📚 Year 8 CCEA Biology: Case Study Practical Drills | Year 8 CCEA 生物：案例分析实战演练
Welcome to a collection of hands‑on case studies designed for Year 8 CCEA Biology. Each practical scenario mirrors a real school laboratory investigation, building your skills in observation, measurement, analysis and drawing conclusions. Work through the cases step by step, just like a scientist.
欢迎来到为 Year 8 CCEA 生物课程设计的一系列动手案例分析。每个实战场景都模拟了真实的学校实验室探究，帮助你锻炼观察、测量、分析和得出结论的能力。像科学家一样一步一步来研究这些案例吧。
1. Case Study 1: Using a Microscope to View Onion Cells | 案例一：显微镜观察洋葱细胞
A student peeled a thin layer of epidermis from an onion scale leaf and prepared a wet mount on a glass slide. A drop of iodine solution was added to stain the cells, making the nucleus more visible.
一位学生从洋葱鳞叶上撕下一层薄薄的表皮，在载玻片上制作了一个临时装片。滴加了一滴碘液来染色，使细胞核更明显。
When viewed under the low‑power objective (×10) and then the high‑ wall, a round nucleus stained dark brown, and a large central vacuole that appeared clear.
在低倍物镜（×10）和高倍物镜（×40）下观察时，排列如砖块的矩形结构变得清晰。每个矩形有明显的细胞壁，一个被染成深褐色的圆形细胞核，以及一个看起来透明的大型中央液泡。
the eyepiece lens (×10) by the objective lens (×40) to give a total magnification of ×400.
学生画了一幅带标注的图，显示了细胞壁、细胞核、细胞质和液泡。放大倍数是由目镜（×10）乘以物镜（×40）计算得出，总放大倍数为 ×400。
Why did the cells not show chloroplasts? Onion epidermal cells are from an underground bulb and do not carry out photosynthesis; therefore, they lack chloroplasts. The iodine stain mainly reacts with starch in the nucleus and cytoplasm, helping to highlight internal parts.
为什么这些细胞没有显示叶绿体？洋葱表皮细胞来自地下鳞茎，不进行光合作用，因此缺少叶绿体。碘液主要与细胞核和细胞质中的淀粉反应，有助于让内部结构更明显。
2. Case Study 2: Testing for Starch in Food Samples | 案例二：测试食物中的淀粉
Five food samples – bread, cheese, potato, apple and rice – were placed on a spotting tile. Two drops of iodine solution were added to each sample, and any colour change was recorded immediately.
将五种食物样品——面包、奶酪、土豆、苹果和米饭——放在点滴板上。向每种样品滴加两滴碘液，并立即记录颜色变化。
Bread, potato and rice turned a blue‑black colour, confirming the presence of starch. Cheese and apple remained the orange‑brown colour of the iodine solution, indicating that starch was absent or below detection level.
面包、土豆和米饭变成了蓝黑色，证实含有淀粉。奶酪和苹果仍保持碘液的橙棕色，说明淀粉不存在或低于检测限。
A control was set up using distilled water and iodine; no colour change occurred, proving that the iodine itself did not react with the tile or water. This ensured that any blue‑black colour came solely from starch in the food.
使用蒸馏水与碘液设置了对照；没有发生颜色变化，证明碘液本身不与点滴板或水反应。这确保任何蓝黑色都仅仅来自食物中的淀粉。
From these results, the student concluded that plant storage organs are rich in starch, whereas animal‑based foods and fruits low in starch do not give a positive test. Repeating the test three times gave consistent results, adding reliability.
根据这些结果，学生得出结论：植物贮藏器官富含淀粉，而动物性食品和淀粉含量低的水果不会产生阳性反应。重复测试三次得到一致结果，增加了可靠性。
3. Case Study 3: Investigating the Effect of Exercise on Heart Rate | 案例三：探究运动对心率的影响
A student measured their resting heart rate by counting radial pulse beats for 30 seconds and multiplying by 2. The resting rate was 72 beats per minute (bpm). They then performed two minutes of step‑ups on a bench and immediately remeasured the pulse.
一位学生通过计数桡动脉脉搏 30 秒再乘以 2 来测量安静心率。安静心率为每分钟 72 次。然后他们在台阶上做两分钟的高抬腿踏步，并立即再次测量脉搏。
Immediately after exercise, the heart rate jumped to 128 bpm. After one minute of rest it fell to 98 bpm, and after five minutes it returned to 74 bpm, close to the resting value. The student recorded the data in a table and drew a line graph of heart rate against time.
