📚 Edexcel Maths 2.2 Biological Molecules: Quantitative Skills and Models | Edexcel 数学 2.2 生物分子：定量技能与模型
In Edexcel A-level Mathematics, contextual problems often use biological molecules as a setting to test core skills such as ratio, proportion, exponential growth and decay, logarithms, differentiation and integration. This article links the biology topic ‘Biological Molecules’ with the mathematical techniques required in Edexcel Maths, focusing on calculations involving concentration, dilution, reaction rates and standard curves.
在 Edexcel A-level 数学中，应用题经常以生物分子为背景，考查比例、指数增长与衰减、对数、微分和积分等核心技能。本文将生物学科中的“生物分子”主题与 Edexcel 数学所需的技术结合起来，重点讲解涉及浓度、稀释、反应速率和标准曲线的计算。
1. Concentration, Moles and Molar Mass | 浓度、摩尔与摩尔质量
In biochemistry, the amount of a substance is often measured in moles. The number of moles n is found by dividing the mass m by the molar mass M, while the concentration c is the number of moles per unit volume V. These two equations are fundamental in Edexcel Maths applied contexts.
在生物化学中，物质的量通常用摩尔表示。摩尔数 n 等于质量 m 除以摩尔质量 M，而浓度 c 是单位体积 V 中的摩尔数。这两个方程是 Edexcel 数学应用题中的基础。
n = m / M and c = n / V
For example, glucose has the formula C₆H₁₂O₆ and a molar mass of 180 g mol⁻¹. If 9 g of glucose is dissolved in 0.5 dm³ of water, the number of moles is n = 9 / 180 = 0.05 mol, and the concentration is c = 0.05 / 0.5 = 0.10 mol dm⁻³.
例如，葡萄糖的分子式为 C₆H₁₂O₆，摩尔质量为 180 g mol⁻¹。如果将 9 g 葡萄糖溶解在 0.5 dm³ 水中，摩尔数 n = 9 / 180 = 0.05 mol，浓度 c = 0.05 / 0.5 = 0.10 mol dm⁻³。
- Always convert volume to dm³ if concentration is in mol dm⁻³.
- 如果浓度单位是 mol dm⁻³，一定要把体积换算成 dm³。
2. Dilution and the Dilution Factor | 稀释与稀释因子
Serial dilution is a common technique when preparing biological molecule solutions. The key mathematical principle is that the amount of solute remains constant before and after dilution, giving the equation c₁V₁ = c₂V₂.
连续稀释是制备生物分子溶液时的常用技术。其核心数学原理是稀释前后溶质的量保持不变，因此得到方程 c₁V₁ = c₂V₂。
c₁V₁ = c₂V₂
If 2.0 cm³ of a 0.50 mol dm⁻³ protein solution is diluted with water to a final volume of 10.0 cm³, the new concentration is c₂ = (0.50 × 2.0) / 10.0 = 0.10 mol dm⁻³. A ten-fold dilution has a dilution factor of 10, often written as 10⁻¹.
如果将 2.0 cm³ 浓度为 0.50 mol dm⁻³ 的蛋白质溶液加水稀释至最终体积 10.0 cm³，新浓度 c₂ = (0.50 × 2.0) / 10.0 = 0.10 mol dm⁻³。十倍稀释的稀释因子为 10，常写作 10⁻¹。
3. Standard Curves and Linear Regression | 标准曲线与线性回归
Spectrophotometry is used to determine the concentration of biological molecules such as DNA or proteins. The absorbance A of a sample is directly proportional to its concentration c, so a calibration graph of A against c gives a straight line through the origin: A = mc.
分光光度法用于测定 DNA 或蛋白质等生物分子的浓度。样品的吸光度 A 与其浓度 c 成正比，因此以 A 对 c 作图得到一条过原点的直线：A = mc。
A = mc + b
In Edexcel Statistics, the least-squares regression line y = a + bx is often used. If a standard curve has equation A = 0.025c + 0.002, and an unknown sample has an absorbance of 0.207, then c = (0.207 – 0.002) / 0.025 = 8.2 units.
在 Edexcel 统计中，常使用最小二乘回归直线 y = a + bx。如果标准曲线方程为 A = 0.025c + 0.002，某未知样品的吸光度为 0.207，则 c = (0.207 – 0.002) / 0.025 = 8.2 单位。
| Concentration c / μg cm⁻³ | Absorbance A |
|---|---|
| 0 | 0.002 |
| 2 | 0.052 |
| 4 | 0.102 |
| 6 | 0.152 |
| 8 | 0.202 |
4. pH and Logarithmic Scales | pH 与对数尺度
The pH scale is a logarithmic measure of hydrogen ion concentration. This topic directly tests Edexcel Pure Maths logarithm skills, including base 10 logarithms and inverse operations.
pH 标度是氢离子浓度的对数度量。这一主题直接考查 Edexcel 纯数学中的对数技能，包括以 10 为底的对数及其逆运算。
pH = -log₁₀[H⁺]
To find the hydrogen ion concentration from pH, the inverse operation is used: [H⁺] = 10⁻ᵖᴴ. For example, if the pH of a solution containing a biological buffer is 4.8, then [H⁺] = 10⁻⁴·⁸ ≈ 1.58 × 10⁻⁵ mol dm⁻³.
