📚 GCSE AQA Biology: Worked Examples | GCSE AQA 生物：典型例题详解
This article presents a series of carefully chosen worked examples that cover the key topics in the GCSE AQA Biology specification. Each example is designed to mirror the style of exam questions, helping you to develop problem-solving skills, apply core concepts, and become familiar with the types of calculations and extended responses you will encounter. By working through these examples and comparing your approach with the model answers, you will strengthen your understanding of biological principles and improve your exam technique.
本文提供了一系列精心挑选的典型例题，覆盖 GCSE AQA 生物大纲中的核心主题。每个例题的设计都模仿真实考题的风格，帮助你培养解决问题的能力、应用核心概念，并熟悉考试中会出现的计算题和扩展写作题型。通过练习这些例题，并将自己的解题过程与标准答案进行对比，你将加深对生物学原理的理解并提高应试技巧。
1. Magnification Calculations | 显微镜放大倍数计算
A student views a plant cell under a light microscope. The image size of the nucleus is measured as 12 mm. The actual size of the nucleus is 6 µm. Calculate the magnification of the image. Show your working. [2 marks]
一名学生在光学显微镜下观察一个植物细胞。细胞核的图像尺寸经测量为 12 mm。细胞核的实际尺寸为 6 µm。请计算该图像的放大倍数。写出你的计算过程。[2分]
Convert all measurements to the same unit. 12 mm = 12 000 µm. Magnification = image size / actual size = 12 000 µm / 6 µm = ×2000. Always check that the units match before dividing; 1 mm = 1000 µm.
将所有测量值转换为相同单位。12 mm = 12 000 µm。放大倍数 = 图像尺寸 / 实际尺寸 = 12 000 µm / 6 µm = ×2000。在相除之前一定要确保单位一致；1 mm = 1000 µm。
nification. Remember that image size is always larger than actual size in a magnified image.
如果题目要求计算实际尺寸，将公式变形：实际尺寸 = 图像尺寸 / 放大倍数。请记住，在放大后的图像中，图像尺寸总是大于实际尺寸。
2. Cell Specialisation and Stem Cells | 细胞分化与干细胞
Explain how the structure of a sperm cell is adapted for its function. [4 marks]
解释精子细胞的结构如何适应其功能。[4分]
The sperm cell has a long tail (flagellum) that provides motility, enabling it to swim towards the egg. The middle section is packed with mitochondria, which release energy through respiration for the tail’s movement. The acrosome in the head contains digestive enzymes that break down the outer layer of the egg, allowing the sperm to penetrate. The nucleus carries the haploid set of chromosomes, which is essential for sexual reproduction and the restoration of the diploid number upon fertilisation.
精子细胞有一条长尾（鞭毛），提供运动能力，使其能游向卵细胞。中部充满了线粒体，线粒体通过呼吸作用释放能量，为尾部运动供能。头部的顶体含有消化酶，可以分解卵细胞的外层，使精子能够穿入。细胞核携带单倍染色体组，这对有性生殖至关重要，受精后能恢复二倍体数目。
Many exam questions ask you to link a specialised cell’s features to its specific job. Always connect the adaptation directly to the function – for example, not just ‘it has many mitochondria’ but ‘it has many mitochondria to release more energy for tail movement’.
许多试题会要求你将特化细胞的特征与其特定功能联系起来。一定要将适应性特征与其功能直接关联——例如，不仅仅写“它有很多线粒体”，而应写“它有很多线粒体，以便为尾部运动释放更多能量”。
Stem cells are undifferentiated cells that can divide to produce more stem cells or differentiate into specialised cells. In plants, meristem tissue contains stem cells that can differentiate into any type of plant cell, allowing plant growth throughout life. In humans, adult stem cells (e.g. in bone marrow) can only differentiate into a limited range of cell types, unlike embryonic stem cells which are pluripotent.