运动后立即测量，心率升至每分钟 128 次。休息一分钟后降至 98 次，五分钟后恢复到 74 次，接近安静值。学生将数据记录在表格中，并绘制了心率随时间变化的线图。
The rapid increase was explained by the muscles needing more oxygen and glucose for respiration during exercise. The heart pumped faster to deliver these materials and remove carbon dioxide. The gradual recovery showed the body's ability to return to a resting state.
可解释为：运动期间肌肉需要更多氧气和葡萄糖进行呼吸作用，心脏搏动加快以输送这些物质并带走二氧化碳。逐渐恢复表明身体有能力回到安静状态。
To improve the investigation, the student suggested keeping the step‑up rhythm constant using a metronome, measuring heart rate immediately and after set intervals, and repeating the trial on three different days for more reliable average data.
为改进探究，学生建议使用节拍器保持踏步节奏一致，在运动后立刻并按固定间隔测量心率，并在三天重复试验以获得更可靠的平均数据。
4. Case Study 4: Measuring Breathing Rate Before and After Activity | 案例四：测量活动前后的呼吸频率
At rest, a student counted 16 breaths per minute by observing chest movements. After running on the spot for two minutes, the breathing rate rose to 31 breaths per minute. Each breath was counted as one complete rise and fall of the chest.
在安静状态下，一位学生通过观察胸部起伏计数得出每分钟 16 次呼吸。原地跑步两分钟后，呼吸频率上升到每分钟 31 次。胸廓一次完整的起伏计为一次呼吸。
The increase in breathing rate is linked to the need to take in more oxygen and expel carbon dioxide produced by increased muscle activity. Deeper breathing also occurs, bringing more air into the lungs with each breath.
呼吸频率的增加与需要摄入更多氧气、排出因肌肉活动增加而产生的二氧化碳有关。呼吸也会加深，每次呼吸将更多空气吸入肺部。
By measuring the volume of air exhaled into a spirometer or a simple balloon, the student could also estimate tidal volume. Preliminary results suggested that both rate and depth of breathing rise during intense exercise to meet energy demands.
通过测量呼入肺活量计或简单气球中的空气体积，学生还可以估算潮气量。初步结果表明，剧烈运动时呼吸频率和深度都会增加，以满足能量需求。
When the student rested, the breathing rate fell slowly. After three minutes, it was down to 20 breaths per minute, showing that the body takes time to clear the oxygen debt and remove excess carbon dioxide.
当学生休息时，呼吸频率缓慢下降。三分钟后降至每分钟 20 次，这表明身体需要时间来偿还氧债并清除多余的二氧化碳。
5. Case Study 5: Photosynthesis in Pondweed – Counting Bubbles | 案例五：水草光合作用——气泡计数
A sprig of pondweed (Elodea) was placed in a boiling tube filled with water and a pinch of sodium hydrogencarbonate to supply carbon dioxide. A lamp was positioned at a fixed distance, and the number of gas bubbles released per minute was counted.
在装满水的沸腾管中放入一枝水草（伊乐藻），并加入少量碳酸氢钠提供二氧化碳。在固定距离处放置一盏灯，并计数每分钟释放的气泡数。
When the lamp was 10 cm away, the pondweed produced 42 bubbles per minute. Moving the lamp to 20 cm reduced the rate to 25 bubbles per minute, and at 30 cm only 11 bubbles per minute were observed. The gas collected in a test tube relit a glowing splint, confirming it was oxygen.
当灯距 10 厘米时，水草每分钟产生 42 个气泡。将灯移至 20 厘米时，速率降至每分钟 25 个气泡；灯距 30 厘米时仅观察到每分钟 11 个气泡。用试管收集的气体使带火星的木条复燃，证实是氧气。
split water molecules and release oxygen. The word equation is: carbon dioxide + water → glucose + oxygen, in the presence of light and chlorophyll.
这个演示表明光照强度影响光合作用的速率。光源越近，可用于分解水分子并释放氧气的光能就越多。文字方程式为：二氧化碳 + 水 → 葡萄糖 + 氧气，条件是有光和叶绿素。
To ensure a fair test, the student kept the temperature constant by using a water bath and used the same piece of pondweed. The number of bubbles was counted three times at each distance to calculate a mean, reducing random error.
