📚 Year 10 SQA Biology: Case Study Practical Drills | SQA 十年级生物案例分析实战演练
Case studies are a powerful way to bridge the gap between textbook theory and real-world biology. In the SQA Year 10 curriculum, you are expected not just to recall facts but to apply your understanding to unfamiliar scenarios, interpret data, and justify conclusions. These practical drills are designed to sharpen your analytical skills across the key areas of cell biology, genetics, ecology, and human physiology. Each case mirrors the style of SQA examination questions, giving you the perfect revision tool.
案例分析是将课本理论与现实生物学连接起来的有效方法。在 SQA 十年级课程中，你不仅要记住知识点，还要能在陌生情境中应用理解、解释数据并论证结论。这些实战演练旨在磨炼你在细胞生物学、遗传学、生态学和人体生理学等关键领域的分析能力。每个案例都模拟了 SQA 考试的题型风格，是完美的复习工具。
1. Genetics Case Study: Sickle Cell Anaemia | 遗传学案例：镰刀型细胞贫血症
Sickle cell anaemia is caused by a mutation in the HBB gene, resulting in an abnormal haemoglobin molecule. The normal allele is HbA, and the mutant allele is HbS. Individuals who are homozygous for HbS (HbS/HbS) produce only the abnormal haemoglobin and suffer from severe anaemia. Heterozygotes (HbA/HbS) are carriers and show the sickle cell trait without major symptoms, while also being resistant to malaria. In parts of West Africa, the frequency of the HbS allele remains high due to this heterozygote advantage.
镰刀型细胞贫血症由 HBB 基因突变引起，导致产生异常的血红蛋白分子。正常等位基因为 HbA，突变等位基因为 HbS。HbS 纯合子（HbS/HbS）只产生异常血红蛋白，患有严重贫血。杂合子（HbA/HbS）是携带者，表现出镰刀细胞特征但无明显症状，同时对疟疾具有抵抗力。在西非部分地区，由于这种杂合子优势，HbS 等位基因频率仍然很高。
In a hypothetical population of 10,000 individuals, the following genotype counts were recorded in a malaria-endemic region:
在一个疟疾流行区，对 10,000 名个体的假设群体记录到如下基因型计数：
| Genotype | Number of individuals |
|---|---|
| HbA/HbA | 7200 |
| HbA/HbS | 2600 |
| HbS/HbS | 200 |
200 + 2600 = 14,400 + 2,600 = 17,000. Frequency of HbA, p = 17,000/20,000 = 0.85. Number of HbS alleles = 2×200 + 2600 = 400 + 2600 = 3,000. Frequency of HbS, q = 3,000/20,000 = 0.15.
首先，计算各等位基因的频率。等位基因总数为 20,000（因为每人有两个等位基因）。HbA 等位基因数 = 2×7200 + 2600 = 14,400 + 2,600 = 17,000。HbA 频率 p = 17,000/20,000 = 0.85。HbS 等位基因数 = 2×200 + 2600 = 400 + 2600 = 3,000。HbS 频率 q = 3,000/20,000 = 0.15。
m mating: expected frequency of HbA/HbA = p² = 0.85² = 0.7225, HbA/HbS = 2pq = 2×0.85×0.15 = 0.255, HbS/HbS = q² = 0.15² = 0.0225.
检验：p + q = 0.85 + 0.15 = 1.00，符合。现在用哈代-温伯格定律预测随机交配下的期望基因型频率：HbA/HbA 期望频率 = p² = 0.85² = 0.7225，HbA/HbS = 2pq = 2×0.85×0.15 = 0.255，HbS/HbS = q² = 0.15² = 0.0225。
re close to expected, suggesting the population is roughly in Hardy-Weinberg equilibrium. However, the slightly lower number of HbS/HbS individuals (200 vs 225) may reflect selection against severely affected homozygotes, while the high heterozygote count is maintained by malaria resistance.
