📚 Case Study Practice in Year 7 OCR Biology | 案例分析实战演练
Welcome to this case study practice for Year 7 OCR Biology. Here you will work through ten real-world scenarios that test your understanding of key biological concepts. Each case study presents an observation or experiment, followed by analysis questions. Read the scenario, think about the science behind it, and then check the explanations. This method will help you apply knowledge to new situations—just like in your exams!
欢迎来到七年级 OCR 生物案例分析实战演练。你将通过十个真实场景来检验对关键生物学概念的理解。每个案例都会呈现一个观察或实验，然后进行分析。阅读情境，思考背后的科学原理，再查看解释。这种方法将帮助你学会把知识应用到新情境中，就像考试一样!
1. Microscope Case Study | 显微镜观察案例
Sarah prepared a thin slice of onion epidermis and placed it on a slide. She added a drop of iodine solution to stain the cells, then viewed the specimen under the microscope using the low-power objective lens. She saw neat rectangular shapes and moved to the high-power lens to observe more detail.
萨拉制作了洋葱表皮的薄片并放在载玻片上。她加了一滴碘液给细胞染色，然后使用低倍物镜在显微镜下观察标本。她看到了整齐的长方形，随后转换到高倍物镜以观察更多细节。
Under high power, Sarah could clearly identify a dark dot in each cell—the nucleus. She noticed that the cells had a firm outer boundary, the cell wall, and a thin inner membrane, the cell membrane. There were no chloroplasts visible, even though the onion is a plant. She then compared this with a cheek cell from her mouth, which was irregular in shape, lacked a cell wall and chloroplasts, but still showed a nucleus and cytoplasm.
在高倍镜下，萨拉可以清楚地看到每个细胞中的深色圆点——细胞核。她注意到细胞有坚硬的外边界，即细胞壁，以及内部薄薄的一层细胞膜。即使洋葱是植物，也没有可见的叶绿体。然后她将其与自己的口腔上皮细胞进行比较，后者形状不规则，缺少细胞壁和叶绿体，但仍显示出细胞核和细胞质。
The case reminds us that plant cells have a cell wall, cell membrane, cytoplasm, nucleus and often chloroplasts (though not in onion epidermis). Animal cells have only cell membrane, cytoplasm and nucleus. The iodine stain makes the nucleus visible by reacting with starch in the cytoplasm.
这个案例提醒我们，植物细胞有细胞壁、细胞膜、细胞质、细胞核，通常还有叶绿体(尽管洋葱表皮细胞没有)。动物细胞只有细胞膜、细胞质和细胞核。碘液通过与细胞质中的淀粉反应，使细胞核可见。
2. Food Chain in a Garden | 花园食物链案例
A gardener noticed that her rose bushes were infested with greenfly (aphids), which suck sap from the stems. Ladybirds were eating the greenfly, and occasionally a blue tit visited to feed on the ladybirds. She wondered what would happen if she sprayed insecticide to kill all the greenfly.
一位园丁发现她的玫瑰丛上爬满了蚜虫，它们吸食茎中的汁液。瓢虫正在捕食蚜虫，偶尔还有蓝山雀飞来以瓢虫为食。她想知道如果喷洒杀虫剂杀死所有蚜虫会发生什么。
s the producer because it makes its own food by photosynthesis. The greenfly is the primary consumer, the ladybird is the secondary consumer, and the blue tit is the tertiary consumer. The arrows show the direction of energy flow.
我们可以将捕食关系表示为一条食物链：玫瑰 → 蚜虫 → 瓢虫 → 蓝山雀。玫瑰是生产者，因为它通过光合作用制造自己的食物。蚜虫是初级消费者，瓢虫是次级消费者，蓝山雀是三级消费者。箭头表示能量流动的方向。
If the insecticide kills all greenfly, the ladybirds lose their food source and their population would decrease. The blue tits would then have fewer ladybirds to eat, so they might leave the garden or starve. This case shows how removing one organism can disrupt the whole food chain.
如果杀虫剂杀死所有蚜虫，瓢虫失去食物来源，其数量会减少。蓝山雀能吃到的瓢虫更少，因此它们可能离开花园或饿死。这个案例说明，移除一种生物会干扰整条食物链。
3. Germination Investigation | 种子发芽实验案例
A class set up an experiment to find out what conditions cress seeds need for germination. They placed seeds in four test tubes: Tube A had moist cotton wool and was left at room temperature (20°C) with air; Tube B had dry cotton wool at room temperature with air; Tube C had moist cotton wool but was placed in a fridge at 4°C; Tube D had moist cotton wool but the seeds were submerged in boiled water with a layer of oil to exclude air.
