📚 Year 11 SQA Biology: Formula & Theorem Quick Reference Handbook | SQA 生物：公式定理速查手册
This quick reference handbook is designed for Year 11 students following the SQA National 5 Biology curriculum. It brings together all the essential mathematical formulae, biological equations, and key scientific theorems you will encounter in the course. From photosynthesis and respiration calculations to genetic ratios and population sampling methods, this article condenses the core quantitative and conceptual relationships into one easy-to-navigate resource. Keep it handy for revision, homework, and exam preparation.
这本速查手册专为学习 SQA National 5 生物课程的 Year 11 学生而编写，汇总了课程中涉及的所有关键数学公式、生物方程和科学定理。从光合作用与呼吸作用计算，到遗传比例和种群取样方法，本文将核心的定量关系与概念性原理浓缩为易于查阅的资源。无论是复习、作业还是备考，这本手册都能助你一臂之力。
1. Photosynthesis Word and Chemical Equations | 光合作用文字与化学方程
rall word equation summarises the inputs and outputs of this vital reaction: carbon dioxide + water → glucose + oxygen. This reaction only proceeds in the presence of light and chlorophyll, which acts as the catalyst and energy absorber within chloroplasts.
光合作用是绿色植物、藻类及某些细菌将光能转化为储存在葡萄糖中的化学能的过程。总文字方程概括了这一重要反应的反应物与产物：二氧化碳 + 水 → 葡萄糖 + 氧气。该反应仅在光和叶绿素存在时才能进行，叶绿素在叶绿体内充当催化剂和能量吸收剂。
6H₂O → C₆H₁₂O₆ + 6O₂. Note that light energy and chlorophyll are written above or below the arrow to indicate that they are necessary conditions rather than reactants. Students should be able to identify carbon dioxide and water as raw materials and glucose and oxygen as the products.
配平的化学方程从分子层面精确描述了光合作用。National 5 课程中常用的配平方程为：6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂。需注意，光能和叶绿素写在箭头上方或下方，以表明它们是必要条件而非反应物。学生应能识别二氧化碳和水为原料，葡萄糖和氧气为产物。
Understanding this equation is fundamental because it links to energy transfer through ecosystems. The glucose produced can be used immediately for respiration, converted into starch for storage, or transformed into cellulose for cell wall construction. The oxygen released is essential for aerobic respiration in almost all living organisms.
理解该方程至关重要，因为它与生态系统的能量传递密切相关。产生的葡萄糖可立即用于呼吸作用、转化为淀粉储存或转化为纤维素构建细胞壁。释放的氧气则是几乎所有生物进行有氧呼吸所必需的。
2. Aerobic and Anaerobic Respiration Equations | 有氧呼吸与无氧呼吸方程
ygen, and anaerobic respiration, which proceeds without oxygen. The word equation for aerobic respiration is: glucose + oxygen → carbon dioxide + water + energy (ATP). The balanced chemical equation is: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy. This reaction releases a large amount of energy, approximately 38 ATP molecules per glucose molecule in ideal conditions.
呼吸作用是细胞从葡萄糖中释放能量的过程。National 5 层面存在两种形式：需氧的有氧呼吸和不需氧的无氧呼吸。有氧呼吸的文字方程为：葡萄糖 + 氧气 → 二氧化碳 + 水 + 能量（ATP）。配平的化学方程为：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 能量。该反应释放大量能量，理想条件下每个葡萄糖分子约产生 38 个 ATP 分子。
mall amount of energy. The word equation is: glucose → lactic acid + energy. This occurs during vigorous exercise when oxygen delivery to muscles is insufficient. The build-up of lactic acid causes muscle fatigue and eventually cramping, and an oxygen debt must be repaid after exercise to oxidise the lactic acid back to harmless compounds.
无氧呼吸在动物细胞与植物或酵母细胞中有所不同。在动物肌肉细胞中，葡萄糖被分解为乳酸和少量能量，文字方程为：葡萄糖 → 乳酸 + 能量。此过程发生在剧烈运动、肌肉供氧不足时。乳酸的积累导致肌肉疲劳甚至痉挛，运动后必须偿还氧债，将乳酸重新氧化为无害物质。
anol + carbon dioxide + energy. This process, known as fermentation, is exploited industrially in brewing and baking. The balanced chemical equation is: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ + energy. Students should be clear that anaerobic respiration releases far less energy than aerobic respiration because glucose is only partially broken down.
在植物和酵母中，无氧呼吸遵循不同途径，产生乙醇和二氧化碳，文字方程为：葡萄糖 → 乙醇 + 二氧化碳 + 能量。这一过程称为发酵，被广泛应用于酿造和烘焙工业。配平的化学方程为：C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ + 能量。学生应明确无氧呼吸释放的能量远少于有氧呼吸，因为葡萄糖仅被部分分解。
3. Enzyme Activity and the Effect of Temperature and pH | 酶活性及温度与 pH 的影响
Enzymes are biological catalysts made of protein that speed up chemical reactions in living organisms without being consumed. The fundamental theorem governing enzyme action is the lock and key model: each enzyme has an active site of a specific shape that is complementary to its substrate. Only when the substrate fits precisely into the active site can an enzyme-substrate complex form and catalysis proceed. This model explains enzyme specificity — each enzyme catalyses only one type of reaction.
酶是由蛋白质构成的生物催化剂，能在不被消耗的情况下加速生物体内的化学反应。支配酶作用的基本定理是锁钥模型：每种酶具有特定形状的活性位点，与其底物互补。只有当底物精确契合活性位点时，才能形成酶-底物复合物并进行催化。该模型解释了酶的专一性——每种酶仅催化一类反应。
Temperature affects enzyme activity according to a predictable pattern. As temperature increases, kinetic energy rises, leading to more frequent successful collisions between enzyme and substrate, so the rate of reaction increases. However, this relationship only holds up to the optimum temperature, typically around 37°C for human enzymes. Beyond this point, the weak hydrogen bonds maintaining the enzyme's tertiary structure begin to break; the active site becomes denatured and permanently loses its complementary shape. The rate of reaction then drops sharply to zero.
温度按照可预测的模式影响酶活性。温度升高，动能增加，酶与底物间的成功碰撞更加频繁，因此反应速率加快。但这种关系仅维持到最适温度（人体酶约为 37°C）。超过该点，维持酶三级结构的弱氢键开始断裂；活性位点发生变性，永久丧失其互补形状，反应速率随即骤降至零。
pH similarly has an optimum value for each enzyme, beyond which activity declines. Most enzymes work best around pH 7, but exceptions exist — pepsin in the stomach has an optimum around pH 2. Either side of the optimum, the altered hydrogen ion concentration disrupts the ionic and hydrogen bonds holding the protein in its precise three-dimensional shape, leading to denaturation. The key mathematical relationship to remember is that denaturation is irreversible: once an enzyme is denatured, its catalytic function cannot be restored by returning to optimum conditions.
pH 同样对每种酶有一个最适值，超出该值活性即下降。多数酶在 pH 7 左右活性最佳，但也有例外——胃蛋白酶的最适 pH 约为 2。在最适值两侧，氢离子浓度的改变会破坏维持蛋白质精确三维结构的离子键和氢键，导致变性。需牢记的关键数学关系是变性不可逆：一旦酶变性，即使恢复至最适条件，其催化功能也无法复原。
4. Calculating Enzyme Reaction Rates | 酶反应速率计算
ven time period. The fundamental formula for calculating rate is straightforward: rate = quantity of product formed (or quantity of substrate used) ÷ time taken. Units must be expressed correctly depending on what is being measured. If the product accumulated is measured in cm³ and time in seconds, the rate will be expressed in cm³/s. Common units in SQA exams include cm³/min, g/s, or arbitrary units per minute.
酶控反应速率衡量的是给定时间内底物转化为产物或产物积累的快慢。计算速率的基本公式很简单：速率 = 生成的产物量（或消耗的底物量）÷ 所用时间。单位必须根据测量对象正确表达。若产物积累量以 cm³ 计量、时间以秒计量，则速率单位为 cm³/s。SQA 考试中常见单位包括 cm³/min、g/s 或任意单位/分钟。
e in y ÷ change in x. At the initial linear portion of a progress curve, the steepness of the slope indicates the initial rate of reaction. As substrate is used up or the enzyme becomes limiting, the curve flattens, indicating a decreasing rate. Students should practise extracting data points from graphs and applying the formula to quantify the rate at different stages.
