📚 Year 11 CAIE Biology: Formula & Theorem Quick-Reference Handbook | Year 11 CAIE 生物：公式定理速查手册
This quick-reference handbook compiles the essential formulas, equations, and key principles required for the CAIE IGCSE Biology syllabus (0610/0970). Mastering these calculations and concepts will help you solve quantitative problems and understand fundamental biological laws effectively. Keep this guide handy for revision and exam preparation.
本速查手册汇编了 CAIE IGCSE 生物学（0610/0970）大纲所要求的基本公式、方程式和关键原理。掌握这些计算和概念，将有效帮助你解决定量问题并理解基础生物学定律。在复习和备考时请随身参考本指南。
1. Magnification & Unit Conversions | 放大倍数与单位换算
g specimen size is Magnification (M) = Image size (I) divided by actual size (A). Always ensure both measurements are in the same unit before applying the formula.
显微镜的总放大倍数是目镜倍数乘以物镜倍数。计算标本尺寸的核心公式为：放大倍数（M）= 图像尺寸（I）÷ 实际尺寸（A）。在代入公式前请务必统一单位。
M = I ÷ A
Common unit conversions you must memorise: 1 cm = 10 mm, 1 mm = 1000 µm, 1 µ be calibrated using a stage micrometer for accurate measurements under high magnification.
必须牢记的常用单位换算：1 厘米 = 10 毫米，1 毫米 = 1000 微米，1 微米 = 1000 纳米。高倍镜下使用目镜测微尺时需用镜台测微尺进行校准，方可获得准确数据。
| Length conversion | ÷ 10 | × 1000 | × 1000 |
| 1 cm | = 10 mm | = 10 000 µm | = 10 000 000 nm |
A = I ÷ M and always state the final answer in the required unit, typically micrometres (µm) for cells and sub-cellular structures.
你可能会被要求根据比例尺计算实际尺寸，或估算细胞直径。可将公式变形为 A = I ÷ M，最终答案务必以题目要求的单位呈现，通常细胞和亚细胞结构以微米（µm）为单位。
2. Surface Area to Volume Ratio | 表面积与体积之比
The surface area : volume ratio (SA:V) explains why cells are microscopic and why organisms require specialised exchange surfaces. As an object increases in size, its volume grows much faster than its surface area, so the SA:V ratio decreases. This limits the rate at which substances can enter or leave a cell by diffusion alone.
表面积与体积之比（SA:V）解释了细胞为何微小以及生物体为何需要特化的交换表面。随着物体增大，体积的增长远快于表面积，因此 SA:V 比值下降，从而限制了仅靠扩散进出细胞的物质交换速率。
SA:V = Surface Area ÷ Volume
6 / L. Notice that a small L gives a large ratio, while a large L gives a small ratio. The same principle applies to spheres and cylinders, making it a universal biological constraint.
对于边长为 L 的立方体：表面积 = 6L²，体积 = L³，因此 SA:V = 6L² / L³ = 6 / L。可见 L 越小，比值越大；L 越大，比值越小。该原理同样适用于球体和圆柱体，是普适的生物学限制因素。
Adaptations such as flattened shapes (e.g., leaves, alveolar walls), microvilli on intestinal epithelial cells, and highly folded membranes all serve to increase the surface area available for exchange without a large increase in volume.
扁平结构（如叶片、肺泡壁）、小肠上皮细胞的微绒毛以及高度折叠的膜，都可在不大幅增加体积的前提下，增大可用于物质交换的表面积。
3. Rate Calculations in Biology | 生物学中的速率计算
ormed (or reactant used) ÷ Time taken. You will encounter rate calculations when investigating enzyme activity, photosynthesis, respiration, and transpiration.
速率衡量生物过程进行的快慢。通用公式为：速率 = 生成产物的量（或消耗反应物的量）÷ 所用时间。在研究酶活性、光合作用、呼吸作用以及蒸腾作用时都会遇到速率计算。
Rate = Change in quantity ÷ Time
nts measuring the appearance of a product (e.g., starch breakdown), rate = change in absorbance or mass ÷ time. In all cases, include appropriate units such as cm³/min, arbitrary units/s, or g/h.
