📚 Pre-U WJEC Biology: Quick-Reference Formula & Theorem Handbook | Pre-U WJEC 生物：公式定理速查手册
Mastering the quantitative side of Pre-U Biology is just as critical as understanding key concepts. This handbook compiles all the essential formulas, equations, and theorems you need to memorise for the WJEC examination, from magnification and surface area to volume ratios, through to Hardy-Weinberg equilibrium and population growth models. Each entry includes a clear definition, the formula itself, worked examples where helpful, and tips for avoiding common pitfalls in the exam. Keep this guide close during revision to ensure you can apply these quantitative tools accurately and confidently under timed conditions.
掌握 Pre-U 生物中的定量内容与理解核心概念同样重要。本手册汇总了 WJEC 考试必须记忆的所有关键公式、方程式与定理，涵盖从放大倍率、表面积与体积比，直至哈迪-温伯格平衡与种群增长模型。每一条目均包含明确定义、公式本身、适当的工作示例以及避免考试常见陷阱的提示。复习期间随身携带本指南，确保你能够限时准确且自信地运用这些量化工具。
1. Magnification Formula | 放大倍率公式
Magnification is the factor by which an image appears larger than the actual specimen. The standard formula relates image size, actual size, and magnification. Ensure all measurements are converted to the same unit before calculation—typically millimetres or micrometres. A common error involves mixing units, which can throw the answer off by a factor of 1000. Remember that 1 mm = 1000 µm, so converting actual size to mm or image size to µm before solving is essential for accuracy.
放大倍率指图像比实际标本放大的倍数。标准公式将图像尺寸、实际尺寸与放大倍率联系在一起。确保所有测量值计算之前转换为相同单位——通常使用毫米或微米。常见错误涉及单位混用，可能导致答案偏差 1000 倍。记住 1 mm = 1000 µm，解题前将实际尺寸转换为毫米或将图像尺寸转换为微米对准确性至关重要。
Magnification = Image size ÷ Actual size
If you measure an image of a cell as 25 mm and the actual cell diameter is 0.05 mm, then Magnification = 25 ÷ 0.05 = ×500. In most Pre-U questions, you will rearrange this equation to find either actual size or magnification. Use a ruler to measure the image directly on the exam paper, and always write the units clearly in your working.
如果你测量的一张细胞图像为 25 mm，实际细胞直径为 0.05 mm，则放大倍率 = 25 ÷ 0.05 = ×500。在大多数 Pre-U 问题中，你需要变换此方程以求出实际尺寸或放大倍率。直接使用直尺在试卷上测量图像，并在计算过程中始终清晰标注单位。
2. Surface Area to Volume Ratio | 表面积与体积比
The surface area to volume ratio (SA:V) is a fundamental concept that explains why cells are microscopic and why larger organisms require specialised exchange surfaces and transport systems. As an organism or cell increases in size, its volume grows proportionally faster than its surface area, decreasing the SA:V ratio. This limits the efficiency of diffusion for nutrient uptake and waste removal. Simple geometric formulas are used to calculate surface area and volume for cubes, spheres, and cylinders.
表面积与体积比 (SA:V) 是一个基础概念，解释了细胞为何微观，以及为何较大生物体需要特化的交换表面与运输系统。随着生物体或细胞尺寸增大，其体积增长比例上快于表面积，从而降低了 SA:V 比。这限制了扩散对营养摄取与废物排出的效率。使用简单几何公式即可计算立方体、球体与圆柱体的表面积和体积。
For a cube of side length L: Surface area = 6L², Volume = L³, therefore SA:V = 6/L. As L doubles, SA:V halves. For a sphere of radius r: Surface area = 4πr², Volume = (4/3)πr³, SA:V = 3/r. You may be asked to calculate these ratios and then explain biological consequences, such as why large active animals need lungs, gills, or circulatory systems while single-celled organisms rely on diffusion alone.
对于边长为 L 的立方体：表面积 = 6L²，体积 = L³，因此 SA:V = 6/L。随 L 加倍，SA:V 减半。对于半径为 r 的球体：表面积 = 4πr²，体积 = (4/3)πr³，SA:V = 3/r。你可能被要求计算这些比率，然后解释生物学后果，比如为何大型活跃动物需要肺、鳃或循环系统，而单细胞生物仅依靠扩散。
3. Fick’s Law of Diffusion | 菲克扩散定律
Fick’s Law describes the rate of diffusion across a membrane or exchange surface. It states that the rate is directly proportional to the surface area and the concentration difference, and inversely proportional to the thickness of the diffusion pathway. This equation is conceptual rather than computational in most Pre-U contexts, but you must be able to explain how structural adaptations maximise diffusion rate by modulating these three factors. Alveoli, villi, and gill lamellae all exemplify these principles.
