📚 Year 9 SQA Biology: Case Study Practical Exercises | Year 9 SQA 生物：案例分析实战演练
Case studies are a powerful way to build your understanding of biology. By working through real-life scenarios, you can practise applying scientific concepts, interpreting data, and refining your investigation skills. This article presents eight practical exercises designed to mirror the types of questions you may encounter in SQA assessments. Each case study breaks down a biological problem and guides you through the thought process required to solve it.
案例分析是加深生物学理解的有效方式。通过研习真实情景，你可以练习运用科学概念、解读数据并提升探究技能。本文提供八个实战演练，模拟你可能在 SQA 测评中遇到的题型。每个案例都会拆解一个生物问题，引导你掌握解题所需的思路。
1. Case Study 1: Investigating Photosynthesis | 案例一：探究光合作用
A Year 9 student wanted to find out how light intensity affects the rate of photosynthesis. She placed some Elodea pondweed in a beaker of water, added sodium hydrogencarbonate as a carbon dioxide source, and positioned a lamp at different distances from the plant. She counted the number of oxygen bubbles released per minute. She recorded her results in a table.
一位九年级学生想知道光照强度如何影响光合作用速率。她把一些伊乐藻放入装有水的烧杯中，加入碳酸氢钠作为二氧化碳来源，并将一盏灯放置在距离植物不同远近的位置。她统计了每分钟释放的氧气气泡数量，并将结果记录在表格中。
The independent variable in this experiment was the distance of the lamp, which changes the light intensity. The dependent variable was the number of bubbles produced per minute, representing the rate of photosynthesis. Controlled variables included water temperature, the amount of sodium hydrogencarbonate, and the length of pondweed.
该实验的自变量是灯的距离，它改变了光照强度。因变量是每分钟产生的气泡数量，代表光合作用速率。控制变量包括水温、碳酸氢钠的用量以及水草的长度。
If the student repeated the experiment with the lamp very close to the plant, she would expect more bubbles. However, if the light intensity becomes too high, the rate might plateau because another factor, such as CO₂ concentration, becomes limiting. This illustrates the principle of limiting factors.
如果学生将灯放在离植物很近的位置重复实验，她会预期有更多气泡。然而，如果光照强度过高，速率可能会趋于平稳，因为另一个因素（如二氧化碳浓度）成为限制因素。这体现了限制因素的原理。
2. Case Study 2: Enzyme Activity and Temperature | 案例二：酶活性与温度
In the lab, a group tested the effect of temperature on the activity of amylase, an enzyme that breaks down starch. They mixed starch solution with amylase at different temperatures (0 °C, 20 °C, 40 °C, 60 °C, and 80 °C). Every 30 seconds, they added a drop of the mixture to iodine solution on a spotting tile. The time taken for the iodine to stop turning blue-black indicated how fast starch was digested.
在实验室中，一组学生测试了温度对淀粉酶活性的影响。淀粉酶是一种分解淀粉的酶。他们将淀粉溶液与淀粉酶在不同温度（0 °C、20 °C、40 °C、60 °C 和 80 °C）下混合，每隔 30 秒取一滴混合物加到点滴板上盛放的碘液中。碘液不再变为蓝黑色所需的时间表示淀粉被消化的速度。
Their data showed that digestion was fastest at 40 °C, while at 0 °C and 80 °C iodine continued to turn blue-black even after several minutes. This is because enzymes have an optimum temperature. At low temperatures molecules move slowly, so collisions are rare. At very high temperatures the enzyme denatures – its active site loses shape and can no longer bind to starch.
他们的数据显示，40 °C 时消化最快，而在 0 °C 和 80 °C 时，即使几分钟后碘液仍会变成蓝黑色。这是因为酶有最适温度。低温下分子运动缓慢，碰撞机会少；而高温下酶会变性——其活性位点形状改变，无法再与淀粉结合。
One student suggested repeating the test at 37 °C, the human body temperature. It is reasonable to predict that amylase would also work well at this temperature, because human amylase functions inside the body at around 37 °C.
有学生建议在 37 °C（人体体温）下重复实验。可以合理预测淀粉酶在此温度下也会表现良好，因为人体淀粉酶在体内约 37 °C 的环境中发挥作用。
3. Case Study 3: Modelling Digestion | 案例三：消化模型
To understand how food is broken down, students designed a model gut using Visking tubing, which acts like a partially permeable membrane. They filled the tubing with a mixture of starch and amylase solution and placed it in a beaker of warm water at 37 °C. After 20 minutes, they tested the water outside the tubing for starch and reducing sugar.
