📚 Year 8 CAIE Biology: Interdisciplinary Integrated Questions Practice | Year 8 CAIE 生物：跨学科综合题型训练
Biology does not exist in isolation. Whether you are measuring the growth of a plant, modelling the flow of blood, or analysing data from a field study, you draw on ideas from chemistry, physics, mathematics and geography. This article provides a series of integrated question sets designed to help Year 8 CAIE learners make those connections confidently.
生物学并不是孤立存在的。无论是测量植物的生长、模拟血液流动，还是分析野外研究的数据，你都需要借助化学、物理、数学和地理等学科的思想。本文提供了一系列跨学科综合题型训练，帮助 Year 8 CAIE 学习者自信地建立这些联系。
1. Photosynthesis and Chemical Equations | 光合作用与化学方程式
Photosynthesis is the process by which green plants use light energy to convert carbon dioxide and water into glucose and oxygen. The balanced chemical equation is often written as: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. Notice how the number of atoms of each element is the same on both sides – this is a key chemical principle.
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。平衡后的化学方程式通常写作：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂。请注意，每种元素的原子数在两边是相同的——这是一条关键的化学原理。
Integrated question
A potted plant is placed on a balance under a bright lamp. The mass of the system slowly increases over several hours. Explain this observation using the photosynthesis equation and the law of conservation of mass. Which gas contributes most to the mass gain?
综合题目
将一盆盆栽植物放在强光灯下的天平上。几小时内，整个系统的质量缓慢增加。请用光合作用方程式和质量守恒定律解释这一现象。哪种气体对质量增加贡献最大？
Answer: The plant takes in carbon dioxide from the air and incorporates its carbon and oxygen atoms into glucose and other organic molecules. The mass of carbon dioxide fixed is greater than the mass of oxygen released, so the total mass of the plant increases. Carbon dioxide is the main contributor.
答案：植物从空气中吸收二氧化碳，并将其碳原子和氧原子固定到葡萄糖等有机物中。固定的二氧化碳质量大于释放的氧气质量，因此植物的总质量增加。二氧化碳是主要的贡献者。
2. Blood Circulation and Pressure | 血液循环与压力
Blood flows around the body through arteries, capillaries and veins. Physics helps us understand how pressure differences drive this circulation. When the heart contracts, it generates high pressure in the arteries; by the time blood reaches the veins, the pressure has dropped significantly. The relationship between pressure, flow rate and vessel diameter can be explored with simple models.
血液通过动脉、毛细血管和静脉在全身流动。物理学帮助我们理解压力差如何驱动循环。当心脏收缩时，在动脉中产生高压；当血液到达静脉时，压力已大幅下降。我们可以用简单的模型探究压力、流速和血管直径之间的关系。
Integrated question
A student models an artery using a flexible tube connected to a pump. When the tube is partially clamped to simulate narrowing, the pressure upstream rises, while the flow rate downstream falls. Why does this happen in the human body if a coronary artery becomes narrowed by fatty deposits?
综合题目
一名学生用连接水泵的软管来模拟动脉。当软管被部分夹紧以模拟变窄时，上游的压力升高，而下游的流速下降。如果冠状动脉因脂肪沉积而变窄，人体内为什么会出现这种情况？
Answer: A narrower artery increases resistance, so the heart must pump with higher pressure to push blood through. The reduced cross-sectional area also decreases the volume of blood that can flow per minute, limiting oxygen delivery to heart muscle cells. This can cause chest pain.
答案：动脉变窄会增加阻力，因此心脏必须以更高的压力泵血。横截面积减小也会降低每分钟能够流过的血量，从而限制向心肌细胞输送氧气，这可能导致胸痛。
3. Ecosystems and Climate | 生态系统与气候
The distribution of biomes such as tropical rainforests, deserts and tundra is largely determined by temperature and precipitation. Biologists work with geographers to interpret climate graphs and predict which organisms can survive in a given region. Long-term climate data also reveal how shifts in temperature affect species migration and flowering times.
热带雨林、沙漠和冻原等生物群系的分布很大程度上由温度和降水决定。生物学家与地理学家合作，解读气候图表，预测哪些生物能够在特定区域生存。长期气候数据还揭示了温度变化如何影响物种迁徙和开花时间。
Integrated question
Study the climate data below. (Table shows average monthly temperature and rainfall for a location.) The vegetation is dominated by drought-resistant shrubs and succulents. Identify the likely biome. Explain how one adaptation of a named succulent helps it survive in this climate.
