📚 Year 10 WJEC Biology: Case Study Practical Drills | Year 10 WJEC 生物：案例分析实战演练
Mastering WJEC Biology at Year 10 is not just about memorising facts—it is about applying concepts to unfamiliar scenarios exactly as the exam demands. This article provides a series of practical case studies covering key topics from cells, enzymes and transport to genetics, ecology and experimental skills. Each section models the type of data interpretation, graph analysis and structured reasoning required by the WJEC specification, helping you build confidence and accuracy when tackling exam‑style questions.
掌握 Year 10 WJEC 生物不仅仅是记忆知识点——更重要的是像考试要求的那样，将概念应用到陌生的情境中。本文提供一系列实战案例分析，覆盖细胞、酶和运输、遗传学、生态学以及实验技能等核心主题。每个小节都模拟了 WJEC 大纲所要求的数据解读、图表分析和结构化推理，帮助你建立信心，提高解答考试类题目的准确性。
1. Command Words in WJEC Exams | WJEC 考试中的指令词
WJEC mark schemes rely heavily on specific command words. Understanding what each one expects is the first step to a structured answer. 'State' asks for a brief, factual reply, whereas 'describe' requires a detailed step‑by‑step account of what happens. 'Explain' goes further: you must give reasons or mechanisms, often linking cause and effect. 'Calculate' demands working and correct units. When you see 'suggest', you are expected to propose a plausible biological explanation even if the exact fact has not been taught—use your scientific reasoning. 'Compare' means noting similarities and differences, while 'evaluate' involves weighing up evidence, advantages and disadvantages before reaching a conclusion.
WJEC 评分方案高度依赖特定的指令词。理解每个词的要求是给出条理化答案的第一步。'State' 要求简短、事实性的回答，而 'describe' 需要逐步详细叙述发生的过程。'Explain' 更进一步：你必须给出原因或机制，常常需要把因果联系起来。'Calculate' 要求写出演算过程和正确的单位。当你看到 'suggest' 时，即使具体事实没学过，也应该运用科学推理提出一个合理的生物学解释。'Compare' 意味着指出相似与差异，而 'evaluate' 则涉及权衡证据、优缺点然后得出结论。
In every case study that follows, pay attention to the command word used. Practise re‑wording the question so that your answer directly addresses its demand. For example, a question asking 'Explain why the rate of enzyme activity decreases above 40 °C' must not simply describe the decrease but must state the cause—the active site changes shape, the enzyme denatures, and the substrate can no longer bind.
在下面的每一个案例分析中，请留意所使用的指令词。练习把问题重新措辞，让你的答案直接满足它的要求。例如，如果问题是 '解释为什么酶活性在 40 °C 以上会下降'，你就不应仅仅描述下降的现象，而必须说出原因——活性部位的形状改变，酶变性，底物无法再结合。
2. Enzyme Activity: Temperature and pH | 酶活性：温度和 pH
Case: A student investigated the effect of temperature on catalase extracted from potato. Five tubes containing hydrogen peroxide (H₂O₂) and catalase were kept at 0 °C, 10 °C, 25 °C, 40 °C and 60 °C. The volume of oxygen produced in 60 seconds was recorded. The results table is shown below.
案例：一名学生研究了温度对土豆中提取的过氧化氢酶活性的影响。五个装有双氧水 (H₂O₂) 和过氧化氢酶的试管分别置于 0 °C、10 °C、25 °C、40 °C 和 60 °C。记录 60 秒内产生的氧气体积。结果表格如下。
Question: 'Describe and explain the pattern shown by the data.'
问题：'描述并解释数据所呈现的变化规律。'
Your answer should first describe the trend: oxygen production is low at 0 °C, rises to a maximum at 40 °C, then falls sharply at 60 °C. Then explain using kinetic theory and denaturation. At low temperatures, molecules have less kinetic energy so collisions are less frequent and fewer enzyme‑substrate complexes form. At the optimum temperature (around 40 °C), the frequency of successful collisions is highest. Above 50–60 °C the enzyme's active site loses its specific shape (denatures) permanently, so the substrate cannot fit and the reaction virtually stops.
