📚 IB & Edexcel Biology: Calculation Practice | IB与Edexcel生物：计算题专项训练
Calculation questions form an essential part of both IB and Edexcel Biology examinations. From microscopy and cell counting to water potential and population genetics, being fluent in these quantitative skills can earn you easy marks and strengthen your analytical thinking. This article brings together the most common calculation types, explains the underlying principles, and walks you through worked examples.
计算题是IB和Edexcel生物考试的重要组成部分。从显微镜测量和细胞计数到水势和群体遗传学，熟练掌握这些定量技能不仅能帮你轻松拿分，还能提升分析思维。本文汇总了最常见的计算题型，解释原理，并带你一步步完成例题。
1. Microscopy Calculations | 显微镜计算
The essential formula is Magnification = Image size ÷ Actual size, often written as M = I / A. Both image size and actual size must be in the same units, so become confident converting between millimetres (mm) and micrometres (µm). Remember: 1 mm = 1000 µm.
核心公式是放大倍数 = 图像尺寸 ÷ 实际尺寸，常写作 M = I / A。图像尺寸和实际尺寸必须使用相同单位，因此要熟练进行毫米 (mm) 和微米 (µm) 之间的转换。请记住：1 mm = 1000 µm。
Example: An image of a root hair measures 75 mm in length. Its actual length is 0.05 mm. Magnification = 75 ÷ 0.05 = 1500×. If the actual length had been given as 50 µm, we could convert image size to µm: 75 mm = 75 000 µm; M = 75 000 ÷ 50 = 1500×.
示例：一张根毛图像的长度为 75 mm，实际长度为 0.05 mm。放大倍数 = 75 ÷ 0.05 = 1500×。如果实际长度给出的是 50 µm，可将图像尺寸换算为 µm：75 mm = 75 000 µm；M = 75 000 ÷ 50 = 1500×。
M = I / A (units must match)
2. Haemocytometer Cell Counting | 血球计数板细胞计数
Haemocytometers contain a grid of known dimensions. The central ruled area typically has a square of 1 mm × 1 mm, covered by a coverslip that leaves a depth of 0.1 mm. The volume above this square is therefore 1 mm × 1 mm × 0.1 mm = 0.1 mm³.
血球计数板上有已知尺寸的网格。通常中央计数区域包含一个 1 mm × 1 mm 的方格，盖上盖玻片后深度为 0.1 mm。因此该方格上方的体积为 1 mm × 1 mm × 0.1 mm = 0.1 mm³。
Because 1 cm³ = 1 mL and 1 cm³ = 1000 mm³, it follows that 1 mm³ = 0.001 mL. Thus 0.1 mm³ = 0.0001 mL. To obtain cells per mL, multiply the average count per 0.1 mm³ by 10 000 ( = 1 / 0.0001). If the sample was diluted, include the dilution factor: cells mL⁻¹ = average count × dilution factor × 10⁴.
由于 1 cm³ = 1 mL，且 1 cm³ = 1000 mm³，可得 1 mm³ = 0.001 mL。所以 0.1 mm³ = 0.0001 mL。要将计数结果换算为每 mL 的细胞数，需将 0.1 mm³ 内的平均计数乘以 10 000（= 1 / 0.0001）。如果样品经过稀释，还需乘以稀释因子：细胞数 mL⁻¹ = 平均计数 × 稀释倍数 × 10⁴。
Worked example: A yeast suspension is diluted 100 times. An average of 48 cells is counted in the central 0.1 mm³ square. Cell concentration = 48 × 100 × 10⁴ = 4.8 × 10⁷ cells mL⁻¹.
计算示例：酵母悬液被稀释了 100 倍。中央 0.1 mm³ 方格内平均计数为 48 个细胞。细胞浓度 = 48 × 100 × 10⁴ = 4.8 × 10⁷ 个 mL⁻¹。
3. Serial Dilutions and Standard Curves | 系列稀释与标准曲线
Serial dilutions reduce a concentrated stock solution stepwise by a constant factor. The dilution equation C₁V₁ = C₂V₂ allows you to calculate the new concentration after adding solvent. For instance, to prepare 10 mL of 0.1 mol L⁻¹ glucose from a 1 mol L⁻¹ stock, V₁ = (0.1 × 10) ÷ 1 = 1 mL; add 9 mL solvent.
