Year 7 Edexcel Biology: Cross-curricular Applied Questions Training | 跨学科综合题型训练

📚 Year 7 Edexcel Biology: Cross-curricular Applied Questions Training | 跨学科综合题型训练

Biology in Year 7 goes beyond memorising facts—it often requires you to use skills from mathematics, physics, chemistry and geography to solve real-world problems. This article guides you through common types of integrated questions, showing how to connect knowledge from different subjects and improve your exam technique.

七年级的生物学习不仅仅是记忆事实——它时常要求你运用数学、物理、化学和地理等多学科技能解决真实世界的问题。本文将带你梳理常见的综合题型，展示如何将不同学科的知识联系起来，提升你的应试技巧。

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1. Microscopy Calculations: Linking Biology and Mathematics | 显微镜计算：生物学与数学的结合

A typical question will ask you to calculate total magnification or the real size of a cell using a scale bar. This requires you to multiply and divide confidently, and to convert between units such as millimetres (mm) and micrometres (µm).

典型的题目会要求你计算总放大倍数或利用比例尺求出细胞的真实大小。这需要你熟练地乘除，并在毫米（mm）和微米（µm）等单位之间进行转换。

The total magnification of a light microscope is found using: total magnification = eyepiece magnification × objective magnification.

光学显微镜的总放大倍数计算公式为：总放大倍数 = 目镜放大倍数 × 物镜放大倍数。

Example 1: If the eyepiece lens is ×10 and the objective lens is ×40, the total magnification is 10 × 40 = ×400.

例题1： 如果目镜是10×，物镜是40×，总放大倍数为10 × 40 = 400×。

To calculate the actual size of a specimen seen under the microscope, use: actual size = image size ÷ magnification. Always check that the units are the same before dividing.

计算在显微镜下看到的标本实际大小时，使用公式：实际大小 = 图像大小 ÷ 放大倍数。在相除之前务必确保单位一致。

Practice question: A student draws a cell that measures 60 mm across at a magnification of ×300. What is the real diameter of the cell in micrometres? (Remember: 1 mm = 1000 µm)

练习题： 某学生在300×放大倍数下绘制了一个直径为60 mm的细胞。该细胞的真实直径是多少微米？（提示：1 mm = 1000 µm）

Solution: Real size = 60 mm ÷ 300 = 0.2 mm = 200 µm. This type of maths frequently appears in Edexcel papers.

解答：实际大小 = 60 mm ÷ 300 = 0.2 mm = 200 µm。这类数学计算在Edexcel试卷中经常出现。

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2. Energy Flow in Food Chains: A Physics Perspective | 食物链中的能量流动：物理学视角

Food chains show the transfer of energy from one organism to another. The arrows in a chain represent the direction of energy flow, not who eats whom. Understanding that energy is passed along but lost as heat helps you answer questions correctly.

食物链展示能量从一个生物体传递到另一个生物体。链中的箭头表示能量流动的方向，而不是谁吃谁。理解能量在传递过程中会以热的形式损耗，有助于你正确作答。

In every food chain, the producer (often a green plant) traps light energy from the Sun during photosynthesis. When a primary consumer eats the plant, only a fraction of the energy is transferred—much is lost through movement, respiration and waste.

在每条食物链中，生产者（通常是绿色植物）通过光合作用捕获来自太阳的光能。当初级消费者吃植物时，只有一小部分能量被传递——大量能量通过运动、呼吸和排泄损耗了。

Exam-style question: Explain why a food chain rarely has more than four or five trophic levels.

考试型问题： 解释为什么一条食物链很少超过四个或五个营养级。

Model answer: There is insufficient energy left at the higher levels to support another layer of consumers because most energy is lost to the surroundings as heat.

标准答案：由于大部分能量以热的形式散失到周围环境中，在较高营养级剩余的能量不足以再支撑一个消费者层级。

Here you are using the physics principle of energy conservation and dissipation while discussing a biological concept.

这里，你在讨论生物学概念的同时，也在运用物理学中能量守恒与耗散的原理。

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3. Photosynthesis Equation: Chemistry in Plants | 光合作用方程式：植物的化学

Photosynthesis is a chemical reaction that takes place inside chloroplasts. You need to know the word equation and be able to use chemical symbols as a way to reinforce the link with chemistry.

