📚 Year 8 OCR Biology Case Study Practice | 八年级 OCR 生物案例分析实战演练
Welcome to this case study practice session, designed especially for Year 8 OCR Biology students. Throughout these ten real-world scenarios, you will apply your knowledge of cells, organ systems, ecology, genetics and more. Each case presents a description, key questions, model answers and detailed explanations. This approach helps you move beyond memorising facts and towards using scientific ideas to solve problems — a skill that is essential for OCR assessments.
欢迎参加本次案例分析实战演练，专为八年级 OCR 生物学学生设计。通过这十个真实情景，你将运用有关细胞、器官系统、生态、遗传等方面的知识。每个案例都包含情景描述、关键问题、参考答案和详细解释。这种方法能帮助你从单纯的记忆事实，转变为运用科学观点解决问题——这也是 OCR 评估所必需的核心技能。
1. Case Study 1: Identifying Cell Structures | 案例一：识别细胞结构
A student prepared a temporary mount of onion epidermal cells and stained it with iodine solution. Under the light microscope at ×400 magnification, she observed neat rectangular cells. Each cell showed a distinct cell wall, a dark-stained nucleus and a large, transparent central vacuole. She then prepared a slide of her own cheek cells using methylene blue and noticed irregularly shaped cells with a nucleus and granular cytoplasm, but no cell wall or large vacuole.
一名学生制作了洋葱表皮细胞的临时装片，并用碘液染色。在400倍光学显微镜下，她观察到排列整齐的长方形细胞。每个细胞都能看到清晰的细胞壁、被染成深色的细胞核以及一个透明的大中央液泡。随后她用亚甲蓝制作了自己的口腔上皮细胞装片，发现细胞形状不规则，有细胞核和颗粒状细胞质，但没有细胞壁和大液泡。
Question: Which three structures are present in both onion and human cheek cells? Why are cell walls and chloroplasts not observed in the cheek cells?
问题：哪三种结构同时存在于洋葱细胞和人口腔上皮细胞中？为什么在口腔上皮细胞中看不到细胞壁和叶绿体？
Answer: Both cell types contain a cell membrane, cytoplasm and a nucleus. Cheek cells lack a cell wall because animal cells do not have cellulose cell walls; they have only a flexible cell membrane. Neither onion epidermal cells nor cheek cells contain chloroplasts — onion cells grow underground and do not photosynthesise, while animal cells are never photosynthetic.
答案：两种细胞都含有细胞膜、细胞质和细胞核。口腔上皮细胞没有细胞壁，因为动物细胞不具有纤维素细胞壁，它们只有有弹性的细胞膜。洋葱表皮细胞和口腔上皮细胞都不含叶绿体——洋葱细胞生长在地下，不进行光合作用，而动物细胞永远无法进行光合作用。
Explanation: The cell wall is a rigid layer outside the cell membrane, made of cellulose in plants. It provides structural support and prevents bursting when water enters. The large central vacuole in mature plant cells stores water and dissolved substances, helping maintain turgor pressure. The nucleus houses DNA and directs protein synthesis. Methylene blue and iodine are stains that bind to different cell components, making the nucleus and cytoplasm more visible under the microscope.
解释：细胞壁是位于细胞膜外的坚硬层，在植物中由纤维素组成。它提供结构支撑，并防止细胞因吸水而胀破。成熟植物细胞中的大中央液泡储存水分和溶解物，帮助维持膨压。细胞核含有 DNA 并指导蛋白质合成。亚甲蓝和碘液是能与不同细胞成分结合的染剂，使细胞核和细胞质在显微镜下更易观察。
2. Case Study 2: Enzyme Action in Digestion | 案例二：消化中的酶作用
Zara chewed a piece of plain bread slowly and noticed that after a couple of minutes it began to taste sweet. Her teacher explained that an enzyme in saliva starts breaking down the starch in bread into smaller sugar molecules. To investigate this, the class set up four test tubes containing starch suspension and treated them as follows: Tube A — saliva at 37°C, Tube B — saliva at 5°C, Tube C — boiled saliva at 37°C, Tube D — water at 37°C. After 15 minutes, they added iodine solution to each tube.
