📚 Year 7 CAIE Biology: Cross-Curricular Practice | Year 7 CAIE 生物：跨学科综合题型训练
In Year 7 CAIE Biology, you will encounter questions that blend biology with mathematics, physics, chemistry, and geography. These cross-curricular tasks train you to think like a scientist, using data, calculations, and logical reasoning to solve real-world problems. This article provides twelve worked examples of integrated question types to build your confidence.
在 Year 7 CAIE 生物课程中，你会遇到融合数学、物理、化学和地理的考题。这类跨学科任务训练你像科学家一样思考，运用数据、计算和逻辑推理解决实际问题。本文提供十二个综合题型的详细示例，帮助你建立信心。
1. Analysing Population Data | 分析种群数据
Ecologists sample populations and use mathematical tools to describe their findings. Calculating the mean, median, and range helps reveal patterns in species abundance.
生态学家对种群进行取样，并用数学工具描述结果。计算平均值、中位数和极差有助于揭示物种丰度的变化规律。
A student counted woodlice under logs in a park on five days: 12, 9, 15, 10, 14. To find the mean, she added 12 + 9 + 15 + 10 + 14 = 60, then divided by 5 to get 12. The median is the middle number when sorted: 9, 10, 12, 14, 15 → 12. The range is 15 – 9 = 6.
一名学生在公园的木头下连续五天统计鼠妇数量：12、9、15、10、14。为求平均值，她总和 60 后除以 5，得到 12。中位数是排序后的中间值：9、10、12、14、15 → 12。极差为 15 − 9 = 6。
- Mean (average): total ÷ number of values
- Median (middle): the centre value when listed in order
- Range (spread): largest – smallest
- 平均值：总和 ÷ 数值个数
- 中位数：按顺序排列后的中间值
- 极差：最大值 − 最小值
Using these statistics allows biologists to compare habitats fairly, even when sample sizes differ.
使用这些统计量，生物学家可以公平地比较不同栖息地，即便样本数量不同。
2. Photosynthesis Rate and Inverse Square Law | 光合作用速率与平方反比关系
The rate of photosynthesis decreases as a plant is moved farther from a light source. This follows a physical pattern: light intensity is inversely proportional to the square of the distance.
当植物远离光源时，光合作用速率会下降。这遵循一个物理规律：光照强度与距离的平方成反比。
If an aquatic plant produces 20 bubbles of oxygen per minute at 10 cm from the lamp, how does the bubble count change at 20 cm? The distance doubles, so light intensity becomes 1/(2²) = 1/4 of the original. Hence, we expect about 20 ÷ 4 = 5 bubbles per minute.
如果一株水生植物在距灯 10 cm 处每分钟产生 20 个氧气泡，那么在 20 cm 处会怎样？距离加倍，光照强度变为原来的 1/(2²) = 1/4。因此预计产生 20 ÷ 4 = 5 个气泡。
| Distance (cm) | Relative light intensity | Expected bubbles/min |
| 10 | 1 | 20 |
| 20 | 1/4 | 5 |
| 30 | 1/9 | ≈ 2.2 |
This exercise links biology with physics and simple ratio calculations, teaching you to predict experimental outcomes.
这个练习将生物与物理及简单比例计算联系起来，让你学会预测实验结果。
3. Energy Transfer in Food Chains | 食物链中的能量传递
Only about 10% of the energy in one trophic level is passed to the next; the rest is lost as heat, movement, and waste. Solving these energy budgets requires percentages and multiplication.
在一个营养级中，只有大约 10% 的能量传递给下一级；其余以热、运动和排泄物形式散失。解决这类能量预算问题需要百分比和乘法。
Suppose grass contains 8000 kJ of energy. A rabbit eats the grass and gains 10% of that energy: 8000 × 0.10 = 800 kJ. A fox then eats the rabbit and receives 10% of the rabbit’s energy: 800 × 0.10 = 80 kJ. You can see why food chains rarely exceed four or five levels.
假设草含有 8000 kJ 能量。兔子吃草获得 10% 的能量：8000 × 0.10 = 800 kJ。狐狸再吃兔子，获得兔子能量的 10%：800 × 0.10 = 80 kJ。这让你明白为什么食物链很少超过四到五级。
Energy at trophic level n = Previous energy × 0.10
第 n 营养级的能量 = 前一级能量 × 0.10
Being able to calculate energy loss improves your understanding of ecosystem efficiency and population sizes.
能计算能量损耗有助于你更好地理解生态系统效率和种群大小。
4. Enzymes and pH in Digestion | 消化酶与酸碱度
Digestive enzymes are proteins that work best at specific pH values. Linking biology with chemistry, you can investigate how acid or alkaline conditions affect reaction rates.
消化酶是一种蛋白质，在特定 pH 值下活性最高。将生物与化学结合起来，你可以探究酸或碱条件如何影响反应速率。
The enzyme pepsin in the stomach works optimally at pH 2, while trypsin in the small intestine prefers pH 8. If you were to test starch breakdown by amylase (optimum pH 7), you might use buffer solutions at pH 4, 7, and 10. The time taken for starch to disappear can be recorded and plotted.
胃蛋白酶在 pH 2 时活性最佳，而小肠中的胰蛋白酶适宜 pH 8。如果要检测淀粉酶分解淀粉（最适 pH 7），可使用 pH 4、7、10 的缓冲液。记录淀粉消失所需的时间并作图。
| pH | Time for starch to break down (min) |
| 4 | 15 |
| 7 | 4 |
| 10 | 12 |
This kind of practical investigation demands careful measurement (science skills) and graph drawing (maths), preparing you for future lab work.
这类实践探究要求仔细测量（科学技能）和绘图（数学），为未来的实验工作做好准备。
5. Microscopy and Scale Calculations | 显微镜与比例计算
Using a microscope involves physics (light, lenses) and mathematics (magnification, actual size). You must convert units and use the formula:
使用显微镜涉及物理（光、透镜）和数学（放大倍数、实际尺寸）。你需要换算单位并运用以下公式：
Magnification = Image size ÷ Actual size
放大倍数 = 图像大小 ÷ 实际大小
An image of a cell measures 30 mm on a drawing. If the magnification is ×400, what is the actual size? Actual size = 30 mm ÷ 400 = 0.075 mm. Convert to micrometres: 0.075 mm × 1000 = 75 µm.
细胞图片在图上尺寸为 30 mm。若放大倍数为 ×400，实际大小是多少？实际大小 = 30 mm ÷ 400 = 0.075 mm。换算成微米：0.075 mm × 1000 = 75 µm。
- 1 mm = 1000 µm
- Always express actual sizes in micrometres for cells.
- 1 mm = 1000 µm
- 细胞的实际大小始终以微米表示。
Such problems develop numerical fluency and unit conversion skills vital for any science qualification.
这类问题培养计算能力和单位换算技巧，对任何科学考试都至关重要。
6. Lung Capacity and Data Analysis | 肺活量与数据分析
Measuring lung volume with a spirometer connects biology to physics (gas volume) and mathematics (data representation). Repeated measurements help assess reliability.
用肺活量计测量肺容积将生物与物理（气体体积）和数学（数据表示）联系起来。重复测量有助于评估数据的可靠性。
Five students recorded their vital capacity (litres): 2.8, 3.2, 2.9, 3.1, 2.7. Calculate the mean, identify any anomalous results, and suggest how exercise might change these values. The mean is (2.8+3.2+2.9+3.1+2.7) ÷ 5 = 14.7 ÷ 5 = 2.94 L. All values lie close to the mean, so no clear anomaly exists. Regular aerobic exercise can increase vital capacity.
五名学生记录了肺活量（升）：2.8, 3.2, 2.9, 3.1, 2.7。计算平均值，识别异常值，并说明运动可能如何改变这些数值。平均值为 (2.8+3.2+2.9+3.1+2.7) ÷ 5 = 2.94 L。所有数值都接近平均值，无明显异常。有规律的有氧运动可提高肺活量。
Constructing a bar chart of class data strengthens your ability to present and interpret biological results clearly.
绘制全班数据的条形图能强化你清晰展示和解读生物实验结果的能力。
7. Temperature and Seed Germination | 温度与种子萌发
Seeds require suitable temperatures to germinate, linking biology to geography (climate zones) and physics (heat). You can design controlled experiments to test the effect of temperature.
种子需要适宜的温度才能萌发，这连接了生物与地理（气候带）和物理（热量）。你可以设计对照实验来测试温度的影响。
Cress seeds were placed at 5°C, 15°C, 25°C, and 35°C. After 48 hours, the percentage of germinated seeds was: 0%, 45%, 90%, 20%. Which is the optimum temperature? 25°C gives the highest germination rate. Explain why 35°C showed a drop—enzymes may denature at high temperatures, a concept from chemistry.
将水芹种子分别放置在 5°C、15°C、25°C 和 35°C 环境中。48 小时后发芽率为：0%、45%、90%、20%。最适温度是多少？25°C 出芽率最高。解释为什么 35°C 下降——酶在高温下可能变性，这是化学概念。
Drawing a line graph of these results lets you interpolate and extrapolate, skills used in geography when studying climate and vegetation.
绘制这些结果的折线图可以让你练习内插和外推，这是地理中研究气候和植被时用到的技能。
8. Genetic Probability with Punnett Squares | 遗传概率与旁纳特方格
Inheritance follows mathematical rules of probability. Using a Punnett square, you can predict the chance of offspring showing a particular trait.
遗传遵循数学上的概率规律。利用旁纳特方格，你可以预测后代出现某一性状的几率。
In pea plants, tall (T) is dominant over dwarf (t). Cross two heterozygous tall plants (Tt × Tt). The Punnett square gives: TT, Tt, tT, tt. The probability of a tall offspring is 3/4 (75%), and dwarf is 1/4 (25%). Expressing this as a ratio 3:1 mimics Mendel’s findings.
在豌豆中，高茎 (T) 对矮茎 (t) 为显性。将两株杂合高茎豌豆 (Tt × Tt) 杂交。旁纳特方格结果为：TT、Tt、tT、tt。高茎后代的概率为 3/4 (75%)，矮茎为 1/4 (25%)。用 3:1 的比例表示，正好符合孟德尔的发现。
Probability of a given genotype = number of squares with that genotype ÷ 4
特定基因型的概率 = 具有该基因型的方格数 ÷ 4
Converting probabilities into percentages and ratios demonstrates how maths deepens your genetic predictions.
将概率转化为百分数和比例，体现数学如何加深你对遗传预测的理解。
9. Blood Cell Counts and Health Diagnosis | 血细胞计数与健康诊断
Doctors use blood test data—red blood cells, white blood cells, and platelets—to diagnose diseases. Interpreting these numbers is a cross-curricular task merging biology with numerical reasoning.
医生利用血液检测数据——红细胞、白细胞和血小板——来诊断疾病。解读这些数字是一个融合生物与数字推理的跨学科任务。
A patient’s blood report shows: red blood cells 4.0 million per µL (normal: 4.5–5.5 million), white blood cells 15,000 per µL (normal: 4,000–11,000). The low red cell count could suggest anaemia, while high white cells indicate possible infection. Calculate the percentage deviation from normal to quantify the abnormality.
某患者的血检报告显示：红细胞 4.0 百万/µL（正常 4.5–5.5 百万），白细胞 15,000/µL（正常 4,000–11,000）。红细胞偏低可能提示贫血，而白细胞偏高表明可能感染。计算偏离正常值的百分比以量化异常程度。
For red cells: (4.0 – 4.5) ÷ 4.5 × 100% ≈ −11.1% (below normal). For white cells: (15000 – 11000) ÷ 11000 × 100% ≈ +36.4% (above normal).
