📚 Case Study Practice for Pre-U Cambridge Biology | 剑桥Pre-U生物：案例分析实战演练
Case study and data-analysis questions form a substantial part of the Cambridge Pre-U Biology examination, requiring you to integrate factual knowledge, quantitative skills and critical evaluation. This article presents ten worked examples spanning ecology, genetics, physiology, evolution and experimental design. Each scenario is structured to mimic the style of exam questions, with step-by-step reasoning and model answers that highlight the techniques examiners look for.
案例分析与数据解读题在剑桥Pre-U生物考试中占很大比重，需要考生整合事实性知识、定量技能和批判性评价。本文呈现十个涵盖生态学、遗传学、生理学、进化与实验设计的解题范例。每个情景均模拟考题风格，附有逐步推理和参考解答，突显阅卷官看重的答题技巧。
1. Ecological Sampling: Comparing Two Habitats | 生态取样：比较两种生境
A student sampled two grassland areas, A (unmanaged meadow) and B (improved pasture), using ten 1 m² quadrats randomly placed in each. The numbers of individuals of four plant species were recorded. The table shows total counts summed over the ten quadrats.
一名学生用随机放置的十个1 m²样方对两块草地A（未开垦草甸）和B（改良牧场）进行取样，记录四种植物个体的数量。表格为十个样方的总和。
| Species | A count | B count |
|---|---|---|
| Poa pratensis | 320 | 550 |
| Trifolium repens | 180 | 160 |
| Ranunculus acris | 95 | 50 |
| Cirsium arvense | 5 | 40 |
To quantify biodiversity, calculate Simpson’s Diversity Index D = 1 – Σ(n/N)² where n = number of individuals of a species, N = total number of individuals of all species. For habitat A: N = 600. Sum of (n/N)² = (320/600)² + (180/600)² + (95/600)² + (5/600)² ≈ 0.2844 + 0.0900 + 0.0251 + 0.00007 = 0.3996. DA = 1 – 0.3996 = 0.6004. For habitat B: N = 800, sum of (n/N)² ≈ (550/800)² + (160/800)² + (50/800)² + (40/800)² = 0.4727 + 0.0400 + 0.0039 + 0.0025 = 0.5191. DB = 1 – 0.5191 = 0.4809. The unmanaged meadow has a higher diversity index, reflecting greater evenness and fewer dominant species.
为量化生物多样性，计算辛普森多样性指数 D = 1 – Σ(n/N)²，其中 n 为某物种个体数，N 为所有物种个体总数。生境A：N = 600。Σ(n/N)² ≈ 0.3996，DA = 0.6004。生境B：N = 800，Σ(n/N)² ≈ 0.5191，DB = 0.4809。未开垦草甸的多样性指数更高，反映其均匀度更高且优势种较少。
In an exam answer you should comment that differences may arise from grazing pressure, fertiliser input in pasture B promoting competitive grasses, while less disturbed meadow A supports more niches. Always link the index to ecological theory.
答题时你应指出差异可能源于放牧压力、牧场B施肥促进竞争性禾草，而干扰较少的草甸A承载更多生态位。务必将指数与生态学理论关联。
2. Genetic Pedigree: Autosomal Recessive Disorder | 遗传系谱：常染色体隐性遗传病
The pedigree shows a rare condition affecting individuals II-3 and III-2. Unaffected parents I-1 and I-2 have three children; only daughter II-3 is affected. II-3 marries an unaffected unrelated man, producing affected son III-2. Determine the mode of inheritance and the probability that III-1 is a carrier.
系谱图显示一种罕见疾病，患者为II-3和III-2。未患病父母I-1和I-2育有三名子女，仅女儿II-3患病。II-3与一名无亲缘关系的未患病男性婚配，生下患病儿子III-2。判断该遗传模式并计算III-1为携带者的概率。
Clues for autosomal recessive inheritance: affected individuals born to unaffected parents (I-1, I-2) imply heterozygous carriers. Both sexes are affected, and the condition skips generations. Since II-3 is affected (homozygous recessive), her partner must be a carrier for III-2 to be affected. The probability that III-1 (unaffected sibling) is a carrier: parents are affected (aa) and carrier (Aa). Their Punnett square yields 1 Aa : 1 aa; however, III-1 is unaffected, so the aa outcome is excluded. Therefore the conditional probability that III-1 is a carrier equals 100% (Aa out of Aa and impossible AA). More precisely: II-3 genotype aa, partner is Aa (must be carrier). Offspring genotypes: 50% Aa, 50% aa. Knowing III-1 is unaffected removes the aa class, leaving only Aa, so carrier probability = 1 (certain). Note: If the partner’s genotype were unknown, one would need to incorporate population carrier frequency; here the pedigree shows he is obligate carrier because he transmitted the recessive allele.
常染色体隐性遗传线索：患病个体由未患病父母（I-1、I-2）所生，表明双亲为杂合携带者。两性皆可患病，疾病隔代出现。因II-3患病（隐性纯合），其配偶必定是携带者才能使III-2患病。III-1（未患病同胞）为携带者的概率：亲本为患病(aa)与携带者(Aa)。旁氏表产生1 Aa : 1 aa；但III-1未患病，故排除aa类别，条件概率为100%（仅Aa）。注意：仅在配偶基因型已知为携带者时成立；系谱已显示其为必然携带者。
3. Enzyme Kinetics: Determining Vmax and Km | 酶动力学：测定Vₘₐₓ与Kₘ
An experiment measured the initial rate (v) of an enzyme-catalysed reaction at different substrate concentrations [S]. Use the data to estimate Vₘₐₓ and Kₘ.
实验测定了不同底物浓度[S]下的酶促反应初始速率(v)。用数据估算Vₘₐₓ和Kₘ。
| [S] (mmol dm⁻³) | v (µmol min⁻¹) |
|---|---|
| 0.5 | 0.75 |
| 1.0 | 1.20 |
| 2.0 | 1.71 |
| 4.0 | 2.18 |
| 8.0 | 2.53 |
Because the curve is hyperbolic, linear transformations help extract parameters. Using the Hanes-Woolf plot ([S]/v against [S]) provides a quick approach, but for a simple estimate you can solve simultaneous equations. Select two well-spaced points, e.g. [S] = 0.5, v = 0.75 and [S] = 4.0, v = 2.18. According to the Michaelis-Menten equation: v = Vₘₐₓ[S] / (Kₘ + [S]). Rearranging: Vₘₐₓ[S] = v(Kₘ + [S]) → Vₘₐₓ[S] = vKₘ + v[S]. Plug in both data sets:
0.5 Vₘₐₓ = 0.75 Kₘ + 0.375 … (1)
4.0 Vₘₐₓ = 2.18 Kₘ + 8.72 … (2)
Multiply (1) by 8: 4.0 Vₘₐₓ = 6.0 Kₘ + 3.0. Equate with (2): 6.0 Kₘ + 3.0 = 2.18 Kₘ + 8.72 → 3.82 Kₘ = 5.72 → Kₘ ≈ 1.50 mmol dm⁻³. Substitute back: 0.5 Vₘₐₓ = 0.75 × 1.50 + 0.375 = 1.50 → Vₘₐₓ = 3.0 µmol min⁻¹. These estimates agree well with more rigorous regression methods.
由于曲线呈双曲线，可用线性转换提取参数。使用汉斯-伍尔夫图（[S]/v 对 [S]）是一种快速方法，但为简化估算，可选取两个间隔较远的点，例如 [S]=0.5, v=0.75 及 [S]=4.0, v=2.18。代入米氏方程，联立求解得 Kₘ ≈ 1.50 mmol dm⁻³，Vₘₐₓ ≈ 3.0 µmol min⁻¹。该估算与更严谨的回归方法一致。
Interpretation: Kₘ indicates the substrate concentration at half Vₘₐₓ; a low Kₘ implies high affinity. Always discuss the limitations of using only two points and the advantage of linearised plots for accuracy.
