GCSE CIE Biology: Typical Exam Questions Explained | GCSE CIE 生物：典型例题详解

📚 GCSE CIE Biology: Typical Exam Questions Explained | GCSE CIE 生物：典型例题详解

In CIE GCSE Biology, understanding core concepts is only half the battle – being able to apply them to exam-style questions is what earns top marks. This article walks you through ten typical questions that appear regularly on CIE papers, breaking down the key points, common pitfalls and model answers. Each question is presented in English and Chinese, with detailed explanations paired in both languages to help you build confidence and accuracy.

在 CIE GCSE 生物考试中，理解核心概念只是成功的一半——能够将这些概念应用到考试题型中才是取得高分的关键。本文带你逐一分析 CIE 试卷中频繁出现的十道典型例题，拆解关键知识点、常见错误和标准答案。每道题均以中英双语呈现，并配有详细的中英对照解析，帮助你提升信心和答题准确度。

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1. Enzyme Activity and Temperature | 酶活性与温度

Question: The graph below shows the activity of an enzyme at different temperatures. At 20°C the activity is low, it peaks at 40°C, and then falls sharply to zero by 60°C. Explain why the enzyme activity decreases above 40°C. (3 marks)

问题：下图显示了某酶在不同温度下的活性。20°C 时活性较低，40°C 时达到峰值，然后急剧下降，60°C 时降为零。解释为什么酶活性在 40°C 以上会下降。（3 分）

Enzymes are proteins with a specific three-dimensional shape. The active site is held in place by weak bonds, including hydrogen bonds. As temperature rises beyond the optimum, these bonds begin to break, causing the active site to lose its complementary shape – this is called denaturation. The substrate can no longer bind, so the rate of reaction drops. At 60°C denaturation is complete and the enzyme is permanently inactive.

酶是具有特定三维结构的蛋白质。活性位点由弱键（包括氢键）维持。当温度超过最适温度时，这些键开始断裂，导致活性位点失去与底物互补的形状——这称为变性。底物无法再结合，因此反应速率下降。60°C 时变性完全，酶永久失活。

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2. Osmosis in Plant Cells | 植物细胞中的渗透作用

Question: A piece of potato is placed in a concentrated sugar solution. After 30 minutes the potato strip becomes soft and floppy. Use your knowledge of osmosis to explain this observation. (4 marks)

问题：将一块土豆放入浓糖溶液中。30 分钟后，土豆条变得柔软松弛。运用渗透作用的知识解释这一现象。（4 分）

The sugar solution has a lower water potential than the potato cells. Water moves by osmosis from a region of higher water potential (inside the cells) to a region of lower water potential (the external solution) through a partially permeable membrane. As water leaves the cells, the vacuole shrinks and the cytoplasm pulls away from the cell wall – the cells become plasmolysed. This loss of turgor pressure makes the tissue soft.

糖溶液的水势低于土豆细胞内的水势。水通过部分透性膜，由水势较高的区域（细胞内）渗透移动到水势较低的区域（外部溶液）。随着水分流失，液泡缩小，细胞质与细胞壁分离——细胞发生质壁分离。膨压的丧失使组织变软。

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3. Limiting Factors of Photosynthesis | 光合作用的限制因素

Question: A farmer grows tomatoes in a glasshouse. In winter she adds extra carbon dioxide and heats the glasshouse. Explain how each of these changes increases the rate of photosynthesis. (4 marks)

问题：一位农民在温室中种植番茄。冬季她额外补充二氧化碳并给温室加温。解释这两种改变如何分别提高光合作用速率。（4 分）

Carbon dioxide is a raw material for photosynthesis; it is fixed in the Calvin cycle to produce glucose. Increasing CO₂ concentration provides more substrate for the enzyme RuBisCO, raising the rate of the light-independent reactions as long as light and temperature are not limiting. Raising the temperature increases the kinetic energy of molecules, so enzyme-catalysed reactions proceed faster. However, the farmer must avoid overheating because enzymes denature at very high temperatures.

