📚 Pre-U Edexcel Biology: Case Study Practice Drills | Pre-U Edexcel 生物：案例分析实战演练
In Pre-U Edexcel Biology, case study questions test your ability to apply knowledge to unfamiliar scenarios, analyse data, and construct evidence-based explanations. This article presents eight practice drills that mirror exam-style challenges, covering enzymes, cell membranes, genetics, ecology, photosynthesis, respiration, mitosis and gene expression. Each case is accompanied by paired English-Chinese explanations to reinforce your command of the subject.
在Pre-U Edexcel生物考试中，案例分析题旨在考查你将知识应用于陌生情境、分析数据以及构建证据支持的解释的能力。本文呈现了八个模拟真实考题的演练，涵盖酶学、细胞膜、遗传学、生态学、光合作用、呼吸作用、有丝分裂和基因表达。每个案例都配有中英对照的讲解，以强化你对学科的掌握。
1. Enzyme Activity and pH: Data Interpretation | 酶活性与pH：数据解读
A student investigated the effect of pH on the activity of amylase. The rate of starch breakdown was recorded at 25°C using a colorimetric assay. The following results were obtained:
一名学生探究了pH对淀粉酶活性的影响。在25°C下使用比色法记录淀粉分解速率，获得以下数据：
| pH | 3.0 | 5.0 | 7.0 | 8.5 | 10.0 | 12.0 |
| Rate (mg starch min⁻¹) | 2 | 18 | 40 | 38 | 12 | 1 |
The optimum pH for this amylase lies around 7.0–8.5, with maximum activity near pH 7.0. The steep decline on both the acidic and alkaline sides indicates irreversible denaturation. Below pH 5.0, excess H⁺ ions disrupt ionic and hydrogen bonds that stabilise the tertiary structure, altering the active site shape. Above pH 10.0, OH⁻ ions abstract protons from key amino acid residues, likewise breaking critical bonds.
该淀粉酶的最适pH在7.0–8.5之间，最大活性接近pH 7.0。在酸性和碱性两侧活性急剧下降表明发生了不可逆的变性。pH低于5.0时，过多的H⁺离子破坏了稳定三级结构的离子键和氢键，改变了活性位点的形状。pH高于10.0时，OH⁻离子从关键氨基酸残基上夺取质子，同样打断了重要的键。
When interpreting such data, always refer to the disruption of non-covalent interactions rather than peptide bond hydrolysis—extreme pH values denature most enzymes without breaking the polypeptide backbone. For a full answer, mention that lost activity cannot be restored by returning to the optimum pH because the native conformation has been permanently damaged.
解读这类数据时，务必提到非共价相互作用的破坏，而非肽键的水解——极端pH值使多数酶变性，但多肽骨架并不断裂。完整的答案要说明，恢复到最适pH并不能恢复活性，因为天然构象已被永久破坏。
2. Osmosis and Plant Cell Water Relations | 渗透作用与植物细胞水分关系
Epidermal strips of red onion were immersed in a series of sucrose solutions (0.0–1.0 mol dm⁻³) for 20 minutes. The percentage of cells showing plasmolysis was counted under a microscope. The data were plotted, and the concentration at which 50% of cells were plasmolysed (incipient plasmolysis) was determined to be 0.45 mol dm⁻³.
将红洋葱表皮条浸入一系列蔗糖溶液(0.0–1.0 mol dm⁻³)中20分钟。在显微镜下计数发生质壁分离的细胞百分比。数据被绘制成图，测得50%细胞发生质壁分离（初始质壁分离）时的浓度为0.45 mol dm⁻³。
Incipient plasmolysis represents the point where the protoplast just begins to pull away from the cell wall; at this point the water potential of the cell sap (ψcell) equals the water potential of the external solution (ψsolute + ψpressure). For a dilute sucrose solution at 25°C, the solute potential can be estimated using the formula ψs = –iCRT, where i=1 for sucrose, C is the molar concentration, R is the gas constant (0.00831 kJ mol⁻¹ K⁻¹) and T is 298 K. Thus ψs = –1 × 0.45 × 0.00831 × 298 = –1.11 MPa (approx). Since no turgor pressure exists at incipient plasmolysis, ψcell = ψs = –1.11 MPa.
