📚 Formula & Theorem Quick Reference Handbook | 公式定理速查手册
This handbook summarises the essential formulas, word equations, and key principles covered in Year 8 Edexcel Biology. Use it as a quick revision companion to remember the core concepts when preparing for assessments.
本手册总结了 Edexcel 八年级生物课程中必须掌握的公式、文字方程式和核心原理，可作为考前快速复习的速查指南，帮助你牢固记忆关键概念。
1. Photosynthesis Word Equation | 光合作用文字方程式
Photosynthesis is the process by which green plants produce glucose and oxygen using light energy, carbon dioxide and water. The word equation must be memorised precisely.
光合作用是绿色植物利用光能、二氧化碳和水生成葡萄糖和氧气的过程。需要准确牢记该文字方程式。
Carbon dioxide + Water → Glucose + Oxygen
二氧化碳 + 水 → 葡萄糖 + 氧气
The reaction takes place in the chloroplasts and requires chlorophyll and sunlight. Light energy is absorbed by chlorophyll and converted into chemical energy stored in glucose.
该反应发生在叶绿体中，需要叶绿素和阳光。光能被叶绿素吸收后转化为化学能，储存在葡萄糖中。
2. Balanced Chemical Equation for Photosynthesis (Symbol) | 光合作用平衡化学方程式（符号）
Although Year 8 mainly uses word equations, the symbol equation provides a clearer picture of the atoms involved. The balanced symbol equation is:
虽然八年级主要使用文字方程式，但符号方程式能更清晰地展示原子种类和数目。平衡后的符号方程式为：
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
This shows that six molecules of carbon dioxide react with six molecules of water to produce one molecule of glucose and six molecules of oxygen. It is essential to note that the atoms are rearranged, not created or destroyed.
其表示6个二氧化碳分子与6个水分子反应，生成1个葡萄糖分子和6个氧气分子。重要的是，原子只是重新组合，并未被创造或消灭。
3. Aerobic Respiration Word Equation | 有氧呼吸文字方程式
Aerobic respiration releases energy from glucose in the presence of oxygen. It occurs continuously in the mitochondria of most cells. The word equation is:
有氧呼吸是在有氧条件下从葡萄糖中释放能量的过程，持续发生在大多数细胞的线粒体中。文字方程式为：
Glucose + Oxygen → Carbon dioxide + Water (+ energy)
葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）
Energy is released in the form of ATP, which powers life processes such as muscle contraction, active transport and maintaining body temperature.
能量以ATP形式释放，为肌肉收缩、主动运输和维持体温等生命活动提供动力。
4. Anaerobic Respiration in Animals | 动物体内的无氧呼吸
When oxygen is insufficient, animal cells switch to anaerobic respiration. This produces less energy and leads to the formation of lactic acid, which can cause muscle fatigue. The word equation is:
当氧气不足时，动物细胞转为无氧呼吸，产生的能量较少，并生成乳酸，后者可能导致肌肉疲劳。其文字方程式为：
Glucose → Lactic acid (+ energy)
葡萄糖 → 乳酸（+ 能量）
Unlike aerobic respiration, this process does not require oxygen and produces no carbon dioxide or water in animals. The lactic acid must be oxidised later when oxygen becomes available, creating an oxygen debt.
与有氧呼吸不同，此过程无需氧气，在动物体内不产生二氧化碳和水。之后氧气充足时，乳酸需被氧化，这就形成了氧债。
5. Anaerobic Respiration in Plants and Yeast | 植物和酵母的无氧呼吸
In plants and microorganisms such as yeast, anaerobic respiration produces ethanol and carbon dioxide instead of lactic acid. This is exploited in baking and brewing. The word equation is:
在植物和酵母等微生物中，无氧呼吸产生的是乙醇和二氧化碳而非乳酸。这被应用于烘焙和酿酒。文字方程式为：
Glucose → Ethanol + Carbon dioxide (+ energy)
葡萄糖 → 乙醇 + 二氧化碳（+ 能量）
Carbon dioxide released causes bread dough to rise, while ethanol contributes to the flavour and alcohol content in beverages. This process is also known as fermentation.
