📚 Year 10 Edexcel Biology: Formula and Theorem Quick Reference Handbook | Year 10 Edexcel 生物：公式定理速查手册
This quick reference handbook compiles the essential formulas, equations, key principles and theorems for Year 10 Edexcel Biology. Use it to revise core quantitative skills, recall important biological laws, and support your exam preparation.
本速查手册汇集了 Edexcel 考试局 Year 10 生物课程的核心公式、方程式、关键原理和定理。利用它来复习重要的定量技能、记住关键的生物学定律，助力你的考试准备。
1. Magnification and Unit Conversion | 放大倍数与单位换算
The magnification of a microscopic image tells you how many times larger the image is than the actual object. The triangle formula helps you rearrange the relationship: Magnification = Image size ÷ Actual size. Always express both sizes in the same unit before calculating.
显微镜图像的放大倍数告诉你图像比实际物体大多少倍。这个三角形公式可以帮你重组关系：放大倍数 = 图像尺寸 ÷ 实际尺寸。计算前一定要将两个尺寸转换为相同的单位。
Magnification = Image size ÷ Actual size
放大倍数 = 图像尺寸 ÷ 实际尺寸
Key unit conversions are essential for microscopy: 1 cm = 10 mm; 1 mm = 1000 µm; 1 µm = 1000 nm. For example, when an image measures 50 mm at 400×, the actual size is 50 mm ÷ 400 = 0.125 mm = 125 µm.
关键的单位换算是显微镜工作必须掌握的：1 厘米 = 10 毫米；1 毫米 = 1000 微米；1 微米 = 1000 纳米。例如，一张图像在 400 倍下测得 50 毫米，实际尺寸为 50 mm ÷ 400 = 0.125 mm = 125 µm。
2. Surface Area to Volume Ratio | 表面积与体积比
As an object or organism gets larger, its volume grows faster than its surface area. The surface area : volume ratio (SA:V) is calculated by dividing surface area by volume. A high SA:V ratio allows rapid diffusion, while a low ratio limits exchange, making specialised transport systems necessary.
物体或生物体变大时，体积的增长速度要快于表面积。表面积与体积之比（SA:V）用表面积除以体积来计算。高的 SA:V 比有利于快速扩散，低的比值则会限制物质交换，因此需要特化的运输系统。
SA:V ratio = Surface area ÷ Volume
SA:V 比值 = 表面积 ÷ 体积
For a 1 cm cube, SA = 6 cm², V = 1 cm³, ratio = 6:1. For a 2 cm cube, SA = 24 cm², V = 8 cm³, ratio = 3:1. This fall explains why larger organisms need lungs, gills and transport systems.
以一个边长 1 cm 的立方体为例，表面积 = 6 cm²，体积 = 1 cm³，比值为 6:1。边长变为 2 cm 时，表面积 = 24 cm²，体积 = 8 cm³，比值降为 3:1。这一变化解释了为什么较大的生物需要肺、鳃和运输系统。
3. Percentage Change in Mass | 质量变化百分比
In osmosis experiments, such as the potato cylinder practical, you calculate the percentage change in mass to compare water uptake or loss. A positive percentage indicates water has entered the tissue; a negative one shows water has left.
在渗逶实验中，例如土豆条实验，通过计算质量变化百分比来比较吸水或失水的情况。正百分比说明水分进入组织，负百分比说明水分流失。
Percentage change = (Final mass − Initial mass) ÷ Initial mass × 100%
变化百分比 = (最终质量 − 初始质量) ÷ 初始质量 × 100%
Always record masses in grams and calculate to at least two decimal places. Plotting percentage change against sucrose concentration lets you estimate the solute potential inside the cells.
记录质量时始终使用克，并至少保留两位小数。将变化百分比对蔗糖浓度作图，可以估算细胞内溶液的溶质势。
4. Rate of Reaction Calculations | 反应速率计算
Reaction rate measures how fast a substrate is used or a product is formed. For enzyme-catalysed reactions, you often find the rate by recording the volume of gas produced or the disappearance of a colour. The average rate is the total change divided by time.
反应速率衡量底物消耗或产物形成的快慢。对于酶促反应，通常通过记录产生的气体体积或颜色消退来计算速率。平均速率等于总变化量除以时间。
Rate = Change in quantity (e.g. volume, mass) ÷ Time
速率 = 变化量（如体积、质量）÷ 时间
When starch is broken down by amylase, you can measure the time taken for the iodine test to stop turning blue-black. A useful proxy rate is then 1 / time (1/t). Units could be s⁻¹.
当淀粉酶分解淀粉时，可以记录碘液不再变为蓝黑色的时间。此时一个实用的替代速率是 1 ÷ 时间（1/t），单位可用 s⁻¹。
5. Population Estimation Techniques | 群体大小估计方法
Ecologists estimate population sizes using quadrats for slow-moving or stationary organisms, and the capture-mark-recapture method for motile animals. The quadrat formula is: Population size = Mean count per quadrat × (Total area ÷ Quadrat area).
生态学家对运动缓慢或不动的生物使用样方法，对活动动物则使用捕捉–标记–再捕捉法。样方法的公式为：群体大小 = 每样方平均数量 × (总面积 ÷ 样方面积)。
Estimating population using quadrats: N = n̅ × (Atotal ÷ Aquadrat)
样方估计法：N = n̅ × (Atotal ÷ Aquadrat)
The Lincoln index for capture-recapture is: N = (M × C) ÷ R, where M is the number initially marked and released, C is the total caught in the second sample, and R is the number of recaptures that were already marked. Assumptions include no migration, no births or deaths between samples, marks are not lost, and marked individuals mix randomly.