一个班级设计实验探究水芹种子发芽需要哪些条件。他们将种子放入四支试管：试管 A 在室温(20°C)下，有湿棉花和空气；试管 B 室温下干棉花和空气；试管 C 湿棉花但放入 4°C 冰箱；试管 D 湿棉花但种子浸没在凉开水中并覆盖一层油以隔绝空气。
After five days, only the seeds in Tube A germinated normally. Tube B showed no germination—water was missing. Tube C seeds remained dormant because the temperature was too low. Tube D seeds did not germinate because oxygen was excluded by the boiled water and oil layer. Thus, seeds require water, warmth and oxygen (air) to germinate. Light is not essential for germination, as cress will germinate in the dark if other conditions are met.
五天后，只有试管 A 的种子正常发芽。试管 B 没有发芽——缺少水分。试管 C 的种子保持休眠，因为温度太低。试管 D 的种子未发芽，因为凉开水和油层排除了氧气。因此，种子发芽需要水分、适宜的温度和氧气(空气)。光照对发芽并非必需，水芹在黑暗中只要其他条件满足也能发芽。
4. Testing a Leaf for Starch | 检验叶片淀粉案例
A pupil wanted to prove that light is needed for photosynthesis. She took a potted geranium plant and placed it in a dark cupboard for 48 hours to destarch the leaves. She then attached a black paper strip to one leaf, covering only a part of it, and exposed the plant to bright light for six hours. After that, she removed the leaf, boiled it in water, then in ethanol to remove chlorophyll, and finally rinsed it in warm water and added iodine solution.
一名学生想证明光合作用需要光。她取了一盆天竺葵，放在暗柜中 48 小时以耗尽叶片中的淀粉。然后在叶片上贴一条黑纸，只遮住部分，再将植株放在强光下照射六小时。之后她摘下叶片，先在水中煮沸，再用酒精煮沸以去除叶绿素，最后用温水漂洗并加碘液。
The part of the leaf that had been exposed to light turned blue-black with iodine, showing that starch was present. The covered part remained pale yellow-brown, indicating no starch. The student correctly concluded that light is necessary for photosynthesis to produce starch. The overall photosynthesis equation is:
叶片见光部分遇碘液变蓝黑色，表明有淀粉存在。遮光部分仍为浅黄褐色，说明无淀粉。学生正确得出结论：光是光合作用制造淀粉所必需的。光合作用总方程式为：
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
二氧化碳 + 水 → 葡萄糖 + 氧气
This classic experiment demonstrates the importance of a control (covered portion) and safe use of ethanol (heated in a water bath). The alkane ethanol removes chlorophyll so the colour change is clearer.
这个经典实验展示了对照组(遮光部分)的重要性以及酒精的安全使用(水浴加热)。酒精去除叶绿素后颜色变化更清晰。
5. Digestion Case Study | 消化系统案例
John ate a cheese sandwich for lunch. He wondered what happens to the bread (high in starch) as it travels through his digestive system. In the mouth, the sandwich is chewed and mixed with saliva, which contains an enzyme called amylase. Amylase starts breaking down starch into smaller sugar molecules like maltose.
约翰午餐吃了一个奶酪三明治。他想知道面包(富含淀粉)经过消化系统时会发生什么。在口腔中，三明治被咀嚼并与含有淀粉酶的唾液混合。淀粉酶开始将淀粉分解成更小的糖分子，如麦芽糖。
The food then travels down the oesophagus to the stomach, where acid and enzymes digest the cheese (protein). The bread, now a bolus, continues to be digested. In the small intestine, more enzymes from the pancreas break down starch completely into glucose, which is small enough to be absorbed through the villi into the bloodstream. The large intestine absorbs water, leaving undigested fibre to be excreted.
食物随后沿食道进入胃，胃酸和酶消化奶酪(蛋白质)。已成食团的面包继续被消化。在小肠中，来自胰腺的更多酶将淀粉完全分解成葡萄糖，后者小到足以通过绒毛吸收进入血液。大肠吸收水分，留下未消化的纤维被排出。
This case illustrates that different food groups are digested in different parts of the gut, and that enzymes are biological catalysts that speed up the breakdown of large molecules. It also highlights the role of the mouth, stomach, small intestine and large intestine in processing a meal.