为确保公平测试，学生用水浴保持温度恒定并使用同一枝水草。在每个距离下计数气泡三次以计算平均值，减少随机误差。
6. Case Study 6: Demonstrating the Need for Light in Photosynthesis | 案例六：证明光合作用需要光
A potted geranium plant was placed in the dark for 48 hours to destarch it. A strip of black paper was clipped onto one leaf, covering part of it. The plant was then exposed to bright sunlight for six hours.
将一盆天竺葵在暗处放置 48 小时以去除叶片中的淀粉。用黑色纸条夹在一片叶子上遮住部分区域，然后将植株放在强光下照射六小时。
After the light exposure, the leaf was boiled in water to soften it, then boiled in ethanol in a water bath to remove chlorophyll. The decolourised leaf was rinsed and tested with iodine solution. Only the area that had received light turned blue‑black; the covered portion remained pale brown.
光照结束后，将叶片在水中煮沸软化，然后在酒精中水浴加热以去除叶绿素。脱色的叶片漂洗后用碘液测试。只有接受光照的区域变成蓝黑色；被遮盖部分仍为浅棕色。
This proved that light is necessary for photosynthesis to produce starch. The dark‑covered region could not manufacture glucose and therefore stored no starch. Ethanol must be heated indirectly in a water bath because it is highly flammable.
这证明了光是光合作用产生淀粉的必要条件。被遮光区域无法制造葡萄糖，因此没有储存淀粉。乙醇必须在水浴中间接加热，因为它高度易燃。
The student considered why the whole plant was initially kept in the dark: to ensure that any starch detected after the experiment had been made during the light exposure, not before. This step makes the test valid.
学生思考了为什么一开始要将整株植物放在暗处：为了确保实验后检测到的任何淀粉都是光照期间制造的，而不是之前就存在的。这一步骤使测试有效。
7. Case Study 7: Modelling the Digestive System with Everyday Objects | 案例七：用日常物品模拟消化系统
Classmates built a physical model to mimic the journey of food through the digestive tract. A plastic bag represented the stomach, tights represented the small intestine, and a paper cup with holes acted as the large intestine.
同学们制作了一个物理模型来模拟食物在消化道中的旅程。塑料袋代表胃，紧身连裤袜代表小肠，一个带孔的纸杯充当大肠。
A slice of bread was torn to represent chewing (mouth), then placed into the plastic bag with a small amount of water and squeezed – like stomach churning. The mixture was then pushed into one leg of the tights, and students squeezed the tights to mimic peristalsis, so the paste moved along.
将一片面包撕碎代表咀嚼（口腔），然后放入装有少量水的塑料袋中挤压——类似胃的搅动。之后将混合物推入连裤袜的一条腿中，学生们挤压袜子模拟蠕动，使糊状物向前移动。
At the end of the tights, much of the water was squeezed out through the mesh, representing absorption of nutrients in the small intestine. The remaining solid material was passed to the paper cup, where more water drained through holes, modelling water absorption in the large intestine. The leftover solid represented faeces.
在袜子末端，大部分水通过网眼被挤出，模拟小肠中营养物质的吸收。剩余的固体物转移到纸杯中，水从孔中排出，模拟大肠吸收水分。剩下的固体代表粪便。
The model helped pupils understand the physical breakdown of food and the progressive absorption of water. It did not show chemical digestion by enzymes, but it effectively illustrated mechanical digestion and movement along the gut.
这个模型帮助学生们理解食物的物理分解以及水分的逐步吸收。它没有展示酶的化学消化，但很好地说明了机械消化和食物沿着肠道的移动。
8. Case Study 8: Investigating Osmosis Using Potato Strips | 案例八：用马铃薯条探究渗透作用
Three potato strips of equal length (5 cm) were cut and weighed. Each was placed into a different solution: distilled water, 0.5 M sugar solution, and 1.0 M sugar solution. After 30 minutes the strips were blotted dry and re‑measured.
切取三根等长（5 厘米）的马铃薯条并称重。将它们分别放入不同溶液中：蒸馏水、0.5 M 蔗糖溶液和 1.0 M 蔗糖溶液。30 分钟后取出吸干表面水分，再次测量。
The potato in distilled water increased in length to 5.4 cm and gained mass, feeling turgid and firm. The potato in 0.5 M sugar solution changed only slightly to 5.1 cm, while the strip in 1.0 M sugar solution shrank to 4.6 cm and became flaccid and soft.