换算为 10,000 人中的期望数目：HbA/HbA = 7225，HbA/HbS = 2550，HbS/HbS = 225。观察值与期望值接近，表明群体大致处于哈代-温伯格平衡。但 HbS/HbS 个体略少（200 对 225），可能反映了对严重患者的负选择，而高数量的杂合子则因疟疾抗性得以维持。
2. Enzyme Activity: Effect of Temperature | 酶活性：温度的影响
Catalase is an enzyme that breaks down hydrogen peroxide into water and oxygen. A student investigated the effect of temperature on catalase activity by measuring the volume of oxygen produced per minute. The results are shown in the table below.
过氧化氢酶能催化过氧化氢分解为水和氧气。一位学生通过测量每分钟产生的氧气体积，研究了温度对过氧化氢酶活性的影响。结果见下表。
| Temperature (°C) | Oxygen produced (cm³ min⁻¹) |
|---|---|
| 10 | 2.0 |
| 20 | 5.0 |
| 30 | 8.5 |
| 40 | 9.0 |
| 50 | 7.0 |
| 60 | 0.5 |
The data show that oxygen production increases as temperature rises up to about 40°C, which is near the optimum temperature for catalase. At this point, the rate reaches a maximum because the enzyme and substrate molecules have high kinetic energy, leading to more frequent successful collisions. Above 40°C, the rate declines sharply because the enzyme begins to denature – the heat breaks the hydrogen bonds maintaining its tertiary structure, changing the shape of the active site so that the substrate no longer fits.
数据表明，氧气产量在温度升至约 40°C 时增加，这接近过氧化氢酶的最适温度。此时速率达到最大，因为酶和底物分子具有较高的动能，导致更频繁的有效碰撞。超过 40°C 后，速率急剧下降，因为酶开始变性——高温破坏了维持其三级结构的氢键，改变了活性部位的形状，使底物不再契合。
Calculate the Q₁₀ value for the temperature range 20– is value indicates a typical enzymatic temperature coefficient, showing the reaction rate nearly doubles with a 10°C rise, as long as the enzyme is not denatured.
计算 20–30°C 范围内的 Q₁₀ 值。20°C 时速率为 5.0 cm³ min⁻¹，30°C 时为 8.5 cm³ min⁻¹。Q₁₀ = (T+10) 时的速率 / T 时的速率 = 8.5/5.0 = 1.7。该值表示典型的酶促温度系数，表明在酶未变性时，温度每升高 10°C 反应速率接近翻倍。
3. Photosynthesis: Light Intensity Experiment | 光合作用：光照强度实验
A student used a sealed flask containing pondweed and a hydrogencarbonate indicator solution to measure the rate of photosynthesis at different light intensities. The indicator turns from red to purple as CO₂ concentration decreases, indicating photosynthesis. The time taken for the indicator to change to a standard purple colour was recorded. The inverse of time (1/t) was used as a measure of photosynthetic rate.
一位学生使用装有水草和碳酸氢盐指示剂的密封烧瓶，测量不同光照强度下的光合作用速率。当 CO₂ 浓度因光合作用下降时，指示剂由红色变为紫色。记录指示剂变为标准紫色所需的时间，并以时间倒数（1/t）作为光合速率的度量。
| Light intensity (arbitrary units) | Time for colour change (s) | Rate (1/t, s⁻¹) |
|---|---|---|
| 2 | 600 | 0.0017 |
| 4 | 300 | 0.0033 |
| 6 | 200 | 0.0050 |
| 8 | 150 | 0.0067 |
| 10 | 150 | 0.0067 |
At low light intensities, the rate of photosynthesis increases proportionally, demonstrating that light was the limiting factor. However, at 8 and 10 units, the rate plateaus, indicating that another factor, most likely the CO₂ concentration or temperature, has become limiting. The light intensity at which the rate levels off is known as the light saturation point.
在低光照强度下，光合速率成比例增加，表明光是限制因子。但当光照强度为 8 和 10 单位时，速率持平，表明另一个因子（最有可能是 CO₂ 浓度或温度）已成为限制因子。速率不再增加时的光照强度称为光饱和点。
If the temperature were increased from 20°C to 25°C, the rate at high light intensity might increase because the enzymes of the Calvin cycle would have more kinetic energy. However, if CO₂ is scarce, no further increase would occur.
