📚 Year 7 SQA Biology: Formula & Theorem Quick Reference | 七年级SQA生物公式定理速查手册
This quick reference handbook brings together every essential formula, equation, and principle you need for Year 7 SQA Biology. Use it to review key calculations and core concepts before assessments.
这本速查手册汇集了七年级SQA生物所有必考的公式、方程式和基本原理，帮助你巩固重点，高效备考。
1. Total Magnification in Microscopy | 显微镜总放大倍率
The total magnification of a light microscope is found by multiplying the magnifying power of the eyepiece lens by the magnifying power of the objective lens. This tells you how much larger a specimen appears compared with its actual size.
光学显微镜的总放大倍率等于目镜放大倍数乘以物镜放大倍数。它表示标本比实际尺寸放大了多少倍。
Total Magnification = Eyepiece Magnification × Objective Magnification
iece lens is 10× and the objective lens is 40×, the total magnification is 10 × 40 = 400×.
示例：若目镜为10×，物镜为40×，则总放大倍率 = 10 × 40 = 400×。
2. Photosynthesis Word Equation | 光合作用文字方程式
Photosynthesis is the process by which green plants make their own food. The word equation summarises the reactants and products, showing that light energy and chlorophyll are essential for the reaction.
光合作用是绿色植物制造自身养分的过程。文字方程式概括了反应物和产物，并表明光能和叶绿体是反应发生的关键条件。
carbon dioxide + water → glucose + oxygen
The reaction only proceeds when chlorophyll absorbs light energy. Glucose is used for respiration, growth and storage, while oxygen is released as a waste gas.
只有当叶绿体吸收光能时，该反应才能进行。葡萄糖用于呼吸、生长和储存，氧气为副产品释放。
3. Aerobic Respiration Word Equation | 有氧呼吸文字方程式
Aerobic respiration takes place in the mitochondria of cells when oxygen is available. It releases a relatively large amount of energy from glucose, producing carbon dioxide and water as waste.
有氧呼吸在氧气充足时发生于细胞的线粒体中。它从葡萄糖释放大量能量，同时产生二氧化碳和水作为废物。
glucose + oxygen → carbon dioxide + water (+ energy)
The energy released is used for muscle contraction, maintaining body temperature, and powering all active processes in living organisms.
释放的能量用于肌肉收缩、维持体温以及驱动所有生命活动。
4. Anaerobic Respiration in Animals | 动物无氧呼吸
When oxygen is in short supply, such as during vigorous exercise, animal cells switch to anaerobic respiration. This process releases less energy and produces lactic acid instead of carbon dioxide and water.
当氧气不足时（如剧烈运动期间），动物细胞会转而进行无氧呼吸。此过程释放较少能量，并产生乳酸而非二氧化碳和水。
glucose → lactic acid (+ a small amount of energy)
Lactic acid build-up causes muscle fatigue and soreness. After exercise, oxygen is needed to break down the lactic acid – this is called oxygen debt.
乳酸积累会导致肌肉疲劳和酸痛。运动后需要氧气分解乳酸，这称为氧债。
5. The 10% Rule of Energy Transfer | 能量传递的10%法则
Only about 10% of the energy stored in one trophic level is passed on to the next trophic level in a food chain. Most energy is lost through movement, heat, respiration and undigested wastes.
在食物链中，每一营养级储存的能量只有大约10%能传递给下一营养级。大部分能量因运动、产热、呼吸和未消化物质而流失。
Energy passed on ≈ 10% of the energy taken in
This principle explains why food chains rarely have more than four or five trophic levels and why the biomass decreases at higher levels.
这一原理解释了为什么食物链很少超过四到五个营养级，并且越高级生物量越少。
6. Calculating Percentage Efficiency of Energy Transfer | 能量传递效率计算
To work out the exact efficiency of energy transfer between two trophic levels, use the percentage efficiency formula. It is often used with data from food chains or pyramids of energy.
要精确计算两个营养级间的能量传递效率，可使用百分比效率公式，常结合食物链或能量金字塔数据使用。
Efficiency (%) = (Energy available to next level / Energy available from previous level) × 100%
om grass and only 500 kJ is stored in its tissues, the efficiency is (500/5000) × 100 = 10%.
例如，一只兔子从草中摄取5000千焦能量，仅500千焦储存于自身组织，则效率为 (500/5000) × 100 = 10%。
7. Percentage Change in Mass (Osmosis Experiments) | 质量变化百分比（渗透实验）
When investigating osmosis using plant tissues such as potato cylinders, you must calculate the percentage change in mass to compare results fairly. This formula accounts for differences in starting mass.
在用土豆条等植物组织研究渗透作用时，必须计算质量变化百分比，以公平比较结果并消除初始质量差异的影响。
Percentage Change in Mass = ((Final Mass – Initial Mass) / Initial Mass) × 100%
A positive value indicates water has entered the cells by osmosis (gain in mass). A negative value means water has left the cells, causing plasmolysis.
正值表示水分通过渗透进入细胞（质量增加）；负值表示水分离开细胞，发生质壁分离。
8. Estimating Population Size Using Quadrats | 样方估算种群数量
Ecologists use quadrats to sample plant or slow-moving animal populations in a habitat. The mean number of organisms per quadrat is scaled up to the whole area to estimate the total population.
生态学家用样方对一个生境中的植物或移动缓慢的动物进行取样，再将每个样方的平均个体数外推至整个区域，估算总种群大小。
Estimated Population = Mean count per quadrat × (Total area / Quadrat area)
aisies and the field area is 500 m², the estimated daisy population is 8 × (500/1) = 4000.
示例：若一个1平方米的样方平均有8朵雏菊，田地面积为500平方米，则估算雏菊数量为 8 × (500/1) = 4000。
9. Pyramids of Numbers and Biomass | 数量金字塔与生物量金字塔
A pyramid of numbers shows the count of organisms at each trophic level, while a pyramid of biomass shows the total mass of living material. Both pyramids usually get smaller towards the top because of energy losses.
数量金字塔显示每个营养级的个体数量，生物量金字塔显示总生物存量。由于能量逐级损失，两种金字塔通常越往上越小。
Exceptions can occur: one large tree may support many insects, giving an inverted pyramid of numbers. However, the pyramid of biomass nearly always has a broad base and a narrow apex.
也有例外：一棵大树可能支撑大量昆虫，形成倒置的数量金字塔；但生物量金字塔几乎总是底宽顶窄。
10. Lock and Key Model of Enzyme Action | 酶作用的锁钥模型
Enzymes are biological catalysts that speed up reactions. The lock and key model describes how an enzyme’s active site has a specific shape that only fits a complementary substrate, just as a key fits a lock.
酶是加速反应的生物催化剂。锁钥模型描述酶活性部位拥有特定形状，只能与互补的底物结合，如同钥匙插入锁孔。
Once the substrate is bound, the reaction occurs and products are released. The enzyme remains unchanged and can be used again. High temperatures or extreme pH can denature the enzyme, changing the shape of the active site so it no longer works.
底物结合后发生反应并释放产物，酶本身不变且可重复使用。高温或极端pH会使酶变性，改变活性部位形状使其失活。
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