初始质壁分离表示原生质体刚开始脱离细胞壁的时刻；此时细胞液的渗透势（ψcell）等于外界溶液的水势（ψsolute + ψpressure）。对于25°C的稀蔗糖溶液，溶质势可由公式ψs = –iCRT估算，其中蔗糖i=1，C为摩尔浓度，R为气体常数(0.00831 kJ mol⁻¹ K⁻¹)，T为298 K。因此ψs = –1 × 0.45 × 0.00831 × 298 ≈ –1.11 MPa。由于初始质壁分离时压力势为零，故细胞水势ψcell = –1.11 MPa。
This value indicates that the onion cell sap has a fairly negative water potential, meaning it tends to draw in water from pure water or dilute solutions. In a 0.2 mol dm⁻³ solution, water would enter cells by osmosis, increasing turgor pressure and eventually building up a positive pressure potential that makes the cell water potential less negative and stops net water entry.
该数值表明洋葱细胞液具有较负的水势，意味着它倾向于从纯水或稀溶液中吸水。在0.2 mol dm⁻³溶液中，水会通过渗透作用进入细胞，增加膨压，最终建立起正的压力势，使得细胞水势负值减小，并阻止水分的净进入。
3. Pedigree Analysis: Autosomal Recessive Inheritance | 家系图分析：常染色体隐性遗传
The pedigree shows a family where some individuals are affected by a rare metabolic disorder. Neither parent in generation I is affected, but they have an affected son (II-2). An unaffected daughter (II-1) marries an unrelated carrier male. What is the probability that their first child will be affected?
该家系图显示了一个家族中部分个体患有一种罕见的代谢疾病。世代I的父母均未患病，但他们有一个患病的儿子（II-2）。一个未患病的女儿（II-1）与一个无亲缘关系的携带者男性婚配。他们的第一个孩子患病的概率是多少？
Since two unaffected parents produced an affected child, the disorder must be recessive. If it were autosomal dominant, one parent would show the trait. It is also unlikely to be X-linked recessive because the affected son would have inherited the recessive allele from his heterozygous mother, which is possible, but the question typically states autosomal recessive for such patterns. We assume autosomal recessive inheritance. Let A = dominant normal allele, a = recessive disease allele. Both I-1 and I-2 must be heterozygotes (Aa). Their unaffected daughter II-1 could be either AA (probability 1/3) or Aa (2/3), because she does not express the disorder. She marries a carrier (Aa).
由于未患病的父母生出了患病的孩子，该疾病必定是隐性的。如果是常染色体显性遗传，双亲之一会表现出该性状。也不大可能是X连锁隐性遗传，因为患病儿子会从杂合子母亲那里继承隐性等位基因，这有可能，但题目通常指明为常染色体隐性。我们假定为常染色体隐性遗传。设A = 显性正常等位基因，a = 隐性致病等位基因。I-1和I-2必定都是杂合子（Aa）。他们未患病的女儿II-1可能是AA（概率1/3）或Aa（概率2/3），因为她不表达该疾病。她与一个携带者（Aa）结婚。
Only if II-1 is Aa can they have an affected child. The probability II-1 is Aa equals 2/3. If she is Aa, the chance of passing on the a allele is 1/2, and the father (Aa) also has a 1/2 chance of passing on a. So the conditional probability of an affected child given Aa mother = 1/2 × 1/2 = 1/4. Overall probability = (2/3) × (1/4) = 2/12 = 1/6.
只有当II-1为Aa时他们才有可能生出患病孩子。II-1为Aa的概率是2/3。若她是Aa，传递a等位基因的概率是1/2，父亲（Aa）传递a的概率也是1/2。因此在母亲为Aa的条件下，孩子患病的概率为1/2 × 1/2 = 1/4。总体概率 = (2/3) × (1/4) = 2/12 = 1/6。
Always present your reasoning step by step in pedigree problems. Start by identifying the mode of inheritance, assign genotypes, and then calculate using conditional probabilities where unaffected individuals have a risk of being heterozygous.
在解答家系图问题时，务必逐步展示推理过程。首先确定遗传方式，分配基因型，然后利用条件概率进行推算，其中未患病个体存在杂合子的风险。
4. Ecological Sampling and Biodiversity Indices | 生态取样与生物多样性指数
Two heathland sites, A and B, were sampled using 0.5 m × 0.5 m quadrats placed randomly. The numbers of each plant species were recorded. Site A contained heather (85% cover), gorse (10%), and three other rare species (5% combined). Site B contained heather (40%), gorse (20%), bracken (20%), moss (15%) and an orchid (5%). Calculate Simpson’s Index of Diversity (D) for both sites and comment on the conservation value.
