📚 SQA Higher Biology Formula & Theorem Quick Reference | SQA 高等生物公式定理速查手册
This quick-reference guide compiles the essential formulas, equations, and key theorems required for the Year 12 SQA Higher Biology course. It focuses on quantitative skills assessed across units – from computing magnification and diversity indices to applying Hardy-Weinberg equilibrium and estimating population size. Each formula is paired with a clear statement of its biological context, helping you move confidently between calculations and conceptual understanding.
本速查手册汇集了 SQA 高等生物（Year 12）课程必备的公式、方程式及核心定理，覆盖从计算放大倍数和多样性指数到应用哈代-温伯格平衡及估算种群数量等各个单元的定量分析要求。每条公式均配有明确的生物学背景说明，帮助你在计算与概念理解之间自如切换。
1. Magnification and Actual Size | 放大倍数与实际尺寸
In microscopy, the relationship between image size, actual size, and magnification is fundamental. The standard equation is: Image size = Actual size × Magnification. Rearranging gives the more commonly used form in practical work: Actual size = Image size ÷ Magnification. Always ensure that image and actual sizes are expressed in the same units; convert to micrometres (µm) where necessary (1 mm = 1000 µm).
在显微镜使用中，图像大小、实际大小与放大倍数之间的关系是基础。标准方程为：图像大小 = 实际大小 × 放大倍数。在实际操作中，更常用的是变形公式：实际大小 = 图像大小 ÷ 放大倍数。必须确保图像大小与实际大小使用相同的单位，必要时转换为微米（µm；1 mm = 1000 µm）。
Actual Size (µm) = (Image Size (mm) × 1000) ÷ Magnification
This calculation is critical for calibrating eyepiece graticules and interpreting cell structures. For example, if a cell image measures 20 mm under a ×400 magnification, the actual length is (20 × 1000) ÷ 400 = 50 µm. Common pitfalls include forgetting to convert millimetres to micrometres, or omitting the division step.
该计算对于校准目镜测微尺和判读细胞结构至关重要。例如，若一个细胞在 400 倍放大下的图像长度为 20 mm，则实际长度为 (20 × 1000) ÷ 400 = 50 µm。常见错误包括忘记将毫米转换为微米，或忽略了除法步骤。
2. Respiratory Quotient (RQ) | 呼吸商
Respiratory Quotient (RQ) reveals which substrate is being metabolised during aerobic respiration. It is defined as the ratio of carbon dioxide produced to oxygen consumed over a given period.
呼吸商（RQ）揭示有氧呼吸期间被代谢的底物类型，定义为一定时间内产生的二氧化碳体积与消耗的氧气体积之比。
RQ = Volume of CO₂ produced ÷ Volume of O₂ consumed
Pure carbohydrate respiration yields an RQ of 1.0 (e.g. C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O), while lipid respiration yields approximately 0.7 due to the lower oxygen content of fatty acids. Protein yields around 0.9. RQ is measured using a respirometer, and values above 1.0 may indicate anaerobic respiration, where CO₂ is produced without O₂ uptake.
纯碳水化合物的呼吸商为 1.0（例如 C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O），而脂质呼吸商约为 0.7，因为脂肪酸的氧含量较低；蛋白质约为 0.9。RQ 通过呼吸计测定，若数值超过 1.0 可能表明发生了无氧呼吸，即产生 CO₂ 却未消耗 O₂。
3. Productivity: GPP and NPP | 生产力：总初级生产力与净初级生产力
In ecosystem energetics, Gross Primary Productivity (GPP) is the total rate at which producers convert solar energy into chemical energy through photosynthesis. A significant portion of this energy is used for plant respiration (R). Net Primary Productivity (NPP) is the energy that remains to form new biomass and is available to the next trophic level.
在生态系统能量学中，总初级生产力（GPP）是生产者通过光合作用将太阳能转化为化学能的总速率。其中相当一部分能量用于植物自身的呼吸消耗（R）。净初级生产力（NPP）则为剩余的能量，用于形成新生物量并可供下一营养级利用。
NPP = GPP − R
All three quantities are usually expressed in units of energy per area per time (kJ m⁻² year⁻¹) or as biomass per area per time (g m⁻² year⁻¹). For example, in a temperate forest, GPP might be 50 kJ m⁻² year⁻¹, respiration 30 kJ m⁻² year⁻¹, leaving an NPP of 20 kJ m⁻² year⁻¹. This relationship underpins pyramids of energy and helps explain why food chains are limited in length.
三者通常用单位面积单位时间的能量（kJ m⁻² year⁻¹）或生物量（g m⁻² year⁻¹）表示。例如，在温带森林中，GPP 可能为 50 kJ m⁻² year⁻¹，呼吸消耗 30 kJ m⁻² year⁻¹，则 NPP 为 20 kJ m⁻² year⁻¹。这一关系是能量金字塔的基础，也有助于解释为什么食物链的长度是有限的。
4. Energy Transfer Efficiency Between Trophic Levels | 营养级间能量传递效率
When energy flows from one trophic level to the next, only a small percentage is transferred. This efficiency can be calculated using either energy content or biomass per unit area.
