📚 IGCSE AQA Biology: Worked Examples Explained | IGCSE AQA 生物：典型例题详解
Welcome to our detailed walkthrough of common IGCSE AQA Biology exam questions. Each worked example is designed to mirror the style and demand of real AQA assessment material, showing you how to apply knowledge, structure answers, and avoid typical mistakes. By studying these examples carefully, you will build confidence in tackling calculations, data interpretation, and extended response questions.
欢迎来到 IGCSE AQA 生物典型例题详解。每个例题都模拟 AQA 真实考题的风格与难度，带你一步步运用知识、组织答案、避开常见错误。认真学习这些题目，你会更有信心应对计算、数据分析及扩展型问答。
1. Worked Example: Magnification | 例题：放大倍数计算
A student draws a plant cell observed under a light microscope. The drawing measures 9.6 cm across. The actual diameter of the cell is 0.12 mm. Calculate the magnification used. Show your working.
一名学生绘制了光学显微镜下的植物细胞，绘图宽度为9.6厘米。该细胞的实际直径为0.12毫米。请计算所使用的放大倍数并展示计算过程。
First, recall the formula: Magnification = Image size / Actual size. Here, image size is 9.6 cm and actual size is 0.12 mm.
首先记住公式：放大倍数 = 图像尺寸 ÷ 实际尺寸。本题中图像尺寸为9.6厘米，实际尺寸为0.12毫米。
Convert both measurements to the same unit. 9.6 cm = 96 mm. So Magnification = 96 mm / 0.12 mm = 800. Alternatively, convert to micrometres: 96 mm = 96,000 µm, 0.12 mm = 120 µm; 96,000 / 120 = 800. The answer is ×800 (or 800 times).
将两个数值统一为相同单位。9.6厘米 = 96毫米。因此放大倍数 = 96毫米 ÷ 0.12毫米 = 800。也可以转换为微米：96毫米 = 96,000微米，0.12毫米 = 120微米；96,000 ÷ 120 = 800。答案为800倍。
In reverse, if the magnification is 400× and the image is 2 cm, actual size = 2 cm / 400 = 0.005 cm = 0.05 mm. Always include units and check that the magnification has no unit.
反过来，如果放大倍数为400倍，图像为2厘米，实际尺寸 = 2厘米 ÷ 400 = 0.005厘米 = 0.05毫米。务必标注单位，并注意放大倍数本身不带单位。
2. Worked Example: Osmosis and Potato Cylinders | 例题：渗透作用与土豆条实验
Five potato cylinders of equal mass were placed in sucrose solutions of different concentrations. After 30 minutes they were blotted dry and reweighed. The percentage change in mass is shown in the table below. Explain the results.
五根等质量的土豆条分别放入不同浓度的蔗糖溶液中。30分钟后吸干表面水分并重新称重，质量变化百分比如下表所示。请解释实验结果。
| Sucrose concentration / mol dm⁻³ | % change in mass |
|---|---|
| 0.0 | +22.5 |
| 0.2 | +10.0 |
| 0.4 | -4.5 |
| 0.6 | -15.2 |
| 0.8 | -24.8 |
Potato cells have a lower water potential than pure water, so water enters by osmosis in dilute solutions, causing an increase in mass. In 0.0 mol dm⁻³ and 0.2 mol dm⁻³ solutions, the mass increased.
土豆细胞的水势比纯水低，因此在稀溶液中水通过渗透进入细胞，使质量增加。在0.0 mol dm⁻³ 和0.2 mol dm⁻³ 溶液中质量都增加了。
As sucrose concentration rises, the external water potential falls. At around 0.4 mol dm⁻³, there is no net change in mass — this is the point where the solution is isotonic to the potato cells. In higher concentrations, water leaves the cells and mass decreases.
随着蔗糖浓度升高，外部水势下降。在大约0.4 mol dm⁻³ 处，质量没有净变化——此时溶液与土豆细胞等渗。浓度更高时，水离开细胞，质量下降。
This practical is often used to estimate the solute concentration inside potato cells. Always mention osmosis, water potential gradients, and the terms hypotonic, hypertonic, isotonic where relevant.
