📚 A-Level Edexcel Biology: Calculation Practice Drills | A-Level Edexcel 生物：计算题专项训练
Calculation questions appear in every Edexcel A-Level Biology paper, from magnifying cells to evaluating statistical significance. This guide walks you through the core calculation types, step-by-step worked examples, and the key formulas you must recall under exam conditions. Strong numerical skills not only secure marks directly but also support accurate interpretation of data and experimental design.
计算题在 Edexcel A-Level 生物试卷中无处不在，从细胞放大测量到统计显著性检验。本指南将带你梳理核心计算类型，通过分步范例和必记公式，帮助你在考试中稳拿分数。扎实的计算能力不仅可以赢得直接分值，还能确保你正确解读实验数据、设计探究方案。
1. Magnification Calculations | 放大倍数计算
Magnification is the ratio between the size of an image and the actual size of the specimen. The fundamental equation is Magnification = Image size ÷ Actual size. Always ensure both measurements use the same units before dividing. Image size is usually given in millimetres or centimetres, while actual size is typically in micrometres.
放大倍数是图像大小与标本实际大小的比值。基本公式为 放大倍数 = 图像大小 ÷ 实际大小。计算前务必统一单位：图像大小常用毫米或厘米给出，而实际大小常以微米表示。
M = I ÷ A or A = I ÷ M
Worked example: An electron micrograph shows a mitochondrion image length of 30 mm. The magnifying power is ×20,000. Calculate the actual length in µm. Step 1: convert image size to µm (30 mm = 30,000 µm). Step 2: actual size = image size / magnification = 30,000 ÷ 20,000 = 1.5 µm.
范例：一张电镜照片中线粒体的图像长度为 30 mm，放大倍数为 ×20,000。计算其实际长度（µm）。步骤一：将图像大小转换为 µm（30 mm = 30,000 µm）。步骤二：实际大小 = 图像大小 ÷ 放大倍数 = 30,000 ÷ 20,000 = 1.5 µm。
2. Actual Size and Unit Conversions | 实际大小与单位换算
Edexcel questions frequently ask you to determine actual size from a diagram equipped with a scale bar. Measure the scale bar length, note the distance it represents, then use the ratio: Actual size of structure = (measured length of structure ÷ measured length of scale bar) × scale bar value. Be ready to convert among millimetres (mm), micrometres (µm) and nanometres (nm).
Edexcel 考题常要求根据带比例尺的示意图求出结构的实际大小。量取比例尺长度，记住其代表的距离，然后使用比值：结构实际大小 = (结构测量长度 ÷ 比例尺测量长度) × 比例尺值。同时要熟练在毫米 (mm)、微米 (µm) 和纳米 (nm) 之间转换。
1 mm = 1000 µm 1 µm = 1000 nm
For example, a chloroplast image has a scale bar of 20 mm representing 5 µm. A granum measures 12 mm on the image. Actual granum length = (12 mm ÷ 20 mm) × 5 µm = 0.6 × 5 µm = 3.0 µm.
例如，叶绿体图像中，比例尺长 20 mm 代表 5 µm。类囊体颗粒在图像上量得 12 mm。颗粒实际长度 = (12 mm ÷ 20 mm) × 5 µm = 0.6 × 5 µm = 3.0 µm。
3. Serial Dilutions and Concentrations | 连续稀释与浓度计算
Serial dilutions involve stepwise dilution of a stock solution, often by a constant factor (e.g., 1 in 10). The final concentration after n steps is Cₙ = C₀ × (dilution factor)ⁿ. You may also be asked to calculate volumes using C₁V₁ = C₂V₂ where a single dilution is made. Recognise when a dilution factor applies: a 10⁻² dilution means the original concentration is multiplied by 0.01.
连续稀释是将储存液按恒定倍数（如 1/10）逐级稀释的过程。经 n 步后的终浓度为 Cₙ = C₀ × (稀释因子)ⁿ。若为单步稀释，可用 C₁V₁ = C₂V₂ 求所需体积。注意识别稀释因子的含义：10⁻² 稀释即原浓度乘以 0.01。
Example: A glucose stock of 2.0 mol dm⁻³ undergoes three 1 in 10 serial dilutions. Final concentration = 2.0 × (0.1)³ = 2.0 × 0.001 = 0.002 mol dm⁻³.
示例：2.0 mol dm⁻³ 的葡萄糖储存液经过三次 1/10 连续稀释。终浓度 = 2.0 × (0.1)³ = 2.0 × 0.001 = 0.002 mol dm⁻³。
4. Probability and Genetic Ratios | 概率与遗传比例
Monohybrid and dihybrid crosses produce expected phenotypic ratios. You must demonstrate the use of Punnett squares and calculate probability of a specific genotype. For a heterozygous cross (Aa × Aa), the probability of a recessive homozygote (aa) is 1/4. When two independent genes interact, apply the product rule: multiply individual probabilities.
