📚 Year 11 CIE Biology Formula & Theorem Quick Reference Handbook | Year 11 CIE 生物公式定理速查手册
Mastering the quantitative and theoretical side of CIE IGCSE Biology is just as important as understanding the concepts themselves. This quick reference handbook brings together all the essential formulas, equations, ratios, and principles that frequently appear in Paper 2, Paper 4, and Paper 6. From magnification calculations to Lincoln Index, each entry is explained with worked examples so you can apply them accurately in your exams.
掌握 CIE IGCSE 生物学的定量与理论部分，与理解概念本身同样关键。这本速查手册汇总了所有经常出现在试卷二、试卷四和试卷六中的基本公式、方程、比率和原理。从放大倍数计算到林肯指数，每个条目都配有具体示例讲解，帮助你在考试中准确运用。
1. Magnification Formula | 放大倍数公式
The magnification of a drawing or micrograph tells you how many times larger the image is compared to the real object. The standard formula is used in microscopy and biological drawing exercises.
绘图或显微照片的放大倍数表示图像比真实物体大多少倍。该标准公式广泛应用于显微镜观察和生物绘图练习中。
Magnification = Image size ÷ Actual size
Always ensure image size and actual size are in the same unit before dividing. It is common to convert mm to µm or nm as needed.
计算前务必确保图像尺寸和实际尺寸单位一致。通常需要将毫米转换为微米或纳米。
Example: An image of a cell measures 50 mm across. The actual cell diameter is 0.025 mm. The magnification is 50 ÷ 0.025 = 2000×. If the image size is given in mm and actual size in µm, convert first (1 mm = 1000 µm).
示例：一个细胞图像直径为 50 mm。实际细胞直径为 0.025 mm。放大倍数为 50 ÷ 0.025 = 2000 倍。如果图像尺寸以毫米为单位，实际尺寸以微米为单位，请先进行单位换算（1 mm = 1000 µm）。
2. Actual Size Calculation | 实际大小计算
When you know the magnification and can measure the image size, rearrange the formula to find the actual size of a specimen. This is particularly useful for estimating cell dimensions from photomicrographs.
当已知放大倍数并可测量图像大小时，运用变形公式即可求出标本的实际大小。此方法对显微照片中细胞尺寸的估算尤为重要。
Actual size = Image size ÷ Magnification
If an image of a chloroplast is 30 mm long and the magnification is 6000×, then actual length = 30 mm ÷ 6000 = 0.005 mm, which is 5 µm after conversion.
如果一个叶绿体图像长 30 mm，放大倍数为 6000 倍，则实际长度 = 30 mm ÷ 6000 = 0.005 mm，换算后为 5 µm。
Remember to express your final answer in an appropriate biological unit, typically micrometres (µm) for cell structures. Show your working clearly on exam papers.
请记住以合适的生物学单位呈现最终答案，细胞结构通常以微米（µm）表示。在试卷上清晰展示计算过程。
3. Balanced Equations for Photosynthesis | 光合作用方程式
Photosynthesis is the process by which green plants produce glucose and oxygen from carbon dioxide and water, using light energy. Both word and symbol equations are required knowledge.
光合作用是绿色植物利用光能，将二氧化碳和水转化为葡萄糖和氧气的过程。需要同时掌握文字方程式和化学符号方程式。
Word equation: carbon dioxide + water → glucose + oxygen (in the presence of light and chlorophyll).
文字方程式：二氧化碳 + 水 → 葡萄糖 + 氧气（在光和叶绿素存在下）。
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
The balanced symbol equation must show six molecules of each input and output, illustrating that matter is conserved. The equation is a foundation for topics like leaf structure and gas exchange.
配平后的化学方程式必须体现出各六个分子的输入与输出，表明物质守恒。该方程式是叶片结构和气体交换等主题的基础。
4. Balanced Equation for Aerobic Respiration | 有氧呼吸方程式
Aerobic respiration releases energy from glucose in the presence of oxygen. The symbolic equation is effectively the reverse of photosynthesis, emphasising the cycling of gases in nature.
有氧呼吸在有氧气参与的条件下从葡萄糖中释放能量。其他学方程式实质上与光合作用互为可逆，体现了自然界中气体的循环。
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O (+ ATP)
While ATP is not a chemical substance that appears in a perfectly balanced equation, it is often included to remind students that energy is released. In CIE exams, the version showing ATP as an output is widely accepted.