此实验常用于估测土豆细胞内的溶质浓度。回答时务必提及渗透、水势梯度，并在相关处使用低渗、高渗和等渗等术语。
3. Worked Example: Enzyme Activity and Temperature | 例题：酶活性与温度
An investigation recorded the time taken for amylase to digest starch at different temperatures. The table gives time and calculated rate (1/time). Plot a graph and determine the optimum temperature.
某项研究记录了不同温度下淀粉酶消化淀粉所需的时间。表格给出了时间和计算出的速率（1/时间）。请绘制图形并确定最适温度。
| Temperature / °C | Time / s | Rate / s⁻¹ |
|---|---|---|
| 10 | 210 | 0.0048 |
| 20 | 90 | 0.0111 |
| 30 | 45 | 0.0222 |
| 40 | 38 | 0.0263 |
| 50 | 62 | 0.0161 |
| 60 | 250 | 0.0040 |
The rate increases from 10 °C to around 40 °C because enzyme and substrate molecules have more kinetic energy and collide more frequently. The peak rate is at approximately 40 °C — the optimum temperature for this amylase.
速率从10 °C升高到约40 °C，因为酶和底物分子具有更多动能，碰撞更频繁。速率峰值约在40 °C——这是该淀粉酶的最适温度。
Above 40 °C the rate falls sharply. The high temperature breaks hydrogen bonds in the enzyme, altering the shape of its active site. The enzyme denatures and can no longer bind the substrate.
40 °C以上，速率急剧下降。高温破坏了酶分子内的氢键，改变了活性部位的形状。酶发生变性，无法再与底物结合。
In an exam, always relate temperature changes to collision frequency and enzyme denaturation. Use data from the table to support your answer.
考试时务必把温度变化与碰撞频率和酶变性联系起来，并用表格数据支撑你的答案。
4. Worked Example: Heart Structure and Blood Flow | 例题：心脏结构与血流
A diagram of the human heart is provided with chambers labelled A, B, C, D. Blood vessels W, X, Y, Z are also indicated. Describe the pathway of blood from the vena cava to the aorta, naming the chambers and valves.
下图为人类心脏示意图，腔室标为A、B、C、D，血管标为W、X、Y、Z。请描述从腔静脉到主动脉的血液路径，说出经过的腔室和瓣膜名称。
Deoxygenated blood enters the right atrium (chamber A) from the vena cavae. When the atrium contracts, blood is pushed through the tricuspid valve into the right ventricle (chamber B).
缺氧血从腔静脉进入右心房（A室）。心房收缩时，血液通过三尖瓣被推入右心室（B室）。
Contraction of the right ventricle forces blood through the semilunar valve into the pulmonary artery (vessel X) and to the lungs. After gas exchange, oxygenated blood returns via the pulmonary vein (vessel Y) to the left atrium (chamber C).
右心室收缩，血液通过半月瓣进入肺动脉（血管X）流向肺部。气体交换后，含氧血经肺静脉（血管Y）返回左心房（C室）。
Blood then moves through the bicuspid (mitral) valve into the left ventricle (chamber D), which generates high pressure to pump oxygenated blood through the aortic semilunar valve into the aorta (vessel Z) and around the body.
随后血液通过二尖瓣进入左心室（D室），左心室产生高压将含氧血经主动脉半月瓣泵入主动脉（血管Z）并送往全身。
Always state that the left ventricle has a thicker muscular wall because it must pump blood much farther. Remember coronary arteries supply the heart muscle itself.
回答时常指出左心室肌肉壁更厚，因为它需要将血液泵得更远。记住冠状动脉为心肌自身供血。
5. Worked Example: Breathing and Gas Exchange | 例题：呼吸与气体交换
The table below shows the approximate composition of inhaled and exhaled air. Explain the differences.
下表显示了吸入气和呼出气的大致组成。请解释差异。
| Gas | Inhaled air / % | Exhaled air / % |
|---|---|---|
| Oxygen | 21 | 16 |
| Carbon dioxide | 0.04 | 4 |
| Nitrogen | 78 | 78 |
| Water vapour | Variable (low) | Saturated |
Oxygen decreases from 21% to 16% because it diffuses from the alveoli into the blood for aerobic respiration in cells. Carbon dioxide, a waste product of respiration, increases from 0.04% to about 4%.
氧气从21%下降至16%，因为它从肺泡扩散到血液中，用于细胞的有氧呼吸。二氧化碳是呼吸作用的废物，从0.04%上升至约4%。
Nitrogen is inert and does not take part in respiration, so its proportion remains unchanged. Exhaled air is saturated with water vapour because the lining of the alveoli is moist and water evaporates into the air.
