📚 Year 8 Edexcel Biology: Case Study Practice | 案例分析实战演练
Welcome to this focused case study revision session designed for Year 8 Edexcel Biology. In the real world, biology is not just about memorising facts – it is about applying them to explain observations, solve problems, and interpret data. The following case studies have been carefully selected to mirror typical exam-style scenario questions. Each case invites you to think like a biologist, linking core concepts such as cells, nutrition, respiration, photosynthesis, ecosystems, and inheritance to everyday phenomena. Work through the cases, read the analysis step by step, and use the bilingual explanations to strengthen both your scientific understanding and your academic language skills.
欢迎参加本次为 Year 8 Edexcel 生物学定制的案例分析强化训练。在现实生活中，生物学不只是记忆事实，更是运用知识解释现象、解决问题和分析数据。以下案例经过精心挑选，模仿典型的考试情境题。每个案例都邀请你像生物学家一样思考，将细胞、营养、呼吸、光合作用、生态系统和遗传等核心概念与日常现象联系起来。请逐步阅读案例与分析，利用中英双语解释加深你的科学理解和学术语言能力。
1. The Shrinking Potato | 土豆条缩小之谜
A student placed identical potato cylinders into three beakers: one with pure water, one with 10% sucrose solution, and one with 25% sucrose solution. After 30 minutes, the potato in pure water became slightly longer and very firm. The potato in 10% sucrose became a little shorter and softer. The potato in 25% sucrose became much shorter and very floppy.
一名学生将相同的土豆条分别放入三个烧杯中：一杯为纯水，一杯为 10% 蔗糖溶液，一杯为 25% 蔗糖溶液。30 分钟后，纯水中的土豆条略微变长且很硬；10% 蔗糖溶液中的土豆条稍变短、变软；25% 蔗糖溶液中的土豆条明显缩短且非常萎软。
The water moved by osmosis. The potato cells contain a certain amount of dissolved substances, so their cytoplasm has a lower water potential than pure water. Water entered the cells in pure water by osmosis, making them turgid. In the sugar solutions, the water potential outside the cells was lower (more concentrated), so water left the cells by osmosis, causing them to become flaccid. The higher the sucrose concentration, the greater the water loss.
水通过渗透作用移动。土豆细胞含有一定量溶解物，因此细胞质的水势比纯水低。纯水中水通过渗透进入细胞，使细胞硬挺。在糖溶液中，细胞外水势较低（浓度更高），水通过渗透离开细胞，导致细胞萎软。蔗糖浓度越高，失水越多。
This case reminds you that osmosis is a special type of diffusion – the net movement of water molecules through a partially permeable membrane from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution). In exam questions, always mention the partially permeable cell membrane and compare water potentials.
这个案例提醒你，渗透是一种特殊的扩散——水分子通过部分透性膜从水势较高区域（稀溶液）向水势较低区域（浓溶液）的净移动。在考试中，务必提到部分透性细胞膜并比较水势。
2. The Sailor with Bleeding Gums | 牙龈出血的水手
A historical report describes sailors on a long sea voyage who had no fresh fruit or vegetables for months. They developed swollen, bleeding gums, their wounds healed slowly, and they felt very weak. The ship’s doctor suspected a dietary deficiency.
一份历史记录描述了一次长途航海中数月没有新鲜水果和蔬菜的水手。他们牙龈肿胀出血，伤口愈合缓慢，并且感到极度虚弱。船医怀疑是饮食缺乏所致。
The symptoms point to a lack of vitamin C (ascorbic acid), which causes scurvy. Vitamin C is needed to make collagen, a protein that helps keep skin, gums, and blood vessels healthy. Without it, capillaries become fragile and bleed easily. Fresh citrus fruits, tomatoes, and green vegetables are rich sources of vitamin C.
这些症状表明缺乏维生素 C（抗坏血酸），导致坏血病。维生素 C 是合成胶原蛋白所必需的，胶原蛋白有助于保持皮肤、牙龈和血管健康。缺乏时，毛细血管变脆，容易出血。新鲜柑橘类水果、西红柿和绿色蔬菜富含维生素 C。
This case links to food tests: vitamin C can be detected using DCPIP solution, which turns from blue to colourless when ascorbic acid is present. A balanced diet must include vitamins and minerals alongside carbohydrates, proteins, and fats to prevent deficiency diseases.
