📚 Year 7 CAIE Biology: Interdisciplinary Exam Practice | Year 7 CAIE 生物：跨学科综合题型训练
Welcome to an integrated revision session designed to sharpen your problem-solving skills by linking biology with mathematics, chemistry, physics and geography. In real scientific investigations, subject boundaries blur. The following ten worked examples each present a Year‑7 style question, a step‑by‑step answer, and an explanation of how disciplines connect. Use them to strengthen both your biological understanding and your ability to think like a true scientist.
欢迎来到这堂跨学科综合复习课，我们将通过把生物与数学、化学、物理、地理联系起来，锤炼你的解题能力。在真实的科学探究中，学科界限是模糊的。下面十个精心设计的例题，每一题都呈现了一道七年级风格的题目、分步解答以及学科交叉的解说。利用它们来夯实你的生物学理解，同时培养真正科学家的思维习惯。
1. Microscopy and Mathematics | 显微镜与数学
Question: A student uses a light microscope with an eyepiece lens marked ×10 and an objective lens marked ×40. She draws a cheek cell that is 0.05 mm wide and wants to know the actual width of the cell in micrometres (µm). The image width in her drawing is 50 mm. Calculate (a) the total magnification, (b) the actual cell width in µm, and (c) the magnification of her drawing.
问题：一名学生使用一台目镜为×10、物镜为×40的光学显微镜。她绘制了一个口腔上皮细胞，绘图中的细胞宽度是50 mm，而实际细胞宽度为0.05 mm。请计算（a）显微镜的总放大倍数，（b）以微米（µm）为单位的实际细胞宽度，以及（c）她绘图的放大倍数。
Answer (a): Total magnification = eyepiece magnification × objective magnification = 10 × 40 = ×400.
答案（a）：总放大倍数 = 目镜放大倍数 × 物镜放大倍数 = 10 × 40 = ×400。
Answer (b): 1 mm = 1000 µm, so 0.05 mm = 0.05 × 1000 = 50 µm. The actual cell width is 50 µm.
答案（b）：1 mm = 1000 µm，故 0.05 mm = 0.05 × 1000 = 50 µm。细胞实际宽度为 50 µm。
Answer (c): Drawing magnification = drawing size ÷ actual size = 50 mm ÷ 0.05 mm = ×1000. Notice this is much larger than the microscope magnification because the drawing is enlarged further by hand. Always convert units to the same before dividing.
答案（c）：绘图放大倍数 = 绘图尺寸 ÷ 实际尺寸 = 50 mm ÷ 0.05 mm = ×1000。注意这比显微镜放大倍数大得多，因为手工绘图又进行了放大。计算前务必将单位统一。
Cross‑disciplinary link: This question blends biology (microscopy) with mathematics (multiplication, unit conversion, ratio). Being able to switch between millimetres and micrometres is a key numeracy skill that appears in many practical exams.
跨学科联系：这道题融合了生物学（显微镜）与数学（乘法、单位换算、比例）。能在毫米和微米之间灵活转换是一项关键的运算技能，在众多实验考试中都会用到。
2. Photosynthesis and Chemical Testing | 光合作用与化学检测
Question: A leaf from a plant kept in sunlight is boiled in water, then in ethanol, and finally rinsed in warm water. Iodine solution is added. The leaf turns blue‑black. (a) What substance is present in the leaf? (b) Why is the leaf boiled in ethanol? (c) Name a safety precaution when heating ethanol.
问题：一片在阳光下放置的叶片先后经过水煮、乙醇煮，再用温水漂洗。滴加碘液后，叶片变为蓝黑色。（a）叶片中存在什么物质？（b）为什么要用乙醇煮叶片？（c）加热乙醇时需采取什么安全措施？
Answer (a): Starch. Iodine solution turns blue‑black in the presence of starch, confirming photosynthesis has occurred.
答案（a）：淀粉。碘液遇淀粉会变成蓝黑色，从而证实光合作用已经发生。
Answer (b): Boiling in ethanol removes chlorophyll, the green pigment, so the colour change with iodine can be seen clearly. Without this step, the green colour masks the blue‑black colour.
答案（b）：乙醇煮沸可去除叶绿素（绿色色素），这样碘液的颜色变化才能清晰可见。不经此步，绿色会遮盖蓝黑色。
Answer (c): Ethanol is flammable, so it must be heated using a hot water bath (beaker of hot water), never directly over a flame.
答案（c）：乙醇易燃，因此必须使用热水浴（烧杯盛热水）加热，绝不可直接用火焰加热。
Cross‑disciplinary link: This classic biology experiment requires understanding of chemical properties (iodine as an indicator, solubility of chlorophyll in ethanol) and laboratory safety, linking biology with chemistry.