综合题目
研究以下气候数据。（表格显示了某地月平均温度和降雨量。）当地植被以耐旱灌木和多肉植物为主。请判断这可能是哪种生物群系，并以一种多肉植物为例，解释它的某项适应如何帮助它在这样的气候中生存。
| Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
| Temp (°C) | 26 | 25 | 23 | 20 | 17 | 15 | 14 | 16 | 18 | 20 | 23 | 25 |
| Rainfall (mm) | 15 | 18 | 22 | 35 | 60 | 85 | 92 | 78 | 50 | 30 | 20 | 14 |
Answer: The location shows warm temperatures year-round and moderate rainfall with a dry summer – this is a Mediterranean biome or a dry shrubland. A succulent like Aloe vera stores water in thick, fleshy leaves, and its waxy cuticle reduces water loss by evaporation during the hot, dry summer months.
答案：该地全年温暖，降雨适中，夏季干燥——这是地中海生物群系或干燥灌木林地。像芦荟这样的多肉植物将水分储存在厚厚的肉质叶片中，其蜡质角质层减少了炎热干燥夏季的水分蒸发。
4. Population Growth Calculations | 种群增长计算
Population size changes when births, deaths, immigration and emigration occur. Biologists use simple equations to calculate population growth rate: r = (birth rate + immigration rate) – (death rate + emigration rate). If r is positive, the population grows. Mathematics skills are essential for plotting growth curves and predicting future numbers.
当出生、死亡、迁入和迁出发生时，种群大小会发生变化。生物学家使用简单的方程来计算种群增长率：r = (出生率 + 迁入率) – (死亡率 + 迁出率)。如果 r 为正，种群就会增长。数学技能对于绘制增长曲线和预测未来数量至关重要。
Integrated question
A population of rabbits on an island has 120 individuals at the start of the year. Over the year, 80 are born, 30 die, 10 rabbits leave and 5 join from a neighbouring island. Calculate the growth rate r and the final population size. Sketch a simple graph showing how the population might change over the next three years if resources remain limited.
综合题目
某岛上一群兔子在年初有 120 只。年内，出生 80 只，死亡 30 只，10 只离开，5 只从邻近岛屿迁入。计算增长率 r 和年终种群数量。如果资源始终有限，画出未来三年种群数量可能如何变化的简图。
Answer: r = (80 + 5) – (30 + 10) = 85 – 40 = 45 individuals. Final number: 120 + 45 = 165. With limited resources, the population may rise initially, then slow as it reaches carrying capacity, giving an S-shaped (sigmoid) curve.
答案：r = (80 + 5) – (30 + 10) = 85 – 40 = 45 只。年终数量：120 + 45 = 165。在资源有限的情况下，种群一开始会上升，然后随着达到环境容纳量而减缓，形成 S 型 (sigmoid) 曲线。
5. Microscopes and Optics | 显微镜与光学
The compound light microscope uses two convex lenses – the eyepiece and the objective – to magnify tiny specimens. Total magnification = eyepiece magnification × objective magnification. Physics optics explains how convex lenses refract light to create enlarged, virtual images. Understanding focal length and resolution helps biologists choose the correct equipment.
复式光学显微镜使用两个凸透镜——目镜和物镜——来放大微小的标本。总放大倍数 = 目镜放大倍数 × 物镜放大倍数。物理光学解释了凸透镜如何折射光线以形成放大的虚像。了解焦距和分辨率有助于生物学家选择合适的设备。
Integrated question
A student uses a 10x eyepiece and a 40x objective to view onion epidermal cells. Calculate the total magnification. If the field of view diameter is 0.4 mm, estimate the length of one cell that spans about 1/8 of the diameter. Suggest why using the 100x oil immersion objective might give a sharper image.
综合题目
一名学生使用 10× 目镜和 40× 物镜观察洋葱表皮细胞。计算总放大倍数。如果视野直径为 0.4 mm，估算一个细胞约占直径的 1/8 时，该细胞的长度。并说明为什么使用 100× 油浸物镜可能会得到更清晰的图像。
Answer: Total magnification = 10 × 40 = 400x. Cell length ≈ 0.4 mm × (1/8) = 0.05 mm = 50 µm. The 100x oil immersion objective has a higher numerical aperture, reducing the loss of light rays due to refraction, which improves resolution and thus sharpness.
