← 返回资讯屏轩教育 · 行业资讯

Year 8 Cambridge Biology: Formula & Theorem Quick Reference | 剑桥八年级生物：公式定理速查手册

2026-07-18 02:22 · 本文部分内容由AI辅助创作

📚 Year 8 Cambridge Biology: Formula & Theorem Quick Reference | 剑桥八年级生物：公式定理速查手册

Welcome to your quick reference handbook for Year 8 Cambridge Biology. This guide compiles the essential formulas and theorems you must know, from magnification and energy efficiency to genetic ratios and physiological equations. Use it to revise key concepts, practise calculations and strengthen your understanding of life processes.

欢迎使用剑桥八年级生物公式定理速查手册。本指南汇集了你必须掌握的基本公式和定理，从放大倍率、能量效率到遗传比例和生理方程式。用它来复习关键概念、练习计算，并加深对生命过程的理解。

1. Magnification Formula | 放大倍率公式

The magnification of a specimen viewed under a microscope or in a photomicrograph is calculated using this relationship:

在显微镜或显微照片中观察标本的放大倍率用以下关系式计算：

Magnification = Image size ÷ Actual size

Image size is the size of the object as seen in the image (for example, in mm), while actual size is the real-life size of the specimen. It is essential to convert both measurements to the same unit before dividing, otherwise the answer will be wrong by a factor of 1000.

图像大小是图像中看到的物体尺寸（例如以毫米计），而实际大小是标本的真实尺寸。在进行除法之前，必须将两个测量值转换为相同的单位，否则答案会相差 1000 倍。

Common unit conversions: 1 mm = 1000 µm; 1 µ ; 1 cm = 10 mm. Always write the unit as a multiplication sign, for example 100×, never as "100 times bigger".

常见单位换算：1 毫米 = 1000 微米；1 微米 = 1000 纳米；1 厘米 = 10 毫米。答案中的单位要写成乘号形式，例如 100×，不要写成"大 100 倍"。

= 5 ÷ 0.05 = 100×. If the two numbers are in different units, convert first: 0.05 mm = 50 µm, so 5000 µm ÷ 50 µm = 100×.

例题：如果图像大小 = 5 毫米，实际大小 = 0.05 毫米，则放大倍率 = 5 ÷ 0.05 = 100 倍。若两个数值单位不同，要先换算：0.05 毫米 = 50 微米，因此 5000 微米 ÷ 50 微米 = 100 倍。

Rearranging the formula is a common exam skill:

公式变形是常见的考试技能：

Actual size = Image size ÷ Magnification

Image size = Actual size × Magnification

Remember that magnification has no units, because it is a ratio of two lengths measured in the same unit.

请记住放大倍率没有单位，因为它是两个用相同单位测量的长度之比。

2. Microscope Total Magnification | 显微镜总放大倍率

A compound light microscope uses two lenses, so the total magnification is the product of the eyepiece and the objective lens:

复式光学显微镜使用两个透镜，因此总放大倍率是目镜与物镜放大倍率的乘积：

Total magnification = Eyepiece magnification × Objective magnification

e 4× objective gives only 40×, which is useful for finding the specimen before zooming in.

例如，10× 目镜配合 40× 物镜，得到 10 × 40 = 400×。换成 4× 物镜则只有 40×，适合在放大之前先找到标本。

There is a trade-off to remember: as magnification increases, the field of view becomes smaller and the depth of field becomes shallower, so less of the specimen is in focus at one time.

需要记住一个权衡关系：放大倍率越高，视野越小，景深越浅，因此同一时间能清晰对焦的标本范围更小。

3. Photosynthesis Equation | 光合作用方程式

Photosynthesis is the process by which green plants and some other organisms use light energy to convert carbon dioxide and water into glucose and oxygen.

