📚 Year 12 CIE Biology: Case Study Practice Walkthrough | Year 12 CIE 生物：案例分析实战演练
Case studies are a powerful way to apply your Year 12 biology knowledge to real exam-style scenarios. In this walkthrough, we will tackle five detailed case studies drawn from the CIE AS syllabus, covering enzyme kinetics, membrane transport, genetic inheritance, immune response, and plant water relations. Each case is broken down step by step, highlighting the key concepts, common pitfalls, and effective strategies for constructing high-scoring answers. Use this as a model for how to approach data interpretation and experimental analysis questions in your own revision.
案例分析是将 Year 12 生物学知识应用到真实考试情境中的有效方法。本实战演练将深入探讨五个来自 CIE AS 大纲的详细案例，涵盖酶动力学、膜运输、遗传继承、免疫应答和植物水分关系。每个案例都逐步拆解，突出关键概念、常见错误以及构建高分答案的有效策略。你可以将其作为复习中如何应对数据分析与实验设计题目的范本。
1. Enzyme Kinetics Case Study: Investigating Trypsin Activity | 酶动力学案例：探究胰蛋白酶活性
ues. The raw data are shown in Table 1. The rate of reaction was calculated as 1/time × 1000.
一位学生研究了 pH 对胰蛋白酶活性的影响。她记录了不同 pH 值下明胶薄膜被消化所需的时间。原始数据见表 1。反应速率按 1/时间 × 1000 计算。
| pH | Time (s) | Rate (1/time × 1000) |
|---|---|---|
| 4.0 | 240 | 4.17 |
| 5.5 | 140 | 7.14 |
| 7.0 | 60 | 16.7 |
| 8.0 | 75 | 13.3 |
| 9.5 | 160 | 6.25 |
Explain why the rate of reaction is calculated as 1/time. (2 marks)
解释为什么反应速率按 1/时间计算。（2 分）
Answer approach: The rate is inversely proportional to time taken; a shorter time indicates a faster rate. Using 1/time gives a measure directly proportional to the rate, making it easier to compare and plot graphs with rate on the y-axis. Multiplying by 1000 simply scales the numbers for convenience.
答案思路：速率与所用时间成反比；时间越短表示速率越快。使用 1/时间可得到与速率成正比的量度，便于比较和在 y 轴上绘制速率图。乘以 1000 仅为缩放数值以便使用。
Describe the effect of pH on trypsin activity shown by the data. (3 marks)
根据数据描述 pH 对胰蛋白酶活性的影响。（3 分）
Answer: Trypsin works best around pH 7.0 (optimum). At pH 4.0 the rate is very low (4.17); as pH rises to 7.0 the rate increases sharply to a maximum of 16.7. Beyond 7.0, the rate declines gradually: 13.3 at pH 8.0 and 6.25 at pH 9.5. The activity is low at extremes because changes in pH alter the charges on amino acid residues in the active site, disrupting hydrogen and ionic bonds and changing the tertiary structure, so the substrate no longer fits.
答案：胰蛋白酶在 pH 约 7.0 时活性最高（最适 pH）。pH 4.0 时速率极低（4.17）；随着 pH 升至 7.0，速率急剧增加至最高 16.7。超过 7.0 后，速率逐渐下降：pH 8.0 时为 13.3，pH 9.5 时为 6.25。极端 pH 下活性很低，因为 pH 变化改变了活性位点氨基酸残基的电荷，破坏了氢键和离子键，改变了三级结构，使底物不再适配。
2. Membrane Transport Case Study: Beetroot Permeability | 膜运输案例：甜菜根透性
A class investigated the effect of temperature on membrane permeability using beetroot cylinders. Beetroot pieces were placed in water baths at temperatures ranging from 10 °C to 70 °C for 30 minutes. The absorbance of the surrounding water was measured using a colorimeter. Higher absorbance indicates more betalain pigment leaked out of the cells.