菲克定律描述了跨膜或交换表面的扩散速率。它阐明速率与表面积和浓度差成正比，与扩散通路厚度成反比。在大多数 Pre-U 语境中，该方程侧重概念而非计算，但你须能解释结构适应如何通过调节这三个因素使扩散速率最大化。肺泡、小肠绒毛和鳃薄片均是这些原理的实例。
Rate of diffusion ∝ (Surface area × Concentration difference) ÷ Thickness of membrane
A thin, moist alveolar wall with a large total surface area and a steep oxygen gradient between alveolar air and blood ensures rapid gas exchange. When answering exam questions, avoid simply stating the law; instead, relate each component to a specific anatomical feature. For example, the extensive capillary network maintains a steep concentration gradient by rapidly carrying oxygen away.
薄而湿润的肺泡壁、巨大的总表面积，加上肺泡气与血液之间陡峭的氧浓度梯度，确保了快速气体交换。回答考试问题时，避免简单陈述该定律；而应将每个组成部分关联到具体的解剖特征。例如，广泛的毛细血管网络通过迅速带走氧气维持了陡峭的浓度梯度。
4. Osmosis and Water Potential | 渗透作用与水势
Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential across a partially permeable membrane. Water potential (Ψ) is measured in kilopascals (kPa) and pure water has a water potential of zero by definition. Solute addition lowers water potential into negative values, while increased pressure raises it. The relationship between solute concentration and water potential can be inferred from calibration curves, and you must be able to determine water potential from experimental data.
渗透作用是水分子通过部分透性膜从水势较高区域向水势较低区域的净移动。水势 (Ψ) 以千帕 (kPa) 为单位，根据定义纯水的水势为零。溶质添加使水势降低为负值，而增加压力则使其升高。溶质浓度与水势之间的关系可从校准曲线推断，并且你需能根据实验数据测定水势。
Ψ = Ψₛ + Ψₚ (solute potential + pressure potential)
In a typical plant tissue practical, you plot percentage change in mass against sucrose concentration, and the point where the line crosses zero change corresponds to the solution where Ψ of the tissue equals Ψ of the external solution. You then convert this concentration to a Ψ value using a provided table. Examiners frequently test your understanding that at incipient plasmolysis, the pressure potential is zero, so Ψ = Ψₛ of the cell contents.
在典型的植物组织实验中，你将质量变化百分比相对蔗糖浓度作图，质量零变化点对应的溶液即为组织 Ψ 等于外部溶液 Ψ 的位置。随后你利用提供的表格将该浓度转换为 Ψ 值。考官常会考察你对初始质壁分离时压力势为零、因而 Ψ 等于细胞内容物 Ψₛ 的理解。
5. Mark-Release-Recapture (Lincoln Index) | 标记-释放-重捕法（林肯指数）
The Lincoln Index estimates the population size of motile species where a direct count is impractical. The method involves capturing a sample, marking them harmlessly, releasing them back to mix randomly with the population, then capturing a second sample after a suitable interval. The ratio of marked to unmarked individuals in the second sample allows estimation of total population size. Several assumptions underpin reliability: marks must not be lost, mortality and emigration must be equal between marked and unmarked individuals, and mixing must be complete.
林肯指数用于估算直接计数不现实的活动物种的种群大小。该方法包括捕获一个样本、无害标记后放回使其与种群随机混合，然后间隔合适时间后捕获第二个样本。第二个样本中标记与未标记个体的比例，可以估算出种群总规模。若干假设支撑其可靠性：标记不可脱落、标记与未标记个体之间的死亡率与迁出率必须相等，混合必须充分。
N = (M × C) ÷ R
Where N = estimated total population, M = number marked in the first sample, C = total caught in the second sample, R = number of marked individuals recaptured in the second sample. For example, if you initially catch and mark 40 woodlice, release them, then later catch 50 woodlice of which 10 are marked, then N = (40 × 50) ÷ 10 = 200. You may also encounter variations that account for population changes, but the basic form suffices for WJEC Pre-U.
