📚 Year 10 Eduqas Biology: Formulas & Principles Quick Reference | Year 10 Eduqas 生物：公式定理速查手册
This quick-reference handbook compiles the most essential formulas, principles and key calculations covered in Year 10 Eduqas GCSE Biology. It brings together magnification, actual-size calculations, unit conversions, rates of reaction, percentage change and ecological efficiency in one place, with worked examples written in the style Eduqas examiners expect. Use it to check your method, your units and the plausibility of your answer before every assessment.
本速查手册汇总了 Year 10 Eduqas GCSE 生物中最关键的公式、定理和重要计算，涵盖放大倍数、实际大小换算、单位换算、反应速率、百分比变化和生态效率等，方便你在备考时快速查阅。所有公式均配有清晰的解释、针对 Eduqas 大纲的示例，以及考官常看的评分点，适合在做题前核对方法与单位。
1. Magnification Formula | 放大倍数公式
e. Before you divide, both sizes must be expressed in the same unit, normally millimetres (mm) or micrometres (µm).
放大倍数表示图像比实际物体放大了多少倍。标准公式为：放大倍数 = 图像大小 ÷ 实际大小。计算前必须把两项大小换算成同一单位，通常使用毫米（mm）或微米（µm）。
Magnification = Image size ÷ Actual size
e answer is written as 1500 ×. Notice that the units cancel, so magnification has no unit.
示例：图像中一个细胞宽 30 mm，而实际细胞宽 0.02 mm，则放大倍数 = 30 ÷ 0.02 = 1500，写作 1500 ×。注意单位在此被约掉，因此放大倍数本身没有单位。
Always express magnification as a number followed by '×' If your calculation gives a tiny number such as 0.07 ×, you have almost certainly divided the wrong way round and swapped the image and actual sizes.
放大倍数应以数字加"×"表示，并检查结果是否合理。光学显微镜一般可达 ×1000 至 ×1500，而电子显微镜照片可高达 10 万倍甚至更高。若算出的倍数极小，比如 0.07 ×，几乎可以肯定你把除法的顺序写反了，颠倒了图像大小与实际大小。
2. Actual Size from Micrographs | 显微图像中的实际大小
en. Rearranging the magnification equation gives Actual size = Image size ÷ Magnification.
Eduqas 考试中非常常见的一种题型是：给出放大倍数和图像上测得的大小，要求算出标本的实际大小。把放大倍数公式变形即可得到：实际大小 = 图像大小 ÷ 放大倍数。
Actual size = Image size ÷ Magnification
taken at ×2000. Actual size = 40 ÷ 2000 = 0.02 mm. Multiplying by 1000 converts this to 20 µm, which is a realistic size for a mitochondrion.
示例：某显微照片的放大倍数为 ×2000，图中线粒体长 40 mm。实际大小 = 40 ÷ 2000 = 0.02 mm，乘以 1000 换算为 20 µm，这正是一般线粒体的真实尺寸范围。
Useful size benchmarks help you judge whether an answer is reasonable: a typical animal cell is 10-30 µm, a bacterium is 1-5 µm, a red blood cell is about 7 µm, and a ribosome is around 20-25 nm. If your calculated cell comes out as 5 m, you have made a conversion slip; always write the unit at every step.
记住一些常见尺寸有助于判断答案是否合理：典型动物细胞为 10-30 µm，细菌为 1-5 µm，红细胞约 7 µm，核糖体约 20-25 nm。如果算出细胞为 5 m 那么大，说明单位换算出错；建议每一步都写出单位。
3. Unit Conversions for Microscopy | 显微计算的单位换算
Most lost marks in magnification questions are conversion errors, not formula errors. Learn the chain centimetres - millimetres - micrometres - nanometres, where each step down multiplies by 1000 for µm and nm, and by 10 from cm to mm.
