📚 Year 10 OCR Biology: Case Study Practice | Year 10 OCR 生物：案例分析实战演练
This article brings together eight typical case studies that you may encounter in the Year 10 OCR Gateway Biology course. Each scenario is designed to reinforce your data analysis, experimental design, and evaluation skills – the very competencies examined through AO2 and AO3 questions. By working through these examples, you will gain confidence in interpreting graphs, calculating rates, justifying conclusions, and applying good scientific practice like CORMS.
本文汇集了八个你在 Year 10 OCR Gateway 生物学课程中可能遇到的典型案例。每个场景都旨在强化你的数据分析、实验设计和评价能力——这正是通过 AO2 和 AO3 题目考查的核心素养。通过这些范例，你将更自信地解读图表、计算速率、论证结论并运用诸如 CORMS 的规范科学实践。
1. Case Study 1: The Effect of pH on Amylase Activity | 案例一：pH 对淀粉酶活性的影响
Scenario: A student investigated how pH affects the rate of starch breakdown by the enzyme amylase. Starch solution and amylase were mixed at pH values ranging from 4 to 9. Every 10 seconds a drop of the mixture was removed and tested with iodine solution. The time taken for the iodine to remain orange‑brown (indicating complete starch digestion) was recorded. The table below shows the results.
情景：一位学生研究了 pH 如何影响淀粉酶分解淀粉的速率。在 pH 4 至 9 的条件下将淀粉溶液和淀粉酶混合。每 10 秒取出一滴混合物用碘液测试，记录碘液保持橙棕色（表明淀粉完全消化）所需的时间。下表为实验结果。
| pH | Time for starch to disappear (s) |
|---|---|
| 4.0 | 180 |
| 5.0 | 95 |
| 6.0 | 55 |
| 7.0 | 38 |
| 8.0 | 60 |
| 9.0 | 135 |
Typical questions: Plot a line graph of pH against time. Describe the trend. Explain why the activity decreases at very high and very low pH. Suggest how the student could improve the reliability of the results.
典型问题：绘制 pH 对应时间的折线图。描述变化趋势。解释为何酶活性在极高和极低 pH 下会下降。提出提高结果可靠性的改进方法。
How to approach: When plotting, put pH on the x‑axis (independent variable) and time on the y‑axis (dependent variable). Note that a shorter time indicates a faster rate, so the rate of reaction is inversely proportional to the time taken. At pH extremes, the enzyme's active site is denatured – its shape changes so that the substrate no longer fits. The optimum pH for human amylase is around 7.0, close to the pH in the mouth and small intestine. To improve reliability, the student should perform repeats at each pH and calculate a mean, identifying any anomalous results. Always identify the control variables: volume of starch, volume of amylase, temperature (use a water bath), concentration of solutions, and the end‑point colour matching.
解题策略：绘图时，将 pH 放在 x 轴（自变量），时间放在 y 轴（因变量）。注意时间越短意味着速率越快，所以反应速率与所用时间成反比。在极端 pH 下，酶的活性位点发生变性——形状改变，底物不再与之契合。人类淀粉酶的最适 pH 在 7.0 左右，接近于口腔和小肠的 pH。为了提高可靠性，学生应在每个 pH 下进行重复实验并计算平均值，识别异常数据。务必识别控制变量：淀粉体积、淀粉酶体积、温度（使用水浴）、溶液浓度以及终点颜色匹配。
2. Case Study 2: Photosynthesis and Light Intensity | 案例二：光合作用与光照强度
Scenario: A class measured the rate of photosynthesis in pondweed (Elodea) by counting oxygen bubbles released per minute at different distances from a lamp. The light intensity was calculated using the inverse square law (intensity ∝ 1/distance²). Results are shown below.
情景：全班通过计算伊乐藻（Elodea）在不同灯距下每分钟释放的氧气气泡数来测量光合作用速率。光强根据平方反比定律（光强 ∝ 1/距离²）计算。结果如下。
| Distance from lamp (cm) | 1/distance² (arbitrary units) | Bubbles per minute |
|---|---|---|
| 10 | 0.0100 | 45 |
| 20 | 0.0025 | 22 |
| 30 | 0.0011 | 13 |
| 40 | 0.0006 | 7 |
| 50 | 0.0004 | 4 |
Typical questions: Plot a graph of bubbles per minute against 1/distance². Explain why the rate levels off at high light intensities. Suggest why the pondweed is placed in a beaker of water. State three limiting factors for photosynthesis.
