📚 Case Study Practice for Year 8 SQA Biology | 八年级SQA生物案例分析实战演练
Case studies are a fantastic way to apply your biological knowledge to real-world scenarios. In this revision guide, we will work through eight extended case studies covering key topics in the Year 8 SQA Biology curriculum. Each case presents a situation, asks you to analyse data or explain phenomena, and then provides model answers so you can check your understanding. By working through these examples, you will sharpen your skills in observation, data interpretation, and scientific reasoning — exactly what you need for your assessments.
案例研究是将生物学知识应用于真实情境的绝佳方式。在这份复习指南中，我们将通过八个拓展案例，涵盖八年级SQA生物课程的核心主题。每个案例都会给出情境，要求你分析数据或解释现象，然后提供参考答案，以便你检验自己的理解。通过逐一演练这些例子，你将提升观察、数据解读与科学推理的能力——这正是你在测评中所需要的。
1. Ecosystem Dynamics: Rabbit and Fox Populations | 生态系统动态：兔与狐的数量
A survey in a Scottish grassland recorded the number of rabbits and foxes over a six-year period. The data are shown in the table below. Study the figures and then answer the questions.
在苏格兰一处草地进行的调查记录了六年里兔子与狐狸的数量，数据见下表。仔细阅读这些数据并回答问题。
| Year | Rabbit population | Fox population |
|---|---|---|
| 2018 | 120 | 15 |
| 2019 | 200 | 22 |
| 2020 | 280 | 35 |
| 2021 | 320 | 50 |
| 2022 | 250 | 48 |
| 2023 | 180 | 40 |
Describe the general relationship between the rabbit and fox populations from 2018 to 2021. Why do you think the rabbit numbers increased first?
请描述2018年至2021年间兔子和狐狸数量之间的大致关系。你认为为什么兔子数量会先增加？
Between 2018 and 2021, the rabbit population rose steadily from 120 to 320, and the fox population also increased from 15 to 50. This shows a predator-prey relationship: as the prey (rabbits) become more abundant, the predator (foxes) have more food, so their numbers also grow. The rabbit population increased first because foxes need time to respond — more food means higher survival and reproduction rates for foxes, but this takes a season or two to show in the numbers.
在2018年至2021年间，兔子数量从120稳步上升到320，狐狸数量也从15增加到50。这体现了捕食者-猎物关系：当猎物（兔子）更加丰富时，捕食者（狐狸）有了更多食物，因此其数量也增长。兔子数量先增加是因为狐狸需要时间做出反应——更多的食物意味着狐狸更高的存活率和繁殖率，但这需要一两个繁殖季节才能在数据中体现出来。
Explain what might have caused the rabbit population to drop after 2021, even though fox numbers remained high.
请解释2021年后兔子数量下降的可能原因，尽管狐狸数量仍然较高。
After 2021, the rabbit population fell from 320 to 180. A likely reason is that the large fox population exerted heavy predation pressure, killing many rabbits. Additionally, with such a high rabbit density, food resources for rabbits — such as grass and herbs — may have become scarce, leading to competition and starvation. Disease could also spread more easily in a dense population. The fox numbers stayed high for a while because they could still find prey, but eventually they began to decline as rabbits became harder to find.
2021年后，兔子数量从320下降到180。可能的原因是那时很高的狐狸数量施加了极大的捕食压力，杀死了许多兔子。另外，兔子密度如此之高，它们所需的食物——如草和草本植物——可能变得稀缺，导致竞争和饥饿。疾病在密集种群中也更容易传播。狐狸数量暂时保持高位是因为它们仍能找到猎物，但最终随着兔子变得难找，狐狸数量也开始下降。
2. Enzyme Action in Digestion | 消化过程中的酶作用
A student investigated the effect of temperature on the breakdown of starch by amylase. Four test tubes were set up with starch solution and amylase at different temperatures, and the time taken for the starch to disappear was recorded using iodine solution.