释放的二氧化碳使面包面团膨胀，乙醇则为饮料带来风味和酒精含量。这一过程也称为发酵。
6. Enzyme Action and the Lock-and-Key Model | 酶的作用与锁钥模型
Enzymes are biological catalysts that speed up chemical reactions without being used up. The ‘lock and key’ model explains how each enzyme has an active site with a specific shape that fits only a particular substrate, like a key fits a lock.
酶是生物催化剂，能加速化学反应而自身不被消耗。”锁钥”模型说明，每种酶具有特定形状的活性位点，只与特定的底物结合，好比一把钥匙开一把锁。
Once the enzyme-substrate complex forms, the reaction occurs and products are released, leaving the enzyme unchanged. Denaturation occurs when high temperature or extreme pH alters the shape of the active site permanently, preventing binding.
酶与底物形成复合物后，发生反应并释放产物，酶本身不变。高温或极端pH会永久改变活性位点的形状，导致酶变性失活。
7. Digestive Enzymes – Key Facts Table | 消化酶关键信息表
The three main types of digestive enzymes break down large insoluble molecules into smaller soluble ones for absorption. The table below summarises their actions.
三大类消化酶将不溶性大分子分解为可溶的小分子以便吸收。下表总结了它们的作用。
| Enzyme / 酶 | Substrate / 底物 | Product(s) / 产物 | Site of Action / 作用部位 |
|---|---|---|---|
| Amylase / 淀粉酶 | Starch / 淀粉 | Maltose / 麦芽糖 | Mouth, small intestine / 口腔、小肠 |
| Protease / 蛋白酶 | Protein / 蛋白质 | Amino acids / 氨基酸 | Stomach, small intestine / 胃、小肠 |
| Lipase / 脂肪酶 | Lipids (fats) / 脂类 | Fatty acids + glycerol / 脂肪酸 + 甘油 | Small intestine / 小肠 |
Bile is not an enzyme but emulsifies fats, breaking large lipid droplets into smaller ones to increase the surface area for lipase action. This is a mechanical, not chemical, breakdown.
胆汁不是酶，但能乳化脂肪，将大脂肪滴分解为小脂肪滴，增大脂肪酶作用的表面积。这属于物理性消化，而非化学性消化。
8. Microscopy Magnification Formula | 显微镜放大倍数公式
When using a light microscope, the total magnification is calculated by multiplying the eyepiece lens magnification by the objective lens magnification. The formula is:
使用光学显微镜时，总放大倍数为目镜放大倍数与物镜放大倍数的乘积。公式为：
Total magnification = Eyepiece magnification × Objective magnification
总放大倍数 = 目镜放大倍数 × 物镜放大倍数
For example, if the eyepiece magnifies ×10 and the objective lens magnifies ×40, the total magnification is ×400. This formula helps when drawing scaled diagrams, though Year 8 mainly focuses on understanding scale rather than calculating actual size.
例如，若目镜为10×、物镜为40×，则总放大倍数为400×。该公式有助于绘制按比例放大的图，不过八年级阶段更侧重理解比例关系，而非计算实际大小。
9. Genetic Inheritance – Monohybrid Cross and Probability | 遗传 – 单基因杂交与概率
A monohybrid cross tracks the inheritance of a single characteristic. Using a Punnett square, we can predict the probability of offspring showing dominant or recessive traits. The basic principle: each parent contributes one allele per trait.
单基因杂交追踪单一性状的遗传。通过庞纳特方格，可以预测后代出现显性或隐性性状的概率。基本原则：每个亲本针对每个性状提供一个等位基因。
If ‘A’ represents a dominant allele and ‘a’ a recessive allele, a cross between two heterozygous parents (Aa × Aa) gives a genotypic ratio of 1 AA : 2 Aa : 1 aa, and a phenotypic ratio of 3 dominant : 1 recessive. Probability of a dominant phenotype is 3/4 or 75%.