光合作用是绿色植物和其他一些生物利用光能将二氧化碳和水转化为葡萄糖和氧气的过程。

Word equation: Carbon dioxide + Water → Glucose + Oxygen

文字方程式：二氧化碳 + 水 → 葡萄糖 + 氧气

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

This reaction takes place in the chloroplasts using chlorophyll and light energy. Without light, photosynthesis stops, because light provides the energy needed to break the bonds in water and carbon dioxide.

该反应在叶绿体中借助叶绿素和光能进行。没有光，光合作用就会停止，因为光提供了断裂水和二氧化碳中化学键所需的能量。

The conditions that control the rate of photosynthesis can be plotted on a graph. The factor in shortest supply is called the limiting factor.

控制光合作用速率的条件可以绘制成图。供应量最少的那个因素称为限制因素。

Factor | 因素Effect on rate | 对速率的影响
Light intensity 光照强度Rate rises with intensity, then plateaus 速率随强度上升后趋于平稳
Carbon dioxide 二氧化碳More CO₂ increases rate up to a limit 增加浓度可提高速率，但有上限
Temperature 温度Rises to an optimum (about 25 °C), then falls 升至最适温度（约 25 °C）后下降
Chlorophyll 叶绿素Variegated leaves photosynthesise faster in green areas 斑叶的绿色区域光合作用更快

The balanced equation conserves atoms: 6 carbon, 12 hydrogen and 18 oxygen atoms appear on both sides. Energy is not conserved as chemical energy alone, because light energy is converted into chemical energy stored in glucose.

配平方程式遵守原子守恒：两边各有 6 个碳原子、12 个氢原子和 18 个氧原子。能量方面并非只以化学能形式守恒，因为光能被转化为储存在葡萄糖中的化学能。

4. Respiration Equations | 呼吸作用方程式

Respiration releases the energy stored in glucose so that cells can carry out life processes such as movement, growth and active transport.

呼吸作用释放储存在葡萄糖中的能量，使细胞能够完成运动、生长和主动运输等生命活动。

Aerobic respiration uses oxygen and releases a large amount of energy per glucose molecule:

有氧呼吸需要氧气，每分子葡萄糖释放大量能量：

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Anaerobic respiration happens when oxygen is in short supply. In animals and humans it produces lactic acid; in yeast and plants it produces ethanol and carbon dioxide.

缺氧时进行无氧呼吸。动物和人体中产生乳酸；酵母和植物中产生乙醇和二氧化碳。

Glucose → Lactic acid | 葡萄糖 → 乳酸

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ | 酵母：葡萄糖 → 乙醇 + 二氧化碳

Note that respiration and photosynthesis are opposites chemically. Respiration works day and night in all living cells, while photosynthesis only occurs in cells containing chlorophyll and only in the light.

注意呼吸作用和光合作用在化学上是相反的过程。呼吸作用在所有活细胞中昼夜不停地进行，而光合作用只发生在含叶绿素的细胞中，并且只在有光的条件下进行。

Comparing energy release helps you explain fatigue: anaerobic respiration releases far less energy per glucose molecule than aerobic respiration, which is why intense exercise cannot be sustained for long without oxygen.

比较能量释放有助于解释疲劳现象：每分子葡萄糖无氧呼吸释放的能量远少于有氧呼吸，这就是剧烈运动离不开氧气、无法长时间持续的原因。

5. Energy Efficiency and Trophic Levels | 能量效率与营养级

Energy is transferred along a food chain from producers to consumers. At each transfer, only part of the energy is passed on because organisms use energy for movement, respiration and growth, and lose energy in waste products.

能量在食物链中由生产者传递给消费者。每一次传递只有部分能量被传递下去，因为生物要把能量用于运动、呼吸和生长，并通过排泄物损失能量。

广告 / Sponsored

牛津AQA国际GCSE生物?OxfordAQA International GCSE Biology (9201): Onl?进口原版
牛津AQA国际GCSE生物?OxfordAQA International GCSE Biology (9201): Onl?进口原版
¥399.0
📱 去京东购买
京东精选

Efficiency (%) = (Energy in next level ÷ Energy in previous level) × 100

Typical transfer efficiency in a food chain is only about 10 per cent. If a producer level contains 20 000 kJ of energy, the herbivores receive about 2000 kJ and the carnivores only about 200 kJ.

