📚 Year 8 WJEC Biology: Interdisciplinary Integrated Question Practice | Year 8 WJEC 生物：跨学科综合题型训练
In Year 8 WJEC Biology, you will often meet questions that combine biology with ideas from chemistry, physics, maths or geography. These interdisciplinary questions test your ability to link different topics together. This article takes you through ten common types of integrated question, with clear examples and worked answers, so you can feel confident tackling them in tests.
在 Year 8 WJEC 生物中，你经常会遇到结合化学、物理、数学或地理知识的题目。这些跨学科题目考察你联系不同主题的能力。本文带你梳理十种常见的综合题型，配有清晰的例子和解答，让你在考试中应对自如。
1. Biology and Chemistry: Enzymes and pH | 生物与化学：酶与pH值
Enzymes are biological catalysts made of protein. Each enzyme has an active site with a shape that fits a specific substrate. The activity of an enzyme depends heavily on pH, because pH can change the shape of the active site.
酶是由蛋白质构成的生物催化剂。每种酶都有一个活性位点，其形状与特定底物契合。酶的活性很大程度上依赖于pH值，因为pH可以改变活性位点的形状。
Typical question: Explain why the enzyme pepsin works best in the stomach at pH 2, but would not work in the small intestine at pH 8.
典型问题：解释为什么胃蛋白酶在胃中 pH 2 时活性最高，但在小肠 pH 8 时不起作用。
Worked answer: Pepsin’s active site has a shape that is perfectly adapted to acidic conditions. At pH 2, the shape is correct and can bind to protein substrates. If the pH rises to 8, the shape of the active site changes (the enzyme denatures) and the substrate no longer fits. Therefore, the reaction cannot be catalysed.
解答：胃蛋白酶的活性位点形状完全适应酸性条件。在 pH 2 时，形状正常，可与蛋白质底物结合。如果 pH 升至 8，活性位点形状改变（酶变性），底物无法契合，从而无法催化反应。
Many digestive enzymes have different optimum pH values. For instance, salivary amylase works best at neutral pH (around 7), while trypsin in the small intestine prefers alkaline conditions (pH 8–9). Being able to link pH to enzyme structure is a key cross-subject skill.
许多消化酶有不同的最适pH。例如，唾液淀粉酶在中性（约 pH 7）时活性最佳，而小肠中的胰蛋白酶偏好碱性条件（pH 8–9）。将pH与酶结构联系起来是一项关键的跨学科技能。
2. Biology and Physics: Diffusion and Temperature | 生物与物理：扩散与温度
Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration. It is a physical process that does not require energy from the cell. According to the particle theory (from physics), raising the temperature gives particles more kinetic energy, so they move faster and spread out more quickly.
扩散是粒子从高浓度区域向低浓度区域的净移动。这是一个物理过程，不需要细胞提供能量。根据物理学中的粒子理论，升高温度给予粒子更多动能，使它们运动更快、扩散得更快。
Typical question: Suggest why an aquatic animal finds it harder to obtain oxygen from warm water than from cold water. Use ideas about diffusion and particle motion.
典型问题：为什么水生动物在温水里比在冷水里更难获得氧气？请用扩散和粒子运动的知识解释。
Worked answer: In warm water, the gas particles have more kinetic energy, so oxygen diffuses rapidly out of the water. However, the solubility of oxygen decreases as temperature rises, meaning less oxygen is dissolved. Furthermore, the animal’s own body processes speed up, so demand for oxygen increases. Taken together, warm water can stress the animal because less oxygen is available, even though diffusion itself is faster.
解答：在温水中，气体粒子动能更大，氧气迅速从水中扩散逸出。但随着温度升高，氧的溶解度下降，水中溶解的氧气更少。同时，动物体内的代谢加快，对氧的需求增加。因此，虽然扩散本身更快，但可用氧气减少，温水会给水生动物带来压力。
In your examination, you might also see questions linking diffusion to surface area, such as: why do cells have folded membranes? The answer combines surface area ideas with the physics of diffusion rate.
