📚 Interdisciplinary Integrated Question Practice for Year 7 AQA Biology | AQA七年级生物跨学科综合题型训练
In Year 7 AQA Biology, questions are not always limited to pure biology. Examiners often design tasks that link what you have learned in biology with ideas from chemistry, physics and mathematics. This article will help you recognise these cross-curricular links and practise answering integrated questions step by step. By the end, you will feel more confident in applying your knowledge across subjects.
在AQA七年级生物中，题目并不总是纯粹的生物学问题。考官经常设计一些任务，把你学到的生物知识与化学、物理和数学概念联系起来。本文会帮助你识别这些跨学科的联系，并逐步练习回答综合题型。学完后，你会更有信心在不同学科之间应用知识。
1. Cells and Diffusion: Linking Biology with Physics | 细胞与扩散：连接生物与物理
All cells need to take in nutrients and remove waste. This happens through the cell membrane by diffusion, which is the movement of particles from a region of higher concentration to a region of lower concentration. Diffusion does not require energy from the cell.
所有细胞都需要吸收营养并排出废物。这通过细胞膜以扩散的方式实现，扩散是粒子从高浓度区域向低浓度区域移动的过程，不需要细胞提供能量。
In physics we learn that particles in a liquid or gas are in constant random motion. The higher the temperature, the faster the particles move. This means that temperature affects the rate of diffusion – a wonderful example of biology and physics working together.
在物理中我们学到，液体或气体中的粒子处于持续的无规则运动中。温度越高，粒子运动越快。这意味着温度会影响扩散速率——这是生物与物理完美结合的一个例子。
Practice question: A student dropped a purple crystal of potassium permanganate into a beaker of cold water and another identical crystal into a beaker of hot water. Describe what the student would observe and explain the difference.
练习题：一位学生将一粒紫色高锰酸钾晶体放入一杯冷水中，将另一粒完全相同的晶体放入一杯热水中。描述学生将观察到的现象并解释差异。
Answer guide: In hot water the purple colour spreads much faster than in cold water. Because the hot water has a higher temperature, the water particles have more kinetic energy and move faster. This increases the rate of diffusion, so the colour spreads more quickly through the liquid.
答题指导：热水中紫色扩散比冷水中快得多。因为热水温度更高，水分子具有更大的动能，运动更快。这使扩散速率增加，因此颜色在液体中扩散得更快。
2. Photosynthesis and Light Intensity | 光合作用与光强度
Photosynthesis is the process by which green plants use light energy to convert carbon dioxide and water into glucose and oxygen. Light is a form of energy, and in physics we can measure light intensity and investigate how distance from a light source affects it.
光合作用是绿色植物利用光能将二氧化碳和水转化为葡萄糖和氧气的过程。光是一种能量形式，在物理中我们可以测量光强度，并研究距离光源的远近如何影响光强度。
In a common biology experiment, pondweed is used to count oxygen bubbles produced at different distances from a lamp. The data allow us to practise graph-drawing skills and to connect light intensity with the rate of photosynthesis.
在一个常见的生物实验中，我们使用伊乐藻来计数在不同距离的灯光下产生的氧气泡数量。这些数据可以用来练习绘制图表，并将光强度与光合作用速率联系起来。
Table of results:
结果表：
| Distance from lamp (cm) / 灯距离 (cm) | Bubbles per minute / 每分钟气泡数 |
|---|---|
| 10 | 24 |
| 20 | 13 |
| 30 | 7 |
| 40 | 4 |
Question: Plot a line graph of bubble rate against distance. Explain why the number of bubbles decreases as the distance increases. Use ideas from both biology and physics in your explanation.
问题：绘制气泡产生速率与距离关系的折线图。解释为什么气泡数随着距离增加而减少。在解释中运用生物和物理两门学科的观点。
Sample answer: As the distance from the lamp increases, the light intensity reaching the pondweed decreases. This is because light spreads out and appears dimmer further away. In biology, a lower light intensity provides less energy for photosynthesis, so the rate of reaction falls and fewer oxygen bubbles are produced per minute.
