IGCSE AQA Biology: Calculation Practice | IGCSE AQA 生物：计算题专项训练

📚 IGCSE AQA Biology: Calculation Practice | IGCSE AQA 生物：计算题专项训练

Calculation questions are a key part of the IGCSE AQA Biology exam. They test your ability to handle numerical data, convert units, and apply formulae in biological contexts. From microscope magnification to population estimates, mastering these skills can make a real difference to your grade. This article provides a structured revision guide covering the most common calculation types you will meet, with worked examples and tips for avoiding common mistakes.

计算题是 IGCSE AQA 生物考试的重要组成部分，考查你处理数值数据、换算单位以及在生物学情境中应用公式的能力。从显微镜放大率到种群数量估算，掌握这些技能能真正提升你的成绩。本文提供一份系统的复习指南，涵盖你将遇到的最常见计算类型，配有演练示例和避免常见错误的技巧。

1. Microscopy: Magnification Formula | 显微镜：放大率公式

The relationship between magnification, image size, and actual size is given by the formula: Magnification = Image size ÷ Actual size. You must be able to rearrange this to find any of the three variables. Remember that ‘image size’ is what you measure with a ruler on a diagram or photomicrograph, and ‘actual size’ is the real size of the specimen. Always convert both to the same unit before calculating.

放大率、图像大小和实际大小之间的关系由公式：放大率 = 图像大小 ÷ 实际大小 给出。你必须能够重新排列公式以求出三个变量中的任意一个。记住，“图像大小”是你用尺子在图或显微照片上量得的长度，而“实际大小”是标本的真实尺寸。在计算前，始终将两者换算为相同单位。

A typical question might give you an image of a cell that measures 50 mm across, and the actual cell diameter is 0.02 mm. Magnification = 50 ÷ 0.02 = 2500×. Alternatively, you may be told the magnification and actual size, and asked for the image size: Image size = Magnification × Actual size. If a 5 μm mitochondrion is viewed at 10 000×, image size = 10 000 × 5 μm = 50 000 μm = 50 mm. An exam favourite is unit conversion: 1 mm = 1000 μm, 1 μm = 1000 nm. Use standard form to express very large or very small numbers clearly, e.g. 0.002 mm = 2 × 10⁻³ mm = 2 μm.

典型题目可能会给出一个细胞图像，测量其宽度为 50 mm，而实际细胞直径为 0.02 mm。放大率 = 50 ÷ 0.02 = 2500×。或者，可能告知放大率和实际大小，要求求图像大小：图像大小 = 放大率 × 实际大小。如果一个 5 μm 的线粒体以 10 000× 放大观察，图像大小 = 10 000 × 5 μm = 50 000 μm = 50 mm。考试中最常见的是单位换算：1 mm = 1000 μm，1 μm = 1000 nm。使用标准形式清晰表示非常大或非常小的数字，例如 0.002 mm = 2 × 10⁻³ mm = 2 μm。

2. Measuring Cells and Organelles from Diagrams | 从图像中测量细胞和细胞器

In the exam, you may be given a printed micrograph with a scale bar. The scale bar represents a known actual length. First, measure the scale bar in mm with your ruler. Then calculate the magnification of the image using the formula: Magnification = measured length of scale bar (mm) ÷ actual length scale bar represents (mm). Once you have the magnification, measure the structure of interest and divide by magnification to find actual size. Alternatively, if no scale bar is given, there may be a magnification stated on the diagram; use it directly.

考试中，可能会提供一张带有比例尺的印刷显微照片。比例尺代表已知的实际长度。首先，用尺子以毫米为单位测量比例尺的长度。然后，使用公式计算图像的放大率：放大率 = 量得的比例尺长度（mm）÷ 比例尺代表的实际长度（mm）。得到放大率后，测量你感兴趣的结构，除以放大率即可求得实际大小。或者，如果没有给出比例尺，图像上可能会注明放大倍数；直接使用即可。

For example, a scale bar labelled ’10 μm’ measures 20 mm on paper. Magnification = 20 mm ÷ 0.01 mm (since 10 μm = 0.01 mm) = 2000×. If a chloroplast in the same image measures 8 mm, its actual length = 8 mm ÷ 2000 = 0.004 mm = 4 μm. Always check your units carefully: converting everything to mm or μm before calculation avoids mistakes. Practice with past papers to get comfortable reading scale bars and using your ruler under time pressure.