为了理解食物如何被分解，学生用赛璐玢透析管（模拟部分透性膜）设计了肠道模型。他们将淀粉和淀粉酶混合液装入透析管，置于 37 °C 的温水烧杯中。20 分钟后，他们检测了管外水中的淀粉和还原糖。
The test for starch using iodine was negative outside the tubing, while Benedict's test showed a brick-red precipitate. This indicates that starch was digested into sugar (maltose) by the enzyme, and the smaller sugar molecules were able to pass through the membrane into the surrounding water. Starch molecules were too large to leave the tubing.
对管外溶液用碘液检测淀粉的结果呈阴性，而本尼迪克特试剂检测显示砖红色沉淀。这表明淀粉被酶消化成糖（麦芽糖），较小的糖分子能够透过膜进入周围的水中。淀粉分子太大，无法穿过透析管。
This model helps explain why the body needs to break down large molecules into smaller ones: only small, soluble molecules can be absorbed through the wall of the small intestine into the blood.
这个模型有助于解释为什么身体需要把大分子分解成小分子：只有可溶性的小分子才能通过小肠壁吸收进入血液。
4. Case Study 4: Respiration in Yeast | 案例四：酵母的呼吸作用
A class investigated respiration by mixing yeast with a glucose solution in a conical flask and attaching a balloon over the opening. The flask was kept in a water bath at 30 °C. After 30 minutes, the balloon had inflated. A second flask with boiled yeast (killed by heat) did not inflate a balloon.
全班同学开展呼吸作用探究，将酵母与葡萄糖溶液在锥形瓶中混合，瓶口套上一个气球。锥形瓶置于 30 °C 水浴中。30 分钟后，气球膨胀。第二个装有煮沸酵母（被高温杀死）的锥形瓶，气球没有膨胀。
The inflation of the balloon indicates the production of carbon dioxide during anaerobic respiration (fermentation). Yeast cells break down glucose to release energy, producing CO₂ and ethanol as waste products. The boiled yeast produced no gas because the enzymes were denatured and respiration could not occur.
气球的膨胀表明无氧呼吸（发酵）过程中产生了二氧化碳。酵母细胞分解葡萄糖释放能量，产生二氧化碳和乙醇作为废物。煮沸的酵母没有产生气体，因为酶已变性，无法进行呼吸作用。
The word equation for fermentation is: glucose → ethanol + carbon dioxide (+ a small amount of energy). In industry, this process is used for baking and brewing.
发酵的文字方程式为：葡萄糖 → 乙醇 + 二氧化碳（+少量能量）。在工业上，这一过程被用于烘焙和酿造。
5. Case Study 5: Food Chains and Energy Transfer | 案例五：食物链与能量传递
l. They determined that the grass contained 10,000 kJ of energy over a given period. The grasshoppers contained 1,000 kJ, the frogs 100 kJ, and the owls only 10 kJ.
生态学家研究了一条草原食物链：草 → 蚱蜢 → 青蛙 → 猫头鹰。他们测定，在一定时期内草含有 10,000 kJ 能量，蚱蜢含有 1,000 kJ，青蛙含有 100 kJ，而猫头鹰仅含有 10 kJ。
This pattern shows that only about 10% of the energy is transferred from one trophic level to the next. The rest is lost through movement, heat, respiration, and undigested materials. This is why food chains rarely have more than four or five trophic levels – there simply is not enough energy to support another level.
这种模式表明，每一营养级只传递了约 10% 的能量给下一级。其余能量通过运动、散热、呼吸和未消化的物质而损失。这就是食物链很少超过四到五个营养级的原因——根本没有足够的能量支撑更多的层级。
Farmers can increase energy efficiency by reducing the number of trophic levels between humans and their food, for example by eating plants directly rather than feeding them to livestock.
农民可以通过减少人类与食物之间的营养级数量来提高能量效率，例如直接食用植物，而不是用植物喂养牲畜。
6. Case Study 6: Osmosis in Potato Strips | 案例六：土豆条的渗透作用
Students placed potato strips of equal length into sugar solutions of different concentrations: 0 M (distilled water), 0.2 M, 0.4 M, 0.6 M, and 0.8 M. After 30 minutes, they measured the length of each strip again. The results showed that strips in 0 M and 0.2 M solutions increased in length, while those in higher concentrations decreased in length.