典型问题：绘制气泡数/分钟对应 1/距离² 的图像。解释为何在较高光强下速率趋于平稳。说明为何将伊乐藻置于盛水烧杯中。列举光合作用的三个限制因素。
How to approach: The graph should be a scatter plot with a line of best fit, and may show a plateau. At low light intensities, light is the limiting factor. At higher intensities, another factor – normally carbon dioxide concentration or temperature – becomes limiting. The beaker of water acts as a heat shield, absorbing infrared radiation from the lamp so that temperature does not become a confounding variable. Control variables include: the species of pondweed, length of stem, volume of sodium hydrogen carbonate solution (providing CO₂), and same lamp. Remember CORMS: Change light intensity, Organism same pondweed, Repeat three times, Measure bubbles per minute, Same – all other factors kept constant.
解题策略：图形应为带最佳拟合线的散点图，并可能出现平台。在低光强下，光是限制因素。光强较高时，另一个因素——通常是二氧化碳浓度或温度——成为限制因素。烧杯中的水充当隔热层，吸收灯的红外辐射，使温度不会成为混杂变量。控制变量包括：伊乐藻种类、茎长度、碳酸氢钠溶液体积（提供 CO₂）以及相同的灯。牢记 CORMS：C（改变）光强，O（生物）相同的伊乐藻，R（重复）三次，M（测量）每分钟气泡数，S（相同）所有其他因素保持恒定。
3. Case Study 3: Osmosis in Potato Cylinders | 案例三：土豆条中的渗透作用
Scenario: A student prepared potato cylinders of equal length and mass and placed them in sucrose solutions ranging from 0.0 mol/dm³ (distilled water) to 1.0 mol/dm³. After 30 minutes, the cylinders were dried and re‑weighed. The percentage change in mass was calculated.
情景：一位学生准备了等长等质量的土豆条，并将其分别放入从 0.0 mol/dm³（蒸馏水）到 1.0 mol/dm³ 的蔗糖溶液中。30 分钟后，将土豆条干燥并重新称重，计算了质量变化百分比。
| Sucrose concentration (mol/dm³) | % change in mass |
|---|---|
| 0.0 | +12.5 |
| 0.2 | +7.0 |
| 0.4 | +2.0 |
| 0.6 | -3.5 |
| 0.8 | -9.0 |
| 1.0 | -14.0 |
Typical questions: Plot a line graph of concentration against percentage change in mass. Determine the concentration inside the potato cells. Explain why mass increases in distilled water but decreases in concentrated sucrose solution.
典型问题：绘制浓度对应质量变化百分比的折线图。确定土豆细胞内部的浓度。解释为何在蒸馏水中质量增加而在浓蔗糖溶液中质量下降。
How to approach: Draw a smooth curve and read the x‑axis value at 0% mass change – this is the isotonic point, where water potential inside and outside the cell is equal. In distilled water, water enters the potato cells by osmosis because the water potential is higher outside (less negative), causing the cells to become turgid and increasing mass. In strong sucrose solution, water leaves the cells; they become flaccid and mass decreases. Always discuss water potential gradient and the partially permeable membrane. To improve accuracy, blot cylinders carefully, use a balance reading to 0.01 g, and control temperature.
解题策略：绘制平滑曲线，在质量变化 0% 处读取 x 轴值——这是等渗点，细胞内外的水势相等。在蒸馏水中，水通过渗透作用进入土豆细胞，因为外部水势更高（负得少），细胞变得硬胀，质量增加。在浓蔗糖溶液中，水离开细胞，细胞变得松软，质量下降。始终讨论水势梯度和部分透性膜。为提高准确性，仔细吸干土豆条，使用精确到 0.01 g 的天平，并控制温度。
4. Case Study 4: Respiration in Yeast – Which Sugar is Best? | 案例四：酵母的呼吸作用——哪种糖最佳？
Scenario: A student set up four boiling tubes, each containing yeast suspension and a different sugar solution: glucose, fructose, sucrose, and lactose. The tubes were placed in a water bath at 30 °C, and the volume of carbon dioxide produced in 10 minutes was measured using a gas syringe or inverted measuring cylinder.