能量从一个营养级流向下一营养级时，只有一小部分被传递。传递效率可用能量含量或单位面积生物量计算。
Efficiency (%) = (Energy in trophic level n ÷ Energy in trophic level n−1) × 100
Typical efficiencies range from 5% to 20%, with 10% often used as a rough average. Losses occur through respiration, egestion, excretion, and non-consumption of parts of the prey. In SQA calculations, you may be asked to determine efficiency from given data tables. For instance, if the producer level contains 10 000 kJ m⁻² year⁻¹ and primary consumers contain 1 500 kJ m⁻² year⁻¹, the transfer efficiency is (1500 ÷ 10 000) × 100 = 15%.
典型的传递效率为 5%–20%，通常粗略取 10% 作为平均数。能量散失通过呼吸、排遗、排泄及未食用部分等途径。在 SQA 计算中，你可能需要根据给定数据表求出效率。例如，若生产者级含有 10 000 kJ m⁻² year⁻¹，初级消费者含有 1 500 kJ m⁻² year⁻¹，则传递效率为 (1500 ÷ 10 000) × 100 = 15%。
5. Mark-Recapture: Estimating Population Size | 标记重捕法：估算种群数量
The Lincoln Index (mark-recapture method) is used to estimate the population size of mobile organisms. A sample is captured, marked, and released; later, a second sample is taken and the proportion of marked individuals recaptured allows estimation.
林肯指数（标记重捕法）用于估算运动能力较强的生物种群数量。先捕获一份样本进行标记后释放；随后采集第二份样本，通过重捕中标记个体的比例进行估算。
N = (M × C) ÷ R
Where N = estimated total population size, M = number marked in the first sample, C = total number captured in the second sample, and R = number of marked individuals recaptured in the second sample. The method assumes that: (i) marks are not lost or overlooked; (ii) marked individuals mix randomly with the rest of the population; (iii) no significant births, deaths, immigration, or emigration occur between samples; and (iv) marking does not affect survival or behaviour.
这里 N = 估算的总种群数量，M = 第一次样本中标记的个体数，C = 第二次取样总个体数，R = 第二次取样中重捕的标记个体数。该方法假设：(i) 标记不丢失、不被忽略；(ii) 标记个体与种群其余个体随机混合；(iii) 两次取样期间无显著的出生、死亡、迁入或迁出；(iv) 标记不影响存活或行为。
6. Hardy-Weinberg Principle | 哈代-温伯格平衡原理
The Hardy-Weinberg principle provides a null model for population genetics. For a gene with two alleles, the allele frequencies in a large, randomly mating population remain constant across generations if no evolutionary influences act. The two fundamental equations are:
哈代-温伯格原理为群体遗传学提供了零假设模型。对于一个具有两个等位基因的基因，在一个大且随机交配的种群中，如果没有进化因素影响，等位基因频率代代保持不变。两个基本方程为：
p + q = 1
p² + 2pq + q² = 1
p is the frequency of the dominant allele, q the frequency of the recessive allele. p² represents the frequency of homozygous dominant individuals, 2pq of heterozygotes, and q² of homozygous recessive individuals. The conditions for Hardy-Weinberg equilibrium are: no mutation, no gene flow (migration), large population size, random mating, and no natural selection. Even slight deviations can be detected using the chi-squared test, helping to infer evolutionary forces.
p 为显性等位基因频率，q 为隐性等位基因频率。p² 代表纯合显性个体频率，2pq 为杂合子频率，q² 为纯合隐性个体频率。哈代-温伯格平衡的条件包括：无突变、无基因流动（迁移）、大种群、随机交配且无自然选择。即便微小的偏差也可通过卡方检验检测，从而推断进化力量的作用。
7. Exponential Population Growth | 指数种群增长
When resources are unlimited, populations with discrete generations can grow exponentially. The number of individuals after a certain number of generations is given by the geometric growth model.
在资源无限的情况下，具有离散世代的种群可呈指数增长。经过若干世代后的个体数可用几何增长模型表示。
Nt = N0 × 2n
Nt is the population size at generation t, N0 is the initial population size, and n is the number of generations. For example, starting with 100 bacteria (N0 = 100) that double every generation, after 5 generations (n = 5) the population is 100 × 2⁵ = 3200. The time for one generation is the generation time (g). In continuous growth with overlapping generations, the exponential model N = N0 ert may be used, but the SQA Higher course typically emphasises the discrete form.
