📚 Typical IGCSE Biology Exam Questions with Detailed Solutions | IGCSE 生物典型例题详解
This article walks through a set of typical IGCSE Biology exam questions, covering core topics from cell structure to ecology. Each example is presented with a detailed solution, highlighting key marking points and common pitfalls. Use these worked examples to sharpen your exam technique and deepen your understanding of the syllabus.
本文精选了一组 IGCSE 生物典型考题，涵盖细胞结构到生态学等核心主题。每道例题都配有详细解答，突出了关键得分点与常见易错处。通过这些典型例题的剖析，帮助你提升解题技巧、加深对考纲内容的理解。
1. Identifying Cell Structures | 识别细胞结构
A student observes a cell under a light microscope and labels three structures: X is a large, spherical organelle near the centre. Y is a small, rod-shaped organelle found in large numbers. Z is the thin outer boundary of the cell. Name X, Y and Z and state one function of each. [3 marks]
一名学生在光学显微镜下观察细胞，标注了三个结构：X 是靠近中央的一个大型球形细胞器；Y 是数量众多的小型杆状细胞器；Z 是细胞外部的薄层边界。请写出 X、Y 和 Z 的名称，并各写出一个功能。[3 分]
X is the nucleus. It contains genetic material (DNA) and controls the activities of the cell. Y is a mitochondrion (plural mitochondria). It is the site of aerobic respiration, releasing energy for cell activities. Z is the cell membrane. It controls the movement of substances into and out of the cell, acting as a selectively permeable barrier.
X 是细胞核。它含有遗传物质（DNA），并控制细胞的活动。Y 是线粒体。它是有氧呼吸的场所，为细胞活动释放能量。Z 是细胞膜。它控制物质进出细胞，起选择性渗透屏障的作用。
2. Osmosis Experiment with Potato | 土豆条渗透实验
Potato cylinders of equal initial mass (5.0 g) were placed in distilled water, 0.5 mol/dm³ sucrose solution and 1.0 mol/dm³ sucrose solution for 30 minutes. After blotting, the masses were measured: distilled water +15%, 0.5 mol/dm³ -6%, 1.0 mol/dm³ -14%. Explain these results in terms of water potential. [4 marks]
将初始质量相等（5.0 g）的土豆条分别放入蒸馏水、0.5 mol/dm³ 蔗糖溶液和 1.0 mol/dm³ 蔗糖溶液中，浸泡 30 分钟后吸干称重：蒸馏水 +15%，0.5 mol/dm³ -6%，1.0 mol/dm³ -14%。请从水势的角度解释这些结果。[4 分]
Distilled water has a higher water potential than the cytoplasm of the potato cells, so water enters the cells by osmosis. The cells become turgid, and the tissue swells, increasing mass. The 0.5 mol/dm³ sucrose solution has a lower water potential than the cells, causing water to leave the cells by osmosis. The cells become flaccid, and the tissue shrinks, decreasing mass. The 1.0 mol/dm³ solution has an even lower water potential, so more water is drawn out, resulting in a greater mass loss. The cell wall prevents the plant cells from bursting in distilled water but allows plasmolysis in concentrated sugar solution.
蒸馏水的水势高于土豆细胞质，水分通过渗透作用进入细胞。细胞变得饱满（质壁分离复原），组织膨胀，质量增加。0.5 mol/dm³ 蔗糖溶液的水势低于细胞，水分通过渗透作用离开细胞。细胞萎蔫，组织收缩，质量减少。1.0 mol/dm³ 溶液水势更低，更多水分被吸出，质量损失更大。细胞壁可防止细胞在蒸馏水中涨破，但在浓糖溶液中可以发生质壁分离。
3. Effect of Temperature on Enzyme Activity | 温度对酶活性的影响
An enzyme-catalysed reaction was carried out at different temperatures. The rate of reaction increased up to 40°C, but above 45°C the rate dropped sharply. Explain these observations. [3 marks]
某酶促反应在不同温度下进行。反应速率在 40°C 之前逐渐增加，但在 45°C 以上急剧下降。请解释观察到的现象。[3 分]
As temperature increases up to the optimum (around 40°C), the kinetic energy of enzyme and substrate molecules increases. They collide more frequently and with more energy, so more enzyme-substrate complexes form and the rate of reaction rises. Above 45°C, the high temperature causes the enzyme to denature. The shape of the enzyme’s active site is permanently altered, so the substrate can no longer fit. The enzyme loses its catalytic function, causing the rate to fall dramatically.
温度升高至最适温度（约 40°C）时，酶和底物分子的动能增加。它们碰撞更频繁且能量更高，形成更多的酶–底物复合物，因此反应速率上升。当温度超过 45°C，高温使酶变性。酶的活性位点形状发生永久性改变，底物无法再与之契合。酶失去催化功能，导致速率急剧下降。
4. Testing a Leaf for Starch | 淀粉叶片测试
Describe the steps you would take to test a variegated leaf for the presence of starch, and explain the purpose of each step. [4 marks]
请描述检测杂色叶片中是否存在淀粉的实验步骤，并解释每一步的目的。[4 分]
- Boil the leaf in water for about 2 minutes to kill the tissue and stop all chemical reactions.