例如，一条标注为“10 μm”的比例尺在纸上量得 20 mm。放大率 = 20 mm ÷ 0.01 mm（因为 10 μm = 0.01 mm）= 2000×。如果同一图像中的叶绿体量得 8 mm，其实际长度 = 8 mm ÷ 2000 = 0.004 mm = 4 μm。始终仔细检查单位：在计算前将所有数值转换为毫米或微米可以避免错误。通过往年真题练习，便于在时间压力下熟练读取比例尺并使用尺子。

3. Estimating the Size of a Subcellular Structure Using the Field of View | 利用视野估算亚细胞结构大小

Another method involves knowing the diameter of the field of view. For example, under low power you may observe a cell that stretches across half the field. If the field diameter is known to be 400 μm, the cell length is roughly 200 μm. To find actual size under higher magnifications, you can use the relationship: diameter of field at low power × magnification of low power = diameter of field at high power × magnification of high power. This is because the field diameter is inversely proportional to magnification.

另一种方法需要知道视野直径。例如，在低倍镜下你可能观察到某个细胞横跨视野的一半。如果已知视野直径为 400 μm，那么细胞长度大约就是 200 μm。要在更高放大倍数下求实际大小，可以使用以下关系：低倍镜视野直径 × 低倍镜放大倍数 = 高倍镜视野直径 × 高倍镜放大倍数。这是因为视野直径与放大倍数成反比。

If using a light microscope with a ×10 eyepiece and a ×40 objective (total 400×), the field diameter is often provided. Suppose the field diameter is 0.4 mm (400 μm) at 100× magnification. Then at 400×, field diameter = (100 × 0.4) ÷ 400 = 0.1 mm = 100 μm. Now count the number of cells or organelles that fit across the field; dividing the field diameter by the number gives average object size. This technique is useful for microbes like yeast or bacteria, where direct measurement is tricky.

如果使用装有×10目镜和×40物镜（总放大倍数 400×）的光学显微镜，视野直径通常会给出。假设在 100× 放大倍数下视野直径为 0.4 mm（400 μm）。那么在 400× 下，视野直径 = (100 × 0.4) ÷ 400 = 0.1 mm = 100 μm。现在数一数横跨视野的细胞或细胞器数量；用视野直径除以数量即为物体平均大小。这种技巧对酵母或细菌等微生物特别有用，因为直接测量很困难。

4. Magnification and Total Magnification in Light Microscopes | 光学显微镜中的放大率和总放大率

Total magnification of a light microscope is the product of the eyepiece lens magnification and the objective lens magnification. For example, an eyepiece of ×10 and an objective of ×40 give a total magnification of 400×. Questions may ask you to calculate the missing lens magnification. If total magnification is 100× and the eyepiece is ×10, the objective must be 100 ÷ 10 = 10×. While this is straightforward, it can be combined with drawing magnification. When you draw a specimen observed under a microscope, your drawing has its own magnification: drawing size ÷ actual size. This is often larger than the microscope magnification, and you must keep the two concepts distinct.

光学显微镜的总放大率是目镜放大倍数与物镜放大倍数的乘积。例如，目镜为×10、物镜为×40，则总放大率为 400×。题目可能会要求你计算缺失的镜头放大倍数。若总放大率为 100×，目镜为×10，则物镜必定为 100 ÷ 10 = 10×。虽然这很简单，但它可能与绘图放大率结合在一起。当你在显微镜下绘制观察到的标本时，你的绘图有自己的放大率：绘图大小 ÷ 实际大小。这通常比显微镜放大率更大，你必须将这两个概念区分清楚。

A typical exam question: A student draws a cell as seen under a 400× microscope. The drawing of the cell is 80 mm long. The actual cell is 0.02 mm long. What is the drawing magnification? Answer: drawing magnification = 80 ÷ 0.02 = 4000×. Notice that the microscope magnification is not directly used in this calculation unless you are asked about the relationship between the two. The image seen through the microscope is already magnified 400×, but the drawing further enlarges that image. This layered magnification can confuse students, so always identify which ‘image’ size you are dealing with.

