📚 Year 11 Cambridge IGCSE Biology: Formula & Theorem Quick Reference | 剑桥11年级IGCSE生物公式定理速查手册
This quick-reference handbook brings together all the essential formulas, equations, and key principles that frequently appear in Year 11 Cambridge IGCSE Biology (0610/0970) quantitative questions. Being confident with these tools will save you time in the exam and reduce simple calculation errors.
本速查手册汇集了11年级剑桥IGCSE生物(0610/0970)定量题中频繁出现的重要公式、方程和核心原理。熟练掌握这些工具，可以帮助你在考试中节省时间，并减少简单的计算失误。
1. Magnification Formula | 放大倍率公式
The magnification of a drawing or micrograph tells you how many times larger the image is compared to the real specimen. The core relationship is: Magnification = Image size ÷ Actual size.
绘图或显微照片的放大倍率表示图像比真实标本大多少倍。核心关系是：放大倍率 = 图像尺寸 ÷ 实际尺寸。
M = I ÷ A
To find the actual size of a cell or organelle when you know the image size and the magnification, rearrange the formula: Actual size = Image size ÷ Magnification.
如果已知图像尺寸和放大倍率，要求细胞或细胞器的实际尺寸，可将公式变形：实际尺寸 = 图像尺寸 ÷ 放大倍率。
All lengths must be converted to the same unit before you do the division. In CIE IGCSE Biology, this almost always means working in millimetres (mm) or micrometres (µm).
在相除之前，所有长度必须换算成相同单位。剑桥IGCSE生物几乎总是要求使用毫米(mm)或微米(µm)进行计算。
2. Unit Conversions for Microscopy | 显微镜单位换算
Clear unit conversions are essential when measuring cells and organelles under the microscope. Memorise these three fundamental relationships.
在显微镜下测量细胞和细胞器时，清晰的单位换算必不可少。牢记以下三个基本关系。
| 1 centimetre (cm) = 10 millimetres (mm) | 1 厘米 (cm) = 10 毫米 (mm) |
| 1 millimetre (mm) = 1000 micrometres (µm) | 1 毫米 (mm) = 1000 微米 (µm) |
| 1 micrometre (µm) = 1000 nanometres (nm) | 1 微米 (µm) = 1000 纳米 (nm) |
When converting from a larger unit to a smaller unit, multiply by the conversion factor. When changing a smaller unit into a larger one, divide rather than multiply.
从大单位转换为小单位时，要乘以换算系数。从小单位转换为大单位时，则用除法而不是乘法。
3. Rate of Reaction in Enzyme Studies | 酶促反应速率
Many IGCSE practicals measure how long it takes for a change to be completed, for example the disappearance of a starch-iodine colour or the time needed for a certain volume of gas to be collected. In these cases, rate is taken as the reciprocal of time.
许多IGCSE实验会测量某一变化完成所需的时间，例如淀粉与碘液颜色的消失，或收集一定体积气体所需的时间。在这些情况下，速率常用时间的倒数来表示。
Rate ∝ 1 ÷ time (s)
If the rate is calculated from a graph showing volume of oxygen produced, the rate at a given point can be found from the gradient: Rate = Change in volume ÷ Change in time.
若根据显示氧气产量的图表来计算速率，某一点的速率可由斜率求得：速率 = 体积变化量 ÷ 时间变化量。
Always state the units of rate clearly, such as cm³/s or s⁻¹.
一定要清晰注明速率的单位，例如 cm³/s 或 s⁻¹。
4. Percentage Change | 百分比变化
In osmosis experiments with potato cylinders or other plant tissues, you frequently need to calculate the percentage change in mass or length. The formula compares the difference to the initial reading.
在使用土豆条或其他植物组织的渗透实验中，经常需要计算质量或长度的百分比变化。该公式将变化量与初始读数进行比较。
Percentage change = (Final value − Initial value) ÷ Initial value × 100%
A negative percentage change indicates a net loss of water by osmosis, while a positive value indicates a net gain. Reporting percentage change rather than absolute change allows a fair comparison between samples of different starting masses.
负百分比变化表示渗透过程净失水，正百分比变化则表示净吸水。报告百分比变化而非绝对变化，可以对不同起始质量的样本进行公平对比。
5. Cardiac Output | 心输出量
Cardiac output describes the volume of blood pumped by the heart per minute. It links two measurable variables: how often the heart beats and how much blood leaves the left ventricle with each beat.
心输出量指心脏每分钟泵出的血液体积。它将两个可测量的变量联系起来：心跳的频率以及每次搏动从左心室射出的血量。
Cardiac output (cm³/min) = Heart rate (beats/min) × Stroke volume (cm³/beat)
Stroke volume is the volume of blood ejected from the left ventricle during one contraction. During exercise, both heart rate and stroke volume increase, leading to a much higher cardiac output to deliver more oxygen to muscles.
每搏输出量是指左心室在一次收缩中射出的血液体积。运动时，心率和每搏输出量都会增加，从而大幅提高心输出量，向肌肉输送更多氧气。
6. Energy Content of Food | 食物能量含量
The energy released when food is burned can be estimated by measuring the temperature rise of a known volume of water. The calculation uses the specific heat capacity of water, approximately 4.2 J/(g °C).
