📚 Year 11 WJEC Biology: Practical Case Study Drills | Year 11 WJEC 生物：案例分析实战演练
WJEC Year 11 Biology challenges you to apply knowledge to unfamiliar scenarios through case study questions. These require careful data interpretation, experimental analysis and logical deduction. This article provides a series of practical drill exercises that mirror the style of real exam questions, building your confidence and refining the skills you need to succeed.
WJEC Year 11 生物考试通过案例分析题考查你在陌生情境中应用知识的能力。这类题目要求仔细解读数据、分析实验并进行逻辑推理。本文提供一系列贴近真题风格的实战演练，帮助你建立信心、打磨应考技巧。
1. Understanding Case Study Questions in WJEC Biology | 理解 WJEC 生物中的案例分析题
A case study question presents a real-world biological scenario, often accompanied by a table of results, a graph or a description of an experiment. Your job is to identify the relevant biological concept, describe patterns in the data, and use scientific reasoning to explain what is happening.
案例分析题呈现一个真实的生物学情境，通常附带一张结果表、一幅图或一段实验描述。你要做的是识别相关的生物学概念，描述数据中的规律，并用科学原理解释观察到的现象。
Examiners routinely test how well you can calculate rates, evaluate experimental design, suggest improvements and apply knowledge from one context to another. Mark schemes reward precise use of data and clear linkage between evidence and conclusion.
考官经常考查你计算速率、评价实验设计、提出改进方法以及在不同情境间迁移知识的能力。评分标准青睐精准引用数据、并将证据与结论清晰联系起来的作答。
Top-performing students always frame their answers around the 'describe – explain – evaluate' structure: state the trend, provide the biological reason, and then comment on reliability or limitations.
高分考生的答题结构总是围绕“描述—解释—评价”：先陈述趋势，再给出生物学原因，最后评价数据可靠性或实验局限性。
2. Interpreting Graphs and Tables with Confidence | 自信解读图表与数据表
Begin every graph question by noting the independent variable (x-axis) and the dependent variable (y-axis), including units. For example, a graph titled 'Effect of temperature on enzyme activity' will have temperature (°C) on the x-axis and rate of reaction (e.g. cm³ O₂/min) on the y-axis.
解答图表题时，先注意自变量（x轴）和因变量（y轴）及其单位。例如，标题为“温度对酶活性的影响”的图，x轴为温度（°C），y轴为反应速率（如 cm³ O₂/min）。
Describe the overall shape: does the line rise, fall, peak at an optimum, or plateau? Then back up your description with plotted numbers, such as 'the rate increased from 2.1 cm³/min at 10 °C to 5.8 cm³/min at 40 °C before dropping sharply at 60 °C'.
描述曲线总体形状：是上升、下降、出现最适峰值，还是趋于平台？然后用图上具体数据支持你的描述，如“速率从 10 °C 时的 2.1 cm³/min 升至 40 °C 的 5.8 cm³/min，随后在 60 °C 急剧下降”。
If you are asked to calculate a rate, use the formula:
rate = change in y ÷ change in x
如需计算速率，使用公式：速率 = y轴变化量 ÷ x轴变化量。
3. Case Study 1 – The Effect of pH on Enzyme Action | 案例1：pH 对酶活性的影响
A student investigated amylase activity by mixing the enzyme with starch solution at pH values from 3 to 11, recording the time taken for starch to be completely digested. The shortest time was at pH 7 (22 seconds), while at pH 4 it was 78 seconds and at pH 10 it was 69 seconds.
某学生通过将淀粉酶与淀粉溶液在不同 pH（3–11）混合，记录淀粉被完全分解所需的时间。最短时间出现在 pH 7（22秒），而 pH 4 时需78秒，pH 10 时需69秒。
Because reaction time is an inverse measure of activity, the fastest rate occurs at the optimum pH – here approximately pH 7. The shape of the active site is maintained by hydrogen and ionic bonds, which are disrupted by excess H⁺ or OH⁻ ions in extreme pH, leading to denaturation.
由于反应时间是酶活性的反向指标，速率最快时即为最适 pH——此处约为 pH 7。活性位点的形状由氢键和离子键维持，极端 pH 下过量的 H⁺ 或 OH⁻ 会破坏这些键，导致酶变性。
When answering, always state that beyond the optimum the enzyme's active site changes shape so the substrate no longer fits, decreasing the rate. Quote the data to illustrate: at pH 7 the time was 22 s, compared with 85 s at pH 11.
作答时务必说明：超过最适 pH 后，酶活性位点形状改变，底物不再契合，因此速率下降。引用数据加以说明：pH 7 时仅需 22 秒，而 pH 11 时需 85 秒。
4. Case Study 2 – Limiting Factors of Photosynthesis | 案例2：光合作用的限制因素
Pondweed was placed in water containing hydrogencarbonate indicator or connected to a gas syringe to measure oxygen production. At 0.04 % CO₂, the oxygen output rose with light intensity but levelled off at 15 cm³/min. When CO₂ was raised to 0.12 %, the rate continued to climb to 28 cm³/min before plateauing.
将水草置于含碳酸氢盐指示剂的水中或连接气体注射器，以测量氧气产量。在 0.04 % CO₂ 下，氧气产量随光强增加，但在 15 cm³/min 时不再上升。将 CO₂ 提升至 0.12 % 后，速率继续攀升至 28 cm³/min 后才趋于平缓。
The first plateau indicates that CO₂ was the limiting factor; once it was raised, the rate increased because the enzyme RuBisCO had more substrate to fix. At the second plateau, either light intensity or temperature must have become limiting.
