📚 Formulas and Principles Quick Reference for CIE Year 10 Biology | Year 10 CIE 生物：公式定理速查手册
Welcome to your essential quick reference guide for CIE Year 10 Biology. This handbook brings together all the key formulas, equations and mathematical principles you need to confidently tackle calculations and data analysis in your IGCSE course. From magnification to energy transfer, each entry is presented with clear worked examples and paired bilingual explanations to reinforce your understanding and exam technique.
欢迎使用这份 CIE Year 10 生物学核心公式定理速查手册。本手册汇集了 IGCSE 课程中所有关键计算公式、化学方程式和数学原理，帮助你自信应对计算题和数据分析题。每个公式都配有清晰的示例和中英双语解释，巩固理解并提升应试技巧。
1. Magnification Formula | 放大倍率公式
ification of a specimen tells you how much larger the image appears compared to the object' ze ÷ Actual size. All three quantities must have the same units, typically millimetres (mm) or micrometres (µm).
放大倍率表示图像相比物体实际尺寸放大了多少倍。计算公式为：放大倍数 = 图像尺寸 ÷ 实际尺寸。三个量必须使用相同的单位，通常为毫米 (mm) 或微米 (µm)。
Magnification = Image size ÷ Actual size
To use the formula successfully, always convert all measurements to the same unit before calculating. A helpful memory aid is the 'MIA' triangle, where M sits at the top, I (image) and A (actual) sit in the bottom corners. Cover the quantity you need to find to reveal the correct operation.
为正确使用该公式，务必先将所有测量值转换为相同单位。一个有效的记忆方法是 "MIA" 三角形，M 在顶部，I (图像) 和 A (实际) 在底部两角。遮住需要求解的量，即可得出正确的运算。
2. Calculating Actual Size from Magnification | 根据放大倍数计算实际尺寸
s: Actual size = Image size ÷ Magnification. This is especially useful for interpreting micrographs and biological diagrams with a scale bar.
当已知图像尺寸和放大倍数时，可以通过公式变形求出物体的实际尺寸。变形公式为：实际尺寸 = 图像尺寸 ÷ 放大倍数。这在解读带有比例尺的显微照片和生物图时特别有用。
Actual size = Image size ÷ Magnification
0 = 0.0125 mm. Convert this to micrometres by multiplying by 1000: 0.0125 mm × 1000 = 12.5 µm. Always express the final answer in the simplest appropriate unit.
例如，某细胞图像长度为 25 mm，放大倍数为 2000，则实际尺寸为 25 ÷ 2000 = 0.0125 mm。转换为微米需乘以 1000：0.0125 mm × 1000 = 12.5 µm。最终答案须用最简明的合适单位表示。
3. Photosynthesis Equation | 光合作用公式
Photosynthesis is the process by which green plants and some other organisms use light energy to synthesise glucose from carbon dioxide and water. The overall word equation summarises this energy conversion, while the balanced chemical equation shows the precise number of molecules involved.
光合作用是绿色植物及某些其他生物利用光能，将二氧化碳和水合成葡萄糖的过程。总文字方程概括了这一能量转换，平衡的化学方程式则精确表示了参与反应的分子数目。
Carbon dioxide + Water → Glucose + Oxygen
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Light energy is absorbed by chlorophyll in chloroplasts. The equation is often shown with 'light' and 'chlorophyll' written above the arrow. Remember that oxygen is released as a by-product, which is essential for aerobic respiration in most living organisms.
光能被叶绿体中的叶绿素吸收。方程常在箭头上方标注“光”和“叶绿素”。记住氧气作为副产物释放，这对大多数生物的有氧呼吸至关重要。
4. Aerobic Respiration Equation | 有氧呼吸公式
Aerobic respiration is the complete breakdown of glucose in the presence of oxygen to release energy for cellular activities. The word and symbol equations are the reverse of photosynthesis, reflecting the complementary relationship between the two processes.
有氧呼吸是葡萄糖在氧气参与下被完全分解、为细胞活动释放能量的过程。文字方程和符号方程与光合作用恰好相反，体现了两个过程的互补关系。
Glucose + Oxygen → Carbon dioxide + Water
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Energy is released in the form of ATP. The equation represents the summary reaction; the actual process involves many enzyme-controlled steps inside mitochondria. Students should be able to write both the word and balanced chemical equations accurately.
能量以 ATP 形式释放。该方程是总反应式；实际过程在线粒体内部经过许多酶控步骤。学生应能准确书写文字方程和平衡的化学方程式。
5. Anaerobic Respiration Equations | 厌氧呼吸公式
When oxygen is absent or in short supply, cells can respire anaerobically to produce a small amount of ATP. The products differ between organisms: yeast produces ethanol and carbon dioxide, while animal muscle cells produce lactic acid.
当缺氧或供氧不足时，细胞可进行厌氧呼吸以产生少量 ATP。不同生物的产物有所差异：酵母产生乙醇和二氧化碳，而动物肌肉细胞产生乳酸。
In yeast: Glucose → Ethanol + Carbon dioxide
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
In animal muscles: Glucose → Lactic acid
C₆H₁₂O₆ → 2C₃H₆O₃
Lactic acid accumulation causes muscle fatigue and cramping. Yeast's anaerobic respiration is exploited in baking and brewing industries. Note that anaerobic respiration yields far less energy per glucose molecule than aerobic respiration.
乳酸积累会导致肌肉疲劳和痉挛。酵母的厌氧呼吸被应用于烘焙和酿酒工业。注意厌氧呼吸每分子葡萄糖所产生的能量远少于有氧呼吸。
6. Enzyme Reaction Rate | 酶促反应速率
e taken for a specific event, such as the colour change when starch has been completely digested.