捕捉–再捕捉的林肯指数公式为：N = (M × C) ÷ R，其中 M 是首次标记释放的数量，C 是第二次捕获的总数，R 是第二次捕获中已标记的个体数。前提假设包括：没有迁移、两次采样间无出生或死亡、标记不脱落、标记个体随机混合。
6. Energy and Biomass Transfer Efficiency | 能量与生物量传递效率
In food chains, only a fraction of the energy or biomass at one trophic level is transferred to the next. The efficiency is expressed as a percentage. Typical ecological efficiency is around 10%, with losses due to respiration, movement, excretion and uneaten parts.
在食物链中，上一个营养级只有一小部分能量或生物量传递到下一个营养级。效率以百分比表示。典型的生态效率约为 10%，能量损失来源于呼吸作用、运动、排泄和未进食的部分。
Efficiency (%) = (Energy/biomass transferred to next level ÷ Energy/biomass available at previous level) × 100%
效率 (%) = (传递到下一级的能量/生物量 ÷ 前一级可用的能量/生物量) × 100%
You may be asked to construct pyramids of biomass or energy and calculate efficiency from given data. Remember to subtract energy lost through waste before working out the amount available to the next level.
考试中可能要求你构建生物量或能量金字塔，并根据数据计算效率。记住在计算可供下一级利用的量之前，要先减去浪费损失的能量。
7. Punnett Squares and Genetic Probability | 庞纳特方格与遗传概率
A Punnett square shows the possible genotypes of offspring from a genetic cross. When both parents are heterozygous (Aa), the expected genotypic ratio is 1 AA : 2 Aa : 1 aa, and the phenotypic ratio for a completely dominant trait is 3 dominant : 1 recessive.
庞纳特方格显示了遗传杂交中后代的可能基因型。当双亲均为杂合子（Aa）时，预期的基因型比例为 1 AA : 2 Aa : 1 aa；对于完全显性性状，表型比例为 3 显性 : 1 隐性。
| A | a | |
| A | AA | Aa |
| a | Aa | aa |
For sex determination, the cross is XX (female) × XY (male). The offspring ratio is 1 XX : 1 XY, meaning equal probability of female and male.
性別決定方面，杂交形式为 XX（雌性）× XY（雄性）。后代比例为 1 XX : 1 XY，代表雌性和雄性的概率相等。
8. Cell Theory and Central Principles | 细胞理论与核心原理
Modern cell theory states three fundamental ideas: (1) All living organisms are composed of one or more cells. (2) The cell is the basic structural and functional unit of all living things. (3) All cells arise from pre-existing cells through division.
现代细胞理论阐述三个基本观点：（1）所有生物体由一个或多个细胞构成。（2）细胞是所有生物体的基本结构和功能单位。（3）所有细胞都通过分裂从原有细胞产生。
These principles underpin all of biology. They explain why bacteria, plants and animals are all subject to the same molecular and organisational logic, and why processes like mitosis and meiosis are essential for growth and reproduction.
这些原理奠定了整个生物学的基础。它们解释了为何细菌、植物和动物都遵循相同的分子与组织逻辑，也揭示了有丝分裂和减数分裂对生长与繁殖为何是必不可少的。
9. Gas Exchange Principles | 气体交换原理
The rate of diffusion across a membrane or tissue depends on surface area, the concentration gradient and the thickness of the exchange surface. This relationship is often written as a proportionality.
通过膜或组织扩散的速率取决于表面积、浓度梯度和交换表面的厚度。这一关系通常用比例式来表示。
Rate of diffusion ∝ (Surface area × Concentration difference) ÷ Thickness of membrane
扩散速率 ∝ (表面积 × 浓度差) ÷ 膜的厚度
Efficient exchange surfaces, such as alveoli, gill filaments and spongy mesophyll in leaves, share common features: large surface area, thin barriers, good blood supply (or ventilation) and a steep concentration gradient maintained by continuous flow.
高效的交换表面——如肺泡、鳃丝和叶片的海绵状叶肉——都有共同特征：表面积大、屏障薄、血液供应充足（或能通风），并通过持续流动保持陡峭的浓度梯度。
10. Photosynthesis and Respiration Equations | 光合作用与呼吸作用方程式
Photosynthesis uses light energy to convert carbon dioxide and water into glucose and oxygen. The balanced symbol equation summarises the chemical changes.
光合作用利用光能将二氧化碳和水转化为葡萄糖和氧气。配平的化学符号方程式概括了其中的化学变化。
Word equation: carbon dioxide + water → glucose + oxygen
文字方程式：二氧化碳 + 水 → 葡萄糖 + 氧气
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Aerobic respiration releases energy from glucose. The same equation is reversed, and energy is produced as ATP.
有氧呼吸从葡萄糖中释放能量。同一方程式反向进行，并以 ATP 的形式产生能量。
Word equation: glucose + oxygen → carbon dioxide + water (+ energy)
文字方程式：葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
11. Interpreting Graphs and Calculating Rate | 图表解读
Published by TutorHao | Year 10 Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导