你的答案应首先描述趋势：产氧量在 0 °C 时很低，在 40 °C 升高到最大值，然后在 60 °C 急剧下降。然后用分子运动理论和变性来解释。低温时分子动能较低，碰撞频率低，形成的酶‑底物复合物较少。在最适温度（约 40 °C）时，有效碰撞频率最高。超过 50–60 °C 后，酶的活性部位永久失去特定形状（变性），底物无法契合，反应几乎停止。
A common WJEC question also tests pH. Remember that extreme pH disrupts the ionic and hydrogen bonds maintaining the tertiary structure of the enzyme, again causing irreversible denaturation. The key phrase is 'change in shape of the active site'.
WJEC 常见的考题也会测试 pH。记住，极端的 pH 会破坏维持酶三级结构的离子键和氢键，同样引起不可逆的变性。关键词是 '活性部位的形状变化'。
3. Osmosis in Plant Tissues | 植物组织中的渗透作用
Case: Potato cylinders of equal length were placed in sucrose solutions of different concentrations (0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm⁻³) for 24 hours. The percentage change in mass was calculated and plotted. The line crosses the x‑axis at 0.45 mol dm⁻³.
案例：将等长的土豆条浸入不同浓度的蔗糖溶液（0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm⁻³）中 24 小时。计算质量变化百分比并绘图。曲线在 0.45 mol dm⁻³ 处穿过 x 轴。
Question: 'Explain why the potato gains mass in 0.0 mol dm⁻³ solution and loses mass in 1.0 mol dm⁻³ solution. What is the significance of the point where the curve crosses the x‑axis?'
问题：'解释为什么土豆在 0.0 mol dm⁻³ 溶液中质量增加，而在 1.0 mol dm⁻³ 溶液中质量减少。曲线与 x 轴交点的意义是什么？'
Answer: In pure water (0.0 mol dm⁻³), the water potential outside the cells is higher (less negative) than inside the cell vacuoles. Water moves by osmosis into the cells, which swell and increase in mass. In 1.0 mol dm⁻³ solution, the external solution has a lower (more negative) water potential; water leaves the cells, causing flaccidity and a loss in mass. The x‑intercept at 0.45 mol dm⁻³ represents the sucrose concentration where the water potential of the solution equals that of the potato cells—there is no net water movement, so mass stays constant. This is an estimate of the solute potential of the cells.
解答：在纯水 (0.0 mol dm⁻³) 中，细胞外的水势高于细胞内液泡的水势（较不负面）。水通过渗透作用进入细胞，细胞膨胀，质量增加。在 1.0 mol dm⁻³ 溶液中，外液水势较低（更负）；水离开细胞，导致质量减少并变得疲软。在 0.45 mol dm⁻³ 处的 x 轴交点代表蔗糖浓度使得溶液水势与土豆细胞水势相等——没有净水移动，因此质量恒定。这是对细胞溶质势的估测。
WJEC markers look for the terms 'water potential', 'net movement' and 'partially permeable membrane'. Always mention that osmosis is a passive process.
WJEC 评分者看重 '水势'、'净移动' 和 '部分通透膜' 这些术语。务必提及渗透是一种被动过程。
4. Limiting Factors of Photosynthesis | 光合作用的限制因素
Case: An experiment measured the rate of oxygen production by pondweed under different light intensities and at two carbon dioxide concentrations (0.04 % and 0.12 %). The graph shows two curves: at low CO₂ the rate plateaus at about 40 % of maximum light intensity; at high CO₂ the rate continues to increase up to full light intensity.
案例：一项实验测量了不同光强度和两种二氧化碳浓度（0.04 % 和 0.12 %）下水草的产氧速率。图表显示两条曲线：在低 CO₂ 条件下，速率在约 40 % 最大光强度时趋于平坦；在高 CO₂ 条件下，速率持续增加直至最大光强度。
Question: 'Identify the limiting factor on the plateau of the low‑CO₂ curve and explain why the high‑CO₂ curve does not plateau.'