酶促反应速率衡量底物转化为产物的快慢。可用简单公式表示为：速率 = 1 ÷ 特定事件所需时间，例如淀粉完全消化时颜色变化的时间。
Rate of reaction = 1 ÷ Time (s⁻¹)
product amount (or substrate used) ÷ Time. Common investigations include the effect of temperature or pH on catalase or amylase activity.
在记录一段时间内可测量变化的实验中，也可用公式：速率 = 产物生成量(或底物消耗量) ÷ 时间。常见探究包括温度或 pH 对过氧化氢酶或淀粉酶活性的影响。
7. Cardiac Output Formula | 心输出量公式
Cardiac output is the volume of blood pumped by one ventricle of the heart per minute. It is an important indicator of cardiovascular fitness and is calculated by multiplying heart rate and stroke volume.
心输出量指心脏一个心室每分钟泵出的血液体积。它是衡量心血管健康的重要指标，由心率与每搏输出量相乘计算得出。
Cardiac output (dm³/min) = Heart rate (beats/min) × Stroke volume (dm³)
Stroke volume is the amount of blood ejected per beat. Ensure that units are consistent: if stroke volume is given in cm³, convert to dm³ by dividing by 1000, or convert the cardiac output accordingly. For example, a resting heart rate of 70 bpm and stroke volume of 0.07 dm³ gives a cardiac output of 4.9 dm³/min.
每搏输出量指每次心跳搏出的血量。应确保单位一致：若每搏输出量以 cm³ 为单位，需除以 1000 转化为 dm³，或相应调整心输出量单位。例如，静息心率 70 bpm，每搏输出量 0.07 dm³，则心输出量为 70 × 0.07 = 4.9 dm³/min。
8. Population Estimation Using Capture-Recapture | 标志重捕法估算种群数量
The Lincoln Index provides a method for estimating the population size of mobile organisms. It involves capturing a sample, marking them, releasing them, and then capturing a second sample. The formula assumes that marked and unmarked individuals mix randomly.
林肯指数提供了一种估算移动生物种群数量的方法。步骤包括捕捉样本、进行标记、释放，然后进行第二次捕捉。该公式假设标记和未标记个体随机混合。
Population size = (M × C) ÷ R
marked individuals recaptured in second sample. This method relies on several assumptions: no births, deaths, immigration or emigration occur between samples, marks are not lost, and marking does not affect survival.
其中 M = 第一次捕捉并标记的数量，C = 第二次捕捉的总数，R = 第二次捕捉中带有标记的个体数。该方法依赖若干假设：两次采样之间无出生、死亡、迁入或迁出，标记不脱落，且标记不影响生存。
9. Surface Area to Volume Ratio | 表面积与体积比
Surface area to volume ratio (SA:V) explains why cells and organisms are often small or have specialised exchange surfaces. As an object increases in size, its volume grows faster than its surface area, reducing the SA:V ratio. This limits diffusion efficiency for heat, gases and nutrients.
表面积与体积比 (SA:V) 解释了为什么细胞和生物体通常很小或具有特化的交换表面。随着物体变大，其体积增长快于表面积，导致 SA:V 比值下降，从而限制了热、气体和营养物质的扩散效率。
For a simple cube with side length s, the surface area is 6s² and volume is s³, giving the ratio:
对于边长为 s 的简单立方体，表面积为 6s²，体积为 s³，比值为：
SA:V = 6s² ÷ s³ = 6 ÷ s
er organisms have larger relative surface area for exchange. This principle underpins adaptations such as villi in the small intestine and alveoli in the lungs.
边长为 1 cm 的立方体 SA:V = 6:1；边长为 2 cm 的立方体 SA:V = 24 ÷ 8 = 3:1。生物体越小，其相对交换表面积越大。这一原理是小肠绒毛和肺泡等适应结构的基础。
10. Genetic Probability in Monohybrid Crosses | 单基因杂交中的遗传概率
Monohybrid inheritance involves a single characteristic controlled by one gene with two alleles. A Punnett square allows you to predict the genotype and phenotype ratios of offspring. Probability is expressed as a fraction, decimal or percentage.
单基因遗传指由一个基因的两个等位基因控制单一性状。借助庞纳特方格可预测子代的基因型和表现型比例。概率可用分数、小数或百分数表示。
plants (Tt × Tt), where T (tall) is dominant over t (dwarf). The resulting genotypes are:
以两株杂合高茎豌豆杂交 (Tt × Tt) 为例，T (高茎) 对 t (矮茎) 为显性。产生的基因型如下：
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
The expected phenotype ratio is 3 tall : 1 dwarf. Thus, the probability of an offspring being tall is 3/4 (75%), and the probability of being dwarf is 1/4 (25%). Chance has no memory; each fertilisation event is independent.
预期的表现型比例为 3 高 : 1 矮。因此，子代出现高茎的概率为 3/4 (75%)，出现矮茎的概率为 1/4 (25%)。概率没有记忆，每次受精都是独立事件。
11. Energy Transfer Efficiency in Food Chains | 食物链中的能量传递效率
Energy is lost at each trophic level through respiration, heat, uneaten parts and excretion. The efficiency of energy transfer between two levels is calculated as a percentage of energy passed on. This explains why food chains rarely exceed four or five trophic levels.
能量在每一营养级通过呼吸、散热、未被摄食部分和排泄而散失。两个营养级之间的能量传递效率以传递的能量百分比来计算。这解释了为什么食物链鲜少超过四至五个营养级。
Efficiency (%) = (Energy at next trophic level ÷ Energy at current trophic level) × 100% Published by TutorHao | Year 10 Biology Revision Series | aleveler.com
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