干细胞是未分化的细胞，它们能分裂产生更多的干细胞，或者分化成特化的细胞。在植物中，分生组织含有干细胞，可以分化为任何类型的植物细胞，使植物能够终身生长。在人体中，成体干细胞（例如骨髓中的干细胞）只能分化为有限类型的细胞，与可分化为多种细胞类型的胚胎多能干细胞不同。
3. Enzyme Action and Temperature | 酶的作用与温度影响
An investigation was carried out into the effect of temperature on the rate of an enzyme-controlled reaction. The table shows the results. At 20 °C, the rate was 0.8 arbitrary units; at 30 °C, 1.5 units; at 40 °C, 2.2 units; at 50 °C, 1.8 units; at 60 °C, 0.2 units. Explain why the rate of reaction decreases above 40 °C. [3 marks]
某实验研究了温度对酶促反应速率的影响。结果如下表所示。在20 °C时，速率为0.8个任意单位；在30 °C时为1.5个单位；在40 °C时为2.2个单位；在50 °C时为1.8个单位；在60 °C时为0.2个单位。请解释为何在40 °C以上反应速率会下降。[3分]
At temperatures above the optimum (around 40 °C for many human enzymes), the enzyme molecules vibrate more vigorously. These increased vibrations break some of the weak bonds (hydrogen and ionic bonds) that hold the enzyme’s tertiary structure in place. The shape of the active site changes, so the substrate is no longer complementary and can no longer bind. The enzyme is denatured, meaning the change is permanent, and the rate of reaction falls sharply.
在高于最适温度（人类许多酶的最适温度约为40 °C）时，酶分子振动加剧。这些增强的振动破坏了维持酶三级结构的弱键（氢键和离子键）。活性部位的形状发生改变，底物不再与之互补，因此无法结合。酶发生了变性，意味着这种改变是永久性的，反应速率急剧下降。
It is important to distinguish between denaturation and simply lower kinetic energy at low temperatures. At low temperatures, the enzyme and substrate have less kinetic energy, so they move more slowly, resulting in fewer successful collisions per second. However, the enzyme is not denatured, and the activity increases again if the temperature is raised. Denaturation is always linked to a permanent change in the active site shape.
区分变性和在低温下动能较低的现象非常重要。在低温下，酶和底物的动能较小，因此它们移动更慢，导致每秒成功碰撞的次数减少。但此时酶并未变性，如果温度回升，酶的活性也会重新升高。变性总是与活性部位形状的永久性改变有关。
4. Digestion and Absorption of Starch | 淀粉的消化与吸收
Describe the digestion of starch in the human digestive system and explain how the products are absorbed. [5 marks]
描述淀粉在人体消化系统中的消化过程，并解释其产物是如何被吸收的。[5分]
Starch digestion begins in the mouth, where salivary amylase breaks starch down into maltose. The food is swallowed and travels to the small intestine. Here, pancreatic amylase continues the breakdown of any remaining starch into maltose. Maltase, an enzyme in the membranes of the epithelial cells lining the small intestine, hydrolyses maltose into glucose. Glucose is absorbed by active transport into the epithelial cells and then diffuses into the blood capillaries. The small intestine’s villi provide a large surface area, and their walls are one cell thick, which reduces the diffusion distance for efficient absorption.
淀粉的消化从口腔开始，唾液淀粉酶将淀粉分解为麦芽糖。食物被吞咽并进入小肠。在这里，胰淀粉酶继续将残余淀粉分解为麦芽糖。小肠上皮细胞膜上的麦芽糖酶将麦芽糖水解为葡萄糖。葡萄糖通过主动运输被小肠上皮细胞吸收，然后扩散进入毛细血管。小肠绒毛提供了很大的表面积，其壁只有一个细胞的厚度，缩短了扩散距离，从而提高了吸收效率。
Questions on digestion often require you to name specific enzymes, their sites of production and action, and the substrate and product. Ensure you specify that the products of digestion (glucose, amino acids, fatty acids and glycerol) are small and soluble, which allows them to be absorbed into the blood.
有关消化的题目常常要求你列出具体酶的名称、产生部位和作用部位，以及底物和产物。务必说明消化的产物（葡萄糖、氨基酸、脂肪酸和甘油）是小分子且可溶的，这使得它们能够被吸收进入血液。
5. Limiting Factors in Photosynthesis | 光合作用的限制因素
A student investigates the effect of light intensity on the rate of photosynthesis in pondweed. The graph of oxygen produced per minute against light intensity shows the rate rising quickly at first and then levelling off. At the plateau, the rate does not increase even when light intensity is increased further. Explain why. [3 marks]
一个学生研究了光强度对水草光合作用速率的影响。以每分钟产生的氧气量对光强度作图，曲线起初快速上升，随后趋于平缓。在平台期，即使继续增加光强度，光合速率也不再提高。请解释原因。[3分]
As light intensity increases, the rate of photosynthesis rises because more energy is available for the light-dependent reactions. Eventually, another factor becomes limiting – usually carbon dioxide concentration or temperature. At the plateau, the enzymes involved in the Calvin cycle (or the light-independent reactions) are working at their maximum capacity, and the rate cannot increase unless the limiting factor is raised. If CO₂ concentration is increased at the plateau, the rate would rise again, provided temperature is not limiting.