氮气是惰性气体，不参与呼吸，因此其比例保持不变。呼出气水蒸气饱和，因为肺泡内壁湿润，水分蒸发进入空气中。
When describing ventilation, mention the diaphragm and intercostal muscles. Inspiration: diaphragm contracts and flattens, external intercostals contract, ribcage moves up and out, volume increases, pressure decreases, air rushes in.
描述通气时要提到膈肌和肋间肌。吸气：膈肌收缩变平，外肋间肌收缩，肋骨上抬外扩，胸腔容积增大，压力降低，空气进入。
6. Worked Example: Photosynthesis Practical | 例题：光合作用实验
A student investigates the effect of light intensity on photosynthesis using an Elodea (pondweed) placed in water. Bubbles of oxygen produced per minute were counted at different distances from a lamp. The results are shown below.
一名学生使用伊乐藻（水生植物）研究光强度对光合作用的影响，记录不同灯距下每分钟产生的氧气气泡数。结果如下。
| Distance / cm | Bubbles per minute |
|---|---|
| 10 | 42 |
| 20 | 25 |
| 30 | 14 |
| 40 | 9 |
As distance increases, light intensity decreases (inverse square law), so the rate of photosynthesis falls. At 10 cm, the rate is highest because more light energy is available to drive the light-dependent reactions.
灯距增加时，光强度降低（光照度反比于距离平方），光合作用速率下降。10 cm处速率最高，因为有更多光能驱动光反应。
Other factors may become limiting at high light intensity, such as carbon dioxide concentration or temperature. In this investigation, the independent variable is distance, and the dependent variable is bubble count per minute.
在高光强下，其他因素如二氧化碳浓度或温度可能会成为限制因子。本实验中，自变量为距离，因变量为每分钟气泡数。
Control variables: same piece of Elodea, same volume of water, same temperature, addition of sodium hydrogencarbonate to provide CO₂, and wait time between readings. Always list at least three control variables in a plan.
控制变量：同一段伊乐藻、等量水、相同温度、加入碳酸氢钠提供CO₂，以及每次读数前等待一定时间。在实验设计中务必列出至少三项控制变量。
7. Worked Example: Monohybrid Inheritance | 例题：单基因遗传
In pea plants, tall stem (T) is dominant over dwarf stem (t). A heterozygous tall plant is crossed with a homozygous dwarf plant. Determine the expected genotypic and phenotypic ratios using a Punnett square.
在豌豆中，高茎（T）对矮茎（t）为显性。一株杂合高茎与一株纯合矮茎杂交。请用庞纳特方格确定预期的基因型比和表现型比。
Parental genotypes: Tt × tt. Gametes from the tall parent: T and t (each 50%). Gametes from dwarf parent: all t. Draw the Punnett square:
亲本基因型：Tt × tt。高茎亲本产生的配子：T 和 t（各占50%）。矮茎亲本产生的配子：全部为t。绘制庞纳特方格：
| T | t | |
| t | Tt | tt |
| t | Tt | tt |
Genotypic ratio: 2 Tt : 2 tt, simplified to 1 Tt : 1 tt. Phenotypic ratio: 1 tall : 1 dwarf (since Tt is tall, tt is dwarf).
基因型比：2 Tt : 2 tt，简化为 1 Tt : 1 tt。表现型比：1高茎 : 1矮茎（Tt表现为高茎，tt表现为矮茎）。
Always define alleles clearly at the start, use uppercase for dominant and lowercase for recessive, and show each step. In an exam, link offspring ratios back to random fertilisation of gametes.
一开始就要清晰地定义等位基因，显性用大写字母，隐性用小写字母，并展示每一步。考试时须将后代比例与配子的随机受精联系起来。
8. Worked Example: Food Chains and Energy Transfer | 例题：食物链与能量传递
A food chain in a field is: grass → grasshoppers → frogs → snakes. The biomass of grass is 2000 kg. Only about 10% of biomass is passed to the next trophic level. Calculate the maximum biomass of snakes.
