📚 Year 8 AQA Biology: Case Study Practical Exercises | 八年级 AQA 生物：案例分析实战演练
This article provides a series of case study practical exercises for Year 8 AQA Biology. Each case study presents a real-world scenario or experimental data, requiring you to apply your knowledge of key biological concepts. Work through each example to build your analytical skills and deepen your understanding of ecosystems, human body systems, cells, photosynthesis, variation, and more.
本文为八年级 AQA 生物课提供一系列案例分析实战演练。每个案例呈现真实情境或实验数据，要求你运用关键生物学概念。逐一练习以培养分析能力，加深对生态系统、人体系统、细胞、光合作用、变异等主题的理解。
1. Pond Food Web | 池塘食物网
In a local pond, the following organisms were found: algae, water fleas, small fish, perch (a predatory fish), and herons. Algae use sunlight to photosynthesise and are the producers. Water fleas feed on algae. Small fish eat water fleas. Perch eat small fish, and herons eat perch.
在当地一个池塘中，发现了下列生物：藻类、水蚤、小鱼、鲈鱼（一种捕食性鱼）和苍鹭。藻类利用阳光进行光合作用，是生产者。水蚤以藻类为食。小鱼吃水蚤。鲈鱼吃小鱼，苍鹭吃鲈鱼。
Question: A chemical spill kills most of the water fleas. Predict and explain the short-term effects on the populations of algae, small fish, and perch.
问题：一次化学品泄漏杀死了大部分水蚤。预测并解释对藻类、小鱼和鲈鱼种群的短期影响。
Answer: With fewer water fleas, fewer algae are eaten, so the algae population increases. Since water fleas are the small fish's main food, small fish numbers will decline due to starvation. Perch rely on small fish for food; their population will also drop as a result. This demonstrates how interconnected organisms are in a food web.
解答：水蚤数量减少意味着被吃掉的藻类变少，因此藻类数量增加。由于水蚤是小鱼的主要食物，小鱼因饥饿而减少。鲈鱼依赖小鱼为食，其数量也会随之下降。这说明食物网中生物之间是如何紧密相连的。
2. Digestion of a Ham Sandwich | 三明治的消化
A student eats a ham sandwich. The bread contains starch, while the ham provides protein and fat. Using your knowledge of enzymes, describe how the sandwich is digested in different parts of the gut.
一名学生吃了一个火腿三明治。面包含有淀粉，火腿提供蛋白质和脂肪。运用你对酶的知识，描述三明治在消化道各部分如何被消化。
In the mouth, salivary amylase begins breaking starch down into smaller sugars like maltose. The food is then swallowed and enters the stomach, where pepsin (a protease) starts digesting the ham's protein into polypeptides. The stomach's acidic environment helps pepsin work. In the small intestine, pancreatic amylase continues starch digestion, trypsin further breaks down proteins, and lipase digests fats into fatty acids and glycerol. Bile from the liver emulsifies fats, increasing the surface area for lipase action. The small, soluble molecules are then absorbed into the bloodstream.
在口腔中，唾液淀粉酶开始将淀粉分解为麦芽糖等较小的糖。食物然后被吞咽进入胃，胃蛋白酶（一种蛋白酶）开始将火腿中的蛋白质分解为多肽。胃中的酸性环境有助于胃蛋白酶工作。在小肠中，胰淀粉酶继续消化淀粉，胰蛋白酶进一步分解蛋白质，脂肪酶将脂肪消化为脂肪酸和甘油。来自肝脏的胆汁乳化脂肪，增大脂肪酶作用的表面积。这些可溶的小分子随后被吸收进入血液。
3. Breathing Rate and Exercise | 运动与呼吸频率
A student measured their breathing rate at rest and after 5 minutes of slow jogging. The results are shown in the table below.
一名学生测量了自己在休息时和慢跑5分钟后的呼吸频率。结果如下表所示。
| Condition | Breathing rate (breaths/min) |
|---|---|
| At rest | 12 |
| After slow jogging | 30 |
Explain why the breathing rate increases during exercise.