- 将叶片在沸水中煮约 2 分钟，杀死组织并停止所有化学反应。
- Place the leaf in a boiling tube of ethanol and heat in a water bath. The ethanol removes chlorophyll, decolourising the leaf.
- 将叶片放入盛有乙醇的沸腾管中，水浴加热。乙醇去除叶绿素，使叶片脱色。
- Rinse the leaf in warm water to remove the ethanol and soften the leaf.
- 用温水冲洗叶片，洗去乙醇并使叶片变软。
- Spread the leaf on a white tile and add a few drops of iodine solution. Areas that turn blue-black contain starch; yellow-brown areas lack starch.
- 将叶片平铺在白色瓷板上，滴加数滴碘液。变为蓝黑色的区域含有淀粉，呈黄褐色的区域不含淀粉。
Safety note: ethanol is flammable, so it must be heated in a water bath away from open flames.
安全提示：乙醇易燃，因此必须使用水浴加热，并远离明火。
5. Food Test for Reducing Sugars | 还原糖食物测试
A student wants to test whether an unknown solution contains a reducing sugar such as glucose. Describe the test and the expected positive result. [2 marks]
某学生想检测一种未知溶液中是否含有还原糖（如葡萄糖）。请描述测试方法及预期阳性结果。[2 分]
Add an equal volume of Benedict’s solution to the sample and heat the mixture in a boiling water bath for 2–3 minutes. A positive result is a colour change from blue to green, yellow or brick-red, depending on the concentration of reducing sugar.
向样品中加入等体积的本尼迪克特溶液，并在沸水浴中加热 2–3 分钟。阳性结果是颜色由蓝变为绿、黄或砖红色，取决于还原糖的浓度。
6. Digestive Enzymes and Their Actions | 消化酶及其作用
Complete the table to show the enzyme, substrate, product and site of action for each digestive process. [4 marks]
完成表格，写出以下消化过程所涉及的酶、底物、产物和作用部位。[4 分]
| Enzyme / 酶 | Substrate / 底物 | Products / 产物 | Site / 作用部位 |
|---|---|---|---|
| Amylase / 淀粉酶 | Starch / 淀粉 | Maltose / 麦芽糖 | Mouth, small intestine / 口腔、小肠 |
| Protease / 蛋白酶 | Protein / 蛋白质 | Amino acids / 氨基酸 | Stomach, small intestine / 胃、小肠 |
| Lipase / 脂肪酶 | Lipids (fats) / 脂类 | Fatty acids and glycerol / 脂肪酸和甘油 | Small intestine / 小肠 |
This table summarises the key enzymes in chemical digestion. Amylase breaks starch into maltose, protease breaks proteins into amino acids, and lipase digests lipids into fatty acids and glycerol. Bile emulsifies fats to increase the surface area for lipase action but does not contain enzymes.
该表总结了化学消化中的关键酶。淀粉酶将淀粉分解为麦芽糖，蛋白酶将蛋白质分解为氨基酸，脂肪酶将脂类消化成脂肪酸和甘油。胆汁乳化脂肪以增大脂肪酶作用的表面积，但胆汁本身不含酶。
7. Structure of the Heart | 心脏结构
On a diagram of the heart, chambers A, B, C and D are labelled. A is the right atrium, B is the right ventricle, C is the left atrium and D is the left ventricle. State one function of the valves between the atria and ventricles, and name the blood vessel that carries blood from the left ventricle to the body. [2 marks]
在心脏示意图上，标注了腔室 A、B、C 和 D。A 为右心房，B 为右心室，C 为左心房，D 为左心室。请写出位于心房和心室之间瓣膜的一个功能，并说出将血液从左心室运往全身的血管名称。[2 分]
The atrioventricular valves (tricuspid on the right, bicuspid/mitral on the left) prevent backflow of blood from the ventricles into the atria when the ventricles contract. The blood vessel that carries oxygenated blood from the left ventricle to the body is the aorta.
房室瓣（右侧三尖瓣、左侧二尖瓣）可防止心室收缩时血液从心室倒流回心房。将含氧血液从左心室输送到全身的血管是主动脉。
8. Aerobic Respiration Equation and Experiments | 有氧呼吸方程式及实验
In a school investigation, germinating seeds were placed in a flask connected to a test tube of limewater. Air was drawn through the apparatus. After several hours the limewater turned milky. Write the balanced word and chemical equations for aerobic respiration, and explain what caused the limewater to change. [3 marks]
在一项学校探究实验中，将萌发种子放入一个连接着石灰水试管的烧瓶里，抽吸空气通过装置。几小时后石灰水变浑浊。请写出有氧呼吸的文字和化学方程式，并解释石灰水变化的原因。[3 分]
Word equation: glucose + oxygen → carbon dioxide + water (+ energy)
文字方程式：葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O (+ energy)
The germinating seeds respire aerobically, producing carbon dioxide. The CO₂ bubbles through the limewater and reacts with calcium hydroxide to form insoluble calcium carbonate, which turns the limewater milky.
