📚 AS CIE Biology: Formula & Theorem Quick Reference Handbook | AS CIE 生物：公式定理速查手册
Welcome to the AS CIE Biology essential revision guide. This handbook compiles the key quantitative relationships, laws, and indices you must master for your examination. From magnification and Fick's law to cardiac output and Simpson's diversity index, every formula is explained with clear definitions and worked examples to ensure you can apply them accurately under timed conditions.
欢迎使用 AS CIE 生物必备复习手册。本手册整理了考试中必须掌握的核心定量关系、定律和指数。从放大倍数和菲克定律到心输出量和辛普森多样性指数，每个公式都配有清晰的定义和计算示例，确保你在限时考试中能够准确运用。
1. Magnification Formula | 放大倍数公式
The magnification formula is fundamental to all microscopy work. It links the size of an image under the microscope to the real size of the specimen. You must be able to rearrange it for any of the three variables.
放大倍数公式是所有显微镜工作的基础。它将显微镜下图像的大小与标本的实际尺寸联系起来。你必须能够对该公式进行变形，求解任意一个变量。
Magnification = Image size ÷ Actual size
Always ensure that image size and actual size are expressed in the same units before performing the calculation. The image size is usually measured in millimetres or centimetres on the micrograph, while the actual size is expressed in micrometres (µm) or nanometres (nm). Convert all lengths to the same unit first, typically micrometres, to avoid errors.
在计算前务必保证图像尺寸和实际尺寸的单位一致。图像尺寸通常在显微照片上以毫米或厘米测量，而实际尺寸以微米 (µm) 或纳米 (nm) 表示。为避免错误，应先将所有长度转换为同一单位，通常转换成微米。
h has a stated magnification of ×8000. Calculate the actual length of the mitochondrion in µm.
示例：一名学生在电子显微照片上测得一个线粒体的长度为 24 mm。该显微照片标注的放大倍数为 ×8000。计算线粒体的实际长度（单位 µm）。
Actual size = Image size ÷ Magnification = 24 mm ÷ 8000 = 0.003 mm. Convert mm to µm: 0.003 mm = 0.003 × 1000 µm = 3 µm. The actual length is 3 µm.
实际尺寸 = 图像尺寸 ÷ 放大倍数 = 24 mm ÷ 8000 = 0.003 mm。将 mm 转换为 µm：0.003 mm × 1000 = 3 µm。实际长度为 3 µm。
2. Unit Conversions in Microscopy | 显微镜单位换算
Precise unit conversion is essential when working with microscopic structures. AS CIE Biology always expects you to express biological dimensions in the most appropriate metric unit. The table below summarises the conversions you must memorise.
在处理显微结构时，精确的单位换算至关重要。AS CIE 生物考试始终要求用最合适的公制单位表示生物尺寸。下表总结了你必须牢记的换算关系。
| Conversion | Equivalent |
| 1 metre (m) | = 1000 mm = 1,000,000 µm |
| 1 millimetre (mm) | = 1000 micrometres (µm) |
| 1 micrometre (µm) | = 1000 nanometres (nm) |
All unit conversions for microscopic work rely on multiplying or dividing by 1000. When moving from a larger unit to a smaller unit (e.g., mm to µm), multiply by 1000. When moving from a smaller unit to a larger unit (e.g., µm to mm), divide by 1000.
显微镜工作的所有单位换算都依赖于乘以或除以 1000。从较大单位转换为较小单位时（例如 mm 转为 µm），乘以 1000；从较小单位转换为较大单位时（例如 µm 转为 mm），除以 1000。
It is particularly common to measure image size in mm, convert to µ cation formula. Keeping a clear record of these steps in your working prevents careless mistakes.
尤其常见的情形是，图像尺寸以 mm 为单位，先转换为 µm（×1000），再代入放大倍数公式。在计算过程中清晰记录这些步骤有助于防止粗心错误。
3. Surface Area to Volume Ratio | 表面积与体积比
A cell's ability to exchange materials with its environment depends on its surface area to volume ratio (SA:V). This ratio decreases as an organism or cell increases in size, placing a fundamental limit on cell dimensions.