系列稀释是将浓贮备液以恒定的倍数逐步稀释。稀释公式 C₁V₁ = C₂V₂ 可用来计算加入溶剂后的新浓度。例如，要由 1 mol L⁻¹ 的葡萄糖贮备液配制 10 mL 0.1 mol L⁻¹ 的溶液，所需 V₁ = (0.1 × 10) ÷ 1 = 1 mL；再加 9 mL 溶剂。
A dilution factor (DF) is often expressed as 1:10 or 10⁻¹. Repeating this step gives a geometric series: 10⁻¹, 10⁻², 10⁻³… Absorbance readings of these dilutions can be used to plot a standard curve of absorbance versus concentration, which then allows the concentration of an unknown sample to be read off the calibration line.
稀释因子常表示为 1:10 或 10⁻¹。重复这一步骤可得到几何级数：10⁻¹、10⁻²、10⁻³……测量这些稀释液的吸光度，可绘制吸光度-浓度标准曲线，随后即可从校准线上读取未知样品的浓度。
4. Water Potential and Osmosis | 水势与渗透作用
Water potential (ψ) determines the direction of water movement. It is the sum of the solute potential (ψₛ) and the pressure potential (ψₚ): ψ = ψₛ + ψₚ. In an open system, ψₚ is often zero, so ψ ≈ ψₛ.
水势 (ψ) 决定了水分子的运动方向。它是溶质势 (ψₛ) 与压力势 (ψₚ) 的总和：ψ = ψₛ + ψₚ。在开放体系中，压力势常为零，因此 ψ ≈ ψₛ。
Solute potential of a solution is calculated using ψₛ = –iCRT, where i is the ionisation constant (1 for sucrose), C is molar concentration (mol L⁻¹), R is the pressure constant (8.314 kPa L mol⁻¹ K⁻¹), and T is absolute temperature in Kelvin (K = °C + 273).
溶液的溶质势可用 ψₛ = –iCRT 计算，其中 i 为电离常数（蔗糖为1），C 为摩尔浓度 (mol L⁻¹)，R 为压力常数 (8.314 kPa L mol⁻¹ K⁻¹)，T 为绝对温度开尔文 (K = °C + 273)。
Example: Calculate the water potential of a 0.2 mol L⁻¹ sucrose solution at 20 °C, assuming ψₚ = 0. ψₛ = –1 × 0.2 × 8.314 × (20 + 273) = –1 × 0.2 × 8.314 × 293 ≈ –487 kPa. Hence ψ ≈ –487 kPa.
示例：计算 20°C 下 0.2 mol L⁻¹ 蔗糖溶液的水势，假设 ψₚ = 0。ψₛ = –1 × 0.2 × 8.314 × (20 + 273) = –1 × 0.2 × 8.314 × 293 ≈ –487 kPa。因此 ψ ≈ –487 kPa。
5. Enzyme Rate Calculations | 酶促反应速率计算
The initial rate of an enzyme‑catalysed reaction is found by drawing a tangent to the progress curve at time zero. Rate = Δproduct / Δtime or Δsubstrate used / Δtime. Common units are cm³ min⁻¹, µmol min⁻¹ or absorbance change per second.
计算酶促反应的初始速率，可在反应进程曲线上于时间为零处绘制切线。速率 = 产物增加量 / 时间间隔 或 底物消耗量 / 时间间隔。常用单位有 cm³ min⁻¹、µmol min⁻¹ 或吸光度每秒的变化。
Worked example: In a catalase experiment, the first 2 minutes of a tangent show a volume increase of 6.0 cm³ of O₂. Initial rate = 6.0 cm³ ÷ 2 min = 3.0 cm³ min⁻¹. If you need to convert to µmol min⁻¹, remember that 1 cm³ of O₂ at room temperature is roughly 40 µmol; 3.0 cm³ ≈ 120 µmol min⁻¹.