示例答案：随着灯距离增加，到达伊乐藻的光强度减弱。这是因为光会扩散开，距离越远光照越暗。在生物学方面，较低的光强度为光合作用提供的能量减少，因此反应速率下降，每分钟产生的氧气泡数量变少。
3. Respiration and the Energy in Food | 呼吸作用与食物中的能量
Respiration is a chemical reaction that releases energy from glucose inside living cells. This energy is used for movement, growth and keeping warm. The energy originally comes from the chemical energy stored in food.
呼吸作用是从活细胞内的葡萄糖中释放能量的化学反应。这些能量用于运动、生长和保持体温。这些能量最初来源于食物中储存的化学能。
alculated using the formula: energy (J) = mass of water (g) × 4.2 × temperature rise (°C).
物理学家可以通过燃烧食物并利用释放的热量加热已知质量的水来测量食物中的能量。传递的能量可以用以下公式计算：能量（焦）= 水的质量（克）× 4.2 × 温度升高（℃）。
Integrated question: A 1 g piece of biscuit was burned and the heat used to warm 20 g of water. The water temperature rose from 20 °C to 50 °C. Calculate the energy released. How does this relate to respiration?
综合题型：燃烧一块1克的饼干，用释放的热量加热20克水。水温从20 °C上升到50 °C。计算释放的能量。这跟呼吸作用有何关系？
Solution: me from the stored chemical energy in the biscuit. During respiration in our body, glucose is broken down to release energy in a controlled way, providing about the same amount of energy for the same mass of glucose.
解答：温度升高 = 50 - 20 = 30 °C。能量 = 20 × 4.2 × 30 = 2520 焦。这些能量来自饼干中储存的化学能。在我们身体的呼吸作用中，葡萄糖被分解，以受控的方式释放能量，相同质量的葡萄糖提供大约相同的能量。
4. Food Chains and Energy Transfer | 食物链与能量传递
A food chain shows how energy passes from one organism to another. Plants convert light energy into chemical energy, but only about 10% of the energy is transferred to the next trophic level. Much of the energy is lost through movement, heat and undigested materials.
食物链展示了能量如何从一个生物体传递到另一个。植物将光能转化为化学能，但只有大约10%的能量传递到下一营养级。大部分能量通过运动、热和未消化的物质而损失。
This ecological rule allows us to use simple percentage calculations from mathematics to estimate energy at each level.
这条生态规律让我们能用数学中的简单百分比计算来估算每一营养级的能量。
Question: A field of grass captured 20 000 kJ of energy from sunlight through photosynthesis. Calculate the energy available to the rabbit that eats the grass and to the fox that eats the rabbit. Assume 10% efficiency at each step.
问题：一片草场通过光合作用从阳光中固定了20 000 kJ能量。计算吃草的兔子，以及吃兔子的狐狸所能获得的能量。假设每一步的效率为10%。
Working: Energy in grass = 20 000 kJ. Energy in rabbit = 20 000 × 10% = 2000 kJ. Energy in fox = 2000 × 10% = 200 kJ. The fox receives only 1% pped by the grass. This explains why most food chains have fewer than five trophic levels.
计算过程：草中的能量 = 20 000 kJ。兔子的能量 = 20 000 × 10% = 2000 kJ。狐狸的能量 = 2000 × 10% = 200 kJ。狐狸只获得了草最初固定能量的1%。这解释了为什么大多数食物链不超过五个营养级。
5. The Skeleton and Levers | 骨骼与杠杆
Your skeleton provides support, but when muscles pull on bones, they also form levers. Physics teaches us that a lever can multiply force or increase speed, depending on the positions of the effort, load and pivot.
你的骨骼提供支撑，而当肌肉牵拉骨骼时，它们也构成了杠杆。物理告诉我们，杠杆可以放大力或者增加速度，这取决于力、负荷和支点的位置。
The forearm works as a third-class lever. The elbow is the pivot, the biceps muscle provides the effort, and the hand holds the load. Because the effort is closer to the pivot than the load, the muscle must produce a larger force to lift a weight.
