📚 Year 10 AQA Biology: Formula & Principle Quick-Reference Handbook | Year 10 AQA 生物：公式定理速查手册
Welcome to your go-to guide for every essential formula, principle and numerical relationship covered in Year 10 AQA Biology. Whether you are calculating magnification, working out rates of reaction or interpreting physiological data, this handbook organises all the must-know quantitative skills into clear, bite-sized sections. Each entry pairs an English explanation with its Chinese counterpart so you can reinforce your subject knowledge bilingually, exactly the way you need for classwork, homework and revision.
欢迎使用这本 Year 10 AQA 生物必备公式、原理和数量关系的速查手册。无论你是在计算放大倍数、求反应速率，还是解读生理数据，本手册都将所有必会的定量技能整理成清晰、精简的小节。每一条内容都配有中英文对照讲解，帮助你在双语环境下巩固学科知识，完美适配课堂练习、作业和备考复习。
1. Magnification Formula | 放大倍数公式
One of the most frequently used equations in cell biology links the size of an image to the real size of the specimen. Magnification = image size ÷ actual size. Always ensure both measurements are in the same unit before dividing. You can rearrange the triangle: Actual size = image size ÷ magnification and Image size = magnification × actual size.
细胞生物学中最常用的公式之一将图像尺寸与标本的真实尺寸联系起来。放大倍数 = 图像尺寸 ÷ 实际尺寸。计算前务必统一单位。你可以通过三角形变形得到：实际尺寸 = 图像尺寸 ÷ 放大倍数，图像尺寸 = 放大倍数 × 实际尺寸。
M = I ÷ A
400, the actual size is 50 mm ÷ 400 = 0.125 mm. Convert to micrometres (×1000) to give 125 µm. Always show your unit conversion clearly.
考试中，如果细胞图像在 400 倍下测得长度为 50 mm，实际尺寸为 50 mm ÷ 400 = 0.125 mm。转换为微米（×1000）即得 125 µm。务必清晰地展示单位换算过程。
| Term | 术语 | Meaning | 含义 |
|---|---|
| Image size | Measured length in a drawing or micrograph | 图或显微照片中测得的长度 |
| Actual size | True length of the specimen | 标本的真实长度 |
| Magnification | How many times larger the image is than the real object | 图像比实物放大的倍数 |
2. Enzyme Activity and Rate Calculations | 酶活性与速率计算
The rate of an enzyme-controlled reaction can be measured by the amount of product formed or substrate used per unit time. Rate = quantity of product ÷ time. A common practical uses amylase breaking down starch, with iodine testing; the time taken for starch to disappear gives a relative rate as 1 ÷ time.
酶促反应的速率可以通过单位时间内产物的生成量或底物的消耗量来衡量。速率 = 产物量 ÷ 时间。一个常见实验用淀粉酶分解淀粉，并用碘液检测；淀粉消失所需时间的倒数（1 ÷ 时间）可作为相对速率。
Rate = 1 / t (when endpoint is fixed)
When plotting a graph, put time on the x‑axis and product formed on the y‑axis. The gradient of the line at any point gives the rate at that moment. Always mention ‘arbitrary units’ if no absolute mass or volume is recorded.
作图时将时间放在 x 轴，产物量放在 y 轴。线上任意一点的斜率代表该时刻的瞬时速率。如果没有记录绝对质量或体积，记得注明“任意单位”。
3. Percentage Change | 百分比变化
To compare two measurements, use the percentage change formula: Percentage change = ((final value – initial value) ÷ initial value) × 100. A negative result indicates a decrease. This is vital in osmosis experiments when measuring the change in mass of potato cylinders.
比较两个测量值时可使用百分比变化公式：百分比变化 = ((最终值 − 初始值) ÷ 初始值) × 100。负值表示下降。这在渗透实验中测量土豆条质量变化时至关重要。
%Δ = ((Vf − Vi) ÷ Vi) × 100
an initial mass of 5.0 g and a final mass of 5.8 g shows a ((5.8 − 5.0) ÷ 5.0) × 100 = +16% change. Explain the direction of water movement based on the sign.
例如，薯条初始质量 5.0 g，最终质量 5.8 g，则 ((5.8 − 5.0) ÷ 5.0) × 100 = +16%。根据正负号解释水分子的净移动方向。
4. Rate of Photosynthesis | 光合作用速率
The rate of photosynthesis can be measured indirectly by the volume of oxygen produced per minute, or by the time taken for a leaf disc to rise in a bicarbonate solution. Rate = 1 ÷ time taken for a standard response, e.g. the time for a leaf disc to float to the surface.