某班级用甜菜根圆柱体探究了温度对膜透性的影响。将甜菜根块置于 10 °C 至 70 °C 不同温度的水浴中 30 分钟。用比色计测量周围水的吸光度。吸光度越高，表示泄漏到细胞外的甜菜红素越多。
| Temperature (°C) | Absorbance (arbitrary units) |
|---|---|
| 10 | 0.12 |
| 20 | 0.15 |
| 30 | 0.22 |
| 40 | 0.35 |
| 50 | 0.58 |
| 60 | 0.85 |
| 70 | 1.20 |
Explain why the absorbance increases with temperature. (4 marks)
解释为什么吸光度随温度升高而增加。（4 分）
Model answer: As temperature rises, the phospholipids in the cell membrane gain kinetic energy, making the bilayer more fluid and increasing permeability. At high temperatures (above 50 °C), membrane proteins denature, opening channels and disrupting the integrity of the membrane. This allows betalain, which is normally confined inside the vacuole and cytoplasm, to leak out, raising the absorbance. At very high temperatures the membrane structure is permanently damaged.
模范答案：随着温度升高，细胞膜中的磷脂分子获得动能，使磷脂双分子层更易流动，通透性增加。在高温下（高于 50 °C），膜蛋白变性，打开通道并破坏膜完整性。这使得通常局限于液泡和细胞质中的甜菜红素泄漏出来，从而提高吸光度。在极高温度下膜结构发生永久损伤。
State two control variables for this investigation. (2 marks)
列出该实验的两个控制变量。（2 分）
Answer: Surface area/volume of beetroot pieces (e.g., using same size borer), duration of incubation (30 minutes), volume of water used, wavelength of colorimeter, and the same batch of beetroot. Keeping these constant ensures any change in absorbance is due to temperature alone.
答案：甜菜根块的面积/体积（例如使用同一打孔器打出相同尺寸）、孵育时间（30 分钟）、用水量、比色计波长以及同一批甜菜根。控制这些变量可确保吸光度的变化仅由温度引起。
3. Genetic Inheritance Case Study: Dihybrid Cross in Fruit Flies | 遗传继承案例：果蝇双因子杂交
In Drosophila melanogaster, the allele for grey body (G) is dominant over black body (g), and the allele for normal wings (W) is dominant over vestigial wings (w). A pure-breeding grey, normal-winged fly was crossed with a black, vestigial-winged fly. The F1 generation were all grey with normal wings. An F1 fly was then test-crossed with a black, vestigial-winged fly. The following offspring were obtained:
在果蝇中，灰身（G）对黑身（g）为显性，正常翅（W）对残翅（w）为显性。一只纯种灰身正常翅果蝇与一只黑身残翅果蝇杂交。F1 代全部为灰身正常翅。然后将一只 F1 代果蝇与黑身残翅果蝇进行测交，获得以下后代：
| Phenotype | Number |
|---|---|
| Grey body, normal wings | 965 |
| Black body, vestigial wings | 944 |
| Grey body, vestigial wings | 206 |
| Black body, normal wings | 185 |
Explain what these results indicate about the linkage between the two genes. (5 marks)
解释这些结果说明了两个基因之间的连锁关系。（5 分）
d black-normal) are present in much smaller numbers. This shows the genes for body colour and wing shape are linked, i.e., they are on the same chromosome. If they were unlinked, test cross would yield a 1:1:1:1 ratio. The recombinant frequency = (206+185)/2300 × 100 ≈ 17%. This indicates the distance between the linked genes; crossing over occurred in 17% eioses, producing new allele combinations. The map distance is approximately 17 map units.