在使用水草进行的光合作用实验中，速率通常表示为每分钟产生的氧气气泡数。在测定产物（如淀粉分解）出现的酶实验中，速率 = 吸光度或质量变化 ÷ 时间。所有情况下都要标明适当单位，例如 cm³/min、相对单位/s 或 g/h。
strate depletion. Draw a tangent to the curve at time zero and use the gradient = Δy / Δx.
你可能还需要从图线的初始线性部分计算速率，此时反应最快且不受底物耗尽限制。在时间零点处绘制曲线的切线，使用斜率 = Δy / Δx 即可求得初始速率。
4. The Q₁₀ Temperature Coefficient | Q₁₀ 温度系数
The Q₁₀ coefficient describes how the rate of an enzyme-controlled reaction changes when the temperature is increased by 10 °C. For most biological reactions under physiological temperatures, Q₁₀ is approximately 2, meaning the rate doubles for each 10 °C rise.
Q₁₀ 系数描述温度每升高 10 °C 时酶控反应速率的变化情况。在生理温度范围内，大多数生物反应的 Q₁₀ 约为 2，即温度每升高 10 °C，反应速率加倍。
Q₁₀ = Rate at (T + 10) °C ÷ Rate at T °C
cantly above 2 can indicate that the enzyme is approaching its optimum temperature; beyond the optimum, enzymes denature and the rate drops rapidly, giving a Q₁₀ far below 1.
例如，某酶促反应在 20 °C 时速率为 1.2 相对单位，在 30 °C 时速率为 2.5 相对单位，则 Q₁₀ = 2.5 / 1.2 ≈ 2.1。若数值显著大于 2，可能表明酶正接近最适温度；超过最适温度后酶会变性，速率急剧下降，此时 Q₁₀ 远低于 1。
Students often confuse Q₁₀ with a simple rate ratio. Remember it only applies for a 10 °C interval. If the interval is different, you must first calculate the rate per 10 °C rise or adjust the formula accordingly. This concept also explains why fever is dangerous and why reptiles become sluggish when cold.
学生常将 Q₁₀ 与简单的速率比混淆。请记住，它的前提是温度变化恰好为 10 °C。若温差不同，需先折算为每 10 °C 的速率变化，或相应调整公式。这一概念也解释了发烧为何危险以及爬行动物遇冷时行动迟缓的原因。
5. Energy Content of Food | 食物中的能量含量
The energy stored in food can be estimated by burning a known mass of the food sample and measuring the temperature rise of a fixed volume of water. This method assumes that all the heat released by combustion is absorbed by the water, though in reality there are significant heat losses.
通过燃烧已知质量的食品样品并测量一定体积水的温升，可以估算食物中所储存的能量。该方法假设燃烧释放的热量全部被水吸收，但实际上存在大量热损失。
Energy (J) = m × c × ΔT
Where m is the mass of water in grams, c is the specific heat capacity of water (4.2 J/g°C), and ΔT is the temperature rise in °C. The result is often expressed as energy per gram of food (J/g) after dividing by the mass of the food sample burned.
式中 m 指水的质量（g），c 为水的比热容（4.2 J/g°C），ΔT 为温升（°C）。所得结果常除以燃烧食品样品的质量，表示为每克食物的能量含量（J/g）。
red to water = 20 × 4.2 × 15 = 1260 J. Energy per gram = 1260 J/g (or 1.26 kJ/g). Remember to list sources of error such as incomplete combustion, heat loss to the air, and evaporation of water.
在典型实验中，你可能会在盛有 20 g 水的沸腾管下燃烧一片 1 g 的薯片，观察到温升为 15 °C。传递给水的能量 = 20 × 4.2 × 15 = 1260 J。每克能量 = 1260 J/g（即 1.26 kJ/g）。记得列举误差来源，如燃烧不充分、热量向空气散失以及水的蒸发。
6. Mendelian Genetics: Laws of Inheritance | 孟德尔遗传定律
Gregor Mendel's experiments with pea plants established two fundamental laws. The Law of Segregation states that each organism possesses two alleles for a trait, which separate during gamete formation, so each gamete carries only one allele. The Law of Independent Assortment states that alleles for different genes are distributed into gametes independently of one another.