该案例与食物测试相关：维生素 C 可用 DCPIP 溶液检测，存在抗坏血酸时溶液由蓝变为无色。均衡饮食必须包含维生素和矿物质以及碳水化合物、蛋白质和脂肪，以预防营养缺乏病。
3. The Model Gut Investigation | 肠道模型探究
A class used a model ‘gut’ – a Visking tubing bag containing starch solution and amylase enzyme, suspended in a water bath at 37 °C. The surrounding water was tested every minute with iodine solution and Benedict’s reagent. Initially, the water outside gave a blue-black colour with iodine, but after 20 minutes no starch was detected outside. However, the outside water did give an orange-red precipitate with Benedict’s reagent after heating.
班级使用“肠道”模型——一个装有淀粉溶液和淀粉酶的透析袋，悬挂在 37 °C 水浴中。周围的水每分钟用碘液和本尼迪克特试剂检测。最初，外部水遇碘液呈蓝黑色，但 20 分钟后外部未检测到淀粉。然而，外部水加热后与本尼迪克特试剂产生橙红色沉淀。
Amylase breaks down starch into smaller sugar molecules, mainly maltose. Starch molecules are too large to pass through the Visking tubing (which mimics the partially permeable membrane of the small intestine), but the smaller sugar molecules can diffuse out. That is why starch was not found outside but reducing sugar was present. The colour change from blue-black to no colour with iodine inside the tubing confirmed starch digestion.
淀粉酶将淀粉分解为较小的糖分子，主要是麦芽糖。淀粉分子太大，无法通过透析袋（模拟小肠的部分透性膜），但较小的糖分子可以扩散出去。因此外部未检出淀粉，却检出了还原糖。透析袋内碘液由蓝黑变为无色，证实了淀粉被消化。
In exams, you should explain why temperature matters: 37 °C is the optimum for human enzymes, providing the best kinetic energy for enzyme-substrate collisions without denaturing the enzyme. This model demonstrates digestion and absorption, two key stages in human nutrition.
考试中，要解释温度的重要性：37 °C 是人体酶的最适温度，提供最佳的酶与底物碰撞动能而不使酶变性。此模型展示消化与吸收两个人体营养的关键阶段。
4. The Athlete’s Breathing Rate | 运动员的呼吸频率
A student measured her breathing rate at rest: 16 breaths per minute. After running on the spot for two minutes, her breathing rate immediately increased to 32 breaths per minute. Even after five minutes of rest, her rate was still 20 breaths per minute.
一名学生测量了安静时的呼吸频率：每分钟 16 次。原地跑步两分钟后，呼吸频率立即升至每分钟 32 次。即使休息五分钟后，频率仍为每分钟 20 次。
During exercise, muscle cells respire more rapidly to release energy for contraction. Aerobic respiration requires oxygen and produces carbon dioxide. The increased breathing rate brings in more oxygen and removes excess CO₂ faster. The equation for aerobic respiration can be summarised as: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O (+ energy). The slower recovery shows that the body still needed extra oxygen to break down lactic acid produced during anaerobic respiration when oxygen delivery lagged behind demand.
运动时，肌肉细胞呼吸加快以释放收缩所需的能量。有氧呼吸需要氧气并产生二氧化碳。呼吸频率增加可更快吸入更多氧气并排出多余的 CO₂。有氧呼吸的方程式可概括为：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O（+ 能量）。恢复较慢表明身体仍需要额外氧气来分解无氧呼吸产生的乳酸，当时供氧滞后于需求。
You should be able to link heart rate and breathing rate to gas exchange and the circulatory system. When answering such case studies, always connect the demand for energy with oxygen supply and waste removal.
你应能将心率和呼吸频率与气体交换和循环系统联系起来。回答此类案例分析时，始终将能量需求与氧气供应及废物清除联系起来。
5. The Bubbling Pondweed | 冒泡的水草
A piece of pondweed was placed in a beaker of water with a lamp at different distances. The number of oxygen bubbles released per minute was counted. At 10 cm, 45 bubbles; at 20 cm, 30 bubbles; at 40 cm, 12 bubbles; and in the dark, 0 bubbles.
将一段水草放入盛水的烧杯中，用不同距离的灯照射。记录每分钟释放的氧气泡数。距离 10 cm 时 45 个气泡；20 cm 时 30 个；40 cm 时 12 个；黑暗中为 0 个气泡。
The bubbles contain oxygen produced during photosynthesis. The rate of photosynthesis decreases as the distance from the light source increases because light intensity decreases. In the dark, no photosynthesis occurs, so no oxygen bubbles are produced. The equation for photosynthesis is: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, showing light energy is essential.
