📚 Year 7 SQA Biology: Case Study Practical Drill | Year 7 SQA 生物：案例分析实战演练
Welcome to this case-study practical drill for Year 7 SQA Biology. In real examinations, you are often given a scenario that tests your ability to apply knowledge, analyse data, and draw conclusions. This article presents a series of realistic biology case studies covering ecosystems, body systems, photosynthesis, disease and more. For each case, we provide a model analysis that demonstrates how to break down a problem, use correct terminology and reach a sound scientific judgement. Studying these worked examples will sharpen your investigative skills and deepen your understanding of key biological concepts.
欢迎来到 Year 7 SQA 生物案例分析实战演练。在实际考试中，你经常需要根据给定情景应用知识、分析数据并得出结论。本文围绕生态系统、人体系统、光合作用、疾病等多个主题，提供一系列真实的生物案例。每个案例都附有示范分析，教你如何拆解问题、使用正确术语并做出严谨的科学判断。学习这些范例可以锻炼你的探究能力，加深对核心生物概念的理解。
1. Pond Food Web Analysis | 池塘食物网分析
A garden pond contains the following organisms: pondweed, microscopic algae, water fleas, small fish, frogs and a grey heron. The food web can be summarised as: pondweed and algae are eaten by water fleas; water fleas are eaten by small fish; small fish are eaten by frogs; frogs are eaten by herons. Sometimes water fleas also feed directly on algae, and small fish can eat both water fleas and algae.
一座花园池塘里生活着以下生物：池塘草、微藻、水蚤、小鱼、青蛙和苍鹭。食物网可以概括为：池塘草和藻类被水蚤取食；水蚤被小鱼取食；小鱼被青蛙取食；青蛙被苍鹭取食。有时水蚤也直接摄食藻类，而小鱼可以同时吃水蚤和藻类。
First, identify the producers in this community. Producers are organisms that can make their own food by photosynthesis, using light energy, carbon dioxide and water. Pondweed and microscopic algae are the only producers because they contain chlorophyll and convert inorganic substances into biomass. All other organisms are consumers; they obtain energy by eating other living things.
首先，找出这个群落中的生产者。生产者是能利用光能、二氧化碳和水通过光合作用制造自身食物的生物。池塘草和微藻是仅有的生产者，因为它们含有叶绿素，能将无机物转化为生物量。其他所有生物都是消费者；它们通过取食其他生物获得能量。
Now, suppose a fungal disease kills most of the frog population. Predict and explain the likely impact on the food web. With fewer frogs, the predation pressure on small fish would decrease, so the small fish population would initially increase. An expanded small fish population would then consume more water fleas, causing the water flea numbers to drop. Meanwhile, pondweed and algae might increase because fewer water fleas are feeding on them. However, after some time the small fish themselves might overgraze and then experience a food shortage, leading to a natural fluctuation. This case shows how removing one predator can cause a cascade of effects throughout the food web.
现在假设一种真菌疾病杀死了大部分青蛙种群。请预测并解释这对食物网可能产生的影响。青蛙数量减少后，小鱼的捕食压力下降，因此小鱼数量初期会增加。增多的鱼群会吃掉更多的水蚤，导致水蚤数量下降。同时，由于取食水蚤的生物减少，池塘草和藻类的数量可能上升。但一段时间后，小鱼自身可能因食物短缺而数量回落，导致自然波动。这个案例说明，移除一种捕食者会在整个食物网中引发一系列连锁效应。
Finally, state the direction of energy flow in this food web. Energy enters the system from the Sun, is captured by producers during photosynthesis, and then transferred from one trophic level to the next as organisms are eaten. At each step, some energy is lost as heat through respiration, so the amount available to top predators is much smaller than the energy stored in producers.
最后，陈述该食物网中的能量流动方向。能量从太阳进入系统，被生产者通过光合作用捕获，然后随着生物被取食从一个营养级传递到下一个营养级。每一步都有部分能量通过呼吸作用以热的形式散失，因此传递给顶级捕食者的能量远少于储存在生产者中的能量。
2. Journey Through the Human Digestive System | 人体消化系统之旅
A student eats a sandwich containing cheese and bread. Describe the physical and chemical digestion that takes place as the meal passes through the human digestive tract. Begin in the mouth, where teeth physically break down the food by chewing. Saliva contains an enzyme called amylase, which starts digesting starch from the bread into smaller sugars. The tongue shapes the chewed food into a bolus, which is swallowed.
