📚 Edexcel Mathematics 2.5: Biological Membranes | 爱德思数学 2.5 生物膜
In Edexcel Mathematics, contextual problems often draw on biological systems such as cell membranes. This article explains how core mathematical techniques can model diffusion, osmosis, surface area effects and membrane potentials, while keeping the focus on examinable mathematics.
在爱德思数学中，情境题常以生物系统（如细胞膜）为背景。本文解释如何用核心数学技巧对扩散、渗透、表面积效应和膜电位进行建模，同时把重点放在可考查的数学内容上。
The biology is only the setting. The marks come from ratio, proportion, linear functions, logarithms, exponential models and graph interpretation, all of which can be tested through membrane-based data.
生物学只是背景。分数来自比例、正反比例、线性函数、对数、指数模型和图像解读，这些都可以通过生物膜数据来考查。
1. Surface Area to Volume Ratio for Cubes | 立方体的表面积与体积比
ce area to volume ratio is therefore 6x² : x³, which simplifies to 6 : x, or SA:V = 6 ÷ x.
对于边长为 x 的立方体细胞，表面积为 6x²，体积为 x³。因此表面积与体积比为 6x² : x³，化简为 6 : x，即 SA:V = 6 ÷ x。
SA:V = 6x² : x³ = 6 : x = 6 ÷ x
As x increases, the ratio 6 ÷ x decreases. A large cube has a smaller surface area relative to its volume, so diffusion across its membrane is less efficient per unit of volume.
当 x 增大时，比值 6 ÷ x 减小。大立方体相对于其体积只有较小的表面积，因此单位体积通过膜的扩散效率较低。
Small cells and folded membranes both increase the effective SA:V ratio. Folding adds membrane area without adding much volume, so the effective ratio becomes larger than the simple cube model suggests.
小细胞和折叠膜都能增大有效的表面积体积比。折叠增加了膜面积而几乎不增加体积，因此有效比值会大于简单立方体模型所给出的结果。
In exam questions, you may be asked to compare two cubes by calculating the ratio for each and then finding the percentage change. Always keep the same units before dividing.
考试题中可能要求你通过分别计算两个立方体的比值，再求百分比变化来比较。相除前务必保持单位一致。
2. Spheres, Membrane Folding and Standard Results | 球体、膜折叠与标准结果
is A = 4πr² and the volume is V = (4 ÷ 3)πr³. Dividing area by volume gives SA:V = 3 ÷ r.
对于半径为 r 的球体，表面积 A = 4πr²，体积 V = (4 ÷ 3)πr³。用面积除以体积得到 SA:V = 3 ÷ r。
SA:V = 4πr² ÷ ((4 ÷ 3)πr³) = 3 ÷ r
This is another standard result to memorise. If the radius doubles, the SA:V ratio halves; if the radius halves, the ratio doubles.
这是另一个需要记住的标准结果。若半径加倍，SA:V 比值减半；若半径减半，比值加倍。
Membrane folding can be modelled by replacing the simple geometric area with an effective area. If a membrane is folded so that its area is multiplied by a factor k, then the effective SA:V becomes k times the original ratio.
膜折叠可以通过用有效面积替代简单几何面积来建模。如果膜折叠后面积乘以因子 k，则有效 SA:V 变为原来比值的 k 倍。
A common exam task is to compare a cube and a sphere of equal volume, or to explain why microvilli increase absorption. Both require the same mathematical idea: area scales differently from volume.
常见考题是比较等体积的立方体和球体，或解释微绒毛为何能增强吸收。两者都需要同一个数学思想：面积与体积的缩放方式不同。
3. Percentage Change and Comparing Cell Sizes | 百分比变化与细胞大小比较
Percentage change is calculated as (new value − old value) ÷ old value × 100% his allows you to compare cells of different sizes without listing every ratio separately.
百分比变化计算为（新值 − 旧值）÷ 旧值 × 100%。对于 SA:V，这让你无需逐一列出所有比值，就能比较不同大小的细胞。
Percentage change = (new − old) ÷ old × 100%
s from 6 ÷ x to 6 ÷ (2x) = 3 ÷ x. The percentage change is (3 ÷ x − 6 ÷ x) ÷ (6 ÷ x) × 100% = −50%.
对于立方体，如果边长从 x 加倍到 2x，SA:V 从 6 ÷ x 变为 6 ÷ (2x) = 3 ÷ x。百分比变化为 (3 ÷ x − 6 ÷ x) ÷ (6 ÷ x) × 100% = −50%。
nal decrease when its radius doubles, because SA:V = 3 ÷ r also depends inversely on size.
负号表示减少。球体半径加倍时也显示相同的比例下降，因为 SA:V = 3 ÷ r 同样与尺寸成反比。
When comparing two cells, avoid subtracting ratios directly unless the question asks for a difference. Most Edexcel mark schemes reward a clear ratio or percentage comparison.
比较两个细胞时，除非题目要求差值，否则不要直接相减比值。多数爱德思评分标准更认可清晰的比值或百分比比较。
4. Fick's Law as Direct and Inverse Proportion | 菲克定律与正比反比
Fick's law states that the diffusion rate across a membrane is directly proportional to the surface area and the concentration gradient, and inversely proportional to the membrane thickness.
菲克定律指出，跨膜扩散速率与表面积和浓度梯度成正比，与膜厚度成反比。
Rate = D × A × ΔC ÷ d
Here D is the diffusion coefficient, A is the surface area, ΔC is the concentration difference and d is the membrane thickness. The symbol Δ means 'change in'.
其中 D 是扩散系数，A 是表面积，ΔC 是浓度差，d 是膜厚度。符号 Δ 表示 "变化量"。
Doubling A doubles the rate; doubling ΔC doubles the rate; doubling d halves the rate. These are direct and inverse proportion relationships, and they can be tested without calculating a numerical answer.
A 加倍则速率加倍；ΔC 加倍则速率加倍；d 加倍则速率减半。这些都是正比例和反比例关系，可以不算出具体数值而直接考查。
A useful proportional form is Rate ∝ hat if A and ΔC are both doubled while d is unchanged, the rate becomes four times larger.
一个有用的比例形式是 Rate ∝ A × ΔC ÷ d。它表明如果 A 和 ΔC 都加倍而 d 不变，速率会变为原来的四倍。
5. Rearranging Fick's Law in Exam Calculations | 考题中重排菲克定律
k you to find the fourth. Rearranging Rate = D × A × ΔC ÷ d gives D = Rate × d ÷ (A × ΔC).
许多题目给出四个变量中的三个，要求求第四个。重排 Rate = D × A × ΔC ÷ d 可得 D = Rate × d ÷ (A × ΔC)。
Rate and A = Rate × d ÷ (D × ΔC). Practise these rearrangements until they are automatic.
同样，d = D × A × ΔC ÷ Rate，A = Rate × d ÷ (D × ΔC)。把这些重排练到自动化。
Unit consistency is essential. If A is in m², d in m and ΔC in mol m⁻³, then D is usually in m² s⁻¹ and the rate is in mol s⁻¹.
单位一致至关重要。如果 A 用 m²，d
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