放大倍数题失分大多不是公式用错，而是单位换算出错。请牢记这条换算链：厘米 - 毫米 - 微米 - 纳米。其中 cm 到 mm 乘 10，mm 到 µm、µm 到 nm 各乘 1000。
| Conversion | 换算 | Relationship | 关系 | Example | 示例 |
| cm → mm | × 10 | 2 cm = 20 mm |
| mm → µm | × 1000 | 0.05 mm = 50 µm |
| µm → nm | × 1000 | 7 µm = 7000 nm |
| µm → mm | ÷ 1000 | 2500 µm = 2.5 mm |
| cm → µm | × 10 000 | 1 cm = 10 000 µm |
When converting, decide whether the number should get bigger or smaller. Moving to a smaller unit gives a bigger number (0.02 mm becomes 20 µm); moving to a larger unit gives a smaller number (2500 µm becomes 2.5 mm). This single check catches most mistakes.
换算时先判断数字应变大还是变小：换成更小的单位，数字变大（0.02 mm 变成 20 µm）；换成更大的单位，数字变小（2500 µm 变成 2.5 mm）。仅凭这一步检查就能发现大多数错误。
4. Total Magnification and Scale Bars | 总放大倍数与标尺
ation × objective magnification. A ×10 eyepiece with a ×40 objective therefore gives ×400.
使用学校显微镜时，总放大倍数等于两个镜头的放大倍数相乘：总放大倍数 = 目镜放大倍数 × 物镜放大倍数。例如 ×10 目镜配 ×40 物镜，总放大倍数为 ×400。
Total magnification = eyepiece × objective
gth of the bar, then compare it with the structure you are measuring. If a bar labelled 10 µm measures 20 mm on the page, then 1 mm of the image represents 0.5 µm. A structure measuring 30 mm therefore has an actual length of 30 × 0.5 = 15 µm.
有些图像不给出放大倍数，而是给出标尺。先量出标尺在纸面上的长度，再与你要测的结构作比较。若标注 10 µm 的标尺在纸面上长 20 mm，则图像上每 1 mm 代表 0.5 µm。一个长度 30 mm 的结构，实际长度即为 30 × 0.5 = 15 µm。
Remember that magnification compares sizes and is dimensionless, while resolution is the ability to distinguish two close points and is a property of the microscope, not of the calculation. A high magnification with poor resolution gives a large but blurry image.
注意：放大倍数只比较大小，没有单位；而分辨力指区分两个相邻点的能力，是显微镜本身的属性，与计算无关。放大倍数高但分辨力差，只能得到又大又模糊的图像。
5. Surface Area to Volume Ratio | 表面积与体积比
me, with both quantities in the same unit, so the ratio itself has units of cm⁻¹, or is written as a ratio such as 6:1.
表面积与体积比（SA:V）解释了细胞为何保持很小、交换表面为何要折叠。计算方法为：SA:V = 表面积 ÷ 体积，两者需用同一单位，因此比值本身单位为 cm⁻¹，也可写成如 6:1 的比式。
SA:V = Surface area ÷ Volume
and volume = 3³ = 27 cm³, so SA:V = 54 ÷ 27 = 2.0. The larger cube has the smaller ratio.
以立方体为例：边长 1 cm 时，表面积 = 6 × 1² = 6 cm²，体积 = 1³ = 1 cm³，SA:V = 6.0；边长 3 cm 时，表面积 = 6 × 3² = 54 cm²，体积 = 3³ = 27 cm³，SA:V = 54 ÷ 27 = 2.0。可见物体越大，比值越小。
This is why single-celled organisms rely on diffusion alone, while larger organisms need exchange surfaces such as villi in the small intestine, alveoli in the lungs, gills in fish and root hair cells in plants, all of which increase surface area without greatly increasing volume.
因此，单细胞生物仅靠扩散即可满足需求，而较大生物必须依靠交换表面，如小肠绒毛、肺泡、鱼鳃和植物根毛细胞。这些结构在不显著增加体积的前提下大幅增加了表面积。
6. Rate of Reaction and Enzyme Activity | 反应速率与酶活性速率
nt of product formed ÷ Time taken. Typical units in Eduqas practicals are cm³ s⁻¹ for gas volume, g min⁻¹ for mass loss, or mol dm⁻³ s⁻¹ for concentration change.