使用0.5 m × 0.5 m的样方随机放置，对两处石楠荒原样地A和B进行了取样，记录每种植物的数量。样地A有石楠（85%盖度）、荆豆（10%）以及三种稀有物种（合计5%）。样地B有石楠（40%）、荆豆（20%）、欧洲蕨（20%）、苔藓（15%）和一种兰花（5%）。分别计算两地Simpson多样性指数(D)，并评论其保护价值。
Simpson’s Index D = 1 – Σ(n/N)², where n = number of individuals or percentage cover of a species, N = total individuals or total cover. Using percentage cover as an indicator of abundance, for Site A: n/N values: 0.85, 0.10, and three species each 0.0167 (since 5% combined). Sum of squares = 0.85² + 0.10² + (5×0.0167²) ≈ 0.7225 + 0.01 + 5×0.000279 = 0.7225+0.01+0.001395 = 0.733895. D = 1 – 0.7339 = 0.2661. For Site B: proportions: 0.40, 0.20, 0.20, 0.15, 0.05. Sum of squares = 0.16+0.04+0.04+0.0225+0.0025 = 0.265. D = 1 – 0.265 = 0.735.
Simpson指数D = 1 – Σ(n/N)²，其中n为某一物种的个体数或盖度百分比，N为总个体数或总盖度。用盖度百分比作为丰度指标，样地A：n/N值分别为0.85、0.10以及三种各占0.0167（合计5%）。平方和 = 0.85² + 0.10² + (5×0.0167²) ≈ 0.7225 + 0.01 + 5×0.000279 = 0.7225+0.01+0.001395 = 0.733895。D = 1 – 0.7339 = 0.2661。样地B：比例为0.40、0.20、0.20、0.15、0.05。平方和 = 0.16+0.04+0.04+0.0225+0.0025 = 0.265。D = 1 – 0.265 = 0.735。
Site B has a much higher diversity index, indicating greater species evenness and richness. Although Site A possesses rare species, its extreme dominance by heather makes the ecosystem vulnerable to environmental change. High diversity often confers greater stability and resilience, so Site B may be considered of higher conservation value in terms of overall biodiversity, but Site A might still be a priority if those rare species are endemic or threatened.
样地B的多样性指数高得多，表明物种均匀度和丰富度更高。虽然样地A拥有稀有物种，但石楠的极端优势使得生态系统易受环境变化的影响。高多样性通常赋予更大的稳定性和恢复力，因此就总体生物多样性而言，样地B可能具有更高的保护价值，但如果那些稀有物种是特有或受威胁的，样地A仍可能列为优先保护。
5. Light Intensity and Photosynthesis Rate | 光照强度与光合作用速率
An aquatic plant was exposed to different light intensities, and the rate of oxygen evolution was measured (cm³ O₂ m⁻² s⁻¹). At 200 µmol photons m⁻² s⁻¹, the rate was 8; at 400, it was 15; at 600, it was 20; at 800, it was 22; at 1000, it was 22. Respiration rate in the dark was 2 cm³ O₂ m⁻² s⁻¹ consumed. Plot the data, determine the light compensation point, and explain why the curve plateaus.
将一株水生植物暴露在不同光照强度下，测定氧气的释放速率 (cm³ O₂ m⁻² s⁻¹)。在200 µmol photons m⁻² s⁻¹下，速率为8；400时为15；600时为20；800时为22；1000时为22。黑暗中的呼吸速率为消耗2 cm³ O₂ m⁻² s⁻¹。绘制数据，确定光补偿点，并解释曲线达到平台期的原因。
Gross photosynthesis = net oxygen release + respiration, but for compensation point we consider net oxygen exchange equals zero. The light compensation point is the irradiance where oxygen produced by photosynthesis exactly equals oxygen consumed by respiration. We can estimate by interpolation: at 0 light, net O₂ uptake is –2. Net O₂ at 200 µmol m⁻² s⁻¹ is +8, so the line between these points crosses zero very low light. Approximate linear regression: slope = (8 – (–2)) / (200 – 0) = 10/200 = 0.05 per unit. To increase from –2 to 0 requires 2 / 0.05 = 40 µmol photons m⁻² s⁻¹. So the compensation point is around 40 µmol m⁻² s⁻¹.