跨学科联系：这个经典的生物实验需要理解化学性质（碘作为指示剂、叶绿素能溶于乙醇）以及实验室安全，将生物与化学紧密结合起来。
3. Food Webs and Geography | 食物网与地理
Question: A tropical rainforest food web contains trees → beetles → frogs → snakes → hawks. (a) Draw this as a food chain. (b) Explain why rainforests can support food chains with many trophic levels. (c) Predict what might happen to frog numbers if a fungal disease kills most beetles.
问题：某一热带雨林的食物网包含：树木 → 甲虫 → 蛙 → 蛇 → 鹰。（a）将此画成一条食物链。（b）解释为什么热带雨林能支持含有多个营养级的食物链。（c）如果真菌病害导致大部分甲虫死亡，预测蛙的数量会如何变化。
Answer (a): Trees → beetles → frogs → snakes → hawks. (Arrows show energy flow.)
答案（a）：树木 → 甲虫 → 蛙 → 蛇 → 鹰。（箭头表示能量流动。）
Answer (b): Tropical rainforests have high temperatures, abundant rainfall and intense sunlight all year round. This allows producers (trees) to photosynthesise rapidly, producing a large biomass. More energy enters the ecosystem, so more trophic levels can be sustained.
答案（b）：热带雨林全年高温、多雨、光照强烈。这使得生产者（树木）能快速进行光合作用，产生巨大的生物量。进入生态系统的能量更多，因此能够支撑更多的营养级。
Answer (c): Frogs feed on beetles, so with fewer beetles available, frogs would have less food. The frog population would likely decrease due to starvation, and snakes and hawks may also be affected.
答案（c）：蛙以甲虫为食，可食用的甲虫减少，蛙的食物来源就减少了。蛙的数量很可能因饥饿而下降，蛇和鹰也会受到影响。
Cross‑disciplinary link: This question connects biology (feeding relationships, populations) with geography (climate and biomes). Understanding the physical environment helps explain biological diversity and pyramid structure.
跨学科联系：这道题将生物学（捕食关系、种群）与地理（气候与生物群系）相结合。理解自然环境有助于解释生物多样性和金字塔结构。
4. Skeleton, Muscles and Levers | 骨骼、肌肉与杠杆
Question: When you lift a book by bending your elbow, the biceps muscle contracts. (a) Identify the fulcrum, effort and load in this lever system. (b) State the class of lever. (c) Explain why the biceps must produce a force greater than the weight of the book.
问题：当你弯曲手臂举起一本书时，肱二头肌收缩。（a）指出这个杠杆系统中的支点、施力点和负荷。（b）说出该杠杆的类型。（c）解释为什么肱二头肌产生的力必须大于书的重量。
Answer (a): Fulcrum = elbow joint. Effort = force from biceps muscle (attached near the top of the forearm). Load = weight of the book in the hand.
答案（a）：支点 = 肘关节。施力 = 来自肱二头肌的力（附着于前臂上端附近）。负荷 = 手中书的重量。
Answer (b): It is a class 3 lever because the effort is between the fulcrum and the load.
答案（b）：这是第三类杠杆，因为施力点位于支点和负荷之间。
Answer (c): In a class 3 lever, the effort is closer to the fulcrum than the load, so the effort arm is shorter. A larger effort force is needed to balance a smaller load, following the principle of moments: effort × effort arm = load × load arm.
答案（c）：在第三类杠杆中，施力点比负荷更靠近支点，因而施力臂较短。根据力矩原理（施力 × 施力臂 = 负荷 × 负荷臂），需要较大的施力来平衡较小的负荷。
Cross‑disciplinary link: The physics of levers and moments explains movement in biology. This integration of biomechanics and physics is essential for understanding how our body performs work.
跨学科联系：杠杆和力矩的物理学原理解释了生物学中的运动。这种生物力学与物理的融合，对于理解人体如何做功至关重要。
5. Breathing Rate and Data Handling | 呼吸速率与数据处理
Question: A student measured her breaths per minute at rest and during exercise. The results: at rest 14, 16, 15; after jogging 22, 25, 24; after sprinting 32, 34, 35. (a) Calculate the mean breathing rate for each condition. (b) Present the means in a bar chart. (c) Explain the biological reason for the trend.
问题：一名学生测量了自己安静时和运动时的每分钟呼吸次数。结果：安静时 14, 16, 15；慢跑后 22, 25, 24；冲刺后 32, 34, 35。（a）计算每种状态下的平均呼吸速率。（b）用条形图表示这些平均值。（c）解释该趋势的生物学原因。
Answer (a): Rest mean = (14+16+15) ÷ 3 = 15 breaths/min. Jogging mean = (22+25+24) ÷ 3 = 23.7, round to 24 breaths/min. Sprinting mean = (32+34+35) ÷ 3 = 33.7, round to 34 breaths/min.