其中 N = 估算总种群数，M = 第一个样本中的标记数量，C = 第二个样本中捕获的总数，R = 第二个样本中重捕的标记个体数。例如，如果你初始捕获标记了 40 只鼠妇，放回后稍后捕获了 50 只，其中 10 只带有标记，则 N = (40 × 50) ÷ 10 = 200。你可能遇到考虑种群变化的变体，但 WJEC Pre-U 使用基本形式即可。
6. Population Growth Rate | 种群增长率
Population growth rate quantifies how a population changes over a given period, accounting for births, deaths, immigration, and emigration. The balance of these four factors determines whether a population increases, decreases, or remains stable. In closed systems like laboratory cultures, immigration and emigration are zero, so the growth rate depends entirely on birth rate versus death rate. Understanding this equation is essential for interpreting population growth curves and discussing factors that limit population size.
种群增长率量化了一个种群在给定时期内的变化，需考虑出生、死亡、迁入与迁出。这四个因素的平衡决定了种群是在增长、缩减还是保持稳定。在实验室培养等封闭系统中，迁入与迁出为零，因此增长率完全取决于出生率与死亡率的对比。理解此方程对解读种群增长曲线和讨论限制种群大小的因素至关重要。
Population growth rate = (Births + Immigration) – (Deaths + Emigration)
When expressed as a per capita rate, r = (B – D) / N, where B is births per time period, D is deaths per time period, and N is the current population size. If r is positive, the population grows; if negative, it declines. Carrying capacity (K) is the maximum population size an environment can sustain, and growth typically slows as N approaches K due to density-dependent factors like competition, disease, and resource depletion.
当以人均率表示时，r = (B – D) / N，其中 B 为每个时间段的出生数，D 为每个时间段的死亡数，N 为当前种群规模。若 r 为正，种群增长；若为负，则下降。环境容纳量 (K) 是环境可维持的最大种群规模，随着 N 趋近 K，由于竞争、疾病、资源枯竭等密度制约因素，增长通常会减缓。
7. Net Primary Production (NPP) | 净初级生产量 (NPP)
Net primary production is the rate at which plants store chemical energy in their biomass after accounting for their own respiratory losses. It represents the energy available to the next trophic level—herbivores and decomposers. Gross primary production (GPP) is the total energy fixed by photosynthesis. The difference between GPP and plant respiration (R) gives NPP. This is a crucial measure in ecosystem energetics and frequently appears in questions involving energy transfer efficiency between trophic levels.
净初级生产量是植物扣除自身呼吸损耗后在生物量中储存化学能的速率。它代表了可供下一个营养级——食草动物与分解者——利用的能量。总初级生产量 (GPP) 是光合作用固定的总能量。GPP 与植物呼吸作用 (R) 之差即为 NPP。这是生态系统能量学中的关键指标，频繁出现在涉及营养级间能量传递效率的问题中。
NPP = GPP – R
All three variables are typically expressed in energy units per unit area per unit time, such as kJ m⁻² year⁻¹. A typical exam question might provide GPP and R values and ask you to calculate NPP, then ask what percentage of GPP is available to primary consumers. Remember that some energy is also lost as heat and in indigestible materials, so only a fraction of NPP is actually ingested and assimilated by herbivores.
这三个变量通常都以单位面积单位时间的能量单位表示，例如 kJ m⁻² year⁻¹。典型考试题可能给出 GPP 和 R 值，要求你计算 NPP，然后询问 GPP 中有多大百分比可供初级消费者利用。请记住，部分能量还以热能和不可消化物质的形式散失，因此 NPP 中仅有一部分被食草动物实际摄入和同化。
8. Exponential and Logistic Growth Models | 指数型与逻辑型增长模型
Exponential growth occurs when resources are unlimited, producing a J-shaped curve. Logistic growth incorporates carrying capacity, producing an S-shaped (sigmoid) curve, which more realistically models most natural populations. The mathematical descriptions allow you to predict population size at any time and to identify the point of maximum growth rate, which is crucial for sustainable harvesting strategies and conservation biology.
指数增长发生在资源无限时，产生 J 形曲线。逻辑增长纳入了环境容纳量，产生 S 形（钟形）曲线，更真实地模拟了大多数自然种群。数学描述使你能预测任何时间的种群规模，并确定最大增长速率点，这对可持续收获策略与保护生物学至关重要。
Exponential: dN/dt = rN | Logistic: dN/dt = rN [(K – N) / K]
For the logistic model, when N is very small, [(K – N) / K] approximates 1, so growth is nearly exponential. When N = K/2, the growth rate is maximal, and this point is the maximum sustainable yield point in fisheries management. When N = K, growth rate equals zero and the population stabilises. You are not required to solve differential equations in the exam, but you should be able to interpret these equations and link them to shape features of growth curves.
