Year 11 Edexcel Biology: Interdisciplinary Integrated Question Practice | 跨学科综合题型训练

📚 Year 11 Edexcel Biology: Interdisciplinary Integrated Question Practice | 跨学科综合题型训练

In the Edexcel Year 11 Biology course, many exam questions require you to link biological concepts with ideas from chemistry, physics, mathematics, and geography. These interdisciplinary questions test your ability to apply knowledge, analyse data, and solve problems across subject boundaries. This revision guide provides worked examples of common question types to help you build confidence and master the skills needed for top marks.

在 Edexcel 十一年级生物课程中，许多考题要求你将生物学概念与化学、物理、数学和地理等学科的思想联系起来。这些跨学科题目考查你运用知识、分析数据和跨学科解决问题的能力。本复习指南提供常见题型的详细例题，帮助你树立信心，掌握获取高分所需的技巧。

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1. Enzyme Activity and Rate Calculations | 酶活性与速率计算

Question: A student investigated the breakdown of hydrogen peroxide by catalase. The volume of oxygen produced was recorded every 10 seconds.

题目：一名学生研究了过氧化氢酶对过氧化氢的分解。每 10 秒记录一次产生的氧气体积。

Time / s	Volume of O₂ / cm³
0	0
10	12
20	20
30	26
40	29
50	30
60	30

Use the data to calculate the mean rate of reaction between 0 and 30 seconds, and explain why the rate changes after 40 seconds.

利用数据计算 0 至 30 秒之间的平均反应速率，并解释 40 秒后速率为何发生变化。

Step 1: Identify the volume change. At 0 s the volume is 0 cm³; at 30 s the volume is 26 cm³. The total volume produced is 26 cm³.

步骤 1：找出体积变化。0 秒时体积为 0 cm³；30 秒时体积为 26 cm³。产生的总体积为 26 cm³。

= change in volume ÷ time taken = 26 cm³ ÷ 30 s = 0.867 cm³/s (to 3 significant figures).

步骤 2：使用速率公式。速率 = 体积变化 ÷ 所用时间 = 26 cm³ ÷ 30 s = 0.867 cm³/s（保留三位有效数字）。

Step 3: Explain the decrease in rate. After 40 s the gradient of the line flattens. The substrate (hydrogen peroxide) concentration decreases, so fewer enzyme–substrate complexes form per second, leading to a lower rate. Eventually the substrate is used up.

步骤 3：解释速率的下降。40 秒后曲线斜率趋于平缓。底物（过氧化氢）浓度降低，因此每秒形成的酶‑底物复合物减少，导致速率下降。最终底物耗尽。

Cross‑ rate = Δy/Δx, which is the same as the gradient formula you use in physics and chemistry.

跨学科联系：该题将生物学与速率的数学计算以及图表解读相结合。你需要使用速率 = Δy/Δx 公式，这与物理和化学中使用的斜率公式相同。

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2. Photosynthesis Equation and Stoichiometry | 光合作用方程式与化学计量

alculate the mass of glucose it could produce. (Relative atomic masses: C = 12, O = 16, H = 1)

题目：光合作用的配平方程式为 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂。一株番茄植物在一个晴天吸收了 88 克二氧化碳。计算它可能产生的葡萄糖质量。（相对原子质量：C = 12，O = 16，H = 1）

Step 1: Find the molar mass of CO₂. Mᵣ(CO₂) = 12 + (16 × 2) = 44 g/mol.

步骤 1：求出 CO₂ 的摩尔质量。Mᵣ(CO₂) = 12 + (16 × 2) = 44 g/mol。

p 2: Calculate the number of moles of CO₂ absorbed. Moles = mass / Mᵣ = 88 ÷ 44 = 2.0 mol.

步骤 2：计算吸收的 CO₂ 的物质的量。物质的量 = 质量 / Mᵣ = 88 ÷ 44 = 2.0 mol。

uation. 6 mol CO₂ produce 1 mol C₆H₁₂O₆. So 2 mol CO₂ produce (2 ÷ 6) = 0.333 mol glucose.

步骤 3：利用方程式中的物质的量之比。6 mol CO₂ 生成 1 mol C₆H₁₂O₆。因此 2 mol CO₂ 生成 (2 ÷ 6) = 0.333 mol 葡萄糖。

ᵣ(glucose) = (12×6) + (1×12) + (16×6) = 180 g/mol. Mass = moles × Mᵣ = 0.333 × 180 = 60 g.

步骤 4：求出葡萄糖的质量。Mᵣ（葡萄糖） = (12×6) + (1×12) + (16×6) = 180 g/mol。质量 = 物质的量 × Mᵣ = 0.333 × 180 = 60 g。

Cross‑disciplinary link: This problem uses chemical equations and stoichiometry, which are central to GCSE Chemistry. You must be able to calculate molar masses, convert between mass and moles, and apply mole ratios.