要点：数量最多的两种表型（灰身正常翅和黑身残翅）为亲本型，而重组型（灰身残翅和黑身正常翅）数量极少。这表明体色和翅形基因是连锁的，即位于同一条染色体上。如果它们不连锁，测交结果应为 1:1:1:1 的比例。重组频率 = (206+185)/2300 × 100 ≈ 17%。这表明连锁基因间的距离；减数分裂中有 17% 发生了交叉互换，产生了新的等位基因组合。基因图距约为 17 个图距单位。
State the genotypes of the F1 fly and the test-cross parent, using correct notation to show linkage. (2 marks)
用正确的符号表示连锁，写出 F1 代果蝇和测交亲本的基因型。（2 分）
mozygous recessive gw/gw. The cross is GW/gw × gw/gw. Parental gametes from F1 are GW and gw; recombinant gametes are Gw and gW, explaining the phenotypes.
答案：F1 代从一个亲本继承了 G W，从另一亲本继承了 g w，因此其基因型为 GW/gw（连锁，顺式排列）。测交亲本为隐性纯合子 gw/gw。杂交为 GW/gw × gw/gw。F1 产生的亲本型配子为 GW 和 gw；重组型配子为 Gw 和 gW，这解释了各表型。
4. Immunology Case Study: Monoclonal Antibody Production | 免疫学案例：单克隆抗体生产
Monoclonal antibodies are used in pregnancy test kits and cancer treatment. The production process involves the fusion of a plasma cell with a myeloma cell. The resulting hybridoma cell divides indefinitely and secretes antibodies specific to the original antigen.
单克隆抗体用于验孕棒和癌症治疗。其生产过程涉及浆细胞与骨髓瘤细胞的融合。生成的杂交瘤细胞可无限分裂，并分泌对原始抗原具有特异性的抗体。
Outline the steps involved in producing monoclonal antibodies targeting human chorionic gonadotropin (hCG). (6 marks)
概述生产针对人绒毛膜促性腺激素（hCG）的单克隆抗体的步骤。（6 分）
Model answer: A mouse is injected with purified hCG. Its B-lymphocytes are activated and produce anti-hCG antibodies. The mouse spleen is removed and plasma cells are isolated. These plasma cells are fused with HGPRT-deficient myeloma cells using polyethylene glycol (or electrical fusion). The mixture is placed in HAT medium; unfused myeloma cells die because they cannot use salvage pathway for nucleotide synthesis, and unfused plasma cells die naturally because they cannot divide indefinitely. Only fused hybridoma cells survive and divide. The hybridoma cells are separated into individual wells and screened for anti-hCG antibody production using ELISA. Positive clones are cultured to produce large quantities of the desired monoclonal antibody.
模范答案：向小鼠注射纯化的 hCG。其 B 淋巴细胞被激活并产生抗 hCG 抗体。取出小鼠脾脏并分离浆细胞。用聚乙二醇（或电融合）将这些浆细胞与 HGPRT 缺陷型骨髓瘤细胞融合。将混合物置于 HAT 培养基中；未融合的骨髓瘤细胞因不能利用补救途径合成核苷酸而死亡，未融合的浆细胞因不能无限分裂自然死亡。只有融合的杂交瘤细胞存活并分裂。将杂交瘤细胞分离到单个孔中，使用 ELISA 筛选产生抗 hCG 抗体的细胞。阳性克隆被培养以大量生产所需的单克隆抗体。
Explain why the use of HAT medium is a key step. (3 marks)
解释为什么使用 HAT 培养基是关键步骤。（3 分）
HAT medium contains aminopterin that blocks the de novo synthesis of nucleotides. Myeloma cells lack HGPRT and cannot use the salvage pathway, so they die. Plasma cells cannot divide in culture and will eventually die. Only hybridomas that inherit the HGPRT enzyme from the plasma cell and the ability to divide from the myeloma cell survive. This provides an efficient selection for fused cells without the need for manual sorting.