速率的含义始终是"单位时间内发生多少变化"：速率 = 生成物量 ÷ 所用时间。Eduqas 实验题中常见单位有气体体积的 cm³ s⁻¹、质量减少的 g min⁻¹，以及浓度变化的 mol dm⁻³ s⁻¹。
Rate = Amount of product ÷ Time
t, the rate is often expressed as Rate = 1 ÷ Time, giving arbitrary units of s⁻¹ that allow fair comparisons between repeats.
示例：40 s 内收集到 20 cm³ 氧气，平均速率 = 20 ÷ 40 = 0.5 cm³ s⁻¹。若实验测量的是到达终点所需的时间，速率常写作：速率 = 1 ÷ 时间，单位为 s⁻¹（任意单位），便于比较不同重复实验。
For enzyme graphs, the initial rate is the steepest part of the curve and is found by drawing a tangent at time zero. Varying temperature or pH beyond the optimum denatures the enzyme by changing the shape of its active site, so the rate falls sharply; increasing substrate concentration raises the rate only until all active sites are occupied.
在酶促反应图像中，初速率对应曲线最陡的部分，可在时间为零处作切线求出。温度或 pH 超出最适范围会使酶变性，活性位点形状改变，速率急剧下降；而增加底物浓度只能提高速率，直到所有活性位点被占据为止。
7. Percentage Change and Percentage Efficiency | 百分比变化与百分比效率
A positive answer is an increase and a negative answer is a decrease; always state which.
百分比变化用于把新数值与原数值比较：百分比变化 = （变化量 ÷ 原值）× 100。结果为正表示增加，为负表示减少，答题时务必说明方向。
Percentage change = (Change ÷ Original) × 100
25 g afterwards. The change is 25 − 20 = 5 g, so percentage change = (5 ÷ 20) × 100 = +25%, a 25% increase caused by water entering by osmosis.
示例：马铃薯条放入蒸馏水前质量为 20 g，之后为 25 g。变化量为 25 − 20 = 5 g，故百分比变化 = （5 ÷ 20）× 100 = +25%，即因渗透吸水而增加了 25%。
ers 20 J of light from 100 J of electrical energy, its efficiency is (20 ÷ 100) × 100 = 20%, with the remaining 80 J lost mainly as heat.
百分比效率把同样的思路用于能量或质量：百分比效率 = （有用输出 ÷ 总输入）× 100。若一盏灯由 100 J 电能中输出 20 J 光能，则效率为 （20 ÷ 100）× 100 = 20%，其余 80 J 主要以热能形式散失。
8. Ecological Efficiency and Energy Transfer | 生态效率与能量传递
level ÷ Energy available at the previous level) × 100. It is always calculated between two adjacent levels.
生态效率衡量能量从一个营养级传递到下一营养级的比例：生态效率 = （传递到下一营养级的能量 ÷ 上一营养级可利用的能量）× 100。计算必须在相邻两个营养级之间进行。
Ecological efficiency = (Energy at level n+1 ÷ Energy at level n) × 100
y consumers contain 5 000 kJ m⁻² yr⁻¹. Ecological efficiency = (5 000 ÷ 50 000) × 100 = 10%, which matches the typical value of about 10% quoted for food chains.
示例：草地的生产者含能量 50 000 kJ m⁻² yr⁻¹，初级消费者含能量 5 000 kJ m⁻² yr⁻¹。生态效率 = （5 000 ÷ 50 000）× 100 = 10%，与食物链中常引用的约 10% 典型值一致。
The remaining 90% is not wasted in the sense of disappearing: it is lost as heat from respiration, used for movement and maintaining body temperature, egested in faeces and urine as undigested material, or left uneaten as roots, bones and fur. This explains why food chains rarely have more than four or five trophic levels.
其余 90% 并非"消失"：一部分以呼吸产热形式散失，一部分用于运动与维持体温，一部分以未消化的物质随粪便和尿液排出，还有一部分如根、骨骼和毛发未被取食。这解释了为何食物链通常不超过四到五个营养级。
9. Biomass, Dry Mass and Pyramids | 生物量、干重与生态金字塔
Biomass is the mass of living material in an organism or trophic level, usually measured as dry mass because water content varies widely. A pyramid of biomass is drawn with bars proportional to the biomass at each level, so it normally narrows towards the top.