跨学科联系：该题运用了化学方程式和化学计量学，这些是 GCSE 化学的核心内容。你必须能够计算摩尔质量、在质量与物质的量之间转换，并应用物质的量之比。

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3. Respiration and Energy Values | 呼吸作用与能量值

Question: The aerobic respiration of glucose releases 2870 kJ of energy per mole. A cereal bar contains 35 g of carbohydrate, assumed to be entirely glucose. (a) Calculate the energy available from the cereal bar in kJ. (b) A 60 kg person uses roughly 8400 kJ per day. What percentage of their daily energy need does one cereal bar provide?

题目：葡萄糖的有氧呼吸每摩尔释放 2870 kJ 能量。一根谷物棒含有 35 克碳水化合物，假设全为葡萄糖。(a) 计算该谷物棒提供的能量（kJ）。(b) 一名 60 千克的人每天大约消耗 8400 kJ。一根谷物棒提供了每日能量需求的百分之多少？

≈ 0.1944 mol. Energy = 0.1944 × 2870 ≈ 558 kJ. (Approx. 560 kJ to 2 significant figures.)

(a) 部分：首先求出葡萄糖的物质的量。Mᵣ(C₆H₁₂O₆) = 180 g/mol。物质的量 = 35 ÷ 180 ≈ 0.1944 mol。能量 = 0.1944 × 2870 ≈ 558 kJ。（约 560 kJ，取两位有效数字。）

Part (b): Percentage = (energy from bar ÷ daily need) × 100 = (558 ÷ 8400) × 100 ≈ 6.6 %.

(b) 部分：百分比 =（谷物棒能量 ÷ 每日需求）× 100 = (558 ÷ 8400) × 100 ≈ 6.6 %。

Cross‑disciplinary link: This question links biology with nutritional science and mathematics. You must handle energy conversions, use the mole concept from chemistry, and perform percentage calculations—skills that also appear in physics and everyday contexts.

跨学科联系：该题将生物学与营养科学及数学联系起来。你必须处理能量转换，运用化学中的物质的量概念，并进行百分比计算——这些技能也出现在物理和日常生活情景中。

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4. Genetics and Probability | 遗传学与概率

Question: Cystic fibrosis (CF) is caused by a recessive allele ‘c’. Two parents are both carriers (Cc). They plan to have three children. Calculate the probability that exactly two of the three children will have cystic fibrosis.

题目：囊性纤维化（CF）由隐性等位基因 ‘c’ 引起。父母双方均为携带者（Cc）。他们计划要三个孩子。计算恰好有两个孩子患囊性纤维化的概率。

c gives: CC, Cc, cC, cc. Probability of CF (cc) = 1/4; probability of not having CF = 3/4.

步骤 1：画出一个孩子的庞纳特方格。杂交 Cc × Cc 产生：CC、Cc、cC、cc。患 CF（cc）的概率 = 1/4；不患 CF 的概率 = 3/4。

C₂ = 3 ways. Probability = ₃C₂ × (1/4)² × (3/4)¹ = 3 × (1/16) × (3/4) = 3 × (3/64) = 9/64.

步骤 2：使用二项式概率。对于三个孩子中恰好两个患 CF，我们需要组合 ₃C₂ = 3 种方式。概率 = ₃C₂ × (1/4)² × (3/4)¹ = 3 × (1/16) × (3/4) = 3 × (3/64) = 9/64。

Step 3: Express as a decimal or percentage. 9/64 ≈ 0.141 or 14.1 %.

步骤 3：以小数或百分比表示。9/64 ≈ 0.141 或 14.1 %。

Cross‑disciplinary link: This genetic problem requires the use of combinatorial probability from mathematics. Being able to move between the Punnett square (biology) and probability calculations (maths) is essential for solving inheritance questions involving multiple offspring.

跨学科联系：这道遗传学题需要运用数学中的组合概率。能够在庞纳特方格（生物学）和概率计算（数学）之间进行转换，对解决涉及多个后代的遗传题目至关重要。

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5. Sampling and Population Estimates | 取样与种群估计

Question: A student used the mark–release–recapture method to estimate the population of woodlice in a garden. On day 1, she caught 40 woodlice, marked them with a dot of non‑toxic paint, and released them. On day 2, she caught 50 woodlice, of which 8 were marked. (a) Calculate the estimated population size. (b) State one assumption made when using this method.

题目：一名学生使用标记‑释放‑重捕法估算花园中鼠妇的种群数量。第一天她捕获了 40 只鼠妇，用无毒颜料做标记后释放。第二天她捕获了 50 只鼠妇，其中 8 只有标记。(a) 计算估算的种群大小。(b) 说出使用该方法时的一条假设。

cond sample = 50, R = number of marked recaptures = 8. N = (40 × 50) / 8 = 2000 / 8 = 250.

(a) 部分：使用林肯指数：估算的种群数量 N = (M × C) / R，其中 M = 第一次样本中标记的数量 = 40，C = 第二次捕获总数 = 50，R = 重捕中有标记的数量 = 8。N = (40 × 50) / 8 = 2000 / 8 = 250。

Part (b): Assumptions include: marked individuals mix randomly with the rest of the population; no immigration, emigration, deaths, or births between samples; the mark does not affect survival or chance of recapture; marked marks are not lost. (Any one)

(b) 部分：假设包括：标记个体与种群其余个体随机混合；两次取样之间没有迁入、迁出、死亡或出生；标记不影响生存或重捕概率；标记不会脱落。（任答一点）

Cross‑disciplinary link: This ecological method applies statistical sampling theory from mathematics. The Lincoln index is an algebraic equation, and you must understand how variables affect the accuracy of the estimate, similar to evaluating data quality in geography fieldwork.