随着光强度的增加，光合速率上升，因为更多的能量可用于光反应。最终，另一种因素成为限制因素——通常是二氧化碳浓度或温度。在平台期，参与卡尔文循环（或暗反应）的酶已处于最大工作效率，除非该限制因素增加，否则速率无法提高。如果在平台期增加二氧化碳浓度，只要温度不是限制因素，速率会再次上升。
You may be asked to interpret graphs with several limiting factors. Remember the principle: at any given time, the rate is limited by the factor that is in shortest supply. When a graph plateaus, identify which factor is now in shortest supply and explain how you could overcome it experimentally.
你可能会遇到要求解读多个限制因素曲线图的题目。请记住这一原则：在任何时候，反应速率都受最缺乏的那个因素的限制。当曲线趋于平缓时，找出此时最缺乏的因素是什么，并解释你可以如何通过实验克服这一限制。
6. Aerobic and Anaerobic Respiration | 有氧呼吸与无氧呼吸
Compare the processes of aerobic and anaerobic respiration in animal cells. In your answer you should refer to the products, the amount of energy released and the site of the reactions. [4 marks]
比较动物细胞中有氧呼吸和无氧呼吸的过程。你的回答应涉及产物、释放的能量量和反应发生的部位。[4分]
Aerobic respiration uses oxygen and fully oxidises glucose to carbon dioxide and water. It takes place in the mitochondria and releases a large amount of energy (approximately 32 ATP molecules per glucose). Anaerobic respiration occurs when oxygen is absent or in short supply. In animal cells, glucose is partially broken down in the cytoplasm to produce lactic acid and a small amount of energy (only 2 ATP molecules). Anaerobic respiration does not produce carbon dioxide and results in an oxygen debt that must be repaid by oxidising lactic acid back to pyruvate or glucose when oxygen is available.
有氧呼吸需要利用氧气，将葡萄糖彻底氧化成二氧化碳和水。此过程发生在线粒体中，释放大量的能量（每分子葡萄糖约产生32个ATP）。无氧呼吸在缺氧或氧气供应不足时发生。在动物细胞中，葡萄糖在细胞质中被不完全分解，产生乳酸和少量能量（仅2个ATP）。无氧呼吸不产生二氧化碳，并导致氧债，待氧气充足时需将乳酸氧化回丙酮酸或葡萄糖来偿还氧债。
When comparing two processes, use a clear structure: state the conditions, location, products and energy yield for each, and then contrast them. In plant and yeast cells, anaerobic respiration produces ethanol and carbon dioxide instead of lactic acid; this is a common point of confusion, so check the context carefully.
在比较两个过程时，请使用清晰的结构：分别陈述每个过程的条件、部位、产物和能量产量，然后进行对比。在植物和酵母细胞中，无氧呼吸产生乙醇和二氧化碳，而不是乳酸；这是常见的混淆点，因此要仔细审题。
7. Reflex Arc and Synapses | 反射弧与突触
A person accidentally touches a hot pan and immediately pulls their hand away. Describe the nervous pathway involved in this reflex action. Explain the role of a synapse in this response. [5 marks]
某人偶然碰到热锅，立即把手缩回。请描述这一反射活动中涉及的神经通路。解释突触在此反应中的作用。[5分]
The stimulus (heat) is detected by receptor cells in the skin. A sensory neurone carries the impulse from the receptors to the spinal cord. In the spinal cord, the sensory neurone passes the impulse to a relay neurone via a synapse. The relay neurone then transmits the impulse to a motor neurone, again across a synapse. The motor neurone carries the impulse to the effector (the biceps muscle), which contracts to pull the hand away. A synapse ensures the impulse travels in one direction only because neurotransmitter is released from the presynaptic neurone and binds to receptors on the postsynaptic neurone; it also allows coordination between multiple neurones.
刺激（热量）被皮肤中的感受器细胞检测到。感觉神经元将神经冲动从感受器传向脊髓。在脊髓中，感觉神经元通过突触将神经冲动传递给中间神经元。中间神经元再通过另一个突触将冲动传递给运动神经元。运动神经元将冲动传至效应器（肱二头肌），肌肉收缩将手拉回。突触确保神经冲动只能单向传递，因为神经递质由突触前神经元释放，与突触后神经元上的受体结合；此外，突触还能协调多个神经元间的信号。
Many candidates lose marks by forgetting to mention the synapse’s role in unidirectionality. You should also be able to explain that a synapse slows down the impulse slightly due to the time taken for neurotransmitter diffusion, but this allows accurate coordination.