解读酶活性图表时，速率通过计算直线斜率求得。对于直线图，斜率 = y 轴变化量 ÷ x 轴变化量。在反应进程曲线的初始线性段，斜率陡度表明初始反应速率。随着底物耗尽或酶成为限制因素，曲线趋于平坦，表明速率下降。学生应练习从图表中提取数据点并应用公式量化不同阶段的速率。
of reaction increases when the temperature is raised by 10°C. It is calculated as: Q₁₀ = rate at (T + 10)°C ÷ rate at T°C. For many biological reactions within the enzyme’s tolerance range, the Q₁₀ is approximately 2, meaning the rate doubles for every 10°C rise. However, this formula only applies below the optimum temperature; above the optimum, denaturation dominates and Q₁₀ becomes meaningless.
与温度对酶活性影响相关的一个特殊公式是 Q₁₀ 温度系数。Q₁₀ 值表示温度每升高 10°C 反应速率增加的倍数。计算公式为：Q₁₀ = (T + 10)°C 时的速率 ÷ T°C 时的速率。对于酶耐受范围内的许多生物反应，Q₁₀ 约为 2，即温度每升高 10°C 速率翻倍。但该公式仅适用于最适温度以下；超过最适温度，变性占主导地位，Q₁₀ 便失去意义。
5. Surface Area to Volume Ratio | 表面积与体积比
pic, why organisms have specialised exchange surfaces, and how organisms adapt to their environments. For a cube-shaped cell of side length L, the formulas are: surface area = 6L² and volume = L³. Therefore, SA:V ratio = 6L² ÷ L³ = 6 ÷ L. As the size of a cell or organism increases, its SA:V ratio decreases.
表面积与体积比（SA:V）是 National 5 生物中最重要的定量概念之一。它解释了细胞为何微小、生物为何具有特化的交换表面以及生物如何适应其环境。对于边长为 L 的立方体细胞，公式为：表面积 = 6L²，体积 = L³。因此，SA:V 比 = 6L² ÷ L³ = 6 ÷ L。随着细胞或生物体积增大，其 SA:V 比减小。
The biological significance of a high SA:V ratio is profound. A high ratio means a large surface area relative to the internal volume, allowing efficient diffusion of oxygen, carbon dioxide, nutrients, and waste products across the cell membrane. This is why actively metabolising cells like those in the alveoli, villi, and root hair cells have flattened or elongated shapes that maximise surface area. Conversely, a low SA:V ratio means the internal volume is too large for the surface to supply adequately by diffusion alone, necessitating specialised transport systems such as the circulatory system in animals and the xylem and phloem in plants.
高 SA:V 比的生物学意义深远。高比值意味着相对于内部体积，表面积较大，可使氧气、二氧化碳、营养物质和废物高效地跨细胞膜扩散。这就是为什么肺泡、绒毛和根毛细胞等代谢活跃的细胞具有扁平或细长形状以最大化表面积。相反，低 SA:V 比意味着内部体积过大，仅靠扩散无法充分供应，因此需要专门的运输系统，如动物的循环系统以及植物的木质部和韧皮部。
Calculations involving SA:V ratios frequently appear in SQA exam questions. Students may be asked to calculate the surface area, volume, and ratio for simple geometric shapes or to predict which of two differently sized cells will absorb nutrients faster. The key principle to remember is: the smaller the organism or structure, the larger its SA:V ratio, and the more efficient its exchange with the environment. This theorem also explains why animals in cold climates tend to be larger (Bergmann's rule) — a lower SA:V ratio reduces heat loss.
涉及 SA:V 比的计算频繁出现在 SQA 考试中。学生可能被要求计算简单几何形状的表面积、体积和比值，或预测两个不同大小的细胞中哪个吸收营养更快。需牢记的关键原理是：生物体或结构越小，其 SA:V 比越大，与环境交换的效率就越高。这一定理也解释了为何寒冷气候中的动物体型往往较大（伯格曼法则）——较低的 SA:V 比可减少热量散失。
6. Genetic Crosses and Monohybrid Ratios | 遗传杂交与单基因杂交比例
Monohybrid inheritance examines how a single characteristic determined by one gene is passed from parents to offspring. The theorem central to this is Mendel’s law of segregation: each individual possesses two alleles for each gene, and these alleles separate during gamete formation so that each gamete carries only one allele. By convention, dominant alleles are represented by a capital letter (e.g., T for tall) and recessive alleles by the lower-case equivalent (e.g., t for dwarf).
