Year 9 CIE Biology: Case Study Practice | Year 9 CIE 生物：案例分析实战演练

📚 Year 9 CIE Biology: Case Study Practice | Year 9 CIE 生物：案例分析实战演练

Case studies are a vital part of Year 9 CIE Biology, helping you apply knowledge to real experiments and data. This article presents ten practice cases covering key topics such as enzymes, osmosis, photosynthesis, genetics, and ecology. For each case, we provide data, questions, and model answers to sharpen your scientific skills. Read the English analysis, then check the Chinese version to reinforce understanding.

案例分析是 Year 9 CIE 生物学习的重要环节，它帮助你运用知识解决真实实验和数据问题。本文提供十个实战案例，覆盖酶、渗透、光合作用、遗传和生态等重点主题。每个案例都给出数据、问题和参考答案，助你提升科学技能。先读英文解析，再看中文版以巩固理解。

1. Enzyme Activity and Temperature – Interpreting Data | 酶活性与温度——数据解读

A student investigated the effect of temperature on the activity of catalase from potato tissue. The catalase broke down hydrogen peroxide, and the volume of oxygen produced in 2 minutes was recorded. The results are shown in the table below.

一名学生研究了温度对马铃薯组织中过氧化氢酶活性的影响。该酶分解过氧化氢，记录2分钟内产生的氧气体积。结果如下表所示。

Temperature / °C	0	10	20	30	40	50	60
Oxygen produced / cm³	2	7	18	32	38	15	3

Question: At which temperature is the rate of reaction highest? Suggest why the activity decreases above 40 °C.

问题：反应速率在哪个温度下最高？请解释为何活性在40°C以上下降。

Answer: The highest oxygen yield occurs at 40 °C, indicating the optimum temperature for catalase. Beyond this point, enzyme molecules lose their three-dimensional shape because the heat breaks the bonds maintaining the active site. The enzyme becomes denatured, and the substrate can no longer fit, so activity falls rapidly.

答案：最高产氧量在40°C，表明这是过氧化氢酶的最适温度。超过该温度，酶分子因热破坏了维持活性部位的键而失去立体构型。酶发生变性，底物无法匹配，活性急剧下降。

Key concept: Enzymes have an optimum temperature; high temperatures cause irreversible denaturation. In experiments, control all other variables such as pH and enzyme concentration to obtain valid results.

核心概念：酶有最适温度；高温导致不可逆变性。实验中需控制其他变量（如pH和酶浓度）以获得有效结果。

2. Osmosis in Potato Strips – Experiment Analysis | 马铃薯条渗透实验——分析

Five potato strips of equal initial mass were placed in sucrose solutions of different concentrations for 30 minutes. The percentage change in mass was calculated. The table shows the data.

五个初始质量相等的马铃薯条分别浸入不同浓度蔗糖溶液中30分钟。计算质量变化百分比，数据如下表。

Sucrose concentration / mol dm⁻³	0.0	0.2	0.4	0.6	0.8
Change in mass / %	+12.5	+4.0	-3.5	-10.2	-15.8

Question: Explain why the strip gained mass in 0.0 mol dm⁻³ solution. What can you conclude about the water potential of the potato tissue?

问题：解释为何马铃薯条在0.0 mol dm⁻³溶液中质量增加。你对马铃薯组织的水势有何结论？

Answer: In pure water (0.0 mol dm⁻³), the water potential outside the potato cells is higher than inside. Water moves by osmosis from a region of higher water potential to lower water potential, entering the cells, so the strip gains mass. The point where mass change is zero lies between 0.2 and 0.4 mol dm⁻³; this suggests the water potential of potato cytoplasm is equivalent to approximately 0.3 mol dm⁻³ sucrose solution.

答案：在纯水中，细胞外水势高于细胞内。水通过渗透从高水势区向低水势区移动，进入细胞，使条质量增加。质量零变化的浓度介于0.2至0.4 mol dm⁻³之间，表明马铃薯细胞质的水势约相当于0.3 mol dm⁻³蔗糖溶液。

3. Testing for Biological Molecules – Qualitative Observations | 生物分子测试——定性观察

A mixture of unknown food substances was tested with four reagents: iodine solution, Benedict’s solution (with heating), biuret reagent, and ethanol emulsion. The observations were recorded.

某未知食物混合物用四种试剂测试：碘液、本尼迪克特试剂（加热）、双缩脲试剂和乙醇乳液。记录观察结果。

Iodine: turned blue-black.
Benedict’s: brick-red precipitate after 5 min heating.
Biuret: purple colour appeared.
Ethanol + water: cloudy white emulsion.

