CIE A-Level Biology Formula & Theorem Quick Reference | CIE 生物公式定理速查手册

📚 CIE A-Level Biology Formula & Theorem Quick Reference | CIE 生物公式定理速查手册

For Year 13 students following the CIE A‑Level Biology (9700) syllabus, quantitative skills and the application of key formulae are essential not only for problem‑solving but also for the analysis and interpretation of experimental data. This quick‑reference handbook brings together the most important equations, statistical tests and biological constants that you are expected to understand, recall and use confidently in the examination. Each entry includes the formula in its standard notation and a concise explanation of when and why it is used.

对于修读 CIE A‑Level 生物 (9700) 课程的 Year 13 学生来说，数据处理能力和关键公式的应用不仅是解决实际问题的基础，也是分析和解读实验数据的核心。这本速查手册汇集了考试要求你理解、记忆并灵活运用的重要方程、统计检验和生物学常数。每个条目均提供了标准记号的公式以及简要说明，帮助你掌握何时及为何使用这些公式。

1. Magnification and Unit Conversions | 放大倍率与单位换算

Magnification is calculated from the ratio of the size of an image to the actual size of the specimen. Always ensure that both measurements are in the same units before performing the calculation. Electron micrographs often use micrometres (µm) or nanometres (nm), so conversions between millimetres, micrometres and nanometres are tested regularly.

放大倍率的计算基于图像尺寸与标本实际尺寸之比。进行计算前，务必保证两者使用相同单位。电子显微照片常使用微米 (µm) 或纳米 (nm)，因此毫米、微米与纳米之间的换算是考试中的常规考点。

Magnification = Image size ÷ Actual size
A = I/M
1 mm = 1000 µm = 1 000 000 nm

2. Hardy–Weinberg Principle | 哈迪-温伯格原理

The Hardy–Weinberg principle provides a mathematical model that predicts allele and genotype frequencies in a population that is not evolving. In CIE examinations you need to use the two equations to estimate the frequency of alleles, carriers and affected individuals, particularly in autosomal recessive conditions.

哈迪-温伯格原理提供了一个数学模型，用于预测在非进化群体中等位基因和基因型频率。在 CIE 考试中，你需要运用这两个方程来估算等位基因频率、携带者频率以及患病个体频率，尤其是在常染色体隐性遗传疾病的分析中。

p + q = 1
p² + 2pq + q² = 1
Where p = frequency of dominant allele, q = frequency of recessive allele

3. Standard Deviation and Standard Error | 标准差与标准误

Standard deviation (s) measures the spread of a set of data around the mean. A larger standard deviation indicates greater variability. The standard error of the mean (SE) describes how far the sample mean is likely to be from the true population mean; it decreases as sample size increases. These values are used to plot error bars on bar charts and to compare the reliability of data sets.

标准差 (s) 衡量一组数据围绕平均值的分散程度。标准差越大，表明变异性越高。均值的标准误 (SE) 描述样本均值与总体真实均值之间可能存在的差距；随着样本量的增加，标准误会减小。这些数值常用于绘制柱状图上的误差棒，并用于比较数据集的可靠性。

s = √[ Σ(x – x̄)² / (n – 1) ]
SE = s / √n
x̄ = sample mean, n = sample size

4. Chi‑squared Test (χ² Test) | 卡方检验

The chi‑squared test is used to determine whether there is a significant difference between observed and expected frequencies in categorical data. In genetics you will test the goodness of fit of experimental ratios to Mendelian ratios; in ecology you might test whether the distribution of a species is associated with a particular habitat factor. You must be able to state the null hypothesis, calculate χ², determine degrees of freedom and compare the calculated value against a critical value at a given probability level.

卡方检验用于判断分类数据中观测频率与期望频率之间是否存在显著差异。在遗传学中，你需要检验实验比率是否符合孟德尔比率；在生态学中，你可能需要检验某一物种的分布是否与特定栖息地因素相关联。你必须能够提出零假设，计算 χ² 值，确定自由度，并将计算值与给定概率水平下的临界值进行比较。

χ² = Σ (O – E)² / E
Degrees of freedom (df) = number of categories – 1

5. Student’s t‑test | 学生t检验

The t‑ ce is statistically significant. The CIE course expects familiarity with the unpaired (two‑sample) t‑ e critical value at p = 0.05 is essential for the analysis of experimental investigations.

t检验比较两组连续数据的均值，以评估它们之间的差异是否具有统计学显著性。CIE 课程要求学生熟悉非配对（双样本）t检验。掌握 t 值的计算方法、确定自由度以及在 p = 0.05 水平下与临界值进行比较并解读结果，对于实验调查的分析至关重要。

t = (x̄₁ – x̄₂) / √s₁²/n₁ + s₂²/n₂
df = n₁ + n₂ – 2 (approximate)

6. Simpson’s Index of Diversity | 辛普森多样性指数

Simpson’s Index measures the biodiversity of a habitat, taking into account both species richness and evenness. A higher index value indicates greater diversity. CIE candidates may be given the formula and required to calculate the index from species count data, or to explain its ecological significance.