答案：总放大倍数 = 10 × 40 = 400×。细胞长度 ≈ 0.4 mm × (1/8) = 0.05 mm = 50 µm。100× 油浸物镜具有更高的数值孔径，减少了因折射造成的光线损失，从而提高了分辨率，因此图像更清晰。
6. Carbon Cycle and Energy Transfer | 碳循环与能量转移
Carbon atoms move between the atmosphere, organisms, oceans and rocks. Photosynthesis fixes carbon dioxide, respiration releases it, and combustion of fossil fuels returns ancient carbon to the air. At each step, energy is transferred – light energy to chemical energy, then to heat energy. This cycle links biology with chemistry and environmental science.
碳原子在大气、生物体、海洋和岩石之间移动。光合作用固定二氧化碳，呼吸作用释放它，而化石燃料的燃烧则将远古的碳放回空气。在每一个步骤中，能量都在转移——光能转变为化学能，再转变为热能。这个循环将生物学与化学和环境科学联系在一起。
Integrated question
Draw a simplified diagram of the carbon cycle including a green plant, a rabbit, a decomposer and a factory. Label the processes: photosynthesis, feeding, respiration, death, decomposition, and combustion. Explain how the energy from the Sun is eventually lost from the ecosystem.
综合题目
画一幅包含绿色植物、兔子、分解者和工厂的碳循环简图。标注过程：光合作用、摄食、呼吸、死亡、分解和燃烧。解释来自太阳的能量最终是如何从生态系统中散失的。
Answer: In the diagram, arrows show carbon moving from air to plant, plant to rabbit, rabbit and plant to decomposer, and decomposer to air; factory burning fuel releases CO₂. Sunlight energy converted to chemical energy in plant tissues is eventually transformed to heat through respiration at each trophic level, and this heat is lost to space.
答案：图中用箭头表示碳从空气到植物、植物到兔子、兔子及植物到分解者、以及分解者到空气的移动；工厂燃烧燃料释放 CO₂。在植物组织中将光能转化而成的化学能，通过每个营养级的呼吸作用最终转化为热能，散失到太空中。
7. Nutrition and Chemical Digestion | 营养与化学消化
Large food molecules – starch, proteins and fats – must be broken down chemically into smaller, soluble units before they can be absorbed. Enzymes such as amylase, proteases and lipases catalyse these hydrolysis reactions. Understanding the pH and temperature optima of these enzymes requires knowledge of both biology and basic chemistry.
大分子的食物——淀粉、蛋白质和脂肪——必须通过化学消化分解为较小的可溶性单位才能被吸收。淀粉酶、蛋白酶和脂肪酶等酶催化这些水解反应。要理解这些酶的最适 pH 和温度，需要生物学和基础化学的知识。
Integrated question
A student mixes amylase solution with starch suspension in a test tube at 37°C. After 10 minutes, the iodine test is negative. State what this tells you about starch digestion. Explain why the same reaction would slow down significantly if the pH were changed from 7 to 2.
综合题目
一名学生将淀粉酶溶液与淀粉悬浊液在 37°C 的试管中混合。10 分钟后，碘液测试结果为阴性。说明这告诉你关于淀粉消化的什么信息。并解释为什么如果将 pH 从 7 调至 2，该反应会明显减慢。
Answer: A negative iodine test means all the starch has been broken down into maltose. Amylase works best at neutral pH (around 7). At pH 2, the enzyme’s active site becomes denatured – its shape is altered – so it can no longer bind to starch molecules, and the rate of reaction decreases sharply.
答案：碘液测试呈阴性意味着所有淀粉都已被分解为麦芽糖。淀粉酶在中性 pH（约 7）时活性最佳。在 pH 2 时，酶的活性部位变性——形状发生改变——因此无法再与淀粉分子结合，反应速率急剧下降。
8. Data Interpretation from Experiments | 实验数据解读
Scientific inquiry often produces numerical data that must be organised in tables and graphs. Biologists look for patterns, calculate means and ranges, and draw conclusions. Interdisciplinarity with mathematics is vital when uncertainties and averages are discussed. A well-constructed graph can reveal relationships that words alone cannot.