蒸馏水中的马铃薯条长度增加到 5.4 厘米，质量增加，变得饱满坚硬。0.5 M 蔗糖溶液中的马铃薯仅略微变化到 5.1 厘米，而 1.0 M 蔗糖溶液中的马铃薯条收缩到 4.6 厘米，变得松软。
These changes are explained by osmosis – the movement of water molecules from a region of higher water concentration to a region of lower water concentration across a partially permeable membrane. Plant cells have such membranes, so water enters when the external solution is more dilute (higher water potential) and leaves when it is more concentrated.
这些变化可以用渗透作用来解释——水分子通过部分透性膜从水浓度较高的区域向水浓度较低的区域运动。植物细胞具有这样的膜，因此当外部溶液较稀（水势较高）时水进入细胞，当外部溶液较浓时水离开细胞。
To improve reliability, the student suggested measuring the strips with a digital calliper to 0.1 mm and calculating percentage change in mass, which allows better comparison between strips of different starting masses.
为提高可靠性，学生建议用数显卡尺测量至 0.1 毫米并计算质量变化百分比，这能更好比较不同初始质量的薯条。
9. Case Study 9: Testing for Carbon Dioxide with Limewater | 案例九：用石灰水检测二氧化碳
A student set up a flask containing marble chips and hydrochloric acid. The gas produced was bubbled through a delivery tube into a test tube of clear limewater. Within seconds, the limewater turned milky (cloudy white).
一位学生搭建了一个装有石灰石碎片和盐酸的锥形瓶。产生的气体通过导管被通入盛有澄清石灰水的试管中。几秒钟内，石灰水变浑浊（乳白色）。
ride + water + carbon dioxide. The CO₂ reacts with calcium hydroxide in the limewater to form insoluble calcium carbonate, causing the cloudiness.
乳白色现象证实了二氧化碳的存在。相关的反应为：碳酸钙 + 盐酸 → 氯化钙 + 水 + 二氧化碳。二氧化碳与石灰水中的氢氧化钙反应，生成不溶的碳酸钙，导致变浑。
This test is commonly used to detect CO₂ in exhaled air. The student blew gently through a straw into fresh limewater; it also turned milky, demonstrating that our breath contains carbon dioxide from respiration.
该测试常用于检测呼出气体中的二氧化碳。学生通过吸管轻轻向新鲜石灰水中吹气，石灰水也变得浑浊，说明我们的呼吸中含有来自呼吸作用的二氧化碳。
An excess of carbon dioxide was bubbled for two minutes, and the milky solution eventually cleared again. This happened because the calcium carbonate reacted further to form soluble calcium hydrogencarbonate, showing the importance of using the test quickly.
持续通入过量二氧化碳两分钟后，浑浊的溶液最终又变澄清。这是因为生成的碳酸钙进一步反应形成了可溶的碳酸氢钙，这表明应快速进行该测试。
10. Case Study 10: Enzyme Action – Saliva and Starch Breakdown | 案例十：酶的作用——唾液与淀粉分解
A student collected 2 ml of saliva in a test tube. Two test tubes were prepared: Tube A contained 5 ml starch solution + 1 ml saliva; Tube B contained 5 ml starch solution + 1 ml distilled water (control). Both tubes were placed in a water bath at 37 °C, body temperature.
一位学生在试管中收集了 2 毫升唾液。准备了两支试管：试管 A 含 5 毫升淀粉溶液 + 1 毫升唾液；试管 B 含 5 毫升淀粉溶液 + 1 毫升蒸馏水（对照）。均放入 37 °C 水浴（体温）。
Every minute, a drop from each tube was tested on a spotting tile with iodine solution. At the start, both samples turned blue‑black, confirming starch. After 5 minutes, Tube A no longer gave a blue‑black colour, while Tube B remained blue‑black throughout the 15‑minute observation period.
每隔一分钟，从每支试管中取一滴液体放在点滴板上用碘液测试。开始时两个样品都变蓝黑色，证实有淀粉。5 分钟后，试管 A 不再产生蓝黑色，而试管 B 在整个 15 分钟观察期内始终保持蓝黑色。
Saliva contains the enzyme amylase, which breaks starch down into maltose, a reducing sugar. Because maltose does not react with iodine, the colour change disappeared. The control showed that without the enzyme, starch does not break down at 37 °C.