中文翻译：碘液——呈蓝黑色；本尼迪克特加热——砖红色沉淀；双缩脲——紫色出现；乙醇+水——乳白色浑浊。

Question: Identify the biological molecules present in the sample. Explain your reasoning.

问题：判断样品中含有哪些生物分子，并说明理由。

Answer: Starch is present because iodine gives a blue-black colour. A reducing sugar such as glucose or maltose is present because Benedict’s solution forms a brick-red precipitate on heating. Protein is present because biuret reagent turns purple. Lipids (fats/oils) are present because the ethanol emulsion test produces a cloudy white suspension. Therefore, the sample contains starch, reducing sugars, protein, and lipid.

答案：淀粉存在，因为碘呈蓝黑色。还原糖（如葡萄糖或麦芽糖）存在，因为本尼迪克特加热生成砖红色沉淀。蛋白质存在，因为双缩脲变紫。脂类存在，因为乙醇乳液试验产生乳白色悬浮。故样品含淀粉、还原糖、蛋白质和脂质。

4. Photosynthesis Rate – Limiting Factors Graph | 光合速率——限制因素图

A pondweed was exposed to different light intensities. The rate of photosynthesis was measured by counting bubbles of oxygen released per minute. Carbon dioxide concentration and temperature were kept constant.

一段水草置于不同光照强度下，通过计数每分钟释放的氧气泡数测量光合速率。二氧化碳浓度和温度保持恒定。

The graph (described here) shows the rate rising linearly with light intensity up to 2000 lux, then levelling off. Question: Identify the limiting factor at low light intensity and explain why the rate plateaus at high light intensity. How would increasing CO₂ concentration affect the plateau?

曲线（此处描述）显示速率随光强在2000 lux内线性上升，之后趋于平稳。问题：指出低光强时的限制因素，解释为何高光强下速率趋于平稳。增加CO₂浓度如何影响曲线平台？

Answer: At low light intensity, light is the limiting factor because there is insufficient energy for the light-dependent reactions. As light intensity increases, the rate rises until another factor becomes limiting. Here, the plateau suggests that CO₂ concentration or temperature has become the limiting factor. If CO₂ concentration is increased, the plateau should rise to a higher level, provided temperature is also adequate.

答案：低光强时光照是限制因素，因为光反应缺乏足够能量。随光强增加速率上升，直至另一因素成为限制。这里平台表明CO₂浓度或温度成为限制因素。若增加CO₂浓度，平台将升高，前提是温度也充足。

5. Diffusion and Surface Area – Agar Cube Investigation | 扩散与表面积——琼脂块探究

Phenolphthalein agar cubes of different sizes (1 cm, 2 cm, 3 cm sides) were placed in hydrochloric acid. The acid diffuses into the cubes, turning them from pink to colourless. The time taken for complete decolorisation was recorded.

将含酚酞的不同大小琼脂块（边长1 cm, 2 cm, 3 cm）放入盐酸中。酸扩散进入块内，使其由粉红变无色。记录完全褪色所需时间。

Cube side / cm	Surface area / cm²	Volume / cm³	SA:V ratio	Time to clear / s
1	6	1	6:1	45
2	24	8	3:1	210
3	54	27	2:1	540

Question: Explain the relationship between surface area to volume ratio (SA:V) and the time for diffusion. Why do large active organisms require transport systems?

问题：解释表面积与体积比（SA:V）与扩散时间的关系。为什么大型活跃生物需要运输系统？

Answer: As the cube size increases, the SA:V ratio decreases. A lower SA:V means less surface area relative to volume for substances to enter, so diffusion to the centre takes much longer. Large organisms cannot rely solely on diffusion to supply oxygen and remove waste; they need a circulatory system and specialized exchange surfaces to overcome the limitation of a small SA:V ratio.

答案：随着琼脂块变大，SA:V比减小。SA:V越小，相对于体积的表面积越少，物质进入中心所需扩散时间更长。大型生物不能仅靠扩散供氧和排废；它们需要循环系统和特化交换表面来克服低SA:V比的限制。

6. Sampling with Quadrats – Estimating Population Size | 样方取样——估算种群大小

Students estimated the population of daisies in a 500 m² field using a 0.25 m² quadrat. They placed the quadrat randomly 10 times and counted the daisies in each quadrat. The counts were: 3, 5, 2, 7, 4, 6, 5, 8, 3, 5.

学生们用0.25 m²样方估算一块500 m²田地中雏菊的种群数量。他们随机放置样方10次，记录每次雏菊数量：3, 5, 2, 7, 4, 6, 5, 8, 3, 5。

Question: Calculate the mean number of daisies per quadrat. Estimate the total population of daisies in the field. Suggest why random sampling is necessary.