尽管 ATP 并非严格配平方程中的化学物质，但常被纳入以提示学生能量的释放。在 CIE 考试中，将 ATP 作为产物的版本也被广泛接受。
Note that anaerobic respiration in yeast and muscle cells produces different products and far less ATP, but the aerobic equation serves as the key reference for energy calculations.
请注意，酵母和肌肉细胞中的无氧呼吸产生不同产物且 ATP 产量少得多，但有氧呼吸方程式是能量计算的关键参考。
5. Respiratory Quotient (RQ) | 呼吸商
The respiratory quotient indicates which substrate is being respired by comparing the volume of carbon dioxide produced to the volume of oxygen consumed over a given time.
呼吸商通过比较特定时间内产生二氧化碳的体积与消耗氧气的体积，揭示正在被呼吸的底物类型。
RQ = Volume of CO₂ released ÷ Volume of O₂ consumed
An RQ of 1.0 typically means carbohydrate is being respired. RQ values near 0.7 suggest lipid metabolism, while RQ around 0.8 indicates protein. The concept often appears in respirometer data analysis questions.
呼吸商为 1.0 通常意味着正在呼吸碳水化合物。RQ 值接近 0.7 表明脂类代谢，约为 0.8 则提示蛋白质。这一概念常出现在呼吸计数据分析题中。
In a practical scenario using a respirometer, soda lime absorbs CO₂, so the reduction in volume of air directly reflects oxygen uptake. Combining this with simultaneous experiments that measure CO₂ allows RQ determination.
在实际使用呼吸计的实验中，钠石灰会吸收 CO₂，因此空气体积的减少直接反映氧气消耗量。将此数据与同时测量 CO₂ 的实验相结合，即可计算出呼吸商。
6. Cardiac Output | 心输出量
Cardiac output is a key measure of heart performance, representing the volume of blood pumped by one ventricle per minute. It links heart rate and stroke volume through a simple multiplication.
心输出量是衡量心脏功能的关键指标，表示一个心室每分钟泵出的血液体积。它通过一个简单的乘法将心率和每搏输出量关联起来。
Cardiac Output (CO) = Heart Rate (HR) × Stroke Volume (SV)
Stroke volume is the amount of blood ejected by a ventricle in one contraction. A typical resting value might be 70 mL, and with a heart rate of 70 bpm, cardiac output is around 4.9 L/min.
每搏输出量是一个心室每次收缩射出的血量。典型的静息值约为 70 mL，配合 70 bpm 的心率，心输出量约为 4.9 L/min。
During exercise, both heart rate and stroke volume rise, dramatically increasing cardiac output. Understanding this equation is vital for topics covering the circulatory system and the effects of activity.
运动时，心率和每搏输出量均上升，使心输出量显著增加。理解这一方程对掌握循环系统和活动影响等主题至关重要。
7. Lincoln Index (Capture-Mark-Recapture) | 林肯指数（标记重捕法）
The Lincoln Index estimates the size of a mobile animal population by capturing, marking, releasing, and recapturing individuals. It assumes random mixing and no significant births, deaths, or migration.
林肯指数通过捕获、标记、释放并再次捕获个体来估算活动动物种群的大小。该方法假设个体随机混合且没有显著的出生、死亡或迁徙。
N = (M × C) ÷ R
Here, N = estimated population size, M = number of individuals captured and marked in the first sample, C = total number captured in the second sample, R = number of marked individuals recaptured in the second sample.
其中，N = 估计种群大小，M = 第一次捕获并标记的个体数，C = 第二次捕获的总个体数，R = 第二次捕获中带标记的个体数。
For example, if 40 woodlice are marked and released, then later 50 are captured of which 10 are marked, N = (40 × 50) ÷ 10 = 200. The reliability depends on the assumptions being met and a large enough sample size.
例如，先标记并释放 40 只潮虫，随后捕获 50 只，其中 10 只带标记，则 N = (40 × 50) ÷ 10 = 200。可靠性取决于假设是否满足以及样本量是否足够大。
8. The 10% Energy Transfer Rule | 10% 能量传递法则
In food chains and food webs, only about 10% of the energy stored in one trophic level is transferred to the next level. The remaining energy is lost through respiration, heat, egestion, and uneaten parts.
在食物链和食物网中，上一个营养级储存的能量只有约 10% 能够传递到下一个营养级。其余能量通过呼吸作用、散热、排遗和未被取食的部分而损失。
Energy available at next level ≈ 10% of energy from previous level
This principle explains why food chains rarely exceed four or five trophic levels. Pyramids of biomass and energy also become narrower at higher levels, reflecting the diminishing energy flow.