若将温度从 20°C 提高到 25°C，高光照下的速率可能会提高，因为卡尔文循环中的酶具有更高的动能。但若 CO₂ 不足，则不会进一步上升。
4. Food Web and Energy Transfer | 食物网与能量传递
In a grassland ecosystem, the following approximate energy values were measured within a simple food chain: grass (producer) - 10,000 kJ m⁻² year⁻¹; grasshoppers (primary consumer) - 1,000 kJ m⁻² year⁻¹; frogs (secondary consumer) - 100 kJ m⁻² year⁻¹; snakes (tertiary consumer) - 10 kJ m⁻² year⁻¹.
在一个草地生态系统中，测定了以下简单食物链中的能量近似值：草（生产者）- 10,000 kJ m⁻² year⁻¹；蚱蜢（初级消费者）- 1,000 kJ m⁻² year⁻¹；青蛙（次级消费者）- 100 kJ m⁻² year⁻¹；蛇（三级消费者）- 10 kJ m⁻² year⁻¹。
one trophic level to the next. From grass to grasshoppers: efficiency = (1,000/10,000)×100% = 10%. From grasshoppers to frogs: (100/1,000)×100% = 10%. From frogs to snakes: (10/100)×100% = 10%. The consistent 10% in ecology; most energy is lost through respiration, heat, egestion, and uneaten material.
计算从一个营养级传递到下一个营养级的能量百分比。从草到蚱蜢：效率 = (1,000/10,000)×100% = 10%。从蚱蜢到青蛙：(100/1,000)×100% = 10%。从青蛙到蛇：(10/100)×100% = 10%。恒定的 10% 效率体现了生态学中的经验法则；大部分能量通过呼吸、散热、排泄和未食用的物质而损耗。
Draw a pyramid of energy for this food chain. The pyramid would have a broad base (grass) and each succeeding bar would be one-tenth the width of the previous one. This shape reflects why food chains rarely exceed four or five trophic levels – there is insufficient energy to support another level.
画出这条食物链的能量金字塔。金字塔的底层（草）宽广，每个后继条形宽度为前一级的十分之一。这种形状反映了为何食物链很少超过四或五个营养级——没有足够的能量支撑更高的营养级。
If a pesticide accumulates in the tissues, biomagnification would cause the highest concentration in the snake, as it consumes multiple frogs. This is an example of how toxic substances concentrate along the food chain.
若某种杀虫剂在组织中积累，生物放大作用会使蛇体内的浓度最高，因为它以大量青蛙为食。这是有毒物质沿食物链浓缩的一个例子。
5. Natural Selection: Peppered Moth | 自然选择：桦尺蛾
The peppered moth (Biston betularia) exists in two forms: a pale, speckled variety (typica) and a dark, melanic form (carbonaria). Before the Industrial Revolution, tree trunks were covered in pale lichens, and the pale moths were well camouflaged. In the mid-19th century, industrial pollution killed lichens and blackened tree trunks with soot, giving the dark moths a selective advantage. Data collected in Manchester showed the following changes in the frequency of melanic moths: in 1848, 2% were carbonaria; by 1895, the figure had risen to 95%.
桦尺蛾（Biston betularia）有两种形态：浅色带斑点的类型（typica）和深色黑化型（carbonaria）。工业革命前，树干上长满浅色地衣，浅色蛾伪装良好。19 世纪中期，工业污染杀死地衣，并用煤烟使树干变黑，深色蛾获得了选择优势。在曼彻斯特收集的数据显示黑化蛾频率发生了如下变化：1848 年，2% 为 carbonaria；到 1895 年，这一数字上升到 95%。
Explain this dramatic change using natural selection. Within the moth population, there was variation in colour. Dark moths were initially rare but better camouflaged on sooty trees, so they were less likely to be eaten by birds. They survived and reproduced more, passing on the alleles for dark colour to their offspring. Over generations, the frequency of the advantageous allele increased, while the pale allele decreased. This is directional selection in response to an environmental change.
用自然选择解释这一显著变化。在蛾类种群中存在颜色变异。深色蛾起初稀少，但在煤烟覆盖的树上伪装得更好，不易被鸟类捕食。它们存活并繁殖
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