扎拉慢慢咀嚼了一块白面包，几分钟后她注意到面包开始有甜味。老师解释说，唾液中的一种酶开始将面包里的淀粉分解成较小的糖分子。为了探究这一点，全班设置了四支装有淀粉悬浮液的试管，分别作如下处理：试管 A——唾液，37°C；试管 B——唾液，5°C；试管 C——煮沸过的唾液，37°C；试管 D——水，37°C。15分钟后，他们向每支试管滴加碘液。
Question: In which tube would the iodine stay yellow-brown, indicating that starch had been digested? Explain why the other tubes would turn blue-black.
问题：哪支试管中的碘液会保持黄褐色，表明淀粉已被消化？解释为什么其他试管会变成蓝黑色。
Answer: Only in Tube A would starch be digested, so the iodine would remain yellow-brown. Tubes B and C would turn blue-black because the enzyme amylase was either too cold to work effectively (B) or denatured by boiling (C). Tube D contained no enzyme at all, so starch was not broken down.
答案：只有试管 A 中的淀粉会被消化，因此碘液保持黄褐色。试管 B 和 C 会变成蓝黑色，因为淀粉酶在管 B 中温度太低无法高效工作，在管 C 中则因煮沸而变性。试管 D 不含任何酶，所以淀粉没有被分解。
Explanation: The enzyme in saliva is salivary amylase. Its optimum temperature is around 37°C, body temperature. At 5°C the enzyme molecules have very little kinetic energy, so collisions with starch are infrequent. Boiling irreversibly changes the shape of the enzyme’s active site, preventing the starch from fitting — this is denaturation. Iodine solution tests for starch: a blue-black colour indicates starch is present, whereas a yellow-brown colour indicates starch has been broken down to maltose or other sugars.
解释：唾液中的酶是唾液淀粉酶。其最适温度约为 37°C，即体温。在 5°C 时，酶分子的动能很小，与淀粉碰撞的频率很低。煮沸会不可逆地改变酶活性部位的形状，使淀粉无法与之契合——这就是变性。碘液用于检验淀粉：蓝黑色表示淀粉存在，黄褐色则表示淀粉已被分解为麦芽糖或其他糖类。
3. Case Study 3: The Journey of a Sandwich | 案例三：三明治的旅程
Tom ate a chicken and lettuce sandwich on wholemeal bread. Describe the path the sandwich takes through Tom’s digestive system and explain how mechanical and chemical digestion work together to break it down into absorbable molecules, naming the enzymes involved.
汤姆吃了一份全麦面包夹鸡肉和生菜的三明治。请描述这份三明治在汤姆消化系统中的路径，并解释机械消化和化学消化如何协同作用，将其分解为可吸收的小分子，同时说出所涉及的酶。
Answer: The sandwich is first chewed in the mouth, where teeth carry out mechanical digestion by grinding the food. Salivary amylase begins the chemical digestion of starch in the bread. The food bolus travels down the oesophagus by peristalsis to the stomach. Here, gastric juices containing hydrochloric acid and protease enzymes (such as pepsin) break down proteins in the chicken. The stomach also churns the food, mixing it with enzymes. The mixture then enters the small intestine, where pancreatic amylase, trypsin (protease) and lipase act on starch, protein and fats respectively. Bile from the liver emulsifies fat droplets, increasing the surface area for lipase. The products — glucose, amino acids and fatty acids/glycerol — are absorbed through the villi into the blood. Undigested fibre and other waste move into the large intestine, where water is reabsorbed, and finally faeces are egested through the anus.
