📚 Year 11 CAIE Biology: Quick Reference Handbook of Formulas & Key Principles | Year 11 CAIE 生物：公式定理速查手册
This article provides a concise yet comprehensive compilation of essential formulas, equations, and biological principles required for the CAIE Year 11 Biology syllabus. Mastering these quantitative and conceptual tools is vital for success in examinations and practical assessments.
本文为CAIE Year 11生物课程提供了一份精炼而全面的必考公式、方程式和生物学原理汇编。掌握这些量化和概念工具对考试及实践评估的成功至关重要。
1. Magnification Calculations | 放大倍率计算
The most fundamental formula in microscopy links image size (I), actual size (A), and magnification (M): M = I / A. Rearranging gives I = M × A and A = I / M. Always ensure I and A are in the exact same unit before dividing.
显微镜学中最基本的公式将图像尺寸（I）、实际尺寸（A）和放大倍率（M）联系起来：M = I / A。变形可得 I = M × A 以及 A = I / M。在相除之前，务必确保 I 和 A 使用完全相同的单位。
For example, if a cell image measures 25 mm on a micrograph and its actual length is 0.05 mm, the magnification is 25 ÷ 0.05 = 500 ×. If a drawing of a chloroplast is 40 mm long at a stated magnification of ×4000, its actual length is 40 / 4000 = 0.01 mm.
例如，某细胞在显微照片中的图像长度为25 mm，实际长度为0.05 mm，则放大倍率为 25 ÷ 0.05 = 500倍。若一个叶绿体的绘图长度为40 mm，标明的放大倍率为×4000，则其实际长度为 40 / 4000 = 0.01 mm。
Magnification = Image size / Actual size M = I ÷ A
2. Unit Conversions for Microscopy | 显微镜单位换算
Biologists routinely convert between millimetres (mm), micrometres (µm), and nanometres (nm). The key relationships are: 1 mm = 1000 µm; 1 µm = 1000 nm. To convert mm to µm, multiply by 1000; to convert µm to nm, multiply by 1000. Going the opposite direction requires division.
生物学家经常需要在毫米（mm）、微米（µm）和纳米（nm）之间进行转换。关键关系为：1 mm = 1000 µm；1 µm = 1000 nm。要将mm转换为µm，乘以1000；将µm转换为nm，同样乘以1000。反向转换则需要除以1000。
| From | To | Operation |
|---|---|---|
| mm | µm | ×1000 |
| µm | nm | ×1000 |
| µm | mm | ÷1000 |
| nm | µm | ÷1000 |
A practical example: an organelle measures 0.002 mm. In µm this is 0.002 × 1000 = 2 µm, and in nm it is 2 × 1000 = 2000 nm. Consistent unit use prevents magnification errors.
实际示例：某细胞器测得长度为0.002 mm。换算成µm为 0.002 × 1000 = 2 µm，再换算成nm则为 2 × 1000 = 2000 nm。统一的单位使用能防止放大倍率计算出错。
3. Estimating Population Size | 种群数量估计
The capture-mark-recapture method (Lincoln Index) estimates animal population size where direct counting is impossible. The formula is: N = (M × C) / R, where N = population estimate, M = number captured and marked in first sample, C = total number captured in second sample, R = number of marked individuals recaptured in second sample.
捕捉-标记-再捕捉法（林肯指数）用于估算无法直接计数的动物种群大小。公式为：N = (M × C) / R，其中 N = 种群估计值，M = 首次样本中捕获并标记的个体数，C = 第二次样本中捕获的总数，R = 第二次样本中带标记的个体数。
Assumptions include: marked organisms mix randomly, no migration, no births or deaths, and marks are not lost. For example, 40 snails are marked and released; next day 50 snails are caught, 10 of which are marked. N = (40 × 50) / 10 = 200 snails.
假设包括：标记个体随机混合、无迁移、无出生或死亡、标记不会脱落。例如，标记并释放40只蜗牛；次日捕获50只，其中10只带标记。N = (40 × 50) / 10 = 200只蜗牛。
Estimated population = (Marked in first sample × Total in second sample) / Marked recaptured
4. Photosynthesis Equation | 光合作用方程式
The overall balanced symbol equation for photosynthesis is essential. Carbon dioxide and water, in the presence of light energy and chlorophyll, produce glucose and oxygen. The equation must be memorised with correct coefficients.
光合作用的总平衡化学方程式是必考内容。二氧化碳和水在光能及叶绿素存在下，生成葡萄糖和氧气。必须记住正确的系数。
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
The rate of photosynthesis can be measured by counting oxygen bubbles produced by an aquatic plant per minute, or by measuring the volume of oxygen collected over time. Rate = number of bubbles / time taken, or rate = volume of O₂ / time.
