📚 Year 12 Cambridge Biology: Case Study Practical Drills | 案例分析实战演练
Case studies bridge the gap between textbook knowledge and real-world biological inquiry. In Cambridge AS and A Level Biology, Paper 2 and Paper 3 often include data-based or scenario-based questions that test your ability to interpret unfamiliar information. This article provides a collection of case study drills, each followed by guided analysis, to strengthen your application skills.
案例研究是连接课本知识与现实生物学探究的桥梁。在剑桥AS和A Level生物考试中，Paper 2和Paper 3经常包含基于数据或情境的问题，考验你解读陌生信息的能力。本文提供了一系列案例研究练习，每个练习都附有引导分析，帮助你强化应用能力。
1. Understanding Case Study Questions | 理解案例分析题
To succeed in case study questions, start by highlighting command terms such as ‘explain’, ‘suggest’, or ‘calculate’. Identify the biological topic from clues in the text, then recall relevant principles. Always link your explanation back to specific data or observations.
要成功应对案例分析题，首先标出指令词，如“解释”、“建议”或“计算”。从文本线索中识别生物主题，然后回顾相关原理。始终将解释与具体数据或观察结果联系起来。
For instance, if a scenario describes ‘a drop in pH reduces enzyme activity’, you must not only state that enzymes denature at extreme pH but connect the change to disruption of ionic and hydrogen bonds in the active site.
例如，如果一个情境描述“pH下降降低了酶活性”，你不仅要说明酶在极端pH下变性，还要将这种变化与活性位点中离子键和氢键的破坏联系起来。
2. Enzyme Kinetics and Inhibitors | 酶动力学与抑制剂
An experiment measured the initial rate of reaction at different substrate concentrations for a purified enzyme, with and without an added inhibitor X. The data are shown in the table below.
一项实验测量了不同底物浓度下纯化酶在有无添加抑制剂X时的初始反应速率。数据如下表所示。
| Substrate concentration (mmol dm⁻³) | Rate without inhibitor (µmol min⁻¹) | Rate with inhibitor X (µmol min⁻¹) |
|---|---|---|
| 0.5 | 0.50 | 0.25 |
| 1.0 | 0.80 | 0.45 |
| 2.0 | 1.20 | 0.75 |
| 4.0 | 1.50 | 1.00 |
| 8.0 | 1.60 | 1.15 |
Analyse the data to determine whether inhibitor X is competitive or non-competitive. With inhibitor X, the maximum rate (Vmax) appears lower (approximately 1.15 vs 1.60 µmol min⁻¹), while the substrate concentration needed to reach half Vmax (Km) appears increased.
分析数据，确定抑制剂X是竞争性还是非竞争性。在有抑制剂X的情况下，最大速率(Vmax)似乎较低（约1.15对1.60 µmol min⁻¹），而达到半Vmax所需的底物浓度(Km)似乎增加了。
However, competitive inhibitors increase Km without affecting Vmax at high substrate concentrations. Here Vmax is reduced, which suggests non-competitive inhibition. A non-competitive inhibitor binds to an allosteric site, changing the enzyme’s shape and reducing the number of functional active sites.
然而，竞争性抑制剂会增加Km而不影响高底物浓度下的Vmax。这里Vmax降低了，这提示是非竞争性抑制。非竞争性抑制剂结合在别构位点，改变酶的形状，减少功能性活性位点的数量。
Therefore, inhibitor X is most likely a non-competitive inhibitor. You could also calculate the Km values from the data to support this conclusion.
因此，抑制剂X最可能是非竞争性抑制剂。你也可以根据数据计算Km值来支持这一结论。
3. Membrane Transport in Action | 膜运输的实际应用
Red blood cells were placed in NaCl solutions of different concentrations. The time taken for haemolysis (rupture) to occur was measured. In a hypotonic solution (0.3% NaCl), cells lysed within 30 seconds. In isotonic (0.9% NaCl), no lysis occurred. In hypertonic (3% NaCl), cells shrank.
将红细胞置于不同浓度的NaCl溶液中。测量溶血（破裂）发生的时间。在低渗溶液(0.3% NaCl)中，细胞在30秒内裂解。在等渗溶液(0.9% NaCl)中，未发生裂解。在高渗溶液(3% NaCl)中，细胞皱缩。
Explain these observations using the concepts of water potential and osmosis. Water moves from a region of higher water potential to lower water potential across a partially permeable membrane.
运用水势和渗透的概念解释这些观察结果。水通过部分透性膜从水势较高的区域向水势较低的区域移动。
In hypotonic solution, the external water potential is higher than the cell interior; water enters, causing swelling and haemolysis. In isotonic, no net movement; in hypertonic, water leaves cells, causing crenation.
在低渗溶液中，外部水势高于细胞内部；水进入细胞，导致膨胀和溶血。在等渗溶液中，没有净移动；在高渗溶液中，水离开细胞，导致皱缩。
This is a classic demonstration of osmotic effects on animal cells, which lack a cell wall to withstand pressure.