光合作用速率可以通过每分钟产生的氧气体积，或者叶圆片在碳酸氢盐溶液中上浮所需时间来间接测量。速率 = 1 ÷ 产生标准反应所用时间，例如叶圆片上浮到液面所需的时间。
Rate of O₂ production (cm³/min) = volume of O₂ ÷ time
When light intensity is limiting, doubling the distance from a lamp reduces light intensity to one quarter (inverse square law). State that beyond a certain concentration of CO₂ or light, the rate plateaus because another factor becomes limiting.
当光照强度为限制因素时，将灯的距离加倍，光强降为原来的四分之一（平方反比定律）。要说明当 CO₂ 浓度或光照超过一定值后，速率不再增加，因为另一个因素成为了限制因素。
5. Balanced Symbol Equations for Photosynthesis and Respiration | 光合作用与呼吸作用的配平化学方程式
These word and symbol equations are fundamental. For photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (light energy and chlorophyll over the arrow). For aerobic respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O (energy released). Recognise them as reverse processes.
以下文字与化学方程式是基础。光合作用：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂（箭头上方需写光能和叶绿体）。有氧呼吸：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O（释放能量）。要认识到两者互为逆过程。
Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Aerobic respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
You must be able to state that glucose is the reactant in respiration and a product in photosynthesis, and that oxygen is produced by photolysis of water during the light-dependent reactions.
你必须能说出葡萄糖是呼吸作用的反应物、光合作用的产物，并知道氧气是由水在光反应阶段光解产生的。
6. Oxygen Debt and Anaerobic Respiration | 氧债与无氧呼吸
During vigorous exercise, muscle cells respire anaerobically, producing lactic acid. Oxygen debt is the extra oxygen needed after exercise to oxidise lactic acid back to glucose or CO₂ and water. In yeast, anaerobic respiration produces ethanol + CO₂: the equation is Glucose → ethanol + carbon dioxide.
剧烈运动时，肌细胞进行无氧呼吸，产生乳酸。氧债是指运动后额外需要的氧气，用于将乳酸氧化回葡萄糖或 CO₂ 和水。在酵母中，无氧呼吸产生乙醇 + 二氧化碳，方程式为：葡萄糖 → 乙醇 + 二氧化碳。
Muscle cell anaerobic respiration: Glucose → lactic acid
Yeast anaerobic respiration: Glucose → ethanol + CO₂
肌细胞无氧呼吸：葡萄糖 → 乳酸
酵母无氧呼吸：葡萄糖 → 乙醇 + CO₂
There is no single numerical formula here, but you need to interpret graphs showing a rise in heart and breathing rate after exercise staying high to repay the oxygen debt.
这里没有单一的数字公式，但你需要解读显示运动后心率和呼吸频率持续升高以偿还氧债的图表。
7. Cardiac Output Formula | 心输出量公式
Cardiac output is the volume of blood pumped by one ventricle per minute. Cardiac output = heart rate × stroke volume bpm, stroke volume 70 cm³, giving cardiac output ≈ 4900 cm³/min. Units must be consistent.
心输出量是指一个心室每分钟泵出的血液体积。心输出量 = 心率 × 每搏输出量。静息时的典型数值：心率 70 bpm，每搏输出量 70 cm³，心输出量约为 4900 cm³/min。单位需保持一致。
CO = HR × SV
During exercise, both heart rate and stroke volume increase, dramatically raising cardiac output. A well-trained athlete may have a higher stroke volume and lower resting heart rate, achieving the same cardiac output more efficiently.
运动时，心率和每搏输出量均上升，心输出量显著增加。训练有素的运动员搏出量更高、静息心率更低，能以更高效率达到相同的心输出量。
8. Body Mass Index (BMI) | 身体质量指数 (BMI)
BMI is an estimate of body fat based on height and mass. BMI = body mass (kg) ÷ height² (m²) MI = 70 ÷ (1.75)² ≈ 22.9 kg/m². AQA contexts often link this to lifestyle and health data.