HAT 培养基含有氨基蝶呤，可阻断核苷酸的从头合成。骨髓瘤细胞缺乏 HGPRT，无法使用补救途径，因而死亡。浆细胞在培养液中不能分裂，最终死亡。只有从浆细胞获得 HGPRT 酶、并从骨髓瘤细胞获得分裂能力的杂交瘤细胞才能存活。这为融合细胞提供了有效的选择，无需手工分选。
5. Plant Water Relations Case Study: Transpiration Pull Measurement | 植物水分关系案例：蒸腾拉力测定
A potometer was set up to measure the rate of water uptake in a leafy shoot under different environmental conditions. The air bubble moved 25 mm in 10 minutes under normal lab conditions. When a fan was placed near the shoot, the bubble moved 58 mm in 10 minutes. When a plastic bag was tied around the leaves, movement was only 6 mm in 10 minutes.
设置蒸腾计，测量不同环境条件下带叶枝条的吸水量。在正常实验室条件下，气泡在 10 分钟内移动 25 mm。当在枝条旁放置风扇时，气泡在 10 分钟内移动 58 mm。当用塑料袋套住叶片时，10 分钟内仅移动 6 mm。
Calculate the rate of water uptake for each condition in mm per minute. (2 marks)
计算每种条件下每分钟的吸水速率（mm/分钟）。（2 分）
Fan: 58/10 = 5.8 mm/min. Bag: 6/10 = 0.6 mm/min. (Always show the calculation and units.)
正常：25/10 = 2.5 mm/min。风扇：58/10 = 5.8 mm/min。塑料袋：6/10 = 0.6 mm/min。（必须写出计算过程和单位。）
Explain why the rate increased with the fan. (3 marks)
解释为什么风扇下速率增加。（3 分）
The fan moves air across the leaf surface, reducing the layer of still, humid air at the boundary layer. This increases the water vapour concentration gradient between the leaf's internal air spaces and the external atmosphere. A steeper diffusion gradient increases the rate of evaporation of water from mesophyll cell walls, which in turn increases transpiration pull and water uptake through the xylem. More rapid loss of water vapour creates a more negative water potential in the leaves, drawing water up faster.
风扇使空气流过叶片表面，减薄了边界层中静止湿润的空气层。这增大了叶片内部空气间隙与外界大气之间的水蒸气浓度梯度。更陡的扩散梯度加快了叶肉细胞壁水分的蒸发速率，进而增加蒸腾拉力和通过木质部的吸水量。更快的水蒸气散失使叶片内水势更负，从而更快地将水拉上来。
Suggest one precaution to ensure the potometer readings are reliable. (2 marks)
提出一项确保蒸腾计读数可靠的预防措施。（2 分）
Ensure the shoot is cut under water to prevent air embolisms entering the xylem; all connections must be airtight using petroleum jelly; allow time for the shoot to acclimatize before recording; use the same shoot and reset the bubble between readings.
确保在水下剪切枝条，以防空气栓塞进入木质部；所有连接处需用凡士林密封；记录前让枝条适应一段时间；使用同一枝条，并在读数间重置气泡。
6. Cell Signalling Case Study: Adrenaline Action on Liver Cells | 细胞信号案例：肾上腺素对肝细胞的作用
Adrenaline binding to its receptor on the surface of liver cells triggers a cascade that results in the conversion of glycogen to glucose. This second messenger system involves G proteins and adenylate cyclase.
肾上腺素与肝细胞表面受体结合，触发级联反应，使糖原转化为葡萄糖。该第二信使系统涉及 G 蛋白和腺苷酸环化酶。
Describe the sequence of events from adrenaline binding to the release of glucose into the blood. (5 marks)
描述从肾上腺素结合到葡萄糖释放进入血液的过程。（5 分）
Adrenaline (first messenger) binds to a specific receptor on the plasma membrane. The receptor changes shape, activating a G protein. The G protein in turn activates adenylate cyclase, which catalyses the conversion of ATP to cyclic AMP (cAMP). cAMP acts as a second messenger and binds to protein kinase A, activating it. Protein kinase A phosphorylates and activates glycogen phosphorylase kinase, which then activates glycogen phosphorylase. This enzyme breaks glycogen down to glucose-1-phosphate, which is converted to glucose-6-phosphate and finally glucose, which exits the cell via facilitated diffusion into the blood.