这是渗透效应对动物细胞影响的经典演示，动物细胞缺乏能承受压力的细胞壁。
4. DNA Replication and Repair | DNA复制与修复
In an experiment similar to Meselson and Stahl, bacteria were grown in a medium containing heavy nitrogen (¹⁵N) then transferred to light nitrogen (¹⁴N) medium. DNA was extracted after zero, one, and two generations and centrifuged.
在一项类似于Meselson和Stahl的实验中，细菌在含有重氮(¹⁵N)的培养基中生长，然后转移到轻氮(¹⁴N)培养基中。在零代、一代和两代后提取DNA并进行离心。
After one generation, a single band of intermediate density appeared, indicating semi-conservative replication. After two generations, both light and intermediate bands were present.
经过一代繁殖后，出现了一条中等密度的单带，表明DNA复制是半保留的。经过两代后，同时出现了轻带和中间带。
Why does this pattern rule out conservative replication? Conservative replication would produce both heavy and light bands after one generation, not an intermediate band.
为什么这个模式排除了全保留复制？全保留复制在一代后会产生重带和轻带，而不是中间带。
The data confirm that each new DNA molecule consists of one old strand and one newly synthesised strand.
数据证实，每个新的DNA分子由一条旧链和一条新合成的链组成。
5. Protein Synthesis and Mutations | 蛋白质合成与突变
A point mutation occurs in the gene for beta-globin, changing the codon from GAG to GUG. This leads to sickle cell anaemia. Refer to the genetic code: GAG codes for glutamate, GUG codes for valine.
β-珠蛋白基因发生点突变，密码子从GAG变为GUG。这导致镰刀型细胞贫血症。请参阅遗传密码表：GAG编码谷氨酸，GUG编码缬氨酸。
Explain how this single base substitution alters the primary structure of haemoglobin and its function. The replacement of a hydrophilic glutamate with hydrophobic valine causes haemoglobin molecules to aggregate under low oxygen conditions.
解释这一单碱基替换如何改变血红蛋白的一级结构及其功能。亲水性谷氨酸被疏水性缬氨酸替换，导致血红蛋白分子在低氧条件下聚集。
This aggregation distorts red blood cells into a sickle shape, impairing oxygen transport and causing blockages in capillaries.
这种聚集使红细胞扭曲成镰刀形，损害氧气运输，并导致毛细血管堵塞。
The example illustrates how a change in genotype can impact phenotype through protein structure.
这个例子说明了基因型的变化如何通过蛋白质结构影响表现型。
6. Cell Division and Cancer | 细胞分裂与癌症
A biopsy from a tumour shows a high mitotic index compared to normal tissue. The mitotic index is (number of cells in mitosis / total number of cells) × 100.
一份肿瘤活检显示与正常组织相比有丝分裂指数很高。有丝分裂指数是（有丝分裂中的细胞数 / 总细胞数）× 100。
Why is a high mitotic index a characteristic of cancerous tissues? Cancer cells have lost normal cell cycle control mechanisms, such as checkpoints, leading to uncontrolled division.
为什么高有丝分裂指数是癌变组织的一个特征？癌细胞失去了正常的细胞周期控制机制，如检查点，导致不受控制的分裂。
A high proportion of cells in mitosis indicates rapid proliferation. Chemotherapy drugs often target rapidly dividing cells, causing side effects in tissues like hair follicles and gut lining.
高比例的有丝分裂细胞表明快速增殖。化疗药物通常靶向快速分裂的细胞，从而导致毛囊和肠道内壁等组织的副作用。
7. Immune Response and Vaccination | 免疫反应与疫苗接种
A graph shows antibody concentration over time after first and second exposure to a pathogen. After first exposure, there is a lag phase of about 10 days, a low peak. After second exposure, a rapid and higher peak is seen.
一张图表显示了初次和再次接触病原体后抗体浓度随时间的变化。初次接触后，大约有10天的滞后期，峰值较低。再次接触后，看到快速且更高的峰值。
Use the concepts of primary and secondary immune responses to explain the graph. The primary response involves activation of naive B and T lymphocytes, taking time for clonal selection and differentiation into plasma and memory cells.
使用初次和再次免疫应答的概念解释图表。初次应答涉及初始B和T淋巴细胞的激活，需要时间进行克隆选择和分化为浆细胞和记忆细胞。
The secondary response is faster and stronger because memory cells persist and rapidly proliferate upon re-exposure, producing high antibody titres.
再次应答更快更强，因为记忆细胞持续存在，并在再次暴露时迅速增殖，产生高滴度抗体。
This principle underlies booster vaccinations to maintain long-term immunity.
这一原理是加强疫苗接种的基础，以维持长期免疫力。
8. Photosynthesis Under Stress | 胁迫下的光合作用
An aquatic plant was exposed to different light intensities, and the rate of photosynthesis was measured by counting oxygen bubbles. At low light intensities, the rate increased linearly; at higher intensities, it plateaued.
将一种水生植物置于不同光强下，通过计数氧气气泡来测量光合作用速率。在低光强下，速率线性增加；在较高光强下，速率趋于平稳。
Explain the shape of the light intensity curve. At low light, light is the limiting factor; the rate
Published by TutorHao | Year 12 Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply