📚 Year 7 WJEC Biology: Cross-Curricular Integrated Question Training | Year 7 WJEC 生物：跨学科综合题型训练
In WJEC Year 7 Biology, you will often meet questions that mix science with maths, geography, chemistry and even physics. These cross-curricular questions test not only your biological knowledge but also your ability to calculate, interpret graphs, use chemical symbols and recognise patterns in data. This article brings together ten key topic areas where biology links with other subjects, helping you practise the skills needed to answer integrated questions confidently.
在 WJEC 七年级生物中，你经常会遇到将科学与数学、地理、化学甚至物理结合起来的题目。这些跨学科问题不仅测试你的生物学知识，还考察你计算、解读图表、使用化学符号和识别数据规律的能力。本文汇集了生物学与其他学科交叉的十个关键主题领域，帮助你练习自信回答综合题所需的技能。
1. Microscopy and Mathematics | 显微镜与数学
ification. For example, a microscope with a ×10 eyepiece and a ×40 objective lens gives a total magnification of ×400. You may also be asked to work out the actual size of a cell if you are given the image size and the magnification. The equation is: actual size = image size ÷ magnification. Remember to keep all measurements in the same units, usually millimetres (mm) or micrometres (µm). 1 mm = 1000 µ and the magnification is ×500, the actual size is 25 ÷ 500 = 0.05 mm, which converts to 50 µm.
计算总放大倍数是基本技能。公式为：总放大倍数 = 目镜放大倍数 × 物镜放大倍数。例如，一个带有×10目镜和×40物镜的显微镜，总放大倍数为×400。你还可能被要求根据给出的图像大小和放大倍数计算细胞的实际大小。公式是：实际大小 = 图像大小 ÷ 放大倍数。记住所有测量值要使用相同单位，通常为毫米(mm)或微米(µm)。1 mm = 1000 µm。如果一个细胞在图像中宽25 mm，放大倍数为×500，则实际大小为25 ÷ 500 = 0.05 mm，换算后为50 µm。
ated questions may also involve scale bars. A scale bar on a micrograph might represent 10 µ length = 20 mm, actual length = 0.01 mm, so magnification = 20 ÷ 0.01 = ×2000. Then use this to find the size of another structure.
综合题还可能涉及比例尺。显微照片上的比例尺可能代表10 µm，在图像上测量为2 cm。你可以先通过将测量长度除以实际长度来计算图像放大倍数。此处图像长度为20 mm，实际长度为0.01 mm，因此放大倍数 = 20 ÷ 0.01 = ×2000。然后用此数值计算另一个结构的大小。
2. Food Chains and Percentage Efficiency | 食物链与能量传递效率
Energy transfer along a food chain is rarely more than 10% ns 5000 kJ of energy, the rabbit would receive approximately 500 kJ. The mathematical skill tested here is calculating percentages. Percentage efficiency = (energy in consumer ÷ energy in producer) × 100%. So for the rabbit, (500 ÷ 5000) × 100% = 10%.
食物链中能量传递的效率很少超过10%。这意味着当兔子吃草时，草中储存的能量只有大约十分之一传递给兔子。如果草含有5000 kJ能量，兔子大约会获得500 kJ。这里考查的数学技能是计算百分比。传递效率 = (消费者能量 ÷ 生产者能量) × 100%。因此兔子的情况为 (500 ÷ 5000) × 100% = 10%。
tween levels. For instance, if a food chain has grass → vole → owl, and grass has 2000 kJ, the vole has 180 kJ, the owl has 15 kJ, the efficiency from grass to vole is (180 ÷ 2000) × 100% = 9%. The large loss is due to respiration, movement, heat and uneaten parts.
题目常提供数据表展示每一营养级的生物量或能量。你需要计算百分比损失或营养级间的效率。例如，一条食物链为草 → 田鼠 → 猫头鹰，草含2000 kJ，田鼠含180 kJ，猫头鹰含15 kJ，则从草到田鼠的效率为(180 ÷ 2000) × 100% = 9%。大量能量损失来自呼吸、运动、热能和未被吃掉的部位。
3. Surface Area to Volume Ratio and Diffusion | 表面积与体积比和扩散
rface area to volume ratio (SA:V). You can calculate SA:V by imagining a cube. For a cube with side length 1 cm, surface area = 6 × (1 cm)² = 6 cm², volume = (1 cm)³ = 1 cm³, so SA:V = 6:1. For a larger cube of side 3 cm, SA = 6 × (3)² = 54 cm², volume = 27 cm³, SA:V = 54:27 = 2:1. As the organism gets larger, the ratio decreases, making diffusion alone insufficient. This is why large organisms need transport systems like the circulatory system.
小的生物如细菌依靠体表的扩散获取氧气和排出废物。扩散速率取决于表面积与体积比(SA:V)。你可以通过想象一个立方体来计算SA:V。对于边长为1 cm的立方体，表面积 = 6 × (1 cm)² = 6 cm²，体积 = (1 cm)³ = 1 cm³，因此SA:V = 6:1。对于边长为3 cm的较大立方体，SA = 6 × (3)² = 54 cm²，体积 = 27 cm³，SA:V = 54:27 = 2:1。随着生物体变大，比值下降，仅靠扩散就不够了。这就是为什么大型生物需要循环系统等运输系统。
Integrated questions ask you to compare ratios and predict which shape is more efficient for exchange. A flat, thin shape, like a leaf or a flatworm, has a higher surface area relative to its volume. Calculations using simple arithmetic and powers (squares and cubes) are common. Always remember units: area in cm², volume in cm³.
综合题会要求你比较比值并预测哪种形状更利于物质交换。扁平而薄的形状，如叶片或扁虫，其表面积相对体积更大。计算常涉及简单的算术和幂（平方与立方）。始终记住单位：面积用cm²，体积用cm³。
4. Photosynthesis and Chemical Equations | 光合作用与化学方程式
nced chemical equation uses symbols: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. Notice that there are 6 carbon atoms, 12 hydrogen atoms and 18 oxygen atoms on each side of the equation. This cross-over with chemistry helps you understand the conservation of atoms during a chemical reaction.
光合作用可以用文字方程式概括：二氧化碳 + 水 → 葡萄糖 + 氧气，在光和叶绿素存在下进行。用化学符号书写的配平方程式为：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂。注意等号两边各有6个碳原子、12个氢原子和18个氧原子。这种与化学的交叉帮助你理解化学反应中的原子守恒。
ced equation and asked to balance it, or to count the number of atoms in reactants and products. For example, what is the total number of hydrogen atoms in 6H₂O? o calculate the mass of glucose produced from a certain mass of CO₂, using the idea that mass is conserved. This links biology with chemical calculations and arithmetic.
考试中你可能会遇到未配平的方程式并被要求进行配平，或者统计反应物和生成物中的原子个数。例如，6H₂O中氢原子的总数是多少？那是 6 × 2 = 12。你还可能被要求利用质量守恒的思想，根据一定质量的CO₂计算生成的葡萄糖质量。这将生物学与化学计算和算术联系起来。
5. Respiration, Energy and Physics Links | 呼吸作用、能量与物理联系
ction has similarities to burning a fuel. In physics, you learn that energy is measured in joules (J) or kilojoules (kJ). When measuring the energy content of food, a simple calorimetry experiment can be done: burning a piece of food under a boiling tube of water and measuring the temperature rise. The energy transferred can be estimated using: energy (J) = mass of water (g) × 4.2 × temperature rise (°C). Biology and physics merge in this investigation.
有氧呼吸利用氧气从葡萄糖中释放能量。文字方程式为：葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）。这个反应与燃烧燃料有相似之处。在物理中你学到能量以焦耳(J)或千焦(kJ)计量。当测量食物中的能量时，可以做简单的量热实验：在盛水的沸腾管下方燃烧一片食物并测量水温上升。传递的能量可以用以下公式估算：能量(J) = 水的质量(g) × 4.2 × 温度上升(°C)。这项研究中生物学与物理融合在了一起。
he temperature of 20 g of water by 15°C, the energy released = 20 × 4.2 × 15 = 1260 J. Energy per gram = 1260 J ÷ 0.5 g = 2520 J/g or 2.52 kJ/g. This works on the assumption that all heat goes to the water, but you should also discuss experimental errors, linking to working scientifically.
题目常要求你通过将总能量除以燃烧的食物质量来计算每克食物所含的能量。如果0.5 g薯片使20 g水升温15°C，则释放的能量 = 20 × 4.2 × 15 = 1260 J。每克能量 = 1260 J ÷ 0.5 g = 2520 J/g 或 2.52 kJ/g。这是基于所有热量都传递给水的假设，但你也要讨论实验误差，这与科学工作方法相联系。
6. Ecosystems and Geographical Distribution | 生态系统与地理分布
The distribution of organisms depends on abiotic factors such as temperature, sunlight, water and soil type. This topic links closely with geography, particularly when studying biomes. In WJEC Year 7, you may be asked to interpret a climate graph (climograph) showing monthly temperature and rainfall for a region, and then predict the type of ecosystem that might exist there. For example, high rainfall and high temperatures all year round suggest a tropical rainforest, while very low rainfall and high summer temperatures suggest a desert.
生物的分布取决于温度、光照、水分和土壤类型等非生物因素。这一主题与地理学紧密相连，尤其是在研究生物群系时。在 WJEC 七年级中，你可能会被要求解读显示某地区月均温和降雨量的气候图表，然后预测那里可能存在的生态系统类型。例如，全年高降雨量和高温意味着热带雨林，而极低的降雨量和夏季高温意味着沙漠。
You could also be asked to draw a line or bar graph from raw data, or to describe a pattern using geographical language, such as 'organisms decrease in number as you move away from the equator'. Understanding how to read map keys and use latitude lines is also a useful cross-curricular skill for explaining global species distribution.
你还可能被要求根据原始数据绘制折线图或柱状图，或者用地理语言描述规律，例如“随着远离赤道，生物数量减少”。懂得阅读地图图例和使用纬线也是解释全球物种分布的一项有用的跨学科技能。
7. Inheritance and Probability | 遗传与概率
ctly with probability and simple fractions in mathematics. For a single gene with two alleles, one dominant (B) and one recessive (b), a cross between two heterozygous parents (Bb × Bb) produces offspring genotypes of BB, Bb, Bb and bb. The probability of a dominant phenotype like brown eyes is 3 out of 4, or 3/4, or 75%. The probability of a recessive phenotype is 1/4 or 25%.
眼睛颜色或卷舌能力等特征的遗传可以用庞纳特方格来模拟。这将生物学与数学中的概率和简单分数直接联系起来。对于一个具有两个等位基因的单基因，一个显性(B)一个隐性(b)，两个杂合亲本交配(Bb × Bb)产生后代的基因型为BB、Bb、Bb和bb。显性表型（如棕色眼睛）的概率为4分之3，即3/4或75%。隐性表型的概率为1/4或25%。
s 4 children, the chance of all four showing the recessive trait is (¼)⁴ = 1/256. However, remember each child' o trace traits, interpreting symbols for male, female, affected and unaffected individuals – a task that requires logical thinking.
考题可以延伸至要求你用比率、百分比表示概率，或者用大样本预测数量。如果一对夫妇有4个孩子，四个都表现出隐性性状的几率是(¼)⁴ = 1/256。但要记住每个孩子的结果是独立的。运用分数和百分比构建了重要的算术技能。你或许还会使用家族谱系图来追踪性状，解读代表男性、女性、患者和正常个体的符号——这是一项需要逻辑思维的任务。
8. Infectious Disease and Data Analysis | 传染病与数据分析
ysing epidemiological data. A table or bar chart might show the number of cases in different regions or over time. You need to be able to describe trends, such as 'cases rose sharply between day 5 and day 10' or 'the region with the lowest vaccination rate had the highest infection rate' 000 people is a common cross-curricular task: (number of cases ÷ total population) × 1000.
