📚 GCSE AQA Biology: Mastering Calculation Questions | GCSE AQA 生物：计算题专项训练
Calculation questions form a vital part of AQA GCSE Biology, appearing across topics from cell biology to ecology. Mastering these calculations requires a clear understanding of formulae, unit conversions, and methodical problem-solving. This guide will walk you through the key calculation types, offering step-by-step strategies and worked examples to boost your confidence for the exam.
计算题是 AQA GCSE 生物考试的重要组成部分，涉及从细胞生物学到生态学的各个主题。掌握这些计算需要清晰地理解公式、单位换算和有条理的解题方法。本指南将带你理解关键的计算类型，提供逐步解题策略和实例，增强你应对考试的信心。
1. Magnification Calculations | 放大倍数计算
Magnification describes how many times larger an image appears compared to the real object. The triangle formula helps you rearrange: cover the quantity you need to find.
放大倍数描述图像比实物大多少倍。三角形公式可帮助你进行公式变形：遮住你需要求的量即可。
Magnification = Image size ÷ Actual size
放大倍数 = 图像尺寸 ÷ 实际尺寸
Image size and actual size must be in the same unit. Usually image size is measured in millimetres (mm) and actual size in micrometres (µm). Always convert all measurements to the smallest unit given before calculating.
图像尺寸和实际尺寸必须使用相同单位。通常图像尺寸以毫米 (mm) 计量，实际尺寸以微米 (µm) 计量。计算前务必将所有测量值统一为最小单位。
| Unit | Equivalent in mm | 单位 | 相当于 mm |
|---|---|---|---|
| 1 cm | 10 mm | 1 厘米 | 10 毫米 |
| 1 mm | 1 mm | 1 毫米 | 1 毫米 |
| 1 µm | 0.001 mm | 1 微米 | 0.001 毫米 |
Worked example: An image of a cell measures 24 mm across. The actual cell diameter is 0.006 mm. Calculate the magnification.
例题：一个细胞图像的直径为 24 mm，实际细胞直径为 0.006 mm。计算放大倍数。
Step 1: Convert both to micrometres. Image size = 24 mm × 1000 = 24,000 µm. Actual size = 0.006 mm × 1000 = 6 µm.
步骤1：统一单位为微米。图像尺寸 = 24 mm × 1000 = 24000 µm。实际尺寸 = 0.006 mm × 1000 = 6 µm。
Step 2: Substitute into formula. Magnification = 24,000 µm ÷ 6 µm = 4000.
步骤2：代入公式。放大倍数 = 24000 µm ÷ 6 µm = 4000。
Answer: The magnification is ×4000.
答案：放大倍数为 ×4000。
2. Calculating Actual Size and Unit Conversions | 实际尺寸计算与单位换算
When you are given magnification and image size, rearrange the formula: Actual size = Image size ÷ Magnification. Unit conversions are often tested together with this skill.
当已知放大倍数和图像尺寸时，变形公式：实际尺寸 = 图像尺寸 ÷ 放大倍数。单位换算常常与此类题目同时考查。
Example: n is 60 mm long at a magnification of ×20,000. Calculate the actual length in micrometres.
例题：一张放大20000倍的线粒体图像长度为 60 mm，计算其实际长度（微米）。
Step 1: Image size in µm = 60 mm × 1000 = 60,000 µm.
步骤1：图像尺寸转换为微米 = 60 mm × 1000 = 60000 µm。
Step 2: Actual size = 60,000 µm ÷ 20,000 = 3 µm.
步骤2：实际尺寸 = 60000 µm ÷ 20000 = 3 µm。
Remember: 1 mm = 1000 µm; 1 cm = 10 mm. Always show your conversion steps to avoid mistakes.
记住：1 mm = 1000 µm；1 cm = 10 mm。要始终展示你的换算步骤以避免错误。
3. Surface Area to Volume Ratio | 表面积与体积比
As an object gets larger, its surface area to volume ratio (SA:V) decreases. This limits the size of cells because diffusion of substances becomes less efficient. To calculate SA:V, work out surface area and volume separately, then write as a ratio, simplifying to 1:n.
随着物体变大，其表面积与体积比（SA:V）会下降。这限制了细胞的大小，因为物质扩散效率会降低。要计算 SA:V，需分别计算出表面积和体积，然后写出比值并简化为 1:n。
For a cube: Surface area = 6 × side²; Volume = side³.
对于立方体：表面积 = 6 × 边长²；体积 = 边长³。
Worked example: 6 × (2 cm)² = 6 × 4 cm² = 24 cm². V = (2 cm)³ = 8 cm³. SA:V = 24 : 8, simplified to 3 : 1.
例题：一个立方体形细胞边长为 2 cm。表面积 SA = 6 × (2 cm)² = 6 × 4 cm² = 24 cm²。体积 V = (2 cm)³ = 8 cm³。SA:V = 24 : 8，简化为 3 : 1。
³, SA:V = 12 : 1. Smaller objects have a much higher SA:V, making exchange more efficient.
