📚 Year 11 OCR Biology: Formulas & Theorems Quick Reference Handbook | Year 11 OCR 生物公式定理速查手册
This handbook gathers the essential formulas, equations and key biological theorems required for Year 11 OCR Biology. Whether you are tackling magnification calculations, analysing energy transfers in ecosystems or revising the fundamental principles of life, this reference guide provides clear statements, worked examples and practical tips to help you master numerical skills and conceptual understanding.
本手册汇集了 Year 11 OCR 生物课程中必备的公式、方程式和核心生物学定理。无论你是在应对显微镜放大倍数计算、分析生态系统的能量传递，还是复习生命的基本原理，这本速查手册都提供了清晰的陈述、实例解析和实用提示，帮助你掌握计算技能和概念理解。
1. The Cell Theory | 细胞学说
All living organisms are composed of one or more cells; the cell is the basic structural and functional unit of life; all cells arise from pre-existing cells by division. This principle underpins the entire study of biology, linking structure to function across all kingdoms of life.
所有生物体由一个或多个细胞组成；细胞是生命的基本结构和功能单位；所有细胞都源自已有细胞的分裂。这条原理支撑着整个生物学的研究，将结构与功能连接到所有生物界。
Viruses are not considered living under the cell theory because they lack cellular organisation and cannot reproduce independently. The theory has been refined by advances in microscopy and biochemistry, but its core statements remain central to OCR exam questions on cell structure and origins.
根据细胞学说，病毒不被视为生物，因为它们缺乏细胞组织且不能独立繁殖。该学说随着显微镜和生物化学的进步而不断完善，但其核心论述仍然是 OCR 考试中关于细胞结构和起源问题的中心内容。
2. Magnification Formula | 放大倍数公式
mage size ÷ Actual size. Rearranging this triangle allows you to calculate any missing value when the other two are known. Both image size and actual size must be in the same units before division.
放大倍数、图像尺寸和实际尺寸之间的关系由公式 放大倍数 = 图像尺寸 ÷ 实际尺寸 给出。通过重排这个三角关系，你可以在已知其中两个量的情况下计算缺失的数值。图像尺寸和实际尺寸必须使用相同单位后再进行除法运算。
M = I ÷ A
size, A = actual size. Typical units: image in millimetres (mm) converted to micrometres (µ age measuring 24 mm, and the actual cell is 0.006 mm, magnification = 24 ÷ 0.006 = 4000 ×.
其中 M = 放大倍数，I = 图像尺寸，A = 实际尺寸。常见单位：图像以毫米（mm）计，需转化为微米（µm）乘以 1000，而实际尺寸常以微米给出。如果显微照片显示细胞图像大小为 24 mm，而实际细胞为 0.006 mm，则放大倍数 = 24 ÷ 0.006 = 4000 ×。
practicals, you will measure images with a ruler and use scale bars to verify your calculations.
请记住：不带单位的放大倍数（如 4000×）只是一个数字。在重排公式时务必展示计算步骤；计算实际尺寸时 A = I ÷ M。在 OCR 实验操作中，你将用直尺测量图像，并使用比例尺来验证你的计算。
3. Actual Size from Scale Bar | 利用比例尺求实际尺寸
ruler and applying: Actual size of object = (measured length of object ÷ measured length of scale bar) × value represented by scale bar.
许多示意图和显微照片包含一个显示已知真实长度的比例尺。实际尺寸可以通过用直尺测量比例尺长度并应用以下公式来确定：物体的实际尺寸 = (测量的物体长度 ÷ 测量的比例尺长度) × 比例尺代表的数值。
Aₒ = (Lₒ ÷ Lₛ) × S
S = stated real size of scale bar. For example, scale bar measures 15 mm and represents 10 µm; an organelle measures 30 mm in the same image. Its actual size = (30 ÷ 15) × 10 µm = 20 µm.
其中 Aₒ = 物体的实际尺寸，Lₒ = 物体的测量长度，Lₛ = 比例尺的测量长度，S = 比例尺标注的真实尺寸。例如，比例尺测量为 15 mm 代表 10 µm；同一图像中某细胞器测量为 30 mm。其实际尺寸 = (30 ÷ 15) × 10 µm = 20 µm。
This method avoids common unit conversion errors and is frequently tested in OCR B1–B3 cell biology contexts. Always check that the object and scale bar are measured in the same unit and convert the final answer to the most appropriate unit (µm or mm).
这种方法可以避免常见的单位转换错误，并在 OCR B1–B3 细胞生物学内容中经常被考查。务必确保物体和比例尺的测量使用相同单位，并将最终答案转换为最合适的单位（µm 或 mm）。
4. Cardiac Output Equation | 心输出量公式
n) = Stroke volume (cm³) × Heart rate (beats per minute). This relationship reveals how the cardiovascular system meets the body's oxygen demands.