细胞与外界交换物质的能力取决于其表面积与体积的比值 (SA:V)。该比值随着生物体或细胞体积的增大而减小，这从根本上限制了细胞的尺寸。
SA:V Ratio = Surface Area ÷ Volume
6L² and volume = L³, so the SA:V = 6/L. The larger L becomes, the smaller the SA:V ratio.
考试可能要求你计算简单形状（如立方体或球体）的 SA:V 比值。对于边长为 L 的立方体，表面积 = 6L²，体积 = L³，因此 SA:V = 6/L。L 越大，SA:V 比值越小。
Adaptations such as flattened leaves, folded internal membranes (cristae of mitochondria, villi in the small intestine), and hollow centres (xylem vessels) all serve to increase surface area without proportionally increasing volume, thereby maximising the exchange rate.
扁平的叶片、折叠的内膜（线粒体嵴、小肠绒毛）以及中空结构（木质部导管）等适应性结构，均可在不按比例增加体积的情况下增大表面积，从而最大限度地提高交换速率。
In the context of transport, a high SA:V ratio is essential for efficient diffusion of oxygen, carbon dioxide, and nutrients. This principle explains why unicellular organisms can rely on diffusion alone while larger organisms require specialised transport systems.
在运输方面，较高的 SA:V 比值对于氧气、二氧化碳和营养物质的高效扩散至关重要。这一原理解释了为什么单细胞生物仅靠扩散就能生存，而较大的生物则需要专门的运输系统。
4. Fick's Law of Diffusion | 菲克扩散定律
Fick's law describes the factors affecting the rate of diffusion of a substance across a membrane or through a tissue. It is widely applicable in gas exchange surfaces, plant leaves, and any biological interface where passive movement occurs.
菲克定律描述了影响物质跨膜或透过组织扩散速率的因素。它广泛应用于气体交换表面、植物叶片以及任何发生被动运动的生物界面。
Rate of diffusion ∝ (Surface area × Concentration difference) ÷ Diffusion distance
The law summarises three key variables: increasing surface area (e.g., the alveoli and villi) increases rate; increasing the concentration gradient increases rate; increasing the diffusion distance (thickness of the membrane or tissue) decreases rate.
该定律概括了三个关键变量：增大表面积（例如肺泡和绒毛）会提高速率；增大浓度梯度会提高速率；增加扩散距离（膜或组织的厚度）会降低速率。
In an exam question, you may be asked to explain structural adaptations using Fick's law. For instance, the alveolar epithelium is extremely thin (short diffusion distance), the lungs are highly branched (large surface area), and breathing and blood flow maintain a steep concentration gradient. These all maximise the rate of gas exchange.
考试题目可能会要求你用菲克定律解释结构上的适应性。例如，肺泡上皮极薄（扩散距离短）、肺高度分支（表面积大），而呼吸和血流维持了陡峭的浓度梯度，这些都最大化了气体交换速率。
Remember that the expression is a proportionality, not an equation with a fixed constant. This means you should describe how manipulation of each factor alters the rate rather than trying to compute exact values.
请记住，该表达式是一个比例关系，而非带有固定常数的等式。这意味着你应该描述每个因素如何影响速率，而不是尝试计算出精确数值。
5. Rate of Enzyme-Controlled Reaction | 酶促反应速率
The rate of enzyme activity is commonly expressed as the reciprocal of the time taken for a measurable change, particularly when quantifying the effect of temperature, pH, or inhibitor concentration.
酶活性的速率通常表示为发生可测量变化所需时间的倒数，尤其用于量化温度、pH 或抑制剂浓度的影响。
Rate = 1 ÷ Time (for a standard end‑point)
If a practical investigation records the time taken for the iodine‑starch colour to disappear, or for a certain mass of product to be formed, the rate can be calculated using this simple relationship. Longer time means slower rate; shorter time means faster rate.
