📚 2.2 Biological Molecules: Exam Practice | 生物：2.2 生物大分子真题精练
This article is a focused revision drill for the ‘Biological Molecules’ topic, commonly found in A-Level Biology Paper 2 or equivalent. We cover the most testable concepts—monomers and polymers, carbohydrates, lipids, proteins, enzymes, nucleic acids, ATP, and water—paired with typical exam-style commentary and pitfall alerts. Each section presents English explanations immediately followed by Chinese translations, making it ideal for bilingual learners aiming for top marks.
本文是针对A-Level生物学“生物大分子”章节的真题精练。内容涵盖单体与多聚体、碳水化合物、脂质、蛋白质、酶、核酸、ATP以及水的核心考点，每部分都配有考试中常见的解析思路和易错警示。每个要点均采用先英文后中文的配对讲解，适合双语学习者冲刺高分。
1. Monomers and Polymers | 单体与多聚体
Most large biological molecules are polymers made of repeating monomer units. Monomers are joined by condensation reactions that remove a water molecule, while polymers are broken down by hydrolysis, which adds water. This principle applies to carbohydrates, proteins, and nucleic acids.
大多数生物大分子是由重复单体组成的多聚体。单体通过缩合反应连接起来，每连接一个单体脱去一分子水；多聚体则通过水解反应断裂，需要加入水分子。这一原理普遍适用于碳水化合物、蛋白质和核酸。
A typical condensation equation for a disaccharide: C₆H₁₂O₆ + C₆H₁₂O₆ → C₁₂H₂₂O₁₁ + H₂O. Notice the product has one fewer H₂O than the sum of the reactants.
典型的双糖缩合反应式：C₆H₁₂O₆ + C₆H₁₂O₆ → C₁₂H₂₂O₁₁ + H₂O。注意产物比两个单体的总和少了一分子水。
2. Carbohydrates: Testing for Reducing Sugars | 碳水化合物：还原糖的检测
The Benedict’s test distinguishes reducing from non‑reducing sugars. All monosaccharides and some disaccharides (e.g., maltose) are reducing sugars because they have a free aldehyde or ketone group that can donate electrons to Cu²⁺, reducing it to Cu⁺ which forms a brick‑red precipitate.
本尼迪克特试剂可区分还原糖和非还原糖。所有单糖和部分双糖（如麦芽糖）是还原糖，因为它们含有游离的醛基或酮基，能将试剂中的Cu²⁺还原成Cu⁺，生成砖红色沉淀。
In the exam, you must state that the mixture must be heated in a water bath at ≥80°C. A green → yellow → orange → brick‑red colour sequence indicates increasing concentration of reducing sugar. Non‑reducing sugars (e.g., sucrose) give a negative result until they are first hydrolysed by boiling with dilute HCl, neutralised, and then tested again.
答题时必须说明需要在水浴中加热至80°C以上。颜色由绿→黄→橙→砖红表明还原糖浓度递增。非还原糖（如蔗糖）直接测试为阴性；必须先与稀盐酸煮沸水解，中和后再进行本尼迪克特检测，才会出现阳性结果。
3. Polysaccharides: Starch, Glycogen and Cellulose | 多糖：淀粉、糖原与纤维素
These glucose polymers differ in structure and function. Use this comparison table to lock in marks for ‘structure–function’ questions.
这三种葡萄糖多聚体在结构和功能上各不相同。利用下表可牢固掌握“结构—功能”类考题。
| Feature | Starch | Glycogen | Cellulose |
|---|---|---|---|
| Monomer | α‑glucose | α‑glucose | β‑glucose |
| Glycosidic bonds | α‑1,4 and α‑1,6 (amylopectin) | α‑1,4 and many α‑1,6 | β‑1,4 |
| Branching | Amylose unbranched; amylopectin branched | Highly branched | Unbranched, straight chains |
| Role | Energy storage in plants | Energy storage in animals | Structural component of plant cell walls |
| Key property | Compact, insoluble, easily hydrolysed | More branched → rapid glucose release | Hydrogen bonds between parallel chains form microfibrils → high tensile strength |
中文释义：单体：α‑葡萄糖（淀粉/糖原），β‑葡萄糖（纤维素）；糖苷键：α‑1,4和α‑1,6（淀粉），高度分支α‑1,6（糖原），β‑1,4（纤维素）；分支程度：直链/支链/高度分支；作用：植物储能、动物储能、植物细胞壁结构；关键性质：淀粉致密不溶易水解；糖原分支更多供快速动员；纤维素链间氢键形成微纤丝，具高抗拉强度。
4. Lipids: Triglycerides and Ester Bonds | 脂质：甘油三酯与酯键
Triglycerides form when one glycerol molecule condenses with three fatty acids. Each condensation creates an ester bond and releases one water molecule. The overall reaction can be summarised:
甘油三酯由一分子甘油与三分子脂肪酸缩合而成。每形成一个酯键脱去一分子水。总反应可表示为：
Glycerol + 3 Fatty acids → Triglyceride + 3H₂O
The ester bond is –COO–. In a diagram, you must be able to circle this linkage and label the reactants. The resulting molecule is hydrophobic and ideal for long‑term energy storage, thermal insulation, and protection of organs.