食物燃烧时释放的能量，可以通过测量已知体积水的温度升高来估算。计算中使用水的比热容，约为 4.2 J/(g °C)。
Energy per gram (J/g) = (Mass of water (g) × 4.2 × Temperature rise (°C)) ÷ Mass of food (g)
This method does not capture all energy released because some heat is lost to the surroundings. Nevertheless, the equation allows you to compare the energy density of different foods, such as a crisp versus a piece of apple.
此方法无法捕捉所有释放的能量，因为部分热量散失到周围环境中。但该方程仍然可以用来比较不同食物的能量密度，例如薯片与一块苹果。
Remember that the volume of water in cm³ is numerically equal to its mass in grams, so 20 cm³ of water has a mass of 20 g.
请记住，水的体积以 cm³ 表示时，数值等于其以克为单位的质量，因此 20 cm³ 水的质量为 20 g。
7. Genetic Probability Rules | 遗传概率法则
When predicting the outcomes of monohybrid crosses, two basic rules of probability are repeatedly used: the product (multiplication) rule and the addition rule.
在预测单基因杂交的结果时，会反复用到两条基本的概率法则：乘法法则和加法法则。
Product rule: The probability that two independent events both occur is the product of their individual probabilities. For example, the chance of inheriting a recessive allele from the mother AND a recessive allele from the father is ½ × ½ = ¼.
乘法法则：两个独立事件同时发生的概率等于各自概率的乘积。例如，同时从母亲和父亲那里遗传隐性等位基因的概率为 ½ × ½ = ¼。
P(A and B) = P(A) × P(B)
Addition rule: The probability that either of two mutually exclusive events occurs is the sum of their individual probabilities. In a Punnett square, if there are two different genotypes that produce the dominant phenotype, you add their probabilities.
加法法则：两个互斥事件中任一事件发生的概率等于各自概率之和。在庞尼特方格中，如果有两种不同的基因型都能产生显性表型，则需要将它们的概率相加。
P(A or B) = P(A) + P(B)
8. Surface Area to Volume Ratio | 表面积与体积比
The surface area to volume ratio (SA:V) explains why cells are so small and why organisms need specialised transport systems. For a cube-shaped cell of side length L, the surface area is 6L² and the volume is L³, giving the ratio.
表面积与体积比 (SA:V) 解释了为什么细胞如此微小，以及生物体为何需要特化的运输系统。对于一个边长为 L 的立方体细胞，表面积为 6L²，体积为 L³，由此得出比值。
SA : V = 6L² ÷ L³ = 6 / L
As the side length L increases, the numerical value of the ratio decreases. A large cube has a smaller SA:V than a small one, which means it has proportionally less surface area to absorb oxygen, nutrients, or remove wastes. This principle underpins the need for circulatory systems and why organisms cannot simply grow larger without special adaptations.
随着边长 L 增大，比值的数值会降低。大立方体的 SA:V 小于小立方体，这意味着它用于吸收氧气、营养或排除废物的表面积相对较少。这一原理是循环系统存在的基础，也解释了为什么生物体若无特殊适应，不能简单变大。
9. Population Growth Rate | 种群增长率
The rate at which a population changes over a unit of time is determined by the balance between additions and removals of individuals. Additions come from births and immigration, while removals are deaths and emigration.
种群在单位时间内变化的速率，取决于个体增加量与减少量之间的平衡。增加来自出生和迁入，减少来自死亡和迁出。
Population growth rate = (Birth rate − Death rate) + (Immigration rate − Emigration rate)
Birth rate and death rate are normally expressed per 1000 individuals per year. If the growth rate is positive, the population is expanding; if negative, the population is declining.
出生率和死亡率通常以每年每千名个体的数量表示。若增长率为正，种群正在扩大；若为负，种群则在下降。
10. Biomass Transfer Efficiency | 生物量传递效率
In a food chain, only a fraction of the biomass consumed at one trophic level is converted into new biomass at the next level. The efficiency of this transfer can be calculated and is a key reason why food chains rarely exceed five trophic levels.
在食物链中，某一营养级所消耗的生物量中，只有一部分会转化为下一营养级的新生物量。这一传递效率可以计算，也是食物链很少超过五个营养级的关键原因。
Efficiency (%) = (Biomass in higher trophic level ÷ Biomass in lower trophic level) × 100
Significant energy is lost through respiration, egestion of undigested materials, and heat. Consequently, the efficiency of biomass transfer is typically around 10%, which means that far more plant material is needed to support a predator than to support a herbivore.
大量能量因呼吸作用、未消化物质的排泄和散热而损失。因此，生物量传递效率通常仅在 10% 左右，这意味着养活一头捕食者所需的植物材料远比养活食草动物多得多。
Applying this formula helps explain why pyramids of biomass almost always narrow dramatically from producers to top carnivores.
应用该公式有助于解释为什么生物量金字塔从生产者到顶级食肉者几乎总是急剧变窄。
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