如果实验记录了碘‑淀粉颜色消失所需的时间，或生成一定质量产物所需的时间，就可以用这个简单关系计算速率。时间越长意味着速率越慢；时间越短意味着速率越快。
adient of the tangent at time zero. Use the formula: Initial rate = (Change in product concentration) ÷ (Change in time).
另外一种情况是，你可能需要根据产物生成量随时间变化的曲线图来确定反应的初始速率。初始速率等于时间零点处切线的斜率。使用公式：初始速率 = (产物浓度的变化量) ÷ (时间的变化量)。
When explaining effects, always link the change in rate to changes in the frequency of successful enzyme‑substrate collisions, or the alteration of the active site's shape at extreme conditions. The reciprocal relationship with time makes it easier to plot comparative rates in laboratory reports.
在解释影响时，务必把速率变化与酶‑底物成功碰撞的频率变化，或在极端条件下活性位点形状的改变联系起来。速率与时间的倒数关系使得在实验报告中绘制速率比较图更为简便。
6. Mitotic Index | 有丝分裂指数
The mitotic index is a parameter used to estimate the proportion of cells in a tissue that are actively dividing. It is particularly useful when studying meristems in plants or tumour growth in animals.
有丝分裂指数是用于估算组织中正在活跃分裂的细胞比例的指标。在研究植物分生组织或动物肿瘤生长时尤其有用。
Mitotic Index = (Number of cells in mitosis ÷ Total number of cells) × 100%
Cells in mitosis can be identified by the appearance of condensed chromosomes during prophase, metaphase, anaphase, and telophase. Interphase cells are not counted in the numerator. You must view a sample of cells under the microscope and count all cells in the field of view.
处于有丝分裂的细胞可通过前期、中期、后期和末期时浓缩染色体的出现来识别。间期细胞不计入分子部分。你必须在显微镜下观察一个细胞样本，并对视野中的所有细胞进行计数。
A high mitotic index suggests rapid tissue growth, such as in root tips or developing embryos. Cancerous tissues also exhibit an elevated mitotic index. This calculation is often combined with a practical where you prepare a root tip squash and count cells in stages of mitosis.
有丝分裂指数高表明组织生长迅速，例如在根尖或发育中的胚胎中。癌变组织也会表现出升高的有丝分裂指数。该计算常与根尖压片实验结合，在观察有丝分裂各阶段细胞时进行计数。
or example, if you count 34 mitotic cells out of 500 cells, mitotic index = (34/500) × 100% = 6.8%.
根据计数的细胞总数，将结果表示为适当的精度。例如，如果在 500 个细胞中计得有 34 个处于有丝分裂，则有丝分裂指数 = (34/500) × 100% = 6.8%。
7. Cardiac Output | 心输出量
Cardiac output is the volume of blood pumped by the left ventricle into the aorta per minute. It is a core concept in the mammalian transport system and allows you to link heart activity to the delivery of oxygen and nutrients to tissues.
心输出量是指左心室每分钟泵入主动脉的血量。它是哺乳动物运输系统的核心概念，使你能够将心脏活动与向组织输送氧气和营养物质联系起来。
Cardiac Output = Heart Rate × Stroke Volume
- Heart rate (HR): number of heart beats per minute (bpm).
- Stroke volume (SV): volume of blood pumped by the left ventricle in one beat, typically about 70 ml in a resting adult human.
- Cardiac output (CO) is therefore measured in millilitres per minute (ml min⁻¹) or litres per minute (L min⁻¹).
- 心率 (HR)：每分钟心跳次数 (bpm)。
- 每搏输出量 (SV)：左心室每次搏动泵出的血量，静息成年人通常约为 70 ml。
- 心输出量 (CO) 因此以毫升每分钟 (ml min⁻¹) 或升每分钟 (L min⁻¹) 为单位。
An increased demand for oxygen, for example during vigorous exercise, can be met by increasing heart rate (chronotropic effect) or increasing stroke volume (inotropic effect). Athletes often have a larger stroke volume and a lower resting heart rate.
对氧气的需求增加时，例如在剧烈运动期间，可通过加快心率（变时效应）或增加每搏输出量（变力效应）来满足。运动员通常每搏输出量较大，静息心率较低。
te the cardiac output in litres per minute. CO = 135 × 80 = 10800 ml min⁻¹ = 10.8 L min⁻¹.