心输出量是指一个心室每分钟泵出的血液体积。它由每搏输出量和心率计算得出：心输出量（cm³/min）= 每搏输出量（cm³）× 心率（每分钟心跳数）。这一关系揭示了心血管系统如何满足身体的氧气需求。
CO = SV × HR
to 21 600 cm³/min. Units: always ensure stroke volume is in cm³ (or mL, as 1 cm³ = 1 mL) and heart rate in bpm.
一名静息状态下的运动员，SV 可能为 70 cm³，HR 为 60 bpm，则 CO = 4200 cm³/min。运动时，SV 可增加至 120 cm³，HR 增至 180 bpm，使 CO 升到 21 600 cm³/min。单位：务必确保每搏输出量使用 cm³（或 mL，因为 1 cm³ = 1 mL），心率使用 bpm。
OCR questions often ask you to calculate one variable when the other two are given, or to explain changes in cardiac output during physical activity. Remember that cardiac output is different for left and right ventricles under normal circumstances; they are equal when averaged over time but may be given for the left ventricle in questions.
OCR 试题通常要求根据另外两个变量计算其中一个变量，或解释体力活动期间心输出量的变化。记住，正常情况下左右心室的心输出量不同；但在长时间平均下是相等的，题目中通常给出左心室的数据。
5. Respiratory Quotient (RQ) | 呼吸商
gen consumed: RQ = CO₂ produced ÷ O₂ consumed. Different substrates yield characteristic RQ values: carbohydrate 1.0, lipid around 0.7, protein approximately 0.9.
呼吸商指示有氧呼吸中利用的底物类型。它被计算为产生的二氧化碳与消耗的氧气的比值：RQ = 产生的 CO₂ ÷ 消耗的 O₂。不同底物产生特征性的 RQ 值：碳水化合物 1.0，脂类约 0.7，蛋白质约 0.9。
RQ = CO₂ produced / O₂ consumed
ced is 0.7 cm³, RQ = 0.7, indicating lipid. Germinating seeds often show an intermediate RQ as they use mixed substrates.
如果呼吸计实验显示使用 0.8 cm³ O₂并产生 0.8 cm³ CO₂，则 RQ = 1.0，表明底物为碳水化合物。如果 O₂ 消耗为 1.0 cm³，CO₂ 产生为 0.7 cm³，则 RQ = 0.7，表明底物为脂类。正在萌发的种子通常显示出中间 RQ 值，因为它们利用混合底物。
RQ is only meaningful for aerobic respiration; anaerobic respiration produces CO₂ without O₂ uptake, making the ratio theoretically infinite. In OCR practicals, you will use simple respirometers with soda lime to absorb CO₂ and measure O₂ uptake, then calculate CO₂ from a parallel setup.
RQ 仅对有氧呼吸有意义；无氧呼吸产生 CO₂ 而不消耗 O₂，使得理论上比值为无穷大。在 OCR 实验操作中，你会使用装有碱石灰的简易呼吸计来吸收 CO₂ 并测量 O₂ 消耗量，然后通过平行装置计算 CO₂ 生成量。
6. Rate of Photosynthesis | 光合作用速率
Photosynthesis rate can be approximated by measuring oxygen production or the time taken for a standard indicator change. A common laboratory method uses the time taken for leaf discs to float as they produce oxygen: Rate ∝ 1 ÷ time taken (s⁻¹). A faster rise indicates a higher rate of photosynthesis.
光合作用速率可以通过测量氧气产出或标准指示剂变化所需的时间来估算。一种常见的实验室方法利用叶圆片上浮的时间，因为它们产生氧气：速率 ∝ 1 ÷ 所用时间（s⁻¹）。上浮越快表明光合作用速率越高。
Rate = 1 / t (where t = time for discs to float or colour change)
min⁻¹. More oxygen bubbles counted per minute also give a direct rate. Always state the unit: e.g. bubbles per minute or s⁻¹.
在碳酸氢盐指示剂实验中，随着 CO₂ 被消耗，指示剂变为紫色。如果颜色变化需要 5 分钟，速率 = 1/5 = 0.2 min⁻¹。每分钟计数的氧气泡数也会给出直接速率。务必注明单位：例如每分钟气泡数或 s⁻¹。
OCR may ask you to interpret graphs of rate against light intensity, carbon dioxide concentration or temperature. Remember that rate increases until another factor becomes limiting. When solving rate calculations, always identify which variable is being measured and keep units consistent.
OCR 可能会要求你解释速率随光照强度、二氧化碳浓度或温度变化的曲线图。牢记速率会增加，直到另一因子成为限制因素。在求解速率计算时，一定要识别被测量的变量并保持单位一致。
7. Energy Transfer Efficiency | 能量传递效率
s, the efficiency of energy transfer between trophic levels is calculated as: Efficiency (% all percentage is transferred because energy is lost through respiration, waste, and heat.