光合速率可通过计数水生植物每分钟产生的氧气气泡数，或测量一段时间内收集的氧气体积来测定。速率 = 气泡数量 / 所用时间，或速率 = O₂ 体积 / 时间。
5. Respiration Equations | 呼吸作用方程式
Aerobic respiration releases a large amount of energy by fully oxidising glucose. The balanced equation is similar to the reverse of photosynthesis. It occurs in the mitochondria of cells when oxygen is present.
有氧呼吸通过完全氧化葡萄糖释放大量能量。其平衡方程式类似于光合作用的逆反应。该过程在细胞内的线粒体中进行，需要氧气存在。
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (ATP)
Anaerobic respiration in yeast produces ethanol and carbon dioxide, while in animal muscles it produces lactic acid. These processes release much less energy and occur without oxygen.
酵母的无氧呼吸产生乙醇和二氧化碳，而动物肌肉的无氧呼吸则产生乳酸。这些过程释放的能量要少得多且不需要氧气。
Yeast: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ + energy
Muscle: C₆H₁₂O₆ → 2C₃H₆O₃ + energy
6. Respiratory Quotient (RQ) | 呼吸商
The respiratory quotient reveals the substrate being respired. It is calculated by dividing the volume of carbon dioxide produced by the volume of oxygen consumed over the same time period. The formula is RQ = CO₂ / O₂.
呼吸商可揭示被呼吸的底物类型。其计算方法为：相同时间内产生的二氧化碳体积除以消耗的氧气体积。公式为 RQ = CO₂ / O₂。
For carbohydrates RQ = 1.0, for lipids RQ ≈ 0.7, and for proteins RQ ≈ 0.8–0.9. Germinating seeds respiring fats show RQ less than 1. Respirometers can measure the gas volumes.
碳水化合物的呼吸商为1.0，脂质约为0.7，蛋白质约为0.8–0.9。萌发种子若以脂肪为呼吸底物，其RQ小于1。呼吸计可用于测量气体体积。
RQ = Volume of CO₂ produced / Volume of O₂ consumed
7. Calculating Energy in Food | 食物能量计算
The energy released by burning food can be calculated using a simple calorimeter. The heat from the burning food raises the temperature of a known mass of water. Energy (in joules) = mass of water (g) × temperature rise (°C) × 4.2 J/g°C.
燃烧食物释放的能量可用简易量热计计算。食物燃烧产生的热量使已知质量的水升温。能量（焦耳）= 水的质量（克） × 温度升高值（°C） × 4.2 J/g°C。
For example, if 20 g of water increases from 22 °C to 45 °C when a 0.5 g piece of food is burned, the energy released is 20 × (45 − 22) × 4.2 = 1932 J. Energy per gram = 1932 / 0.5 = 3864 J/g. The specific heat capacity 4.2 J/g°C is standard for water.
例如，燃烧0.5克食物使20克水的温度从22°C升至45°C，则释放的能量 = 20 × (45 − 22) × 4.2 = 1932 J。每克能量 = 1932 / 0.5 = 3864 J/g。水的比热容标准值为4.2 J/g°C。
Energy (J) = mass of water (g) × ΔT (°C) × 4.2
8. Cardiac Output | 心输出量
Cardiac output (CO) is the volume of blood pumped by the heart per minute. It is calculated as heart rate (HR) multiplied by stroke volume (SV). CO = HR × SV. Heart rate is beats per minute, stroke volume is mL per beat, yielding CO in mL/min, often converted to L/min.
心输出量（CO）是指心脏每分钟泵出的血液体积。计算公式为心率（HR）乘以每搏输出量（SV）。CO = HR × SV。心率单位为次/分钟，每搏输出量单位为毫升/次，所得CO单位为 mL/min，常转换为 L/min。
Typical resting values for an adult: HR = 70 bpm, SV = 70 mL/beat, CO = 4900 mL/min or 4.9 L/min. During exercise, both HR and SV can rise, dramatically increasing CO to deliver more oxygen to muscles.
成年人典型的静息值：HR = 70 bpm，SV = 70 mL/beat，CO = 4900 mL/min 即 4.9 L/min。运动时，心率和每搏输出量均可升高，显著增加心输出量为肌肉输送更多氧气。
Cardiac output (mL/min) = Heart rate (bpm) × Stroke volume (mL/beat)
9. Body Mass Index (BMI) | 身体质量指数
BMI assesses whether a person’s mass is healthy for their height. The formula is: BMI = mass (kg) / [height (m)]². It provides a numerical indicator used to classify underweight, normal, overweight, and obese categories.