解释为什么运动时呼吸频率会增加。
During exercise, muscles contract more frequently and require more energy. This energy is released through aerobic respiration, which uses oxygen and glucose to produce ATP, carbon dioxide, and water. The word equation is:
运动时肌肉收缩更频繁，需要更多能量。这种能量通过有氧呼吸释放，该过程利用氧气和葡萄糖产生 ATP、二氧化碳和水。文字方程式为：
glucose + oxygen → carbon dioxide + water (+ energy)
葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）
To supply more oxygen and remove the increased carbon dioxide, the breathing rate and heart rate both rise. This ensures the blood carries enough oxygen to the muscle cells and eliminates waste CO₂ efficiently.
为了提供更多氧气并清除增多的二氧化碳，呼吸频率和心率都会升高。这确保血液携带足够的氧气到达肌肉细胞，并有效清除废物 CO₂。
4. Onion Cell Observation | 洋葱细胞观察
ctive lens. They measured the length of one cell in the drawing as 2.5 cm. Calculate the actual length of the cell and describe its visible structures.
一名学生制作了洋葱表皮装片，用 ×10 目镜和 ×40 物镜观察。他们测量绘图中一个细胞的长度为 2.5 厘米。计算该细胞的实际长度，并描述可见的结构。
Total magnification = eyepiece magnification × objective magnification = 10 × 40 = ×400.
总放大倍数 = 目镜放大倍数 × 物镜放大倍数 = 10 × 40 = ×400。
Actual size = image size ÷ magnification = 2.5 cm ÷ 400 = 0.00625 cm = 0.0625 mm = 62.5 µm.
实际大小 = 图像大小 ÷ 放大倍数 = 2.5 cm ÷ 400 = 0.00625 cm = 0.0625 mm = 62.5 µm。
Under the microscope, the student should observe a rectangular cell with a clear cell wall, a nucleus visible as a darker spot, and a large central vacuole. The cytoplasm may appear granular. No chloroplasts are seen in onion epidermis cells because they grow underground and do not photosynthesise.
在显微镜下，学生应观察到矩形细胞，有明显的细胞壁，可以看见作为深色点的细胞核，以及一个大型中央液泡。细胞质可能呈颗粒状。洋葱表皮细胞中看不到叶绿体，因为它们生长在地下，不进行光合作用。
5. Photosynthesis Experiment | 光合作用实验
A student placed a piece of Elodea pondweed in a beaker of water and counted oxygen bubbles released at different distances from a lamp. The results are recorded below.
一名学生将一根伊乐藻放入盛水的烧杯中，计数在不同距离的灯光下释放的氧气泡数。结果记录如下。
| Distance from lamp (cm) | Bubbles per minute |
|---|---|
| 10 | 35 |
| 20 | 18 |
| 30 | 8 |
Analyse the relationship between light intensity and the rate of photosynthesis. Include the balanced symbol equation for photosynthesis.
分析光照强度与光合作用速率之间的关系。写出光合作用的平衡化学方程式。
The data show that as the lamp is moved closer (higher light intensity), the rate of bubbling increases dramatically. This is because light provides the energy needed to drive the photosynthesis reaction. The balanced equation is:
数据表明，当灯移得更近（光照强度增加）时，冒泡速率显著增加。这是因为光提供了驱动光合作用反应所需的能量。平衡方程式为：
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Light energy is absorbed by chlorophyll in the chloroplasts to convert carbon dioxide and water into glucose and oxygen. At greater distances, less light energy reaches the plant, so the rate of photosynthesis decreases.
叶绿体中的叶绿素吸收光能，将二氧化碳和水转化为葡萄糖和氧气。距离更远时，到达植物的光能减少，因此光合作用速率下降。
6. Variation in Sunflowers | 向日葵的变异
A class measured the heights of 20 sunflower plants (in cm): 120, 135, 140, 125, 130, 138, 142, 128, 136, 133, 127, 131, 137, 129, 134, 126, 130, 139, 132, 128. Plot a frequency diagram and determine whether this is continuous or discontinuous variation. Explain the causes of such variation.
一个班级测量了 20 株向日葵的高度（厘米）：120、135、140、125、130、138、142、128、136、133、127、131、137、129、134、126、130、139、132、128。绘制频率分布图，判断这是连续变异还是不连续变异。解释这种变异的原因。
Grouping the data into intervals of 5 cm (120-124, 125-129, etc.) shows a normal distribution with most plants around 130-134 cm. This is continuous variation – the characteristic can take any value within a range and is influenced by both genes and the environment. In sunflowers, height depends on multiple genes (polygenic inheritance) as well as factors such as light, water, and mineral availability. Continuous variation produces a bell-shaped curve.