一个典型的考试题：一名学生绘制了一个在 400× 显微镜下观察到的细胞。绘图中的细胞长 80 mm，实际细胞长 0.02 mm。绘图放大率是多少？答案：绘图放大率 = 80 ÷ 0.02 = 4000×。注意，除非题目询问两者间的关系，显微镜放大率在此计算中不直接使用。通过显微镜看到的图像已经放大 400×，而绘图又进一步放大了该图像。这种分层的放大率可能会让学生感到困惑，因此始终要明确你所处理的是哪个“图像”尺寸。

5. Estimating Population Size: Capture-Mark-Recapture | 估算种群大小：捕获-标记-重捕法

The formula is: Population size = (Number in first sample × Number in second sample) ÷ Number of marked individuals recaptured. This assumes that marked individuals mix randomly with the rest of the population, no migration, no births or deaths between sampling, and that marking does not affect survival or behaviour. In the exam, you may be given data and asked to calculate this estimate, as well as discuss the assumptions and their effect on the accuracy of the estimate.

公式为：种群大小 =（第一次样本数量 × 第二次样本数量）÷ 重捕到的标记个体数量。该方法假设标记个体与种群其余部分随机混合，在两次抽样之间无迁移、无出生或死亡，且标记不影响个体存活或行为。在考试中，可能会给你数据并要求计算这一估计值，还可能讨论假设及其对估计准确性的影响。

For example: In a pond, 40 water boatmen are caught, marked, and released. Later, a second sample of 50 is caught, and 10 of them are found to be marked. Estimated population size = (40 × 50) ÷ 10 = 200. If the marking made them more visible to predators, recapture rate would drop, leading to an overestimate. Questions often ask you to evaluate the reliability of the estimate and suggest improvements, such as using a larger sample or repeating the recapture several times. This is an application of the Lincoln Index, widely used in ecology.

例如：在一个池塘中，首次捕获 40 只水黾，标记后放回。后来，第二次捕获 50 只，其中 10 只有标记。估算的种群大小 = (40 × 50) ÷ 10 = 200。如果标记使它们更容易被天敌发现，重捕率会下降，导致估算值偏高。题目常常要求你评估估算值的可靠度并提出改进建议，例如使用更大的样本或多次重复重捕。这是林肯指数的一种应用，广泛用于生态学中。

6. Calculating Biomass and Efficiency of Energy Transfer | 计算生物量和能量传递效率

Biomass is the mass of living material, often measured as dry mass (after removing water). You may be asked to calculate biomass of a trophic level from given data, or the efficiency of energy transfer between trophic levels. Efficiency (%) = (Biomass at higher trophic level ÷ Biomass at lower trophic level) × 100. Alternatively, it can be based on energy content per unit area per year (kJ m⁻² year⁻¹). Typical values are around 10%, but you will be given the numbers to plug into the formula.

生物量是活体材料的质量，通常以干重（除去水分后）测量。你可能会被要求根据给定数据计算某一营养级的生物量，或计算营养级之间能量传递的效率。效率（%）=（较高营养级的生物量 ÷ 较低营养级的生物量）× 100。或者，也可以基于单位面积年能量含量（kJ m⁻² 年⁻¹）计算。典型值约为 10%，但会给出数据让你代入公式。

Example: In a food chain, the biomass of grass is 20 000 kJ m⁻², and the biomass of rabbits feeding on it is 1 800 kJ m⁻². Efficiency = (1 800 ÷ 20 000) × 100 = 9%. You should be able to explain why energy is lost between trophic levels – respiration, excretion, uneaten parts, and heat. Calculations sometimes involve converting wet mass to dry mass by using a given percentage, or scaling up from a small sample to the whole ecosystem. For instance, if the dry mass of grass in a 1 m² quadrat is 0.5 kg, then in a 1000 m² field the total dry mass is 500 kg. Combine this with energy content per kg to find total energy stored.

示例：在一条食物链中，草的生物量为 20 000 kJ m⁻²，而以其为食的兔子的生物量为 1 800 kJ m⁻²。效率 = (1 800 ÷ 20 000) × 100 = 9%。你应该能够解释能量在营养级之间损失的原因——呼吸作用、排泄、未被取食的部分以及散热。计算有时涉及使用给定的百分比将湿重转换为干重，或从一个小样方放大到整个生态系统。例如，如果一个 1 m² 样方中的草干重为 0.5 kg，那么在一个 1000 m² 的田地中总干重为 500 kg。将此与每千克的能量含量结合，可得出储存的总能量。

7. Rate of Enzyme-Controlled Reactions | 酶促反应速率

The rate of an enzyme reaction can be calculated from the change in product formed or substrate used per unit time. Rate = Change ÷ Time. You may be asked to interpret graphs of product concentration against time, or calculate the initial rate by drawing a tangent at time zero. If data is given as a table, calculate the rate for each interval and observe how it changes. For example, if the volume of oxygen produced in an enzyme-catalysed breakdown of hydrogen peroxide is 36 cm³ after 120 seconds, the average rate = 36 ÷ 120 = 0.3 cm³/s. Units are important; you might need to convert minutes to seconds or vice versa.

