粉和淀粉酶的试管用不同 pH 缓冲液处理，记录淀粉完全消化所需的时间（碘液不再变蓝黑），并以 1/时间 计算反应速率。
| pH | Time (s) | Rate (s⁻¹) |
|---|---|---|
| 4 | 120 | 0.0083 |
| 5 | 60 | 0.0167 |
| 6 | 30 | 0.0333 |
| 7 | 20 | 0.0500 |
| 8 | 25 | 0.0400 |
| 9 | 45 | 0.0222 |
(a) Using the data, state the optimum pH for amylase and explain your choice.
(a) 根据数据，说出淀粉酶的最适 pH 并解释你的选择。
The optimum pH is 7 because the rate of reaction is highest (0.0500 s⁻¹) at this pH. The enzyme works fastest when its active site has the best complementary shape for the substrate.
最适 pH 为 7，因为在该 pH 下反应速率最高（0.0500 s⁻¹）。酶的活性部位与底物形状最互补时，酶工作最快。
(b) Explain why the rate decreases at pH 9.
(b) 解释为何在 pH 9 时速率下降。
At pH 9, the high alkalinity disrupts the ionic and hydrogen bonds that hold the enzyme’s tertiary structure. The active site changes shape (denatures), so the starch substrate no longer fits, reducing the rate.
在 pH 9 时，强碱性破坏了维持酶三级结构的离子键和氢键，活性部位形状改变（变性），淀粉底物不再契合，速率下降。
3. Genetics: Monohybrid Cross and Punnett Square | 遗传学：单基因杂交与庞纳特方格
In pea plants, the allele for tall stems (T) is dominant to the allele for dwarf stems (t). A heterozygous tall plant is crossed with a dwarf plant. Predict the expected phenotype ratio of the offspring.
在豌豆中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。将一株杂合高茎植株与一株矮茎植株杂交，预测后代的表型比例。
Step 1: Write the parental genotypes. Heterozygous tall = Tt, dwarf = tt.
步骤一：写出亲本基因型。杂合高茎 = Tt，矮茎 = tt。
Step 2: Determine gametes. Tt produces gametes T and t; tt produces only t.
步骤二：确定配子。Tt 产生配子 T 和 t；tt 只产生配子 t。
Step 3: Draw the Punnett square.
步骤三：画出庞纳特方格。
| t | t | |
| T | Tt | Tt |
| t | tt | tt |
Step 4: Offspring genotypes are Tt and tt in equal proportions. Phenotypes: Tt = tall, tt = dwarf. Phenotype ratio = tall : dwarf = 1 : 1.
步骤四：后代基因型为 Tt 和 tt，各占一半。表型：Tt 为高茎，tt 为矮茎。表型比例 高茎 : 矮茎 = 1 : 1。
Key concept: A test cross with a homozygous recessive individual reveals the unknown genotype of a dominant phenotype.
核心概念：与隐性纯合子进行的测交可以揭示显性表型个体的未知基因型。
4. Moles and Mass Calculations in Reactions | 化学反应中的摩尔与质量计算
Magnesium burns in oxygen to form magnesium oxide. The balanced equation is:
镁在氧气中燃烧生成氧化镁，配平方程式为：
2Mg + O₂ → 2MgO
Calculate the mass of magnesium oxide produced when 2.4 g of magnesium reacts completely with excess oxygen. (Relative atomic masses: Mg = 24, O = 16)
计算 2.4 g 镁与过量氧气完全反应后生成的氧化镁质量。（相对原子质量：Mg = 24，O = 16）
Step 1: Calculate moles of Mg. n(Mg) = mass / molar mass = 2.4 g / 24 g/mol = 0.10 mol.
步骤一：计算镁的物质的量。n(Mg) = 质量 ÷ 摩尔质量 = 2.4 g ÷ 24 g/mol = 0.10 mol。
Step 2: Use the mole ratio from the equation. 2 moles Mg produce 2 moles MgO, so moles of MgO = moles of Mg = 0.10 mol.