问题：计算每个样方雏菊的平均数。估算整块田地雏菊的总数。说明为何必须随机取样。

Answer: Mean = sum of counts ÷ number of quadrats = (3+5+2+7+4+6+5+8+3+5) ÷ 10 = 48 ÷ 10 = 4.8 daisies per quadrat. Each quadrat covers 0.25 m², so total number of quadrats that fit in 500 m² is 500 ÷ 0.25 = 2000. Estimated total population = mean per quadrat × total quadrats = 4.8 × 2000 = 9600 daisies. Random sampling avoids bias and ensures the sample is representative of the whole area.

答案：平均数 = 总数 ÷ 10 = 48 ÷ 10 = 4.8 棵/样方。样方面积0.25 m²，500 m²中样方数 = 500 ÷ 0.25 = 2000。估计总数 = 4.8 × 2000 = 9600 棵雏菊。随机取样避免偏差，确保样本代表整个区域。

7. The Human Circulatory System – Heart Rate and Exercise | 人体循环系统——心率与运动

A student measured her resting heart rate (72 bpm) and then ran for 3 minutes. After exercise, she recorded her heart rate every minute for 10 minutes: 160, 148, 134, 120, 110, 102, 96, 88, 82, 76.

一位学生测量静息心率（72次/分），然后跑3分钟。运动后每分钟记录心率：160, 148, 134, 120, 110, 102, 96, 88, 82, 76。

Question: Explain why the heart rate increases during exercise. How long does it take for her heart rate to return to near resting level? What does recovery time indicate about fitness?

问题：解释运动时心率为何上升。她的心率恢复到接近静息水平需要多长时间？恢复时间如何反映体能？

Answer: During exercise, muscles need more oxygen and glucose for increased respiration; they produce more carbon dioxide. The heart pumps faster to deliver oxygen and remove waste. Her heart rate returns to 76 bpm at minute 10, near resting 72 bpm, so recovery takes about 10 minutes. A shorter recovery time usually indicates better cardiovascular fitness because the heart and lungs work more efficiently.

答案：运动时肌肉需要更多氧气和葡萄糖进行呼吸，产生更多二氧化碳。心脏加速泵血以输氧和排废。心率在第10分钟恢复至76次/分，接近静息72次/分，故恢复约需10分钟。恢复时间较短通常表明心血管更健康，因为心肺工作效率更高。

8. Plant Transport – Transpiration Pull Case | 植物运输——蒸腾拉力案例

A potometer was used to measure the rate of water uptake by a leafy shoot under different conditions: still air, moving air (fan), and high humidity (plastic bag). The air bubble moved 12 mm, 28 mm, and 5 mm respectively in 10 minutes. The capillary tube diameter was 1.0 mm (cross-sectional area = π × (0.5 mm)² = 0.785 mm²).

使用蒸腾计测量不同条件下枝条的吸水速率：静止空气、流动空气（风扇）和高湿度（塑料袋）。10分钟内气泡分别移动12 mm, 28 mm, 5 mm。毛细管内径1.0 mm (截面积 = π × (0.5 mm)² = 0.785 mm²)。

Question: Calculate the water uptake rate in mm³ per minute for moving air. Explain the effect of wind and humidity on transpiration.

问题：计算流动空气条件下每分钟吸水速率（mm³/min）。解释风和湿度对蒸腾作用的影响。

Answer: Volume moved = cross-sectional area × distance = 0.785 mm² × 28 mm = 21.98 mm³ in 10 min. Rate = 21.98 ÷ 10 = 2.198 mm³/min. Wind increases transpiration because it removes the humid air layer around stomata, steepening the water vapour concentration gradient. High humidity reduces transpiration because the gradient is smaller, so diffusion of water vapour out of the leaf slows.

答案：移动体积 = 截面积 × 距离 = 0.785 mm² × 28 mm = 21.98 mm³/10 min。速率 = 2.198 mm³/min。风加快蒸腾，因为它吹走气孔周围湿空气层，增大水汽浓度梯度。高湿度则降低蒸腾，因为梯度变小，水汽扩散减慢。

9. Genetics Problems – Monohybrid Crosses | 遗传学问题——单基因杂交

In pea plants, the allele for tall stem (T) is dominant over the allele for dwarf stem (t). A homozygous tall plant is crossed with a dwarf plant.

豌豆中，高茎等位基因（T）对矮茎等位基因（t）显性。一株纯合高茎与一株矮茎杂交。

Question: Determine the genotype and phenotype ratios of the F1 generation. Then, show the outcome of crossing two F1 plants. Use a Punnett square.