一位学生研究了温度对淀粉酶分解淀粉的影响。他在不同温度下设置了四支装有淀粉溶液和淀粉酶的试管，并用碘液记录了淀粉消失所需的时间。
| Temperature (°C) | Time for starch disappearance (seconds) |
|---|---|
| 0 | No change after 600 s |
| 20 | 240 |
| 37 | 60 |
| 60 | No change after 600 s |
At which temperature did the amylase work fastest? Explain your answer in terms of enzyme activity.
在哪个温度下淀粉酶工作得最快？请从酶活性的角度解释你的答案。
The amylase worked fastest at 37°C, as the starch disappeared in only 60 seconds. Enzymes are proteins that catalyse reactions; their activity increases with temperature up to an optimum point because molecules move faster and collide more often. For human amylase, the optimum is near body temperature, 37°C. At 0°C the enzyme was too cold to function effectively, and at 60°C the enzyme was denatured — its shape changed permanently — so it could no longer bind to starch.
淀粉酶在37°C下工作得最快，淀粉仅用60秒就消失了。酶是催化反应的蛋白质；在达到最适温度之前，其活性随温度升高而增加，因为分子运动加快、碰撞更频繁。对于人体淀粉酶，最适温度接近体温37°C。在0°C时，酶太冷而无法有效发挥作用；而在60°C时，酶已经变性——其形状永久改变——因此无法再与淀粉结合。
3. Plant Tropisms: Investigating Phototropism | 植物向性：探究向光性
A class grew oat seedlings in three boxes. Box A had light from above only; Box B had a light source from the right side; Box C was kept in complete darkness. After five days, the direction of shoot growth was recorded.
一个班级在三个盒子中培育燕麦幼苗。A盒只用上方光照；B盒从右侧给光；C盒完全黑暗。五天后记录幼苗的生长方向。
What results would you expect in each box? Explain the role of auxin in causing the observed growth pattern in Box B.
你预期每个盒子中的结果会如何？请解释生长素在造成B盒中观察到的生长模式中的作用。
In Box A, shoots would grow straight up, showing even growth all around. In Box B, shoots would bend towards the light on the right side. In Box C, shoots would grow straight up but would be pale and etiolated because of the lack of light. In Box B, the light from the right causes auxin (a plant hormone) to accumulate on the shaded left side of the shoot tips. Auxin stimulates cell elongation, so the cells on the left side grow longer than those on the right, making the shoot bend towards the light. This is positive phototropism and helps the plant capture more light for photosynthesis.
A盒中的幼苗会笔直向上生长，表现为四周均匀生长。B盒中的幼苗会向右侧的光源弯曲。C盒中的幼苗也会笔直向上生长，但由于缺少光照会显得苍白、徒长。在B盒中，来自右侧的光导致生长素（一种植物激素）在芽尖背光的左侧积累。生长素促进细胞伸长，因此左侧的细胞比右侧长得更长，使幼苗向光源弯曲。这是正向光性，有助于植物为光合作用获取更多光照。
4. Microorganisms and Food Preservation | 微生物与食品保存
Four slices of bread were treated differently and left in sealed bags for one week: Slice 1 was dry; Slice 2 was moistened with water; Slice 3 was moistened and kept in a fridge at 4°C; Slice 4 was moistened and sprinkled with vinegar. The amount of mould growth was compared.
四片面包经过不同处理后密封在袋中放置一周：切片1保持干燥；切片2用水湿润；切片3湿润后置于4°C冰箱中；切片4湿润后洒上醋。比较霉菌的生长量。
Predict the order of mould growth from most to least. Explain how each condition affects microbial growth.
请预测从最多到最少的霉菌生长顺序。解释每种条件如何影响微生物生长。
Most mould: Slice 2 (moist, warm room temperature) > Slice 3 (moist but cold) > Slice 4 (moist with vinegar) > Slice 1 (dry). Microorganisms need water to grow, so the dry slice had almost no mould. Warmth also speeds up enzyme reactions in microbes, so the room-temperature moist slice grew mould fastest. The fridge slowed growth because low temperatures reduce microbial activity. Vinegar is acidic, and most moulds and bacteria cannot grow well in acidic conditions; the vinegar acted as a natural preservative. This links to food preservation methods: drying, refrigeration, and pickling.