要从 pH 求氢离子浓度，使用逆运算：[H⁺] = 10⁻ᵖᴴ。例如，如果含有生物缓冲液的溶液 pH 为 4.8，则 [H⁺] = 10⁻⁴·⁸ ≈ 1.58 × 10⁻⁵ mol dm⁻³。
- Remember that a difference of 1 pH unit means a ten-fold change in [H⁺].
- 记住，pH 相差 1 个单位意味着 [H⁺] 相差十倍。
5. Exponential Growth of Bacterial Cultures | 细菌培养的指数增长
Biological molecules such as nutrients affect the growth rate of bacterial cultures. When resources are unlimited, the number of cells N follows exponential growth with time t. The continuous growth model is N(t) = N₀eᵏᵗ, where N₀ is the initial number and k is the growth rate constant.
营养物质等生物分子会影响细菌培养物的生长速率。当资源无限时，细胞数量 N 随时间 t 呈指数增长。连续增长模型为 N(t) = N₀eᵏᵗ，其中 N₀ 是初始数量，k 是增长速率常数。
N(t) = N₀eᵏᵗ
If a culture starts with 500 cells and doubles every 3 hours, the growth rate is k = ln 2 / 3 ≈ 0.231 h⁻¹. The number after 10 hours is N = 500e^(0.231×10) ≈ 500 × 10.08 ≈ 5040 cells.
如果培养物从 500 个细胞开始，每 3 小时翻一番，则增长速率 k = ln 2 / 3 ≈ 0.231 h⁻¹。10 小时后的数量为 N = 500e^(0.231×10) ≈ 500 × 10.08 ≈ 5040 个细胞。
6. Exponential Decay and Half-Life of Biological Molecules | 生物分子的指数衰减与半衰期
Many biological molecules degrade over time, especially radioactive tracers or unstable proteins. The amount remaining follows exponential decay, and the half-life t₁/₂ is the time taken for the quantity to halve.
许多生物分子会随时间降解，尤其是放射性示踪剂或不稳定蛋白质。剩余量遵循指数衰减，半衰期 t₁/₂ 是数量减半所需的时间。
M(t) = M₀e⁻ᵏᵗ and t₁/₂ = ln 2 / k
Suppose a labelled amino acid has a decay constant k = 0.05 h⁻¹. Its half-life is t₁/₂ = 0.693 / 0.05 = 13.86 h. To find the time for 90% decay, solve e⁻⁰·⁰⁵ᵗ = 0.10, giving t = -ln(0.10) / 0.05 ≈ 46.1 h.
假设一种标记氨基酸的衰减常数 k = 0.05 h⁻¹，其半衰期 t₁/₂ = 0.693 / 0.05 = 13.86 h。要求衰减 90% 所需的时间，解方程 e⁻⁰·⁰⁵ᵗ = 0.10，得 t = -ln(0.10) / 0.05 ≈ 46.1 h。
7. Reaction Rates and the Michaelis-Menten Equation | 反应速率与米氏方程
Enzyme-catalysed reactions involving biological molecules often follow the Michaelis-Menten model. The initial rate v depends on substrate concentration [S] through a rational function, which is a rich source of Pure Maths questions on asymptotes and limits.
涉及生物分子的酶催化反应通常遵循米氏模型。初始速率 v 通过有理函数依赖于底物浓度 [S]，这是纯数学中关于渐近线和极限的丰富题目来源。
v = V_max[S] / (K_m + [S])
As [S] becomes very large, the denominator approaches [S], so v tends to V_max, the maximum rate. The Michaelis constant K_m is the substrate concentration at which v = V_max / 2, giving a simple algebraic substitution problem.
当 [S] 非常大时，分母趋近于 [S]，因此 v 趋近于最大速率 V_max。米氏常数 K_m 是当 v = V_max / 2 时的底物浓度，这是一个简单的代数代入问题。
8. Differentiation: Rate of Change of Concentration | 微分：浓度变化率
In Edexcel Pure Maths, differentiation is used to model the rate of change of a substrate concentration. For a first-order reaction, the rate of change of [S] is proportional to [S] itself, written as d[S]/dt = -k[S].
在 Edexcel 纯数学中，微分用于建立底物浓度变化率的模型。对于一级反应，[S] 的变化率与 [S] 本身成正比，写作 d[S]/dt = -k[S]。
d[S]/dt = -k[S]
The solution to this differential equation is [S] = [S]₀e⁻ᵏᵗ, which links back to exponential decay. The derivative gives the instantaneous rate of reaction at any time t, a skill frequently tested in Edexcel Mechanics and Pure papers.