考试中还可能看到将扩散与表面积联系起来的题目，例如：为什么细胞有折叠的膜？答案需结合表面积概念与扩散速率的物理原理。
3. Biology and Mathematics: Surface Area to Volume Ratio | 生物与数学：表面积与体积比
Cells need to exchange substances with their surroundings. The surface area to volume ratio (SA:V) affects how quickly this exchange can happen. As an object gets larger, its volume increases faster than its surface area, so the SA:V ratio decreases.
细胞需要与环境交换物质。表面积与体积比（SA:V）影响交换速度。物体变大时，体积增长快于表面积，因此 SA:V 比下降。
You can calculate SA:V for a cube using the formula:
SA:V = 6s² ÷ s³ = 6/s
可以用公式计算立方体的 SA:V：
SA:V = 6s² ÷ s³ = 6/s
Example: A cube with side length 1 cm has SA = 6 × 1² = 6 cm², V = 1³ = 1 cm³, so SA:V = 6:1. A cube with side 2 cm has SA = 24 cm², V = 8 cm³, giving a ratio of 3:1.
例子：边长1 cm的立方体，SA = 6×1² = 6 cm²，V = 1³ = 1 cm³，因此 SA:V = 6:1。边长2 cm的立方体，SA = 24 cm²，V = 8 cm³，比例为 3:1。
Typical question: Explain why most cells are very small, using calculations of surface area to volume ratio.
典型问题：运用表面积与体积比的计算解释为什么大多数细胞非常小。
Worked answer: A small cell has a high SA:V ratio, meaning plenty of surface area for diffusion of oxygen and nutrients. If a cell grew too large, its SA:V ratio would drop, and diffusion would not be fast enough to supply the centre. The cell could not survive. That is why large organisms are made from many small cells rather than one giant cell.
解答：小细胞具有高 SA:V 比，意味着有足够的表面积用于氧气和营养物质的扩散。如果细胞长得太大，SA:V 比下降，扩散速度不足以供应中心，细胞将无法存活。因此，大型生物由许多小细胞构成，而非一个巨型细胞。
4. Biology and Geography: Food Chains and Energy Flow | 生物与地理：食物链与能量流动
In an ecosystem, energy enters through photosynthesis by producers. When a herbivore eats a plant, only about 10% of the energy is transferred to the next trophic level. The rest is lost as heat, used in respiration, or remains in uneaten parts.
在生态系统中，能量通过生产者的光合作用进入。当植食动物取食植物时，仅约10%的能量传递到下一营养级。其余以热量、呼吸消耗或未食用部分的形式流失。
Typical question: A producer contains 5000 kJ of energy. Use the 10% rule to calculate how much energy reaches the secondary consumer. Explain why food chains rarely have more than four trophic levels.
典型问题：某生产者含有5000 kJ能量。运用10%法则，计算有多少能量到达次级消费者。并解释为什么食物链很少超过四个营养级。
Worked answer: Producer: 5000 kJ → Primary consumer gets 10% = 500 kJ → Secondary consumer gets 10% of 500 kJ = 50 kJ. By the time you reach the tertiary consumer, only 5 kJ remains. With so little energy, it cannot support another level. This links to geography because maps of biomes show that areas with high primary productivity (like rainforests) can support slightly longer food chains, but the 10% rule still limits length.
解答：生产者：5000 kJ → 初级消费者获得10%，即500 kJ → 次级消费者获得500 kJ的10%，即50 kJ。到了三级消费者时仅剩5 kJ。能量太少，无法支撑下一级。这与地理学有关，因为生物群落分布图显示初级生产力高的区域（如雨林）可支撑稍长的食物链，但10%法则仍然限制其长度。
Interdisciplinary questions often ask you to construct pyramids of numbers or biomass and link them to energy efficiency. You may also need to interpret geographical data, such as climate graphs affecting food chains.
跨学科题目常要求你构建数量金字塔或生物量金字塔，并将其与能量效率联系起来。你还可能需要解读地理数据，如气候图表对食物链的影响。
5. Biology and History: The Discovery of Cells | 生物与历史：细胞的发现
The cell is the basic unit of life, but this idea was only established in the 19th century. In 1665, Robert Hooke observed dead cork cells under a simple microscope and named them ‘cells’. Later, Antonie van Leeuwenhoek observed living cells, including bacteria.