一名学生吃了一个含有奶酪和面包的三明治。请描述这顿餐食经过人体消化道时所发生的物理消化和化学消化。从口腔开始：牙齿通过咀嚼将食物物理磨碎。唾液中含有一种叫做淀粉酶的酶，开始将面包中的淀粉分解成较小的糖。舌把咀嚼过的食物塑造成食团，然后吞咽。
In the stomach, the sandwich encounters gastric juice containing hydrochloric acid and a protease enzyme. The acid kills many bacteria unharmed by chewing and creates the acidic conditions that the protease needs to work. The protease begins the chemical digestion of protein from the cheese, breaking it down into shorter chains of amino acids. The muscular walls of the stomach churn the food, continuing physical digestion and producing a semi-liquid mixture called chyme.
在胃里，三明治遇到含有盐酸和一种蛋白酶的胃液。盐酸能杀死咀嚼中未被消灭的许多细菌，并为蛋白酶提供所需的酸性环境。蛋白酶开始化学消化奶酪中的蛋白质，将其分解为较短的氨基酸链。胃的肌肉壁搅动食物，继续物理消化，产生一种叫做食糜的半流体混合物。
The chyme is then released slowly into the small intestine. Here, digestive juices from the pancreas and the intestinal wall supply more enzymes, including lipases that digest fats from the cheese, and proteases and amylases that complete the breakdown of protein and starch. The inner surface of the small intestine is folded into millions of tiny finger-like villi, which massively increase the surface area for absorption. Digested nutrients pass into the bloodstream and are carried to cells throughout the body. Undigested material continues into the large intestine, where water is reabsorbed, and the remaining waste is formed into faeces for excretion.
食糜随后缓慢释放进入小肠。在这里，来自胰腺和肠壁的消化液提供更多的酶，包括消化奶酪中脂肪的脂肪酶，以及完成蛋白质和淀粉分解的蛋白酶和淀粉酶。小肠内表面折叠成数百万个微小的指状绒毛，极大地增加了吸收的表面积。已消化的营养物质进入血液，并被输送到全身细胞。未消化的物质继续进入大肠，在那里水分被重吸收，剩余废物形成粪便排出体外。
3. Photosynthesis Investigation Using Pondweed | 用池塘草探究光合作用
A scientist places freshly cut pieces of pondweed (Elodea) into a beaker of water and counts the oxygen bubbles released per minute under different light conditions. The results are recorded in the table below.
一名科学家将刚剪下的伊乐藻放入一杯水中，并记录在不同光照条件下每分钟释放的氧气泡数量。结果记录在下表中。
| Condition | 条件 | Bubbles per minute | 每分钟气泡数 |
|---|---|---|---|
| Complete darkness | 全黑暗 | 0 | 0 |
| Low light (lamp 30 cm away) | 低光照（灯距30 cm） | 5 | 5 |
| Bright light (lamp 10 cm away) | 强光照（灯距10 cm） | 18 | 18 |
Analyse the data and explain what the bubbles represent. The bubbles are oxygen gas, a waste product of the light-dependent stage of photosynthesis. In complete darkness, no light energy is available, so photosynthesis does not occur. As light intensity increases, the rate of photosynthesis rises, producing more oxygen. The relationship is approximately proportional over this range. This investigation confirms that light is a necessary factor for the photosynthesis reaction.
分析数据并解释这些气泡代表什么。这些气泡是氧气，即光合作用光反应阶段的废物。在全黑暗条件下，没有可利用的光能，因此光合作用不发生。随着光照强度增加，光合作用速率上升，产生更多氧气。在此范围内，二者的关系大致成正比。这一探究证实，光是光合作用反应的必要因素。
glucose + oxygen, in the presence of light and chlorophyll. The balanced chemical equation can be written as:
光合作用的完整文字表达式是：二氧化碳 + 水 → 葡萄糖 + 氧气，需要光和叶绿素。配平的化学方程式可写作：
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
If the scientist added a sodium hydrogencarbonate solution to the water, more carbon dioxide would be available, and the rate of bubbling would increase even further, showing that carbon dioxide concentration is another limiting factor.
如果科学家向水中加入碳酸氢钠溶液，二氧化碳将更加充足，气泡释放速率会进一步提高，这表明二氧化碳浓度是另一个限制因素。
4. Analysing the Human Heart and Circulation | 分析人体心脏与循环
A diagram of the human heart shows four chambers: left atrium, left ventricle, right atrium and right ventricle. Blood from the body enters the right atrium through the vena cava, then passes to the right ventricle, which pumps it to the lungs via the pulmonary artery. In the lungs, blood picks up oxygen and releases carbon dioxide. Oxygenated blood returns to the left atrium through the pulmonary vein, moves into the left ventricle, and is then pumped out to the body through the aorta.
