📚 GCSE CIE Biology: Formula Summary Handbook | GCSE CIE 生物：公式汇总手册
This handbook brings together the essential equations and calculations you are expected to use in the CIE GCSE Biology course. Mastering these formulas will help you tackle data-based questions, practical work assessments, and core concepts with confidence.
本手册汇集了 CIE GCSE 生物学课程中你必须掌握的基本公式和计算方法。熟练运用这些公式，有助于你自信应对数据分析题、实验评估以及核心概念考查。
1. Magnification Formula | 放大倍数公式
Magnification (M) tells you how many times larger an image appears compared to the real size of the object. The relationship is given by the equation:
放大倍数 (M) 表示图像比物体的实际尺寸大多少倍。它们的关系用如下公式表示：
M = I ÷ A
where M stands for magnification, I is the image size, and A is the actual size of the specimen. All measurements must be in the same unit before using the formula.
M = I ÷ A
其中 M 为放大倍数，I 为图像大小，A 为样品的实际大小。使用公式前，必须将所有测量值换算为同一单位。
2. Actual Size Calculation | 实际大小计算
When you know the magnification and have measured the image size, you can find the actual size of a cell or organelle by rearranging the magnification formula:
如果已知放大倍数并已测量图像大小，通过变形放大倍数公式即可求出细胞或细胞器的实际大小：
A = I ÷ M
Always remember to state the actual size with the correct unit, usually micrometres (µm) or millimetres (mm). This calculation is frequently tested in questions that provide a micrograph and a scale bar.
A = I ÷ M
请务必用正确的单位记录实际大小，通常为微米 (µm) 或毫米 (mm)。提供显微照片和比例尺的考题中，该计算频繁出现。
3. Unit Conversions | 单位换算
Before using any size formula, you may need to convert between millimetres, micrometres and nanometres. The table below shows the common conversions required in the biology exam:
在使用任何大小公式之前，你可能需要在毫米、微米和纳米之间进行换算。下表展示了生物考试中常见的单位换算：
| 1 cm = 10 mm | 1 mm = 1000 µm |
| 1 µm = 1000 nm | 1 m = 1000 mm |
For example, to express 0.02 mm in micrometres, multiply by 1000 to get 20 µm. Correct unit handling is essential to gain full marks in calculation questions.
例如，将 0.02 mm 表示为微米时，乘以 1000 得到 20 µm。正确的单位处理是计算题获取满分的关键。
4. Surface Area to Volume Ratio | 表面积与体积比
The surface area to volume ratio (SA:V) is a key concept explaining why cells are microscopic and why organisms need specialised exchange surfaces. It can be calculated using simple geometry:
表面积与体积比 (SA:V) 是解释细胞为什么微小、生物体为什么需要特化交换表面的重要概念。该比值可通过简单几何公式计算：
SA:V = Surface area ÷ Volume
For a cube-shaped cell, surface area = 6 × side² and volume = side³. As a cell grows, its volume increases much faster than its surface area, so the SA:V ratio falls. This limits the efficiency of diffusion.
SA:V = 表面积 ÷ 体积
对于一个立方体细胞，表面积 = 6 × 边长²，体积 = 边长³。随着细胞长大，其体积增大速度远超表面积，因此 SA:V 比值下降，这限制了扩散效率。
5. Food Energy Content | 食物能量含量
In a calorimetry experiment, the energy released by burning a food sample is absorbed by a known mass of water. The energy content per gram is found using the formula:
在量热法实验中，燃烧食物样品释放的能量被已知质量的水吸收。每克食物的能量含量用以下公式计算：
Energy (J/g) = (mass of water × 4.2 × temperature rise) ÷ mass of food sample
Mass of water is in grams, temperature rise in °C, and the specific heat capacity of water is 4.2 J/g/°C. The mass of the food sample should be recorded before burning and the rise in water temperature measured accurately with a thermometer.