细胞是生命的基本单位，但这一观点直到19世纪才确立。1665年，罗伯特·胡克在简易显微镜下观察到死去的软木细胞，并命名为“细胞”。后来，安东尼·范·列文虎克观察到了活细胞，包括细菌。
Typical question: Evaluate how improvements in microscope technology helped scientists develop the cell theory. Include historical examples.
典型问题：评价显微镜技术的改进如何帮助科学家发展细胞学说。请结合历史实例。
Worked answer: Early microscopes had low magnification and blurry images. Hooke could only see cell walls, not internal structures. By the 1830s, improved lenses allowed Schleiden and Schwann to see nuclei and cytoplasm, leading them to propose that all plants and animals are made of cells. This demonstrates that scientific progress often depends on technological advances. Asking ‘how’ a discovery was made brings history into biology.
解答：早期显微镜放大率低、图像模糊。胡克只能看见细胞壁，看不到内部结构。到19世纪30年代，透镜改进使施莱登和施旺能观察到细胞核和细胞质，从而提出所有动植物都由细胞构成。这表明科学进步常依赖技术发展。探究发现的过程将历史融入了生物学。
Understanding the timeline also helps you appreciate that scientific knowledge builds over time, which is a valuable cross-curricular perspective.
了解时间线还有助于你认识到科学知识是逐步积累的，这是一个有价值的跨学科视角。
6. Biology and Technology: Microscopes and Magnification | 生物与技术：显微镜与放大率
A light microscope uses lenses to magnify specimens. The total magnification is calculated by multiplying the eyepiece lens magnification by the objective lens magnification. To work out the real size of a specimen, you use:
Actual size = Image size ÷ Magnification
光学显微镜使用透镜放大标本。总放大率 = 目镜放大率 × 物镜放大率。计算标本实际大小时，使用：
实际大小 = 图像大小 ÷ 放大率
Typical question: A student views an onion cell using a ×10 eyepiece and a ×40 objective lens. The cell image measures 50 mm across. Calculate the actual size of the cell in micrometres (1 mm = 1000 μm).
典型问题：学生使用×10目镜和×40物镜观察洋葱细胞。图像中的细胞宽度为50 mm。计算该细胞的实际大小（微米），1 mm = 1000 μm。
Worked answer: Total magnification = 10 × 40 = 400×. Actual size = 50 mm ÷ 400 = 0.125 mm. Convert to micrometres: 0.125 mm × 1000 = 125 μm. This interdisciplinary problem combines biology, technology and mathematics, and practising unit conversions is essential for accuracy.
解答：总放大率 = 10 × 40 = 400×。实际大小 = 50 mm ÷ 400 = 0.125 mm。转换为微米：0.125 mm × 1000 = 125 μm。这道题融合了生物、技术和数学，单位换算是确保准确的关键。
7. Biology and Health: Nutrition Labels and Energy | 生物与健康：营养标签与能量
Food provides nutrients: carbohydrates, proteins, fats, vitamins, minerals and fibre. Energy content is measured in kilojoules (kJ) or kilocalories (kcal). On nutrition labels, you often see the amount per 100 g or per serving.
食物提供营养素：碳水化合物、蛋白质、脂肪、维生素、矿物质和膳食纤维。能量含量以千焦（kJ）或千卡（kcal）为单位。营养标签上通常标明每100克或每份的含量。
Typical question: A cereal bar label shows: Fat 8 g, Carbohydrate 30 g, Protein 5 g. Given that fat provides about 38 kJ/g and carbohydrate and protein each provide 17 kJ/g, estimate the total energy in the bar.
典型问题：某谷物棒标签显示：脂肪8克，碳水化合物30克，蛋白质5克。已知脂肪约提供38 kJ/g，碳水化合物和蛋白质各提供17 kJ/g，估算该谷物棒的总能量。
Worked answer: Energy from fat = 8 g × 38 kJ/g = 304 kJ. Energy from carbohydrate = 30 g × 17 kJ/g = 510 kJ. Energy from protein = 5 g × 17 kJ/g = 85 kJ. Total = 304 + 510 + 85 = 899 kJ. This type of calculation helps you make informed choices about diet and understand the biology of energy balance.
解答：脂肪供能 = 8 g × 38 kJ/g = 304 kJ。碳水化合物供能 = 30 g × 17 kJ/g = 510 kJ。蛋白质供能 =
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