Nt 为第 t 代时的种群大小，N0 为初始种群大小，n 为世代数。例如，初始 100 个细菌（N0 = 100），每代翻倍，经过 5 代（n = 5）种群数目为 100 × 2⁵ = 3200。完成一代所需的时间为世代时间（g）。对于世代重叠的连续增长，也可使用指数模型 N = N0 ert，但 SQA Higher 课程通常强调离散形式。
8. Mitotic Index | 有丝分裂指数
The mitotic index is a measure of proliferative activity in a tissue. It is calculated by viewing cells under a light microscope and counting those showing visible chromosomes (prophase, metaphase, anaphase, telophase) relative to the total cells in the field of view.
有丝分裂指数衡量组织中细胞的增殖活性。通过在光学显微镜下观察细胞，计数可见染色体的细胞（前期、中期、后期、末期）占视野内总细胞的比例。
Mitotic Index = (Number of cells in mitosis ÷ Total number of cells) × 100%
A high mitotic index indicates rapid growth, such as in root tips, shoot apices, or cancerous tissues. In cancer diagnosis, an elevated mitotic index can suggest malignancy. For practical assessment, you may be required to identify stages of mitosis in a root-tip squash and calculate the mitotic index to compare different regions or treatments.
有丝分裂指数高表明生长迅速，如根尖、茎尖或癌组织。在癌症诊断中，有丝分裂指数升高可能提示恶性。在实验评估中，你可能需要从根尖压片中识别有丝分裂各期并计算有丝分裂指数，以比较不同区域或处理。
9. Chi-Squared Test (χ²) | 卡方（χ²）检验
The chi-squared (χ²) test is a goodness-of-fit statistical test used to compare observed results with expected theoretical ratios – often in genetics or population studies. It determines whether any deviation is due to chance or a significant factor.
卡方（χ²）检验是一种拟合优度统计检验，用于比较观察结果与期望理论比值——常见于遗传学或种群研究。它判断偏差是否由偶然造成，或是存在显著因素。
χ² = Σ (O − E)² ÷ E
O = observed value, E = expected value. The sum is taken over all categories. The calculated χ² is compared against critical values in a χ² distribution table, typically at the 0.05 probability level, considering the degrees of freedom (df = number of categories − 1). If χ²calc > χ²crit, the null hypothesis is rejected. For example, in a monohybrid cross expecting a 3:1 ratio, if observed numbers of 95 round and 25 wrinkled seeds are compared with expected 90 and 30, χ² = (95−90)²/90 + (25−30)²/30 = 0.28 + 0.83 = 1.11. With df = 1 and critical value 3.84, the null hypothesis is not rejected.
O = 观察值，E = 期望值。对所有类别求和。计算出的 χ² 值与卡方分布表中的临界值比较，通常采用 0.05 概率水平，并考虑自由度（df = 类别数 − 1）。若 χ²计算 > χ²临界，则拒绝零假设。例如，在预期 3:1 的单基因杂合子杂交中，若观察到 95 个圆粒和 25 个皱粒种子，期望值为 90 和 30，则 χ² = (95−90)²/90 + (25−30)²/30 = 0.28 + 0.83 = 1.11。自由度为 1，临界值 3.84，因此不拒绝零假设。
10. Simpson’s Diversity Index | 辛普森多样性指数
Simpson’s Diversity Index (D) quantifies the biodiversity of a habitat, taking into account both species richness and evenness. A higher value indicates greater diversity.
辛普森多样性指数（D）定量描述栖息地的生物多样性，同时考虑物种丰富度和均匀度。指数值越高，多样性越丰富。
D = 1 − Σ (n / N)²
where n = total number of organisms of a particular species, and N = total number of organisms of all species. The term (n/N) represents the proportion of the community made up by that species. For instance, in a sample containing 50 daisies, 20 buttercups, and 10 clovers (N=80): Σ (n/N)² = (50/80)² + (20/80)² + (10/80)² = 0.3906 + 0.0625 + 0.0156 = 0.4687, so D = 1 − 0.4687 = 0.5313. This value can be compared between habitats; those with more even distribution of abundant species yield a higher D. Some SQA contexts may use the reciprocal form 1/Σ(n/N)², but the complement form is standard in Higher.
其中 n = 某一特定物种的总个体数，N = 所有物种的总个体数。(n/N) 表示该物种在整个群落中所占的比例。例如，一份样本包含 50 朵雏菊、20 株毛茛和 10 株三叶草（N = 80）：Σ (n/N)² = (50/80)² + (20/80)² + (10/80)² = 0.3906 + 0.0625 + 0.0156 = 0.4687，因此 D = 1 − 0.4687 = 0.5313。该值可在不同栖息地间比较；物种数量分布越均匀，D 值越高。尽管有些 SQA 背景可能使用倒数形式 1/Σ(n/N)²，但互补形式是 Higher 课程的标准。
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