Edexcel Maths 6.4 Cloning Biotechnology | Edexcel 数学 6.4 克隆生物技术

📚 Edexcel Maths 6.4 Cloning Biotechnology | Edexcel 数学 6.4 克隆生物技术

In this revision unit we explore the mathematical tools that appear under the Edexcel Maths 6.4 theme of cloning and biotechnology. Although the biological ideas provide the context, the marks are awarded for correct equation use, graph interpretation, probability calculations and statistical conclusions.

本复习单元聚焦于 Edexcel 数学 6.4 克隆与生物技术主题中常见的数学工具。虽然生物概念提供背景，但分数取决于正确使用方程、解读图像、计算概率和得出统计结论。

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1. Modelling Cell Growth | 细胞生长建模

A clone is a population of genetically identical cells, so its growth under ideal conditions can be modelled by simple iterative sequences. For Edexcel Maths, you are expected to translate a worded rule into a recurrence relation of the form uₙ₊₁ = k uₙ.

克隆群是一群遗传相同的细胞，因此在理想条件下其增长可以用简单的迭代数列建模。在 Edexcel 数学中，你需要将文字规则转化为形如 uₙ₊₁ = k uₙ 的递推关系。

If one cell divides into two every 20 minutes, then k = 2 and uₙ₊₁ = 2uₙ. The sequence is geometric, so the nth term is:

如果一个细胞每 20 分钟分裂为两个，那么 k = 2 且 uₙ₊₁ = 2uₙ。该数列是等比数列，因此第 n 项为：

uₙ = u₀ × 2ⁿ

Here u₀ is the initial number of cells and n is the number of divisions. This basic model is the foundation for most cloning and biotechnology growth questions.

这里 u₀ 是初始细胞数量，n 是分裂次数。这个基本模型是大多数克隆与生物技术增长问题的基础。

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2. Exponential Growth Curves | 指数增长曲线

When the number of divisions is large, discrete growth can be approximated by the continuous exponential model. You should be able to sketch and interpret graphs of y = A eˣ or y = A eᵏᵗ.

当分裂次数很多时，离散增长可以用连续指数模型近似。你应当能够绘制并解释 y = A eˣ 或 y = A eᵏᵗ 的图像。

For cloning, the cell count N at time t is often written as N(t) = N₀ eᵏᵗ. The constant k is the growth rate per unit time, and N₀ is the starting population.

在克隆问题中，t 时刻的细胞数量 N 通常写作 N(t) = N₀ eᵏᵗ。常数 k 是单位时间增长率，N₀ 是初始种群。

Key exam skills include finding k from two data points and predicting future population size. If N(0) = 100 and N(3) = 800, then:

关键考试技能包括根据两个数据点求 k 并预测未来种群大小。如果 N(0) = 100 且 N(3) = 800，那么：

800 = 100 e³ᵏ ⇒ e³ᵏ = 8 ⇒ 3k = ln 8 ⇒ k = (ln 8)/3

Always show this substitution step clearly; Edexcel marking rewards method, not just the final number.

务必清晰展示这个代入步骤；Edexcel 评分奖励方法，而不仅仅是最终数字。

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3. Logarithms and Doubling Time | 对数与倍增时间

Doubling time is the period needed for a clone population to double. In biotechnology, this tells researchers how quickly a culture becomes large enough for industrial use.

倍增时间是克隆种群翻倍所需的时间。在生物技术中，这告诉研究人员培养物多快能增长到工业使用所需的量。

From the equation N = N₀ × 2^(t/d), taking logs gives ln N = ln N₀ + (t/d) ln 2. Rearranging leads to a linear form, so you can find d from a log-linear graph.

根据方程 N = N₀ × 2^(t/d)，取对数得到 ln N = ln N₀ + (t/d) ln 2。重新整理后可得到线性形式，因此你可以从对数-线性图中求出 d。

A common Edexcel question gives a table of N against t and asks you to plot ln N on the y-axis. The gradient of the straight line equals k or (ln 2)/d depending on the base used.

常见的 Edexcel 题目会给出 N 对 t 的数据表，要求你以 ln N 为纵轴作图。直线的斜率根据使用的底数等于 k 或 (ln 2)/d。

Remember log laws: ln(a × b) = ln a + ln b and ln(aᵖ) = p ln a. These are essential for simplifying exponential growth equations.

记住对数法则：ln(a × b) = ln a + ln b 和 ln(aᵖ) = p ln a。这些对于化简指数增长方程至关重要。

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4. Logistic Growth and Carrying Capacity | 逻辑斯蒂增长与承载容量

Unlimited exponential growth is unrealistic because nutrients and space run out. Biotechnology models therefore use the logistic equation, which introduces a maximum carrying capacity L.

无限制的指数增长并不现实，因为营养物质和空间会耗尽。因此生物技术模型使用逻辑斯蒂方程，其中引入了最大承载容量 L。

A standard logistic model is dN/dt = rN(1 – N/L). At the start, when N is small, growth is almost exponential; as N approaches L, the growth rate slows to zero.

标准逻辑斯蒂模型为 dN/dt = rN(1 – N/L)。在开始时，当 N 很小时，增长几乎是指数式的；当 N 接近 L 时，增长率减慢至零。

In A level Maths, you are unlikely to solve this differential equation analytically, but you must interpret its shape: a sigmoid or S-shaped curve. The steepest point occurs at N = L/2.

在 A level 数学中，你不太可能需要解析求解这个微分方程，但必须解释其形状：S 形曲线。最陡点出现在 N = L/2。

Questions may ask you to estimate L from a graph or match data to the correct model. Use the plateau of the curve as an estimate of carrying capacity.

题目可能要求你从图像估算 L，或将数据与正确模型匹配。将曲线的平台段用作承载容量的估计值。

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5. Probability in Cloning | 克隆中的概率

Cloning and genetic modification involve uncertain outcomes, so probability is central to Edexcel Maths 6.4 questions. A typical problem gives the probability that a single cell survives transformation.

克隆和基因修饰涉及不确定的结果，因此概率是 Edexcel 数学 6.4 题目的核心。典型问题会给出单个细胞在转化后存活下来的概率。

Use tree diagrams when events happen in sequence. For example, if a cell survives with probability 0.8 and then divides successfully with probability 0.7, the probability of both is 0.8 × 0.7 = 0.56.

当事件顺次发生时，使用树状图。例如，如果一个细胞存活概率为 0.8，然后成功分裂的概率为 0.7，那么两者都发生的概率为 0.8 × 0.7 = 0.56。

For independent events, multiply probabilities; for mutually exclusive outcomes, add probabilities. Edexcel candidates often

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