科学探究常常产生需要以表格和图表形式整理的数值数据。生物学家寻找规律、计算平均值和范围并得出结论。在讨论不确定性和平均值时，与数学的跨学科联系至关重要。一张构建良好的图表可以揭示仅靠文字无法表达的关系。
Integrated question
An experiment measured the rate of oxygen produced by pondweed at different light intensities. Results:
| Light intensity (lux) | 200 | 400 | 600 | 800 | 1000 |
| Oxygen rate (cm³/min) | 0.5 | 1.0 | 1.4 | 1.6 | 1.7 |
Plot a line graph. Explain why the curve flattens. Using your knowledge of limiting factors, predict what would happen if the concentration of CO₂ were increased while light intensity remained at 1000 lux.
综合题目
一项实验测量了不同光照强度下池塘草产生的氧气速率。结果如上表。绘制折线图。解释曲线为何趋于平缓。运用限制因子的知识，预测如果二氧化碳浓度增加而光照强度维持在 1000 lux 时会发生什么。
Answer: The graph initially rises steeply then plateaus because light is no longer the limiting factor; something else (possibly CO₂ or temperature) now limits photosynthesis. If CO₂ concentration were increased, the rate of oxygen production would rise further until CO₂ itself becomes limiting or another factor intervenes.
答案：图表起初急剧上升然后趋于平缓，因为光照不再是限制因子；别的因素（可能是 CO₂ 或温度）限制了光合作用。如果增加 CO₂ 浓度，氧气产生速率会进一步上升，直到 CO₂ 自身成为限制因子或另一个因素介入。
9. Diffusion and Temperature Effects | 扩散与温度效应
Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration. The kinetic particle model from chemistry tells us that particles have more kinetic energy at higher temperatures, moving faster and spreading out more quickly. This explains why warm conditions speed up biological processes such as gas exchange.
扩散是粒子从较高浓度区域向较低浓度区域的净移动。化学中的动力学粒子模型告诉我们，温度越高，粒子的动能越大，移动速度越快，扩散得也越快。这就解释了为什么温暖的环境会加速气体交换等生物过程。
Integrated question
A drop of food colouring is added to a beaker of cold water (5°C) and another to a beaker of warm water (40°C). The colour spreads throughout the warm water in half the time. Explain these results using particle theory. Relate this to why mammals and birds maintain a high body temperature.
综合题目
将一滴食用色素分别滴入一杯冷水（5°C）和一杯温水（40°C）中。颜色在温水中扩散至整杯所需的时间是冷水的一半。用粒子理论解释这些结果。并联系说明为什么哺乳动物和鸟类需要维持较高的体温。
Answer: At higher temperatures, water molecules move faster and collide more frequently with the dye particles, causing the dye to spread more rapidly. Mammals and birds keep a constant high body temperature, which ensures that reactions such as the diffusion of oxygen into cells occur quickly enough to support their high metabolic rates.
答案：温度较高时，水分子运动更快，与染料粒子碰撞频率更高，使得染料扩散得更快。哺乳动物和鸟类保持恒定的高体温，这样就能保证氧气扩散到细胞等反应足够迅速地发生，以支持它们较高的新陈代谢率。
10. Enzymes and pH | 酶与pH
Every enzyme has an optimum pH at which it works fastest. Pepsin, a protease in the stomach, is most active at pH 2, while trypsin in the small intestine works best at pH 8. The chemical environment directly shapes the shape of the enzyme’s active site through the concentration of hydrogen ions (H⁺). Too acidic or too alkaline conditions denature the enzyme.
每种酶都有一个使其工作最快的最适 pH。胃蛋白酶是一种胃内的蛋白酶，在 pH 2 时活性最高；而小肠中的胰蛋白酶则在 pH 8 时作用最佳。化学环境通过氢离子（H⁺）的浓度直接影响酶活性部位的形状。过酸或过碱的环境都会使酶变性。
Integrated question
A biology student tests the activity of catalase (found in liver) at pH 4, 7 and 10 by measuring the volume of oxygen produced from hydrogen peroxide in one minute. The results: pH 4 – 8 cm³; pH 7 – 22 cm³; pH 10 – 9 cm³. Plot a bar chart. State the optimum pH from this data. Explain why the activity falls at pH 10.