二氧化碳是光合作用的原料；它在卡尔文循环中被固定以合成葡萄糖。提高 CO₂ 浓度可为酶 RuBisCO 提供更多底物，在光照和温度不受限制的情况下，能提升暗反应速率。升温增加了分子的动能，使得酶促反应更快进行。但农民必须避免过热，因为酶在过高温下会变性。

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4. Comparing Aerobic and Anaerobic Respiration | 有氧呼吸与无氧呼吸的比较

Question: Compare aerobic and anaerobic respiration in animal cells. In your answer, include the products and the relative yield of ATP. (5 marks)

问题：比较动物细胞中有氧呼吸和无氧呼吸，包括产物和 ATP 的相对产量。（5 分）

Aerobic respiration uses oxygen and fully oxidises glucose to carbon dioxide and water. It produces a large amount of ATP – around 36-38 molecules per glucose. Anaerobic respiration occurs when oxygen is absent; in animal cells it converts glucose into lactic acid. It yields only 2 ATP molecules per glucose. Both processes begin with glycolysis in the cytoplasm, but only aerobic respiration continues with the Krebs cycle and the electron transport chain inside mitochondria.

有氧呼吸利用氧气并将葡萄糖完全氧化为二氧化碳和水。它产生大量 ATP——每个葡萄糖分子约 36-38 个。无氧呼吸在缺氧时进行；在动物细胞中将葡萄糖转化为乳酸，每个葡萄糖仅产生 2 个 ATP。两种过程都始于细胞质中的糖酵解，但只有有氧呼吸会在线粒体内继续进行克雷布斯循环和电子传递链。

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5. The Double Circulatory System | 双循环系统

Question: Describe the advantages of a double circulatory system in mammals. (3 marks)

问题：描述哺乳动物双循环系统的优势。（3 分）

Blood passes through the heart twice in one complete circuit: the right side pumps deoxygenated blood to the lungs (pulmonary circulation), and the left side pumps oxygenated blood to the rest of the body (systemic circulation). This separation allows the systemic circulation to operate at a much higher pressure, delivering oxygen and nutrients quickly to respiring tissues. Low pressure in the pulmonary circulation protects the delicate lung capillaries and allows efficient gas exchange without damage.

血液在一次完整的循环中两次经过心脏：右侧将缺氧血泵至肺部（肺循环），左侧将富氧血泵至全身（体循环）。这种分离使得体循环能够在较高压力下运行，快速向呼吸组织输送氧气和营养。肺循环的低压保护了脆弱的肺毛细血管，能够在无损的情况下进行高效的气体交换。

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6. The Transpiration Stream | 蒸腾流

Question: Explain how water moves from the roots to the leaves of a plant. Use the terms transpiration pull, cohesion and adhesion in your answer. (5 marks)

问题：解释水如何从植物的根部移动到叶片。请在你的答案中使用“蒸腾拉力”、“内聚力”和“附着力”。（5 分）

Water evaporates from the mesophyll cells into the air spaces of the leaf and diffuses out through the stomata – this is transpiration. The loss of water creates a negative pressure (tension) at the leaf surface, which pulls more water up the xylem – the transpiration pull. Water molecules are cohesive due to hydrogen bonding, so the column of water in the xylem stays unbroken. Adhesion of water molecules to the xylem walls also helps to pull the water column upwards against gravity. This continuous stream delivers water and dissolved mineral ions from the roots.

水从叶肉细胞蒸发进入叶片气腔，并通过气孔扩散出去——这就是蒸腾作用。水分的流失在叶片表面产生负压（张力），从而将更多的水通过木质部向上拉——蒸腾拉力。水分子由于氢键而具有内聚力，木质部中的水柱因此保持连续不断。水分子对木质部管壁的附着力也有助于将水柱克服重力向上拉。这一连续的水流将水和溶解的矿质离子从根部输送到叶片。

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7. The Reflex Arc | 反射弧

Question: A person accidentally touches a hot pan and immediately pulls their hand away. Describe the pathway of the nerve impulse from the receptor to the effector. (4 marks)

问题：一个人不小心碰到了热锅，手立即缩回。描述神经冲动从感受器到效应器的路径。（4 分）

The heat stimulus is detected by a thermoreceptor in the skin. A sensory neurone carries the impulse to the spinal cord. Inside the spinal cord, the sensory neurone synapses with a relay neurone (also called an internuncial neurone). The relay neurone then passes the impulse to a motor neurone. The motor neurone transmits the impulse to an effector – the biceps muscle, which contracts to pull the hand away. This pathway is a reflex arc; it does not involve the conscious brain, allowing a rapid, automatic response.

高温刺激由皮肤中的温度感受器检测。感觉神经元将冲动传入脊髓。在脊髓内，感觉神经元与中间神经元（也称联络神经元）形成突触。中间神经元再将冲动传递给运动神经元。运动神经元将冲动传至效应器——肱二头肌，使其收缩，手缩回。这条路径就是反射弧；它不经过有意识的大脑参与，从而实现快速、自动的反应。

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8. Hormonal Control of Blood Glucose | 血糖的激素调控

Question: After eating a meal rich in carbohydrates, blood glucose concentration rises. Explain how the hormones insulin and glucagon restore the glucose level to normal. (5 marks)

问题：吃完富含碳水化合物的饭后，血糖浓度升高。解释胰岛素和胰高血糖素如何将血糖恢复至正常水平。（5 分）

High blood glucose is detected by the β cells of the islets of Langerhans in the pancreas. They release insulin into the blood. Insulin stimulates the liver and muscle cells to take up glucose and convert it into glycogen for storage; it also increases cellular respiration. As a result, blood glucose falls. When blood glucose drops too low, α cells in the pancreas secrete glucagon. Glucagon prompts the liver to break down glycogen back into glucose and release it into the blood. This negative feedback loop keeps the glucose concentration within a narrow range.

高血糖被胰腺胰岛中的 β 细胞检测到，它们向血液中释放胰岛素。胰岛素刺激肝细胞和肌肉细胞摄取葡萄糖并将其转化为糖原储存；它还能促进细胞呼吸。结果血糖下降。当血糖过低时，胰腺中的 α 细胞分泌胰高血糖素。胰高血糖素促使肝脏将糖原分解回葡萄糖并释放到血液中。这一负反馈回路使血糖浓度维持在狭窄的范围内。

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9. Monohybrid Inheritance | 单基因杂交遗传

Question: In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). A homozygous tall plant is crossed with a short plant. Use a genetic diagram to predict the phenotype ratio of the F2 generation when two F1 offspring are crossed. (4 marks)

问题：在豌豆中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。将一株纯合高茎与一株矮茎杂交。使用遗传图解预测，当两株 F1 后代杂交时 F2 代的表型比例。（4 分）

etes: T and t from each parent. A Punnett square shows offspring genotypes: TT, Tt, Tt, tt. The phenotype ratio is 3 tall : 1 short. This demonstrates Mendel' ecause each parent contributes one allele, and the recessive allele can be inherited from both parents.

亲本杂交：TT × tt → F1 全部为 Tt（高）。F1 杂交：Tt × Tt。可能的配子：每个亲本产生 T 和 t。庞纳特方格显示后代基因型：TT, Tt, Tt, tt。表型比例为 3 高 : 1 矮。这体现了孟德尔的分离定律；隐性性状在 F2 代重新出现，因为每个亲本提供一个等位基因，隐性等位基因可以从双亲遗传。

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10. Energy Flow in Food Chains | 食物链中的能量流动

Question: In a grassland ecosystem, only about 10% of the energy is transferred from one trophic level to the next. Explain why this percentage is so low. (4 marks)

问题：在草原生态系统中，能量从一个营养级传递到下一个营养级只有大约 10%。解释为什么这个比例这么低。（4 分）

Not all of the organism at one trophic level is consumed by the next. Significant energy is lost as metabolic heat in respiration, which is used for movement, growth and maintaining body temperature. Undigested materials such as cellulose are lost in faeces (egestion). Energy is also used in excretion. By the time energy is assimilated into new biomass, only a small fraction remains for the next consumer. This energy loss limits the length of food chains.

某一营养级的生物并不会全部被下一级消耗。大量能量在呼吸作用中以代谢热的形式散失，这些能量用于运动、生长和维持体温。未消化的物质如纤维素随粪便排出（排遗）。能量还用于排泄。等到能量同化成新的生物量时，只有一小部分可留给下一级消费者。这种能量损失限制了食物链的长度。

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