食物链中典型的传递效率只有约 10%。如果生产者层级含 20 000 千焦能量，食草动物大约获得 2000 千焦，食肉动物仅约 200 千焦。

This explains why food chains rarely have more than four or five trophic levels, and why a diet based on plants can support many more people than a diet based on meat.

这解释了为什么食物链很少超过四到五个营养级，也解释了为什么以植物为主的膳食能供养的人口远多于以肉类为主的膳食。

6. Surface Area to Volume Ratio | 表面积与体积比

The surface area to volume ratio (SA:V) explains why cells are small and why exchange surfaces in organisms are folded or flattened.

表面积与体积比（SA:V）解释了为什么细胞很小，以及为什么生物体内的交换表面呈折叠或扁平状。

SA:V = Surface area ÷ Volume

2 × 2 = 8 cm³, so SA:V = 24 ÷ 8 = 3 cm⁻¹. For a cube of side 1 cm, SA:V = 6 ÷ 1 = 6 cm⁻¹.

边长为 2 厘米的立方体，表面积 = 6 × (2 × 2) = 24 平方厘米，体积 = 2 × 2 × 2 = 8 立方厘米，所以 SA:V = 24 ÷ 8 = 3 cm⁻¹。边长为 1 厘米的立方体，SA:V = 6 ÷ 1 = 6 cm⁻¹。

The rule is simple: as an object gets bigger, its volume grows faster than its surface area, so the SA:V ratio falls. A low ratio means diffusion cannot supply the interior quickly enough.

规律很简单：物体越大，体积增长快于表面积，所以 SA:V 比值下降。比值低意味着扩散无法及时为内部供应物质。

Structure | 结构Adaptation | 适应特征
Villi in small intestine 小肠绒毛Finger-like folds increase surface area 指状褶皱增大表面积
Alveoli in lungs 肺泡Millions of tiny sacs with thin walls 数百万薄壁小囊
Root hair cells 根毛细胞Long thin extension absorbs water and ions 细长突起吸收水和无机盐
Red blood cells 红细胞Biconcave shape, no nucleus 双凹圆盘状，无细胞核

7. Rate Calculations in Biology | 生物学中的速率计算

Many practical investigations in Year 8 ask you to calculate a rate from measured data. The general formula is the same for enzyme activity, gas production and cooling curves.

八年级的许多实验探究要求你根据测量数据计算速率。无论是酶活性、气体产生还是降温曲线，通用公式都相同。

Rate = Change in quantity ÷ Time taken

速率 = 变化量 ÷ 所用时间

Common units include cm³/min for gas volume, g/min for mass loss, and °C/min for temperature change. Always convert time to seconds or minutes consistently before dividing.

常见单位包括气体体积的 cm³/min、质量损失的 g/min、温度变化的 °C/min。做除法前，务必把时间统一换算为秒或分钟。

24 ÷ 4 = 6 cm³/min. If the question asks for cm³ per second, divide 24 ÷ 240 = 0.1 cm³/s.

例题：一根池塘草在 4 分钟内释放 24 立方厘米氧气，则速率 = 24 ÷ 4 = 6 cm³/min。若题目要求每秒多少立方厘米，则 24 ÷ 240 = 0.1 cm³/s。

To compare two results fairly, calculate the rate over the steepest, straightest part of the graph rather than the whole curve. The initial rate is usually the fastest because substrates are plentiful.

要公平比较两组结果，应在图像最陡、最直的部分计算速率，而不是整条曲线。初始速率通常最快，因为底物充足。

8. Genetic Ratios and Punnett Squares | 遗传比例与庞尼特方格

A Punnett square predicts the possible genotypes of offspring from two parents. Capital letters represent dominant alleles and lower-case letters represent recessive alleles.