人体心脏示意图显示了四个腔室：左心房、左心室、右心房和右心室。来自身体的血液通过腔静脉进入右心房，然后进入右心室，右心室将血液经肺动脉泵送到肺部。在肺部，血液吸收氧气并释放二氧化碳。含氧血通过肺静脉返回左心房，进入左心室，再经主动脉泵送到全身。
Compare the thickness of the muscular walls of the left and right ventricles. The left ventricle has a much thicker wall because it must pump blood all around the body, a relatively high-pressure circuit. In contrast, the right ventricle only needs to send blood to the nearby lungs, a shorter distance requiring less force. This structural difference is an adaptation that makes the double circulatory system efficient.
比较左心室和右心室肌肉壁的厚度。左心室壁厚得多，因为它需要将血液泵送至全身，这是一个相对高压的循环。相比之下，右心室只需将血液送到附近的肺部，距离较短，所需力量较小。这一结构差异是一种适应性，使双循环系统高效运转。
If a child's heart had a small hole between the left and right ventricles (a septal defect), predict what would happen to oxygen delivery. Some oxygen-rich blood from the left ventricle would leak into the right ventricle and be sent to the lungs unnecessarily, reducing the volume of oxygenated blood reaching the body. Over time, the heart works harder and may enlarge. This scenario illustrates how a single structural change can affect the entire circulatory performance.
如果一名儿童心脏左右心室之间存在一个小孔（间隔缺损），请预测氧气输送会发生什么变化。来自左心室的部分富氧血液会漏入右心室，被不必要地送往肺部，从而减少到达身体的富氧血液量。时间一长，心脏会更加费力工作并可能扩大。这个情景说明了单一结构改变如何影响整个循环系统的性能。
5. Bioaccumulation of DDT in a Food Chain | DDT 在食物链中的生物积累
Farmers sprayed a lake with the pesticide DDT to control mosquitoes. Even though the concentration in water was very low, scientists later measured DDT levels in the tissues of organisms at different trophic levels. The data are shown below.
农民向一个湖泊喷洒杀虫剂 DDT 以控制蚊虫。尽管水体中浓度极低，科学家随后测量了不同营养级生物组织中的 DDT 水平。数据如下。
| Organism | 生物 | DDT concentration (ppm) | DDT 浓度（ppm） |
|---|---|---|---|
| Water | 水 | 0.0003 | 0.0003 |
| Phytoplankton | 浮游植物 | 0.04 | 0.04 |
| Zooplankton | 浮游动物 | 0.5 | 0.5 |
| Small fish | 小鱼 | 2.0 | 2.0 |
| Large fish | 大鱼 | 5.0 | 5.0 |
| Fish-eating bird | 食鱼鸟 | 25.0 | 25.0 |
Explain the pattern: DDT is a persistent chemical that does not break down easily in the environment. At each trophic level, organisms consume many individuals from the level below, so the DDT stored in each prey animal accumulates in the consumer's body. Over time, the concentration increases dramatically at the top of the food chain — a phenomenon called biomagnification. Top predators such as fish-eating birds therefore suffer the greatest toxic effects, including thinning of eggshells and reproductive failure.
解释这一模式：DDT 是一种持久性化学物质，在环境中不易分解。在每一个营养级，生物个体都会捕食大量下一级的生物，因此每一头猎物体内储存的 DDT 都累积在捕食者体内。随时间推移，食物链顶端的浓度急剧升高——这一现象称为生物放大作用。因此，像食鱼鸟这样的顶级捕食者会遭受最明显的毒性影响，包括蛋壳变薄和繁殖失败。
What lesson does this case teach about the use of pesticides? It shows that even a small amount of a toxin can become concentrated enough to threaten top carnivores and disrupt ecosystem balance. This is why scientists must assess the long-term effects of chemicals before they are released into the environment.
该案例关于杀虫剂使用教会我们什么？它表明，即使极小剂量的毒素也能浓缩到足以威胁顶级食肉动物并破坏生态系统平衡的程度。这就是科学家必须在化学物质被释放到环境之前评估其长期影响的原因。
6. Antibiotic Resistance in Bacteria | 细菌的抗生素耐药性
A 14-year-old girl developed a sore throat caused by Streptococcus bacteria. The doctor prescribed a course of the antibiotic penicillin. After three days, the girl’s symptoms had not improved, and a laboratory test showed that some of the bacteria had become resistant to penicillin. Explain how this resistance developed.