对于逻辑模型，当 N 很小时，[(K – N) / K] 约等于 1，因此增长近乎指数型。当 N = K/2 时，增长率达到最大，该点即为渔业管理中的最大可持续产量点。当 N = K 时，增长率等于零，种群稳定。考试不要求你解微分方程，但你应能解读这些方程并将其与生长曲线的形状特征联系起来。
9. Hardy-Weinberg Principle | 哈迪-温伯格原理
The Hardy-Weinberg principle states that allele and genotype frequencies within a large, randomly mating population remain constant from generation to generation in the absence of evolutionary influences. It provides a null hypothesis against which evolution can be detected. The principle applies only when five conditions are met: no mutation, random mating, no natural selection, extremely large population size, and no gene flow. Calculating allele and genotype frequencies using the two equations is a core Pre-U skill.
哈迪-温伯格原理阐明，在一个缺乏进化影响的随机交配大种群中，等位基因与基因型频率世代间保持恒定。它提供了一个无效假设，进化可借此被检测到。该原理仅在满足五个条件时适用：无突变、随机交配、无自然选择、极大种群规模、无基因流动。运用两个方程计算等位基因与基因型频率是 Pre-U 的核心技能。
p + q = 1 | p² + 2pq + q² = 1
Where p = frequency of the dominant allele, q = frequency of the recessive allele. In the second equation, p² represents the frequency of homozygous dominant, 2pq the heterozygous, and q² the homozygous recessive genotype. A typical problem provides the frequency of the recessive phenotype (which equals q²), and you must calculate q, then p, then genotype frequencies. Remember that the square root of q² gives q, and p = 1 – q. Always check your values—p and q must sum to 1.
其中 p = 显性等位基因频率，q = 隐性等位基因频率。在第二个方程中，p² 表示纯合显性基因型频率，2pq 为杂合子，q² 为纯合隐性。典型问题会给出隐性表型频率（等于 q²），你必须计算 q，然后是 p，再计算基因型频率。记住 q² 的平方根等于 q，而 p = 1 – q。务必检查数值——p 与 q 必须总和为 1。
10. The Chi-Squared Test | 卡方检验
The chi-squared (χ²) test determines whether observed categorical data deviate significantly from expected values. In genetics, it tests whether experimental ratios match Mendelian predictions. In ecology, it can test whether species are uniformly distributed or associated. The test calculates a χ² value and compares it to a critical value from a distribution table at a chosen probability level (typically p = 0.05) with appropriate degrees of freedom. A χ² value larger than the critical value indicates a significant difference, rejecting the null hypothesis.
卡方 (χ²) 检验决定观察到的分类数据是否显著偏离预期值。在遗传学中，它检验实验比率是否符合孟德尔预测。在生态学中，它可检验物种种群分布是否均匀或存在关联。该检验计算 χ² 值，并将其与选定概率水平（通常 p = 0.05）下、适当自由度的分布表临界值进行比较。χ² 值大于临界值表明存在显著差异，拒绝无效假设。
χ² = Σ [(O – E)² ÷ E]
Where O = observed frequency, E = expected frequency. Degrees of freedom typically equal the number of categories minus one, but in tests of association in contingency tables, df = (rows – 1) × (columns – 1). You will be provided with a critical value table in the exam, but you need to know how to state clear null and alternative hypotheses, calculate χ² step-by-step, interpret the result, and write a biological conclusion. Common pitfalls include forgetting to use actual numbers and not proportions, and miscalculating degrees of freedom.
其中 O = 观察频数，E = 预期频数。自由度通常等于类别数减一，但在列联表关联性检验中，df = (行数 – 1) × (列数 – 1)。考试会提供临界值表，但你需知道如何陈述清晰的无效与备择假设、逐步计算 χ²、解释结果以及写出生物学结论。常见错误包括忘记使用实际频数而非比例，以及错误计算自由度。
11. pH and the Hydrogen Ion Concentration | pH 与氢离子浓度
pH is a logarithmic measure of hydrogen ion concentration, fundamental to understanding enzyme activity, buffering in blood, and transport across membranes. A small change in pH represents a large change in H⁺ concentration due to the logarithmic scale—each unit change represents a tenfold difference. The Pre-U specification expects you to understand the relationship quantitatively, linking pH to the likelihood of enzyme denaturation and the importance of maintaining stable pH for biological processes.
pH 是氢离子浓度的对数度量，对理解酶活性、血液缓冲作用及跨膜运输至关重要。由于使用对数标度，pH 的微小变化即代表氢离子浓度的巨大变化——每个单位变化代表十倍的差异。Pre-U 教学大纲期望你定量地理解这一关系，将 pH 与酶变性可能以及维持生物过程稳定 pH 的重要性相联系。
pH = –log₁₀[H⁺]
A solution with a hydrogen ion concentration of 1×10⁻⁷ mol dm⁻³ has a pH of 7. If [H⁺] increases to 1×10⁻⁶ mol dm⁻³, the pH drops to 6—a tenfold increase in hydrogen ion concentration. You should be comfortable converting between [H⁺] and pH. Additionally, understand the carbonic acid-bicarbonate buffer system that maintains blood pH around 7.4: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻.
