📚 Year 11 Eduqas Biology: Quick Reference Formula & Theorems Handbook | Year 11 Eduqas 生物：公式定理速查手册
This handbook brings together all the essential formulas, equations and key principles you need to master for the Eduqas GCSE Biology examination. Use these concise summaries to check your understanding and to support your revision for the Year 11 assessments.
本手册汇集了你在 Eduqas GCSE 生物考试中需要掌握的所有核心公式、方程式和关键原理。使用这些简明摘要来检验你的理解，并为 Year 11 评估的复习提供支持。
1. Magnification Formula | 放大率公式
The magnification of a biological specimen is calculated by dividing the image size by the real size of the object. This formula is essential when using microscopes and interpreting micrographs. Remember to convert all measurements to the same unit before calculating.
生物标本的放大率通过图像大小除以物体的实际大小来计算。该公式在使用显微镜和解读显微照片时至关重要。计算前记得将所有测量值转换为相同单位。
Magnification = Image size ÷ Actual size
放大率 = 图像大小 ÷ 实际大小
- Image size is measured with a ruler on a micrograph; actual size is often given or can be found using a stage micrometer.
- 用直尺在显微照片上测量图像大小；实际大小通常已给出，或可通过载台测微尺得出。
2. Cardiac Output Equation | 心输出量公式
Cardiac output is the volume of blood pumped by the heart per minute. It is determined by the heart rate (beats per minute) and the stroke volume (volume of blood pumped per beat). This relationship helps explain how exercise or disease affects circulation.
心输出量是心脏每分钟泵出的血液体积。它由心率（每分钟心跳次数）和每搏输出量（每搏泵出的血量）决定。这一关系有助于解释运动或疾病如何影响循环。
Cardiac output = Heart rate × Stroke volume
心输出量 = 心率 × 每搏输出量
- Units: cardiac output in cm³/min or L/min; heart rate in beats/min (bpm); stroke volume in cm³ or L.
- 单位：心输出量以 cm³/min 或 L/min 表示；心率单位为每分钟次数 (bpm)；每搏输出量单位为 cm³ 或 L。
3. Body Mass Index (BMI) | 体质指数 (BMI)
BMI is a simple index used to classify whether an adult has a healthy body mass for their height. It is widely used to assess risks associated with being underweight, overweight or obese.
BMI 是一个用于判断成年人体重相对身高是否健康的简单指数，广泛用于评估体重过轻、超重或肥胖的相关风险。
BMI = Body mass (kg) ÷ (Height in metres)²
BMI = 体重 (kg) ÷ (身高以米为单位)²
- r example, a person weighing 70 kg and 1.75 m tall: BMI = 70 ÷ (1.75 × 1.75) = 22.9 kg/m².
- 例如，体重 70 kg、身高 1.75 m 的人：BMI = 70 ÷ (1.75 × 1.75) = 22.9 kg/m²。
4. Energy Transfer Efficiency in Ecosystems | 生态系统的能量传递效率
In a food chain, only a fraction of the energy is transferred from one trophic level to the next. The efficiency of energy transfer is calculated to understand how much biomass can be supported at higher levels.
在食物链中，只有一部分能量从一个营养级传递到下一个营养级。计算能量传递效率有助于理解更高营养级能够维持多少生物量。
Efficiency (%) = (Energy transferred to biomass ÷ Total energy intake) × 100
效率 (%) = (传递至生物质的能量 ÷ 总摄入能量) × 100
- Energy is lost through respiration, egestion, excretion and as heat. Typically efficiency is around 10% between trophic levels.
- 能量通过呼吸作用、排遗、排泄和散热损失。营养级之间的效率通常约为 10%。
5. Photosynthesis Word and Chemical Equation | 光合作用文字与化学方程式
Photosynthesis is the process by which plants, algae and some bacteria convert light energy into chemical energy, producing glucose and oxygen. The balanced chemical equation summarises the reactants and products.
光合作用是植物、藻类和某些细菌将光能转化为化学能、生成葡萄糖和氧气的过程。配平的化学方程式概括了反应物和产物。
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
carbon dioxide + water → glucose + oxygen
二氧化碳 + 水 → 葡萄糖 + 氧气
- Light energy is absorbed by chlorophyll and is essential for the reaction to proceed.
- 光能被叶绿素吸收，是反应进行的必要条件。
6. Aerobic Respiration Equation | 有氧呼吸方程式
Aerobic respiration releases energy from glucose in the presence of oxygen. It is the most efficient way to produce ATP, the energy currency of cells, and takes place in mitochondria.