BMI用于评估一个人的体重相对其身高是否健康。公式为：BMI = 体重（kg）/ [身高（m）]²。该数值用于划分体重过轻、正常、超重和肥胖等类别。
A person weighing 65 kg with a height of 1.75 m has a BMI = 65 / (1.75 × 1.75) = 65 / 3.0625 ≈ 21.2 kg/m², which falls into the normal range.
一位体重65 kg、身高1.75 m的人，BMI = 65 / (1.75 × 1.75) = 65 / 3.0625 ≈ 21.2 kg/m²，属于正常范围。
| BMI Range (kg/m²) | Classification |
|---|---|
| < 18.5 | Underweight |
| 18.5 – 24.9 | Normal weight |
| 25.0 – 29.9 | Overweight |
| ≥ 30.0 | Obese |
10. Genetic Crosses and Ratios | 遗传杂交与比例
Monohybrid inheritance follows Mendel’s law of segregation. A genetic cross between two heterozygous parents (e.g., Bb × Bb) produces offspring with a genotype ratio of 1 BB : 2 Bb : 1 bb and a phenotype ratio of 3 dominant : 1 recessive, assuming complete dominance.
单基因遗传遵循孟德尔分离定律。两个杂合子亲本（例如 Bb × Bb）之间的杂交会产生基因型比为 1 BB : 2 Bb : 1 bb 的后代，若为完全显性，则表型比为 3显性 : 1隐性。
A test cross (Bb × bb) yields a 1 Bb : 1 bb ratio, producing a 1:1 phenotype ratio. Punnett squares systematically predict these outcomes. Understanding these ratios is essential for solving inheritance problems.
测交（Bb × bb）可产生 1 Bb : 1 bb 的比例，表型比为 1:1。庞纳特方格可系统性地预测这些结果。理解这些比例对于解决遗传学问题至关重要。
Sex determination in humans involves the XX (female) and XY (male) system. A cross between a mother (XX) and father (XY) produces a 1 XX : 1 XY ratio, giving a 50% probability of a male or female child.
人类的性别决定涉及XX（女性）和XY（男性）系统。母亲（XX）与父亲（XY）的杂交产生 1 XX : 1 XY 的比例，即生男生女的概率各为50%。
11. Principles of Natural Selection | 自然选择原理
Natural selection is a key mechanism of evolution, not a mathematical formula, but its underlying principles are fundamental theorems in biology. Variation exists within a population; organisms produce more offspring than can survive; there is competition for resources; individuals with traits better adapted to the environment are more likely to survive and reproduce; these advantageous traits are passed to the next generation.
自然选择是进化的关键机制，并非数学公式，但其基本原理是生物学中的根本性定理。种群内存在变异；生物产生的后代数量超过环境承载力；个体间存在资源竞争；具有更适应环境性状的个体更可能存活并繁殖；这些有利性状会传递给下一代。
Over many generations, the frequency of advantageous alleles increases, leading to adaptations and possibly speciation. Classic examples include antibiotic resistance in bacteria and Darwin’s finches’ beak shapes.
经过多代以后，有利等位基因的频率增加，导致适应特征的形成，并可能导致物种形成。经典例子包括细菌的抗生素耐药性和达尔文雀的喙形变化。
12. Enzyme Activity and Rate Calculations | 酶活性与速率计算
The rate of an enzyme-catalysed reaction can be determined by measuring the speed of product appearance or substrate disappearance. A common school laboratory approach uses the time taken for a starch-iodine colour to disappear. Rate is then calculated as 1 / time (s⁻¹).
酶催化反应的速率可通过测量产物出现或底物消失的快慢来确定。学校实验室中常见的做法是记录淀粉-碘混合物的蓝色消失所需的时间。然后计算速率：速率 = 1 / 时间（秒⁻¹）。
For example, if amylase breaks down starch completely in 120 seconds at pH 7, the rate is 1/120 = 0.0083 s⁻¹. More generally, rate = amount of product formed / time taken. The effect of temperature and pH on enzyme activity follows a characteristic bell-shaped curve with an optimum.
例如，若在pH 7下，淀粉酶将淀粉完全分解耗时120秒，则速率为 1/120 = 0.0083 s⁻¹。更一般地，速率 = 生成的产物量 / 所用时间。温度和pH对酶活性的影响遵循一条具有最适点的钟形曲线。
Reaction rate = 1 / time (for a defined endpoint) or Rate = Δ[Product] / time
Students should be able to plot results and identify the initial rate and the leveling off of the curve as substrate is used up or enzyme denatures.
学生应能绘制结果曲线，并识别初始速率以及因底物耗尽或酶变性导致的曲线平台期。
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