Typical Example Questions Explained for IGCSE OCR Science | IGCSE OCR 科学：典型例题详解

📚 Typical Example Questions Explained for IGCSE OCR Science | IGCSE OCR 科学：典型例题详解

In IGCSE OCR Science (Combined or Separate), exam success relies on mastering typical question styles. This article walks you through worked examples from Biology, Chemistry, and Physics, highlighting the key steps, common pitfalls, and examiner expectations. Understanding these will boost your confidence and help you achieve top marks.

在IGCSE OCR科学（综合或分科）考试中，掌握典型题目风格是成功的关键。本文详解生物、化学和物理的例题，展示关键步骤、常见错误和考官预期。掌握这些将提升你的信心，助你取得高分。

1. Enzyme Activity Graph Interpretation | 酶活性图解析

Enzyme-catalysed reactions show a characteristic rate vs temperature curve. At low temperatures, the kinetic energy of molecules is low, leading to fewer successful collisions between enzyme and substrate. The rate of reaction is slow.

酶催化反应呈现典型的反应速率-温度曲线。在低温下，分子动能较低，酶与底物之间的成功碰撞较少，因此反应速率缓慢。

As the temperature rises, kinetic energy increases, so more enzyme-substrate complexes form per second. The rate increases up to an optimum temperature, usually around 37 °C for human enzymes.

随着温度升高，动能增加，每秒形成的酶-底物复合物增多。反应速率上升直至最适温度，通常人体酶的最适温度在 37 °C 左右。

Beyond the optimum, the enzyme begins to denature. The weak bonds maintaining the tertiary structure break, the active site loses its specific shape, and the substrate can no longer bind. The reaction rate drops sharply to zero.

超过最适温度后，酶开始变性。维持三级结构的弱键断裂，活性中心失去特定形状，底物无法结合，反应速率急剧下降至零。

Exam questions often provide a graph and ask you to explain the shape. Always link the science: increased collisions before the optimum, denaturation after.

考试题常给出曲线并要求解释其形状。务必联系科学原理：最适温度前碰撞增加，之后变性。

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2. Genetics: Punnett Squares and Phenotypic Ratios | 遗传学：庞纳特方格与表型比例

Consider a monohybrid cross for cystic fibrosis, caused by a recessive allele f. Two heterozygous parents (Ff) mate. Set up a Punnett square to predict offspring genotypes.

以囊性纤维化为例，它是一种隐性等位基因 f 导致的疾病。两个杂合亲本（Ff）婚配，构建庞纳特方格预测子代基因型。

	F	f
F	FF	Ff
f	Ff	ff

The resulting genotypic ratio is 1 FF : 2 Ff : 1 ff. The phenotypic ratio is 3 unaffected : 1 affected. The probability of an affected child is 25%.

基因型比例是 1 FF : 2 Ff : 1 ff。表型比例为 3 正常 : 1 患病。患病孩子的概率是 25%。

Always state your ratio in the simplest form and label the phenotypes clearly. If the question asks for a percentage, multiply the probability by 100.

始终用最简形式表示比例并清楚标注表型。若题目要求百分比，将概率乘以 100。

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3. Mole Calculations Using Formula Triangles | 使用公式三角形的摩尔计算

A typical question: Calculate the mass of sodium hydroxide (NaOH) needed to prepare 250 cm³ of a 0.100 mol/dm³ solution. Molar mass M(NaOH) = 40 g/mol.

典型题目：计算配制 250 cm³ 浓度为 0.100 mol/dm³ 的氢氧化钠溶液所需的质量。摩尔质量 M(NaOH) = 40 g/mol。

First, convert the volume to dm³: V = 250 cm³ ÷ 1000 = 0.250 dm³. Use the relationship c = n / V to find the amount: n = c × V = 0.100 × 0.250 = 0.0250 mol.

首先将体积换算为 dm³：V = 250 cm³ ÷ 1000 = 0.250 dm³。利用关系式 c = n / V 求物质的量：n = c × V = 0.100 × 0.250 = 0.0250 mol。

n = c × V

Then, mass m = n × M = 0.0250 mol × 40 g/mol = 1.00 g. Always show the units and work step by step.

然后，质量 m = n × M = 0.0250 mol × 40 g/mol = 1.00 g。务必写出单位并逐步展示计算过程。

A common error is forgetting to convert cm³ to dm³, leading to an answer 1000 times too large. Use the conversion 1000 cm³ = 1 dm³.

常见错误是忘记将 cm³ 转换为 dm³，导致答案为正确值的 1000 倍。牢记 1000 cm³ = 1 dm³。

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4. Electrolysis of Aqueous Sodium Chloride: Half Equations | 氯化钠水溶液电解：半方程式

When aqueous NaCl is electrolysed using inert electrodes, several ions are present: Na⁺, Cl⁻, H⁺ and OH⁻. At the cathode, hydrogen is produced because H⁺ is reduced more easily than Na⁺.

用惰性电极电解氯化钠水溶液时，存在 Na⁺、Cl⁻、H⁺ 和 OH⁻ 等离子。在阴极产生氢气，因为 H⁺ 比 Na⁺ 更容易得电子而被还原。

The cathode half equation: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (or 2H⁺ + 2e⁻ → H₂). At the anode, chlorine is produced: 2Cl⁻ → Cl₂ + 2e⁻.

阴极半方程式：2H₂O + 2e⁻ → H₂ + 2OH⁻（或 2H⁺ + 2e⁻ → H₂）。在阳极产生氯气：2Cl⁻ → Cl₂ + 2e⁻。

Sodium ions remain in solution, forming sodium hydroxide with the OH⁻ left behind. The overall reaction is: 2NaCl + 2H₂O → H₂ + Cl₂ + 2NaOH.