有氧呼吸在氧气的参与下从葡萄糖释放能量。它是产生 ATP（细胞的能量通货）效率最高的方式，并在线粒体中进行。
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (ATP)
glucose + oxygen → carbon dioxide + water + energy
葡萄糖 + 氧气 → 二氧化碳 + 水 + 能量
- This reaction is essentially the reverse of photosynthesis, releasing the energy stored in glucose.
- 此反应基本上是光合作用的逆过程，释放储存在葡萄糖中的能量。
7. Anaerobic Respiration Equations | 无氧呼吸方程式
When oxygen is insufficient, cells can respire anaerobically to produce a small amount of ATP. In animals, lactic acid is formed; in yeast and plants, ethanol and carbon dioxide are produced.
当氧气不足时，细胞可进行无氧呼吸产生少量 ATP。动物体内形成乳酸；酵母和植物中生成乙醇和二氧化碳。
In animals / 动物中：
C₆H₁₂O₆ → 2C₃H₆O₃ + energy (ATP)
glucose → lactic acid + energy
葡萄糖 → 乳酸 + 能量
In yeast and plants / 酵母和植物中：
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ + energy (ATP)
glucose → ethanol + carbon dioxide + energy
葡萄糖 → 乙醇 + 二氧化碳 + 能量
8. Rate of Enzyme-Controlled Reactions | 酶促反应速率
The rate of an enzyme-catalysed reaction can be determined by measuring the amount of product formed or substrate consumed per unit time. A simple calculation involves the reciprocal of the time taken for a given change to occur.
酶催化反应速率可通过测量单位时间生成的产物量或消耗的底物量来确定。一种简单计算涉及给定变化所需时间的倒数。
Rate = 1 ÷ Time (or change in quantity ÷ time)
速率 = 1 ÷ 时间（或改变量 ÷ 时间）
- If starch is broken down by amylase, the rate can be expressed as 1 ÷ time taken for the iodine test to no longer turn blue-black.
- 若淀粉被淀粉酶分解，速率可表示为 1 ÷ 碘液不再变蓝黑色所需的时间。
9. Fick's Law for Diffusion | 扩散的菲克定律
The rate of diffusion across a membrane depends on the surface area, the concentration gradient and the thickness (distance) of the exchange surface. Fick's law brings these factors together in a proportional relationship.
跨膜扩散速率取决于表面积、浓度梯度和交换表面的厚度（距离）。菲克定律将这些因素综合在一个比例关系中。
Rate of diffusion ∝ (Surface area × Concentration difference) ÷ Diffusion distance
扩散速率 ∝ (表面积 × 浓度差) ÷ 扩散距离
- Adaptations of gas exchange surfaces, such as large alveoli area and thin epithelium, increase diffusion rate according to this law.
- 气体交换表面的适应特征（如巨大肺泡表面积和薄上皮）依照该定律提高扩散速率。
10. Osmosis and Water Potential | 渗透作用与水势
Osmosis is the net movement of water molecules across a partially permeable membrane from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution). Water potential is usually measured in pressure units and pure water has a water potential of zero.
渗透作用是水分子通过选择性透过膜从较高水势（稀溶液）向较低水势（浓溶液）的净移动。水势通常以压力单位衡量，纯水的水势为零。
- No formal equation is required at GCSE, but the concept is vital for understanding turgor in plants and lysis/crenation in animal cells.
- GCSE 不要求正式方程式，但该概念对于理解植物中的膨压和动物细胞中的溶血、皱缩至关重要。
11. Monohybrid Inheritance and Phenotypic Ratios | 单基因杂交与表现型比例
ratio for a heterozygous cross (e.g. Aa × Aa) is 3:1 for dominant to recessive phenotypes.
当一个具有两个等位基因的基因控制某一性状时，我们可以用庞尼特方格预测杂交结果。杂合子杂交（如 Aa × Aa）的经典孟德尔比例是显性表现型与隐性表现型之比为 3:1。
- mozygous dominant crossed with any; 1:1 if test cross (Aa × aa); 3:1 if two heterozygotes.
- 预期比例：若纯合显性与任何杂交，为 1:0；测交（Aa × aa）为 1:1；两个杂合子为 3:1。
12. Population Density and Sampling | 种群密度与取样
Estimating population size in a habitat often involves sampling using quadrats. The abundance can be calculated as density per unit area or as percentage cover. The mean number per quadrat is used to scale up to the total area.
估算栖息地种群大小常通过使用样方取样。多度可表示为单位面积密度或百分比覆盖度。用每个样方的平均数量乘上总面积进行推算。
Estimated population = (Mean number per quadrat × Total area) ÷ Area of quadrat
估计种群数量 = (每个样方平均个体数 × 总面积) ÷ 样方面积
- Alternatively, percentage cover = (area covered by species ÷ total area of quadrat) × 100.
- 或者，覆盖度百分比 = (物种覆盖面积 ÷ 样方总面积) × 100。
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