若同一个细胞边长为 0.5 cm，则 SA = 1.5 cm²，V = 0.125 cm³，SA:V = 12 : 1。较小的物体具有高得多的 SA:V，使物质交换更高效。
4. Estimating Population Size Using Quadrats | 使用样方法估算种群数量
For sessile or slow-moving organisms, quadrats are used to sample an area. The population estimate is then calculated using the mean per quadrat and the total area.
对固着或移动缓慢的生物，使用样方对一片区域进行取样。然后利用每个样方的平均数量与总面积来估算种群数量。
Estimated population = Mean number per quadrat × (Total area ÷ Area of one quadrat)
估算种群数量 = 每样方平均个体数 × (总面积 ÷ 单个样方面积)
Example: ld area is 200 m². Population estimate = 4.2 × (200 ÷ 0.25) = 4.2 × 800 = 3360 dandelions.
例题：10个 0.5 m × 0.5 m 的样方（面积均为 0.25 m²）得到蒲公英平均数为 4.2。场地总面积为 200 m²。种群估算 = 4.2 × (200 ÷ 0.25) = 4.2 × 800 = 3360 株蒲公英。
Place quadrats randomly using random coordinates to get a representative sample.
使用随机坐标放置样方以获得代表性样本。
5. Capture-Recapture Method | 标记重捕法
For mobile animals, the Lincoln index estimates population size:
对于移动性动物，林肯指数估算种群大小：
on size = (Number in first sample × Number in second sample) ÷ Number of marked recaptured
种群数量 = (第一次样本个体数 × 第二次样本个体数) ÷ 重捕到的标记个体数
Example: 0 woodlice are caught, of which 8 are marked. Population = (30 × 40) ÷ 8 = 1200 ÷ 8 = 150.
例题：首次捕捉 30 只鼠妇，标记后释放。之后捕捉 40 只，其中 8 只带标记。种群数量 = (30 × 40) ÷ 8 = 1200 ÷ 8 = 150。
Assumptions: no births, deaths, immigration or emigration between samples; marking does not affect survival; marked individuals mix evenly. Be ready to evaluate validity when assumptions are broken.
假设条件：两次采样之间无出生、死亡、迁入或迁出；标记不影响存活率；标记个体均匀混合。当假设被破坏时，要做好评估可靠性的准备。
6. Energy Transfer Efficiency | 能量传递效率
In a food chain, only a fraction of energy is transferred from one trophic level to the next. Efficiency is calculated using:
在食物链中，只有一小部分能量从一个营养级传递到下一个。其效率计算如下：
Efficiency (%) = (Energy transferred to next level ÷ Energy available at previous level) × 100%
效率 (%) = (传递至下一级的能量 ÷ 上一级可用的能量) × 100%
Example: 20,000 kJ of energy; primary consumers obtain 2,000 kJ. Efficiency = (2000 ÷ 20000) × 100% = 10%.
例题：植物含有 20000 kJ 能量；初级消费者获得 2000 kJ。效率 = (2000 ÷ 20000) × 100% = 10%。
You may also be given biomass figures; apply the same formula. Always present your answer to two significant figures if the question requests.
也可能给的是生物量数据；应用相同公式。如题目要求，结果通常保留两位有效数字。
7. Percentage Change Calculations | 百分比变化计算
Percentage change is used extensively in osmosis, growth and population studies. The formula is:
百分比变化广泛应用于渗透作用、生长和种群研究中。其公式为：
Percentage change = (Final value - Initial value) ÷ Initial value × 100%
百分比变化 = (终值 - 初值) ÷ 初值 × 100%
A negative result indicates a decrease. For example, a potato chip's mass drops from 5.0 g to 4.2 g in sugar solution. % change = (4.2 - 5.0) ÷ 5.0 × 100% = -16%. This implies water loss by osmosis.
负值表示减少。例如，土豆片在糖溶液中质量从 5.0 g 降至 4.2 g。百分比变化 = (4.2 - 5.0) ÷ 5.0 × 100% = -16%。这表明渗透失水。
Always include the negative sign where appropriate and ensure you divide by the initial value, not the final one.
需要时务必写上负号，并确保除以初值而非终值。
8. Genetic Probability | 遗传概率
Monohybrid inheritance questions often ask for the probability of an offspring having a certain genotype or phenotype. Use a Punnett square and express the answer as a fraction, percentage or ratio.
单基因遗传题常会问及子代具有某种基因型或表型的概率。使用庞纳特方格，答案可以表示为分数、百分数或比例。
Example: ort. The Punnett square yields TT, Tt, Tt, tt. The probability of a tall plant = 3/4 or 75%. Probability of short plant = 1/4 or 25%.
例题：两株杂合高茎豌豆 (Tt) 杂交。等位基因 T 为
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