跨学科联系：该生态学方法应用了数学中的统计抽样理论。林肯指数是一个代数方程，你必须理解变量如何影响估算的准确性，这类似于地理野外考察中对数据质量的评估。

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6. Nerve Impulses and Electrical Potentials | 神经冲动与电位

Question: The graph below shows the changes in membrane potential during an action potential in a neurone. The resting potential is –70 mV. The peak is at +40 mV. The whole event lasts 5 ms. Calculate the rate of depolarisation (in mV/ms) from the resting potential to the peak.

题目：下图显示了神经元动作电位期间膜电位的变化。静息电位为 –70 mV。峰值在 +40 mV。整个过程持续 5 ms。计算从静息电位到峰值的去极化速率（单位为 mV/ms）。

Step 1: Determine the total voltage change. From –70 mV to +40 mV, the change = 40 – (–70) = 110 mV.

步骤 1：确定总的电压变化。从 –70 mV 到 +40 mV，变化 = 40 – (–70) = 110 mV。

risation. From the graph, the rising phase (depolarisation) typically takes about 1 ms. We' me 1 ms for the voltage to rise. Rate = voltage change / time = 110 mV / 1 ms = 110 mV/ms.

步骤 2：使用去极化所需时间。从图中，上升阶段（去极化）通常约需 1 ms。我们假设电压升高需时 1 ms。速率 = 电压变化 / 时间 = 110 mV / 1 ms = 110 mV/ms。

110/5 = 22 mV/ms, which is not the depolarisation rate. Always identify the correct phase.

步骤 3：如果计算中错误地使用了整个动作电位持续时间（5 ms），将得到 110/5 = 22 mV/ms，这不是去极化速率。务必找准正确的阶段。

Cross‑disciplinary link: This question applies the physics concept of potential difference (voltage) and rate of change. The calculation mirrors the gradient determination of a voltage–time graph, a skill practised in electricity topics in physics.

跨学科联系：该题运用了物理中的电位差（电压）和变化率概念。该计算与电压‑时间图像的斜率确定方法相似，这是物理电学主题中练习过的技能。

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7. The Eye and Optics | 眼睛与光学

Question: Explain how the eye focuses light from a distant object and why a person with short‑sight (myopia) uses a diverging (concave) lens for correction. Use the terms cornea, lens, retina, focal point, and diverging light.

题目：解释眼睛如何聚焦来自远处物体的光线，以及为什么近视患者用发散（凹）透镜进行矫正。使用角膜、晶状体、视网膜、焦点和发散光线这些术语。

Explanation: Light from a distant object arrives in nearly parallel rays. The cornea and lens together refract the light to focus it on the retina, forming a sharp image. In a short‑sighted eye, the eyeball is too long or the lens is too powerful, so the light is focused in front of the retina, causing a blurred image. A concave lens diverges the incoming parallel rays slightly before they enter the eye, effectively moving the focal point backwards onto the retina.

解释：来自远处物体的光线以近乎平行射入。角膜和晶状体共同折射光线，将其聚焦在视网膜上，形成清晰图像。在近视眼中，眼球过长或晶状体屈光力过强，光线聚焦在视网膜前方，导致图像模糊。凹透镜在光线进入眼睛前使平行光线微微发散，从而将焦点后移，恰好落在视网膜上。

Cross‑disciplinary link: This topic directly overlaps with physics—optics. You need to understand how convex and concave lenses change the path of light rays, and how focal length relates to the power of a lens. Ray diagrams from physics are used to justify the biological correction.

跨学科联系：该主题与物理光学直接重叠。你需要理解凸透镜和凹透镜如何改变光线路径，以及焦距与透镜度数之间的关系。物理中的光线图可用于证明生物学矫正的原理。

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8. Carbon Cycle and Numerical Models | 碳循环与数值模型

Question: The table shows the mass of carbon stored in different global reservoirs.

Reservoir	Carbon stored (Gt)
Atmosphere	830
Terrestrial biomass	560
Ocean surface waters	900
Fossil fuels	4000

If humans burned enough fossil fuel to release an extra 200 Gt of carbon into the atmosphere, calculate the percentage increase in the atmospheric carbon reservoir. Suggest two effects this might have on ecosystems.

如果人类燃烧化石燃料，额外向大气释放 200 Gt 碳，计算大气碳储库增加的百分比。列举这可能对生态系统产生的两个影响。

e increase = (extra carbon / original atmospheric carbon) × 100 = (200 / 830) × 100 ≈ 24.1 %.

增加百分比 =（额外碳量 / 原始大气碳量）× 100 = (200 / 830) × 100 ≈ 24

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