这个案例说明不同食物类别在消化道不同部位被消化，酶是加速大分子分解的生物催化剂。它还突显了口腔、胃、小肠和大肠在处理一顿饭中的作用。
6. Breathing Rate and Exercise | 呼吸速率与运动案例
During a PE lesson, students measured their breathing rate at rest and then immediately after running for three minutes. At rest, Abdi counted 15 breaths per minute. After exercise, his breathing rate rose to 36 breaths per minute. He also noticed he was breathing more deeply.
在一节体育课上，学生们测量了安静时和跑步三分钟后的呼吸速率。安静时阿卜迪测得每分钟 15 次呼吸。运动后，他的呼吸速率升至每分钟 36 次。他还注意到呼吸变得更深了。
The increase in breathing rate supplies more oxygen to the muscles. The muscles need oxygen for aerobic respiration, which releases energy from glucose. The word equation for aerobic respiration is:
呼吸速率加快为肌肉提供更多氧气。肌肉需要氧气进行有氧呼吸，从葡萄糖中释放能量。有氧呼吸的文字方程式为：
Glucose + Oxygen → Carbon dioxide + Water (+ energy)
葡萄糖 + 氧气 → 二氧化碳 + 水 (+ 能量)
While exercising, the muscles produce extra carbon dioxide, which must be exhaled. Deep breathing helps remove CO₂ quickly. This explains why both breathing rate and depth increase. After exercise, the breathing rate gradually returns to normal as the oxygen debt is repaid.
运动时肌肉产生额外的二氧化碳，必须呼出。深呼吸有助于快速排出 CO₂。这就解释了为什么呼吸速率和深度都会增加。运动后，随着氧债的偿还，呼吸速率逐渐恢复正常。
7. Classifying Organisms | 生物分类案例
A Year 7 class was given pictures of six organisms: a frog, a snake, an eagle, a dog, a goldfish and a butterfly. They were asked to group them into vertebrates and invertebrates, and then into more specific groups. All organisms except the butterfly have a backbone, so they are vertebrates. The butterfly is an invertebrate, and it belongs to the arthropods (insects).
一个七年级班级拿到了六种生物的图片：青蛙、蛇、鹰、狗、金鱼和蝴蝶。他们被要求将其分为脊椎动物和无脊椎动物，然后再分为更具体的类群。除蝴蝶外所有动物都有脊柱，因此是脊椎动物。蝴蝶是无脊椎动物，属于节肢动物(昆虫)。
The class then used a simple key to separate the vertebrates. The frog has moist skin and lays eggs without shells, so it is an amphibian. The snake has dry scaly skin and lays leathery eggs, making it a reptile. The eagle has feathers and a beak—it is a bird. The dog has fur and produces milk for its young, so it is a mammal. The goldfish has gills and swims in water; it is a fish. The butterfly has six legs, a three-part body and wings, so it is an insect.
然后班级用一个简单的检索表将脊椎动物分开。青蛙皮肤湿润、产无壳卵，是两栖动物。蛇皮肤干燥有鳞、产革质卵，是爬行动物。鹰有羽毛和喙，是鸟类。狗有皮毛并哺乳幼崽，是哺乳动物。金鱼有鳃且在水中游泳，是鱼类。蝴蝶有六条腿、身体三部分和翅膀，是昆虫。
This case demonstrates how classification is based on observable features and how a dichotomous key helps identify organisms. Understanding classification allows biologists to study biodiversity and evolutionary relationships.
这个案例展示了分类如何基于可观察的特征，以及二叉检索表如何帮助鉴定生物。理解分类使生物学家能够研究生物多样性和进化关系。
8. Pond Ecosystem Sampling | 池塘生态系统取样案例
A group of students used pond-dipping nets to collect organisms from a local pond. They found water fleas (Daphnia), damselfly nymphs, tadpoles, water boatmen and pondweed. They then drew a food web: Pondweed is the producer; water fleas and tadpoles eat pondweed; damselfly nymphs and water boatmen eat water fleas and small tadpoles. The students wondered what would happen if the pond became polluted with fertiliser.