在生态系统中，营养级之间的能量传递效率计算如下：效率 (%) = (下一营养级生物量中的能量 ÷ 上一营养级生物量中的能量) × 100。只有很小百分比被传递，因为能量通过呼吸、排泄物和热能形式损耗。
E (%) = (Eₙ ÷ Eₚ) × 100
contain 20 000 kJ and herbivores contain 2 000 kJ, efficiency = (2000 ÷ 20000) × 100 = 10%. Typical efficiencies are around 10%, though they can be lower.
其中 Eₙ = 下一营养级生物量中的能量，Eₚ = 上一营养级的能量。例如，如果植物含有 20 000 kJ，而食草动物含有 2 000 kJ，则效率 = (2000 ÷ 20000) × 100 = 10%。典型的效率约为 10%，但可能更低。
OCR often supplies data in tables of biomass or energy content. You may be required to calculate efficiency or explain why it is not 100%. Losses occur as metabolic heat, undigested material egested and urea excreted. Understanding this equation helps when drawing pyramids of biomass or energy.
OCR 时常提供生物量或能量含量的数据表格。你可能需要计算效率或解释为什么它不是 100%。能量损耗以代谢热、未消化的排出物和排出的尿素等形式发生。理解这个公式有助于绘制生物量金字塔或能量金字塔。
8. Mark-Release-Recapture (Lincoln Index) | 标记重捕法（林肯指数）
ample) ÷ Number of marked individuals recaptured. This method assumes the population is closed, marking does not affect survival, and mixing is random.
为了估算移动物种的种群大小，生态学家使用林肯指数：种群大小 = (第一次取样数量 × 第二次取样数量) ÷ 重捕的标记个体数。该方法假设种群是封闭的，标记不影响生存，且个体充分混合。
N = (M × C) / R
iduals in second sample. If 30 woodlice are marked and released, and later 25 woodlice are caught including 5 marked, N = (30 × 25) / 5 = 150.
其中 N = 估计的总种群大小，M = 第一次捕捉并标记的数量，C = 第二次捕捉的总数，R = 第二次捕捉中带标记的数量。如果标记并释放 30 只鼠妇，随后捕获 25 只其中 5 只有标记，则 N = (30 × 25) / 5 = 150。
OCR fieldwork questions require you to evaluate the assumptions. If marking makes individuals more visible to predators, R decreases, leading to overestimation. Alternatively, if marks are lost, R is underestimated and N overestimated. Always suggest improvements to method when asked.
OCR 的野外调查题要求你评估这些假设。如果标记使个体更容易被捕食者发现，R 会减小，导致种群被高估。反之，如果标记脱落，R 被低估而 N 被高估。当被提问时，务必对方法提出改进建议。
9. The Theory of Natural Selection | 自然选择理论
Evolution by natural selection is the central theorem explaining the diversity of life. It states that individuals within a population show genetic variation, they produce more offspring than the environment can support, there is a struggle for survival, and those with advantageous alleles are more likely to survive and reproduce, passing on the beneficial traits to the next generation.
自然选择进化是解释生命多样性的核心定理。它指出，一个种群内的个体表现出遗传变异，它们产生的后代数量超过环境所能支持的程度，为生存而斗争，而具有有利等位基因的个体更有可能存活并繁殖，将有利性状传递给下一代。
Over generations, the frequency of these advantageous alleles increases, leading to the adaptation of the population to its environment. OCR expects you to apply this theorem to examples like antibiotic resistance in bacteria, beak shape in finches, or fur colour in pocket mice. The theorem requires no calculation but provides the logical framework for interpreting genetic change.
经过多代，这些有利等位基因的频率增加，导致种群对其环境的适应。OCR 期望你将这一定理应用于诸如细菌抗生素耐药性、雀鸟喙形或鼠类毛色等实例。这个定理无需计算，但为解释遗传变化提供了逻辑框架。
Key terms: variation (from mutation and sexual reproduction), competition, differential survival, and heritability. A common pitfall is stating that organisms 'wish' to evolve or that they adapt during their lifetime; instead, selection acts on pre-existing variation in populations.
关键术语：变异（源于突变和有性生殖）、竞争、差异性存活和遗传力。一个常见误区是说生物“希望”进化或它们在一生中适应；实际上，自然选择作用于种群中预先存在的变异。
10. Enzyme Activity Rate | 酶活性速率
The rate of an enzyme‑ ntity of product formed ÷ Time taken. Alternatively, in iodine-starch investigations, Rate ∝ 1 ÷ time taken for the colour to change.