情景：一位学生准备四支大试管，每支装有酵母悬液和不同的糖溶液：葡萄糖、果糖、蔗糖和乳糖。试管置于 30 °C 水浴中，10 分钟内产生的二氧化碳体积通过气体注射器或倒置量筒测量。
| Sugar | Volume of CO₂ (cm³) |
|---|---|
| Glucose | 28 |
| Fructose | 25 |
| Sucrose | 18 |
| Lactose | 5 |
Typical questions: Draw a bar chart to display the results. Explain why glucose produces the most CO₂. Why might sucrose and lactose give lower yields? Suggest why a control tube with no sugar is necessary.
典型问题：绘制条形图展示结果。解释为何葡萄糖产生的 CO₂ 最多。为什么蔗糖和乳糖的产率较低？建议为什么需要不含糖的对照管。
How to approach: Glucose is a monosaccharide and can be used directly in glycolysis – t first secrete invertase to break it down. Lactose is a disaccharide of glucose and galactose; many yeast strains lack the enzyme lactase, so it is poorly respired. A control containing only yeast and water (no sugar) proves that any CO₂ produced comes from the added sugar, not from the yeast' n dioxide (+ energy). Control variables: concentration of sugars, volume of yeast, temperature, and time.
解题策略：葡萄糖是单糖，可直接用于糖酵解——酵母无氧呼吸的第一步。蔗糖是由葡萄糖和果糖组成的二糖；酵母须先分泌蔗糖酶将其分解。乳糖是葡萄糖和半乳糖的二糖；许多酵母菌株缺乏乳糖酶，因而发酵效果差。只含酵母和水的对照管可以证明任何产生的 CO₂ 都来自添加的糖，而非酵母自身的储备。务必提到酵母无氧呼吸的文字方程式：葡萄糖 → 乙醇 + 二氧化碳（+ 能量）。控制变量：糖浓度、酵母体积、温度和时间。
5. Case Study 5: Smoking and Lung Cancer – Interpreting Epidemiological Data | 案例五：吸烟与肺癌——解读流行病学数据
Scenario: A graph shows the lung cancer mortality rates per 100 000 people in a country between 1950 and 2010, comparing male smokers, male non‑smokers, female smokers and female non‑smokers. The smoker lines are much higher, and the male smoker line rises sharply after 1970, while female smoker rates increase later.
情景：一张图表显示了 1950 年至 2010 年间某国每 10 万人中肺癌死亡率，比较了男性吸烟者、男性不吸烟者、女性吸烟者和女性不吸烟者。吸烟者曲线远高于非吸烟者，男性吸烟者曲线在 1970 年后急剧上升，而女性吸烟者率上升较晚。
Typical questions: Describe the trends shown. Calculate the percentage increase in male smoker mortality from 1960 to 2000. Explain why this data alone does not prove that smoking causes lung cancer. Suggest other factors that should be considered.
典型问题：描述呈现的趋势。计算 1960 年至 2000 年男性吸烟者死亡率的增长百分比。解释为何单靠这些数据不能证明吸烟导致肺癌。提出应考虑的其他因素。
How to approach: ns in tobacco smoke (e.g., tar, benzene), identification of mutations in lung cells, and controlled experiments where confounding variables are eliminated. The delayed rise in female lung cancer parallels the later uptake of smoking among women. Calculate percentage change: ((new - old) / old) × 100% occupational exposure to asbestos, and air pollution. Correlation does not equal causation – this is a key evaluative point.
解题策略：你必须表明这种关系是相关关系。要确立因果关系，需要其他证据：动物研究显示烟草烟雾中的致癌物（如焦油、苯）、肺细胞中突变的发现，以及消除混杂变量的对照实验。女性肺癌死亡率较晚上升与女性开始吸烟较晚的现象一致。计算百分比变化：（（新 - 旧）/ 旧）× 100%。讨论混杂变量，如饮酒、职业石棉接触和空气污染。相关不等于因果——这是一个关键的评价点。
6. Case Study 6: Quadrat Sampling and Population Estimation | 案例六：样方取样与种群估算
Scenario: andom samples were taken. The number of daisies per quadrat were: 3, 0, 2, 1, 5, 0, 4, 1, 2, 0, 6, 1, 0, 3, 2, 0, 4, 1, 0, 3. The student also counted daisies along a transect line from a shaded area to an open sunny area.
情景：一位学生想要估算一片 500 平方米操场上雏菊的数量。使用一个 0.5 米 × 0.5 米的样方，随机取样 20 次。每个样方中的雏菊数为：3, 0, 2, 1, 5, 0, 4, 1, 2, 0, 6, 1, 0, 3, 2, 0, 4, 1, 0, 3。该学生还沿着从荫蔽处到开敞向阳处的样线对雏菊进行了计数。
Typical questions: Calculate the mean number of daisies per quadrat. Estimate the total number of daisies in the field. Explain why random sampling is used. Describe how a transect can show the effect of light intensity on daisy distribution. Evaluate the technique.