答案：三明治首先在口腔中被咀嚼，牙齿通过研磨进行机械消化。唾液淀粉酶开始对面包中的淀粉进行化学消化。食团通过食道的蠕动进入胃。胃液中含有盐酸和蛋白酶（如胃蛋白酶），分解鸡肉中的蛋白质。胃还会搅拌食物，使其与酶充分混合。混合物随后进入小肠，在这里胰淀粉酶、胰蛋白酶（一种蛋白酶）和脂肪酶分别作用于淀粉、蛋白质和脂肪。来自肝脏的胆汁将脂肪滴乳化，增大了脂肪酶作用的表面积。产物——葡萄糖、氨基酸和脂肪酸/甘油——通过小肠绒毛被吸收到血液中。未被消化的纤维素和其他废物进入大肠，在这里水分被重吸收，最后粪便通过肛门排出。
Explanation: Mechanical digestion (chewing, churning) increases the surface area of food, allowing enzymes to work more efficiently. Each enzyme is specific to a substrate: amylase breaks starch into maltose; proteases break proteins into amino acids; lipase breaks fats into fatty acids and glycerol. The villi in the small intestine provide a huge surface area for absorption, and their walls are only one cell thick, reducing the diffusion distance. The whole process is an example of an organ system working together to provide the body with essential nutrients.
解释：机械消化（咀嚼、搅拌）增大了食物的表面积，使酶能更高效地工作。每种酶对底物具有专一性：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；脂肪酶将脂肪分解为脂肪酸和甘油。小肠中的绒毛提供了巨大的吸收表面积，且其壁只有一层细胞厚，缩短了扩散距离。整个过程是器官系统协同工作为身体提供必需营养物质的范例。
4. Case Study 4: Effects of Exercise on Breathing Rate | 案例四：运动对呼吸频率的影响
Sami measured his breathing rate at rest by counting the number of breaths he took in one minute. He recorded 16 breaths per minute. After running on the spot for three minutes, he immediately measured his breathing rate again and found it had increased to 34 breaths per minute. He repeated the measurement at one-minute intervals until his breathing rate returned to resting level.
萨米测量了自己安静时的呼吸频率，记录下一分钟的呼吸次数为 16 次/分钟。在原地跑步三分钟后，他立即再次测量呼吸频率，发现增加到 34 次/分钟。之后他每隔一分钟测量一次，直到呼吸频率恢复到安静水平。
Question: Why does breathing rate increase during exercise? What happens to the body’s demand for oxygen and removal of carbon dioxide?
问题：运动时呼吸频率为什么会增加？身体对氧气的需求和对二氧化碳的清除会发生什么变化？
Answer: During exercise, muscle cells contract more frequently and carry out aerobic respiration at a faster rate to release more energy. This process uses oxygen and produces carbon dioxide. The increased breathing rate brings more oxygen into the lungs and removes the excess carbon dioxide produced. The brain’s breathing centre detects the rise in blood carbon dioxide levels and sends signals to the diaphragm and intercostal muscles to contract more rapidly, increasing the breathing rate and depth.
答案：运动时，肌细胞收缩更频繁，以更快的速率进行有氧呼吸，释放更多能量。这一过程消耗氧气并产生二氧化碳。呼吸频率增加能使更多氧气进入肺部，并清除产生的多余二氧化碳。脑部的呼吸中枢检测到血液中二氧化碳浓度升高，就会向膈肌和肋间肌发送信号，使其收缩更快，从而加快呼吸频率和深度。
Explanation: Aerobic respiration can be summarised by the word equation: glucose + oxygen → carbon dioxide + water (+ energy). The mitochondria in muscle cells are the sites of this reaction. After exercise, the breathing rate stays elevated for some time while the body repays an ‘oxygen debt’ — the extra oxygen needed to break down lactic acid that may have accumulated during anaerobic respiration. This investigation demonstrates how the respiratory and circulatory systems coordinate to maintain homeostasis.