前臂是第三类杠杆。肘是支点，肱二头肌提供动力，手拿重物。由于动力点比负荷离支点更近，肌肉必须产生更大的力才能提起重物。
Numerical task: Suppose your hand holds a 10 N weight. The distance from the elbow to the hand is 35 cm, and the biceps tendon attaches 5 cm from the elbow. Calculate the force the biceps must provide. (Ignore arm mass.)
数学任务：假设你的手拿着一个10 N的重物。肘到手的距离为35 cm，肱二头肌肌腱附着在离肘5 cm处。计算肱二头肌需要提供多大的力。（忽略手臂质量。）
Solution: rt: effort × 5 cm = 350 N cm, so effort = 70 N. The muscle force is seven times the weight! Biology and physics work together to let us move.
解答：负荷产生的力矩 = 10 N × 35 cm = 350 N cm。这必须由动力平衡：动力 × 5 cm = 350 N cm，所以动力 = 70 N。肌肉力量是重物的七倍！生物与物理协同工作，使我们能够运动。
6. Breathing and Pressure Changes | 呼吸与气压变化
When you breathe in, your diaphragm flattens and your ribcage moves upward and outward. This increases the volume inside your thorax, which causes the pressure inside to drop below atmospheric pressure, drawing air into your lungs.
吸气时，膈肌变平，胸腔向上向外扩张。这增大了胸腔内的容积，导致内部气压低于大气压，于是空气被吸入肺中。
The physics idea here is the relationship between gas volume and pressure: when volume increases, pressure decreases, provided the temperature is constant. This explains the mechanics of breathing without needing a pump.
这里的物理原理是气体体积与压强的关系：在温度不变的条件下，体积增大，压强减小。这就解释了呼吸的力学原理，不需要泵。
Question: A student says, 'Air rushes into my lungs because my lungs are pulling the air in.' Explain why this statement is not quite correct, using scientific ideas about pressure.
问题：一位学生说：“空气冲进我的肺是因为肺把空气拉进来了。”利用有关气压的科学观点，解释这个说法为什么不太正确。
Model answer: The lungs themselves do not actively pull air in. When the diaphragm and intercostal muscles contract, the chest volume increases, reducing the pressure inside the lungs. The higher outside atmospheric pressure then pushes air into the lungs along a pressure gradient. It is a push, not a pull.
参考答案：肺本身并不主动把空气拉进来。当膈肌和肋间肌收缩时，胸腔容积增大，肺内气压降低。较高的外部大气压沿着气压梯度将空气推入肺中。这是一种推力，而不是拉力。
7. Enzymes and Reaction Rates | 酶与反应速率
Enzymes are biological catalysts that speed up chemical reactions in living organisms. Their activity is affected by temperature – another place where biology meets chemistry and physics.
酶是生物催化剂，可加速生物体内的化学反应。它们的活性受温度影响——这是生物与化学和物理相遇的另一个地方。
In the lab, you can measure enzyme activity by recording the time taken for starch to break down at different temperatures. Rates can be calculated as 1/time, a skill borrowed from mathematics.
在实验室中，你可以通过记录不同温度下淀粉分解所需的时间来测量酶活性。速率可以计算为1/时间，这需要用到数学技巧。
| Temperature (°C) 温度(℃) | Time for starch to disappear (s) 淀粉消失时间(秒) |
|---|---|
| 10 | 150 |
| 20 | 75 |
| 30 | 38 |
| 40 | 20 |
| 50 | 45 |
| 60 | 120 |
Task: Calculate the rate (1/time) for each temperature. Plot a graph of rate against temperature. Describe the shape and explain it in terms of enzyme action and particle theory.
任务：计算每个温度下的速率（1/时间）。绘制速率随温度变化的图表。描述图形并运用酶作用和粒子理论解释它。
Explanation: Rate increases up to about 40 °C because particles gain kinetic energy, collide more frequently, and more enzyme-substrate complexes form. Above the optimum temperature, enzymes are denatured and the active site changes shape
Published by TutorHao | Year 7 Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图