若 ‘A’ 代表显性等位基因，’a’ 代表隐性等位基因，两个杂合亲本杂交（Aa × Aa），后代基因型比例为 1 AA : 2 Aa : 1 aa，表型比例为3显性 : 1隐性。显性表型出现的概率为3/4，即75%。
The inheritance of sex is determined by sex chromosomes: females (XX) and males (XY). A Punnett square cross between XX and XY shows a 1:1 ratio of female to male offspring.
性别由性染色体决定：女性为XX，男性为XY。用庞纳特方格进行XX与XY的杂交，后代表现出雌雄1:1的比例。
10. Food Tests – Reagents and Expected Results | 食物检测 – 试剂与预期结果
Identifying biological molecules in food samples is a core practical skill. The following summarises the main tests.
鉴定食物样品中的生物分子是一项核心实验技能。下表总结了主要检测方法。
| Molecule / 分子 | Reagent / 试剂 | Positive Result / 阳性结果 |
|---|---|---|
| Starch / 淀粉 | Iodine solution / 碘液 | Blue-black / 蓝黑色 |
| Reducing sugar / 还原糖 | Benedict’s solution (heat) / 本尼迪克特试剂（加热） | Green → yellow → brick-red / 绿色→黄色→砖红色 |
| Protein / 蛋白质 | Biuret solution (copper sulfate + sodium hydroxide) / 双缩脲试剂 | Purple/violet / 紫色 |
| Lipids (fats) / 脂类 | Ethanol then water (emulsion test) / 乙醇然后加水（乳浊液测试） | Cloudy white emulsion / 乳白色浑浊 |
It is important to use appropriate safety precautions, as some reagents (e.g., Benedict’s, Biuret) are irritant and require heating carefully in a water bath.
务必采取适当的安全防护，因为某些试剂（如本尼迪克特试剂、双缩脲试剂）具有刺激性，且需要在水浴中小心加热。
11. Energy in Food and Calorimetry Principle | 食物中的能量与热量测量原理
The energy content of food can be estimated by burning a sample and measuring the temperature rise of a known volume of water. The basic principle: energy released by food = energy absorbed by water.
食物所含能量可通过燃烧样品并测量定体积水温升高来估算。基本原理：食物释放的能量 = 水吸收的能量。
Energy (J) = mass of water (g) × 4.2 J/g°C × temperature rise (°C)
能量（焦耳）= 水的质量（g）× 4.2 J/g°C × 温度升高值（°C）
This is a simplified version; students should understand that not all energy is transferred efficiently due to heat loss to the surroundings, so the calculated value is an estimate. The value 4.2 J/g°C is the specific heat capacity of water.
这是一个简化公式；学生应明白由于热量散失到周围环境中，能量传递并非完全有效，因此计算结果为估值。4.2 J/g°C 是水的比热容。
12. Ecological Key Terms and Energy Transfer Rule | 生态关键术语与能量传递法则
In a food chain, energy passes from one trophic level to the next. On average, only about 10% of the energy is transferred; the rest is lost as heat, through movement or in undigested waste. This 10% rule explains why food chains rarely exceed four or five trophic levels.
在食物链中，能量从一个营养级传递到下一级。平均只有约10%的能量被传递，其余的能量以热量、运动消耗或未消化废物等形式散失。这一“10%法则”解释了为什么食物链通常不超过4～5个营养级。
Important terms: producer (makes its own food, e.g. plant), primary consumer (eats producer), secondary consumer (eats primary consumer), decomposer (breaks down dead matter, returning nutrients to soil).
重要术语：生产者（自养生物，如植物）、初级消费者（以生产者为食）、次级消费者（以初级消费者为食）、分解者（分解死物，使养分回归土壤）。
Biomass is the total mass of living material at a trophic level. Pyramids of biomass are usually drawn to scale and show the decrease in biomass at higher trophic levels.
生物量是指某一营养级上活物质的总质量。生物量金字塔通常按比例绘制，呈现出随营养级升高生物量递减的趋势。
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