问题：'指出低 CO₂ 曲线平台上是什么限制因素，并解释为什么高 CO₂ 曲线没有出现平台。'
On the low‑CO₂ curve, once a certain light intensity is reached, carbon dioxide concentration becomes the limiting factor—no further rise in light intensity can increase the rate because there is insufficient CO₂ for the Calvin cycle. The high‑CO₂ curve does not plateau because light intensity remains the limiting factor across the range tested; enough CO₂ is available, so the rate responds to every increase in light until some other factor, such as temperature, becomes limiting at very high light.
在低 CO₂ 曲线上，一旦达到一定光强度，二氧化碳浓度就成了限制因素——进一步增加光强度无法提高速率，因为卡尔文循环缺乏足够的 CO₂。高 CO₂ 曲线没有出现平台是因为在整个测试范围内光强度始终是限制因素；有充足的 CO₂，所以速率会随着每一次光照强度的增加而提高，直到在极高光强下其他因素（如温度）变成限制因素。
Remember to link the limiting factor to the relevant stage of photosynthesis: light for the light‑dependent reaction, CO₂ for the Calvin cycle, and temperature affecting enzyme activity (rubisco). In the WJEC exam, you may be asked to 'suggest how the rate could be increased further'.
记住要把限制因素与光合作用的相关阶段联系起来：光照对应于光依赖反应，CO₂ 对应于卡尔文循环，温度影响酶活性（rubisco）。在 WJEC 考试中，你可能被要求 '建议如何进一步提高速率'。
5. Aerobic vs Anaerobic Respiration | 有氧呼吸与无氧呼吸
Case: Yeast was added to glucose solution in a closed flask. The carbon dioxide produced was collected over time. In another setup, oxygen was bubbled continuously. The total CO₂ produced after 30 minutes was compared with the theoretical yield from aerobic respiration.
案例：将酵母加入密闭烧瓶的葡萄糖溶液中，收集随时间产生的二氧化碳。在另一套装置中，持续通入氧气。比较 30 分钟后产生的总 CO₂ 量与有氧呼吸的理论产量。
Question: 'Explain why less CO₂ is produced in the absence of oxygen and name the additional product in yeast.'
问题：'解释为什么在无氧条件下产生的 CO₂ 较少，并说出酵母中的另一种产物。'
In the presence of oxygen, yeast respires aerobically, fully oxidising glucose to CO₂ and water, releasing a large amount of energy (up to 38 ATP per glucose). In the absence of oxygen, yeast switches to anaerobic respiration (fermentation), where glucose is only partially broken down to ethanol and CO₂, releasing far less energy. The reduced ATP yield means less glucose is metabolised for the same energy demand, and the CO₂ output per glucose is lower. The additional product is ethanol (C₂H₅OH).
在有氧条件下，酵母进行有氧呼吸，将葡萄糖完全氧化为 CO₂ 和水，释放大量能量（每个葡萄糖最多 38 个 ATP）。在无氧条件下，酵母转为无氧呼吸（发酵），葡萄糖仅被部分分解为乙醇和 CO₂，释放的能量少得多。较低的 ATP 产量意味着满足相同能量需求时代谢的葡萄糖较少，且每个葡萄糖产生的 CO₂ 更少。另一种产物是乙醇 (C₂H₅OH)。
(+ energy). For anaerobic in yeast: glucose → ethanol + carbon dioxide (+ energy). In animals, lactic acid is produced instead of ethanol.
WJEC 往届试卷常要求写出文字方程或简单的配平化学式。有氧呼吸：葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）。酵母无氧呼吸：葡萄糖 → 乙醇 + 二氧化碳（+ 能量）。在动物中，产生乳酸而非乙醇。
6. Sampling Ecosystems with Quadrats | 使用样方采样生态系统
Case: Students compared the abundance of daisies in two fields—one regularly mown, one left to grow wild. They placed ten 0.5 m × 0.5 were 3.2 in the mown field and 12.8 in the wild field. The area of each field was 2000 m².