BMI 是通过身高和体重估算体脂的指标。BMI = 体重 (kg) ÷ 身高² (m²)。一位体重 70 kg、身高 1.75 m 的成人，BMI = 70 ÷ (1.75)² ≈ 22.9 kg/m²。AQA 常将此与生活方式和健康数据联系起来。
BMI = m ÷ h²
Remember that BMI does not distinguish between muscle and fat; very muscular individuals may be classified as overweight. Interpret data rather than memorising categories, though the typical ‘healthy’ range is 18.5 – 24.9 kg/m².
记住 BMI 无法区分肌肉与脂肪；肌肉非常发达的人可能被归为超重。重点是解读数据，而非死记分类，尽管典型的“健康”范围是 18.5 – 24.9 kg/m²。
9. Surface Area to Volume Ratio | 表面积与体积比
nisms need transport systems. The ratio is calculated by dividing surface area by volume. For a cube of side length 2 cm, surface area = 24 cm², volume = 8 cm³, SA:V = 3:1.
随着生物体或细胞尺寸增大，其表面积与体积比下降。这限制了扩散效率，并解释为什么较大的生物需要运输系统。该比例通过表面积除以体积求得。对于一个边长 2 cm 的立方体，表面积 = 24 cm²，体积 = 8 cm³，SA:V = 3:1。
SA:V = Surface area ÷ Volume
You should be able to interpret experimental data on diffusion in agar cubes of different sizes, explaining why the smallest cube turns colourless fastest throughout.
你应该能够解读不同大小琼脂块的扩散实验数据，并解释为什么最小的立方体整体变色最快。
10. Diffusion and Osmosis Principles | 扩散与渗透原理
While not strictly formulas, these principles underlie many numerical applications. Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration down a gradient. Factors affecting rate include temperature, concentration gradient and surface area, modelled by Fick's Law: Rate ∝ (surface area × concentration difference) ÷ membrane thickness.
虽然不完全是公式，但这些原理是许多数值应用的基础。扩散是粒子从高浓度区域向低浓度区域的净移动。影响速率的因素包括温度、浓度梯度和表面积，可用菲克定律描述：速率 ∝ (表面积 × 浓度差) ÷ 膜厚度。
Rate ∝ (SA × ΔC) ÷ d
Osmosis is a special case of diffusion involving water across a partially permeable membrane. Water potential drives the net movement; a dilute solution has a higher water potential than a concentrated one. You must use percentage change calculations to analyse results of osmosis experiments.
渗透是扩散的一种特殊情况，指水分子跨过半透膜的运动。水势驱动净移动；稀溶液的水势高于浓溶液。你必须使用百分比变化计算来分析渗透实验结果。
11. Energy in Food (Calorimetry) | 食物中的能量 (量热法)
The energy content of food can be estimated by burning a sample and measuring the temperature rise of a known volume of water. = mass of water (g) × specific heat capacity of water (4.2 J/g°C) × temperature rise (°C). Subsequently, find energy per gram: Energy per gram = energy released ÷ mass of food burnt.
食物的能量含量可通过燃烧样品、测量已知体积水温的升高来估算。释放的能量 (J) = 水的质量 (g) × 水的比热容 (4.2 J/g°C) × 升高的温度 (°C)。随后求出每克能量：每克能量 = 释放的能量 ÷ 燃烧的食物质量。
Q = m × c × ΔT
In typical classroom calorimetry with a crisp, energy loss to the surroundings is significant, so results are underestimates. AQA expects you to identify sources of error and suggest improvements, such as using a draught shield.
在常见的教室薯片量热实验中，能量散失到环境中很显著，因此结果是低估的。AQA 要求你识别误差来源并提出改进建议，如使用挡风罩。
12. Quadrat Sampling and Population Estimation | 样方采样与种群估算
For estimating population size in a habitat, use a quadrat. = (mean number of organisms per quadrat ÷ area of one quadrat) × total area of the habitat. You can also calculate percentage cover and frequency. Always use random sampling unless the aim is to study zonation along a transect.
估算栖息地种群大小时使用样方。估算种群数量 = (每个样方的平均生物数 ÷ 单个样方面积) × 栖息地总面积。你也可以计算百分比覆盖度和出现频率。除非是要研究样带上的带状分布，否则都应采用随机取样。
N = (n̄ ÷ a) × A
= 48 × 500 = 24 000. Interpret results cautiously, acknowledging sample size limitations.
例如，若每个 0.25 m² 样方内雏菊的平均数量为 12，田地面积为 500 m²，则估算种群数量为 (12 ÷ 0.25) × 500 = 48 × 500 = 24 000。解读结果时要谨慎，承认样本量的局限性。
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