酯键为 –COO– 。在结构图中必须能圈出该键并标注反应物。生成的甘油三酯疏水，适合长期储能、隔热和保护器官。
Phospholipids replace one fatty acid with a phosphate‑containing group. This makes the head hydrophilic and the tail hydrophobic, allowing them to form a bilayer in water—the foundation of cell membranes.
磷脂将一个脂肪酸替换为含磷酸基团，使头部亲水、尾部疏水，能在水中形成双分子层，这是细胞膜的基础。
5. Lipid Tests: Emulsion and Translucent Spot | 脂质检测：乳化和透明斑点
The emulsion test is the standard assay for lipids. Shake the sample with absolute ethanol, then pour the mixture into water. A milky‑white emulsion confirms lipids. The translucent spot test on filter paper is a simpler alternative but is less specific.
脂质的标准检测是乳化测试：将样品与无水乙醇振荡，再将混合液倒入水中，若出现乳白色乳化层即含脂质。滤纸透明斑点测试更简便，但特异性较低。
Exam markers frequently penalise students for mixing ethanol with water before shaking—the ethanol must dissolve the lipid first. Also, remember to cite safety precautions when using ethanol (flammable, use no naked flame).
阅卷人常扣分点：先加水再振荡——乙醇必须先溶解脂质。此外务必提及乙醇易燃的安全注意事项。
6. Amino Acids and the Peptide Bond | 氨基酸与肽键
Every amino acid has a central carbon bonded to an amino group (–NH₂), a carboxyl group (–COOH), a hydrogen atom, and a variable R group. Two amino acids undergo a condensation reaction between the –COOH of one and the –NH₂ of the other, forming a peptide bond (–CONH–) and releasing water.
每个氨基酸有一个中心碳原子，分别连接氨基(–NH₂)、羧基(–COOH)、一个氢原子和可变的R基。两分子氨基酸缩合时，一方的羧基与另一方的氨基脱去水，形成肽键(–CONH–)。
Amino acid₁ + Amino acid₂ → Dipeptide + H₂O
The peptide bond is planar and has partial double‑bond character, restricting rotation. In exam diagrams, be ready to identify the peptide link and count residues.
肽键为平面结构，具部分双键性质，限制了旋转。考试中的结构图要求能识别肽键并计数氨基酸残基。
7. Protein Structure: Four Levels | 蛋白质的四级结构
Proteins have four structural levels. Primary: linear sequence of amino acids. Secondary: local folding into α‑helices or β‑pleated sheets, stabilised by hydrogen bonds along the backbone. Tertiary: overall 3‑D folding driven by interactions between R groups—disulfide bridges, ionic bonds, hydrophobic interactions, and hydrogen bonds. Quaternary: assembly of two or more polypeptide chains (e.g., haemoglobin with four subunits).
蛋白质有四级结构。一级：氨基酸线性序列。二级：局部折叠成α‑螺旋或β‑折叠，由主链上的氢键维持。三级：整体三维折叠，依赖R基间的二硫键、离子键、疏水作用和氢键。四级：两条或多条肽链的组合（如血红蛋白四个亚基）。
A common exam trick is to ask why a change in primary structure affects the protein’s function. Answer must link altered amino acid sequence → changed R group interactions → different tertiary folding → altered shape of active site / binding site → loss of function.
常见考题：为何一级结构改变会影响功能？必须回答：氨基酸序列改变→R基相互作用变化→三级折叠改变→活性位点/结合位点形状变化→功能丧失。
8. Enzymes: Models and Inhibition | 酶：作用模型与抑制
Enzymes are globular proteins that lower activation energy. The induced‑fit model says the active site changes shape slightly as the substrate binds, applying stress to bonds. Exam answers should contrast this with the outdated lock‑and‑key model.
酶是球状蛋白，能降低活化能。诱导契合模型认为底物结合时活性部位略微改变形状，对化学键施加张力。答题时应与过时的锁钥模型进行对比。
Competitive inhibitors resemble the substrate and bind to the active site, competing for it. This effect can be overcome by increasing substrate concentration. Non‑competitive inhibitors bind to an allosteric site, altering the enzyme’s shape irrespective of substrate concentration. Use Vmax and Km graphs to distinguish them.
