📚 Case Study Practice Drill | 案例分析实战演练
Case studies are an essential part of the Cambridge Year 10 Biology assessment, requiring you to apply knowledge to unfamiliar scenarios. This drill will guide you through diverse examples, from enzyme experiments to ecological data analysis, helping you build the skills needed to interpret data, draw conclusions, and explain biological principles with precision.
案例分析是剑桥Year 10生物评估的重要组成部分，要求你将知识应用到陌生情景中。本次实战演练将通过酶实验、生态数据分析等多样化案例，帮助你培养解读数据、得出结论并精准解释生物学原理的能力。
1. How to Tackle a Case Study | 如何应对案例分析
Start by reading the stem carefully, underlining key terms like 'optimum', 'limiting factor', or 'concentration gradient'.
首先仔细阅读题干，划出“最适”、“限制因子”、“浓度梯度”等关键术语。
Identify the type of data provided — tables, line graphs, or diagrams — and note the units on both axes.
识别所提供的数据类型——表格、折线图或示意图——并关注坐标轴的单位。
Link the observations to core concepts: for example, a drop in rate above 40°C nearly always points to enzyme denaturation.
将观察到的现象与核心概念联系起来：例如，速率在40°C以上下降几乎总是指向酶变性。
Structure your answer with a clear claim, supporting evidence from the data, and a biological explanation.
用清晰的论点、数据中的支持性证据和生物学解释来组织你的答案。
When a calculation is required, show all working and check that the unit conversion is correct.
当需要计算时，展示所有步骤并确保单位换算正确。
2. Case Study: Enzyme Activity and Temperature | 案例2：酶活性与温度
An investigation recorded the rate of product formation for catalase at different temperatures.
一项研究记录了过氧化氢酶在不同温度下的产物生成速率。
| Temperature / ℃ | Rate / mg min⁻¹ |
|---|---|
| 10 | 1.2 |
| 20 | 3.5 |
| 30 | 6.8 |
| 40 | 9.1 |
| 50 | 4.3 |
| 60 | 0.0 |
Question (a): State the optimum temperature for this enzyme.
问题(a)：说出该酶的最适温度。
The optimum temperature is 40°C, because the rate of product formation peaks at 9.1 mg min⁻¹.
最适温度为40°C，因为产物生成速率达到峰值9.1 mg min⁻¹。
Question (b): Explain the change in rate between 40°C and 60°C.
问题(b)：解释40°C至60°C之间速率的变化。
Above 40°C, the increased kinetic energy breaks the hydrogen bonds and other interactions holding the enzyme’s tertiary structure.
40°C以上时，增加的动能打破了维持酶三级结构的氢键和其他相互作用。
The active site loses its complementary shape, so the substrate can no longer bind — the enzyme is denatured irreversibly.
活性位点失去了互补形状，底物无法再结合——酶发生了不可逆变性。
3. Case Study: Osmosis in Potato Strips | 案例3：土豆条的渗透作用
Potato strips of equal mass were placed in sucrose solutions of different concentrations for 24 hours.
将质量相等的土豆条放入不同浓度的蔗糖溶液中浸泡24小时。
| Sucrose concentration / mol dm⁻³ | % change in mass |
|---|---|
| 0.0 | +18.5 |
| 0.2 | +8.9 |
| 0.4 | -0.5 |
| 0.6 | -11.2 |
| 0.8 | -20.3 |
Which sucrose solution is closest to the water potential of the potato cells? Explain why.
哪种蔗糖溶液最接近土豆细胞的水势？解释原因。
The 0.4 mol dm⁻³ solution gives a negligible mass change (-0.5%), indicating it is nearly isotonic.
0.4 mol dm⁻³溶液导致的质量变化可忽略不计（-0.5%），表明其近乎等渗。
At this concentration, there is no net movement of water by osmosis, so the water potentials inside and outside the cells are equal.
在此浓度下，渗透作用无净水移动，因此细胞内外水势相等。
4. Case Study: Photosynthesis and Light Intensity | 案例4：光合作用与光照强度
A pond weed was exposed to increasing light intensity, and the rate of oxygen bubble production was recorded.