在研究流感等病毒或霍乱等细菌性疾病的传播时，生物学常通过分析流行病学数据与地理学和数学相遇。表格或柱状图可能展示不同地区或不同时间的病例数。你需要能够描述趋势，如“病例在第5天至第10天之间急剧上升”或“疫苗接种率最低的地区感染率最高”。计算每1000人的感染率是一项常见的跨学科任务：(病例数 ÷ 总人口) × 1000。
You could also be given a line graph comparing two treatments and asked to find which one leads to a faster drop in symptoms. Drawing a smooth curve of best fit and reading values from axes are skills from your mathematics lessons. In addition, mapping disease outbreaks on a world map uses geographical skills. Always label your axes when drawing graphs and use a suitable scale. Cross-curricular questions often ask you to suggest reasons for a pattern, linking to environmental or social factors – this is where geographical knowledge really helps.
你也可能拿到一个比较两种治疗方案的折线图，并被要求找出哪一种能更快减轻症状。绘制一条平滑的最佳拟合曲线并从坐标轴上读取数值是数学课上的技能。此外，在世界地图上标示疾病爆发情况要用到地理技能。绘制图表时务必标注坐标轴并使用合适的尺度。综合题常会要求你为某个规律提供理由，并将其与环境或社会因素联系起来——此时地理知识确实很有帮助。
9. Experimental Design and Working Scientifically | 实验设计与科学探究
A good experiment tests only one independent variable at a time while keeping all other variables constant. This is a key principle linking biology with scientific thinking and mathematics. For example, when investigating how light intensity affects the rate of photosynthesis, you might change the distance of a lamp from a plant while measuring the number of oxygen bubbles produced per minute. The distance is the independent variable, the bubble count is the dependent variable, and variables such as temperature, CO₂ concentration and the type of plant must be controlled. Recognising and describing these variables is a common exam requirement.
一个好的实验每次只测试一个自变量，而保持其他所有变量不变。这是将生物学与科学思维及数学联系起来的关键原则。例如，在研究光照强度如何影响光合作用速率时，你可以改变灯与植物的距离，同时测量每分钟产生的氧气泡数量。距离是自变量，气泡数量是因变量，而温度、CO₂浓度和植物种类等变量必须保持不变。识别并描述这些变量是常见的考试要求。
You must also present results in a clear table, with headings that include the quantity and the unit, like 'Distance (cm)' and 'Bubbles per minute'. Plotting a graph with the independent variable on the x-axis is standard practice. Questions may ask you to identify anomalies in data and suggest improvements to the method, measuring a precise volume of gas with a syringe rather than counting bubbles, thus integrating practical skills and mathematical accuracy.
你还必须用清晰表格展示结果，表头应写明数量和单位，例如“距离 (cm)”和“每分钟气泡数”。将自变量标在x轴上绘制图表是标准做法。题目可能会要求你找出数据中的异常值并提出方法改进，例如用注射器精确测量气体体积而不是数气泡，从而整合实践技能与数学的精确性。
10. Nutrition, Food Labels and Daily Values | 营养、食品标签与每日数值
gy (kJ, kcal), as well as percentages of Reference Intake (RI). For instance, a cereal bar may contain 8 g of fat, and the label states this is 11% is links biology with real-life maths and health education. You may be asked to compare two products and decide which is healthier based on their fat, sugar, salt and fibre contents per 100 g.
阅读食品标签涉及理解质量单位(g, mg)和能量单位(kJ, kcal)，以及参考摄入量(RI)的百分比。例如，一条谷物棒含有8 g脂肪，标签标明这是RI的11%。你可以算出脂肪的RI大约是 8 ÷ 0.11 ≈ 73 g。这将生物学与现实生活中的数学及健康教育联系起来。你可能会被要求比较两种产品，并根据每100 g的脂肪、糖分、盐分和纤维含量判断哪一种更健康。
Some questions involve proportional reasoning: if a biscuit provides 5% ur daily calcium needs and contains 40 mg of calcium, what is the recommended daily amount? Here, 5% = 40 mg, so 100% and teeth. Simple calculations with percentages and ratios are embedded in these biological contexts.
有些问题涉及比例推理：如果一块饼干为你提供每日钙需量的5%，并含有40 mg钙，那么每日推荐量是多少？这里，5% = 40 mg，故100% = 40 mg × 20 = 800 mg。你还可能将这些知识与均衡饮食主题相联系，解释为什么我们需要纤维促进消化，或需要钙来构建强壮的骨骼和牙齿。百分比和比率的简单计算就嵌入在这些生物学情境中。
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