单基因杂交遗传研究由单个基因决定的单一性状如何从亲代传递给子代。其核心定理是孟德尔分离定律：每个个体每个基因拥有两个等位基因，这些等位基因在配子形成时分离，使每个配子仅携带一个等位基因。按惯例，显性等位基因用大写字母表示（如 T 表示高茎），隐性等位基因用对应小写字母表示（如 t 表示矮茎）。
e expected genotypic ratio is 1 TT : 2 Tt : 1 tt. The corresponding phenotypic ratio, assuming complete dominance, is 3 dominant : 1 recessive. In a test cross between a heterozygous individual and a homozygous recessive (Tt × tt), the predicted ratio is 1 Tt : 1 tt, giving a 1:1 phenotypic ratio. These are the two most important ratios to memorise for National 5.
使用庞纳特方格可预测子代的基因型比例和表型比例。在两个杂合亲本的杂交中（Tt × Tt），预期的基因型比例为 1 TT : 2 Tt : 1 tt。假设完全显性，相应的表型比例为 3 显性 : 1 隐性。在杂合个体与纯合隐性的测交中（Tt × tt），预期比例为 1 Tt : 1 tt，表型比例为 1:1。这是 National 5 必须记住的两个最重要比例。
The concept of homozygous and heterozygous is critical. Homozygous means having two identical alleles (TT or tt); heterozygous means having two different alleles (Tt). The phenotype is the observable characteristic, while the genotype is the genetic constitution. When interpreting family pedigree diagrams, students must be able to deduce genotypes of individuals and predict probabilities of offspring inheriting particular traits. The probability of an offspring inheriting a specific genotype or phenotype can be expressed as a fraction, decimal, or percentage.
纯合与杂合的概念至关重要。纯合指携带两个相同等位基因（TT 或 tt）；杂合指携带两个不同等位基因（Tt）。表型是可观察的特征，基因型则是遗传组成。解读家族谱系图时，学生必须能够推断个体基因型并预测子代遗传特定性状的概率。子代遗传特定基因型或表型的概率可用分数、小数或百分比表示。
7. Population Sampling and Abundance Estimation | 种群取样与丰富度估算
y or slow-moving organisms, a quadrat is used. The fundamental formula for estimated population size is: estimated total population = (mean number of individuals per quadrat ÷ area of one quadrat) × total area of the habitat. The mean per quadrat is calculated by summing the counts from all quadrats placed and dividing by the number of quadrats sampled. Random placement of quadrats using random number coordinates avoids sampling bias and ensures the estimate is representative.
生态学家使用取样技术估算种群大小，因为逐一计数通常不可行。对于静止或缓慢移动的生物，使用样方。估算种群大小的基本公式为：估算总种群数 =（每个样方平均个体数 ÷ 单个样方面积）× 栖息地总面积。每个样方的平均数通过将所有样方计数求和并除以取样样方数计算得出。使用随机数坐标随机放置样方可避免取样偏差，确保估算结果具有代表性。
For mobile animals, the capture-mark-recapture (CMR) method — also called the Lincoln Index — ÷ R, where N is the estimated total population size, M is the number of individuals captured and marked on the first occasion, C is the total number captured on the second occasion, and R is the number of marked individuals recaptured in the second sample. This formula assumes that marked individuals mix randomly back into the population, that marks are not lost or overlooked, and that the population is closed with no births, deaths, or migrations between sampling events.
对于可移动的动物，采用捕捉-标记-重捕法（CMR）——也称 Lincoln 指数。公式为：N = (M × C) ÷ R，其中 N 为估算的种群总数，M 为首次捕捉并标记的个体数，C 为第二次捕捉的总数，R 为第二次样本中重捕的已标记个体数。该公式假设已标记个体随机混入种群、标记未丢失或遗漏、且种群在两次取样之间保持封闭，无出生、死亡或迁移。
Limitations of the Lincoln Index must be understood. If marked individuals become more visible to predators, R will be smaller than expected, leading to an overestimate of N. If marks wash off or are shed, R will be underestimated, again inflating N. If marked individuals learn to avoid traps, R will be lower, producing an overestimate. Students should be able to evaluate the reliability of population estimates and suggest improvements such as increasing the number of samples, ensuring marks are harmless and durable, and timing the second capture appropriately.