解释：有氧呼吸可用文字方程式概括：葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）。肌细胞的线粒体是这一反应的场所。运动后，呼吸频率会在一段时间内保持较高水平，此时身体正在偿还“氧债”——即分解可能因无氧呼吸积累的乳酸所需的额外氧气。这一探究展示了呼吸系统和循环系统如何协调以维持稳态。
5. Case Study 5: Photosynthesis and Light Intensity | 案例五：光合作用与光照强度
A group of Year 8 students used Canadian pondweed (Elodea) to investigate the effect of light intensity on the rate of photosynthesis. They placed a piece of Elodea in a beaker of water and positioned a lamp at different distances from the beaker. They counted the number of oxygen bubbles released from the cut stem per minute. Their results are shown in a table: at 10 cm, 45 bubbles/min; at 20 cm, 25 bubbles/min; at 30 cm, 12 bubbles/min; at 40 cm, 4 bubbles/min.
一组八年级学生使用加拿大水蕴草（伊乐藻）探究光强对光合作用速率的影响。他们将一段伊乐藻放入盛有水的烧杯中，并将一盏灯放在离烧杯不同距离的位置，然后计算从切口茎两端释放的氧气泡数量（个/分钟）。结果如下：距离 10 cm，45 气泡/分钟；20 cm，25 气泡/分钟；30 cm，12 气泡/分钟；40 cm，4 气泡/分钟。
Question: Explain why the number of bubbles decreases as the distance of the lamp increases. What other factor must be kept constant for the results to be valid?
问题：解释为什么气泡数量随着灯的距离增加而减少。为了使结果有效，必须保持哪个其他因素不变？
Answer: As the lamp is moved further away, the light intensity reaching the plant decreases. Light provides the energy needed for the light‑dependent reactions of photosynthesis, so less light means fewer oxygen molecules are produced. A factor that must be kept constant is temperature — if the water warms up because of the close lamp, the rate of photosynthesis could be affected by temperature rather than light. Similarly, carbon dioxide concentration and the amount of Elodea must be the same in each trial.
答案：当灯移远时，到达植物的光照强度降低。光为光合作用的光依赖反应提供所需能量，因此光越少，产生的氧分子就越少。必须保持不变的因素是温度——如果水因灯靠近而变暖，光合作用速率可能受温度影响而非光照。同样，二氧化碳浓度和伊乐藻的用量在每次试验中也必须相同。
Explanation: Photosynthesis uses light energy to convert carbon dioxide and water into glucose and oxygen. The overall equation is: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. The oxygen bubbles are a direct by‑product of the light reactions. Students should also note that sodium hydrogencarbonate is often added to the water to ensure a constant supply of carbon dioxide. At very high light intensities, the rate will eventually level off because other factors, such as CO₂ concentration or temperature, become limiting.
解释：光合作用利用光能将二氧化碳和水转化为葡萄糖和氧气。总方程式为：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂。氧气泡是光反应的直接副产物。学生还需注意，常在水中加入碳酸氢钠以确保稳定的二氧化碳供应。在很高的光强下，速率最终会趋向平稳，因为其他因素，如 CO₂ 浓度或温度，成为限制因子。
6. Case Study 6: Food Chains and Trophic Levels | 案例六：食物链与营养级
In a meadow ecosystem, the following food chain is observed: grass → grasshopper → frog → snake → hawk. During a wet summer, a fungal disease killed a large proportion of the frog population. Scientists monitored the populations of the other organisms over the next two years.
在一处草甸生态系统中，观察到以下食物链：草 → 蚱蜢 → 青蛙 → 蛇 → 鹰。在一个多雨的夏季，一种真菌疾病导致大量青蛙死亡。科学家在随后的两年中监测了其他生物的数量。
Question: Predict and explain the likely changes in the populations of grasshoppers and hawks after the decline of the frogs.
问题：预测并解释青蛙减少后，蚱蜢和鹰的数量可能发生的变化。
Answer: With fewer frogs preying on them, the grasshopper population is likely to increase because they experience less predation. More grasshoppers will eat more grass, possibly reducing the grass population. Hawks, on the other hand, depend on snakes as a food source, but the snake population will also decline because snakes rely on frogs for food. With fewer snakes available, hawks may find less prey and their population could decrease, or they may switch to alternative prey if available.
答案：由于捕食蚱蜢的青蛙减少，蚱
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