📚 Year 9 WJEC Biology: Case Study Practice Drill | Year 9 WJEC 生物：案例分析实战演练
In WJEC Year 9 Biology, case study questions test your ability to apply scientific knowledge to real-world scenarios. This article provides a series of practice cases with guided solutions to sharpen your analytical skills.
在WJEC九年级生物学中，案例分析题考查你将科学知识应用于真实场景的能力。本文提供一系列带指导方案的练习题，帮助你提升分析能力。
1. Photosynthesis Rate and Light Intensity | 光合作用速率与光照强度
Case: A student placed a piece of pondweed in a beaker of water with a lamp at distances of 10 cm, 20 cm, 30 cm, and 40 cm. Bubbles produced per minute were counted. Results: 45, 22, 10, 5 bubbles per minute.
案例：一名学生将水草放在装有水的烧杯中，光源分别置于10厘米、20厘米、30厘米和40厘米处，记录每分钟产生的气泡数。结果：每分钟45、22、10、5个气泡。
Explain the trend shown by the data. Why does bubble count decrease as the lamp moves further away? What controlled variables are important here?
解释数据所显示的趋势。为什么气泡数随着灯距离增加而减少？这里哪些控制变量很重要？
Analysis: As distance increases, light intensity decreases. Photosynthesis requires light energy, so the rate falls. The inverse square law explains this rapid drop. Controlled variables include temperature, CO₂ concentration, and plant health.
分析：随着距离增加，光照强度减弱。光合作用需要光能，因此速率下降。平方反比定律解释了这种快速减少。控制变量包括温度、二氧化碳浓度和水草健康状况。
2. Enzyme Activity and Temperature | 酶活性与温度
Case: Amylase enzyme was added to starch solutions at 0°C, 20°C, 40°C, 60°C, and 80°C. Iodine tests were done every minute. Time for starch to completely disappear: no change at 0°C, 8 min at 20°C, 3 min at 40°C, 12 min at 60°C, no change at 80°C.
案例：将淀粉酶分别加入到0°C、20°C、40°C、60°C和80°C的淀粉溶液中，每隔一分钟用碘液测试。淀粉完全消失所需时间：0°C无变化，20°C需8分钟，40°C需3分钟，60°C需12分钟，80°C无变化。
Identify the optimum temperature. Explain why the reaction is slow at low temperatures and why it stops at high temperatures.
确定最适温度。解释为什么低温下反应慢，以及为什么高温下反应停止。
Analysis: Optimum is 40°C. At low temperatures, molecules move slowly, reducing collisions between enzyme and substrate. At 80°C, the enzyme denatures – its active site changes shape permanently, so the substrate no longer fits.
分析：最适温度为40°C。低温时分子运动缓慢，酶与底物的碰撞减少。80°C时酶变性——其活性位点永久改变形状，底物不再匹配。
3. Osmosis in Potato Strips | 马铃薯条的渗透作用
Case: Potato strips were placed in sucrose solutions of 0%, 10%, 20%, and 30%. After 30 minutes, percentage change in mass: +15% (0%), +2% (10%), -8% (20%), -18% (30%).
案例：将马铃薯条放入0%、10%、20%、30%的蔗糖溶液中。30分钟后，质量变化百分比为：+15%（0%）、+2%（10%）、-8%（20%）、-18%（30%）。
Which solution is closest to the water potential inside the potato cells? Explain the mass changes.
哪种溶液最接近马铃薯细胞内的水势？解释质量变化。
Analysis: The 10% solution shows very little change, indicating it is near isotonic. In 0% (hypotonic), water enters cells by osmosis, making them turgid and increasing mass. In 20% and 30% (hypertonic), water leaves cells, causing plasmolysis and mass loss.
分析：10%溶液变化很小，表明接近等渗。在0%溶液中，水通过渗透进入细胞，使之变得硬挺，质量增加。在20%和30%溶液中，水从细胞流出，导致质壁分离，质量下降。
4. Food Chains and Bioaccumulation | 食物链与生物富集
Case: In a marine ecosystem, DDT concentration was measured: phytoplankton 0.04 ppm, zooplankton 0.23 ppm, small fish 2.07 ppm, large fish 19.2 ppm, seabirds 124 ppm.
案例：在一个海洋生态系统中，测量DDT浓度：浮游植物0.04 ppm，浮游动物0.23 ppm，小鱼2.07 ppm，大鱼19.2 ppm，海鸟124 ppm。
Describe the pattern and explain why the concentration increases through the food chain.
描述规律，并解释为什么浓度沿食物链递增。
Analysis: The concentration increases at each trophic level. This is bioaccumulation. DDT is fat-soluble and not easily broken down, so it accumulates
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