📚 Case Study Masterclass: Biology in Action | 案例分析实战演练：生物学在行动
Welcome to the Year 7 Cambridge Biology case study masterclass. Here, we apply your scientific knowledge to real-world scenarios, experiments, and data. This will sharpen your analytical skills and prepare you for Checkpoint assessments. Each case study is carefully designed to reflect the key topics: cells, life processes, ecosystems, human physiology, plant biology, and more. As you work through them, think like a scientist – observe, question, and explain.
欢迎来到 Year 7 Cambridge 生物案例分析大师班。在这里，我们将你的科学知识应用到真实情境、实验和数据中。这将锻炼你的分析能力，并为 Checkpoint 评估做好准备。每一个案例都是围绕关键主题精心设计的：细胞、生命过程、生态系统、人体生理、植物生物学等。当你逐一分析时，请像科学家一样思考——观察、提问并解释。
1. Pond Water Mystery | 池塘水之谜
Sara collected a drop of pond water and observed it under a microscope. She saw many tiny creatures moving around. She wondered if these were living organisms. What evidence could she look for to prove they are alive? (Hint: Remember MRS GREN.)
萨拉采集了一滴池塘水并在显微镜下观察。她看到许多微小的生物在移动。她想知道这些是否是生物。她可以寻找哪些证据来证明它们是活的？（提示：记住 MRS GREN。）
To decide if something is living, we use the seven life processes summarized by MRS GREN: Movement, Respiration, Sensitivity, Growth, Reproduction, Excretion, and Nutrition. Sara could observe whether the creatures move on their own, grow, or reproduce over time. She might also test if they respond to a gentle light change (sensitivity). Living things must respire to release energy from food, so eventually she could link nutrition to energy use. Even a single-celled organism like an amoeba carries out all seven processes.
要判断某物是否是有生命的，我们使用由 MRS GREN 概括的生命七过程：运动、呼吸、感应、生长、繁殖、排泄和营养。萨拉可以观察这些生物是否自主移动，是否随着时间生长或繁殖。她也可以测试它们是否对轻微的光线变化有反应（感应）。生物必须通过呼吸作用从食物中释放能量，因此最终她可以把营养与能量使用联系起来。即使是像变形虫这样的单细胞生物，也会进行全部七个过程。
2. Heart Rate Investigation | 心率调查
Tom measured his resting pulse: 70 beats per minute (bpm). After running on the spot for 2 minutes, his pulse shot up to 120 bpm. Five minutes after exercise, it had fallen to 85 bpm. Explain why his pulse increased during exercise and what the recovery rate tells us about his fitness.
汤姆测量了他的静息脉搏：每分钟70次（bpm）。原地跑步2分钟后，他的脉搏飙升至120 bpm。运动停止5分钟后，脉搏回落到85 bpm。请解释为什么运动时他的脉搏会加快，以及恢复速率说明了什么关于他的健康状况。
Muscles need more energy during exercise, so they undergo aerobic respiration at a faster rate. Respiration requires oxygen and glucose, and produces carbon dioxide as waste. The heart beats faster to pump oxygen-rich blood to the working muscles more rapidly. The elevated pulse after exercise is partly because the body is repaying an oxygen debt and clearing lactate. A quicker return toward resting heart rate indicates good cardiovascular fitness; Tom’s recovery (from 120 to 85 bpm in 5 minutes) suggests a reasonably fit heart and efficient circulation.
肌肉在运动时需要更多能量，因此它们以更快的速率进行有氧呼吸。呼吸作用需要氧气和葡萄糖，并产生二氧化碳作为废物。心脏加快跳动，以便更快地将富含氧气的血液泵送到工作的肌肉。运动后脉搏依然偏高，部分原因是身体正在偿还氧债并清除乳酸。较快恢复到静息心率表明良好的心血管健康状况；汤姆的恢复情况（5分钟内从120降至85 bpm）说明他有一颗相当健康的心脏和高效的循环系统。
3. What Do Seeds Need to Germinate? | 种子萌发需要什么？
A student set up four test tubes with cress seeds. Tube A: seeds on damp cotton wool at room temperature, in light. Tube B: seeds on dry cotton wool at room temperature, in light. Tube C: seeds on damp cotton wool in a fridge (4 °C), in light. Tube D: seeds on damp cotton wool at room temperature, in a dark box. After three days, only the seeds in Tube A and Tube D germinated. Analyse the results and explain what conditions are essential for germination.