综合题目
一名生物专业的学生在 pH 4、7 和 10 下测试肝脏中过氧化氢酶的活性，测量一分钟内由过氧化氢产生的氧气体积。结果：pH 4 – 8 cm³；pH 7 – 22 cm³；pH 10 – 9 cm³。绘制条形图。根据数据指出最适 pH。解释为何在 pH 10 时活性下降。
Answer: The bar chart shows the highest column at pH 7, so the optimum pH is 7. At pH 10, the alkaline environment alters the bonds that hold the enzyme’s three-dimensional shape; the active site is no longer complementary to the substrate, so fewer enzyme-substrate complexes form and activity decreases.
答案：条形图显示 pH 7 的柱状最高，因此最适 pH 为 7。在 pH 10 时，碱性环境改变了维持酶三维结构的化学键；活性部位不再与底物契合，因此形成的酶-底物复合物减少，活性下降。
11. Interactions in Food Webs | 食物网中的相互作用
Food webs display the feeding relationships between organisms in an ecosystem. They are built on energy flow, which obeys the laws of physics – namely, that energy is neither created nor destroyed, only transferred. At each trophic level, around 90% of energy is lost as heat, explaining why food chains rarely exceed four or five levels. Mathematics can be used to construct pyramids of numbers and biomass.
食物网展示了生态系统中生物之间的摄食关系。它们建立在能量流动的基础上，而能量流动遵循物理定律——即能量既不能被创造也不能被消灭，只能被转移。在每一个营养级，大约有 90% 的能量以热量的形式散失，这就解释了为什么食物链很少超过四到五个层级。数学可以用来构建数量金字塔和生物量金字塔。
Integrated question
A simple wetland food web: algae → water fleas → small fish → heron. If the algae contain 20 000 kJ of energy, estimate the energy available to the heron, assuming a 10% transfer efficiency at each stage. Show your calculations. State one reason why the actual figure might be even lower.
综合题目
一个简单的湿地食物网：藻类 → 水蚤 → 小鱼 → 鹭。假设藻类含有 20 000 kJ 能量，在每个阶段传递效率为 10% 的情况下，估算鹭可获得的能量。展示计算过程。说明实际数值可能更低的一个原因。
Answer: Energy to water fleas = 20 000 × 10% = 2 000 kJ; to small fish = 2 000 × 10% = 200 kJ; to heron = 200 × 10% = 20 kJ. The actual figure could be lower because not all algae are eaten; some energy is used in movement and other life processes; or the heron might expend energy catching the fish.
答案：传递给水蚤的能量 = 20 000 × 10% = 2 000 kJ；传递给小鱼 = 2 000 × 10% = 200 kJ；传递给鹭 = 200 × 10% = 20 kJ。实际数值可能更低，因为并非所有藻类都被取食；部分能量用于运动和其他生命活动；或者鹭在捕鱼时也会消耗能量。
12. Active Transport and Ion Concentrations | 主动运输与离子浓度
Active transport is the movement of molecules or ions across a cell membrane against a concentration gradient, using energy from respiration. This process is vital in root hair cells taking in nitrate ions from the soil, even when soil concentrations are lower. It ties together biology, chemistry (ion gradients) and physics (energy transformation from glucose to ATP).
主动运输是指分子或离子利用呼吸作用提供的能量，逆浓度梯度跨细胞膜移动。这一过程对根毛细胞从土壤中吸收硝酸根离子至关重要，即使土壤中的浓度更低。它将生物学、化学（离子梯度）和物理学（从葡萄糖到 ATP 的能量转换）紧密联系起来。
Integrated question
A root hair cell has a higher concentration of potassium ions (K⁺) inside than in the surrounding soil water. Yet potassium continues to enter. Explain how this is possible. Why does the process stop if the root is deprived of oxygen? Link your answer to the energy source for active transport.
综合题目
根毛细胞内的钾离子（K⁺）浓度高于周围土壤水中的浓度，但钾离子仍然继续进入细胞。解释这如何可能。为什么如果根部缺氧，这个过程就会停止？请将你的答案与主动运输的能量来源联系起来。
Answer: Potassium ions are moved into the cell by active transport through specific carrier proteins. These proteins change shape to pump the ions uphill, using ATP released from respiration. Without oxygen, aerobic respiration stops, ATP supply falls, so the carrier proteins lack the energy to operate, and potassium uptake ceases.
答案：钾离子通过特定载体蛋白以主动运输的方式进入细胞。这些蛋白利用呼吸作用释放的 ATP 改变形状，将离子逆浓度泵入。缺氧时，有氧呼吸停止，ATP 供应下降，载体蛋白缺乏运作所需的能量，钾离子的吸收便停止。
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