氢离子浓度为 1×10⁻⁷ mol dm⁻³ 的溶液 pH 为 7。如果 [H⁺] 升高到 1×10⁻⁶ mol dm⁻³，pH 将降至 6——氢离子浓度增加了十倍。你应能熟练在 [H⁺] 与 pH 之间进行换算。此外，要理解维持血液 pH 约 7.4 的碳酸-碳酸氢盐缓冲系统：CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻。
12. Cardiac Output Formula | 心输出量公式
Cardiac output is the total volume of blood pumped by a ventricle per minute. It is the product of heart rate (beats per minute) and stroke volume (volume of blood ejected per beat). This formula is crucial for explaining how the cardiovascular system meets varying metabolic demands during rest and exercise. Factors influencing cardiac output are central to circulatory physiology and appear regularly in Pre-U questions about homeostasis, exercise, and cardiovascular disease.
心输出量是一个心室每分钟泵出的血液总量。它是心率（每分钟搏动次数）与每搏输出量（每次搏动射血量）的乘积。此公式对于解释心血管系统如何满足休息和运动期间变化不定的代谢需求至关重要。影响心输出量的因素是循环生理学的核心，经常出现在 Pre-U 关于稳态、运动与心血管疾病的问题中。
Cardiac output = Heart rate × Stroke volume
At rest, a typical cardiac output is about 5 dm³ min⁻¹ (heart rate 70 bpm × stroke volume ~70 cm³). During intense exercise, cardiac output can increase to over 20 dm³ min⁻¹ in trained athletes through increases in both heart rate and stroke volume. Stroke volume depends on venous return (Starling’s Law), the force of myocardial contraction, and afterload. Understanding how sympathetic and parasympathetic nerves, adrenaline, and intrinsic mechanisms regulate cardiac output is linked directly to this formula.
静息时，典型心输出量约为 5 dm³ min⁻¹（心率 70 bpm × 每搏输出量约 70 cm³）。剧烈运动时，经训练运动员的心输出量可增至 20 dm³ min⁻¹ 以上，通过提升心率与每搏输出量共同实现。每搏输出量取决于静脉回流（斯塔林定律）、心肌收缩力以及后负荷。理解交感与副交感神经、肾上腺素及内在机制如何调节心输出量，与此公式直接关联。
13. Pulmonary Ventilation Rate | 肺通气量公式
Pulmonary ventilation rate quantifies the volume of air moved in and out of the lungs per minute. It is determined by the tidal volume (the volume of air per breath) and the ventilation rate (breaths per minute). This measure reflects the overall effectiveness of gas exchange and is tightly regulated to maintain blood oxygen and carbon dioxide levels. Spirometer traces and calculations based on them are standard practical elements in the Pre-U syllabus.
肺通气量量化了每分钟进出肺部的空气量。它由潮气量（每次呼吸的空气量）和通气率（每分钟呼吸次数）决定。该指标反映了气体交换的整体效能，并受到严格调节以维持血液氧与二氧化碳水平。肺活量计图迹及其相关计算是 Pre-U 教学大纲中的标准实践要素。
Pulmonary ventilation = Tidal volume × Ventilation rate
Tidal volume at rest is typically around 0.5 dm³, and ventilation rate about 12 breaths per minute, so pulmonary ventilation = 6 dm³ min⁻¹. This can rise dramatically during exercise. You may also need to calculate other lung volumes from spirometer data: vital capacity, inspiratory reserve volume, expiratory reserve volume, and residual volume. Remember that the total lung capacity = vital capacity + residual volume, and be prepared to interpret traces showing changes during exercise or disease.
静息时潮气量通常约为 0.5 dm³，通气率约每分钟 12 次，故肺通气量 = 6 dm³ min⁻¹。运动期间此值可显著增加。你或许还需根据肺活量计数据计算其他肺容量：肺活量、补吸气量、补呼气量和残气量。记住肺总容量 = 肺活量 + 残气量，并准备好解读显示运动或疾病期间变化的图迹。
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