一名学生用独行菜种子设置了四支试管。试管A：种子放在潮湿的棉花上，室温，有光照。试管B：种子放在干燥的棉花上，室温，有光照。试管C：种子放在潮湿的棉花上，置于冰箱（4 °C），有光照。试管D：种子放在潮湿的棉花上，室温，但在暗箱中。三天后，只有试管A和试管D中的种子萌发了。分析结果并解释萌发需要哪些条件。
Tube B lacked water, which is vital to activate enzymes inside the seed and to soften the seed coat so the embryo can swell. Tube C was too cold; enzymes that drive germination work very slowly at low temperatures, delaying growth. Since Tube D germinated in the dark, light is not a requirement for germination. The essential conditions are therefore water (to start metabolic reactions), warmth (for enzyme action), and oxygen (for aerobic respiration). Light only becomes important later, after leaves emerge for photosynthesis.
试管B缺少水，而水对于激活种子内部的酶以及软化种皮使胚能膨胀至关重要。试管C温度太低；驱动萌发的酶在低温下工作非常缓慢，从而延迟了生长。既然试管D在黑暗中萌发了，说明光不是萌发的必要条件。因此，必需的条件是水（启动代谢反应）、温暖（供酶发挥作用）和氧气（用于有氧呼吸）。光只有在叶子长出后进行光合作用时才变得重要。
4. Analysing a Food Diary | 分析饮食日记
Lucy recorded her food intake for one day: Breakfast – sugary cereal with milk; Lunch – white bread sandwich with chocolate spread; Dinner – chips and fried chicken; Snacks – crisps and a cola drink. She often feels tired and catches colds frequently. Identify the food groups missing from her diet and suggest improvements to help her feel healthier.
露西记录了她一天的饮食：早餐——加牛奶的含糖麦片；午餐——涂巧克力酱的白面包三明治；晚餐——薯条和炸鸡；零食——薯片和可乐。她经常感到疲劳并频繁感冒。请找出她膳食中缺少的食物种类，并提出改善建议以帮助她更健康。
Lucy’s diet is rich in fats, sugars, and refined carbohydrates but very low in fresh fruits, vegetables, and wholegrains. She is likely missing vitamin C (found in citrus fruits and peppers), which supports the immune system and could reduce her frequent colds. She also lacks iron (from leafy greens and red meat) needed for healthy red blood cells, which may explain her tiredness. Fibre from vegetables, fruit, and wholemeal bread is almost absent, putting her at risk for digestive problems. Better choices: swap sugary cereal for porridge with berries, replace white bread with wholemeal, include a side salad with dinner, snack on fruit or yogurt, and drink water instead of cola. A balanced diet must contain carbohydrates, proteins, fats, vitamins, minerals, fibre, and water in the right proportions.
露西的饮食富含脂肪、糖和精制碳水化合物，但新鲜水果、蔬菜和全谷物严重不足。她很可能缺乏维生素C（存在于柑橘类水果和甜椒中），维生素C支持免疫系统，可减少感冒频发。她还缺乏铁（来自绿叶蔬菜和红肉），铁是健康红细胞所必需的，这或许可以解释她的疲劳。来自蔬菜、水果和全麦面包的膳食纤维几乎完全缺失，这使她面临消化问题的风险。更好的选择：将含糖麦片换成加浆果的燕麦粥，用全麦面包替代白面包，晚餐配一份沙拉，用水果或酸奶作为零食，喝水替代可乐。均衡膳食必须按适当比例包含碳水化合物、蛋白质、脂肪、维生素、矿物质、纤维和水。
5. The DDT Crisis in a Food Chain | 食物链中的DDT危机
In a lake ecosystem, DDT concentrations were measured in different organisms: water plants – 0.04 parts per million (ppm); small fish – 0.5 ppm; large fish – 2 ppm; osprey (a bird of prey) – 25 ppm. Explain why the osprey had the highest concentration and name the process involved. Predict how this might affect the osprey population.
