Case Study Practice for Year 11 CCEA Biology | Year 11 CCEA 生物：案例分析实战演练

📚 Case Study Practice for Year 11 CCEA Biology | Year 11 CCEA 生物：案例分析实战演练

In CCEA GCSE Biology, case study questions require you to interpret data, analyse experimental results and apply your knowledge to new contexts. This article provides ten carefully structured case studies drawn from key Year 11 topics. Each case is broken down into background information, experimental data and guided analysis to help you build confidence for exam-style questions.

在 CCEA GCSE 生物考试中，案例分析题要求你解读数据、分析实验结果并将知识应用于新情境。本文精选十个紧扣 Year 11 核心主题的案例，每个案例都拆解为背景信息、实验数据和引导式分析，帮助你从容应对考试风格的题目。

1. Osmosis in Potato Cylinders | 土豆条的渗透作用

A student investigated osmosis by placing potato cylinders of equal mass into different sucrose solutions. The initial mass of each cylinder was 5.0 g. After 30 minutes, the cylinders were blotted dry and reweighed. The percentage change in mass was calculated.

一名学生将质量相等的土豆条放入不同浓度的蔗糖溶液中，探究渗透作用。每个土豆条初始质量为 5.0 g，浸泡 30 分钟后吸干表面水分再次称重，并计算质量变化百分比。

Sucrose concentration / mol dm⁻³	Final mass / g	Change in mass / %
0.0	5.4	+8.0
0.2	5.2	+4.0
0.4	5.0	0.0
0.6	4.7	−6.0
0.8	4.4	−12.0

The data show a trend: as sucrose concentration rose, the percentage mass change became negative, indicating a net loss of water. At 0.4 mol dm⁻³, the mass remained unchanged, suggesting that the solution was isotonic to the potato cells.

数据显示：随着蔗糖浓度升高，质量变化百分比变为负值，表明水分子净流失。在 0.4 mol dm⁻³ 时质量不变，说明该溶液与土豆细胞等渗。

Question: Explain why the cylinders in pure water gained mass. Answer: Pure water has a higher water potential than the cell contents. Water moved into the cells by osmosis, down the water potential gradient, causing the mass to increase.

问题：解释为什么纯水中的土豆条质量增加。答案：纯水的水势高于细胞内容物，水分子沿水势梯度经渗透作用进入细胞，导致质量增加。

Question: Why is there no net change at 0.4 mol dm⁻³? Answer: At this concentration, the water potential inside and outside the cells is equal. There is no net movement of water, so the mass stays the same.

问题：为什么在 0.4 mol dm⁻³ 时没有净变化？答案：在此浓度下，细胞内外的水势相等，水分子没有净移动，因此质量不变。

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2. Enzyme Activity: Amylase Digests Starch | 酶活性：淀粉酶消化淀粉

A class investigated the effect of temperature on amylase action. Starch suspension was mixed with amylase at different temperatures, and a drop of the mixture was tested with iodine solution every 30 seconds. The time taken for the iodine to remain yellow-brown (showing no starch) was recorded.

一个班级探究了温度对淀粉酶活性的影响。在不同温度下将淀粉悬浊液与淀粉酶混合，每隔 30 秒取一滴混合液用碘液测试，记录碘液不再变蓝黑（表明淀粉消失）所需的时间。

Temperature / °C	Time for starch to disappear / seconds
0	240
20	120
35	45
45	200
60	No change

The fastest digestion occurred at 35°C. Below this, enzyme activity was slower because molecules had less kinetic energy. Above 45°C, the enzyme began to denature; the active site changed shape and could no longer bind starch effectively. At 60°C, the enzyme was completely denatured.

最快的消化速率出现在 35°C。低于此温度时，酶活性因分子动能较低而减慢。高于 45°C 时，酶开始变性，活性位点形状改变，无法有效结合淀粉。在 60°C 时酶完全变性。

Question: Why is 35°C described as the optimum temperature? Answer: It is the temperature at which the enzyme works fastest, producing the shortest digestion time without denaturation.

问题：为什么 35°C 被称为最适温度？答案：这是酶作用最快的温度，在不变性的前提下实现了最短的消化时间。

Question: Explain why no digestion occurred at 60°C. Answer: The high temperature broke the bonds maintaining the enzyme's tertiary structure, permanently altering the active site so that the substrate could not fit.

问题：解释 60°C 时为何未发生消化。答案：高温破坏了维持酶三级结构的键，永久性地改变了活性位点，底物无法匹配。

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3. Photosynthesis and Light Intensity | 光合作用与光照强度

A student used an aquatic plant (Elodea) to measure the rate of photosynthesis. The plant was placed in a beaker of water with sodium hydrogencarbonate as a CO₂ source. A lamp was moved to different distances to vary light intensity, and the number of oxygen bubbles produced per minute was counted.