肾上腺素（第一信使）与质膜上的特异性受体结合。受体变构，激活 G 蛋白。G 蛋白继而激活腺苷酸环化酶，该酶催化 ATP 转化为环腺苷酸（cAMP）。cAMP 作为第二信使，与蛋白激酶 A 结合并激活之。蛋白激酶 A 磷酸化并激活糖原磷酸化酶激酶，后者再激活糖原磷酸化酶。该酶将糖原分解为 1-磷酸葡萄糖，再转变为 6-磷酸葡萄糖，最终生成葡萄糖，通过易化扩散离开细胞进入血液。
Explain how this signaling cascade leads to amplification of the signal. (3 marks)
解释此信号级联如何导致信号放大。（3 分）
One adrenaline molecule can activate multiple G proteins, each of which can activate several adenylate cyclase enzymes. Each adenylate cyclase produces many cAMP molecules, and each cAMP activates one protein kinase A molecule which in turn activates many enzyme molecules downstream. At each step the number of active molecules increases exponentially, so a tiny initial signal results in a large release of glucose. This cascade ensures a rapid and massive response to stress.
一个肾上腺素分子可激活多个 G 蛋白，每个 G 蛋白又可激活数个腺苷酸环化酶。每个腺苷酸环化酶生成大量 cAMP 分子，每个 cAMP 激活一个蛋白激酶 A，后者又激活下游许多酶分子。每一步活化的分子数量呈指数增长，因此微小的初始信号即可导致大量葡萄糖释放。这种级联放大了信号，确保对压力作出迅速而剧烈的反应。
7. Infectious Disease Case Study: Tuberculosis Transmission & Control | 传染病案例：肺结核的传播与控制
Mycobacterium tuberculosis is transmitted via airborne droplets. The bacteria can survive inside macrophages by preventing lysosome fusion. CD4+ T cells and interferon gamma are vital in the immune response, forming granulomas to wall off the bacteria.
结核分枝杆菌通过飞沫传播。该菌可通过阻止溶酶体融合而在巨噬细胞内存活。CD4⁺ T 细胞和 γ 干扰素在免疫应答中至关重要，形成肉芽肿以隔离细菌。
Explain why the BCG vaccine is less effective in some geographic regions. (4 marks)
解释为什么卡介苗（BCG）在某些地理区域效果较差。（4 分）
BCG contains a live attenuated strain of Mycobacterium bovis. Its efficacy varies due to several factors: exposure to environmental non-tuberculous mycobacteria may prime the immune system differently, sometimes interfering with protective response; genetic variations in human populations affect immune recognition; different strains of the BCG vaccine have lost some immunogenic antigens after years of subculturing. In tropical regions, co-infections with helminths skew the immune response towards a Th2 profile, reducing the Th1 response needed to clear mycobacteria. Additionally, latent TB infections are more common, and the vaccine is less effective against reactivation of latent TB.
BCG 含有牛分枝杆菌的减毒活菌株。其效果因地区而异，原因有多种：暴露于环境中的非结核分枝杆菌可能以不同方式启动免疫系统，有时干扰保护性应答；人群的遗传变异影响免疫识别；不同菌株的 BCG 疫苗经过多年传代丢失了一些免疫原性抗原。在热带地区，蠕虫共感染使免疫应答偏向 Th2 型，削弱了清除分枝杆菌所需的 Th1 应答。此外，潜伏性结核感染更常见，疫苗对潜伏结核的再激活效果较差。
Describe how a granuloma forms and its role in TB infection. (3 marks)
描述肉芽肿的形成及其在结核感染中的作用。（3 分）
When macrophages infected with M. tuberculosis present antigens on MHC class II, CD4⁺ T cells are activated and secrete interferon gamma. This activates nearby macrophages to increase bactericidal activity and epithelioid cell formation. Macrophages fuse to form multinucleated giant cells. Fibroblasts are stimulated to lay down collagen, creating a tough, caseous centre with a fibrous capsule. The granuloma contains the bacteria, preventing dissemination, but may also provide a niche for dormant bacteria that can reactivate later.