第一个平台表明 CO₂ 是限制因素；提高 CO₂ 后速率提升，因为 RuBisCO 酶获得了更多可固定的底物。在第二个平台，光强或温度必然成为了新的限制因素。
Always identify which factor is limiting at a particular point on the graph and explain why the curve changes slope. The key equation to recall is:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
要始终指出曲线上某一点哪个因素在起限制作用，并解释曲线斜率为何改变。需要回顾的关键方程式为：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
5. Case Study 3 – Heart Rate Response to Exercise | 案例3：运动对心率的影响
A student measured their pulse at rest (70 bpm), immediately after jogging for 5 minutes (122 bpm), and at one-minute intervals during recovery. The recovery curve showed a steep drop in the first two minutes followed by a gradual return to resting level after eight minutes.
某学生测量了自己静息时的心率（70 bpm）、慢跑5分钟后即刻的心率（122 bpm）以及恢复期每分钟的心率。恢复曲线显示前两分钟心率急剧下降，随后在八分钟时逐渐回到静息水平。
During exercise, skeletal muscles respire aerobically (and some anaerobically), generating CO₂ and lactic acid. To supply more O₂ and remove waste, the heart beats faster. The steepness of recovery indicates cardiovascular fitness; a trained athlete's heart rate returns to resting much sooner.
运动时，骨骼肌进行有氧呼吸（以及部分无氧呼吸），产生 CO₂ 和乳酸。为提供更多 O₂ 并清除废物，心跳加快。恢复曲线的陡峭程度反映心血管健康水平；训练有素的运动员心率恢复更快。
Percentage increase can be calculated as (122 - 70) / 70 × 100 = 74.3 % nking them to physiological principles is often a key requirement of case study questions.
可计算心率增加百分比：(122 – 70) / 70 × 100 = 74.3%。掌握这类计算并将其与生理学原理联系起来，往往是案例分析题的关键要求。
6. Case Study 4 – Pedigree Analysis for an Inherited Disorder | 案例4：遗传病的系谱分析
A pedigree chart shows two unaffected parents who have three children: two unaffected and one affected by cystic fibrosis. Cystic fibrosis is caused by a recessive allele (f), while the normal allele is dominant (F). The affected child must have the genotype ff.
一系谱图显示两个未患病、但育有三名子女的父母：两名未患病，一名患有囊性纤维化。囊性纤维化由隐性等位基因 (f) 引起，正常等位基因 (F) 为显性。患病孩子的基因型必为 ff。
Since each parent contributed one allele, both parents must be heterozygous (Ff). Using a Punnett square, the expected offspring genotypes are 25 % FF, 50 % Ff and 25 % ff. If the unaffected children are examined, the probability that one is a carrier (Ff) is 2/3, because the ff genotype is excluded.
由于每个亲本提供一个等位基因，父母两人必然都是杂合子 (Ff)。通过旁氏表可推出后代基因型比例为：25 % FF，50 % Ff，25 % ff。若考虑未患病的子女，其为携带者 (Ff) 的概率为 2/3，因为 ff 基因型已被排除。
Case study questions may extend this to sex-linked traits, asking you to explain why males are more frequently affected by disorders such as haemophilia.
案例分析题可能延伸到伴性遗传，要求你解释为何男性更易患血友病这类疾病。
7. Case Study 5 – Sampling Populations in an Ecosystem | 案例5：生态系统中的种群取样
er. They calculated a mean of 8 limpets per quadrat. The survey area measured 20 m × 30 m.
在岩石海岸，学生沿一条样线随机放置 0.5 m × 0.5 m 样方于 10 个坐标点，统计帽贝数量并评估海藻覆盖度。每个样方平均有 8 只帽贝，调查区域面积为 20 m × 30 m。
0 = 19,200 limpets. Random sampling avoids bias, while a transect reveals patterns of zonation related to tidal exposure.
估算总数：区域可容纳样方数 = (20×30) / (0.5×0.5) = 2400。估计总种群数量 = 8 × 2400 = 19200 只帽贝。随机取样避免偏差，而样线则能揭示与潮汐暴露相关的分层分布模式。
Abiotic factors – such as wave action, salinity and light – explain why barnacles might dominate the upper shore while seaweeds thrive lower down. Use the sampling data to support such ecological conclusions.
非生物因素——如波浪作用、盐度和光照——解释了为何藤壶在潮间带上部占优势而海藻在较低处繁茂。使用取样数据支持这类生态学结论。
8. Case Study 6 – Osmosis and Water Potential in Potato Cylinders | 案例6：土豆圆柱体中的渗透作用与水势
Equal-sized potato cylinders were weighed, placed in a range of sucrose solutions (0.0 to 1.0 mol/dm³) and re-weighed after 30 minutes. The percentage change in mass was plotted against concentration.
将等大的土豆圆柱体称重后，放入一系列蔗糖溶液（0.0 至 1.0 mol/dm³），30 分钟后再次称重，绘制质量变化百分比对浓度的曲线。
In pure water (0.0 mol/dm³), the potato gained mass (+12.3 %) because the water potential inside the cells was lower than that of the surrounding water, so water entered by osmosis. At 0.4 mol/dm³ the mass change was approximately 0 %, indicating that the water potential of the potato cells equalled that of the solution – the isotonic point.
在纯水（0.0 mol/dm³）中，土豆质量增加（+12.3%），因为细胞的水势低于周围水，于是水通过渗透进入。在 0.4 mol/dm³ 时质量变化约为 0%，表明土豆细胞的水势与溶液相等——等渗点。
Above 0.4 mol/dm³ the potato lost mass as water moved out of the cells (plasmolysis). Calculate percentage change with
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