生物量指生物体或某一营养级中活物质的质量，通常以干重测量，因为含水量差异很大。生物量金字塔的条形高度与该营养级的生物量成正比，因此通常向上逐渐变窄。
A pyramid of numbers can look very different and may even be inverted: one oak tree supports thousands of insects, so the bottom bar is a single organism while the bar above it is huge. Exam answers should state that a pyramid of biomass is a better guide to energy flow than a pyramid of numbers.
数量金字塔的形状可能完全不同，甚至倒置：一棵橡树可养活成千上万只昆虫，因此最底层只有一个生物体，而上一层数量巨大。答题时应指出：与数量金字塔相比，生物量金字塔更能反映能量流动的实际情况。
ember to keep units identical (g m⁻², kg m⁻² or kJ m⁻²) and to state the time period over which the measurement was taken.
若要比较两个营养级之间的生物量传递，可用同样的百分比方法：生物量传递率 = （第 n+1 级生物量 ÷ 第 n 级生物量）× 100。注意单位必须一致（g m⁻²、kg m⁻² 或 kJ m⁻²），并说明测量所覆盖的时间范围。
10. Photosynthesis and Respiration Equations | 光合作用与呼吸作用方程
These two word and symbol equations underpin almost every energy topic in Year 10, and they are frequently worth a mark each. Photosynthesis requires light and chlorophyll and stores energy in glucose.
这两个文字方程与符号方程是 Year 10 能量相关主题的基础，往往各占一分。光合作用需要光和叶绿素，并把能量储存于葡萄糖中。
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
ight energy absorbed by chlorophyll. The rate can be measured by counting oxygen bubbles, by the rate of carbon dioxide uptake, or by the rate of biomass increase in pondweed.
用文字表述：二氧化碳 + 水 → 葡萄糖 + 氧气，光能被叶绿素吸收。可通过计数氧气气泡、测量二氧化碳吸收速率或测定水草生物量增加速率来测量光合速率。
eleasing energy for muscle contraction, active transport and maintaining body temperature. Anaerobic respiration in animal cells forms lactic acid, while in yeast it forms ethanol and carbon dioxide, a reaction used in bread making and brewing.
有氧呼吸的文字表述：葡萄糖 + 氧气 → 二氧化碳 + 水，释放的能量用于肌肉收缩、主动运输和维持体温。动物细胞的无氧呼吸产生乳酸，而酵母的无氧呼吸产生乙醇和二氧化碳，这一反应被用于面包发酵和酿酒。
11. Genetics: Probability and Expected Ratios | 遗传学：概率与预期比例
Genetic crosses use the same mathematics as probability. A Punnett square shows all possible combinations of alleles from two parents, and the ratio of phenotypes in the square gives the probability of each outcome.
遗传杂交所用的数学与概率完全一致。庞纳特方格（Punnett square）列出双亲配子的所有可能组合，方格中表现型的比例即为各结果的概率。
ratio 1 AA : 2 Aa : 1 aa, which is a phenotype ratio of 3 dominant : 1 recessive. Each offspring therefore has a 25%, or ¼, chance of showing the recessive phenotype.
两个杂合子杂交（Aa × Aa）的基因型比例为 1 AA : 2 Aa : 1 aa，表现型比例为 3 显性 : 1 隐性。因此每个后代出现隐性表现型的概率为 25%，即 ¼。
Expected number = Probability × Total number of offspring
s, the expected number of recessive individuals is 0.25 × 400 = 100. Remember that this is an expectation, and actual numbers may differ due to chance, especially in small samples.
若要把概率放大到真实群体，可用：预期数量 = 概率 × 后代总数。若 Aa × Aa 杂交产生 400 个后代，预期隐性个体数 = 0.25 × 400 = 100。注意这只是期望值，实际数量可能因随机性而不同，样本越小偏差越明显。
12. Exam Technique and Formula Summary | 考试技巧与公式汇总
In calculation questions, examiners award marks for the correct formula, correct substitution and a correct answer with units, so always write every stage down. Underlining the two values you have been given is a quick way to decide which version of a formula you need.
在计算题中，考官通常按"公式正确、代入正确、答案与单位正确"给分，因此务必写出每一步。把题干给出的两个数值划线标出，是快速判断该用哪条公式的好方法。
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