孟德尔通过豌豆实验建立了两个基本定律。分离定律指出，每个生物体的某一性状由两个等位基因控制，它们在配子形成时彼此分离，使得每个配子只携带一个等位基因。自由组合定律指出，不同基因的等位基因在形成配子时彼此独立地分配。
Monohybrid ratio (heterozygous × heterozygous): 3 : 1
pic ratio for a completely dominant trait is 3 dominant : 1 recessive. For a dihybrid cross (e.g., RrYy × RrYy) where the two genes are unlinked, the expected phenotype ratio is 9 : 3 : 3 : 1.
使用庞纳特方格可以预测后代的基因型和表现型。当双亲均为杂合子（如 Bb × Bb）时，完全显性性状的预期表现型比例为 3 显性 : 1 隐性。对于双基因杂交（如 RrYy × RrYy），若两基因不连锁，预期表现型比例为 9 : 3 : 3 : 1。
chance of a child being both male and affected by a recessive X-linked trait is (½) × (¼) = ⅛. Understand co-dominance and sex linkage as variations on Mendelian patterns.
利用乘法法则可求出特定组合的出现概率：将各独立事件的概率相乘。例如，一个孩子同时为男性且患有 X 连锁隐性遗传病的概率为 (½) × (¼) = ⅛。同时要理解共显性和伴性遗传作为孟德尔模式的变体。
7. Cardiac Output | 心输出量
Cardiac output is the volume of blood pumped by each ventricle per minute. It depends on two factors: the heart rate (number of beats per minute) and the stroke volume (volume of blood pumped out of the left ventricle per beat). This relationship is a key quantitative concept in the transport chapter.
心输出量是每个心室每分钟泵出的血液体积。它取决于两个因素：心率（每分钟搏动次数）和每搏输出量（左心室每次收缩泵出的血量）。该关系是运输章节中重要的定量概念。
Cardiac output = Stroke volume × Heart rate
ring vigorous exercise, heart rate may rise to 180 bpm and stroke volume to 120 mL, increasing cardiac output to over 20 L/min.
成年人静息时的典型数值为：心率约 70 bpm，每搏输出量约 70 mL，心输出量约为 70 × 70 = 4900 mL/min（4.9 L/min）。剧烈运动时，心率可升至 180 bpm，每搏输出量增至 120 mL，心输出量可达 20 L/min 以上。
You may be asked to calculate one variable given the other two, or to interpret heart rate graphs before, during, and after exercise. Remember that stroke volume also increases due to greater venous return stretching the heart muscle (Starling's law), but this mechanism is not always required for a simple calculation.
题目可能给出其中两个变量让你求第三个，或要求解读运动前、中、后的心率变化图。注意，回心血量增多会拉伸心肌，使每搏输出量也增大（斯塔林定律），但简单计算时不一定要求涉及该机制。
8. Population Growth & Sigmoid Curve | 种群增长与S型曲线
Populations in a new environment often show exponential growth at first, followed by a slowing growth rate and eventually a stable plateau around the carrying capacity. This pattern is described by a sigmoid (S-shaped) curve and is the basis of ecological balance.
新环境中的种群最初常呈指数增长，随后增长率减缓，最终在环境容纳量附近稳定下来，形成平台期。这一模式可用 S 型曲线描述，是生态平衡的基础。
Population growth rate = (Birth rate + Immigration) – (Death rate + Emigration)
ces are abundant and the birth rate greatly exceeds the death rate; (3) Deceleration phase, where limiting factors such as food shortage, accumulation of waste, and disease slow the growth; (4) Stationary phase, where birth rate ≈ death rate and the population oscillates around the carrying capacity.