很多考生因忘记提及突触在单向传递中的作用而失分。你还应能解释，由于神经递质扩散需要时间，突触会略微减缓神经冲动传递的速度，但这有助于实现精确的协调。
8. Control of Blood Glucose | 血糖浓度的调节
Explain how the hormones insulin and glucagon work together to maintain a constant blood glucose concentration when a person eats a meal rich in carbohydrates. [6 marks]
解释当一个人吃下一顿富含碳水化合物的食物后，激素胰岛素和胰高血糖素如何共同作用以维持血糖浓度的恒定。[6分]
After a carbohydrate-rich meal, glucose is absorbed from the small intestine into the blood, causing blood glucose concentration to rise. The pancreas detects this rise and the beta cells in the islets of Langerhans secrete insulin into the blood. Insulin stimulates the liver and muscle cells to take up glucose and convert it into glycogen for storage. It also promotes the use of glucose in respiration. As a result, blood glucose falls back to normal. If blood glucose falls below normal, the alpha cells in the pancreas secrete glucagon. Glucagon stimulates the liver to break down glycogen back into glucose and release it into the blood. This negative feedback loop keeps the concentration within a narrow range.
吃完富含碳水化合物的食物后，葡萄糖从小肠吸收进入血液，导致血糖浓度升高。胰腺检测到这一升高，胰岛中的β细胞分泌胰岛素进入血液。胰岛素刺激肝细胞和肌肉细胞摄取葡萄糖，并将其转化为糖原储存起来。胰岛素还促进葡萄糖在呼吸作用中的利用。于是血糖浓度回落到正常水平。如果血糖降至正常值以下，胰腺中的α细胞分泌胰高血糖素。胰高血糖素刺激肝脏将糖原分解为葡萄糖并释放入血。这一负反馈循环使血糖浓度维持在狭窄范围内。
s) → hormonal response → effect on blood glucose → return to normal. Mentioning that glycogen is a storage polymer in liver and muscles will also gain credit.
在 AQA 的扩展题中，你必须展示对负反馈的理解。始终按照以下顺序描述：刺激→感受器（胰腺）→激素响应→对血糖的影响→恢复至正常水平。提到糖原是肝脏和肌肉中的储存多糖也能得分。
9. DNA, Genes and Inheritance of Cystic Fibrosis | DNA、基因与囊性纤维化的遗传
Cystic fibrosis is a recessive genetic disorder caused by a faulty allele of the CFTR gene. Two parents who do not have the disorder have a child with cystic fibrosis. Using a genetic diagram, explain how this is possible and state the probability of their next child having cystic fibrosis. [5 marks]
囊性纤维化是一种隐性遗传病，由 CFTR 基因的一个缺陷等位基因引起。一对没有患病的父母生出了一个患有囊性纤维化的孩子。请用遗传图解说明这如何可能，并说明他们下一个孩子患病的概率。[5分]
Since cystic fibrosis is recessive, a person must inherit two copies of the faulty allele (ff) to have the disorder. The parents are phenotypically normal, so they could be either homozygous dominant (FF) or heterozygous (Ff). Because they produced an affected child (ff), both parents must be carriers – they have the genotype Ff. A Punnett square shows the possible combinations of alleles their children could inherit: FF, Ff, fF and ff. Therefore, there is a 1 in 4 (0.25 or 25%) chance that any child will be homozygous ff and develop cystic fibrosis. The probability remains 25% for each pregnancy, regardless of previous children.
由于囊性纤维化是隐性遗传病，一个人必须遗传到两个缺陷等位基因（ff）才会患病。父母表型正常，所以他们可能是纯合显性（FF）或杂合子（Ff）。因为他们已经生出一个患病孩子（ff），那么父母双方必定都是携带者，其基因型均为 Ff。庞纳特方格显示他们孩子可能获得的等位基因组合为：FF、Ff、fF 和 ff。因此，任何一个孩子为纯合 ff 并患病的概率是四分之一（0.25 或 25%）。每一次怀孕的患病概率仍然是 25%，与之前是否已有患病子女无关。
Genetic diagrams must be clearly labelled with parental genotypes, gametes, and offspring genotypes and phenotypes. Do not forget to state the probability explicitly in the format requested (fraction, decimal or percentage). It is also wise to note that being a carrier (Ff) does not affect the person’s health but still passes on the faulty allele.