案例：学生们比较了两个田地里雏菊的丰度——一块定期修剪，一块任其野生。他们在每块田中随机放置十个 0.5 m × 0.5 m 的样方，计算雏菊的莲座数量。修剪田每样方平均数为 3.2，野生田为 12.8。每块田面积均为 2000 m²。
Question: 'Calculate the estimated total daisy population in each field and explain why the method must be random.'
问题：'计算每块田中雏菊的估计总数，并解释为什么方法必须是随机的。'
To estimate the population, first find the mean per m². Quadrat area = 0.5 0.5 = 0.25 m². In the mown field: density = 3.2 ÷ 0.25 = 12.8 plants per m². Total = 12.8 rdinates) to avoid bias and give a representative sample. If quadrats were placed only in areas that looked rich in daisies, the estimate would be unrepresentative.
要估算总数量，先求每平方米的平均数。样方面积 = 0.5 × 0.5 = 0.25 m²。修剪田：密度 = 3.2 ÷ 0.25 = 12.8 株每平方米，总数 = 12.8 × 2000 = 25,600。野生田：密度 = 12.8 ÷ 0.25 = 51.2 每平方米，总数 = 102,400。方法必须使用随机定位（例如随机数坐标）以避免偏差并获得代表性样本。如果样方只放在看起来雏菊丰富的区域，估算将不具代表性。
Other key terms: abundance, distribution, systematic sampling (e.g., along a transect) is used when investigating change across an environmental gradient. WJEC may also ask about using percentage cover for plants like grass where individual counts are difficult.
其他关键术语：丰度、分布，当研究环境梯度变化时使用系统采样（如沿线样条）。WJEC 也可能要求对难以计数个体的植物（如草）使用覆盖度百分比。
7. The Reflex Arc: A Quick Response | 反射弧：快速反应
Case: A person accidentally touches a hot pan and immediately withdraws their hand. WJEC questions often ask you to label a diagram of the reflex arc and explain why the response is so fast.
案例：一个人不小心碰到热锅，立即缩手。WJEC 问题常要求标记反射弧示意图并解释反应为何如此迅速。
Question: 'Describe the pathway of a reflex impulse from the stimulus to the response, naming all three neurones involved.'
问题：'描述从刺激到反应的反射冲动通路，并说出所涉及的三种神经元。'
Stimulus (heat) is detected by receptors in the skin, which generate an impulse. The impulse travels along a sensory neurone to the spinal cord. Inside the spinal cord, the impulse passes across a synapse to a relay neurone (also called an interneuron), and then via another synapse to a motor neurone. The motor neurone carries the impulse to an effector—the biceps muscle—which contracts and pulls the hand away. There is no conscious thought involved because the pathway bypasses the brain; this makes the response extremely rapid, protecting the body from harm.
刺激（热）被皮肤中的感受器侦测到，进而产生神经冲动。冲动沿感觉神经元传到脊髓。在脊髓内部，冲动通过突触传递给中继神经元（也称中间神经元），然后再经另一个突触传给运动神经元。运动神经元将冲动传至效应器——肱二头肌——肌肉收缩将手拉开。整个过程不涉及意识思考，因为通路绕过了大脑；这使反应极快，保护身体免受伤害。
The WJEC specification also emphasises the role of synapses in ensuring unidirectional flow and the importance of the gap (synaptic cleft) where chemical neurotransmitters diffuse.
WJEC 大纲也强调突触在确保单方向传递中的作用，以及神经递质化学物质扩散的间隙（突触间隙）的重要性。
8. Hormonal Control of Blood Glucose | 血糖的激素调控
Case: A healthy individual's blood glucose concentration was measured during an oral glucose tolerance test. After drinking a glucose solution, blood glucose rose from 5.5 mmol dm⁻³ to 9.5 mmol dm⁻³, then returned to normal within two hours.
案例：在一次口服葡萄糖耐量试验中，测量了一个健康个体的血糖浓度。饮用葡萄糖溶液后，血糖从 5.5 mmol dm⁻³ 上升到 9.5 mmol dm⁻³，然后在两小时内恢复正常。
Question: 'Explain how hormones restore blood glucose to its set point after the glucose drink.'