酶反应速率可以通过单位时间内生成的产物量或消耗的底物量的变化来计算。速率 = 变化量 ÷ 时间。你可能需要分析产物浓度对时间的图像，或通过在时间为零处画切线来计算初始速率。如果数据以表格形式给出，可计算每个时间间隔的速率并观察其变化。例如，如果在过氧化氢酶催化的过氧化氢分解中，120 秒后产生了 36 cm³ 的氧气，则平均速率 = 36 ÷ 120 = 0.3 cm³/s。单位很重要；你可能需要将分钟换算为秒，或反之。

Calculation of rate is also required in photosynthesis experiments measuring oxygen production or carbon dioxide uptake per unit time per unit mass of leaf. Rate = change in gas volume or concentration ÷ (time × mass). The units might be cm³ min⁻¹ g⁻¹. Always show your working, and be prepared to evaluate the limitations of the method, such as gas leaks or temperature fluctuations. A common source of error is forgetting to subtract the control rate from the experimental rate, especially in respiration experiments using a respirometer.

在测量单位时间单位叶质量产氧量或二氧化碳吸收量的光合作用实验中，也需要计算速率。速率 = 气体体积或浓度的变化量 ÷（时间 × 质量），单位可能为 cm³ min⁻¹ g⁻¹。始终展示你的计算过程，并准备好评价方法的局限性，例如气体泄漏或温度波动。一个常见的错误来源是忘记从实验速率中减去对照组速率，尤其是在使用呼吸计测量呼吸作用的实验中。

8. Surface Area to Volume Ratio | 表面积与体积之比

As an organism or cell increases in size, its volume grows faster than its surface area. You may be asked to calculate the surface area : volume ratio for a cube or sphere. For a cube of side length L, surface area = 6L², volume = L³, so the ratio is 6/L. The units are often omitted, but you can express the ratio as a number to 1. For example, a cube of side 2 cm has surface area = 6 × 2² = 24 cm², volume = 2³ = 8 cm³, SA:V = 24 : 8 = 3 : 1. A smaller cube of side 1 cm has SA:V = 6 : 1. As size increases, SA:V decreases.

当一个生物体或细胞体积增大时，其体积的增长快于表面积的增长。你可能需要计算一个立方体或球体的表面积与体积之比。对于边长为 L 的立方体，表面积 = 6L²，体积 = L³，因此比值为 6/L。单位通常省略，但你可以将比值表示为一个数字比 1。例如，一个边长为 2 cm 的立方体，表面积 = 6 × 2² = 24 cm²，体积 = 2³ = 8 cm³，SA:V = 24 : 8 = 3 : 1。一个边长为 1 cm 的较小立方体，SA:V = 6 : 1。随着尺寸增大，SA:V 下降。

This concept explains why large organisms need transport systems (like circulatory system) and why cells are small. The calculation is simple but often embedded in questions about adaptations. You might also need to calculate the surface area and volume of a cylinder or a sphere using given formulae, but usually the cube model is preferred for simplicity. When comparing organisms, calculate the ratio and link it to rate of heat loss or rate of diffusion. A high SA:V means rapid exchange with the environment but also fast heat loss, which is why small mammals have high metabolic rates.

这个概念解释了为什么大型生物需要运输系统（如循环系统）以及为什么细胞很小。计算虽简单，但常常嵌入关于适应性的问题中。你可能还需要使用给定公式计算圆柱体或球体的表面积和体积，但通常为简便起见偏好使用立方体模型。在比较生物体时，计算比值并将其与热量散失速率或扩散速率联系起来。高 SA:V 意味着与环境间的物质交换迅速，但也伴随快速的热量散失，这就是小型哺乳动物代谢率较高的原因。

9. Probability and Genetic Ratios | 概率与遗传比例

In monohybrid crosses, you need to calculate the expected ratio of offspring genotypes and phenotypes. Using a Punnett square, you determine the probability of each combination. For a cross between two heterozygous individuals (Aa × Aa), the genotype ratio is 1 AA : 2 Aa : 1 aa, and if ‘A’ is dominant, the phenotype ratio is 3 dominant : 1 recessive. You can also calculate probabilities as fractions or percentages. The chance of an offspring being homozygous recessive is 1/4 or 25%. Questions may ask for the probability that a couple’s next child will have a particular genetic disease, assuming Mendelian inheritance.