答案（a）：安静时平均值 = (14+16+15) ÷ 3 = 15 次/分。慢跑后 = (22+25+24) ÷ 3 = 23.7，约 24 次/分。冲刺后 = (32+34+35) ÷ 3 = 33.7，约 34 次/分。
Answer (b): Draw a bar chart with ‘Condition’ on the x‑axis (Rest, Jogging, Sprinting) and ‘Mean breathing rate (breaths/min)’ on the y‑axis. Plot bars of height 15, 24, 34 respectively. Label axes and give a title.
答案（b）：绘制条形图，x 轴为“状态”（安静、慢跑、冲刺），y 轴为“平均呼吸速率（次/分）”。分别画出高度为 15、24、34 的柱状图。标注坐标轴并添加标题。
Answer (c): During exercise, muscles contract more and need more energy. Cells increase aerobic respiration, which requires oxygen and produces carbon dioxide. Breathing rate rises to take in more oxygen and remove extra CO₂. After sprinting, oxygen demand is highest, hence the steepest increase.
答案（c）：运动时肌肉收缩增加，需要更多能量。细胞有氧呼吸增强，该过程需要氧气并产生二氧化碳。呼吸速率加快以摄入更多氧气并排出多余的二氧化碳。冲刺后需氧量最高，因此增幅最大。
Cross‑disciplinary link: This task combines biology (gas exchange, respiration) with mathematics (calculating means, drawing bar charts) and physical education (exercise intensity). Handling data and graphing are core scientific skills.
跨学科联系：该任务将生物学（气体交换、呼吸作用）与数学（计算平均值、绘制条形图）以及体育（运动强度）相结合。数据处理和图表绘制是核心科学技能。
6. Diffusion, Surface Area and Mathematics | 扩散、表面积与数学
Question: Two agar cubes stained with phenolphthalein, Cube A (1 cm × 1 cm × 1 cm) and Cube B (2 cm × 2 cm × 2 cm), are placed in hydrochloric acid. The acid diffuses in and turns the cubes colourless. (a) Calculate the surface area to volume ratio (SA:V) for each cube. (b) Predict which cube will become completely colourless first. (c) Explain why cells are usually microscopic.
问题：两块用酚酞染色的琼脂块，方块 A (1 cm × 1 cm × 1 cm) 和方块 B (2 cm × 2 cm × 2 cm)，放入盐酸中。酸会扩散进去并使琼脂变为无色。（a）计算每块琼脂的表面积与体积之比 (SA:V)。（b）预测哪一块会最先完全变为无色。（c）解释为什么细胞通常是微小的。
Answer (a): Cube A: surface area = 6 × (1×1) = 6 cm², volume = 1×1×1 = 1 cm³, SA:V = 6:1. Cube B: surface area = 6 × (2×2) = 24 cm², volume = 2×2×2 = 8 cm³, SA:V = 24:8 = 3:1.
答案（a）：方块 A：表面积 = 6 × (1×1) = 6 cm²，体积 = 1×1×1 = 1 cm³，SA:V = 6:1。方块 B：表面积 = 6 × (2×2) = 24 cm²，体积 = 8 cm³，SA:V = 24:8 = 3:1。
Answer (b): Cube A will turn colourless first because it has a larger surface area relative to its volume. This means a greater proportion of its interior is close to the surface, allowing faster diffusion of acid to the centre.
答案（b）：方块 A 会先变为无色，因为它相对于体积具有更大的表面积。这意味着其内部更大比例靠近表面，酸能更快地扩散到中心。
Answer (c): Cells need to exchange substances (oxygen, nutrients, waste) quickly by diffusion. A high SA:V ratio, as seen in small cubes, makes diffusion efficient. If a cell were very large, its centre would not receive enough materials. Therefore, most cells are microscopic to maintain a high SA:V.
答案（c）：细胞需要通过扩散快速交换物质（氧气、养料、废物）。小的方块具有高 SA:V 比，这使得扩散高效。如果细胞很大，其中心就无法获得足够物质。因此，大多数细胞都很微小，以维持高 SA:V 比。
Cross‑disciplinary link: This investigation uses geometry and ratio (mathematics) to explain a fundamental biological concept – the relationship between size and efficiency of transport. It models how physical constraints shape living organisms.