在一个湖泊生态系统中，测量了不同生物体内的DDT浓度：水生植物——0.04 ppm；小鱼——0.5 ppm；大鱼——2 ppm；鱼鹰（一种猛禽）——25 ppm。解释为什么鱼鹰体内的浓度最高，并指出所涉及的过程名称。预测这可能对鱼鹰种群产生什么影响。
The process is called biomagnification (or bioaccumulation). DDT is a persistent pesticide that is not easily broken down; it gets stored in the fatty tissues of organisms. When small fish eat many contaminated water plants, the DDT accumulates in their bodies. When large fish eat many small fish, the chemical becomes even more concentrated. At the top of the food chain, the osprey consumes many large fish, resulting in the highest DDT load. High levels of DDT can weaken eggshells, causing them to break during incubation. This reduces the number of chicks that hatch, leading to a decline in the osprey population over time.
这一过程称为生物放大（或生物积累）。DDT是一种持久性杀虫剂，不易分解；它储存在生物体的脂肪组织中。当小鱼吃了许多受污染的水生植物时，DDT在其体内积累。当大鱼吃掉许多小鱼时，化学物质变得更加浓缩。在食物链顶端，鱼鹰吃了许多大鱼，导致其体内DDT负担最重。高浓度的DDT会使蛋壳变薄，导致在孵化过程中破裂。这降低了雏鸟的孵化数量，随着时间的推移会导致鱼鹰种群数量下降。
6. Plants Bending Towards Light | 植物弯向光生长
A student placed a potted plant on a windowsill. After one week, the stem had bent strongly toward the window. Another identical plant was rotated a quarter turn each day and grew almost perfectly straight. Explain how plants detect light direction and why the first plant bent. Name the plant hormone responsible and describe how it works.
一名学生将一盆植物放在窗台上。一周后，茎强烈地弯向窗户一侧。另一盆相同的植物每天旋转四分之一圈，结果几乎笔直生长。解释植物如何探测光线方向，以及为什么第一盆植物弯曲了。说出负责的植物激素名称并描述其作用方式。
This growth response is called phototropism – specifically positive phototropism in shoots. The key hormone is auxin. When light shines from one side, auxin produced in the shoot tip moves to the shaded side. Higher auxin concentration on the shaded side causes the cells there to elongate faster than the cells on the lit side. This unequal growth pushes the stem to bend towards the light, placing leaves in the best position for photosynthesis. Rotating the plant distributes auxin evenly around the stem, so all sides grow at similar rates, keeping the shoot straight.
这种生长反应称为向光性——具体来说，是茎的正向光性。关键的激素是生长素。当光线从一侧照射时，茎尖产生的生长素会移动到背光侧。背光侧较高的生长素浓度导致该侧细胞比向光侧细胞伸长得更快。这种不均匀的生长推动茎弯向光源，使叶片处于进行光合作用的最佳位置。旋转植物使生长素均匀分布在茎的周围，因此各侧生长速率相近，保持茎干直立。
7. The Wilting Plant Mystery | 枯萎植物之谜
A healthy potted geranium was accidentally left on a sunny windowsill without water for two days. Its leaves drooped and felt soft. After the plant was watered generously, the leaves became firm again within a few hours. Explain, in terms of cell structure and water movement, why the plant wilted and how it recovered.