遗传图解必须清楚标出亲本基因型、配子、以及子代的基因型和表型。不要忘记明确陈述概率，并按照题目要求的格式（分数、小数或百分数）表示。还应指出，作为携带者（Ff）并不影响个人健康，但仍会传递缺陷等位基因。
10. Natural Selection and Antibiotic Resistance | 自然选择与抗生素耐药性
Explain how the use of antibiotics has led to the development of antibiotic-resistant strains of bacteria, such as MRSA. Use the theory of natural selection in your answer. [5 marks]
解释抗生素的使用如何导致了耐抗生素细菌菌株（如 MRSA）的产生。在你的回答中运用自然选择理论。[5分]
Within a bacterial population, there is genetic variation caused by random mutations. Some of these mutations give certain bacteria resistance to a specific antibiotic. When antibiotics are used, the non-resistant bacteria are killed, but the resistant ones survive. These resistant bacteria reproduce rapidly by binary fission, passing on the resistance allele to their offspring. Over time, the frequency of the resistance allele in the population increases, and the strain becomes largely resistant. This is natural selection: the antibiotic acts as the selection pressure, and the resistant bacteria have a selective advantage, meaning they are better adapted to the environment and more likely to survive and reproduce.
在细菌种群中存在由随机突变引起的遗传变异。其中一些突变使某些细菌对特定抗生素产生耐药性。当使用抗生素时，不具耐药性的细菌被杀死，但耐药菌存活下来。这些耐药菌通过二分裂快速繁殖，将耐药等位基因传递给后代。随着时间的推移，该种群中耐药等位基因的频率增加，菌株变得大部分耐药。这就是自然选择：抗生素充当选择压力，而耐药菌拥有选择优势，即它们更适应环境，更有可能存活和繁殖。
heritance of advantageous allele → change in population. Avoid Lamarckian language such as ‘bacteria become immune’ or ‘they develop resistance because they need it’.
为了获得满分，你必须使用正确的顺序：变异→选择压力（抗生素）→差异性存活→繁殖→优势等位基因的遗传→种群变化。避免使用拉马克式语言，例如“细菌变免疫了”或“因为它们需要耐药性所以就产生耐药性”。
11. Carbon Cycle and Decomposition | 碳循环与分解作用
Describe how carbon is cycled between the atmosphere, plants and animals. Explain the role of microorganisms in this cycle. [6 marks]
描述碳在大气、植物和动物之间如何循环。解释微生物在这个循环中的作用。[6分]
Carbon dioxide in the atmosphere is taken in by plants during photosynthesis, in which carbon is incorporated into glucose and other organic compounds. Plants use these compounds to build biomass. Animals obtain carbon by feeding on plants (or on other animals), digesting the organic molecules and using them for growth and respiration. Both plants and animals return carbon dioxide to the atmosphere through aerobic respiration. When organisms die, decomposers (bacteria and fungi) break down their dead remains and waste products. During decomposition, these microorganisms respire, releasing carbon dioxide back into the atmosphere. If conditions are anaerobic, decomposition may be incomplete, and fossil fuels can be formed, locking carbon away for millions of years. Burning fossil fuels (combustion) returns this stored carbon to the atmosphere as carbon dioxide.
大气中的二氧化碳通过光合作用被植物吸收，碳以葡萄糖和其他有机化合物的形式被固定下来。植物利用这些化合物构建生物量。动物通过摄食植物（或其他动物）获得碳，消化有机分子并将其用于生长和呼吸。植物和动物都通过有氧呼吸将二氧化碳释放回大气。当生物体死亡后，分解者（细菌和真菌）分解它们的遗体和废物。在分解过程中，这些微生物进行呼吸作用，将二氧化碳释放回大气。如果处于缺氧条件下，分解可能不彻底，从而形成化石燃料，将碳封存数百万年。化石燃料的燃烧（即燃烧作用）会将这些储存的碳以二氧化碳的形式返回到大气中。
You often need to close the loop by stating that the carbon cycle maintains the balance of carbon dioxide in the atmosphere and that the processes of photosynthesis and respiration are the main drivers. Remember that microorganisms are crucial in returning nutrients to the soil and carbon to the air, making them essential for ecosystem stability.
通常你需要通过说明碳循环维持大气中二氧化碳的平衡来形成闭环，并指出光合作用和呼吸作用是主要的驱动过程。请记住，微生物在将营养物归还土壤、将碳归还大气方面至关重要，这使它们对于生态系统的稳定不可或缺。
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