将水草暴露于逐渐增强的光照强度下，记录氧气气泡的产生速率。
| Light intensity / lux | Bubble rate / min⁻¹ |
|---|---|
| 200 | 5 |
| 400 | 14 |
| 600 | 22 |
| 800 | 23 |
| 1000 | 23 |
Identify the light intensity at which light stops being the limiting factor, and suggest another factor that may be limiting above this point.
找出光照强度在何处不再是限制因子，并提出在该点以上可能限制速率的另一个因子。
Light stops being the limiting factor around 600 lux, because beyond this the rate plateaus at 23 bubbles per minute.
光照强度在约600 lux时不再是限制因子，因为此后速率稳定在每分钟23个气泡。
A likely limiting factor above 600 lux is carbon dioxide concentration; if CO₂ is fixed, the Calvin cycle cannot proceed faster even with abundant light.
600 lux以上可能的限制因子是二氧化碳浓度；若CO₂固定量不变，即使光能充足，卡尔文循环也无法更快进行。
5. Case Study: Energy Transfer in a Food Chain | 案例5：食物链中的能量传递
A grassland food chain is shown with energy values at each trophic level:
以下草地食物链标出了每个营养级的能量值：
m⁻² yr⁻¹) → Grasshopper (760 kJ m⁻² yr⁻¹) → Frog (83 kJ m⁻² yr⁻¹) → Snake (9 kJ m⁻² yr⁻¹)
Calculate the percentage of energy transferred from the grasshopper to the frog, and explain two reasons for the large loss of energy.
计算从蚱蜢到青蛙的能量传递百分比，并解释能量大量损失的两个原因。
sfer efficiency = (energy in frog / energy in grasshopper) × 100 = (83 / 760) × 100 = 10.9%.
传递效率 = (青蛙能量 / 蚱蜢能量) × 100 = (83 / 760) × 100 = 10.9%。
One reason is that grasshoppers lose much energy through respiration for movement and maintaining body temperature.
原因之一是蚱蜢通过呼吸作用为运动和维持体温消耗了大量能量。
Another reason is that not all parts of the grasshopper are consumed or assimilated; some material is egested as faeces.
另一个原因是并非蚱蜢的所有部分都被摄取或同化；部分物质作为粪便被排出。
6. Case Study: Monohybrid Cross and Pedigree | 案例6：单基因杂交与家系图
The pedigree shows inheritance of a recessive genetic disorder. Affected individuals are shaded.
家系图显示了一种隐性遗传病的遗传方式。患病个体用阴影表示。
If the mother (generation I) is heterozygous and the father is unaffected, what is the probability that their second child will be a carrier?
若母亲（第Ⅰ代）为杂合子而父亲未患病，他们的第二个孩子是携带者的概率是多少？
Let 'A' be the normal allele and 'a' the recessive disease allele. Father’s genotype must be AA (unaffected, no family history). Mother is Aa.
设“A”为正常等位基因，“a”为隐性致病等位基因。父亲基因型必须是AA（未患病，无家族史）。母亲为Aa。
A Punnett square of AA × Aa gives 50% AA and 50% Aa, so the probability that a child is a carrier (Aa) is ½ or 50%.
AA × Aa的庞纳特方格显示50%为AA，50%为Aa，因此孩子是携带者（Aa）的概率为½或50%。
7. Case Study: Vaccination and Herd Immunity | 案例7：疫苗接种与群体免疫
A school with 800 students reports 12 cases of measles. Only 65% of students have received two doses of the MMR vaccine.
一所拥有800名学生的学校报告了12例麻疹病例。只有65%的学生接种了两剂MMR疫苗。
te the minimum vaccination coverage needed to achieve herd immunity for measles (R₀ ≈ 15).
解释疫情发生的原因，并计算实现麻疹群体免疫所需的最低疫苗覆盖率（R₀ ≈ 15）。
Herd immunity threshold = 1 - (1 / R₀) = 1 - (1/15) = 0.933 or 93.3%.