在单基因杂交中，你需要计算子代基因型和表现型的预期比例。使用庞纳特方格，你可以确定每种组合的概率。对于两个杂合个体（Aa × Aa）的杂交，基因型比例为 1 AA : 2 Aa : 1 aa，若“A”为显性，则表现型比例为 3 显性 : 1 隐性。你也可以用分数或百分比表示概率。子代为隐性纯合子的几率为 1/4 或 25%。题目可能会要求计算一对夫妇的下一个孩子患某种特定遗传病的概率，并假设符合孟德尔遗传规律。

These calculations are straightforward but you must interpret the question carefully. For sex-linked traits, the ratios differ between males and females. Example: A carrier female for colour blindness (XᴺXⁿ) and a normal male (XᴺY). Sons have a 1/2 chance of being colour blind; daughters have a 0 chance of being colour blind but a 1/2 chance of being carriers. You should express probability clearly, like ‘1 in 2’ or ‘50%’. Combining probabilities for two independent events requires multiplication, e.g., chance of having two affected children in succession = 1/4 × 1/4 = 1/16. This often appears in pedigree analysis questions.

这些计算并不复杂，但你必须仔细解读题意。对于伴性遗传性状，男女之间的比例有所不同。示例：一个红绿色盲携带者女性（XᴺXⁿ）与一个正常男性（XᴺY）。儿子有 1/2 的概率患色盲；女儿患色盲的概率为 0，但有 1/2 的概率为携带者。你应该清晰地表达概率，例如“1/2”或“50%”。组合两个独立事件的概率需要乘法运算，例如，连续生育两个患病孩子的概率 = 1/4 × 1/4 = 1/16。这经常出现在系谱分析题中。

10. Calculating Percentage Change and Interpreting Data | 计算百分比变化并解读数据

Graph interpretation in biology often requires you to calculate a percentage change between two values. Percentage change = ((Final value – Initial value) ÷ Initial value) × 100. A negative change indicates a decrease. This skill is commonly tested with data from osmosis experiments, where you calculate percentage change in mass of potato cylinders in different sucrose solutions. Plotting the percentage change on a graph helps identify the isotonic point where no net water movement occurs.

生物学中的图表解读常要求你计算两个数值之间的百分比变化。百分比变化 =（（终值 – 初值）÷ 初值）× 100。负变化表示减少。这项技能通常通过渗透实验的数据进行测试，你需要计算马铃薯条在不同蔗糖溶液中质量的百分比变化。在图上标绘百分比变化有助于确定等渗点，即无净水移动的点。

Example: A potato chip of initial mass 5.0 g is placed in a solution; after 30 minutes its mass is 5.4 g. Percentage change = ((5.4 – 5.0) ÷ 5.0) × 100 = +8%. If you are comparing several concentrations, you can use the line of best fit to estimate the concentration where percentage change would be zero (the solute concentration inside the potato cells). Calculations may also involve interpreting gradients – the steeper the line, the faster the rate of change. Always check the scale of axes and show your working clearly.

示例：一个初始质量为 5.0 g 的马铃薯条放入溶液中，30 分钟后其质量为 5.4 g。百分比变化 =（（5.4 – 5.0）÷ 5.0）× 100 = +8%。如果你要比较多个浓度，可以使用最佳拟合线来估计百分比变化为零时的浓度（即马铃薯细胞内部的溶质浓度）。计算还可能涉及解读斜率——线越陡，说明变化速率越快。始终检查坐标轴的刻度并清晰地展示计算过程。

11. Dilution Calculations in Microbiology | 微生物学中的稀释计算

When investigating the effect of antibiotics or disinfectants, or when counting bacterial colonies, serial dilutions are used to get a countable number of colonies (typically 30–300 CFU per plate). You may need to calculate the number of viable bacteria in the original culture. First, find the dilution factor. For example, if 1 cm³ of original culture is added to 9 cm³ of sterile water, that is a 10⁻¹ dilution. Then 1 cm³ of that is added to another 9 cm³ to make a 10⁻² dilution, and so on. If 0.1 cm³ of a 10⁻³ dilution yields 50 colonies, then the number of bacteria per cm³ of original culture = (number of colonies × dilution factor) ÷ volume plated. Here: (50 × 10³) ÷ 0.1 = 500 000 or 5 × 10⁵ CFU per cm³.