能量 (J/g) = (水的质量 × 4.2 × 温度升高值) ÷ 食物样品质量
水的质量用克表示，温度升高值用 °C 表示，水的比热容为 4.2 J/g/°C。燃烧前需记录食物样品质量，并用温度计准确测量水温的上升值。
6. Rate of Enzyme Reactions | 酶促反应速率
The rate of an enzyme-controlled reaction can be determined by measuring how quickly a product appears or a substrate disappears. Two common expressions are:
酶促反应速率可通过测量产物出现的快慢或底物消失的快慢来确定。两种常用表达式如下：
Rate = Amount of product formed ÷ time
速率 = 生成的产物量 ÷ 时间
When a colour change indicates the end point, such as the breakdown of starch by amylase, the rate is often given as the reciprocal of time:
当颜色变化标记终点时，例如淀粉被淀粉酶分解，速率常以时间的倒数表示：
Rate (s⁻¹) = 1 ÷ time (s)
In typical investigations, the time taken for the iodine solution to stop turning blue‑black is recorded, and 1/time is used to compare rates at different pH or temperature conditions.
速率 (s⁻¹) = 1 ÷ 时间 (s)
在典型实验中，记录碘液不再变为蓝黑色所需的时间，并用 1/时间 来比较不同 pH 或温度条件下的反应速率。
7. Rate of Photosynthesis | 光合作用速率
The rate of photosynthesis is often measured by counting the number of oxygen bubbles produced by an aquatic plant like pondweed per unit time, or by collecting the gas volume:
光合作用速率通常通过计算水生植物（如水蕴草）在单位时间内产生的氧气泡数，或收集气体体积来测定：
Rate = Number of O₂ bubbles ÷ time
速率 = O₂ 气泡数 ÷ 时间
or using the volume of gas collected:
或者使用收集的气体体积：
Rate = Volume of O₂ (mm³ or cm³) ÷ time (min)
Keeping light intensity, CO₂ concentration and temperature controlled while changing one variable allows you to investigate the factors that limit photosynthesis.
速率 = O₂ 体积 (mm³ 或 cm³) ÷ 时间 (min)
控制光照强度、CO₂ 浓度和温度不变，仅改变一个变量，即可探究限制光合作用的因素。
8. Rate of Transpiration | 蒸腾速率
A potometer measures the water uptake by a plant, which closely reflects the transpiration rate. The distance the air bubble moves in the capillary tube is recorded over time:
蒸腾计测量植株的吸水量，该值能较好地反映蒸腾速率。记录毛细管内气泡移动的距离和时间：
Transpiration rate = Distance moved by bubble ÷ time
Units may be mm/min or cm/min. The rate can be used to compare how environmental conditions, such as wind speed, humidity, light intensity or temperature, affect transpiration.
蒸腾速率 = 气泡移动的距离 ÷ 时间
单位可为 mm/min 或 cm/min。该速率可用于比较风速、湿度、光照强度或温度等环境条件对蒸腾作用的影响。
9. Percentage Change in Mass (Osmosis) | 质量变化百分比（渗透）
In osmosis experiments with potato cylinders or similar tissues, the change in mass is calculated as a percentage to allow fair comparison between samples of different original masses:
在用土豆条或类似组织进行的渗透实验中，为公平比较不同初始质量的样品，需计算质量变化百分比：
% change = ((final mass − initial mass) ÷ initial mass) × 100
A positive percentage indicates water has entered the tissue by osmosis (the solution was hypotonic), while a negative percentage means water was lost (hypertonic solution). Plotting % change against concentration helps identify the water potential equilibrium point.
质量变化百分比 = ((终质量 − 初始质量) ÷ 初始质量) × 100
正值表示水通过渗透进入组织（低渗溶液），负值表示水流失（高渗溶液）。绘制质量变化百分比与溶液浓度的关系图，有助于确定水势平衡点。
10. Efficiency of Biomass Transfer | 生物量传递效率
In food chains, only a small portion of the biomass consumed by a trophic level is converted into new biomass in the next level. The efficiency of biomass transfer is calculated as follows:
在食物链中，某一营养级消耗的生物量只有一小部分转化为下一个营养级的新生物量。生物量传递效率按如下公式计算：
Efficiency (%) = (Biomass transferred to the next level ÷ Biomass available at the previous level) × 100
Typical efficiencies range around 10 %, explaining why biomass pyramids usually narrow towards higher trophic levels and why most food chains are short.
效率 (%) = (传递给下一级的生物量 ÷ 上一级可用的生物量) × 100
典型传递效率约为 10%，这就解释了为什么生物量金字塔通常向高营养级收窄，以及为什么大多数食物链都很短。
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