这些气泡含有光合作用产生的氧气。随着光源距离增加，光合作用速率下降，因为光强降低。在黑暗中，没有光合作用，因此不产生氧气泡。光合作用方程式为：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂，表明光能必不可少。
Exam questions might ask you to identify the independent variable (light intensity), dependent variable (number of bubbles), and control variables (temperature, CO₂ concentration, type of pondweed). This classic experiment tests how a limiting factor affects the rate of photosynthesis.
考试可能要求你识别自变量（光强）、因变量（气泡数量）和控制变量（温度、CO₂ 浓度、水草种类）。这个经典实验测试限制因素如何影响光合作用速率。
6. The Disappearing Mice | 消失的老鼠
In a grassland food web, owls hunt mice, and mice eat seeds from grass and grains. One year, a disease killed many owls. Scientists observed that the mouse population first increased rapidly, then after a few months began to decline. At the same time, the vegetation cover decreased.
在草原食物网中，猫头鹰捕食老鼠，老鼠以草籽和谷物为食。某一年，一场疾病杀死了许多猫头鹰。科学家观察到老鼠数量先是迅速增加，几个月后又开始下降。同时植被覆盖度减少。
Owls are predators of mice. When owl numbers dropped, fewer mice were eaten, so the mouse population boomed. As more mice fed on seeds, the grass and grain plants were overgrazed, reducing their biomass. This created food shortage for the mice, causing their population to crash. This is a classic predator-prey cycle and demonstrates interdependence within an ecosystem.
猫头鹰是老鼠的捕食者。猫头鹰数量下降，被吃掉的老鼠减少，老鼠数量激增。随着更多老鼠取食种子，草和谷物被过度啃食，生物量下降。这造成老鼠食物短缺，种群数量暴跌。这是经典的捕食者-猎物周期，展示了生态系统内的相互依赖。
You must be able to interpret graphs of predator-prey relationships and explain why populations change over time. Always refer to competition for resources and feeding relationships.
你必须能够解读捕食者-猎物关系图并解释种群数量随时间变化的原因。始终提及资源竞争和摄食关系。
7. The Earlobe Family Tree | 耳垂家族树
The diagram shows a family pedigree for attached (A) and free (F) earlobes. Free earlobes are dominant. In the first generation, the father has attached earlobes and the mother has free earlobes. They have three children: two sons with free earlobes, and one daughter with attached earlobes.
家系图显示了附着耳垂 (A) 和游离耳垂 (F) 的遗传。游离耳垂为显性。第一代父亲为附着耳垂，母亲为游离耳垂。他们有三个孩子：两个儿子为游离耳垂，一个女儿为附着耳垂。
Let F represent the free earlobe allele (dominant) and a represent the attached allele (recessive). The father with attached earlobes must be aa. The mother with free earlobes could be Ff or FF, but because she had a daughter with attached earlobes (aa), she must be heterozygous Ff. The free-earlobe sons are Ff, inheriting F from mother and a from father. This case demonstrates how recessive traits can skip a generation and how parents’ genotypes can be deduced from offspring phenotypes.
设 F 代表游离耳垂等位基因（显性），a 代表附着耳垂等位基因（隐性）。附着耳垂的父亲基因型必为 aa。游离耳垂的母亲可能是 Ff 或 FF，但因为她生有一个附着耳垂的女儿 (aa)，所以必定是杂合子 Ff。游离耳垂的儿子为 Ff，从母亲获得 F，从父亲获得 a。此案例演示了隐性性状如何隔代出现，以及如何从子代表现型推断亲代基因型。
In Year 8, you should be comfortable using simple Punnett squares and genetic diagrams to predict ratios. This case helps you practise interpreting family pedigree charts, a common exam skill.
在 Year 8 阶段，你应熟练使用简单的旁氏表（Punnett square）和遗传图解预测比例。这个案例帮助你练习解读家族谱系图，这是一项常见的考试技能。
8. The Variegated Leaf Starch Test | 彩叶淀粉测试
A variegated geranium leaf with green and white patches was exposed to sunlight for several hours. It was then boiled in water, warmed in ethanol to remove chlorophyll, and tested with iodine solution. The green areas turned blue-black; the white areas remained pale brown/yellow.