在研究抗生素或消毒剂的效果，或计数细菌菌落时，会采用连续稀释法以获得可计数的菌落数（通常每平板 30–300 CFU）。你可能需要计算原始培养液中的活菌数量。首先，确定稀释倍数。例如，若将 1 cm³ 原始培养液加入 9 cm³ 无菌水中，得到 10⁻¹ 稀释液。然后取 1 cm³ 该稀释液加入另外 9 cm³ 水中制得 10⁻² 稀释液，以此类推。如果取 0.1 cm³ 的 10⁻³ 稀释液涂布后长出 50 个菌落，那么原始培养液每 cm³ 的细菌数 =（菌落数 × 稀释倍数）÷ 涂布体积。本例中：(50 × 10³) ÷ 0.1 = 500 000 即 5 × 10⁵ CFU/cm³。

Questions may also involve calculating the number of bacteria after a period of growth, given the doubling time. If a starting population of 100 bacteria doubles every 30 minutes, how many will there be after 3 hours? 3 hours = 6 doubling periods. Population = starting number × 2ⁿ = 100 × 2⁶ = 100 × 64 = 6400. This exponential growth calculation can also be applied to yeast populations. Always check if the question asks for the total number or the increase. Be precise with standard form – it saves time and reduces errors with large numbers.

题目还可能涉及给定倍增时间，计算经过一段时间生长后的细菌数量。如果初始 100 个细菌每 30 分钟分裂一次，3 小时后有多少？3 小时 = 6 个倍增周期。种群数量 = 初始数量 × 2ⁿ = 100 × 2⁶ = 100 × 64 = 6400。这种指数增长计算也可应用于酵母种群。始终注意题目要求的是总数还是增加的数量。精确使用标准形式——这能节省时间并减少处理大数时的错误。

12. Converting Between Units and Using Standard Form | 单位换算与标准形式的使用

A significant stumbling block in many calculation questions is unit conversion. In biology, you constantly move between mm, μm, nm, as well as between grams and kilograms, cm³ and dm³, and seconds and minutes. A clear understanding of metric prefixes is vital: milli (10⁻³), micro (10⁻⁶), nano (10⁻⁹), kilo (10³). You should be able to quickly convert. For example, 0.5 mm = 500 μm, 1.2 kg = 1200 g, 0.08 dm³ = 80 cm³. Use proportion if the question gives you a rate, e.g., if oxygen uptake is 0.25 cm³ per minute per gram, then for 5 g over 10 minutes, total oxygen = 0.25 × 5 × 10 = 12.5 cm³.

许多计算题中的一大绊脚石是单位换算。在生物学中，你经常需要在 mm、μm、nm 之间切换，还有克与千克、cm³ 与 dm³、秒与分钟之间。清楚理解公制词头至关重要：毫（10⁻³）、微（10⁻⁶）、纳（10⁻⁹）、千（10³）。你应该能够快速换算。例如，0.5 mm = 500 μm，1.2 kg = 1200 g，0.08 dm³ = 80 cm³。如果题目给出速率，可使用比例计算，例如，若氧气摄取率为每克每分钟 0.25 cm³，那么 5 克组织在 10 分钟内的总氧气量 = 0.25 × 5 × 10 = 12.5 cm³。

Standard form (scientific notation) allows you to handle very large and very small numbers without confusion. A number is written as A × 10ⁿ, where A is between 1 and 10. Practice this: 0.0034 = 3.4 × 10⁻³; 25000 = 2.5 × 10⁴. When multiplying, add the exponents; when dividing, subtract. In biology, this is especially useful when comparing sizes of organelles, cells, and molecules. For example, a mitochondrion is about 1 × 10⁻⁶ m and a ribosome about 2 × 10⁻⁸ m. The mitochondrion is 50 times larger (10⁻⁶ ÷ 10⁻⁸ = 10²). Mastery of these numerical skills gives you confidence and speed in the exam.

标准形式（科学计数法）使你能够无混淆地处理非常大和非常小的数字。一个数写作 A × 10ⁿ，其中 A 在 1 到 10 之间。练习：0.0034 = 3.4 × 10⁻³；25000 = 2.5 × 10⁴。相乘时指数相加；相除时指数相减。在生物学中，这在比较细胞器、细胞和分子的大小时特别有用。例如，一个线粒体大约为 1 × 10⁻⁶ m，一个核糖体大约为 2 × 10⁻⁸ m。线粒体是核糖体的 50 倍大（10⁻⁶ ÷ 10⁻⁸ = 10²）。掌握这些数值技能能让你在考试中从容且快速。

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