唾液中含有淀粉酶，能将淀粉分解为麦芽糖（一种还原糖）。由于麦芽糖不与碘反应，颜色变化消失。对照组表明，没有酶的情况下淀粉在 37 °C 时不会分解。
To confirm that maltose was produced, the student could carry out a Benedict's test on the solution from Tube A. A green/orange‑red precipitate after heating would indicate the presence of reducing sugar, giving stronger evidence of enzymatic breakdown.
为了确认产生了麦芽糖，学生可以对试管 A 的溶液进行本尼迪克特测试。加热后出现绿色/橙红色沉淀将表明还原糖的存在，为酶分解提供更有力的证据。
11. Case Study 11: Seed Germination – What Conditions Are Needed? | 案例十一：种子萌发需要什么条件？
Four sets of cress seeds were set up:
- Set 1: on damp cotton wool at room temperature (light, water, oxygen, warmth).
- Set 2: on dry cotton wool at room temperature (light, oxygen, warmth, no water).
- Set 3: on damp cotton wool in a fridge at 4 °C (light, water, oxygen, cold).
- Set 4: on damp cotton wool in a sealed jar with an oxygen absorber (light, water, warmth, no oxygen).
准备了四组水芹种子：
- 第 1 组：常温下放在湿棉絮上（光、水、氧气、温暖）。
- 第 2 组：常温下放在干棉絮上（光、氧气、温暖、无水）。
- 第 3 组：在 4 °C 冰箱中放在湿棉絮上（光、水、氧气、低温）。
- 第 4 组：放在密封罐中湿棉絮上并加入除氧剂（光、水、温暖、无氧）。
After three days, only Set 1 seeds had germinated, showing tiny roots and shoots. Seeds in Sets 2, 3 and 4 showed no germination. This demonstrated that water, a suitable temperature and oxygen are all essential for germination; light was not needed for cress seeds, as they had been kept in the light.
三天后，只有第 1 组种子萌发，长出了微小的根和芽。第 2、3、4 组均未萌发。这表明水、适宜的温度和氧气都是萌发所必需的；水芹种子萌发不需要光，因为它们一直处于光照下。
Set 3 showed that low temperatures slow down enzyme reactions required for growth. Set 4 highlighted the need for oxygen for aerobic respiration to provide energy for the developing embryo. This practical gives strong evidence for the three key germination factors.
第 3 组显示低温会减慢生长所需的酶反应。第 4 组强调了有氧呼吸为发育中的胚提供能量需要氧气。该实践为三个关键萌发因素提供了有力的证据。
12. Case Study 12: Using a Model Lung to Show Breathing Mechanism | 案例十二：用肺模型展示呼吸机制
Students constructed a model lung using a plastic bell jar, a Y‑shaped glass tube, two small balloons and a rubber sheet sealed across the bottom. The balloons represented the lungs, the glass tube the trachea and bronchi, and the rubber sheet the diaphragm.
学生们用塑料钟罩、Y 形玻璃管、两个小气球和一块密封在底部的橡胶薄片制作了一个肺模型。气球代表肺，玻璃管代表气管和支气管，橡胶片代表膈肌。
When the rubber sheet was pulled downwards, the volume inside the jar increased, the pressure decreased, and the balloons inflated, showing inhalation. Pushing the sheet upwards reduced the volume, increased the pressure, and deflated the balloons – exhalation.
当将橡胶片向下拉时，钟罩内容积增大，压力减小，气球充气膨胀，显示吸气。向上推薄片则容积减小，压力增大，气球排气缩小——呼气。
This model beautifully illustrates that breathing is caused by pressure changes in the thorax. The diaphragm does not push air into the lungs directly; it expands the chest cavity, causing air to rush in because of the pressure difference. The model correctly shows that the ribs can also move to change volume.
这个模型很好地说明呼吸是由胸腔内的压力变化引起的。膈肌并非直接将空气推入肺部；它扩大胸廓，由于压力差空气被吸入。模型也正确地显示肋骨可以移动以改变容积。
A limitation is that the bell jar wall is rigid, unlike the rib cage which can move slightly. Still, the model allows students to visualise the mechanics of ventilation and understand why a punctured lung or diaphragm injury can impair breathing.
局限性在于钟罩壁是刚性的，不像可以微动的胸廓。尽管如此，该模型帮助学生直观了解通气机制，并理解肺或膈肌损伤为何会影响呼吸。
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