一片带有绿色和白色斑块的彩叶天竺葵叶片在阳光下照射数小时。然后沸水煮过，温热乙醇除去叶绿素，再用碘液测试。绿色区域变成蓝黑色；白色区域仍为浅棕色/黄色。
Photosynthesis produces glucose, which is stored as starch. The green areas contain chlorophyll, which captures light energy needed for photosynthesis. The white areas lack chlorophyll, so no starch was produced there. This investigation provides evidence that chlorophyll is essential for photosynthesis. The ethanol step removes the green colour so that the iodine’s colour change is visible.
光合作用产生葡萄糖，以淀粉形式储存。绿色区域含有叶绿素，能捕获光合作用所需的光能。白色区域缺少叶绿素，因此没有产生淀粉。这个探究提供了叶绿素是光合作用必需条件的证据。乙醇步骤去除了绿色，使碘液颜色变化清晰可见。
You may be asked to explain why boiling ethanol is used instead of water: ethanol dissolves chlorophyll but is flammable, so it must be heated indirectly in a water bath. Safety and fair testing are key aspects of such experimental case studies.
你可能被问到为什么使用沸腾乙醇而不是水：乙醇溶解叶绿素，但易燃，因此必须水浴间接加热。安全和公平测试是此类实验案例分析的关键方面。
9. The Sweet Taste of Chewing Bread | 咀嚼面包的甜味
A student chewed a piece of plain white bread for two minutes without swallowing. She noticed that the bread began to taste slightly sweet. She spat out the bolus and added a few drops of iodine solution; the result was a much lighter brown colour compared to unchewed bread tested with iodine.
一名学生咀嚼一片纯白面包两分钟但不吞咽。她注意到面包开始有淡淡甜味。她吐出食团，加入几滴碘液；与未咀嚼面包的碘液测试结果相比，颜色呈浅棕色。
Bread contains starch, which is a polysaccharide. Saliva contains the enzyme salivary amylase, which starts digesting starch into maltose, a disaccharide sugar. Maltose tastes sweet. The iodine test for starch becomes weaker (lighter brown rather than blue-black) because some starch has been broken down. This is a simple model of chemical digestion that begins in the mouth.
面包含有淀粉，一种多糖。唾液中含有唾液淀粉酶，开始将淀粉分解为麦芽糖，一种二糖。麦芽糖有甜味。淀粉的碘液测试变弱（浅棕色而非蓝黑色），因为部分淀粉已被分解。这是从口腔开始的化学消化简单模型。
In case study questions, you should link the enzyme to its substrate and product, and note that mechanical digestion (chewing) also increases surface area, aiding enzymatic action. This real-life observation makes a memorable connection to digestive system studies.
在案例分析题中，你应将酶与其底物和产物联系起来，并注意物理消化（咀嚼）也增加了表面积，有助于酶的作用。这一生活观察为消化系统的学习建立了难忘的联系。
10. The Polluted Pond Survey | 污染池塘调查
A biologist sampled two ponds. Pond X had crystal clear water with stonefly nymphs, mayfly nymphs, and freshwater shrimps. Pond Y had cloudy water, green algae scum, and only rat-tailed maggots, sludge worms, and bloodworms. The biologist recorded the numbers of each species and concluded that Pond Y was heavily polluted.
一位生物学家对两个池塘采样。池塘 X 水质清澈，有石蝇稚虫、蜉蝣稚虫和淡水虾。池塘 Y 水体浑浊，有绿色藻类浮渣，只发现鼠尾蛆、污泥蠕虫和红虫。生物学家记录各种生物数量，得出结论：池塘 Y 污染严重。
Some aquatic invertebrates are very sensitive to oxygen levels and pollution. Stonefly and mayfly nymphs require high oxygen concentrations and are intolerant of pollution; they are indicator species for clean water. Rat-tailed maggots and sludge worms can survive in low-oxygen, polluted water. The absence of sensitive species and presence of tolerant species in Pond Y strongly suggests organic pollution, which depletes dissolved oxygen as bacteria decompose the waste.
一些水生无脊椎动物对氧气水平和污染非常敏感。石蝇和蜉蝣稚虫需要高浓度溶解氧，不耐受污染；它们是清洁水体的指示物种。鼠尾蛆和污泥蠕虫能在低氧污染的污水中生存。池塘 Y 缺乏敏感物种且耐受物种存在，强烈表明有机污染，因为细菌分解废物时消耗溶解氧。
This case emphasises the use of biological indicators and environmental data to assess ecosystem health. You may be asked to interpret tables of species abundance and infer the level of pollution.
此案例强调运用生物指标和环境数据评估生态系统健康。你可能被要求解读物种丰度表并推断污染程度。
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