问题：确定F1代的基因型和表现型比。然后展示两株F1杂交的结果，使用庞纳特方格。

Answer: Parental genotypes: TT × tt. F1 genotypes all Tt, phenotype all tall. F1 cross: Tt × Tt. Punnett square gives genotype ratio 1 TT : 2 Tt : 1 tt, and phenotype ratio 3 tall : 1 dwarf. This illustrates Mendel’s law of segregation.

答案：亲本基因型：TT × tt。F1基因型全为Tt，表现型全为高茎。F1杂交：Tt × Tt，庞纳特方格得基因型比1 TT : 2 Tt : 1 tt，表现型比为3高茎 : 1矮茎。这体现了孟德尔分离定律。

10. Microscope Skills – Drawing and Magnification Calculations | 显微镜技能——绘图与放大倍数计算

A student observed a plant cell under a microscope with an eyepiece magnification of ×10 and an objective lens of ×40. She measured the image length of a chloroplast using a ruler and found it to be 52 mm. The actual length is 13 µm.

一学生用目镜10×、物镜40×的显微镜观察植物细胞。她用尺测得叶绿体图像长度为52 mm，实际长度为13 µm。

Question: Calculate the total magnification. Verify the actual length using the magnification formula. (Magnification = Image size ÷ Actual size)

问题：计算总放大倍数。使用放大公式验证实际长度。（放大倍数 = 图像大小 ÷ 实际大小）

Answer: Total magnification = eyepiece × objective = 10 × 40 = 400. Using the formula: 400 = Image size / Actual size. Image size = 52 mm = 52 000 µm. So Actual size = 52 000 ÷ 400 = 130 µm. There is a discrepancy with the given 13 µm, suggesting a unit error in the question; the correct actual size should be 130 µm. Check units carefully: converting all lengths to the same unit is essential.

答案：总放大 = 10 × 40 = 400。运用公式：400 = 图像大小/实际大小。图像大小52 mm = 52 000 µm，实际大小 = 52 000 ÷ 400 = 130 µm。题目给出13 µm有误，应为130 µm。单位换算至关重要，必须保持单位一致。

11. Respiration and Fermentation – Yeast Balloon Experiment | 呼吸与发酵——酵母气球实验

A mixture of yeast, sugar solution, and warm water was placed in a flask with a balloon over the neck. The balloon inflated over 30 minutes. Another flask with boiled yeast and sugar did not inflate.

将酵母、糖液和温水放入烧瓶，瓶口气球套住。30分钟内气球膨胀。另一瓶装煮沸酵母和糖，气球不膨胀。

Question: What gas caused the balloon to inflate? Why did the boiled yeast fail to produce gas? Write a word equation for anaerobic fermentation in yeast.

问题：哪种气体使气球膨胀？为何煮沸酵母不产气？写出酵母无氧发酵的文字方程式。

Answer: Carbon dioxide (CO₂) inflated the balloon. Boiling denatures the enzymes in yeast, so fermentation cannot occur. Word equation: Glucose → ethanol + carbon dioxide (+ a small amount of energy). This shows that yeast can respire anaerobically, producing ethanol and CO₂.

答案：二氧化碳（CO₂）使气球膨胀。煮沸使酵母酶变性，发酵无法进行。文字方程式：葡萄糖 → 乙醇 + 二氧化碳（+少量能量）。这表明酵母可无氧呼吸，生成乙醇和CO₂。

12. Ecology and Food Chains – Energy Transfer Calculation | 生态与食物链——能量传递计算

In a simple food chain, grass (producer) contains 25 000 kJ of energy. Rabbits (primary consumer) gain 2 500 kJ by eating the grass, and foxes (secondary consumer) obtain 250 kJ.

在简单食物链中，草（生产者）含25 000 kJ能量，兔（初级消费者）通过食草获得2 500 kJ，狐狸（次级消费者）获得250 kJ。

Question: Calculate the percentage of energy transferred from grass to rabbits and from rabbits to foxes. Explain why only about 10% of energy passes to the next trophic level.

问题：计算从草到兔以及从兔到狐狸的能量传递百分比。解释为何只有约10%的能量传递到下一营养级。

Answer: Grass to rabbits: (2500 ÷ 25000) × 100 = 10%. Rabbits to foxes: (250 ÷ 2500) × 100 = 10%. Only about 10% is transferred because most energy is lost through respiration, movement, heat, and undigested material (egested as faeces). This limits the length of food chains.

答案：草→兔：(2500 ÷ 25000) × 100 = 10%。兔→狐狸：(250 ÷ 2500) × 100 = 10%。仅约10%传递，因为大部分能量通过呼吸、运动、散热及未消化物质（粪便）损失。这限制了食物链的长度。

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Year 9 CIE Biology: Case Study Practice | Year 9 CIE 生物：案例分析实战演练