霉菌最多的是切片2（湿润、温暖室温）> 切片3（湿润但寒冷）> 切片4（湿润并洒醋）> 切片1（干燥）。微生物生长需要水分，因此干燥切片几乎没有长霉。温暖还会加快微生物的酶反应，所以室温下的湿润切片霉菌生长最快。冰箱减缓了生长，因为低温降低了微生物活性。醋是酸性的，大多数霉菌和细菌在酸性条件下无法良好生长；醋起到了天然防腐剂的作用。这与食品保存方法相联系：干燥、冷藏和腌制。
5. Cellular Respiration and Exercise | 细胞呼吸与运动
A person ran on a treadmill for 20 minutes. The breath volume and heart rate were measured every 5 minutes. The results are shown:
一个人在跑步机上跑了20分钟。每5分钟测量一次呼吸量和心率。结果如下：
| Time (min) | Breath volume (L/min) | Heart rate (bpm) |
|---|---|---|
| 0 (rest) | 6 | 70 |
| 5 | 23 | 120 |
| 10 | 28 | 145 |
| 15 | 30 | 155 |
| 20 | 32 | 162 |
Explain why breath volume and heart rate both increase during exercise. Use your knowledge of cellular respiration and oxygen demand.
请解释为什么运动时呼吸量和心率都会增加。运用你关于细胞呼吸和需氧量的知识。
During exercise, muscles contract more often and need more energy. This energy comes from respiration, which requires oxygen and glucose. The equation for aerobic respiration is:
glucose + oxygen → carbon dioxide + water (+ energy)
To supply more oxygen to the muscles, the heart beats faster to pump oxygenated blood more quickly, and breathing rate and depth increase to take in more oxygen and remove the extra carbon dioxide produced. Therefore, both breath volume and heart rate rise to meet the increased demand for aerobic respiration.运动时，肌肉收缩更频繁，需要更多能量。这些能量来自呼吸作用，需要氧气和葡萄糖。有氧呼吸的方程式为：
葡萄糖+氧气 → 二氧化碳+水（+能量）
为了向肌肉输送更多氧气，心脏加快跳动以更快地泵送含氧血，同时呼吸的频率和深度增加，以吸入更多氧气并排出产生的额外二氧化碳。因此，呼吸量和心率双双上升，以满足有氧呼吸增加的需求。6. Inheritance of Traits | 性状的遗传
In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). A gardener crossed a homozygous tall plant (TT) with a short plant (tt). The offspring in the first generation (F₁) were all tall. Then two F₁ plants were crossed to produce the second generation (F₂).
在豌豆植株中，高茎等位基因（T）对矮茎等位基因（t）为显性。一位园丁将一株纯合高茎植株（TT）与一株矮茎植株（tt）杂交。子一代（F₁）全部为高茎。然后两株F₁植株杂交产生了子二代（F₂）。
What are the possible genotypes and phenotypes in the F₂ generation? Use a Punnett square to explain the ratio you expect.
F₂代可能有哪些基因型和表现型？用庞纳特方格解释你预期的比例。
The F₁ plants are all heterozygous (Tt) because they inherit a T from the tall parent and a t from the short parent. Crossing two Tt plants gives the following Punnett square:
F₁植株全部为杂合子（Tt），因为它们从高茎亲本遗传了一个T，从矮茎亲本遗传了一个t。两株Tt植株杂交得到以下庞纳特方格：
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
The genotypes are 1 TT : 2 Tt : 1 tt. Because T is dominant, both TT and Tt plants will be tall, while tt plants will be short. Therefore, the expected phenotype ratio is 3 tall : 1 short.