红细胞：(4.0 – 4.5) ÷ 4.5 × 100% ≈ −11.1%（低于正常）。白细胞：(15000 – 11000) ÷ 11000 × 100% ≈ +36.4%（高于正常）。
This type of analysis trains you to handle clinical data logically, combining biology with percent change calculations.
这种分析训练你逻辑地处理临床数据，将生物与百分比变化计算结合起来。
10. Carbon Cycle and Chemical Equations | 碳循环与化学方程式
The carbon cycle involves biological processes (photosynthesis, respiration) and chemical reactions. Representing them with word and symbol equations sharpens your chemical literacy.
碳循环涉及生物过程（光合作用、呼吸作用）和化学反应。用文字及符号方程式表示它们可以提升你的化学素养。
Photosynthesis word equation: carbon dioxide + water → glucose + oxygen. Symbol equation: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (light energy over the arrow). Respiration is the reverse: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy.
光合作用文字方程式：二氧化碳 + 水 → 葡萄糖 + 氧气。符号方程式：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂（箭头上方注明光能）。呼吸作用则相反：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 能量。
Combustion of fossil fuels adds extra CO₂: CH₄ + 2O₂ → CO₂ + 2H₂O (for methane). Balancing these equations practices chemical bookkeeping and reinforces the link between human activity and atmospheric carbon.
化石燃料燃烧额外排放 CO₂：CH₄ + 2O₂ → CO₂ + 2H₂O（以甲烷为例）。配平这些方程式锻炼了化学计量，并强化人类活动与大气碳含量之间的联系。
11. Osmosis and Concentration Gradients | 渗透作用与浓度梯度
Osmosis is the net movement of water molecules across a partially permeable membrane from a region of high water potential to low water potential. Understanding this requires concepts from both biology and chemistry (solution concentrations).
渗透作用是水分子通过半透膜从高水势区域向低水势区域的净移动。理解这个过程需要生物和化学（溶液浓度）两方面的知识。
A potato cylinder is placed in a 0.5 M sugar solution. It loses mass because water moves out of the potato cells into the more concentrated solution. If you place a similar cylinder in distilled water (0 M), it gains mass. Plotting percentage change in mass against concentration creates a calibration curve you might see in physics or chemistry labs.
将土豆条放入 0.5 M 蔗糖溶液中，它会失重，因为水从土豆细胞流向更高浓度的溶液。若放入蒸馏水 (0 M) 中，它会增重。绘制质量变化百分比-浓度图会得到一条标准曲线，这在物理或化学实验中也很常见。
% change in mass = [(final mass – initial mass) ÷ initial mass] × 100%
质量变化 % = [(最终质量 − 初始质量) ÷ 初始质量] × 100%
This classic experiment blends practical measurement with graphical analysis, preparing you for interdisciplinary assessments.
这个经典实验将实际测量与图形分析融合，为你应对跨学科评估做好准备。
12. Ecological Pyramids and Mathematical Shapes | 生态金字塔与数学图形
Pyramids of numbers, biomass, and energy are graphical summaries of ecosystem structure. Drawing them to scale uses mathematical skills in proportion and area.
数量金字塔、生物量金字塔和能量金字塔是生态系统结构的图示摘要。按比例绘制它们需要用到比例和面积的数学技能。
Given: grass 1000 kJ, snails 100 kJ, thrushes 10 kJ. Draw a pyramid of energy as a stacked bar chart. The height of each bar should reflect the energy value, but more importantly, the area should be roughly proportional. Discuss why the pyramid shape is always upright for energy but can be inverted for numbers (e.g., one tree supporting thousands of insects).
已知：草 1000 kJ，蜗牛 100 kJ，鸫鸟 10 kJ。以堆叠条形图绘制能量金字塔。每一条的高度应反映能量值，但更重要的是面积应大致成比例。讨论为何能量金字塔总是正立，而数量金字塔可能倒置（例如一棵树支持成千上万只昆虫）。
These interpretations require you to connect biological principles with data-presentation techniques learned in mathematics and geography.
这些解释要求你将生物学原理与数学和地理中学习的数据展示技术联系起来。
Published by TutorHao | Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导