计算示例：在过氧化氢酶实验中，前 2 分钟切线段的 O₂ 体积增加了 6.0 cm³。初始速率 = 6.0 cm³ ÷ 2 min = 3.0 cm³ min⁻¹。若需换算为 µmol min⁻¹，可记住室温下 1 cm³ O₂ 约含 40 µmol；3.0 cm³ ≈ 120 µmol min⁻¹。
6. Respiratory Quotient (RQ) | 呼吸商
RQ = CO₂ produced / O₂ consumed. It indicates the respiratory substrate being used (e.g. carbohydrate ≈ 1.0, lipid ≈ 0.7). Using a respirometer containing KOH (which absorbs CO₂), the volume decrease reflects O₂ consumption. A control tube without KOH gives the net gas change, which is O₂ consumed minus CO₂ produced.
RQ = 产生的 CO₂ / 消耗的 O₂。该值可指示被利用的呼吸底物类型（如糖类约为 1.0，脂质约为 0.7）。使用含有 KOH（吸收 CO₂）的呼吸计，体积减少量即反映 O₂ 的消耗量。不装 KOH 的对照管则给出净气体变化，即耗 O₂ 量与产 CO₂ 量之差。
To find CO₂ produced: let X = O₂ consumed (with KOH), Y = net gas change (without KOH). Then CO₂ produced = X – Y. RQ = (X – Y) / X. Example: a respirometer shows a drop of 4.2 mm in the KOH tube and a drop of 1.8 mm in the plain tube. CO₂ produced = 4.2 – 1.8 = 2.4 mm; RQ = 2.4 / 4.2 ≈ 0.57, typical for lipids.
计算 CO₂ 产生量：令 X = 有 KOH 管中 O₂ 消耗量，Y = 无 KOH 管中净气体变化。则 CO₂ 产生量 = X – Y。RQ = (X – Y) / X。示例：呼吸计中 KOH 管液柱下降 4.2 mm，无 KOH 管下降 1.8 mm。CO₂ 产生量 = 4.2 – 1.8 = 2.4 mm；RQ = 2.4 / 4.2 ≈ 0.57，符合脂质特征。
7. Population Size Estimation – Lincoln Index | 种群大小估算 – 林肯指数
For motile organisms, mark-release-recapture uses Lincoln Index: N = (M × C) / R, where M = number captured and marked initially, C = total number captured in the second sample, R = number of marked individuals recaptured in the second sample.
对于能动的生物，标记重捕法使用林肯指数：N = (M × C) / R，其中 M 为初次捕获并标记的个体数，C 为第二次捕获的总个体数，R 为第二次捕获中带有标记的个体数。
Assumptions: marks do not affect survival or catchability; marked individuals mix randomly; no immigration, emigration, births or deaths between samples; marks are not lost.
假定的前提条件：标记不会影响存活率或被捕捉的概率；标记个体随机混合；两次取样之间没有迁入、迁出、出生或死亡；标记不会脱落。
Example: 30 beetles are marked and released. A few days later 40 beetles are captured, of which 10 are marked. Estimated N = (30 × 40) / 10 = 120 beetles.
示例：标记并释放了 30 只甲虫。几天后捕获 40 只甲虫，其中 10 只有标记。估算 N = (30 × 40) / 10 = 120 只。
8. Simpson’s Diversity Index | 辛普森多样性指数
Simpson’s index D = 1 – Σ (n / N)² quantifies biodiversity by considering both species richness and evenness. n = number of individuals of a particular species, N = total number of individuals of all species. A value close to 1 indicates high diversity.
辛普森指数 D = 1 – Σ (n / N)² 通过同时考量物种丰富度和均匀度来量化生物多样性。n 为某一物种的个体数，N 为所有物种的个体总数。数值越接近 1，表示多样性越高。
Example:
示例：
| Species | Count (n) | n/N | (n/N)² |
|---|---|---|---|
| A | 20 | 0.20 | 0.0400 |
| B | 30 | 0.30 | 0.0900 |
| C | 40 | 0.40 | 0.1600 |
| D | 10 | 0.10 | 0.0100 |
| Total | N = 100 | Σ = 0.3000 |
D = 1 – 0.3000 = 0.700.
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