基因型比例为1 TT : 2 Tt : 1 tt。由于T是显性，TT和Tt植株都将是高茎，而tt植株为矮茎。因此预期的表现型比例为3高茎 : 1矮茎。
7. Disease Transmission and Prevention | 疾病传播与预防
A village experienced an outbreak of a bacterial disease spread through contaminated water. Public health officials advised boiling water before drinking and covering wells. After two weeks, the number of new cases dropped sharply.
一个村庄暴发了一种通过受污染水源传播的细菌性疾病。公共卫生官员建议饮用前将水煮沸并遮盖水井。两周后，新发病例数急剧下降。
Explain how boiling water and covering wells help reduce the spread of the disease. Why are antibiotics effective against bacteria but not viruses?
解释煮沸水、遮盖水井如何帮助减少疾病传播。为什么抗生素对细菌有效而对病毒无效？
Boiling water kills bacteria because the high temperature denatures their enzymes and damages cell structures. Covering wells prevents bacteria from entering the water supply, interrupting the transmission route. These are examples of preventing infection by breaking the chain of infection. Antibiotics work by targeting specific bacterial processes, such as cell wall synthesis or protein production, without harming human cells. Viruses are not living cells; they have no metabolic processes of their own and hijack host cells to replicate. Since antibiotics target bacterial features, they are useless against viruses. This is why doctors do not prescribe antibiotics for viral illnesses like the common cold.
煮沸水可杀灭细菌，因为高温使细菌的酶变性并破坏细胞结构。遮盖水井可以防止细菌进入供水系统，切断了传播途径。这些都是通过打破感染链来预防感染的例子。抗生素针对的是细菌特有的过程，如细胞壁合成或蛋白质生产，而不伤害人体细胞。病毒不是活细胞；它们没有自身的代谢过程，而是劫持宿主细胞进行复制。由于抗生素攻击的是细菌的结构特点，因此对病毒无效。这就是医生不会为普通感冒等病毒性疾病开具抗生素的原因。
8. Photosynthesis and Light Intensity | 光合作用与光照强度
An experiment used aquatic plants to measure the rate of photosynthesis. The number of oxygen bubbles produced per minute was counted at different distances from a lamp. The closer the lamp, the higher the light intensity.
一项实验利用水生植物测量光合作用速率。在不同距离放置灯，统计每分钟产生的氧气气泡数。灯越近，光照强度越高。
| Distance from lamp (cm) | Bubbles per minute |
|---|---|
| 10 | 45 |
| 20 | 35 |
| 30 | 22 |
| 40 | 10 |
| 50 | 3 |
What pattern do you see in the data? Explain why the rate of photosynthesis changes with light intensity, and predict what might happen if another factor, such as carbon dioxide concentration, became limiting.
你在数据中看到了什么规律？解释为什么光合作用速率随光照强度变化，并预测如果另一个因子（如二氧化碳浓度）成为限制因子，可能会发生什么。
As the distance from the lamp increases (meaning lower light intensity), the number of oxygen bubbles per minute decreases. This shows that light intensity directly affects the rate of photosynthesis: more light provides more energy for the light-dependent reactions, so more oxygen is produced. However, if light intensity became very high, the rate would no longer increase because another factor, such as available carbon dioxide or temperature, would become limiting. For example, if CO₂ concentration is too low, the Calvin cycle cannot proceed quickly enough, and the rate of photosynthesis will level off regardless of how bright the light is. This is called a limiting factor.
随着灯距增加（即光照强度降低），每分钟产生的氧气泡数量减少。这表明光照强度直接影响光合作用速率：更多光照为光反应提供了更多能量，因此产生更多氧气。然而，如果光照强度非常高，速率不再增加，因为另一个因子（如可用的二氧化碳浓度或温度）会成为限制因子。例如，如果CO₂浓度太低，卡尔文循环无法足够快地进行，无论光线多强，光合作用速率都会趋于平缓。这称为限制因子。
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