光合作用是在叶绿体内发生的一种化学反应。你需要掌握其文字方程式，并能使用化学符号进行表达，以强化与化学学科的联系。

The balanced word equation for photosynthesis is:

光合作用的文字方程式为：

carbon dioxide + water → glucose + oxygen

二氧化碳 + 水 → 葡萄糖 + 氧气

Using chemical formulas, this becomes:

使用化学式表示为：

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Notice that light energy and chlorophyll are required but are not used up in the reaction—this mirrors the concept of a catalyst in chemistry.

请注意，光能和叶绿素是必需的，但它们并未在反应中被消耗——这与化学中催化剂的概念相似。

Integrated question: A plant produces 180 g of glucose. Calculate the mass of carbon dioxide used, given that the relative formula masses are: CO₂ = 44, C₆H₁₂O₆ = 180.

综合题： 某植物产生了180 g葡萄糖。计算所消耗的二氧化碳质量，已知相对分子质量：CO₂ = 44，C₆H₁₂O₆ = 180。

Method: From the equation, 6CO₂ (mass 6×44=264) yield 1 glucose (180). So 180 g glucose needs 264 g CO₂. This connects chemistry calculations directly with plant biology.

解法：根据方程式，6CO₂（质量6×44=264）生成1个葡萄糖（180）。因此180 g葡萄糖需要264 g CO₂。这直接将化学计算与植物生物学联系起来。

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4. Levers in the Human Arm: Biology Meets Physics | 人体手臂中的杠杆：生物学与物理学的碰撞

Your musculoskeletal system works according to the laws of physics. The action of the biceps muscle lifting the forearm is a classic example of a third-class lever, where the effort is between the fulcrum and the load.

你的肌肉骨骼系统根据物理定律运作。肱二头肌提起前臂的动作就是第三类杠杆的经典例子：施力点位于支点和负载之间。

In the arm, the elbow joint acts as the fulcrum, the biceps muscle provides the effort, and the weight in the hand is the load. This arrangement allows a large range of movement but requires more force to lift a load.

在手臂中，肘关节充当支点，肱二头肌提供动力，手中的重量是负载。这种布局允许较大幅度的运动，但提升负载时需要更大的力。

Cross-curricular question: Explain, using the lever principle, why it is easier to lift a heavy object if you keep it close to your body.

跨学科问题： 利用杠杆原理解释，为什么将重物靠近身体时更容易提起。

When the load is closer to the fulcrum (elbow), the distance from fulcrum to load decreases, reducing the moment of the load. The same effort from the biceps therefore produces a larger turning effect, making the lift feel easier. This is a direct application of moments in physics.

当负载靠近支点（肘关节）时，支点到负载的距离减小，负载的力矩变小。因此，肱二头肌施以相同的力能产生更大的转动效果，使提起动作感觉更轻松。这是物理中力矩的直接应用。

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5. Breathing Mechanism and Gas Pressure | 呼吸机制与气体压强

Inhalation and exhalation rely on changing the volume of the chest cavity to create pressure differences—exactly the physics concept of pressure and volume in gases.

吸气和呼气依赖于改变胸腔容积以产生压力差——这正是气体压强与体积的物理学概念。

When the diaphragm contracts and moves downwards, the volume inside the thorax increases. According to the gas laws, increasing volume decreases pressure. The pressure inside the lungs becomes lower than atmospheric pressure, so air rushes in.

当横膈膜收缩并向下移动时，胸腔内的容积增大。根据气体定律，体积增大导致压力降低。肺内气压变得低于大气压，于是空气涌入。

Exhalation occurs when the diaphragm relaxes, the rib cage falls, and the volume decreases, raising the pressure and pushing air out. This is essentially Boyle’s law in action inside your body.

呼气时，横膈膜放松，肋骨下降，容积减小，压力升高，将空气推出。这本质上就是波义耳定律在你体内的实际运用。

Typical Year 7 question: Describe what happens to the diaphragm and rib cage when we breathe in, and explain why air enters the lungs.

七年级典型问题： 描述我们吸气时横膈膜和肋骨的变化，并解释为什么空气会进入肺部。

Answer should mention the increase in volume, decrease in pressure, and the movement of air from higher to lower pressure. Connecting physics with gas exchange systems makes your explanation more precise.