典型问题：计算每个样方中雏菊的平均数量。估算场地内雏菊的总数。解释为何使用随机取样。描述样线如何反映光照强度对雏菊分布的影响。评价该技术。
How to approach: 2000 × 1.9 = 3800 daisies. Random sampling removes bias; use a random number generator to generate coordinates. A transect systematically samples along an environmental gradient; count daisies in a quadrat placed at regular intervals. The technique can be improved by taking more samples, ensuring quadrat placement is consistent, and identifying plants accurately. In an exam, always link to the idea that distribution is affected by abiotic factors such as light, water, and pH of the soil.
解题策略：将雏菊计数求和（38）并除以 20，得平均每个样方 1.9 株。样方面积 = 0.5 × 0.5 = 0.25 平方米。总种群估算值 = （500 ÷ 0.25）× 1.9 = 2000 × 1.9 = 3800 株雏菊。随机取样消除偏差；使用随机数生成器产生坐标。样线沿环境梯度系统取样；每隔固定间隔用样方计数雏菊。该技术可通过增加样本数量、确保样方放置一致以及准确识别植物来改进。考试时务必联系到分布受非生物因素（如光照、水分和土壤 pH）影响的观点。
7. Case Study 7: Antibiotic Effectiveness – Inhibition Zones | 案例七：抗生素的效力——抑菌圈
Scenario: A student spread a lawn of E. coli bacteria on an agar plate. Four filter paper discs, each soaked in a different antibiotic (A, B, C, and D), were placed on the agar. After incubation for 48 hours at 37 °C, the clear zones around each disc were measured. A control disc soaked in sterile water showed no inhibition.
情景：一位学生在琼脂平板上涂布了一层大肠杆菌。四张分别浸泡了不同抗生素（A、B、C 和 D）的滤纸片放在琼脂上。在 37 °C 培养 48 小时后，测量了每个纸片周围的透明圈。用无菌水浸泡的对照纸片未显示抑制。
| Antibiotic | Diameter of inhibition zone (mm) |
|---|---|
| A | 18 |
| B | 25 |
| C | 12 |
| D | 0 |
Typical questions: Which antibiotic should a doctor prescribe based on this data? a disc soaked in water acts as a control. Why must the Petri dish be incubated upside down? otic B (using πr², where π = 3.14). Discuss why it is important to use aseptic techniques.
典型问题：根据数据，医生应该开具哪种抗生素？解释为何用无菌水浸泡的纸片作为对照。为什么培养皿必须倒置培养？计算抗生素 B 抑菌圈的面积（使用 πr²，π = 3.14）。讨论使用无菌技术的重要性。
How to approach: e to the antibiotic, not the disc material or water. Incubating upside down prevents condensation from dripping onto the agar and spreading bacteria. Area of inhibition zone for B: radius = 12.5 mm, area = 3.14 × (12.5)² ≈ 490.6 mm². Aseptic techniques (flaming necks of bottles, using sterile forceps, working near a Bunsen burner) prevent contamination with unwanted microorganisms which could compete with the E. coli or pose a safety risk. Always mention that for accurate comparison, the concentration and volume of antibiotic on each disc must be the same.
解题策略：抗生素 B 的抑菌圈最大（25 mm），因此它对这一菌株最有效。水纸片证明任何抑制都源于抗生素，而非纸片材质或水。倒置培养可防止冷凝水落在琼脂上并散播细菌。B 的抑菌圈面积：半径 = 12.5 mm，面积 = 3.14 × (12.5)² ≈ 490.6 mm²。无菌技术（灼烧瓶口、使用无菌镊子、在本生灯旁操作）可防止不必要的微生物污染，它们可能与大肠杆菌竞争或造成安全风险。一定要提到，为了准确比较，每个纸片上抗生素的浓度和体积必须相同。
8. Case Study 8: Interpreting a Pedigree for Cystic Fibrosis | 案例八：囊性纤维化家系图的解读
Scenario: The diagram shows a pedigree chart for a family. Cystic fibrosis (CF) is a recessive genetic disorder caused by a faulty allele of the CFTR gene. In the pedigree, affected individuals are shaded. Two unaffected parents (gener
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