钠离子留在溶液中，与剩余的 OH⁻ 形成氢氧化钠。总反应为：2NaCl + 2H₂O → H₂ + Cl₂ + 2NaOH。

Questions often ask why sodium is not formed. You must explain the reactivity series: H⁺ is a weaker reducing agent than Na⁺, so it discharges preferentially.

题目常问为何没有生成钠。你必须根据活动性顺序解释：H⁺ 的氧化性比 Na⁺ 弱，因此优先放电。

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5. Circuit Calculations: Ohm’s Law and Resistance | 电路计算：欧姆定律与电阻

Consider a circuit with a 12 V battery and two resistors in series: R₁ = 4 Ω and R₂ = 6 Ω. Find the total resistance and the current.

考虑一个由 12 V 电池和两个串联电阻（R₁ = 4 Ω，R₂ = 6 Ω）组成的电路。求总电阻和电流。

For series: R_total = R₁ + R₂ = 4 + 6 = 10 Ω. Then using Ohm’s Law, I = V / R_total = 12 V / 10 Ω = 1.2 A.

串联时：R_total = R₁ + R₂ = 4 + 6 = 10 Ω。然后由欧姆定律 I = V / R_total = 12 V / 10 Ω = 1.2 A。

V = I × R

Now connect the same resistors in parallel. Use the reciprocal formula: 1/R_total = 1/4 + 1/6 = 3/12 + 2/12 = 5/12. So R_total = 12/5 = 2.4 Ω.

现在把这两个电阻并联。使用倒数公式：1/R_total = 1/4 + 1/6 = 3/12 + 2/12 = 5/12。所以 R_total = 12/5 = 2.4 Ω。

The current from the battery is now I = 12 V / 2.4 Ω = 5 A. Show the parallel formula and the step of taking the reciprocal clearly.

此时从电池流出的电流为 I = 12 V / 2.4 Ω = 5 A。需清晰展示并联公式及取倒数的步骤。

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6. Velocity-Time Graphs and Acceleration | 速度-时间图与加速度

A velocity-time graph for a car shows: stage A – acceleration from rest to 20 m/s in 10 s; stage B – constant velocity of 20 m/s for 20 s; stage C – deceleration to rest in 5 s. Calculate the total distance travelled.

某汽车的速度-时间图显示：A 阶段 – 从静止匀加速到 20 m/s，用时 10 s；B 阶段 – 以 20 m/s 匀速行驶 20 s；C 阶段 – 匀减速至静止，用时 5 s。计算总行驶距离。

Acceleration a = (v – u) / t = (20 – 0) / 10 = 2 m/s². Distance in stage A is the area under the graph: area of triangle = ½ × base × height = ½ × 10 s × 20 m/s = 100 m.

加速度 a = (v – u) / t = (20 – 0) / 10 = 2 m/s²。A 阶段距离为图下面积：三角形面积 = ½ × 底 × 高 = ½ × 10 s × 20 m/s = 100 m。

Stage B distance = rectangle area = 20 s × 20 m/s = 400 m. Stage C distance = triangle = ½ × 5 s × 20 m/s = 50 m. Total distance = 100 + 400 + 50 = 550 m.

B 阶段距离 = 矩形面积 = 20 s × 20 m/s = 400 m。C 阶段距离 = 三角形 = ½ × 5 s × 20 m/s = 50 m。总距离 = 100 + 400 + 50 = 550 m。

Always use the area method for distance in v-t graphs, even when the graph is irregular – count squares or use trapezium formulae.

在 v-t 图中求距离始终使用面积法，即使图形不规则——可数网格或使用梯形公式。

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7. Data Analysis: Drawing Lines of Best Fit | 数据分析：绘制最佳拟合线

In a practical on Hooke’s Law, a student records force (F) and extension (e). After plotting the points, they must draw a line of best fit – a single straight line that passes as close to all points as possible, not dot-to-dot.

在胡克定律实验中，学生记录力（F）与伸长量（e）。描点后必须画出最佳拟合线——一条尽可能靠近所有点的单一直线，而非逐点连接。

If one point lies far off the line, it may be anomalous. Circle it and do not use it when drawing the line. The gradient of the line gives the spring constant k (since F = k e), so k = ΔF / Δe.

如果某个点明显偏离直线，可能是异常点。把它圈起来，画线时不使用它。直线的斜率给出弹簧常数 k（因为 F = k e），所以 k = ΔF / Δe。

Always use a sharp pencil and a ruler. Label the axes with quantities and units, and choose an appropriate scale that uses at least half the graph grid.

始终用削尖的铅笔和直尺作图。标明坐标轴物理量和单位，选取合适的标度使图形至少占网格纸的一半。

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8. Planning Investigations: Variables and Controls | 实验设计：变量与控制

Design an experiment to investigate how temperature affects the rate of reaction between magnesium ribbon and dilute hydrochloric acid. The independent variable is temperature; the dependent variable is the time taken for the magnesium to disappear (or volume of gas produced per minute).

设计实验探究温度如何影响镁条与稀盐酸的反应速率。自变量是温度；因变量是镁条消失所需的时间（或每分钟产生气体的体积）。

Control variables: length (mass) of magnesium ribbon, concentration and volume of acid, surface area of magnesium (same ribbon width). Use a water bath to reach and maintain the desired temperatures: 20 °C, 30 °C, 40 °C, 50 °C, 60 °C.

控制变量：镁条的长度（

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