单基因和双基因杂交产生预期的表型比例。你必须能运用庞纳特方格并计算特定基因型的概率。对于杂合子杂交 (Aa × Aa)，隐性纯合子 (aa) 的概率为 1/4。当两个独立基因互作时，使用乘法法则：将各事件概率相乘。
Probability of A and B = P(A) × P(B)
For instance, the chance of getting an aaBB offspring from a cross AaBb × AaBb: P(aa) = 1/4, P(BB) = 1/4, combined probability = 1/4 × 1/4 = 1/16.
例如，由 AaBb × AaBb 获得 aaBB 后代的概率：P(aa) = 1/4, P(BB) = 1/4，合并概率 = 1/4 × 1/4 = 1/16。
5. Chi-Squared Test | 卡方检验 (χ²)
The chi-squared test compares observed results with expected outcomes to decide if differences are due to chance. The statistic is χ² = Σ((O – E)² ÷ E). Degrees of freedom (df) = number of categories minus 1. Compare your χ² value with the critical value at p = 0.05; if greater, reject the null hypothesis.
卡方检验将观察值与理论预期进行比较，判断差异是否由偶然造成。公式为 χ² = Σ((O – E)² ÷ E)。自由度 (df) = 类别数 – 1。将计算得到的 χ² 值与 p=0.05 下的临界值比较；若大于临界值，则拒绝无效假设。
Example: In a genetic cross, you expect 90 tall and 30 dwarf plants. Observed: 100 tall, 20 dwarf. χ² = (100-90)²/90 + (20-30)²/30 = (100/90) + (100/30) ≈ 1.11 + 3.33 = 4.44. With df=1, critical value is 3.84; since 4.44 > 3.84, the difference is significant.
范例：遗传杂交预期出现 90 高茎 30 矮茎，实际观察为 100 高茎 20 矮茎。χ² = (100-90)²/90 + (20-30)²/30 = (100/90) + (100/30) ≈ 1.11 + 3.33 = 4.44。自由度 1 时临界值为 3.84；4.44 > 3.84，差异显著。
6. Mark-Release-Recapture | 标记重捕法估算种群
The Lincoln index estimates population size (N) for motile organisms: N = (M × C) ÷ R, where M = number initially marked and released, C = total captured in second sample, R = number of marked individuals recaptured. Assumptions include random mixing, no migration, and marks not affecting survival.
林肯指数用于估算活动生物的种群大小 (N)：N = (M × C) ÷ R，其中 M 为初次标记并释放数，C 为第二次捕获总数，R 为第二次捕获中带标记个体数。假设包括随机混合、无迁入迁出、标记不影响存活。
A student marks 40 woodlice, releases them, and later collects 50, of which 10 are marked. N = (40 × 50) ÷ 10 = 2000 ÷ 10 = 200 woodlice.
某学生标记 40 只鼠妇并释放，随后捕捉 50 只，其中 10 只带有标记。N = (40 × 50) ÷ 10 = 2000 ÷ 10 = 200 只。
7. Energy Transfer and Trophic Efficiency | 能量传递与营养级效率
Ecological efficiency between trophic levels is calculated as Efficiency (%) = (Energy in biomass at higher level ÷ Energy in biomass at lower level) × 100. You may also see net production = gross production – respiratory losses. Always check units (kJ m⁻² yr⁻¹) and convert if needed.
营养级间生态效率计算公式为 效率 (%) = (上一级生物量能量 ÷ 下一级生物量能量) × 100。也会涉及净生产量 = 总生产量 – 呼吸损失。注意单位（通常为 kJ m⁻² yr⁻¹）并在必要时换算。
Efficiency = (Eᵤₚₚₑᵣ / Eₗₒwₑᵣ) × 100%
Example: A field of grass absorbs 2,000,000 kJ of sunlight per m² per year. The grass biomass contains 18,000 kJ m⁻² yr⁻¹. Efficiency of photosynthesis = (18,000 ÷ 2,000,000) × 100 = 0.9%.
范例：草地每平方米每年吸收阳光 2,000,000 kJ，草生物量含能 18,000 kJ m⁻² yr⁻¹。光合作用效率 = (18,000 ÷ 2,000,000) × 100 = 0.9%。
8. Respiratory Quotient (RQ) | 呼吸商
RQ indicates the substrate being respired: RQ = CO₂ produced ÷ O₂ consumed. An RQ of 1.0 suggests carbohydrate; 0.7 suggests lipids; around 0.9 indicates protein. In respirometer experiments, you may calculate O₂ uptake from manometer fluid movement and then derive RQ.
呼吸商揭示正在被氧化的呼吸底物：RQ = 产生的 CO₂ ÷ 消耗的 O₂。RQ 为 1.0 提示碳水化合物，0.7 提示脂肪，约 0.9 提示蛋白质。在呼吸计实验中，你可能需要从检压液移动计算 O₂ 吸收量，再求出 RQ。
A germinating seed consumes 0.42 cm³ O₂ and releases 0.36 cm³ CO₂ per minute. RQ = 0.36 ÷ 0.42 = 0.857, suggesting mixed lipids and protein.