庞尼特方格用于预测两个亲本后代的可能基因型。大写字母代表显性等位基因，小写字母代表隐性等位基因。

ohybrid cross between two heterozygotes (Tt × Tt), the Punnett square gives these results:

两个杂合子杂交（Tt × Tt）的庞尼特方格结果如下：

Tt
TTTTt
tTttt

Genotype ratio 1 : 2 : 1 | Phenotype ratio 3 : 1

nt : recessive = 3 : 1). These are probabilities, not guarantees, so a small sample may not match them exactly.

基因型比例显示实际的等位基因组合（TT : Tt : tt = 1 : 2 : 1），而表现型比例显示可见性状（显性 : 隐性 = 3 : 1）。这些是概率而非必然结果，因此小样本的实测比例未必完全吻合。

en two homozygous parents that differ (TT × tt) gives 100 per cent heterozygous offspring.

其他需要记住的比例：杂合子与隐性纯合子测交（Tt × tt）得到 1 : 1。两个不同的纯合亲本杂交（TT × tt）得到 100% 杂合后代。

9. Population Estimation Using Quadrats | 利用样方估算种群数量

Ecologists cannot count every organism in a large area, so they sample using quadrats and then scale up the result.

生态学家无法数清大片区域内的每一个生物，因此使用样方取样，再把结果按比例放大。

timated population = (Total individuals counted ÷ Total area sampled) × Total habitat area

daisies per m². In a field of 250 m², the estimated population = 4.8 × 250 = 1200 daisies.

例题：在十个 1 平方米的样方中共数到 48 株雏菊，则密度为 48 ÷ 10 = 4.8 株/平方米。在 250 平方米的草地中，估算种群数量 = 4.8 × 250 = 1200 株雏菊。

For a valid estimate, quadrats must be placed randomly, for example by using a random number generator to produce coordinates. Sampling along a line (a transect) is used instead when investigating how a population changes across a habitat gradient.

要使估算有效，样方必须随机放置，例如用随机数生成器产生坐标。若要研究种群如何随栖息地梯度变化，则应改用沿直线取样（样带法）。

10. Units, Conversions and Measuring Accuracy | 单位、换算与测量精确度

Accurate units are worth marks in every Cambridge Biology paper. Use the same unit throughout a calculation, and state units clearly in your final answer.

在每一份剑桥生物试卷中，使用准确的单位都能得分。计算过程中要始终使用同一单位，并在最终答案中清楚写出单位。

Quantity | 物理量Unit | 单位
Length 长度mm, µm, nm
Volume 体积cm³, dm³, ml
Mass 质量g, kg, mg
Time 时间s, min
Temperature 温度°C

Percentage change is another formula that appears frequently, especially when comparing growth or mass in experiments:

百分比变化是另一个高频公式，尤其在比较实验中的生长量或质量时：

Percentage change = (Change ÷ Original value) × 100

8 cm to 10 cm, the change is 2 cm and the percentage change = (2 ÷ 8) × 100 = 25 per cent.

例如，一株幼苗从 8 厘米长到 10 厘米，变化量为 2 厘米，百分比变化 = (2 ÷ 8) × 100 = 25%。

Finally, remember that readings always carry uncertainty. A ruler marked in millimetres is read to the nearest 0.5 mm, and repeating a measurement three times and taking a mean reduces the effect of random error.

最后，请记住读数总带有不确定度。以毫米为刻度的尺子只能估读到 0.5 毫米；重复测量三次取平均值可以减少随机误差的影响。

11. Quick Revision Checklist | 快速复习清单

Use this checklist before a test to make sure every formula in this handbook can be recalled and applied without hesitation.

考试前使用这份清单，确保本手册中的每个公式都能毫不犹豫地回忆并应用。