学生们将等长的土豆条放入不同浓度的蔗糖溶液中：0 mol/L（蒸馏水）、0.2 mol/L、0.4 mol/L、0.6 mol/L 和 0.8 mol/L。30 分钟后，他们再次测量每根土豆条的长度。结果显示，在 0 mol/L 和 0.2 mol/L 溶液中的土豆条变长，而较高浓度溶液中的土豆条则变短。
The increase in length in dilute solutions occurred because water moved into the potato cells by osmosis, from a region of high water concentration (the solution) to a region of lower water concentration (inside the cells). In concentrated sugar solutions, water moved out of the cells, causing the strips to shrink.
在稀溶液中长度增加，是因为水通过渗透作用进入土豆细胞，从水浓度高的区域（溶液）移动到水浓度低的区域（细胞内部）。在浓糖溶液中，水从细胞中移出，导致土豆条缩短。
If a potato strip did not change in length, the solution was isotonic – the water concentration inside and outside the cells was equal.
如果某根土豆条长度没有变化，说明该溶液是等渗的——细胞内外水浓度相等。
7. Case Study 7: Simple Inheritance Using Punnett Squares | 案例七：使用庞纳特方格的简单遗传
Tongue-rolling in humans is controlled by a single gene with two alleles. The allele for rolling (R) is dominant over the allele for non-rolling (r). A heterozygous roller (Rr) and a non-roller (rr) plan to have a child. A genetic counselor draws a Punnett square to predict the probability of the child being a roller.
人类卷舌能力由一对等位基因控制，卷舌基因（R）对非卷舌基因（r）为显性。一个杂合卷舌者（Rr）和一个非卷舌者（rr）准备要一个孩子。遗传顾问画出一个庞纳特方格来预测孩子会卷舌的概率。
The Punnett square has Rr parent gametes (R and r) on one side and rr parent gametes (r and r) on the other. The possible combinations are: Rr, Rr, rr, rr. Therefore, there is a 50% chance (2 out of 4) that the child will be a tongue-roller, and a 50% chance the child will be a non-roller.
庞纳特方格一侧为 Rr 亲本的配子（R 和 r），另一侧为 rr 亲本的配子（r 和 r）。可能的组合为：Rr、Rr、rr、rr。因此，孩子有 50%（4 个中有 2 个）的概率是卷舌者，50% 的概率是非卷舌者。
This case illustrates that a recessive trait can appear in offspring even if both parents do not express it, as long as both carry the recessive allele.
这个案例说明，只要父母双方都携带隐性等位基因，即便他们没有表现出隐性性状，隐性性状也可能在孩子身上出现。
8. Case Study 8: Ecosystem Survey Using Quadrats | 案例八：使用样方进行生态系统调查
A team of Year 9 students studied the distribution of daisies on the school field. They placed a 0.5 m x 0.5 m quadrat at 10 random locations and counted the number of daisy plants inside each quadrat. They repeated the survey in an area near a path and in a less trampled area. The data are shown in the table below.
一组九年级学生调查了学校草地上雏菊的分布。他们在 10 个随机位置放置 0.5 m × 0.5 m 的样方，计数每个样方内雏菊植株的数量。他们还在一条小径附近的区域和较少踩踏的区域重复了调查。数据见下表。
| Quadrat number | Near path | Open grassland |
|---|---|---|
| 1 | 3 | 12 |
| 2 | 5 | 15 |
| 3 | 4 | 14 |
| 4 | 2 | 18 |
| 5 | 6 | 11 |
| 6 | 3 | 13 |
| 7 | 4 | 16 |
| 8 | 5 | 10 |
| 9 | 2 | 17 |
| 10 | 4 | 14 |
From the data, students calculated the mean number of daisies per quadrat: 3.8 near the path and 14.0 in the open grassland. This suggests that trampling reduces daisy abundance. The difference could be due to soil compaction, reduced light interception, or direct damage to plants.
根据数据，学生计算出每个样方雏菊的平均数量：近路径处为 3.8，开阔草地为 14.0。这表明踩踏减少了雏菊的数量。差异可能由于土壤压实、光照截获减少或植株直接受损造成。
Random sampling using quadrats is essential to avoid bias. A large number of quadrats increases reliability. The results could be used to inform decisions about protecting certain grassland areas from foot traffic.
使用样方进行随机取样对避免偏差至关重要。大量样方可以提高可靠性。这些结果可用于制定保护某些草地区域免受踩踏的决策。
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