这一原理解释了食物链很少超过四到五个营养级的原因。生物量金字塔和能量金字塔在高级别也变得更窄，反映了逐渐减少的能量流动。
In exam diagrams, you might be asked to calculate the energy passed from producers to primary consumers. If producers contain 50 000 kJ, primary consumers typically receive around 5000 kJ.
在考试图表题中，你可能被要求计算从生产者传递到初级消费者的能量。如果生产者含有 50 000 kJ，初级消费者一般只能获得约 5000 kJ。
9. Fick’s Law of Diffusion | 菲克扩散定律
Fick’s Law describes the factors influencing the rate of diffusion across a membrane or surface. It is fundamental to understanding gas exchange in lungs, gills, and leaves.
菲克定律描述了影响跨膜或跨表面扩散速率的因素。它是理解肺、鳃和叶片中气体交换的基础。
Rate of diffusion ∝ (Surface area × Concentration difference) ÷ Thickness of membrane
To maximise diffusion, organisms have evolved large surface areas, short diffusion distances, and steep concentration gradients. Examples include the many alveoli in the lungs, the flat shape of leaves, and the rich blood supply maintaining gradients.
为了最大化扩散效率，生物体进化出巨大的表面积、较短的扩散距离以及陡峭的浓度梯度。例如，肺内众多的肺泡、扁平的叶片形态以及维持梯度的丰富血液供应。
In exam questions, Fick’s Law helps justify structural adaptations. A thin alveolar wall (one cell thick) reduces diffusion distance; dense capillary networks maintain concentration gradients by constantly carrying gases away.
在考试题中，菲克定律可以合理解释结构适应性。肺泡壁仅有一个细胞的厚度，缩短了扩散距离；密集的毛细血管网络通过不断运走气体来维持浓度梯度。
10. Temperature Coefficient (Q₁₀) | 温度系数
Q₁₀ measures how much the rate of a biological reaction changes with a 10 °C increase in temperature. It is widely used in enzyme-controlled reactions and respiration studies.
Q₁₀ 衡量生物反应速率随温度每升高 10 °C 而发生的变化倍数。该系数广泛用于酶控反应和呼吸作用研究。
Q₁₀ = Rate at (T + 10 °C) ÷ Rate at T °C
For many enzyme reactions, Q₁₀ is around 2, meaning the rate doubles with a 10 °C rise, up to an optimum temperature. Beyond the optimum, denaturation causes rates to plummet.
对于许多酶促反应，Q₁₀ 约等于 2，即温度每上升 10 °C，反应速率加倍（直到最适温度）。超过最适温度后，酶变性会导致速率急剧下降。
If a reaction rate is 5 arbitrary units at 20 °C and 12 units at 30 °C, Q₁₀ = 12 ÷ 5 = 2.4. Students must be careful to subtract the baseline when calculating absolute rates from graphs.
如果 20 °C 时反应速率为 5 个单位，30 °C 时为 12 个单位，则 Q₁₀ = 12 ÷ 5 = 2.4。学生应注意，从曲线图计算绝对速率时须扣除基线值。
11. Surface Area to Volume Ratio (SA:V) | 表面积与体积比
As an organism or cell increases in size, its volume grows faster than its surface area. The SA:V ratio dictates how effectively substances can be exchanged with the environment.
当生物体或细胞增大时，其体积的增长速度快于表面积。SA:V 比率决定了物质与环境交换的有效程度。
SA:V = Surface area ÷ Volume
A small cube with side length 1 cm has a surface area of 6 cm² and volume of 1 cm³, giving an SA:V ratio of 6:1. A larger 3 cm cube has SA = 54 cm² and volume = 27 cm³, so SA:V drops to 2:1.
一个边长为 1 cm 的小立方体，表面积为 6 cm²，体积为 1 cm³，SA:V 比率为 6:1。一个较大的 3 cm 立方体，SA = 54 cm²，体积 = 27 cm³，SA:V 比率降至 2:1。
The high SA:V of single-celled organisms enables direct diffusion; larger multicellular organisms need specialised exchange surfaces and transport systems to compensate for their low SA:V ratios.
单细胞生物的高 SA:V 比率使其能够直接通过扩散进行交换；较大的多细胞生物则需要特化的交换表面和运输系统来弥补过低的 SA:V 比率。
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