答案应提到容积增大、压力降低，以及空气从高压区向低压区的流动。将物理学与气体交换系统联系起来，能使你的解释更加精确。

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6. How We Sense the World: Light and Sound Physics | 我们如何感知世界：光与声的物理学

The eye and the ear are sophisticated biological sensors that respond to physical stimuli—light waves and sound waves. Applying your physics knowledge of how these waves behave helps you understand vision and hearing.

眼睛和耳朵是精密的生物传感器，它们对物理刺激——光波和声波——作出反应。运用你在物理中学到的关于这些波的行为的知识，有助于理解视觉和听觉。

In the eye, the cornea and lens refract (bend) light rays to focus them on the retina. This process follows the rules of refraction you study in physics. The retina converts light energy into electrical impulses, which the brain interprets as images.

在眼中，角膜和晶状体折射（弯曲）光线，使其聚焦在视网膜上。这一过程遵循你在物理中学到的折射规律。视网膜将光能转化为电脉冲，大脑将其解读为图象。

Similarly, sound waves are collected by the outer ear and strike the eardrum, causing it to vibrate. These vibrations are then amplified and transmitted to the inner ear. Understanding the pitch and loudness of sound in physics helps you appreciate how the ear detects different sounds.

类似地，声波被外耳收集并撞击鼓膜，使其振动。这些振动随后被放大并传至内耳。理解物理中声音的音高和响度，能帮你领悟耳朵如何探测不同的声音。

Integrated task: Draw a ray diagram to show how the lens in the eye forms an image on the retina. Explain whether the image is real or virtual. This blends optics from physics with the biology of the eye.

综合任务： 绘制光路图，展示眼中的晶状体如何在视网膜上成像。解释该像是实像还是虚像。这将物理的光学知识与眼睛的生物学结合起来。

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7. Pollination and Climate: Geography in Action | 传粉与气候：地理学在行动

Different plants have flowers adapted for wind or insect pollination. The structure of these flowers is closely linked to the climate and geographical location in which the plant lives.

不同的植物有适应风媒或虫媒传粉的花朵。这些花的结构与植物所处的气候和地理位置密切相关。

Wind-pollinated flowers, such as grasses, often have small, dull petals and large, exposed stamens that shed pollen easily into the air. These are common in open, windy habitats, like grasslands and temperate regions where insect populations may be lower.

风媒花（如禾本科植物）通常花瓣小而暗淡，雄蕊大而外露，容易将花粉散发到空气中。这类花在开阔多风的环境中常见，例如草原和昆虫较少的温带地区。

Insect-pollinated flowers tend to have large, brightly coloured petals, scent and nectar to attract bees or butterflies. Such flowers thrive in warmer, tropical climates where insect activity is high throughout the year, highlighting a clear link with geography and climate.

虫媒花往往具有大型、鲜艳的花瓣、香气和花蜜，以吸引蜜蜂或蝴蝶。这类花在昆虫全年活跃的温暖热带气候中生长繁盛，突显了与地理和气候的明显联系。

Exam connection: You could be asked to suggest why a plant from a windy coastal area is likely to be wind-pollinated. Use geographical reasoning about wind speed and insect scarcity.

考试联系： 你可能会被问到，为什么一个来自多风海岸地区的植物很可能是风媒传粉的。应从风速和昆虫稀少的地理角度进行推理。

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8. Interpreting Biological Data: Graphs and Charts | 解读生物学数据：图表技能

Biology assessments often include tables, line graphs, bar charts and scatter diagrams. Your mathematical skills in reading data, calculating means and identifying trends are essential.

生物学测评经常包含表格、折线图、条形图和散点图。你读取数据、计算平均值和识别趋势的数学技能至关重要。

For example, an experiment on seed germination at different temperatures might produce a bar chart. You need to accurately label axes, choose an appropriate scale, and describe the pattern. The ability to distinguish between continuous and categoric variables comes from your mathematics training.

例如，不同温度下种子萌发的实验可能产生一个条形图。你需要准确标注坐标轴、选择合适的刻度，并描述变化规律。区分连续变量与分类变量的能力来自你的数学训练。

Practice scenario: The table below shows the number of dandelions counted in a 1 m² quadrat on different days in April.