萌发种子进行有氧呼吸，产生二氧化碳。CO₂ 气泡通过石灰水，与氢氧化钙反应生成不溶性碳酸钙，使石灰水变浑浊。
9. Kidney Filtration and Reabsorption | 肾脏过滤与重吸收
Describe how the nephron filters blood and then reabsorbs useful substances. Name the process that occurs in the glomerulus and state two substances that are reabsorbed in the proximal convoluted tubule. [4 marks]
描述肾单位如何过滤血液并重吸收有用物质。写出肾小球中发生的过程名称，并说出在近曲小管中被重吸收的两种物质。[4 分]
In the glomerulus, high blood pressure forces small molecules such as water, glucose, urea and salts out of the blood into the Bowman’s capsule. This is called ultrafiltration. Larger molecules like proteins and blood cells remain in the blood. As the filtrate passes along the nephron, useful substances are reabsorbed back into the blood. In the proximal convoluted tubule, all glucose and most of the water and salts are reabsorbed by active transport and osmosis. In the loop of Henle and collecting duct, further water reabsorption occurs under hormonal control.
在肾小球中，高动脉血压将水、葡萄糖、尿素和盐等小分子物质从血液压入鲍曼囊，这一过程称为超滤。较大的分子如蛋白质和血细胞保留在血液中。当滤液沿肾单位流动时，有用物质被重吸收回血液。在近曲小管，所有葡萄糖以及大部分水和盐通过主动转运和渗透作用被重吸收。在髓袢和集合管中，在激素调控下进一步重吸收水。
10. Monohybrid Inheritance in Flowers | 单基因花色遗传
In pea plants, the allele for red flowers (R) is dominant to the allele for white flowers (r). A heterozygous red-flowered plant is crossed with a white-flowered plant. Using a genetic diagram, predict the ratio of phenotypes in the offspring. [3 marks]
在豌豆植物中，红花等位基因（R）对白花等位基因（r）为显性。将一株杂合红花植株与一株白花植株杂交。请用遗传图解预测后代的表现型比率。[3 分]
Parental genotypes: Rr (heterozygous red) × rr (white). Gametes from Rr: R and r. Gametes from rr: all r. Offspring genotypes: 50% Rr (red) and 50% rr (white). The phenotypic ratio is 1 red : 1 white.
亲本基因型：Rr（杂合红花）× rr（白花）。Rr 产生的配子：R 和 r。rr 产生的配子：全部为 r。后代基因型：50% Rr（红花）和 50% rr（白花）。表现型比例为 1 红花 : 1 白花。
11. Evolution of Antibiotic Resistance | 抗生素耐药性进化
Explain how a population of bacteria can become resistant to an antibiotic, using the principles of natural selection. [4 marks]
请运用自然选择的原理，解释一个细菌种群如何对抗生素产生耐药性。[4 分]
Within a bacterial population, there is genetic variation. A few individuals may possess a mutation that makes them less susceptible to an antibiotic. When the antibiotic is applied, most non-resistant bacteria are killed, but the resistant ones survive. These resistant bacteria reproduce, passing the resistance gene to their offspring. Over many generations, the frequency of the resistance allele increases in the population, so the antibiotic becomes ineffective. This is an example of evolution by natural selection.
在细菌种群中存在遗传变异。少数个体可能携带一个突变，使其对抗生素不那么敏感。当使用抗生素时，大多数非耐药细菌被杀死，但耐药细菌存活下来。这些耐药细菌繁殖，将耐药基因传递给后代。经过多代，耐药等位基因的频率在种群中增加，因此抗生素变得无效。这是自然选择进化的一个例子。
12. Energy Flow and Pyramids of Biomass | 能量流动与生物量金字塔
A pyramid of biomass for a grassland food chain is drawn, showing grass → grasshopper → frog → snake. Explain why the biomass decreases at each successive trophic level. [3 marks]
草地食物链的生物量金字塔绘制为：草 → 蚱蜢 → 青蛙 → 蛇。请解释为什么生物量在每一个连续的营养级上会减少。[3 分]
At each trophic level, organisms use energy for life processes such as respiration, movement and growth. Much of the energy is lost as heat to the environment. Only a small proportion of the biomass consumed is converted into new tissue and becomes available to the next trophic level. Additionally, not all parts of the prey are eaten or digested; some material is lost in faeces. As a result, the biomass becomes progressively smaller from producers to top consumers.
在每个营养级，生物体将能量用于呼吸、运动、生长等生命过程。大量能量以热的形式散失到环境中。只有一小部分被消费的生物量转化为新的组织，并可供下一营养级使用。此外，并非猎物所有部分都被吃掉或消化，部分物质随粪便排出。因此，从生产者到顶级消费者的生物量逐渐减少。
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