📚 Year 11 OCR Biology: Quick Check Formulas & Principles | 年11 OCR生物：公式定理速查手册
This concise reference gathers all the essential formulas, calculation methods, and key principles you need for Year 11 OCR Biology. From magnification to biomass transfer efficiency, each entry is presented with a clear equation, a worked example, and the biological context. Whether you are revising for an end-of-topic test or preparing for your GCSE examinations, use this handbook to strengthen your numerical confidence and conceptual understanding.
这份速查手册汇集了年11 OCR生物学中所有必背公式、计算方法和核心原理。从放大率到生物量传递效率，每个条目都配有清晰的方程、计算示例和生物学情境。无论你是在准备单元测验还是GCSE大考，都可以用这本手册提升计算信心和概念理解。
1. Magnification Formula | 放大率公式
Magnification describes how many times larger an image appears compared with the real object. The formula relates image size, actual size, and magnification. Always convert all measurements to the same unit (usually millimetres or micrometres) before substituting into the equation.
放大率描述图像比实际物体放大了多少倍。公式将图像大小、实际大小和放大率联系起来。代入方程前，务必把所有测量值换算为相同单位（通常用毫米或微米）。
Magnification = Image Size ÷ Actual Size
- Image size is measured with a ruler on a micrograph or drawing.
- Actual size is the real length of the specimen.
- 图像大小用直尺在显微照片或绘图上测量。
- 实际大小指标本的真实长度。
Worked example: on a micrograph. The actual diameter of the cell is 0.006 mm. Magnification = 6 ÷ 0.006 = ×1000.
计算示例：显微照片上一个红细胞长 6 mm，细胞实际直径为 0.006 mm。放大率 = 6 ÷ 0.006 = ×1000。
You can also rearrange the formula to find actual size: Actual Size = Image Size ÷ Magnification.
你也可以变形公式求出实际大小：实际大小 = 图像大小 ÷ 放大率。
2. Rate of Enzyme-Controlled Reactions | 酶控反应速率
The rate of an enzyme-catalysed reaction is often determined by measuring how much product appears, or how much substrate disappears, per unit time. In class experiments, you may record the time taken for a standard change (e.g. starch breakdown detected by iodine) and use the reciprocal.
酶催化反应的速率通常通过测量单位时间内产物的生成量或底物的消耗量来确定。在课堂实验中，你可以记录标准变化所需的时间（例如用碘液检测淀粉分解），并取倒数。
Rate of Reaction = 1 ÷ Time (s)
If you measure product concentration directly, use:
若直接测量产物浓度，使用：
Rate = Change in Amount of Product ÷ Time
Worked example: A potato catalase disc produces 10 cm³ of oxygen in 50 s. Rate = 10 ÷ 50 = 0.2 cm³/s. Alternatively, the time for iodine to stop turning blue-black is 40 s, so rate = 1/40 = 0.025 s⁻¹.
计算示例：一片土豆过氧化氢酶在 50 s 内产生 10 cm³ 氧气。速率 = 10 ÷ 50 = 0.2 cm³/s。或者，碘液不再变蓝黑色的时间为 40 s，则速率 = 1/40 = 0.025 s⁻¹。
Remember to control pH, temperature, enzyme concentration and substrate concentration when investigating a single factor.
在研究单一因素时，记得控制 pH、温度、酶浓度和底物浓度不变。
3. Rate of Photosynthesis | 光合作用速率
The rate of photosynthesis can be estimated by measuring oxygen production (e.g. counting bubbles from pondweed) or the uptake of carbon dioxide over a set time. Light intensity, carbon dioxide concentration and temperature all affect the rate.
光合作用速率可通过测量氧气产量（如计数水草的气泡）或一定时间内的二氧化碳吸收量来估算。光照强度、二氧化碳浓度和温度都会影响速率。
Rate = Volume of O₂ Produced ÷ Time
In practice, you often use the reciprocal of the time taken for a leaf disc to rise or for a fixed number of bubbles to appear.
实际操作中，常使用叶圆片上浮所需时间或固定气泡数出现时间的倒数。
Rate ∝ 1 ÷ Time (when a standard change is measured)
Worked example: Elodea produces 35 bubbles in 5 minutes. Rate = 35 ÷ 5 = 7 bubbles/min. If the time for 10 bubbles to appear decreases from 80 s to 40 s when a lamp is moved closer, the rate roughly doubles.