将数据按 5 厘米分组（120–124、125–129 等）后，显示出大多数植株集中在 130–134 厘米的正态分布。这是连续变异——该特征可以在一个范围内取任意值，受基因和环境共同影响。在向日葵中，高度取决于多个基因（多基因遗传）以及光照、水分和矿物质等因素。连续变异产生钟形曲线。
7. Drug Absorption Simulation | 药物吸收模拟
To model how digested food is absorbed in the small intestine, a student set up a Visking tubing containing a mixture of starch and glucose solution. The tubing was placed in a boiling tube of water. At the start, the water outside the tubing was tested with iodine solution (which remained yellow-brown) and Benedict's reagent (which remained blue). After 30 minutes, the outside water turned blue-black with iodine and gave an orange-red precipitate with Benedict's reagent after heating. Explain this outcome and link it to intestinal absorption.
为了模拟消化后的食物如何在小肠被吸收，一名学生设置了装有淀粉和葡萄糖混合液的Visking 透析管。透析管放入一个装有水的试管中。实验开始时，管外的水用碘液测试（保持黄棕色），用本尼迪特试剂测试（加热后保持蓝色）。30 分钟后，管外的水遇碘液变为蓝黑色，用本尼迪特试剂加热后产生橙红色沉淀。解释这一结果并将其与肠道吸收联系起来。
The Visking tubing has tiny pores that represent the selectively permeable membrane of the small intestine. Glucose molecules are small enough to diffuse through the pores into the surrounding water, while the much larger starch molecules cannot move out. The blue-black colour with iodine proves starch is still inside the tubing (starch present), but the outside water tests positive for reducing sugar with Benedict's reagent, showing glucose has diffused out. This is exactly how the small intestine absorbs small nutrient molecules like glucose and amino acids into the blood, while larger undigested particles remain in the gut.
Visking 透析管具有微小孔隙，代表小肠的选择性渗透膜。葡萄糖分子足够小，可以通过孔隙扩散到周围水中，而大得多的淀粉分子无法通过。碘液变蓝黑色证明淀粉仍在管内（有淀粉），但外部水用本尼迪特试剂检测出还原糖呈阳性，表明葡萄糖已经扩散出来。这正是小肠将葡萄糖、氨基酸等小分子营养物质吸收进入血液的方式，而未消化的大颗粒则留在肠道中。
8. The Carbon Cycle Mystery | 碳循环谜题
A lake ecosystem was studied over a year. Scientists observed that during summer, the CO₂ concentration in the water decreased during the day, while at night it increased. In autumn, dead leaves entering the lake led to a rise in decomposer activity and CO₂ levels. Additionally, nearby factories burning fossil fuels raised atmospheric CO₂, which dissolved into the lake. Construct a simple carbon cycle for the lake and explain how human activities might unbalance it.
研究人员对一个湖泊生态系统进行了为期一年的研究。他们观察到，夏季白天水中 CO₂ 浓度下降，夜晚则上升。秋季，落入湖中的枯叶使分解者活动增加，CO₂ 水平上升。此外，附近工厂燃烧化石燃料提高了大气 CO₂ 浓度，并溶入湖水中。构建该湖的简单碳循环，并解释人类活动可能如何使其失衡。
plants, fish, decomposers) returns CO₂ (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O); decomposition of dead matter releases CO₂; and combustion of fossil fuels adds extra CO₂. Normally, these processes are balanced. However, human activities such as burning fossil fuels and deforestation increase CO₂ input, causing more carbon to dissolve in the lake. This can make the water more acidic and affect organisms like shelled molluscs. Extra CO₂ also intensifies the greenhouse effect, altering temperatures and further disturbing the cycle.
湖泊碳循环包括：水生植物和藻类的光合作用吸收 CO₂（6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂）；所有生物（植物、鱼、分解者）的呼吸作用释放 CO₂（C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O）；死亡物质的分解释放 CO₂；以及化石燃料的燃烧额外增加 CO₂。正常情况下这些过程是平衡的。然而，燃烧化石燃料和砍伐森林等人类活动增加了 CO₂ 输入，导致更多碳溶入湖中。这会使水体变酸，影响带壳的软体动物等生物。额外的 CO₂ 还会加剧温室效应，改变气温，进一步扰乱循环。
Published by TutorHao | Biology Revision Series | aleveler.com
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