练习情境： 下表显示了四月不同日期在1 m²样方中统计的蒲公英数量。

Day	Number of dandelions
日期	蒲公英数量
1	5
8	12
15	18
22	20
29	22

Plot the data on a line graph. Describe the relationship between day and dandelion numbers, and suggest a biological reason for the trend, such as increasing day length or temperature. This task draws on geographical awareness of seasons and mathematical graphing.

把这些数据绘制成折线图。描述天数与蒲公英数量之间的关系，并推测造成这一趋势的生物学原因，如日照时间变长或温度升高。这项任务利用了季节变动的地理意识和数学作图能力。

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9. Inheritance Patterns: Predicting with Probability | 遗传模式：用概率进行预测

Simple monohybrid inheritance can be modelled using Punnett squares, which are essentially probability grids. Year 7 students often face questions about the likelihood of inheriting certain characteristics.

简单的单因子遗传可以用庞尼特方格进行建模，这些方格本质上是概率网格。七年级学生常常遇到关于遗传某种特征的可能性的问题。

Consider the trait of tongue rolling, where rolling (R) is dominant over non-rolling (r). When two heterozygous parents (Rr) are crossed, the possible genotypes of offspring are RR, Rr, and rr.

以卷舌为例，卷舌（R）对非卷舌（r）为显性。当两个杂合子亲本（Rr）杂交时，子代可能的基因型为RR、Rr和rr。

The Punnett square shows the ratio 1 RR : 2 Rr : 1 rr, giving a 3 in 4 (75%) chance of a roller and a 1 in 4 (25%) chance of a non-roller. This is a direct use of mathematical probability.

庞尼特方格显示比例为1 RR : 2 Rr : 1 rr，因而卷舌者的概率为3/4（75%），非卷舌者的概率为1/4（25%）。这是对数学概率的直接运用。

Exam-type question: A couple both have the genotype Rr. They have four children. Does this guarantee that exactly three will be rollers? Explain, linking your answer to probability and the fact that each fertilisation is an independent event.

考试型问题： 一对夫妇的基因型均为Rr。他们生了四个孩子。这是否保证恰好有三个孩子会卷舌？请解释，并将你的答案与概率和每次受精是独立事件这一事实联系起来。

Answer: No, because each child has an independent 75% chance; the 3:1 ratio is a theoretical expectation for a large number of offspring, not an exact prediction for a small family. This reasoning comes from understanding probability theory.

答案：不能，因为每个孩子都有独立的75%几率；3:1的比例是对大量后代的理论预期，而不是对小家庭的确切预测。这一推理来源于对概率论的理解。

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10. Adaptations to Environment: Geography and Survival | 环境适应：地理学与生存

Organisms develop structural, behavioural or physiological adaptations that help them survive in their habitats. Many questions require you to explain how a feature is suited to a particular climate or terrain, which draws on geographical knowledge.

生物体演化出结构、行为或生理上的适应，以帮助它们在栖息地生存。许多问题要求你解释某个特征如何适应特定的气候或地形，这就借助了地理知识。

Think of a camel living in a hot desert. It has long eyelashes to keep out sand, wide feet to stop sinking into the sand, and the ability to store fat in its hump as a metabolic water source. These adaptations only make sense when you appreciate the extreme temperatures and scarcity of water in a desert biome.

想象生活在炎热沙漠中的骆驼。它长有长睫毛阻挡风沙，宽阔的脚防止陷入沙中，并能将脂肪储存在驼峰中作为代谢水源。只有当你了解沙漠生物群系的极端温度和水分稀缺后，这些适应特性才讲得通。

Similarly, a polar bear’s thick white fur provides camouflage and insulation. Explaining this involves understanding the climate of the Arctic and how white fur reduces heat loss by radiation—a concept linked to physics and geography.

同样，北极熊厚厚的白毛提供伪装和隔热。解释这一点需要了解北极的气候，以及白毛如何通过辐射减少热量损失——这是与物理和地理相关的概念。

Cross-curricular task: Choose an organism from a tropical rainforest and from a tundra. For each, list three adaptations and explain how they are linked to the temperature, rainfall and seasonal light of that biome. This exercise combines biology, geography, and even some earth science.

跨学科任务： 选择一种热带雨林生物和一种冻原生物。分别为每一种列出三项适应特征，并解释它们如何与该生物群系的温度、降雨量和季节性光照联系起来。这项练习结合了生物学、地理学，甚至一些地球科学。

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