跨学科联系：这项探究利用几何与比例（数学）来解释一个基本的生物学概念——尺寸与运输效率之间的关系。它模拟了物理约束如何塑造生物体的形态。
7. Ecology Sampling and Statistics | 生态学取样与统计
Question: A class uses a 0.5 m × 0.5 m quadrat to sample daisies on the school field. They throw the quadrat randomly five times and count daisies: 3, 5, 0, 4, 2. (a) Calculate the mean number of daisies per quadrat. (b) Estimate the total number of daisies in the whole field measuring 40 m × 30 m. (c) Why is random sampling important?
问题：一个班级使用 0.5 m × 0.5 m 的样方在学校草坪上抽样调查雏菊。他们随机抛掷样方五次，记录雏菊数量：3, 5, 0, 4, 2。（a）计算每个样方雏菊的平均数量。（b）估算整个场地（40 m × 30 m）的雏菊总数。（c）为什么随机取样很重要？
Answer (a): Mean = (3+5+0+4+2) ÷ 5 = 14 ÷ 5 = 2.8 daisies per quadrat.
答案（a）：平均值 = (3+5+0+4+2) ÷ 5 = 14 ÷ 5 = 2.8 株/样方。
Answer (b): Area of one quadrat = 0.5 × 0.5 = 0.25 m². Area of whole field = 40 × 30 = 1200 m². Number of quadrats that would fit = 1200 ÷ 0.25 = 4800. Estimated total daisies = 4800 × 2.8 = 13 440 daisies.
答案（b）：一个样方面积 = 0.5 × 0.5 = 0.25 m²。整个场地的面积 = 1200 m²。可容纳的样方数 = 1200 ÷ 0.25 = 4800。估算雏菊总数 = 4800 × 2.8 = 13 440 株。
Answer (c): Random sampling avoids bias. If we only placed the quadrat where daisies look abundant, the estimate would be too high. Random throws give each area an equal chance of being selected, making the average more representative.
答案（c）：随机取样可以避免偏差。如果只把样方放在雏菊看似茂盛的地方，估算值就会偏高。随机抛掷使每个区域被选中的机会均等，这样得到的平均值才更具代表性。
Cross‑disciplinary link: This fieldwork uses mathematical statistics (mean, multiplication, area) to answer an ecological question. It demonstrates how biologists use sampling and estimation to study populations without counting every organism.
跨学科联系：这项野外调查利用数学统计（平均值、乘法、面积）来回答生态学问题。它展示了生物学家如何利用抽样和估算来研究种群，而无需计数每一个生物体。
8. Digestion and Enzyme Chemistry | 消化与酶化学
Question: A test tube containing starch solution and amylase is kept at 37 °C. Every minute, a drop is tested with iodine. After 5 minutes, the iodine remains orange‑brown. (a) What does the orange‑brown colour indicate? (b) What is the role of amylase? (c) If the tube were cooled to 5 °C, predict the result after 5 minutes and explain why.
问题：一支试管中装有淀粉溶液和淀粉酶，在 37 °C 下保温。每分钟取一滴液体用碘液检测。5 分钟后，碘液保持橙棕色。（a）橙棕色表明什么？（b）淀粉酶的作用是什么？（c）如果将试管冷却至 5 °C，预测 5 分钟后的结果并解释原因。
Answer (a): The orange‑brown colour indicates that no starch remains; it has all been broken down. Iodine is orange‑brown in the absence of starch and blue‑black when starch is present.
答案（a）：橙棕色表明不再有淀粉残留，淀粉已全被分解。碘液在没有淀粉时为橙棕色，有淀粉时呈蓝黑色。
Answer (b): Amylase is an enzyme that breaks down starch (a polysaccharide) into smaller sugars such as maltose. It acts as a biological catalyst, speeding up the digestion of starch.
答案（b）：淀粉酶是一种酶，能将淀粉（多糖）分解为较小的糖，如麦芽糖。它起到生物催化剂的作用，加速淀粉的消化。
Answer (c): At 5 °C, the iodine test would still show a blue‑black colour after 5 minutes. Enzymes work best at an optimum temperature (around 37 °C for human amylase). At very low temperatures, enzyme molecules move slowly and have much less kinetic energy, so they collide with starch molecules less frequently and the reaction rate drops dramatically.
答案（c）：在 5 °C 时，5 分钟后碘液检测仍会显示蓝黑色。酶在最适温度下活性最高（人体淀粉酶约 37 °C）。在极低温度下，酶分子运动缓慢，动能小得多，与淀粉分子的碰撞频率大大降低，反应速率显著下降。
Cross‑disciplinary link: This connects biology (digestion, enzyme action) with chemistry (chemical indicators, effect of temperature on reaction rate). Understanding the particulate nature of matter and kinetic energy from chemistry explains the observed biological phenomenon.