当感染了结核杆菌的巨噬细胞在 MHC II 类分子上呈递抗原后，CD4⁺ T 细胞被激活并分泌 γ 干扰素。这激活邻近的巨噬细胞，增强杀菌活性并形成上皮样细胞。巨噬细胞融合形成多核巨细胞。成纤维细胞受刺激沉积胶原蛋白，形成含有干酪样中心的坚韧纤维化囊。肉芽肿将细菌包裹其中，防止扩散，但也可能为休眠细菌提供藏身之所，日后可能再激活。
8. Ecological Case Study: Mark-Release-Recapture for Population Estimation | 生态学案例：标记重捕法估算种群数量
Students carried out a mark-release-recapture study on woodlice in a small woodland. On day one, they captured 243 woodlice, marked them with a dot of non-toxic paint, and released them. Two days later, they recaptured 180 woodlice, of which 27 were marked.
学生们对一片小林地中的潮虫进行了标记重捕研究。第一天，他们捕获 243 只潮虫，用无毒颜料点上标记后放回。两天后，他们重捕了 180 只潮虫，其中 27 只带有标记。
Estimate the population size using the Lincoln index. Show your working. (3 marks)
使用林肯指数估算种群数量。写出计算过程。（3 分）
d sample (27). N = (243 × 180)/27 = 43740/27 = 1620. So the estimated population is 1620 woodlice. (Allow formula, substitution, correct answer.)
林肯指数：N = (M × C)/R，其中 M 为初次标记数（243），C 为第二次样本总数（180），R 为第二次样本中标记个体数（27）。N = (243 × 180)/27 = 43740/27 = 1620。因此估算种群数量为 1620 只潮虫。（包括公式、代入和正确答案。）
Discuss two assumptions of this method and how they may have been violated in this investigation. (4 marks)
讨论该方法的两个假设，以及本次调查中它们可能如何被违反。（4 分）
Assumption 1: Marks are not lost between samples. If the paint wore off or the woodlice moulted, the marked number would be underestimated, causing overestimation of population. Assumption 2: Marking does not affect survival or behaviour. If marked woodlice were more conspicuous to predators, they would be predated more, reducing R and overestimating N. Other assumptions: closed population (no migration, births, deaths); if some woodlice moved out of the area or died between sampling, the estimate would be biased. Students should state that the time interval should be short enough to assume a closed population.
假设 1：两次取样间标记不会丢失。如果颜料脱落或潮虫蜕皮，标记数将被低估，导致种群高估。假设 2：标记不影响存活率或行为。如果标记的潮虫更容易被捕食者注意，它们被捕食更多，R 减小，导致 N 高估。其他假设：封闭种群（无迁移、出生、死亡）；如果两次取样间有些潮虫迁出或死亡，估算会产生偏差。学生应说明时间间隔应足够短，以假定为封闭种群。
9. Experimental Design Case Study: Testing the Effect of Light on Chloroplast Activity | 实验设计案例：检验光对叶绿体活性的影响
A student wanted to measure the rate of the light-dependent reactions of photosynthesis using isolated chloroplasts and DCPIP as an artificial electron acceptor. DCPIP is blue when oxidised and colourless when reduced. The reduction rate can be followed by measuring absorbance at 620 nm over time.
一位学生想利用离体叶绿体和 DCPIP 作为人工电子受体，测量光合作用光反应速率。DCPIP 氧化态为蓝色，还原态为无色。还原速率可通过随时间测量 620 nm 处吸光度来追踪。
Design an investigation to determine the effect of light intensity on the rate of DCPIP reduction. Include details of controls, independent and dependent variable, and how you would present results. (6 marks)
设计一个探究光强对 DCPIP 还原速率影响的实验。包括对照组、自变量、因变量以及如何呈现结果的详细说明。（6 分）
Independent variable: light intensity (achieved by varying distance of lamp from the tube). Dependent variable: rate of DCPIP reduction, measured as change in absorbance per unit time (ΔA/min). Control: a tube with all contents wrapped in aluminium foil to block all light; this accounts for any reduction not caused by the light-dependent reactions. Also a boiled chloroplast control to confirm enzymes are required.