一盆健康的天竺葵偶然被遗忘在阳光充足的窗台上两天没浇水。它的叶子下垂且摸上去很软。大量浇水后几小时内，叶子又变得坚挺了。从细胞结构和水分移动的角度解释，植物为什么会枯萎以及它是如何恢复的。
Plant cells are normally turgid because their large central vacuole is full of water, pushing the cytoplasm against the rigid cell wall. This turgor pressure gives stems and leaves their firmness. When water is scarce, cells lose water by osmosis; the vacuole shrinks and the cytoplasm pulls away from the cell wall – a state called plasmolysis. The loss of turgor causes the plant to wilt. Watering restores turgidity because water enters the roots by osmosis, travels up through xylem vessels, and moves into leaf cells. The vacuoles refill, the cytoplasm presses against the cell wall again, and the plant becomes upright and firm.
植物细胞通常是挺直的，因为它们巨大的中央液泡充满了水，把细胞质推压到坚硬的细胞壁上。这种膨压赋予了茎和叶坚挺的状态。当缺水时，细胞通过渗透作用失去水分；液泡缩小，细胞质从细胞壁分离——这种状态称为质壁分离。膨压的丧失导致植物枯萎。浇水恢复挺直状态的原因是：水通过渗透作用进入根部，向上经由木质部导管运输，然后进入叶细胞。液泡重新充满，细胞质再次推压细胞壁，植株恢复挺立和坚挺。
8. Mouldy Bread Experiment | 发霉的面包实验
Four slices of bread were treated differently: Slice 1 – kept dry at room temperature; Slice 2 – moistened and kept at 25 °C; Slice 3 – moistened and kept in a fridge at 4 °C; Slice 4 – moistened and sealed in a plastic bag with no air, kept at 25 °C. After five days, only Slice 2 showed visible mould growth. Explain the results and state what conditions fungi need for growth. Suggest a method to prevent food spoilage using this knowledge.
四片面包被以不同方式处理：切片1——在室温下保持干燥；切片2——湿润并在25 °C下存放；切片3——湿润并在4 °C的冰箱中存放；切片4——湿润并密封在无空气的塑料袋中，25 °C存放。五天后，只有切片2出现了肉眼可见的霉菌。解释实验结果，并说明真菌生长需要哪些条件。基于此知识提出一种防止食物腐败的方法。
Mould is a type of fungus. For active growth, fungi require moisture, warmth, and oxygen. Slice 1 lacked moisture, so fungal spores could not germinate. Slice 3 was placed in a fridge; low temperatures slow down enzyme reactions inside the mould cells, preventing visible growth within the five days. Slice 4 was sealed from oxygen, so the mould could not carry out aerobic respiration to grow. Slice 2 provided all three conditions – moisture, suitable warmth, and oxygen – allowing spores to germinate and hyphae to spread. To prevent food spoilage, we can exploit these needs: keep food dry (e.g., crisps), store in a refrigerator, or use airtight packaging to restrict oxygen.
霉菌是一种真菌。为了活跃地生长，真菌需要水分、温暖和氧气。切片1缺少水分，因此真菌孢子无法萌发。切片3被放在冰箱里；低温减缓了霉菌细胞内酶的反应速度，在五天内阻止了可见的生长。切片4与氧气隔绝，因此霉菌无法进行有氧呼吸来生长。切片2提供了全部三个条件——水分、适宜的温度和氧气——使孢子能够萌发，菌丝得以扩展。为了防止食物腐败，我们可以利用这些需求：保持食物干燥（如薯片），放入冰箱储存，或使用密封包装限制氧气供应。
9. Investigating Lung Capacity | 肺活量探究
Three students of the same age but different activity levels measured their peak flow (a measure of how fast air can be blown out). John, a swimmer: 450 L/min; Mia, who plays video games most days: 320 L/min; Ahmed, who enjoys cycling: 400 L/min. Suggest a possible explanation for the differences. Why do physically active people often have greater lung capacity?