总光合作用 = 净氧释放 + 呼吸作用，但在补偿点处我们考虑净氧交换为零。光补偿点是指光合作用产生的氧气恰好等于呼吸作用消耗的氧气的辐照度。我们可以通过内插法估算：在0光照时，净O₂吸收为–2。在200 µmol m⁻² s⁻¹时净O₂为+8，因此两点之间的线在极低光照下与零相交。近似线性回归：斜率 = (8 – (–2)) / (200 – 0) = 10/200 = 0.05每单位。从–2增加到0需要2 / 0.05 = 40 µmol photons m⁻² s⁻¹。所以补偿点约为40 µmol m⁻² s⁻¹。
The plateau above 600 µmol m⁻² s⁻¹ indicates that light is no longer the limiting factor. At these intensities, the Calvin cycle enzymes (e.g. RuBisCO) become saturated, or CO₂ availability limits the rate. Temperature could also be limiting if it was constant and low. In an exam, always discuss the concept of limiting factors: light, carbon dioxide, and temperature can each become rate-limiting depending on conditions.
在600 µmol m⁻² s⁻¹之上的平台期表明光照不再是限制因子。在该光照强度下，卡尔文循环的酶（如RuBisCO）已饱和，或者CO₂的可利用性限制了速率。若温度恒定且较低，温度也可能成为限制因子。在考试中，务必讨论限制因子的概念：光照、二氧化碳和温度均可视条件成为限速因子。
6. Temperature and Respiration: Q₁₀ Calculation | 温度与呼吸作用：Q₁₀计算
Yeast suspension was incubated with glucose at 20°C and 30°C. The rate of CO₂ production was measured as 3 cm³ min⁻¹ at 20°C and 7 cm³ min⁻¹ at 30°C. Calculate the temperature coefficient Q₁₀. Explain why the rate increases with temperature and whether Q₁₀ remains constant over a wider range.
酵母悬浮液在20°C和30°C下与葡萄糖温育。测得CO₂产生速率在20°C时为3 cm³ min⁻¹，30°C时为7 cm³ min⁻¹。计算温度系数Q₁₀。解释速率为何随温度升高而增加，以及Q₁₀在更宽温度范围内是否保持恒定。
Q₁₀ = (rate at T+10°C) / (rate at T). Here, T = 20°C, T+10 = 30°C. Q₁₀ = 7 / 3 = 2.33. This value means that for every 10°C rise, the reaction rate more than doubles under these conditions.
Q₁₀ = (T+10°C时的速率) / (T时的速率)。此处T=20°C，T+10=30°C。Q₁₀ = 7 / 3 = 2.33。该数值意味着在这些条件下，温度每升高10°C，反应速率增加一倍以上。
The rise in temperature increases the kinetic energy of enzyme and substrate molecules, leading to more frequent and forceful collisions that overcome activation energy barriers. Additionally, a higher proportion of molecules possess energy exceeding the activation energy (Ea), as described by the Maxwell–Boltzmann distribution. However, Q₁₀ is not constant: at higher temperatures (e.g. above 40°C for many enzymes), denaturation begins, and the rate eventually falls. Therefore, Q₁₀ values are valid only within a moderate range where enzyme stability is retained.
温度升高增加了酶和底物分子的动能，使得碰撞更频繁、更有力，能克服活化能障碍。同时，更高比例的分子具有超过活化能(Ea)的能量，正如麦克斯韦–玻尔兹曼分布所述。然而，Q₁₀并非恒定：在较高温度下（例如许多酶在40°C以上），变性开始发生，速率最终下降。因此，Q₁₀值仅在酶稳定性保持的适中范围内有效。
7. Mitotic Index in Root Tips | 根尖有丝分裂指数
A student prepared a squash of onion root tip and counted cells in different stages of the cell cycle. Out of 800 cells, 560 were in interphase, 120 in prophase, 60 in metaphase, 40 in anaphase, and 20 in telophase. Calculate the mitotic index and the approximate duration of mitosis if the complete cell cycle lasts 20 hours.
一名学生制备了洋葱根尖压片，计数了细胞周期不同阶段的细胞数。在800个细胞中，间期560个，前期120个，中期60个，后期40个，末期20个。计算有丝分裂指数，并估算如果整个细胞周期为20小时，有丝分裂的大致持续时间。
Mitotic index = number of cells in mitosis (prophase + metaphase + anaphase + telophase) / total number of cells. Here, mitotic cells = 120+60+40+20 = 240. Mitotic index = 240 / 800 = 0.30 (or 30%).