这个微分方程的解是 [S] = [S]₀e⁻ᵏᵗ，这与指数衰减联系了起来。导数给出任意时间 t 的瞬时反应速率，这是 Edexcel 力学和纯数试卷中常考查的技能。
9. Integration and Total Product Formed | 积分与产物总量
To find the total amount of product formed over a time interval, the rate function must be integrated. This applies definite integration to a biological molecule reaction context.
要求在一段时间内生成产物的总量，必须对速率函数进行积分。这把定积分应用到生物分子反应情境中。
P = ∫ v(t) dt
If the rate of product formation is v(t) = 2e⁻⁰·¹ᵗ mol dm⁻³ min⁻¹, the total product formed from t = 0 to t = 10 minutes is P = ∫₀¹⁰ 2e⁻⁰·¹ᵗ dt = [-20e⁻⁰·¹ᵗ]₀¹⁰ = 20(1 – e⁻¹) ≈ 12.64 mol dm⁻³.
如果产物生成速率为 v(t) = 2e⁻⁰·¹ᵗ mol dm⁻³ min⁻¹，从 t = 0 到 t = 10 分钟生成的总产物为 P = ∫₀¹⁰ 2e⁻⁰·¹ᵗ dt = [-20e⁻⁰·¹ᵗ]₀¹⁰ = 20(1 – e⁻¹) ≈ 12.64 mol dm⁻³。
10. Proportions and Percentage Yield in Biochemistry | 生化中的比例与百分产率
Stoichiometric calculations with biological molecules require proportional reasoning. The percentage yield compares the actual yield to the theoretical yield, a common Edexcel Maths application of percentages and ratios.
生物分子的化学计量计算需要比例推理。百分产率将实际产量与理论产量进行比较，这是 Edexcel 数学中百分比和比例的常见应用。
Percentage yield = (actual yield / theoretical yield) × 100%
If a reaction synthesising a dipeptide has a theoretical yield of 0.80 g but only 0.56 g is produced, the percentage yield is (0.56 / 0.80) × 100% = 70%. Scaling up to larger batches involves direct proportion: to produce 5.6 g, the starting mass must be multiplied by 10.
如果合成二肽的反应理论产量为 0.80 g，但只得到 0.56 g，百分产率为 (0.56 / 0.80) × 100% = 70%。扩大到更大批次涉及正比例：要生产 5.6 g，起始质量必须乘以 10。
11. Statistical Analysis of Experimental Data | 实验数据的统计分析
Biological molecule experiments always involve repeated measurements. Edexcel Statistics requires calculating the mean, standard deviation and confidence intervals to assess reliability. The mean x̄ of n readings is Σx / n, and the sample standard deviation s is √(Σ(x – x̄)² / (n – 1)).
生物分子实验总是涉及重复测量。Edexcel 统计要求计算平均值、标准差和置信区间以评估可靠性。n 次读数的平均值 x̄ 为 Σx / n，样本标准差 s 为 √(Σ(x – x̄)² / (n – 1))。
x̄ = Σx / n and s = √(Σ(x – x̄)² / (n – 1))
For five absorbance readings: 0.21, 0.23, 0.22, 0.24, 0.20, the mean is 0.22 and the sample standard deviation is about 0.0158. Error bars on graphs usually represent ±1 standard deviation.
对于五次吸光度读数：0.21、0.23、0.22、0.24、0.20，平均值为 0.22，样本标准差约为 0.0158。图上的误差线通常表示 ±1 个标准差。
12. Exam-Style Problem Solving | 真题风格问题求解
A typical Edexcel Maths question might combine several of these skills. For example: The absorbance A of a starch-iodine complex is given by A = 0.04c, where c is concentration in mg cm⁻³. A reaction reduces c according to c(t) = 20e⁻⁰·²ᵗ. Find the rate of change of absorbance when t = 5 minutes.
一道典型的 Edexcel 数学题可能综合考查以上多种技能。例如：淀粉-碘络合物的吸光度 A 由 A = 0.04c 给出，其中 c 为浓度，单位 mg cm⁻³。某反应使 c 按 c(t) = 20e⁻⁰·²ᵗ 减小。求 t = 5 分钟时吸光度的变化率。
Using the chain rule, dA/dt = 0.04 × dc/dt = 0.04 × -4e⁻⁰·²ᵗ = -0.16e⁻⁰·²ᵗ. At t = 5, dA/dt = -0.16e⁻¹ ≈ -0.0589 absorbance units per minute.
使用链式法则，dA/dt = 0.04 × dc/dt = 0.04 × -4e⁻⁰·²ᵗ = -0.16e⁻⁰·²ᵗ。当 t = 5 时，dA/dt = -0.16e⁻¹ ≈ -0.0589 吸光度单位每分钟。
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