三名同年龄但运动量不同的学生测量了他们的峰值流速（衡量吹出空气速度的指标）。约翰，一名游泳运动员：450 L/min；米娅，大部分时间玩电子游戏：320 L/min；艾哈迈德，喜欢骑自行车：400 L/min。为这些差异提出可能的解释。为什么经常进行体育锻炼的人通常肺活量更大？
Lung capacity and the strength of respiratory muscles (diaphragm and intercostal muscles) improve with regular aerobic exercise. Swimmers and cyclists repeatedly train their lungs to take in more oxygen and exhale carbon dioxide efficiently. This increases the volume of air that can be moved in and out of the lungs and the speed of exhalation. John’s high peak flow reflects strong respiratory muscles and healthy lung tissue built through swimming. Ahmed’s is also above average due to cycling. Mia’s sedentary lifestyle means her respiratory muscles are not challenged, so her lung function is comparatively lower. Regular exercise also increases the number of capillaries around alveoli, enhancing gas exchange and overall stamina.
肺活量和呼吸肌（膈肌和肋间肌）的力量会通过定期的有氧运动得到提高。游泳运动员和骑自行车的人反复训练肺部，使其能够更有效地吸入更多氧气并呼出二氧化碳。这增加了进出肺部的空气量以及呼气的速度。约翰较高的峰值流速反映了通过游泳锻炼出的强壮呼吸肌和健康肺部组织。艾哈迈德由于骑自行车，数值也高于平均水平。米娅久坐不动的生活方式意味着她的呼吸肌未受到挑战，因此她的肺功能相对较低。定期锻炼还会增加肺泡周围毛细血管的数量，增强气体交换和整体耐力。
10. Classification Challenge at the Zoo | 动物园的分类挑战
A zookeeper gave visitors a set of pictures of five vertebrates: an eagle, a frog, a shark, a kangaroo, and a lizard. Using a simple dichotomous key, they had to sort the animals into their classes. The first question asked: ‘Does the animal have feathers?’ If yes, it is a bird. If no, it goes to the next question: ‘Does it have hair or fur?’ etc. Predict how the animals would be classified and list one key feature of each class.
动物园管理员给游客提供了五种脊椎动物的图片：鹰、青蛙、鲨鱼、袋鼠和蜥蜴。他们需要使用一个简单的二分式检索表，将这些动物分为各自的纲。第一个问题是：“动物有羽毛吗？”如果有，它就是鸟类。如果没有，则进入下一个问题：“它有毛发或皮毛吗？”等等。预测这些动物将如何被分类，并列出每个纲的一个关键特征。
Using a dichotomous key: The eagle has feathers, so it is a bird (Aves) – key feature: feathers and lay hard-shelled eggs. Remaining animals lack feathers. Kangaroo has hair/fur and produces milk, so it is a mammal (Mammalia) – key feature: hair and mammary glands. The frog has smooth, moist skin and begins life in water, so it is an amphibian (Amphibia) – key feature: permeable skin and life cycle involves metamorphosis. The shark has gills and a skeleton made of cartilage, placing it as a cartilaginous fish (Chondrichthyes), a class of fish – key feature: cartilage skeleton and gills. The lizard has dry, scaly skin and lays leathery eggs on land, making it a reptile (Reptilia) – key feature: scales and lungs for breathing. A dichotomous key always splits organisms into two groups at each step based on observable features.
使用二分式检索表：鹰有羽毛，所以它是鸟类（鸟纲）——关键特征：有羽毛，产硬壳卵。其余的动物没有羽毛。袋鼠有毛发/皮毛并且分泌乳汁，所以它是哺乳动物（哺乳纲）——关键特征：毛发和乳腺。青蛙有光滑湿润的皮肤，生命始于水中，所以它是两栖动物（两栖纲）——关键特征：可渗透的皮肤及生命周期包括变态发育。鲨鱼有鳃且骨骼由软骨构成，属于软骨鱼纲，是鱼类的一个纲——关键特征：软骨骨骼和鳃。蜥蜴有干燥、带鳞片的皮肤并在陆地上产有韧性的卵，因此它是爬行动物（爬行纲）——关键特征：鳞片和用于呼吸的肺。二分式检索表在每一步都根据可观察的特征将生物分为两个组。
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