一名学生用水生植物（伊乐藻）测量光合作用速率。将植物放入盛有水且添加碳酸氢钠（CO₂ 来源）的烧杯中，移动台灯改变光照强度，记录每分钟产生的氧气气泡数。

Distance of lamp / cm	Bubbles per minute
5	40
10	35
20	20
30	10
40	5

As the lamp moved further away, light intensity decreased, and the rate of photosynthesis dropped. This is because light provides the energy for the light-dependent reactions. When light is limiting, less ATP and reduced NADP are produced, so the Calvin cycle slows down.

随着台灯距离增加，光照强度降低，光合作用速率下降。这是因为光为光反应提供能量；当光照成为限制因素时，产生的 ATP 和还原型 NADP 减少，导致卡尔文循环减慢。

Question: Why was sodium hydrogencarbonate added? Answer: It provided a constant supply of carbon dioxide, ensuring that CO₂ concentration did not become a limiting factor.

问题：为何加入碳酸氢钠？答案：它提供了稳定的二氧化碳来源，确保 CO₂ 浓度不成为限制因素。

Question: Predict the result if the temperature were increased to 35°C while the lamp remained at 40 cm. Answer: The bubble count might increase slightly because a higher temperature would speed up enzyme reactions in the Calvin cycle, provided other factors are not limiting.

问题：若将温度升至 35°C 而台灯仍保持在 40 cm 处，预测结果。答案：气泡数可能略有增加，因为较高的温度会加快卡尔文循环中的酶促反应，前提是其他因素不成为限制。

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4. The Effect of Exercise on Heart Rate | 运动对心率的影响

A subject's resting heart rate was measured at 70 beats per minute (bpm). They then performed step-ups on a bench for 2 minutes. Heart rate was recorded every minute during a 10-minute recovery period.

受试者静息心率为每分钟 70 次（bpm），随后在台阶上进行 2 分钟踏阶运动。在 10 分钟的恢复期内，每分钟记录一次心率。

Time after exercise / min	Heart rate / bpm
0 (immediately)	135
1	120
3	95
5	80
10	70

Heart rate spiked immediately after exercise to deliver more oxygen and glucose to the muscles and to remove carbon dioxide more rapidly. As the oxygen debt was repaid, heart rate gradually returned to normal. The recovery time indicates the level of cardiovascular fitness.

运动后心率立即飙升，以便向肌肉输送更多氧气和葡萄糖，并更快地清除二氧化碳。随着氧债得到偿还，心率逐渐恢复正常。恢复时间反映了心血管健康水平。

Question: Explain why heart rate remained elevated for several minutes after exercise. Answer: The muscles had built up an oxygen debt. Extra oxygen was needed to oxidise lactic acid, so heart rate stayed high to maintain increased blood flow.

问题：解释运动后心率为何在几分钟内仍保持升高。答案：肌肉积累了氧债，需要额外氧气来氧化乳酸，因此心率保持较高以维持增大的血流量。

Question: How could this experiment be modified to test the effect of exercise duration? Answer: Repeat the procedure with different exercise durations (e.g. 1, 2, 5 minutes) and compare recovery heart rates, keeping all other variables constant.

问题：如何修改本实验以测试运动时长的影响？答案：用不同的运动时长（如 1 分钟、2 分钟、5 分钟）重复实验，并比较恢复期心率，同时保持其他变量不变。

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5. Anaerobic Respiration in Yeast | 酵母的无氧呼吸

A student set up a fermentation tube containing yeast suspended in a glucose solution. The tube was placed in a water bath at 30°C, and the volume of carbon dioxide produced was measured every 5 minutes using a gas syringe.

一名学生搭建了发酵装置，试管内装有悬浮在葡萄糖溶液中的酵母。将试管置于 30°C 水浴中，用气体注射器每隔 5 分钟测量产生的二氧化碳体积。

Time / min	Volume of CO₂ / cm³
0	0
5	4
10	9
15	14
20	18

action can be summarised as: glucose → ethanol + CO₂. This process releases much less energy than aerobic respiration.

二氧化碳体积稳步增加，因为酵母细胞进行无氧呼吸，将葡萄糖转化为乙醇和二氧化碳。反应可概括为：葡萄糖 → 乙醇 + 二氧化碳。该过程释放的能量远少于有氧呼吸。

Question: Why was the experiment conducted at 30°C? Answer: 30°C is close to the optimum temperature for yeast enzymes, ensuring a measurable rate of fermentation without denaturation.

问题：为何实验在 30°C 下进行？答案：30°C 接近酵母酶的最适温度，可在不变性的情况下获得可测量的发酵速率。

Question: Suggest why the rate of CO₂ production slowed towards 20 minutes. Answer: The glucose concentration may have decreased, or ethanol accumulation may have inhibited the yeast enzymes, reducing the reaction rate.