群体免疫阈值 = 1 - (1 / R₀) = 1 - (1/15) = 0.933 即93.3%。
With only 65% vaccinated, many susceptible individuals allow the virus to spread easily; the coverage is far below the required 93%.
疫苗接种率仅为65%，众多易感个体使得病毒易于传播；覆盖率远低于所需的93%。
8. Case Study: Transpiration and Potometer Readings | 案例8：蒸腾作用与蒸腾计读数
A bubble potometer tracked the distance moved by an air bubble in 10 minutes under different conditions.
气泡蒸腾计记录了不同条件下10分钟内气泡移动的距离。
| Condition | Distance / mm |
|---|---|
| Still air, room light | 24 |
| Fan, room light | 58 |
| Fan, bright light | 82 |
| Misted leaves, fan | 31 |
Explain the results using knowledge of factors affecting transpiration rate.
利用影响蒸腾速率的因素知识解释这些结果。
The fan removes the boundary layer of humid air, steepening the water vapour concentration gradient; hence the rate rises from 24 mm to 58 mm.
风扇去除了潮湿空气的边界层，加大了水蒸气浓度梯度；因此速率从24 mm升至58 mm。
Bright light opens stomata wider for photosynthesis, adding to the effect of wind and raising the distance to 82 mm.
强光照使气孔为光合作用而张大，叠加风的影响，使移动距离升至82 mm。
Misting saturates the air around the leaves, reducing the concentration gradient, so the rate drops sharply to 31 mm.
喷雾使叶片周围空气饱和，减小了浓度梯度，因此速率骤降至31 mm。
9. Case Study: Anaerobic Respiration in Yeast | 案例9：酵母的无氧呼吸
A student mixed yeast with different sugar solutions and measured CO₂ production over 30 minutes.
一名学生将酵母与不同糖溶液混合，测量30分钟内的CO₂产量。
| Sugar | CO₂ volume / cm³ |
|---|---|
| Glucose | 48 |
| Sucrose | 42 |
| Lactose | 5 |
| No sugar (control) | 2 |
Explain why glucose gives the highest fermentation rate and why lactose shows very little CO₂.
解释为何葡萄糖产生最高发酵速率，而乳糖产生的CO₂极少。
Yeast can directly phosphorylate glucose to begin glycolysis, so ATP is generated quickly and CO₂ release is high.
酵母可直接磷酸化葡萄糖以启动糖酵解，因此快速生成ATP，CO₂释放量大。
Lactose is a disaccharide that must be hydrolysed by lactase, which most laboratory yeast strains do not produce in sufficient amounts; hence fermentation is minimal.
乳糖是一种必须由乳糖酶水解的二糖，而大多数实验室酵母菌株产生的乳糖酶不足；因此发酵极少。
10. Case Study: Kidney Dialysis Data Analysis | 案例10：肾脏透析数据分析
A table shows the concentration of substances in blood entering and leaving a dialysis machine.
表格显示了血液进入和离开透析机时的物质浓度。
| Substance | Inflow blood / g dm⁻³ | Outflow blood / g dm⁻³ |
|---|---|---|
| Urea | 1.95 | 0.32 |
| Glucose | 1.10 | 1.08 |
| Plasma protein | 70.0 | 69.5 |
Explain how the dialysis membrane achieves these changes, and why protein concentration remains almost unchanged.
解释透析膜如何实现这些变化，以及为何蛋白质浓度几乎不变。
Urea diffuses down its concentration gradient from the blood into the dialysis fluid, which contains no urea, so its concentration drops sharply.
尿素顺着浓度梯度从血液扩散进入不含尿素的透析液中，因此其浓度急剧下降。
Glucose concentration remains similar because the dialysis fluid contains glucose at normal plasma level, preventing net movement.
葡萄糖浓度保持相似，因为透析液中含有与正常血浆水平相当的葡萄糖，阻止了净移动。
Plasma proteins are too large to pass through the partially permeable membrane, so they are retained in the blood.
血浆蛋白分子太大，无法通过部分通透膜，因此被保留在血液中。
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