S 型曲线的四个时期为：(1) 延滞期，个体适应环境，繁殖缓慢；(2) 指数期（对数期），资源充足，出生率远大于死亡率；(3) 减速期，食物短缺、废物积累、疾病等限制因素使增长减速；(4) 稳定期，出生率 ≈ 死亡率，种群数量在容纳量上下波动。
In CAIE exam questions, you might be presented with a graph and asked to identify each phase or to suggest reasons for the shape. The sigmoid model applies to yeast cultures in a flask, bacteria on an agar plate, or rabbits in a field. Human population growth, however, has not yet shown a clear stationary phase globally.
CAIE 考题可能会给出一张图表，让你辨认各阶段或解释曲线形状的原因。S 型模型适用于烧瓶中的酵母菌群、琼脂平板上的细菌或田野中兔子的种群增长。不过，全球人口增长迄今尚未出现明显的稳定期。
9. Energy Transfer in Food Chains | 食物链中的能量传递
Only a small percentage of the energy in one trophic level is transferred to the next. Most energy is lost through respiration, movement, heat, and uneaten parts (e.g., bones, roots). This principle limits the length of food chains to usually no more than 4–5 trophic levels.
营养级中只有少量能量传递到下一级。大部分能量通过呼吸作用、运动、散热以及未被取食的部分（如骨骼、根系）而消耗。该原理将食物链的长度通常限制在 4–5 个营养级以内。
transfer = (Energy in new biomass at next level ÷ Energy available at current level) × 100%
On average, only about 10% of the energy is passed on, although the actual value can vary widely (5–20%). You may be given a pyramid of energy or a table of kJ per square metre per year and asked to calculate the transfer efficiency between specific trophic levels. Always show the full working and express the answer as a percentage to one decimal place.
平均而言，只有约 10% 的能量被传递下去，实际值可在 5%–20% 之间波动。题目可能给出能量金字塔或每平方米每年千焦的表格，要求计算特定营养级之间的传递效率。务必展示完整计算过程，并以百分比表示，保留一位小数。
Understand that energy transfer is not the same as biomass transfer; some biomass is dead organic matter that becomes food for decomposers. The 10% rule explains why large carnivores are rare and why vegetarian diets require fewer resources than meat-based diets at a societal level.
需理解能量传递与生物量传递不同；部分生物量以死有机质形式成为分解者的食物。10% 法则解释了为何大型食肉动物数量稀少，以及在社会层面素食比肉食所需资源更少的原因。
10. Transpiration & Potometer Calculations | 蒸腾作用与蒸腾计计算
uptake = Distance moved by air bubble × Cross-sectional area of capillary tube ÷ Time. Often you simply record distance the bubble travels in a fixed time.
蒸腾计测量带叶枝条吸水的速率，可间接反映蒸腾速率。基本公式为：吸水速率 = 气泡移动距离 × 毛细管横截面积 ÷ 时间。通常只需记录气泡在固定时间内移动的距离。
Transpiration rate ≈ Volume of water taken up ÷ Time
Environmental factors that increase transpiration rate include higher light intensity (stomata open), higher temperature (increases rate of evaporation), higher wind speed (removes humid air), and lower humidity (steeper diffusion gradient). You may be asked to plot a graph of bubble movement under these conditions and calculate the rate from the gradient of the linear portion.
增加蒸腾速率的环境因素包括：较强光照（气孔张开）、较高温度（加速蒸发）、较高风速（带走湿气）以及较低湿度（增大扩散梯度）。你可能需要根据这些条件绘制气泡移动图，并从线性部分的斜率计算速率。
s and to maintain turgor. When interpreting graphs of weight loss in a leafy shoot over time, the rate of transpiration = mass lost ÷ duration, assuming any mass change is due to water evaporation.
计算时需注意，蒸腾计测量的是吸水速率而非失水速率；少量水分用于光合作用和维持膨压。在解读带叶枝条随时间失重图时，假设质量变化均由水分蒸发引起，则蒸腾速率 = 失去的质量 ÷ 时长。
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