必须理解 Lincoln 指数的局限性。若已标记个体更容易被捕食者发现，R 将小于预期，导致对 N 的高估。若标记被冲掉或脱落，R 将被低估，同样会夸大 N。若已标记个体学会避开陷阱，R 将偏低，导致高估。学生应能评估种群估算结果的可靠性，并提出改进措施，如增加样本数量、确保标记无害且持久、合理安排二次捕捉时间。
8. Microorganism Growth and Binary Fission Calculations | 微生物生长与二分裂计算
der optimal conditions, this division can occur at a constant rate, leading to exponential growth. The fundamental formula for predicting bacterial numbers after a given time is: final number of cells = initial number of cells × 2ⁿ, where n is the number of generations (or divisions) that have occurred. The variable n is calculated as: n = total time elapsed ÷ generation time (time for one complete division cycle).
细菌等微生物通过二分裂进行无性繁殖，一个细胞分裂为两个基因相同的子细胞。在最佳条件下，这种分裂以恒定速率发生，导致指数式增长。预测给定时间后细菌数量的基本公式为：最终细胞数 = 初始细胞数 × 2ⁿ，其中 n 为已发生的世代数（即分裂次数）。变量 n 的计算方法为：n = 已过总时间 ÷ 世代时间（完成一个完整分裂周期所需时间）。
9. The final number of cells would be 1 × 2⁹ = 512 cells. This exponential growth cannot continue indefinitely because resources such as nutrients and space become limiting, and toxic waste products accumulate. In practice, bacterial growth in a closed culture follows a sigmoid (S-shaped) curve with four distinct phases: lag phase, exponential (log) phase, stationary phase, and death phase.
例如，若一个世代时间为 20 分钟的单个细菌生长 3 小时（180 分钟），世代数为 180 ÷ 20 = 9。最终细胞数将为 1 × 2⁹ = 512 个细胞。这种指数式增长无法无限持续，因为营养和空间等资源将变得有限，有毒废物也会积累。实际上，封闭培养中的细菌生长遵循 S 形曲线，分为四个明显阶段：延迟期、指数（对数）期、稳定期和死亡期。
s + immigration) − (deaths + emigration) per unit time. For a bacterial colony in a closed system, immigration and emigration are zero, so growth rate simplifies to births − deaths. When birth rate exceeds death rate, the population grows; when they are equal, the population is in stationary phase; when death rate exceeds birth rate, the population declines.
相关的数学概念是种群变化率的计算。种群增长率的公式为：增长率 =（出生数 + 迁入数）−（死亡数 + 迁出数）/ 单位时间。对于封闭系统中的细菌菌落，迁入和迁出为零，因此增长率简化为出生数 − 死亡数。当出生率超过死亡率时，种群增长；两者相等时，种群处于稳定期；当死亡率超过出生率时，种群数量下降。
9. Energy Transfer and Biomass Pyramids | 能量传递与生物质金字塔
Energy flows through ecosystems from producers to consumers, with a fundamental loss at each trophic level. The theorem governing this flow is the ten percent rule: on average, only about 10% of the energy stored in one trophic level is transferred to the next. The remaining 90% is lost through respiration, movement, heat production, egestion of undigested material, and excretion of metabolic wastes. This huge energy loss explains why food chains rarely exceed four or five trophic levels — there is simply insufficient energy to sustain higher-level consumers.
能量通过生态系统从生产者流向消费者，每个营养级均存在固有的能量损失。支配这一流动的定理是十分之一定律：平均而言，一个营养级储存的能量中仅约 10% 传递至下一营养级。其余 90% 通过呼吸作用、运动、产热、未消化物质的排遗和代谢废物的排泄而损失。这种巨大的能量损失解释了为何食物链很少超过四到五个营养级——根本不足以为更高级消费者提供能量。
f energy transfer between trophic levels can be calculated using the formula: efficiency (% evel) × 100. Alternatively, using biomass measured in grams per square metre: efficiency (% s values and asked to calculate percentage efficiency or explain why efficiency is so low.