步骤二：利用方程式的摩尔比。2 mol Mg 生成 2 mol MgO，因此 MgO 的物质的量 = 0.10 mol。
Step 3: Molar mass of MgO = 24 + 16 = 40 g/mol.
步骤三：MgO 的摩尔质量 = 24 + 16 = 40 g/mol。
Step 4: Mass of MgO = moles × molar mass = 0.10 mol × 40 g/mol = 4.0 g.
步骤四：MgO 的质量 = 物质的量 × 摩尔质量 = 0.10 mol × 40 g/mol = 4.0 g。
Common pitfall: Remember that the mass of oxygen reactant is not simply added — you must work through moles and the balanced equation.
常见错误：不要直接将氧气的质量加进去，必须通过物质的量和配平方程式来计算。
5. Electrolysis of Aqueous Sodium Chloride | 氯化钠溶液的电解
Concentrated aqueous sodium chloride is electrolysed using inert electrodes. Describe the products formed at each electrode and name the substance left in solution.
用惰性电极电解浓氯化钠溶液。描述每个电极上的生成物，并说出留在溶液中的物质名称。
At the cathode (negative electrode): Hydrogen gas is produced. The electrolyte contains H⁺ ions from water and Na⁺ ions. Hydrogen ions are discharged more readily than sodium ions: 2H⁺ + 2e⁻ → H₂. The competing reduction is Na⁺ + e⁻ → Na, but hydrogen is produced because it is lower in the reactivity series.
在阴极（负极）：产生氢气。电解液中含有来自水的 H⁺ 和 Na⁺，氢离子比钠离子更易放电：2H⁺ + 2e⁻ → H₂。竞争反应为 Na⁺ + e⁻ → Na，但由于氢在活动性顺序中更低，优先放电。
At the anode (positive electrode): Chlorine gas is produced. Chloride ions are oxidised: 2Cl⁻ → Cl₂ + 2e⁻. Although OH⁻ ions from water are also present, in concentrated NaCl solution Cl⁻ is preferentially discharged because its concentration is high.
在阳极（正极）：产生氯气。氯离子被氧化：2Cl⁻ → Cl₂ + 2e⁻。尽管水中还有 OH⁻，但在浓氯化钠溶液中，由于 Cl⁻ 浓度高，优先放电。
The ions left in solution are Na⁺ and OH⁻, so sodium hydroxide (NaOH) is formed.
留在溶液中的离子是 Na⁺ 和 OH⁻，因此生成氢氧化钠 (NaOH)。
Summary: Cathode → hydrogen, Anode → chlorine, Solution → sodium hydroxide.
总结：阴极 → 氢气，阳极 → 氯气，溶液 → 氢氧化钠。
6. Ohm’s Law and Resistors in Series | 欧姆定律与串联电阻
A circuit consists of a 12 V battery and two resistors of 4 Ω and 6 Ω connected in series. Calculate:
一个电路由 12 V 电池和两个串联的电阻（4 Ω 和 6 Ω）组成。计算：
(a) the total resistance in the circuit.
(a) 电路的总电阻。
For series resistors, Rₜ = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω. (Using Unicode 下标: Rₜ = R₁ + R₂)
串联电阻的总阻值 Rₜ = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω。
(b) the current flowing through the circuit.
(b) 流过电路的电流。
Using Ohm’s Law: I = V / Rₜ = 12 V / 10 Ω = 1.2 A.
根据欧姆定律：I = V / Rₜ = 12 V ÷ 10 Ω = 1.2 A。
(c) the potential difference across each resistor.
(c) 每个电阻两端的电压。
V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V. V₂ = I × R₂ = 1.2 A × 6 Ω = 7.2 V. Check: 4.8 V + 7.2 V = 12 V, matching the battery.
V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V；V₂ = 1.2 A × 6 Ω = 7.2 V。验证：4.8 V + 7.2 V = 12 V，等于电池电压。
Quick tip: In series, current is the same everywhere; the p.d. splits in proportion to resistance.
快速提示：串联电路中电流处处相等；电压按电阻比例分配。
7. Kinetic Energy and Gravitational Potential Energy | 动能与重力势能
Published by TutorHao | GCSE Science Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply