📚 GCSE OCR Biology: Typical Exam Questions Explained | GCSE OCR 生物：典型例题详解
Working through typical exam-style questions is one of the most effective ways to consolidate key GCSE OCR Biology concepts and develop the analytical skills required for top grades. This article presents ten worked examples spanning cell biology, enzymes, transport, physiology, genetics, ecology and biotechnology. Each example is broken down step by step, with clear explanations provided in both English and Chinese to support bilingual learners on the aleveler.com platform.
通过典型考试题型的练习，是巩固 GCSE OCR 生物学核心概念、培养高分所需分析能力的高效方法之一。本文精选十道例题，涵盖细胞生物学、酶、运输、生理、遗传、生态和生物技术等领域。每道题都配以逐步解析，并用中英双语提供清晰解释，旨在支持 aleveler.com 平台上的双语学习者。
1. Cell Structure and Magnification | 细胞结构与放大倍数
Question (a): A student views a plant cell under a light microscope and measures the length of the cell image as 60 mm. The actual length of the cell is 0.15 mm. Calculate the magnification used.
问题 (a)： 一名学生用光学显微镜观察植物细胞，测得图像中细胞长度为 60 mm。该细胞实际长度为 0.15 mm。计算所使用的放大倍数。
Magnification = image size ÷ actual size = 60 ÷ 0.15. 60 divided by 0.15 equals 400, so the magnification is ×400.
放大倍数 = 图像大小 ÷ 实际大小 = 60 ÷ 0.15。60 除以 0.15 等于 400，因此放大倍数为 ×400。
Question (b): Explain why it is important to mount the plant tissue in water and use a coverslip.
问题 (b)： 解释为什么需要将植物组织置于水中并加盖盖玻片。
Mounting in water prevents the specimen from drying out, keeps cells turgid and allows light to pass through for a clearer image. Lowering the coverslip at an angle reduces the trapping of air bubbles, which would obstruct the view.
用水封片可防止标本干燥，保持细胞膨胀，并让光线透过以获得更清晰的图像。倾斜放下盖玻片可以减少气泡的混入，避免干扰观察。
Question (c): Identify two organelles that would be present in this plant cell but absent in an animal cell.
问题 (c)： 指出该植物细胞中存在的、动物细胞所没有的两个细胞器。
Plant cells have a permanent vacuole, chloroplasts and a cellulose cell wall. Any two of these are acceptable, e.g. chloroplasts and a permanent vacuole.
植物细胞具有大液泡、叶绿体和纤维素细胞壁。写出其中任意两个即可得分，例如叶绿体和永久液泡。
2. Enzyme Activity and Temperature | 酶活性与温度
Question: An experiment measured the time taken for amylase to break down starch at different temperatures. The results are shown in the table.
问题： 一项实验测量了不同温度下淀粉酶分解淀粉所需的时间。结果如下表所示。
| Temperature / °C | Time / s |
|---|---|
| 10 | 95 |
| 20 | 60 |
| 30 | 35 |
| 40 | 25 |
| 50 | 50 |
| 60 | No breakdown after 300 s |
(a) Describe the trend shown by the data between 10 °C and 40 °C. (b) Explain why the time taken increases above 40 °C. (c) Calculate the rate of reaction at 30 °C in per second, assuming the starch breakdown is complete. Use rate = 1 / time.
(a) 描述 10 °C 到 40 °C 之间数据呈现的趋势。(b) 解释 40 °C 以上所需时间增加的原因。(c) 假设淀粉完全分解，计算 30 °C 时的反应速率，单位：每秒。使用公式 速率 = 1 / 时间。
(a) As temperature rises from 10 °C to 40 °C, the time taken decreases, meaning the reaction gets faster. This is because enzyme and substrate particles have more kinetic energy and collide more frequently, forming more enzyme-substrate complexes.
(a) 随着温度从 10 °C 升高到 40 °C，所需时间减少，即反应速率加快。这是因为酶与底物粒子具有更高的动能，碰撞频率增加，形成更多的酶-底物复合物。
(b) Above the optimum (around 40 °C), the enzyme’s active site begins to lose its specific shape due to denaturation. The substrate no longer fits, so fewer enzyme-substrate complexes form and the reaction slows. At 60 °C the enzyme is completely denatured.
(b) 超出最适温度（约 40 °C）后，酶活性位点因变性而开始失去特定形状。底物无法契合，酶-底物复合物减少，反应减慢。在 60 °C 时酶已完全变性。
(c) Rate at 30 °C = 1 ÷ 35 s = 0.0286 s⁻¹ (accept 0.029 s⁻¹). The unit can be stated as ‘per second’.
(c) 30 °C 时的速率 = 1 ÷ 35 s = 0.0286 s⁻¹（0.029 s⁻¹ 可接受）。单位可表达为“每秒”。
3. Investigating Osmosis in Potato Tissue | 马铃薯组织渗透作用探究
Question: Strips of potato were blotted dry, weighed and placed in sucrose solutions of different concentrations for 30 minutes. The percentage change in mass was calculated.
问题： 将马铃薯条吸干水分并称重，放入不同浓度的蔗糖溶液中 30 分钟。计算质量变化百分比。
| Sucrose concentration / mol dm⁻³ | % change in mass |
|---|---|
| 0.0 | +12.5 |
| 0.2 | +5.0 |
| 0.4 | -0.2 |
| 0.6 | -8.4 |
| 0.8 | -15.0 |
(a) Determine the sucrose concentration that has the same water potential as the potato cells. (b) Explain why the mass decreased in 0.8 mol dm⁻³ sucrose solution. (c) Suggest why it was necessary to blot the potato strips dry before weighing.
(a) 确定与马铃薯细胞水势相同的蔗糖浓度。(b) 解释在 0.8 mol dm⁻³ 蔗糖溶液中质量减少的原因。(c) 说明称量前吸干马铃薯条表面的水分为何必要。
(a) The solution with approximately 0.4 mol dm⁻³ sucrose shows a negligible change (-0.2%), so its water potential is almost equal to that of the potato cells. This is the point where there is no net movement of water.
(a) 约 0.4 mol dm⁻³ 的蔗糖溶液质量变化极小（-0.2%），其水势与马铃薯细胞水势几乎相等。此点无净水移动。
(b) The 0.8 mol dm⁻³ solution has a lower water potential (more concentrated) than the cell contents. Water moves out of the cells by osmosis, causing the cells to lose turgor and the tissue mass to decrease.
(b) 0.8 mol dm⁻³ 溶液的水势低于细胞液（更浓），水分通过渗透作用从细胞中流失，细胞失去膨压，组织质量下降。
(c) Blotting removes excess surface water so that only the mass of the potato tissue itself is measured, preventing errors from water that has not entered the cells.
(c) 吸干可去除表面多余水分，确保仅测量马铃薯组织自身的质量，避免未进入细胞的水分造成误差。
4. Heart Structure and Double Circulation | 心脏结构与双循环
Question: The diagram shows the human heart. (a) Name the chamber labelled X that receives blood from the vena cava. (b) Explain why the wall of the left ventricle is thicker than the wall of the right ventricle. (c) Describe the advantage of having a double circulatory system.
问题： 图示为人体心脏。(a) 写出标有 X、接受来自腔静脉血液的腔室名称。(b) 解释左心室壁比右心室壁厚的原因。(c) 描述拥有双循环系统的优势。
(a) The chamber that receives deoxygenated blood from the vena cava is the right atrium.
(a) 接受来自腔静脉缺氧血的腔室是右心房。
(b) The left ventricle must pump blood all around the body, so it needs to generate a much higher pressure. The right ventricle only pumps blood a short distance to the lungs, so its muscle wall is thinner.
(b) 左心室需将血液泵送至全身各处，必须产生较高的压力。右心室仅将血液泵至距离较短的肺部，因此其肌肉壁较薄。
(c) A double circulation means blood passes through the heart twice for each complete circuit: once to the lungs and once to the body. This ensures that oxygenated blood is delivered to body tissues at high pressure, which increases the rate of oxygen and glucose supply for respiration, and also keeps deoxygenated blood and oxygenated blood separate.
(c) 双循环指血液在每个完整循环中两次经过心脏：一次去往肺，一次去往身体。这可保证含氧血在高压下被输送至身体组织，提高氧气和葡萄糖供应速率以支持呼吸作用，同时把缺氧血和含氧血分开。
5. Anaerobic Respiration in Yeast | 酵母的无氧呼吸
Question: Yeast is used in bread-making. The equation for anaerobic respiration in yeast is: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. (a) Name the two products shown in the equation. (b) A dough mixture produces 48 cm³ of carbon dioxide in 20 minutes at 28 °C. Calculate the rate of carbon dioxide production in cm³ per minute. (c) Suggest and explain why bread dough is often left in a warm place to rise.
问题： 酵母用于面包制作，其无氧呼吸方程式为：C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂。(a) 写出方程式中显示的两个产物名称。(b) 一个面团混合物在 28 °C 下 20 分钟产生 48 cm³ 二氧化碳，计算二氧化碳的产率，单位 cm³/min。(c) 说明为何做面包时常把面团放在温暖处发酵，并作出解释。
(a) Ethanol (alcohol) and carbon dioxide.
(a) 乙醇（酒精）和二氧化碳。
(b) Rate = volume ÷ time = 48 cm³ ÷ 20 min = 2.4 cm³/min.
(b) 速率 = 体积 ÷ 时间 = 48 cm³ ÷ 20 min = 2.4 cm³/min。
(c) Yeast enzymes work faster at warm temperatures (around 25–35 °C) because particles have more kinetic energy, increasing successful collisions. Faster anaerobic respiration produces more carbon dioxide gas, which makes the dough rise more quickly. If it is too hot, enzymes denature; if too cold, enzymes are inactive.
(c) 酵母酶在温暖温度下（约 25–35 °C）活性更高，因为粒子动能更大，有效碰撞增加。无氧呼吸加快可产生更多二氧化碳，使面团快速膨胀。温度过高酶会变性，过低则酶活性受到抑制。
6. Photosynthesis and Limiting Factors | 光合作用与限制因子
Question: A student investigates the effect of light intensity on the rate of photosynthesis in pondweed. The rate is measured by counting oxygen bubbles per minute. Results:
问题： 一位学生研究光照强度对水草光合作用速率的影响，通过数气泡数测量速率。结果如下：
| Light intensity / arbitrary units | Bubbles per min |
|---|---|
| 2 | 8 |
| 4 | 15 |
| 6 | 22 |
| 8 | 25 |
| 10 | 26 |
(a) Identify the limiting factor between 2 and 6 units. (b) Explain why the rate levels off above 8 units. (c) Suggest how the curve would change if extra carbon dioxide were supplied.
(a) 指出在 2 至 6 单位之间，哪个是限制因子。(b) 解释为何 8 单位以上速率趋于平稳。(c) 推测如果额外提供二氧化碳，曲线会如何变化。
(a) Between 2 and 6 units, as light intensity increases, the rate of photosynthesis rises proportionally. Therefore light intensity is the limiting factor.
(a) 在 2 到 6 单位之间，光合速率随光照强度成比例上升，因此限制因子是光照强度。
(b) Above 8 units, light intensity is no longer limiting. Another factor, such as carbon dioxide concentration or temperature, becomes limiting. The rate cannot increase further until that factor is increased.
(b) 高于 8 单位时，光照强度不再是限制因素。其他因子如二氧化碳浓度或温度成为限制，速率无法再增加，除非提高该因子的水平。
(c) Adding extra carbon dioxide would increase the rate of photosynthesis at high light intensities, so the curve would plateau at a higher level. At low light intensities the curve would not change significantly because light would still be limiting.
(c) 添加额外的二氧化碳会在高光照强度下增加光合速率，因此曲线将在更高的水平上趋于平稳。在低光照强度下曲线变化不大，因为光照仍是限制因素。
7. Monohybrid Inheritance | 单基因遗传
Question: In guinea pigs, black coat colour (B) is dominant to brown (b). A heterozygous black male is crossed with a heterozygous black female.
问题： 在豚鼠中，黑色毛 (B) 对棕色 (b) 为显性。一只杂合黑色雄性与一只杂合黑色雌性交配。
(a) Complete the Punnett square to show the possible genotypes of the offspring.
(a) 完成庞纳特方格，写出后代的可能基因型。
| B | b | |
| B | BB | Bb |
| b | Bb | bb |
Gametes from parents: B, b and B, b. The offspring genotypes are BB, Bb, Bb and bb.
亲本配子：B、b 和 B、b。后代基因型为 BB、Bb、Bb 和 bb。
(b) State the probability that an offspring will be brown. (c) Explain the difference between a dominant and a recessive allele.
(b) 写出后代为棕色的概率。(c) 解释显性等位基因与隐性等位基因的区别。
(b) Brown is bb. There is 1 bb out of 4 possibilities, so probability = 1/4 or 25%.
(b) 棕色基因型为 bb。4 种可能性中有 1 个 bb，因此概率为 1/4 或 25%。
(c) A dominant allele is always expressed in the phenotype even if only one copy is present, while a recessive allele is only expressed when two copies are present (in the homozygous state) and the dominant allele is absent.
(c) 显性等位基因只要存在一个拷贝就在表型中表达；隐性等位基因则仅当存在两个拷贝（纯合状态）且无显性等位基因时才表达。
8. The Carbon Cycle | 碳循环
Question: Carbon is cycled between the atmosphere, living organisms and the physical environment. (a) Name the process by which plants remove carbon dioxide from the air. (b) Explain how carbon in dead plant material can be returned to the atmosphere. (c) State one way human activity is increasing atmospheric carbon dioxide and explain its impact.
问题： 碳在大气、生物体和环境中循环。(a) 写出植物从空气中吸收二氧化碳的过程名称。(b) 解释枯死植物中的碳如何回到大气中。(c) 指出人类活动增加大气二氧化碳的一种方式，并解释其影响。
(a) Photosynthesis. Plants fix carbon dioxide into glucose, which is then used to make other organic compounds.
(a) 光合作用。植物将二氧化碳固定为葡萄糖，进而合成其他有机化合物。
(b) Decomposers such as bacteria and fungi break down dead plant material by respiration, releasing carbon dioxide back into the atmosphere. Combustion of plant material also releases CO₂.
(b) 细菌和真菌等分解者通过呼吸作用分解死物质，将二氧化碳释放回大气。植物材料的燃烧也会释放 CO₂。
(c) Burning fossil fuels releases large quantities of carbon dioxide that had been locked away for millions of years, increasing the greenhouse effect and contributing to global warming.
(c) 燃烧化石燃料会释放大量储存了数百万年的二氧化碳，增强温室效应，导致全球变暖。
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