辛普森指数用于衡量栖息地的生物多样性，同时考虑了物种丰富度和均匀度。指数值越高，表明多样性越高。CIE 考生可能会被要求根据物种数量数据计算该指数，或解释其生态学意义。

D = 1 – Σ(n/N)²
or D = 1 – [ Σ n(n – 1) / N(N – 1) ]
= total number of individuals of a particular species, N = total number of all individuals

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7. Lincoln Index (Capture‑Mark‑Recapture) | 林肯指数（标记重捕法）

The Lincoln Index estimates the size of a mobile animal population. It assumes that the population is closed, marking does not affect survival, and marked individuals mix evenly with the rest of the population. Limitations of the method and ethical considerations are commonly assessed.

林肯指数用于估算移动动物种群的大小。该方法假设种群是封闭的，标记不影响生存，且被标记个体与其余个体均匀混合。该方法的局限性和伦理考量通常是评估内容。

N = (M × C) / R
M = number caught and marked in first sample
C = total number caught in second sample
R = number of marked individuals recaptured in second sample

8. Respiratory Quotient (RQ) | 呼吸商

Respiratory quotient is the ratio of carbon dioxide produced to oxygen consumed during respiration. RQ values provide information about the respiratory substrate being oxidised: carbohydrate gives approximately 1.0, lipid about 0.7, and protein roughly 0.9. RQ is measured using a respirometer, and calculations often involve correcting for temperature and pressure.

呼吸商是呼吸过程中产生的二氧化碳与消耗的氧气之比。RQ 值能够提供关于被氧化的呼吸底物的信息：碳水化合物约为 1.0，脂质约为 0.7，蛋白质约为 0.9。RQ 通过呼吸计进行测量，计算时常需要校正温度与压力。

RQ = Volume (or moles) of CO₂ produced ÷ Volume (or moles) of O₂ consumed

9. Net Primary Productivity (NPP) | 净初级生产力

NPP represents the energy or biomass available to the next trophic level after autotrophs have met their own respiratory demands. It is a fundamental concept in energy flow through ecosystems. You must be able to calculate NPP from given values of gross primary productivity (GPP) and respiration (R), and interpret productivity data.

NPP 代表自养生物满足自身呼吸需求后，可供下一个营养级使用的能量或生物量。这是生态系统中能量流动的基本概念。你必须能够根据给定总初级生产力 (GPP) 和呼吸消耗 (R) 的数值计算 NPP，并解读生产力数据。

NPP = GPP – R
Units: kJ m⁻² yr⁻¹ or g m⁻² yr⁻¹

10. Efficiency of Energy Transfer | 能量传递效率

The efficiency with which energy is transferred between trophic levels is typically low, often around 10%. Calculations of efficiency help explain why food chains rarely exceed four or five trophic levels and why pyramids of energy are always upright. Use the appropriate units and give your answer as a percentage.

各营养级之间能量传递的效率通常很低，一般在 10% 左右。计算能量传递效率有助于解释为什么食物链很少超过四到五个营养级，以及为什么能量金字塔总是呈正立形态。请使用正确的单位，并以百分比形式给出答案。

Efficiency (%) = (Energy available to next level / Energy available to previous level) × 100

11. Rate of Reaction (Enzyme‑catalysed) | 反应速率（酶催化反应）

When investigating enzyme kinetics, the initial rate of reaction can be calculated from the change in product concentration (or substrate depletion) over a fixed time interval. In CIE practical‑based questions you may plot a progress curve and draw a tangent at time zero to determine the initial rate. Understanding the derivation of 1/time as a measure of rate is also tested.

在研究酶动力学时，反应的初始速率可以通过某一时间段内产物浓度的变化（或底物的减少量）计算得出。在 CIE 实验类题目中，你可能需要绘制反应进程曲线，并在时间为零时作切线以确定初始速率。理解用 1/时间 作为速率度量的推导过程也会被考查。

Rate = Change in product concentration / Time
or Rate ∝ 1 / time (for a fixed endpoint assay)

12. Water Potential (Ψ) | 水势

Water potential determines the direction of water movement across a partially permeable membrane. In A2 plant physiology, you must be able to state the water potential equation and explain how solute potential (Ψₛ) and pressure potential (Ψₚ) contribute to the overall water potential. The units are typically kilopascals (kPa), where pure water has a water potential of zero.

水势决定了水通过部分透膜的移动方向。在 A2 植物生理学中，你必须能够写出水势方程并解释溶质势 (Ψₛ) 和压力势 (Ψₚ) 是如何影响总体水势的。单位通常为千帕 (kPa)，纯水的水势为零。

Ψ = Ψₛ + Ψₚ
e i = ionisation constant, C = concentration, R = pressure constant, T = temperature in K)

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