某田间的食物链为：草 → 蚱蜢 → 青蛙 → 蛇。草的生物量为 2000 kg。仅约10%的生物量传递给下一营养级。计算蛇的最大生物量。
Energy transfer between trophic levels is inefficient due to respiration, movement, egestion of waste, and uneaten parts. Using the 10% rule:
营养级之间的能量传递效率不高，因为呼吸、运动、未消化排泄物及未被取食的部分都会损失。按10%法则计算：
Biomass of grasshoppers = 10% of 2000 kg = 200 kg. Biomass of frogs = 10% of 200 kg = 20 kg. Biomass of snakes = 10% of 20 kg = 2 kg.
蚱蜢的生物量 = 2000 kg 的10% = 200 kg。青蛙的生物量 = 200 kg 的10% = 20 kg。蛇的生物量 = 20 kg 的10% = 2 kg。
This explains why food chains rarely have more than five trophic levels — insufficient energy remains to support another level. Pyramids of biomass usually show decreasing mass at higher levels.
这解释了为什么食物链很少超过五个营养级——剩余能量不足以支撑下一级。生物量金字塔通常显示营养级越高，生物量越小。
9. Worked Example: Natural Selection – Peppered Moth | 例题：自然选择——桦尺蛾
In industrial Britain, the proportion of dark peppered moths increased from 1% to over 90% in polluted areas. Use natural selection to explain this change.
在工业时期的英国，污染区深色桦尺蛾的比例从1%上升到90%以上。请用自然选择解释这一变化。
Before industrial pollution, tree trunks were covered in pale lichens, so light-coloured moths were camouflaged and survived predation. Dark moths were easily seen and eaten.
工业污染前，树干上覆盖着浅色地衣，因此浅色蛾子伪装良好，得以幸存；深色蛾子容易被发现并被捕食。
Pollution killed lichens and darkened tree trunks with soot. Now dark moths became better camouflaged, while light moths were more visible to birds. The dark moths survived, reproduced, and passed on their alleles for dark colour.
污染导致地衣死亡，煤烟使树干变黑。此时深色蛾子伪装更好，浅色蛾子更容易被鸟类发现。深色蛾子存活并繁殖，将深色等位基因传递下去。
Over generations, the frequency of the dark allele increased — a typical example of natural selection driven by a change in the environment. This is directional selection.
经过多代，深色等位基因的频率升高——这是由环境变化驱动的自然选择典型例子，属于定向选择。
In an exam, use the steps: variation exists, a selection pressure is present, some variants have a survival advantage, they reproduce more, and advantageous alleles increase in the population over time.
考试中请按步骤回答：存在变异，存在选择压力，某些变异具有生存优势，它们繁殖更多，有利等位基因在种群中的频率随时间增加。
10. Worked Example: Disease Transmission and Prevention | 例题：疾病传播与预防
An outbreak of a bacterial disease occurred in a village. Scientists surveyed the number of cases before and after implementing three measures: chlorination of water, vaccination, and isolating infected individuals. Interpret the bar chart data.
某村庄爆发了一种细菌性疾病。科学家在实施三项措施前后调查了病例数量：饮用水氯消毒、疫苗接种和隔离感染者。请解读条形图数据（数据见下表）。
| Measure | Cases before | Cases after |
|---|---|---|
| Chlorination | 85 | 18 |
| Vaccination | 72 | 9 |
| Isolation | 63 | 12 |
All three measures significantly reduced the number of disease cases. Chlorination killed bacteria in water, cutting off transmission through contaminated drinking water. Vaccination stimulated the immune system to produce memory cells and antibodies, providing active immunity.
三项措施都显著降低了病例数。氯消毒杀死水中细菌，切断了受污染饮用水的传播途径。疫苗接种刺激免疫系统产生记忆细胞和抗体，提供了主动免疫。
Isolation prevented infected individuals from passing the pathogen to susceptible hosts, reducing direct person-to-person transmission. The combined effect is often greater than any single measure alone.
隔离防止感染者将病原体传给易感宿主，减少了直接人际传播。多种措施联合使用常比单一措施效果更好。
Always link control methods to the pathogen’s mode of transmission. For bacterial infections, antibiotics can also be used, but resistant strains are a growing concern. Vaccines often protect the whole community through herd immunity.
回答时务必将控制措施与病原体传播方式联系起来。细菌感染也可使用抗生素，但耐药菌株日益令人担忧。疫苗常通过群体免疫保护整个社区。
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