一组学生用捞网从当地池塘中收集生物。他们发现了水蚤、豆娘稚虫、蝌蚪、水船虫和水草。他们随后画出了食物网：水草是生产者；水蚤和蝌蚪吃水草；豆娘稚虫和水船虫吃水蚤和小蝌蚪。学生们想知道如果池塘被化肥污染会发生什么。
Fertiliser entering the pond would cause an algal bloom, blocking light and killing the pondweed when the algae die and decompose, using up oxygen. Without pondweed, the producers disappear, so consumers like water fleas and tadpoles starve. This would collapse the food web. The students learnt that sampling with nets can reveal the variety of life in a habitat, and that ecosystems are fragile.
流入池塘的化肥会导致藻华，藻类大量繁殖阻挡光线，死亡分解时消耗氧气，从而杀死水草。没有了水草，生产者消失，像水蚤和蝌蚪这样的消费者就会饿死。这将导致食物网崩溃。学生们了解到用捞网取样可以揭示栖息地中的生物多样性，而生态系统是脆弱的。
9. Coloured Flower Experiment | 有色花实验案例
An investigation placed white carnation stems in a vase of water coloured with red food dye. After a few hours, the petals started to show red streaks, and after a full day the petals had turned pink entirely. When the stem was cut, the student observed that the tubes in the centre (xylem) were stained red.
一项实验将白色康乃馨的茎插入染有红色食用色素的水中。几小时后花瓣开始出现红色条纹，一整天后花瓣完全变成了粉红色。当切开茎部时，学生观察到中央管道(木质部)被染成了红色。
The dye moved up the stem with the water because water and dissolved minerals are transported through narrow tubes called xylem vessels. The process relies on transpiration—water evaporates from the leaves, pulling more water up from the roots. The dye molecules were carried along with this water stream and collected in the petals, making them change colour. This demonstrates the role of the stem in transporting substances to all parts of the plant.
染料随水分沿茎上升，因为水和溶解的矿物质通过称为木质部导管的细管运输。该过程依赖蒸腾作用——水分从叶片蒸发，将更多水从根部拉上来。染料分子随这股水流被带走并积聚在花瓣中，使其变色。这证明了茎在将物质输送到植物体各部位中的作用。
If the stem's outer bark (phloem) were removed, the dye would still travel up, showing that water transport occurs in the xylem, not the phloem. The phloem mainly transports food (sugars) downwards. Year 7 students can remember that the stem contains two transport systems working together.
如果剥掉茎的外层树皮(韧皮部)，染料仍会向上运输，说明水分运输发生在木质部而非韧皮部。韧皮部主要向下运输养料(糖)。七年级学生可以记住，茎中含有两个协同工作的运输系统。
10. Hand Washing Experiment | 洗手实验案例
A class tested the effectiveness of hand washing by pressing unwashed fingers onto a sterile nutrient agar plate, then washing hands thoroughly with soap and pressing the same fingers onto a second plate. The plates were sealed and incubated at 25°C for two days. The 'unwashed' plate grew many colonies of different colours and sizes; the 'washed' plate showed very few colonies.
一个班级测试洗手的效果，将未洗的手指按压在无菌营养琼脂平板上，然后用肥皂彻底洗手后，将相同手指按压在第二个平板上。密封平板并在 25°C 条件下培养两天。'未洗手'平板上长出了许多不同颜色和大小的菌落；'洗手'平板上菌落极少。
Microorganisms, including bacteria and fungi, are present on our skin. Washing with soap removes most of them, reducing the risk of infection. The colonies seen on agar are populations of millions of microorganisms that originated from a single cell. This case study highlights the importance of good hygiene, such as washing hands before eating and after using the toilet, to prevent the spread of disease.
微生物，包括细菌和真菌，存在于我们的皮肤上。用肥皂洗手能去除大部分微生物，降低感染风险。琼脂上看到的菌落是由单个细胞繁殖而来的百万微生物种群。这个案例强调了良好卫生习惯的重要性，比如饭前便后洗手，以预防疾病传播。
The class also discussed control variables: both plates used the same type of agar, the fingers were from the same person, and incubation temperature and time were identical. This makes it a fair test. The only difference was the washing treatment, so any difference in colony growth must be due to hand washing.
班级还讨论了控制变量：两个平板使用同类型琼脂，手指来自同一人，培养温度和时间完全相同。这使其成为公平测试。唯一的区别是洗手处理，因此菌落生长的任何差异必定归因于洗手。
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