解读：Kₘ 代表半最大速率时的底物浓度；低 Kₘ 意味着亲和力高。要讨论仅用两点估算的局限性，以及线性化作图提升准确度的优势。
4. Population Genetics: Hardy-Weinberg Equilibrium Test | 群体遗传学：哈代-温伯格平衡检验
A sample of 500 individuals from a large, randomly mating population was typed for the MN blood group, yielding: MM = 165, MN = 250, NN = 85. Determine whether the population is in Hardy-Weinberg equilibrium using a chi-squared test.
从一个大随机交配种群中抽取500人进行MN血型分型，结果：MM=165，MN=250，NN=85。用卡方检验判断该群体是否处于哈代-温伯格平衡。
First calculate allele frequencies. Total alleles = 1000. Frequency of M = p = (2×165 + 250) / 1000 = 580/1000 = 0.58. Frequency of N = q = 1 – 0.58 = 0.42. Expected genotype frequencies under H-W: MM = p² = 0.3364, expected number = 0.3364×500 = 168.2; MN = 2pq = 2×0.58×0.42 = 0.4872, expected = 243.6; NN = q² = 0.1764, expected = 88.2. Chi-squared: χ² = Σ(O-E)²/E = (165-168.2)²/168.2 + (250-243.6)²/243.6 + (85-88.2)²/88.2 ≈ 0.0609 + 0.1682 + 0.1161 = 0.3452. Degrees of freedom: 3 genotypes – 1 – 1 = 1 (for estimating p from data). Critical value at p=0.05 is 3.84. Since 0.345 < 3.84, do not reject H₀; the population is in H-W equilibrium.
先计算等位基因频率：M频率 p=0.58，N频率 q=0.42。哈-温预期基因型数：MM=168.2，MN=243.6，NN=88.2。χ² = Σ(O-E)²/E = 0.3452，自由度1，临界值3.84。因0.345<3.84，不拒绝原假设，群体处于平衡。
In your answer, comment that deviations could indicate natural selection, non-random mating or gene flow. Always state the null hypothesis and mention that the test is valid only if assumptions (large population, random mating, no mutation/migration) are met.
答题时应指出偏离可能暗示自然选择、非随机交配或基因流。务必陈述零假设，并说明只有在满足大种群、随机交配、无突变/迁移等假设时检验才有效。
5. Osmosis and Water Potential in Potato Strips | 渗透与土豆条水势
Strips of potato tissue were immersed in sucrose solutions of varying molarity for 2 hours. The percentage change in mass was recorded. Estimate the water potential of the potato cells assuming the temperature was 25 °C.
土豆条在不同摩尔浓度的蔗糖溶液中浸泡2小时，记录质量变化百分比。假设温度为25°C，估算土豆细胞的水势。
| Sucrose / mol dm⁻³ | % mass change |
|---|---|
| 0.0 | +18.5 |
| 0.2 | +9.3 |
| 0.4 | -1.2 |
| 0.6 | -12.8 |
| 0.8 | -23.1 |
The isotonic point (zero mass change) lies between 0.2 and 0.4 mol dm⁻³. By interpolation, the graph crosses the x‑axis at approximately 0.36 mol dm⁻³. The solute potential Ψs of a solution is given by Ψs = -iCRT, where i = 1 for sucrose (does not dissociate), C = concentration (mol dm⁻³), R = 0.0831 dm³ bar mol⁻¹ K⁻¹, T = 298 K. Thus Ψs = -1 × 0.36 × 0.0831 × 298 ≈ -8.9 bar. At incipient plasmolysis the cell Ψ equals the solution Ψ, so the water potential of potato cells is approximately -8.9 bar. In a Pre‑U answer, mention that the pressure potential is zero at the isotonic point and that the solute potential dominates.
等渗点（质量变化为零）约在0.36 mol dm⁻³。蔗糖溶液溶质势 Ψs = -iCRT = -0.36 × 0.0831 × 298 ≈ -8.9 bar。此时细胞水势等于溶液水势，故土豆细胞水势约为-8.9 bar。答题时须提及在等渗点压力势为零，以溶质势为主。
6. Transpiration Rate Using a Potometer | 蒸腾速率测定：气泡蒸腾计
A bubble potometer was used to measure the rate of transpiration in a leafy shoot under different conditions. The capillary tube had a cross‑sectional area of 0.8 mm². Record the volume of water absorbed in 10 minutes and calculate the rate.
使用气泡蒸腾计测定带叶枝条在不同条件下的蒸腾速率。毛细管横截面积为0.8 mm²。记录10分钟内吸水体积并计算速率。
| Condition | Bubble distance / mm |
|---|---|
| Still air, 20 °C | 12 |
| Fan (wind), 20 °C | 38 |
| Humid chamber, 20 °C | 5 |
Volume moved = distance × cross‑sectional area. For still air: 12 mm × 0.8 mm² = 9.6 mm³ = 0.0096 cm³ in 10 min. Rate = 0.00096 cm³ min⁻¹. Wind increases the rate because it removes the boundary layer of water vapour, steepening the concentration gradient. Humidity reduces the gradient, lowering transpiration. Always convert units consistently. Discuss the assumptions: water uptake approximates transpiration, but a tiny fraction is used in metabolism.
吸水体积 = 距离 × 横截面积。静风条件：12 mm × 0.8 mm² = 9.6 mm³ = 0.0096 cm³，10 min内。速率 = 0.00096 cm³ min⁻¹。风加速蒸腾因吹散水蒸气边界层，增大浓度梯度；高湿度减小梯度，降低蒸腾。答题时讨论假设：吸水量约等于蒸腾量，但少量用于代谢。
7. Interpreting Clinical Trial Data: Risk Ratio | 临床试验数据解读：风险比
A double-blind trial tested a new vaccine against a viral disease. 2000 volunteers received the vaccine and 2000 received a placebo. After one season, 15 in the vaccine group and 85 in the placebo group developed the disease. Calculate the relative risk (RR) and interpret.
一项双盲试验测试某新型病毒疫苗。2000名志愿者接种疫苗，2000名接受安慰剂。一个流行季后，疫苗组15人患病，安慰剂组85人患病。计算相对风险(RR)并解读。
Set up a 2×2 table. Risk in vaccinated = 15/2000 = 0.0075; risk in placebo = 85/2000 = 0.0425. Relative risk = 0.0075/0.0425 = 0.176. Vaccine efficacy = (1 – RR)×100% = 82.4%. This means the vaccine reduces the risk of disease by about 82%. A confidence interval should accompany the estimate; if it does not cross 1, the result is statistically significant. Discuss that RR < 1 indicates a protective effect. Also mention absolute risk reduction (0.0425 - 0.0075 = 0.035) to convey the clinical impact.
疫苗组风险=0.0075，安慰剂组=0.0425，RR=0.176，疫苗效力=82.4%。疫苗将患病风险降低约82%。若置信区间不跨越1，则结果具统计学显著性。RR<1表示保护作用。同时提及绝对风险降低(0.035)，以传达临床效果。
8. Phylogenetic Tree Construction from DNA Sequences | 从DNA序列构建系统发育树
Four species (P, Q, R, S) were sequenced for a homologous region of mitochondrial DNA. The number of nucleotide differences is shown. Construct a simple phylogenetic tree using the unweighted pair-group method (UPGMA).
对四个物种(P, Q, R, S)的线粒体DNA同源区域测序，碱基差异数如下。用不加权分组平均法(UPGMA)构建简易系统发育树。
| P | Q | R | S | |
|---|---|---|---|---|
| P | 0 | 4 | 16 | 17 |
| Q | 0 | 15 | 16 | |
| R | 0 | 6 | ||
| S | Published by TutorHao | Pre-U Biology Revision Series | aleveler.com |
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