酶控反应的速率可以通过单位时间内产物的生成量或底物的消耗量来衡量。常见表达式：速率 = 生成的产物量 ÷ 所用时间。此外，在碘-淀粉实验中，速率 ∝ 1 ÷ 颜色发生变化所需的时间。
Rate = Δ[Product] / Δt
seconds, Rate = 1/120 ≈ 0.0083 s⁻¹. OCR requires you to calculate rates from tables or graphs and to explain the effect of temperature and pH on rate by considering enzyme denaturation and kinetic energy.
如果在 5 分钟内产生 10 mg 麦芽糖，速率 = 10 ÷ 5 = 2 mg/min。当使用定性碘检测时，如果蓝黑色在 120 秒内消退，速率 = 1/120 ≈ 0.0083 s⁻¹。OCR 要求你根据表格或图表计算速率，并通过考虑酶变性和动能来解释温度和 pH 对速率的影响。
Always label axes and provide units in derived rates. At optimum temperature, rate is highest; beyond the optimum, the rate decreases sharply as the active site denatures. For pH, extreme values disrupt ionic bonds, reducing rate. This theorem-like understanding of enzyme specificity is essential alongside numerical calculations.
始终标注坐标轴并在导出速率中提供单位。在最适温度下，速率最高；超过最适温度，由于活性部位变性，速率急剧下降。就 pH 而言，极端值会破坏离子键，降低速率。这种关于酶特异性的定理性理解与数值计算同样重要。
11. Surface Area to Volume Ratio Theorem | 表面积与体积比定理
sed exchange surfaces and transport systems, while single-celled organisms can rely on diffusion alone. The relationship can be expressed as SA:V = surface area ÷ volume.
随着生物体或细胞尺寸增大，其表面积与体积比会减小。这一原理解释了为什么大型生物需要特化的交换表面和运输系统，而单细胞生物可以仅靠扩散生存。这种关系可表示为 SA:V = 表面积 ÷ 体积。
SA:V = SA / V
. Smaller organisms have a higher SA:V, which facilitates rapid gas exchange and waste removal. OCR exams may ask you to calculate SA:V ratios and relate them to adaptations such as alveoli, villi and root hairs.
对于一个边长为 1 cm 的立方体，表面积 = 6 cm²，体积 = 1 cm³，SA:V = 6:1。对于 3 cm 的立方体，SA = 54 cm²，V = 27 cm³，SA:V = 2:1。较小的生物具有较高的 SA:V，这有利于快速气体交换和废物排除。OCR 考试可能会要求你计算 SA:V 比值，并将它们与肺泡、绒毛和根毛等适应性结构联系起来。
The theorem is not a single formula but a unifying concept: metabolic demand is proportional to volume, while supply depends on surface area. Thus, large active organisms have evolved internal transport networks like the circulatory system and structures with folded surfaces to increase functional SA.
这个定理并非一个简单的公式，而是一个统一的概念：代谢需求与体积成正比，而物质供应取决于表面积。因此，大型活跃的生物进化出了内部运输网络，比如循环系统，以及具有折叠表面的结构以增加功能性表面积。
12. Probability and Genetic Ratios (Mendelian Inheritance) | 概率和遗传比例（孟德尔遗传）
) are crossed, and the alleles segregate, the expected genotypic ratio is 1 BB : 2 Bb : 1 bb, and the phenotypic ratio for complete dominance is 3 dominant : 1 recessive. These ratios are based on the laws of segregation and independent assortment.
尽管不是代数意义上的公式，孟德尔遗传遵循可预测的比例。当两个杂合个体（例如 Bb × Bb）杂交，等位基因分离时，预期的基因型比例为 1 BB : 2 Bb : 1 bb，而在完全显性情况下，表型比例为 3 显性 : 1 隐性。这些比例基于分离定律和自由组合定律。
ge. For each offspring from a Bb × Bb cross, probability of dominant phenotype = 3/4 or 75% s, useful in dihybrid crosses (e.g. 9 : 3 : 3 : 1 ratio from two heterozygous gene pairs).
概率可以用分数或百分比表示。对于 Bb × Bb 杂交的每个后代，显性表型的概率 = 3/4 即 75%。乘积法则指出，独立事件共同发生的概率是它们各自概率的乘积，这在双基因杂交中很有用（例如来自两对杂合基因的 9 : 3 : 3 : 1 比例）。
OCR expects you to use Punnett squares to determine outcomes and interpret pedigree charts. Understanding these ratios allows you to predict the likelihood of genetic disorders. Even without a numerical 'theorem', these principles function as mathematical foundations of inheritance.
OCR 期望你使用庞尼特方格确定杂交结果并解释系谱图。理解这些比例使你能够预测遗传疾病的概率。即使没有数字形式的“定理”，这些原则作为遗传的数学基础发挥重要作用。
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