某萌发种子每分钟消耗 0.42 cm³ O₂ 并释放 0.36 cm³ CO₂。RQ = 0.36 ÷ 0.42 = 0.857，提示混合脂类和蛋白质供能。
9. Cardiac Output Calculation | 心输出量计算
Cardiac output (CO) is the volume of blood pumped per minute: CO (dm³ min⁻¹) = Heart rate (bpm) × Stroke volume (dm³). Stroke volume is the volume ejected per beat. You may need to interpret data from an echocardiogram or calculate changes during exercise.
心输出量 (CO) 是每分钟心脏泵出的血液体积：心输出量 (dm³ min⁻¹) = 心率 (bpm) × 每搏输出量 (dm³)。每搏输出量是每次搏动射血量。需要能解读心电图数据或计算运动时的变化。
At rest, a student has a heart rate of 70 bpm and stroke volume of 0.07 dm³. CO = 70 × 0.07 = 4.9 dm³ min⁻¹. During exercise, heart rate rises to 130 bpm and stroke volume to 0.11 dm³; CO becomes 130 × 0.11 = 14.3 dm³ min⁻¹.
静息时，某学生心率 70 次/分，每搏输出量 0.07 dm³。CO = 70 × 0.07 = 4.9 dm³ min⁻¹。运动时心率升至 130 次/分、每搏输出量 0.11 dm³，CO = 130 × 0.11 = 14.3 dm³ min⁻¹。
10. Standard Deviation and Standard Error | 标准差与标准误
Standard deviation (s) measures spread around the mean. The formula for sample SD: s = √(Σ(x – x̄)² ÷ (n – 1)). Standard error of the mean (SE) = s ÷ √n. SE is used to plot error bars and judge overlap between data sets. If error bars do not overlap, the differences are likely significant.
标准差 (s) 衡量数据围绕均值的离散程度。样本标准差公式为 s = √(Σ(x – x̄)² ÷ (n – 1))。均值的标准误 (SE) = s ÷ √n。标准误用于绘制误差棒并判断数据组间重叠：若误差棒不重叠，差异很可能显著。
Data: 5, 6, 7, 8, 9. Mean x̄ = 7. Deviations: -2, -1, 0, 1, 2. Sum of squares = 4+1+0+1+4=10. n-1=4. s = √(10/4) = √2.5 ≈ 1.58. SE = 1.58 ÷ √5 ≈ 0.71.
数据：5, 6, 7, 8, 9。均值 x̄ = 7。离差：-2, -1, 0, 1, 2。平方和 4+1+0+1+4=10。n-1=4，s = √(10/4) = √2.5 ≈ 1.58。SE = 1.58 ÷ √5 ≈ 0.71。
11. Simpson’s Index of Diversity | 辛普森多样性指数
Biodiversity can be quantified using Simpson’s index: D = 1 – (Σ(n ÷ N)²), where n = total individuals of a particular species, N = total individuals of all species. High D (close to 1) indicates high diversity. Edexcel also accepts the reciprocal form, but the ‘1 minus’ version is standard for comparison.
生物多样性可用辛普森指数量化：D = 1 – (Σ(n ÷ N)²)，n 为某一物种的个体总数，N 为所有物种个体总数。D 值高（接近 1）代表多样性高。Edexcel 也接受倒数形式，但 “1 减” 版本更常用于比较。
A pond sample: 20 tadpoles, 15 water boatmen, 10 pond snails, 5 dragonfly nymphs. N = 50. Σ(n/N)² = (20/50)² + (15/50)² + (10/50)² + (5/50)² = 0.16+0.09+0.04+0.01 = 0.30. D = 1 – 0.30 = 0.70.
池塘样本：蝌蚪 20，划蝽 15，椎实螺 10，蜻蜓稚虫 5。N = 50。Σ(n/N)² = (20/50)² + (15/50)² + (10/50)² + (5/50)² = 0.16+0.09+0.04+0.01 = 0.30。D = 1 – 0.30 = 0.70。
12. Rate of Reaction and Enzyme Kinetics | 反应速率与酶动力学
To calculate initial reaction rate from a progress curve, draw a tangent at time zero and find its gradient: Rate = change in product ÷ change in time. Alternatively, if the reaction is linear initially, use Rate = 1 ÷ time taken to reach a set end point (e.g., disappearance of substrate). Units are usually concentration per unit time (e.g., mol dm⁻³ s⁻¹).
要从反应进程曲线求初始速率，在时间为零处作切线并计算斜率：速率 = 产物变化量 ÷ 时间变化量。若初始阶段为直线，也可用速率 = 1 ÷ 达到某终点的时间（如底物消失的时间）。单位通常为浓度每单位时间 (mol dm⁻³ s⁻¹)。
An enzyme-catalysed reaction produces 0.6 µmol of product in the first 20 seconds. The initial linear rate = 0.6 µmol ÷ 20 s = 0.03 µmol s⁻¹.
某酶促反应在最初 20 秒生成 0.6 µmol 产物。初始线性速率 = 0.6 µmol ÷ 20 s = 0.03 µmol s⁻¹。
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