典型考题：某人的心率为 135 bpm，每搏输出量为 80 ml。计算心输出量，以升每分钟表示。CO = 135 × 80 = 10800 ml min⁻¹ = 10.8 L min⁻¹。
8. Percentage Change | 百分比变化
Biologists routinely use percentage change to compare a measured quantity before and after an experimental treatment. This standardises results and allows direct comparison between different groups or variables.
生物学家经常使用百分比变化来比较实验处理前后测得的量。这使结果标准化，并允许不同组别或变量之间直接比较。
Percentage Change = [(Final value − Initial value) ÷ Initial value] × 100%
A negative value indicates a percentage decrease, while a positive value indicates a percentage increase. In AS practical work this is frequently applied to change in mass of potato cylinders in an osmosis investigation, or change in length of a piece of tissue.
负值表示百分比减少，正值表示百分比增加。在 AS 实验操作中，这通常应用于渗透作用研究中马铃薯条的质量变化，或组织片段的长度变化。
It is crucial to use the initial value as the denominator so that all changes are expressed relative to the starting condition. Mistakes most often arise when students divide by the final value instead.
必须将初始值作为分母，这样所有变化都相对于起始条件来表示。最常见的错误是学生除以了最终值。
g after being placed in a salt solution. Percentage change = [(10.8 − 12.0) ÷ 12.0] × 100% = (−1.2 ÷ 12.0) × 100% = −10.0%. The negative sign indicates a loss in mass.
示例：一马铃薯条初始质量为 12.0 g，放入盐溶液中后最终质量为 10.8 g。百分比变化 = [(10.8 − 12.0) ÷ 12.0] × 100% = (−1.2 ÷ 12.0) × 100% = −10.0%。负号表明质量减少。
9. Simpson's Index of Diversity | 辛普森多样性指数
Biodiversity can be quantified using Simpson's Index of Diversity (D). This index takes into account both the number of species present (species richness) and the relative abundance of each species (evenness). It is a required calculation in the CIE AS syllabus.
生物多样性可使用辛普森多样性指数 (D) 来量化。该指数同时考虑了存在的物种数量（物种丰富度）和每个物种的相对丰度（均匀度）。它是 CIE AS 大纲中要求的计算内容。
D = 1 − [Σ n(n−1) ÷ N(N−1)]
number of individuals of all species. The sum (Σ) is taken across all species in the community. The value of D ranges between 0 (low diversity) and 1 (high diversity).
其中 n = 某一特定物种的个体总数，N = 所有物种的个体总数。求和符号 (Σ) 涵盖群落中的所有物种。D 值介于 0（低多样性）和 1（高多样性）之间。
In examinations you may be given a table of species counts and asked to calculate D in steps. First compute N, then for each species compute n(n−1), sum these, and finally apply the formula. This can be time-consuming, so practise methodically.
考试中你可能会看到一张物种数量表，并被要求分步计算 D。首先计算 N，然后对每个物种计算 n(n−1)，将其求和，最后代入公式。这可能很耗时，因此需要有步骤地练习。
10 + 132 + 6 = 348. N(N−1) = 30×29 = 870. D = 1 − (348/870) = 1 − 0.4 = 0.6. This indicates moderate diversity.
示例：一个样本中有 15 只潮虫（物种 A）、12 只蜘蛛（物种 B）和 3 条蜈蚣（物种 C）。N = 30。n(n−1) 总和 = 15×14 + 12×11 + 3×2 = 210 + 132 + 6 = 348。N(N−1) = 30×29 = 870。D = 1 − (348/870) = 1 − 0.4 = 0.6。这表明多样性中等。
A high D value is often associated with a stable, resilient ecosystem. Comparisons of D before and after an environmental disturbance can be used to evaluate the impact of human activity on biodiversity.
高 D 值通常与稳定、有韧性的生态系统相关。比较环境受到扰动前后的 D 值，可用于评估人类活动对生物多样性的影响。
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