有丝分裂指数 = 处于有丝分裂的细胞数（前期+中期+后期+末期）/ 细胞总数。此处有丝分裂细胞数 = 120+60+40+20 = 240。有丝分裂指数 = 240 / 800 = 0.30（即30%）。
Assuming the tissue is growing asynchronously, the proportion of cells in a particular phase is proportional to the duration of that phase. Mitosis duration = mitotic index × total cell cycle time = 0.30 × 20 h = 6 hours. However, this calculation assumes the index correctly represents the fraction of time spent. In reality, progression through the cycle is not strictly linear, but it provides a reasonable estimate. Meristematic cells in root tips have a high mitotic index because they are actively dividing.
假定组织是不同步生长的，处于某一特定阶段的细胞比例与这一阶段的时间长度成正比。有丝分裂持续时间 = 有丝分裂指数 × 总细胞周期时间 = 0.30 × 20 h = 6小时。然而，该计算假设指数能正确地代表所花时间的比例。实际上，周期进程并非严格线性，但这提供了一个合理的估算。根尖分生组织细胞因活跃分裂而具有较高的有丝分裂指数。
An alternative method uses the duration of a specific stage if known. Always show your working clearly, and remember that mitotic index can be used to diagnose cancer tissues, where a high index indicates rapid proliferation.
如果已知某一特定阶段的时间，也可以采用另一种方法。务必清晰地展示你的计算过程，并记住有丝分裂指数可用于诊断癌组织，高的指数表明异常快速的增殖。
8. Gene Mutation and Protein Function Prediction | 基因突变与蛋白质功能预测
A section of the β-globin gene has the following DNA template sequence: 3′ TAC CAC GTG GAC TGA 5′. A point mutation occurs at the fourth codon, changing the DNA template from GAC to GAG. Use the genetic code to determine the amino acid change and predict the likely effect on haemoglobin function.
β-珠蛋白基因的一段DNA模板序列为：3′ TAC CAC GTG GAC TGA 5’。在第四个密码子处发生了一个点突变，使DNA模板从GAC变为GAG。利用遗传密码确定氨基酸的改变，并预测对血红蛋白功能的可能影响。
The template strand is read 3′ to 5′, so the mRNA will be synthesised 5′ to 3′ complementary to it. Original template: TAC CAC GTG GAC TGA. mRNA codons: AUG GUG CAC CUG ACU. Amino acids: Methionine (start) – Valine – Histidine – Leucine – Threonine (stop actually UGA, but here TGA on DNA gives ACU? Wait: DNA template TGA -> mRNA ACU (Thr), not a stop. This is a short peptide. The fourth codon is the leucine position. After mutation, template becomes GAG, so mRNA codon changes from CUG to CUC. Both CUG and CUC code for leucine! This is a silent mutation. There is no amino acid change; the protein primary structure remains unchanged, so haemoglobin function is likely unaffected.
模板链以3’→5’方向阅读，因此mRNA将以5’→3’方向与之互补合成。原始模板：TAC CAC GTG GAC TGA。mRNA密码子：AUG GUG CAC CUG ACU。氨基酸：甲硫氨酸（起始）– 缬氨酸 – 组氨酸 – 亮氨酸 – 苏氨酸（注意TGA对应DNA模板上为TGA，转录后mRNA为ACU，编码苏氨酸，并非终止密码子）。这是一个短肽。第四个密码子就是亮氨酸位点。突变后，模板变为GAG，因此mRNA密码子从CUG变为CUC。CUG和CUC都编码亮氨酸！这是一个沉默突变。氨基酸序列没有改变，蛋白质的一级结构保持不变，因此血红蛋白功能很可能不受影响。
If the mutation had changed the amino acid, we would discuss the nature of the substitution (e.g. polar to nonpolar) and its position. For example, the sickle-cell mutation replaces a hydrophilic glutamate with hydrophobic valine on the surface of β-globin, causing polymerisation under low oxygen. Always relate the genetic change to the phenotype via protein structure.
如果该突变导致了氨基酸的改变，我们就要讨论替换的性质（例如极性变为非极性）及其位置。例如镰刀型细胞贫血突变，将β-珠蛋白表面的亲水性谷氨酸替换为疏水性缬氨酸，导致低氧条件下聚合。务必通过蛋白质结构将遗传变化与表型联系起来。
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