竞争性抑制剂与底物结构相似，占据活性位点，增加底物浓度可克服。非竞争性抑制剂结合变构位点，改变酶的形状，与底物浓度无关。可利用 Vmax 和 Km 图线区分两者。
9. Nucleic Acids: DNA versus RNA | 核酸：DNA 与 RNA 对比
DNA contains deoxyribose, is double‑stranded with complementary base pairing (A‑T, C‑G) and stores genetic information. RNA contains ribose, is usually single‑stranded, uses uracil instead of thymine, and plays roles in protein synthesis (mRNA, tRNA, rRNA).
DNA含脱氧核糖，双链结构，碱基互补配对(A‑T, C‑G)，储存遗传信息。RNA含核糖，通常单链，用尿嘧啶替代胸腺嘧啶，在蛋白质合成中发挥作用（mRNA、tRNA、rRNA）。
A common question asks for two structural differences between DNA and RNA. Provide: (1) DNA has deoxyribose, RNA has ribose; (2) DNA is double‑stranded, RNA is single‑stranded (or DNA uses thymine, RNA uses uracil). Also ensure you can label the 3′ and 5′ ends on a nucleotide diagram.
常考题：列出DNA与RNA的两个结构差异。可答：(1) DNA含脱氧核糖，RNA含核糖；(2) DNA为双链，RNA为单链（或DNA含胸腺嘧啶，RNA含尿嘧啶）。同时要能在核苷酸图上标注3’端和5’端。
10. ATP: Structure and Energy Currency | ATP：结构与能量货币
Adenosine triphosphate (ATP) consists of adenine, ribose, and three phosphate groups. Energy is released when the terminal phosphate bond is hydrolysed: ATP → ADP + Pᵢ (+ 30.6 kJ mol⁻¹ under standard conditions). ATP is the immediate energy source for metabolic reactions, active transport, and muscle contraction.
三磷酸腺苷(ATP)由腺嘌呤、核糖和三个磷酸基团构成。末端磷酸键水解时释放能量：ATP → ADP + Pᵢ（标准条件下约释放30.6 kJ mol⁻¹）。ATP是代谢反应、主动运输和肌肉收缩的直接供能物质。
Exam tip: Do not call ATP a ‘long‑term energy store’—it is a short‑term, immediate donor. Also link ATP to phosphorylation, which often activates enzymes or changes protein shape.
考试提醒：不要将ATP称为“长期储能物质”——它是即时供体。同时将ATP与磷酸化联系起来，磷酸化常激活酶或改变蛋白质形状。
11. Water: Properties and Biological Roles | 水：特性与生物学作用
Water’s polarity and hydrogen bonding give it unique properties: high specific heat capacity (thermal buffer), high latent heat of vaporisation (cooling via sweating), cohesion and surface tension (transpiration stream), excellent solvent (transport medium), and lower density as ice (insulation of aquatic life). For each property, be prepared to state the biological consequence.
水的极性和氢键赋予其独特性质：高比热容（温度缓冲）、高汽化潜热（出汗冷却）、内聚力与表面张力（蒸腾流）、优良溶剂（运输介质）、冰的密度比水小（水体绝缘）。针对每个特性，都要能说出相应的生物学意义。
In exam answers, connect water’s dipole nature to its ability to surround and dissolve charged ions and polar molecules such as glucose and amino acids. This is central to transport in blood and xylem.
考试中需由水的偶极本质联系到它能水合离子和溶解极性的葡萄糖、氨基酸等，这是血液和木质部运输的核心。
12. Exam Pitfalls and Key Reminders | 常见错误与重点提醒
Pitfall 1: Confusing the terms ‘condensation’ and ‘hydrolysis’ in a water‑balance context. Condensation = water released, hydrolysis = water consumed. Draw the arrow accordingly.
陷阱一：搞混缩合和水解与水分子的关系。缩合释放水，水解消耗水。务必正确标注反应箭头上的水分子。
Pitfall 2: Calling starch a ‘single’ polymer—remember starch is a mixture of amylose and amylopectin. Amylose coils, amylopectin branches. Both are polymers of α‑glucose.
陷阱二：称淀粉为“一种”多聚体——淀粉是直链淀粉和支链淀粉的混合物。直链淀粉螺旋状，支链淀粉有分支。两者均为α‑葡萄糖的多聚体。
Pitfall 3: Saying ‘enzymes are killed by high temperature’. They are denatured—the peptide bonds do not break, but the tertiary structure unfolds and active site shape is lost. Use ‘denatured’.
陷阱三：说“酶被高温杀死”。应使用“变性”——肽键未断裂，但三级结构展开、活性位点形状改变。始终用术语“denatured”。
Pitfall 4: For phospholipids, forgetting to mention the hydrophobic barrier role. Always link structure (hydrophilic head outward, hydrophobic tails inward) to selective permeability of the membrane.
陷阱四：磷脂题型中未提及疏水屏障功能。务必将结构（亲水头朝外、疏水尾朝内）与膜的选择透过性联系在一起。
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