问题：'解释饮用葡萄糖后，激素是如何将血糖恢复至设定值的。'
The rise in blood glucose stimulates the pancreas to release the hormone insulin. Insulin travels in the bloodstream and binds to receptors on liver and muscle cells, causing them to take up more glucose. It also activates enzymes that convert glucose into glycogen for storage (glycogenesis) and increases the rate of respiration. As a result, blood glucose falls back to normal. If blood glucose drops too low, the pancreas secretes glucagon, which promotes the breakdown of glycogen into glucose (glycogenolysis) and its release into the blood. This negative feedback loop maintains homeostasis.
血糖升高刺激胰腺释放激素胰岛素。胰岛素随着血液流动，与肝细胞和肌肉细胞上的受体结合，使它们摄取更多的葡萄糖。它还能激活将葡萄糖转化为糖原储存的酶（糖原生成），并提高呼吸速率。结果血糖下降回正常水平。如果血糖降得太低，胰腺分泌胰高血糖素，促进糖原分解为葡萄糖（糖原分解）并将其释放到血液中。这一负反馈回路维持着稳态。
WJEC questions may ask for target organs—liver and muscles. They rarely require details of second messengers at Year 10, but knowing that insulin is a protein hormone that does not enter cells but triggers a cascade inside is useful for top marks.
WJEC 问题可能会问到靶器官——肝脏和肌肉。Year 10 很少要求第二信使的细节，但了解胰岛素是一种蛋白激素，不进入细胞而是触发细胞内级联反应，对获得高分很有帮助。
9. Pedigree Analysis and Genetic Disorders | 系谱分析与遗传病
Case: The pedigree diagram shows the inheritance of cystic fibrosis (CF) in a family. CF is a recessive disorder caused by a faulty allele (f) of the CFTR gene; the normal allele is F. Two unaffected parents have a child with CF. Use the diagram to deduce the genotypes of the parents and the affected child.
案例：系谱图显示了一个家族中囊性纤维化 (CF) 的遗传情况。CF 是一种隐性遗传病，由 CFTR 基因的一个缺陷等位基因 (f) 引起；正常等位基因为 F。一对未患病的父母生了一个患 CF 的孩子。请利用系谱图推断父母和患病孩子的基因型。
Question: 'Explain why two heterozygous parents can produce a child with CF, and calculate the probability.'
问题：'解释为什么两个杂合父母能生出患有 CF 的孩子，并计算概率。'
heterozygous Ff), they can each pass on the f allele. A Punnett square shows the combination: Ff × Ff produces genotypes FF, Ff, Ff and ff—so there is a 25 % The pedigree showing unaffected parents with an affected child confirms they are carriers.
孩子要表现出隐性遗传病，必须遗传两个隐性等位基因 (ff)。如果父母均未患病但都是携带者（杂合子 Ff），他们每人都有可能传递 f 等位基因。庞纳特方格显示：Ff × Ff 产生基因型 FF、Ff、Ff 和 ff——因此孩子有 25 %（1/4）的概率遗传到 ff 并患 CF。系谱中出现未患病父母生患病孩子，就证实了他们是携带者。
WJEC expects you to use correct symbols: capital for dominant, lower case for recessive, and often to denote sex chromosomes in X‑linked disorders. Be ready to interpret pedigrees and explain the evidence for recessive, dominant or sex‑linked inheritance.
WJEC 期望你使用正确的符号：大写字母表示显性，小写表示隐性，在 X 连锁遗传病中还要常标记性染色体。务必准备好解读系谱，并解释隐性和显性或伴性遗传的证据。
10. Natural Selection: Peppered Moth | 自然选择：桦尺蛾
Case: Before the Industrial Revolution, light‑coloured peppered moths (Biston betularia) were camouflaged on lichen‑covered trees, while dark mutants were easily seen by birds. During the industrial period, soot blackened the trees, and the dark form became more common. In recent years, with cleaner air, the light form is increasing again.
案例：工业革命前，浅色桦
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