一名 14 岁女孩因链球菌引起咽喉痛。医生开了一个疗程的抗生素青霉素。三天后，女孩症状未见改善，实验室检测显示部分细菌已对青霉素产生耐药性。试解释这种耐药性是如何出现的。
Within any bacterial population, a few individuals may carry a random gene mutation that gives them resistance to an antibiotic. When the patient started taking penicillin, the antibiotic killed the sensitive, non-resistant bacteria, but the resistant ones survived. With less competition, the resistant bacteria multiplied rapidly. Because the girl may have stopped taking the medicine early or missed doses, the remaining resistant population was able to thrive and cause a persistent infection.
在任何细菌种群中，少数个体可能携带随机基因突变，使其对抗生素具有抵抗力。当病人开始服用青霉素时，抗生素杀死了敏感的、无抗性的细菌，但耐药菌存活了下来。由于竞争减弱，耐药菌迅速繁殖。由于女孩可能提前停药或漏服，存活的耐药菌群得以增殖并导致持续感染。
This case illustrates natural selection in action. The antibiotic was the selection pressure, and resistant bacteria had a selective advantage. To slow the spread of resistance, doctors advise patients to always complete the full course of antibiotics, even if they feel better, to ensure that as many bacteria as possible are killed. Moreover, antibiotics should never be used for viral infections, and the use of antibiotics in farming must be carefully controlled.
这一案例体现了自然选择的作用。抗生素构成了选择压力，而耐药菌具有选择优势。为了减缓耐药性的蔓延，医生建议患者即使感觉好转也应完成整个抗生素疗程，确保尽可能多地杀死细菌。此外，绝不能用抗生素治疗病毒感染，并在农业中严格控制抗生素的使用。
7. Adaptation of a Cactus to a Hot, Dry Habitat | 仙人掌对干热生境的适应
The saguaro cactus lives in the Sonoran Desert, where temperatures can exceed 40 °C and rainfall is scarce. Its survival depends on a set of remarkable adaptations. Describe and explain three key structural adaptations of a cactus.
巨人柱仙人掌生活在索诺兰沙漠，那里气温可超过 40 °C，降雨稀少。它的生存依赖一系列非凡的适应特征。请描述并解释仙人掌的三个关键结构适应。
First, the leaves have been reduced to sharp spines. This dramatically reduces the surface area available for transpiration, minimising water loss. The spines also protect the cactus from herbivores that might seek the stored water. Second, the stem is thick, fleshy and covered with a waxy cuticle. This stem stores large quantities of water and the waxy surface reduces evaporation. The green stem carries out photosynthesis since the leaves are modified. Third, the root system is shallow but extends over a wide area, enabling the cactus to absorb even a small amount of rainfall quickly. These adaptations work together to allow the cactus to thrive in an extreme environment where many other plants would perish.
首先，叶片退化成尖刺。这大大减少了蒸腾作用的表面积，最大限度减少水分流失。刺还能保护仙人掌免遭寻找储水的食草动物侵害。其次，茎厚实多肉并覆盖着一层蜡质角质层。该茎能储存大量水分，蜡质表面可减少蒸发。绿色的茎进行光合作用，因为叶片已经特化。第三，根系虽浅但分布广泛，使仙人掌能迅速吸收哪怕少量的降雨。这些适应共同作用，让仙人掌在许多其他植物无法生活的极端环境中茁壮成长。
How might a prolonged drought still threaten a cactus population? Even adapted plants have limits. If drought lasts for several years, the stored water may become exhausted, and young seedlings that have not yet developed a large water reserve are especially vulnerable. This case reminds us that adaptations are not perfect solutions but compromises that work under a defined range of conditions.
长期干旱仍会怎样威胁仙人掌种群？即便是适应力强的植物也有极限。如果干旱持续数年，储存的水分可能耗尽，且尚未建立充足水储量的幼苗尤其脆弱。这个案例提醒我们，适应并非完美的解决方案，而是在特定条件范围内有效的一种折衷。
8. The Role of Bees in Pollination and Food Security | 蜜蜂在授粉和粮食安全中的作用
In an apple orchard, farmers rely on honeybees to transfer pollen between flowers. Bees collect nectar and pollen as food; while visiting blossoms, pollen grains stick to their hairy bodies and are carried to other flowers, enabling fertilisation. Recently, beekeepers observed a steep decline in bee colonies, a problem partly linked to pesticide use and habitat loss.
在一座苹果园里，果农依赖蜜蜂在花朵间传播花粉。蜜蜂采集花蜜和花粉作为食物；访花时，花粉粒沾附在其多毛的身体上并被带到其他花朵，从而实现受精。近期，养蜂人观察到蜂群大量
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