计算示例：伊乐藻在 5 分钟内产生 35 个气泡。速率 = 35 ÷ 5 = 7 气泡/分钟。当灯靠近时，出现 10 个气泡的时间从 80 s 降至 40 s，速率大致翻倍。
Graphing rate against light intensity reveals a plateau where another factor (e.g. CO₂) becomes limiting.
绘制速率–光照强度曲线时，会出现平台期，表明另一因素（如 CO₂）成为限制因子。
4. Respiratory Quotient (RQ) | 呼吸商（RQ）
The respiratory quotient (RQ) indicates which substrate is being respired by a cell or organism. It compares the volume of carbon dioxide produced to the volume of oxygen consumed.
呼吸商（RQ）指示细胞或生物体正在呼吸哪种底物。它比较产生的二氧化碳体积与消耗的氧气体积。
RQ = Volume of CO₂ Produced ÷ Volume of O₂ Consumed
- RQ for carbohydrate (glucose) ≈ 1.0
- RQ for lipid ≈ 0.7
- RQ for protein ≈ 0.9
- 碳水化合物（葡萄糖）的 RQ ≈ 1.0
- 脂质的 RQ ≈ 0.7
- 蛋白质的 RQ ≈ 0.9
Worked example: ter shows a mouse consumes 50 cm³ O₂ and produces 40 cm³ CO₂ in 10 minutes. RQ = 40 ÷ 50 = 0.8, suggesting a mixed diet with some lipid.
计算示例：呼吸计显示一只小鼠在 10 分钟内消耗 50 cm³ O₂，产生 40 cm³ CO₂。RQ = 40 ÷ 50 = 0.8，提示混合饮食中含有部分脂质。
RQ values help ecologists infer energy sources in seeds or hibernating animals.
RQ 值可帮助生态学家推断种子或冬眠动物的能量来源。
5. Population Density | 种群密度
Population density is the number of individuals of a species per unit area (or volume, for aquatic habitats). It is a fundamental measure in ecology for comparing habitats or monitoring changes over time.
种群密度是指单位面积（或水生栖息地的单位体积）内某物种的个体数量。它是生态学中比较生境或监测时间变化的基本量度。
Population Density = Total Number of Individuals ÷ Area (m²)
When using quadrats, you often calculate a mean number of individuals per quadrat, then scale up:
使用样方时，通常先计算每个样方的平均个体数，再放大：
Estimated Population = (Mean per Quadrat) × (Total Area ÷ Quadrat Area)
Worked example: 7, 3, 8, 5, 6, 4, 2. Mean = 50 ÷ 10 = 5.0. Estimated daisy population = 5.0 × (500 ÷ 1) = 2500.
计算示例：一个 500 m² 的草地中，十个 1 m² 样方的雏菊数量分别为 4, 6, 5, 7, 3, 8, 5, 6, 4, 2。平均值 = 50 ÷ 10 = 5.0。估计雏菊种群 = 5.0 × (500 ÷ 1) = 2500 株。
Sampling must be random, or along a transect for zonation studies, to avoid bias.
取样必须随机，或沿样带进行以研究带状分布，从而避免偏差。
6. Biomass & Energy Transfer Efficiency | 生物量与能量传递效率
Biomass is the dry mass of living material in an organism or in a trophic level. Energy is passed up food chains, but much is lost through respiration, egestion and uneaten parts. Efficiency calculations show how much is transferred.
生物量是指生物体或某一营养级中活物质的干重。能量沿食物链传递时会因呼吸、排遗和未被取食而大量损耗。效率计算可显示传递了多少。
Efficiency (%) = (Biomass or Energy in Higher Level ÷ Biomass or Energy in Lower Level) × 100
Biomass is often given in g/m² or kg/m². Energy content can be measured by burning material in a calorimeter.
生物量常以 g/m² 或 kg/m² 给出。能量含量可通过在热量计中燃烧材料来测量。
Worked example for biomass efficiency: A field of grass has a biomass of 15 000 kg. Rabbits feeding on the grass gain 900 kg of biomass. Efficiency = (900 ÷ 15 000) × 100 = 6%.