Method: Isolate chloroplasts from spinach leaves by grinding in cold buffer and centrifuging. Add a fixed volume of chloroplast suspension to tubes containing DCPIP, buffer, and an electron acceptor (if needed). Place tubes at different distances from a lamp (e.g., 10 cm, 20 cm, 30 cm, 40 cm, 50 cm). Immediately measure absorbance at 620 nm at time 0 and then every 2 minutes for 10 minutes, keeping the tubes in the light between readings. Record absorbance values and calculate the initial rate of decolorisation (slope). Plot rate (y-axis) against light intensity (x-axis where 1/d² indicates relative intensity).
自变量：光强度（通过改变灯距管子的距离实现）。因变量：DCPIP 还原速率，以单位时间内吸光度的变化（ΔA/min）衡量。对照组：一支包有铝箔完全遮光的试管；此组可排除不依赖光反应的还原。还包括煮沸的叶绿体对照，确认需要酶的参与。
方法：用冷缓冲液研磨菠菜叶并离心分离叶绿体。向含有 DCPIP、缓冲液及电子受体的试管中加入固定体积的叶绿体悬液。将试管置于距灯不同距离处（如 10 cm、20 cm、30 cm、40 cm、50 cm）。立即在 0 时刻测量 620 nm 吸光度，此后每 2 分钟测一次，共 10 分钟，测量间隙保持试管在光照下。记录吸光度值并计算褪色初始速率（斜率）。以速率（y 轴）对光强度（x 轴，用 1/d² 表示相对光强）作图。
Presentation: A graph of rate vs. 1/d² will show a directly proportional relationship if light intensity limits the reaction, but the line will plateau when another factor (e.g., DCPIP or chloroplast concentration) becomes limiting. The tube wrapped in foil should show no change in absorbance; any change would indicate background reduction.
呈现：速率对 1/d² 作图，若光强为限制因素，应呈正比关系；但当其他因素（如 DCPIP 或叶绿体浓度）成为限制时，曲线会趋于平台。包有铝箔的试管吸光度应无变化，如有变化则提示存在非特异性还原。
10. Data Analysis Case Study: Interpreting a Calorimeter Trace for Respiration | 数据分析案例：解读呼吸量热计曲线
h an internal diameter of 1.0 mm. A control tube with dry, dead peas showed no movement. (Volume = πr² × distance moved; π = 3.14.)
使用呼吸计测量 25 °C 下萌发豌豆的耗氧量。在 5 分钟内，毛细管中的带色液滴移动了 4.8 cm，毛细管内径为 1.0 mm。含干燥死豌豆的对照管无移动。（体积 = πr² × 移动距离；π = 3.14。）
Calculate the rate of oxygen consumption in cm³ per minute. (3 marks)
计算耗氧率，单位 cm³/分钟。（3 分）
768 cm³. Rate per minute = 0.03768 / 5 = 0.00754 cm³ min⁻¹. Expressed correctly in standard form: 7.54 × 10⁻³ cm³ min⁻¹. (Accept 0.0075 cm³ min⁻¹.)
半径 = 0.5 mm = 0.05 cm。截面积 = π × (0.05)² = 3.14 × 0.0025 = 0.00785 cm²。5 分钟内消耗体积 = 面积 × 距离 = 0.00785 × 4.8 = 0.03768 cm³。每分钟速率 = 0.03768 / 5 = 0.00754 cm³ min⁻¹。可用标准形式表示：7.54 × 10⁻³ cm³ min⁻¹。（接受 0.0075 cm³ min⁻¹。）
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