营养级间的能量传递效率可用公式计算：效率（%）=（较高营养级纳入生物质的能量 ÷ 较低营养级可用的能量）× 100。或者，使用以克/平方米为单位的生物质：效率（%）=（较高营养级生物质 ÷ 较低营养级生物质）× 100。在 SQA 考试题中，学生经常被提供能量或生物质数据表，要求计算百分比效率或解释效率为何如此低。
Pyramids of biomass and pyramids of energy visually represent these relationships. A pyramid of biomass usually, but not always, has a broad base and narrows upwards. An important exception occurs in aquatic ecosystems where phytoplankton reproduce so rapidly that their biomass at any one moment may be less than that of the zooplankton consuming them, creating an inverted pyramid of biomass. Pyramids of energy are always upright because energy is inevitably lost at each transfer, reinforcing the second law of thermodynamics in a biological context.
生物质金字塔和能量金字塔可直观地呈现这些关系。生物质金字塔通常（但不总是）基部宽阔且向上逐渐变窄。一个重要的例外发生在水生生态系统中：浮游植物繁殖极其迅速，使其任一时刻的生物质可能小于以它们为食的浮游动物的生物质，从而形成倒置的生物质金字塔。能量金字塔则始终为正立形状，因为能量在每次传递中必然损失，这从生物学角度印证了热力学第二定律。
10. Osmosis and Water Potential | 渗透作用与水势
Osmosis is the net movement of water molecules from a region of high water concentration (high water potential) to a region of low water concentration (low water potential) through a selectively permeable membrane. It is a passive process that requires no cellular energy. The direction and extent of water movement are determined by the water potential gradient. A dilute solution has a high water potential, while a concentrated solution has a low water potential because the solute particles effectively dilute the water, reducing the number of free water molecules available to move.
渗透作用是水分子通过选择透过性膜，从水浓度高（水势高）的区域向水浓度低（水势低）的区域净移动的过程。这是一种被动过程，无需消耗细胞能量。水移动的方向和程度由水势梯度决定。稀溶液水势高，浓溶液水势低，因为溶质粒子实际上稀释了水，减少了可供移动的自由水分子数量。
The effect of osmosis on animal cells placed in solutions of different water potentials is dramatic. In a hypotonic solution (higher water potential outside the cell), water enters by osmosis and the cell swells; animal cells lacking a cell wall may eventually burst (lysis). In a hypertonic solution (lower water potential outside), water leaves the cell and it shrinks (crenation). In an isotonic solution, water movement in and out is balanced and the cell remains unchanged. These terms are essential vocabulary for answering osmosis-related exam questions.
渗透作用对置于不同水势溶液中的动物细胞产生显著影响。在低渗溶液（细胞外水势较高）中，水通过渗透进入，细胞膨胀；缺乏细胞壁的动物细胞最终可能破裂（溶胞）。在高渗溶液（细胞外水势较低）中，水从细胞中流失，细胞皱缩（皱缩）。在等渗溶液中，水的进出保持平衡，细胞维持原状。这些术语是回答渗透作用相关考题的核心词汇。
Plant cells behave differently due to the presence of a rigid cell wall. In a hypotonic solution, water enters the plant cell by osmosis, causing the vacuole to swell and push the cytoplasm against the cell wall. The cell becomes turgid but does not burst because the inelastic cell wall exerts an equal and opposite pressure. Turgor pressure is crucial for maintaining the rigidity of non-woody plant stems and leaves. In a hypertonic solution, water leaves the plant cell; the vacuole shrinks and the plasma membrane pulls away from the cell wall — a condition called plasmolysis. A plasmolysed cell is flaccid and the plant wilts.
植物细胞因存在坚硬的细胞壁而表现不同。在低渗溶液中，水通过渗透进入植物细胞，使液泡膨胀并将细胞质推向细胞壁。细胞变得饱满（硬挺），但因无弹性的细胞壁施加相等且相反的力而不会破裂。膨压对于维持非木本植物茎和叶的刚性至关重要。在高渗溶液中，水从植物细胞中流失；液泡缩小，质膜从细胞壁上脱离——这一状态称为质壁分离。质壁分离的细胞松弛，植株萎蔫。
11. Calculating Percentage Change and Correct Units | 计算百分比变化与正确单位
ivity, or growth. The formula is: percentage change = (final value − initial value) ÷ initial value × 100. A positive result indicates an increase; a negative result indicates a decrease. When osmotic experiments measure the change in mass or length of potato cylinders in different sugar concentrations, percentage change is preferred over absolute change because it accounts for differences in the initial size of the samples, allowing fair comparisons.
许多 SQA National 5 生物题目要求计算百分比变化，尤其在分析渗透作用、酶活性或生长的实验数据时。公式为：百分比变化 =（终值 − 初值）÷ 初值 × 100。正值表示增加，负值表示减少。当渗透实验测量土豆条在不同糖浓度中质量或长度的变化时，百分比变化优于绝对变化，因为它考虑了样本初始大小的差异，便于进行公平比较。
Correct use of units is non-negotiable in SQA Biology. The International System of Units (SI) is standard: length in millimetres (mm), mass in grams (g) or kilograms (kg), volume in cubic centimetres (cm³) or litres (L), time in seconds (s) or minutes (min), and temperature in degrees Celsius (°C). When reading from measuring instruments such as rulers, measuring cylinders, or thermometers, students must record measurements to the appropriate number of decimal places based on the instrument's precision. For a ruler marked in millimetres, lengths should be given to the nearest 0.5 mm or 1 mm depending on the question instruction.
SQA 生物中正确使用单位不可含糊。国际单位制（SI）为标准：长度用毫米（mm），质量用克（g）或千克（kg），体积用立方厘米（cm³）或升（L），时间用秒（s）或分（min），温度用摄氏度（°C）。使用直尺、量筒或温度计等测量仪器读数时，学生必须根据仪器精度将测量值记录到适当的小数位数。对于以毫米为刻度的直尺，长度应根据题目要求记录到最接近的 0.5 mm 或 1 mm。
A final mathematical skill frequently tested is the construction and interpretation of graphs. Students should be proficient in plotting line graphs with time or concentration on the x-axis (independent variable) and the measured outcome on the y-axis (dependent variable). Axes must be labelled clearly with the variable name and unit, scales must be linear and cover at least half the graph paper, and data points must be plotted accurately with a sharp pencil. When multiple lines are plotted on one set of axes, each line must be clearly distinguishable, and a key provided.
最后一项常考数学技能是图表的绘制与解读。学生应熟练掌握绘制折线图，将时间或浓度置于 x 轴（自变量），将测量结果置于 y 轴（因变量）。坐标轴必须清晰标注变量名和单位，刻度必须线性且占据至少一半图纸，数据点必须用削尖铅笔准确描点。当多组数据在同一组坐标轴上绘制时，每条线必须清晰可辨，并附上图例。
12. The Overall Energy Equation for Photosynthesis versus Respiration | 光合作用与呼吸作用的总能量方程对比
It is crucial for National 5 students to recognise that photosynthesis and aerobic respiration are essentially opposite processes in terms of their overall chemical equations. Photosynthesis consumes carbon dioxide and water, storing light energy as chemical energy in glucose and releasing oxygen as a byproduct. Aerobic respiration consumes oxygen and glucose, releasing the stored chemical energy for cellular work and producing carbon dioxide and water as waste. This cyclical relationship maintains the balance of oxygen and carbon dioxide in the atmosphere and underpins the flow of energy through virtually all ecosystems.
National 5 学生必须认识到，光合作用与有氧呼吸就其总化学方程而言基本上是相反的过程。光合作用消耗二氧化碳和水，将光能以化学能形式储存在葡萄糖中，并释放氧气作为副产物。有氧呼吸消耗氧气和葡萄糖，释放储存的化学能供细胞工作，并产生二氧化碳和水作为废物。这种循环关系维持了大气中氧气和二氧化碳的平衡，并支撑着几乎所有生态系统的能量流动。
Comparing the energy changes quantitatively: photosynthesis is endothermic (ΔH positive), absorbing approximately 2800 kJ of solar energy per mole of glucose synthesised. Aerobic respiration is exothermic (ΔH negative), releasing approximately 2880 kJ per mole of glucose oxidised. The slight discrepancy is due to the inefficiency of energy transfer and the fact that some energy remains bound in ATP molecules rather than being released as heat. Students should be able to identify which process is occurring by examining whether carbon dioxide and oxygen levels are increasing or decreasing in a given scenario.
定量比较能量变化：光合作用是吸能过程（ΔH 为正），每合成一摩尔葡萄糖约吸收 2800 kJ 的太阳能。有氧呼吸是放能过程（ΔH 为负），每氧化一摩尔葡萄糖约释放 2880 kJ 能量。微小差异源于能量传递的低效性以及部分能量仍结合在 ATP 分子中而非以热能形式释放。学生应能通过
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