跨学科联系：这道题将生物学（消化、酶的作用）与化学（化学指示剂、温度对反应速率的影响）联系起来。理解物质的微粒性质和化学中的动能，可以解释所观察到的生物学现象。
9. Osmosis in Plant Cells and Percentage Change | 植物细胞的渗透作用与百分比变化
Question: Identical potato cylinders are weighed and placed in different sugar solutions for 30 minutes. The results are shown in the table.
| Sugar solution (mol/dm³) | Initial mass (g) | Final mass (g) |
|---|---|---|
| 0.0 | 5.0 | 5.8 |
| 0.4 | 5.0 | 4.2 |
| 0.8 | 5.0 | 3.6 |
(a) Calculate the percentage change in mass for the potato in 0.0 mol/dm³ solution. (b) Explain why the mass increased in distilled water (0.0 mol/dm³). (c) Why did the mass decrease in 0.8 mol/dm³ solution?
问题：相同的土豆圆柱体分别称重后放入不同浓度的蔗糖溶液中，30 分钟后结果如上表。（a）计算 0.0 mol/dm³ 溶液中土豆的质量变化百分比。（b）解释为什么在蒸馏水（0.0 mol/dm³）中质量会增加。（c）为什么在 0.8 mol/dm³ 溶液中质量会减少？
Answer (a): Percentage change = ((final mass − initial mass) ÷ initial mass) × 100% = ((5.8 − 5.0) ÷ 5.0) × 100% = (0.8 ÷ 5.0) × 100% = +16%.
答案（a）：质量变化百分比 = ((5.8 – 5.0) ÷ 5.0) × 100% = (0.8 ÷ 5.0) × 100% = +16%。
Answer (b): The water potential inside potato cells is lower than that of pure water (0.0 mol/dm³). Therefore, water moves by osmosis from the region of higher water potential (outside) into the potato cells where water potential is lower. The cells swell and gain mass.
答案（b）：土豆细胞内的水势低于纯水（0.0 mol/dm³）。因此，水通过渗透作用从水势较高的外部区域进入水势较低的土豆细胞内。细胞吸水膨胀，质量增加。
Answer (c): The 0.8 mol/dm³ sugar solution has a lower water potential than the potato cells. As a result, water leaves the potato cells by osmosis and moves into the surrounding solution. Cells become flaccid and the mass decreases.
答案（c）：0.8 mol/dm³ 蔗糖溶液的水势低于土豆细胞。因此，水通过渗透作用离开土豆细胞，进入周围溶液。细胞变得质壁分离，质量减少。
Cross‑disciplinary link: This experiment demands careful use of mathematics (calculating percentage change) and an understanding of concentration gradients, a concept shared by chemistry and physics, to interpret observations in a living system.
跨学科联系：这项实验要求熟练运用数学（计算百分比变化）并理解浓度梯度（化学和物理共有的概念），以解释生命系统中的观察结果。
10. Genetics, Probability and Simulation | 遗传学、概率与模拟
Question: In pea plants, tall (T) is dominant over short (t). Two heterozygous tall plants (Tt) are crossed. (a) Draw a Punnett square to show the possible genotypes of the offspring. (b) What is the probability of obtaining a short offspring? (c) To simulate this, a student tosses two coins 50 times, recording two heads as TT, one head and one tail as Tt, and two tails as tt. Explain how this model represents the genetic cross.
问题：在豌豆中，高茎（T）对矮茎（t）为显性。两株杂合高茎豌豆（Tt）杂交。（a）画出庞纳特方格，显示子代可能的基因型。（b）得到矮茎后代的概率是多少？（c）为模拟此过程，一名学生抛掷两枚硬币 50 次，记录两个正面为 TT，一正一反为 Tt，两个反面为 tt。解释这个模型如何代表遗传杂交。
Answer (a): Punnett square: gametes across top T and t, down side T and t. Combinations: TT, Tt, Tt, tt. So genotypes are TT, Tt, Tt and tt.
答案（a）：庞纳特方格：配子置于上方 T 和 t，侧方 T 和 t。组合为：TT, Tt, Tt, tt。
Answer (b): Only tt gives short plants. Out of 4 equally likely boxes, 1 is tt. Therefore, probability = 1/4 or 25%.
答案（b）：只有 tt 表现为矮茎。4 个均等可能的方格中，1 个是 tt。因此概率 = 1/4 即 25%。
Answer (c): Each coin represents one parent, and each side a possible allele (head = T, tail = t). Tossing the two coins together randomly sorts the alleles, just like independent assortment during gamete formation.
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