生物量效率计算示例：一片草的生物量为 15 000 kg。以草为食的野兔积累了 900 kg 生物量。效率 = (900 ÷ 15 000) × 100 = 6%。
Worked example for energy efficiency: y; small fish feeding on them contain 120 kJ/m². Transfer efficiency = (120 ÷ 800) × 100 = 15%.
能量效率计算示例：浮游动物含能量 800 kJ/m²，以它们为食的小鱼含 120 kJ/m²。传递效率 = (120 ÷ 800) × 100 = 15%。
7. Surface Area to Volume Ratio | 表面积与体积之比
As an object increases in size, its surface area to volume ratio (SA:V) decreases. This principle explains why large multicellular organisms need specialised exchange surfaces (lungs, gills, leaves) and transport systems.
物体增大时，其表面积与体积之比（SA:V）减小。这一原理可以解释为什么大型多细胞生物需要特化的交换表面（肺、鳃、叶片）和运输系统。
SA:V = Total Surface Area ÷ Volume
For a cube of side length L:
对于边长为 L 的立方体：
Surface Area = 6L², Volume = L³, so SA:V = 6 ÷ L
Worked example: ler cube has a higher SA:V, so it exchanges materials more efficiently by diffusion alone.
计算示例：边长为 1 cm 的立方体 SA = 6 cm²，V = 1 cm³，SA:V = 6:1。边长为 3 cm 的立方体 SA = 54 cm²，V = 27 cm³，SA:V = 2:1。较小的立方体 SA:V 更高，故仅靠扩散交换物质的效率更高。
8. Genetic Crosses & Probability | 遗传杂交与概率
Monohybrid inheritance follows predictable ratios when one gene controls a characteristic. Probabilities can be expressed as fractions, percentages, or ratios. A Punnett square models the random fusion of gametes.
当某性状由单个基因控制时，单基因遗传遵循可预测的比例。概率可用分数、百分数或比例表示。庞纳特方格可模拟配子的随机融合。
cross between two heterozygous parents (Bb × Bb), where B is dominant and b is recessive:
两个杂合亲本杂交（Bb × Bb），B 显性，b 隐性：
- Genotypes: ¼ BB, ½ Bb, ¼ bb
- Phenotype ratio: 3 dominant : 1 recessive
- Probability of recessive offspring = ¼ or 25%
- 基因型：¼ BB，½ Bb，¼ bb
- 表型比例：3 显性 : 1 隐性
- 隐性后代概率 = ¼ 或 25%
Worked example: Two plants heterozygous for tallness (Tt) are crossed. The chance a seedling is short (tt) is 1 in 4, or 0.25 (25%). Over 200 seeds, you expect roughly 50 short plants.
计算示例：两株杂合高茎植株（Tt）杂交。幼苗为矮茎（tt）的概率为 1/4，即 0.25（25%）。在 200 粒种子中，预期约有 50 株矮茎。
Sex determination uses the same principle: XX female, XY male. The probability of a male child is ½ (50%).
性别决定遵循相同原理：XX 女性，XY 男性。生男孩的概率为 ½（50%）。
9. Natural Selection Principles | 自然选择原理
Natural selection is the process by which organisms better adapted to their environment tend to survive and produce more offspring. It is not a single formula but a chain of observable facts that lead to evolution.
自然选择是指那些更适应环境的生物体往往会存活并产生更多后代的过程。它不是一个公式，而是一条条可观察的事实链，最终导致进化。
The principles can be summarised as a series of statements:
原理可概括为一系列陈述：
- There is variation within any population. / 任何种群内部都存在变异。
- Many of these variations are heritable. / 许多变异是可遗传的。
- Populations produce more offspring than the environment can support, leading to a struggle for survival. / 种群产生的后代数量超过环境所能支持的范围，从而引发生存斗争。
- Individuals with advantageous alleles are more likely to survive and reproduce. / 携带有利等位基因的个体更可能存活和繁殖。
- These favourable alleles increase in frequency over generations. / 这些有利等位基因的频率在世代中逐步增加。
Worked example: Antibiotic-resistant bacteria: a mutation gives one bacterium resistance. When antibiotics are used, sensitive bacteria die, resistant ones multiply, and over time the population becomes resistant.