问题：推测为何 20 分钟左右 CO₂ 产生速率减慢。答案：葡萄糖浓度可能降低，或乙醇积累抑制了酵母酶，降低了反应速率。

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6. Antibiotic Effectiveness: Zone of Inhibition | 抗生素效果：抑菌圈

Agar plates were spread with a lawn of bacteria. Paper discs soaked in different antibiotics (A, B, C and a control with sterile water) were placed on the surface. After 24 hours of incubation at 25°C, the diameter of the clear zone around each disc was measured.

琼脂平板上均匀涂布一层细菌，再将浸泡过不同抗生素（A、B、C）及无菌水对照的纸片置于表面。在 25°C 培养 24 小时后，测量各纸片周围清晰区域（抑菌圈）的直径。

Antibiotic disc	Diameter of inhibition zone / mm
A	18
B	0
C	10
Sterile water	0

Antibiotic A produced the largest inhibition zone, indicating the greatest effectiveness against this bacterial strain. Antibiotic B showed no zone, so the bacteria were resistant. The control confirmed that any inhibition was due to the antibiotic, not the disc or water.

抗生素 A 产生了最大的抑菌圈，说明对该菌株最有效。抗生素 B 无抑菌圈，表明细菌对此抗药。对照证实抑菌效果来自抗生素，而非纸片或水。

Question: Why is it important to use a sterile water disc? Answer: It acts as a control, showing that the solvent or disc material alone does not inhibit bacterial growth.

问题：为何要使用无菌水纸片？答案：作为对照，表明溶剂或纸片材料本身不会抑制细菌生长。

Question: Explain one way the method could be improved to make the data more reliable. Answer: Replicate each antibiotic on multiple plates and calculate a mean inhibition zone, or use a standardised inoculum to ensure uniform bacterial density.

问题：提出一种改进方法以提高数据可靠性。答案：在多个平板上重复每种抗生素并计算平均抑菌圈，或使用标准化接种物以确保细菌密度均匀。

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7. Genetics: Blood Type Inheritance | 遗传学：血型遗传

Blood type in humans is controlled by a single gene with three alleles: Iᴬ, Iᴮ and Iᴼ. Iᴬ and Iᴮ are codominant, and both are dominant over Iᴼ. A mother with blood type A (genotype IᴬIᴼ) and a father with blood type B (genotype IᴮIᴼ) plan to have children.

人类的血型由单基因控制，该基因有三个等位基因：Iᴬ、Iᴮ 和 Iᴼ。Iᴬ 和 Iᴮ 为共显性，两者均对 Iᴼ 显性。一位 A 型血母亲（基因型 IᴬIᴼ）和一位 B 型血父亲（基因型 IᴮIᴼ）计划生育后代。

A Punnett square can be drawn to predict the probabilities:

可绘制庞尼特方格预测概率：

	Iᴮ	Iᴼ
Iᴬ	IᴬIᴮ (AB)	IᴬIᴼ (A)
Iᴼ	IᴮIᴼ (B)	IᴼIᴼ (O)

The possible genotypes for the children are IᴬIᴮ (type AB), IᴬIᴼ (type A), IᴮIᴼ (type B) and IᴼIᴼ (type O), each with a 25% chance.

后代可能的基因型为 IᴬIᴮ（AB 型）、IᴬIᴼ（A 型）、IᴮIᴼ（B 型）和 IᴼIᴼ（O 型），每种概率为 25%。

Question: One child has blood type O. Explain how this genotype could be expressed even though neither parent is type O. Answer: Both parents are heterozygous, carrying one recessive Iᴼ allele. The child inherited the Iᴼ allele from each parent, resulting in the homozygous recessive genotype IᴼIᴼ.

问题：一个孩子血型为 O 型，而双亲均非 O 型，请解释该基因型如何出现。答案：父母均为杂合子，各携带一个隐性 Iᴼ 等位基因。孩子从双方各继承一个 Iᴼ，形成隐性纯合基因型 IᴼIᴼ。

Question: What is the probability that the next child will be type AB? Answer: 0.25 or 1 in 4 chance, because only the IᴬIᴮ combination results in AB blood type.

问题：下一个孩子为 AB 血型的概率是多少？答案：0.25 或 1/4 的概率，因为只有 IᴬIᴮ 组合才会产生 AB 血型。

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8. Sampling Populations Using Quadrats | 样方调查法估算种群

An ecologist aimed to estimate the abundance of daisies in a 500 m² school field. Ten 0.25 m² quadrats were placed randomly using a random number generator. The number of daisies in each quadrat was recorded.

Published by TutorHao | Year 11 Biology Revision Series | aleveler.com

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