示例：抗生素耐药细菌：某个突变使细菌获得耐药性。使用抗生素时，敏感菌死亡，耐药菌增殖，久而久之种群变得耐药。
10. Osmosis and Water Potential (Conceptual) | 渗透与水势（概念）
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 expressed in pressure units (kPa) and pure water has a water potential of 0 kPa; all solutions have negative water potentials.
渗透是水分子通过选择性半透膜从水势较高的区域向水势较低的区域净移动。水势（ψ）以压强单位（kPa）表示，纯水的水势为 0 kPa；所有溶液的水势均为负值。
Net movement stops when ψ outside = ψ inside (equilibrium)
The water potential of a cell is the sum of its solute potential (ψₛ) and pressure potential (ψₚ). You are not required to calculate these at GCSE, but you must interpret graphs showing % change in mass of plant tissue dipped in different concentrations.
细胞水势等于溶质势（ψₛ）与压力势（ψₚ）之和。GCSE不要求计算，但你必须会解读植物组织在不同浓度溶液中质量变化百分比的曲线图。
Worked interpretation: A potato cylinder gains mass in a solution of low solute concentration because water enters by osmosis (the cells have a lower water potential than the solution). It loses mass in a concentrated solution because water leaves the cells.
解读示例：马铃薯条在低浓度溶液中质量增加，因为水通过渗透进入细胞（细胞水势低于溶液）。在高浓度溶液中质量减少，因为水离开细胞。
11. Biodiversity Indices (Simpson's Index) | 生物多样性指数（辛普森指数）
Biodiversity can be measured using Simpson’s Index of Diversity, which takes into account both species richness (number of species) and evenness (relative abundance). A higher value indicates greater diversity.
生物多样性可以用辛普森多样性指数来衡量，它同时考虑物种丰富度（物种数）和均匀度（相对丰度）。指数值越高，多样性越大。
D = 1 - Σ (n/N)²
mula D = 1/Σ (n/N)² is used, but the OCR specification typically uses the complement form.
其中 n = 某一物种的个体总数，N = 所有物种的个体总数。有时公式为 D = 1/Σ (n/N)²，但OCR大纲通常使用补数形式。
Worked example: (n/N)² values: A (30/50)² = 0.36, B (10/50)² = 0.04, C = 0.04. Sum = 0.44. D = 1 - 0.44 = 0.56. Another site with equal numbers of each would give D closer to 1, indicating higher evenness.
计算示例：某林地样方包含：物种 A 30，物种 B 10，物种 C 10。总个体数 N = 50。(n/N)² 值：A (30/50)² = 0.36，B (10/50)² = 0.04，C = 0.04。总和 = 0.44。D = 1 - 0.44 = 0.56。另一个各样方数量相等的位点 D 值更接近 1，表明均匀度更高。
12. Dilution Calculations in Serial Dilutions | 连续稀释的计算
Serial dilutions are used to create a range of concentrations of a solute, often when investigating the effect of concentration on enzyme activity, osmosis, or microbial growth. Each dilution reduces the concentration by a constant factor.
连续稀释用于配制一系列溶质浓度，常用于研究浓度对酶活性、渗透或微生物生长的影响。每次稀释都将浓度降低固定倍数。
Concentration after dilution = Original Concentration × (Volume of Stock ÷ Total Volume)
Alternatively, if you perform a 1 in 10 serial dilution repeatedly, the concentrations decrease by powers of 10.
或者，如果你反复进行 1/10 的连续稀释，浓度将以 10 的幂次递减。
Worked example: ake 1 cm³ of it and add 9 cm³ of distilled water. The new concentration = 2.0 × (1 ÷ 10) = 0.2 M ke 1 cm³ of this 0.2 M solution and add 9 cm³ water, the next concentration is 0.2 × 0.1 = 0.02 M.
计算示例：你有 2.0 M 蔗糖溶液。取 1 cm³ 溶液加入 9 cm³ 蒸馏水。新浓度 = 2.0 × (1 ÷ 10) = 0.2 M。若再取 1 cm³ 的 0.2 M 溶液加 9 cm³ 水，下一个浓度为 0.2 × 0.1 = 0.02 M。
Always label tubes clearly and use the same unit for all volumes.
务必清楚地标记试管，并对所有体积使用相同的单位。
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