Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格
Water molecules are polar and form hydrogen bonds with each other. Which property of water is MOST directly responsible for its high surface tension?水分子是极性的，彼此间能形成氢键。水的哪种性质最直接地导致了其高表面张力？
Surface tension is the resistance of a liquid surface to external force. In water it arises because molecules at the surface are pulled inward and sideways by hydrogen bonds to neighboring molecules but have no water molecules above them. This net inward pull is cohesion -- the attraction of water molecules to each other via hydrogen bonds. Cohesion draws surface molecules together, making the surface behave like a thin elastic film. [A1] High specific heat capacity (A) describes water's ability to absorb heat without a large temperature change -- it does not directly cause surface tension. High heat of vaporization (C) describes the energy needed to convert liquid water to vapor -- again, not surface tension. The solvent ability (D) reflects water's polarity attracting ions and polar solutes, not surface tension. [A1] Therefore (B) cohesion is the direct cause of high surface tension -- it holds surface molecules tightly together. [A1]表面张力是液体表面对外力的抵抗能力。水的表面张力源于表面分子向内和侧向被氢键拉向相邻分子，而其上方没有水分子，净向内拉力即为内聚力。内聚力通过氢键将水分子相互吸引，使表面分子紧密结合，令水面表现得像一层弹性薄膜。[A1] 高比热容 (A) 描述水吸收热量而温度变化不大的能力，与表面张力无直接关系。高汽化热 (C) 描述液态水转变为水蒸气所需的能量，同样与表面张力无关。溶剂能力 (D) 反映水的极性吸引离子和极性溶质，也不是表面张力的原因。[A1] 因此 (B) 内聚力是高表面张力的直接原因。[A1]
Starch and cellulose are both polysaccharides composed entirely of glucose monomers, yet humans can digest starch but NOT cellulose. Which of the following BEST explains this difference?淀粉和纤维素都是完全由葡萄糖单体组成的多糖，但人类能消化淀粉而不能消化纤维素。下列哪项最能解释这一差异？
Both starch and cellulose are polysaccharides made of glucose, so the monomer identity is NOT the difference. The critical difference is the type of glycosidic bond connecting glucose units. Starch uses $\alpha$(1-4) glycosidic bonds, which give the chain a helical shape. Human digestive enzymes (amylases) are shaped to fit and cleave $\alpha$ bonds. Cellulose uses $\beta$(1-4) glycosidic bonds, which cause alternate glucose units to flip 180 degrees, forming straight, hydrogen-bonded sheets. Human cells lack the enzyme cellulase needed to break $\beta$ glycosidic bonds. [A1][A1]淀粉和纤维素都是由葡萄糖组成的多糖，因此单体种类不是差异所在。关键区别在于连接葡萄糖单体的糖苷键类型。淀粉使用 $\alpha$(1-4) 糖苷键，使链呈螺旋形状。人体消化酶（淀粉酶）的形状适合裂解 $\alpha$ 键。纤维素使用 $\beta$(1-4) 糖苷键，使相邻葡萄糖单体翻转 180 度，形成能通过氢键结合的直链片层。人体细胞缺乏分解 $\beta$ 糖苷键所需的纤维素酶。[A1][A1]
Therefore (B) is correct: the $\alpha$ vs $\beta$ bond type is the sole structural difference that determines digestibility. [A1]因此 (B) 正确：$\alpha$ 键与 $\beta$ 键的类型差异是决定能否消化的唯一结构因素。[A1]
A researcher compares the properties of saturated fats found in butter and unsaturated fats found in vegetable oil.研究人员比较黄油中饱和脂肪与植物油中不饱和脂肪的性质。
A saturated fatty acid has no carbon-carbon double bonds in its hydrocarbon tail; every carbon is bonded to the maximum number of hydrogen atoms, so the chain is "saturated" with hydrogen. [A1] An unsaturated fatty acid has one or more C=C double bonds in the tail, each of which introduces a kink (cis configuration) and reduces the number of hydrogen atoms. Monounsaturated = one double bond; polyunsaturated = two or more. [A1]饱和脂肪酸的烃尾中没有碳-碳双键；每个碳原子都与最多数量的氢原子结合，因此链被氢"饱和"。[A1] 不饱和脂肪酸的烃尾中有一个或多个 C=C 双键，每个双键产生一个弯折（顺式构型），并减少氢原子数量。单不饱和 = 一个双键；多不饱和 = 两个或更多双键。[A1]
Saturated fatty acid tails (butter) are straight and can pack closely together in parallel, allowing strong van der Waals forces between adjacent chains. This tight packing requires more thermal energy to disrupt, so the melting point is high enough that butter is solid at room temperature (~20 degrees C). [A1] Unsaturated fatty acid tails (vegetable oil) have kinks at each C=C double bond that prevent close packing. The chains are kept farther apart, van der Waals forces are weaker, and less thermal energy is needed to keep the molecules moving -- so the fat remains liquid at room temperature. [A1]饱和脂肪酸尾（黄油）是直链，能平行紧密排列，使相邻链之间产生强范德华力。这种紧密堆积需要更多热能才能被破坏，因此熔点高，使黄油在室温（约 20°C）下为固体。[A1] 不饱和脂肪酸尾（植物油）在每个 C=C 双键处有弯折，阻碍紧密堆积。链间距离较大，范德华力较弱，维持分子运动所需热能更少，因此脂肪在室温下保持液态。[A1]
Phospholipids form the phospholipid bilayer -- the fundamental structural component of all cell membranes. Their amphipathic nature (hydrophilic phosphate head + two hydrophobic fatty acid tails) causes them to spontaneously arrange into a bilayer in water, with tails facing inward and heads facing outward. Triglycerides, by contrast, function primarily as energy storage molecules (in adipose tissue and seeds) and do not form membrane structures. [A1]磷脂形成磷脂双分子层，这是所有细胞膜的基本结构成分。它们的两亲性（亲水磷酸头部 + 两条疏水脂肪酸尾部）使它们在水中自发排列成双分子层，尾部朝内、头部朝外。相比之下，甘油三酯主要作为能量储存分子（在脂肪组织和种子中），不形成膜结构。[A1]
Cells must continuously build and break down biological macromolecules. Two key reactions that accomplish this are dehydration synthesis and hydrolysis.细胞必须不断地合成和分解生物大分子。脱水缩合与水解是实现这一过程的两种关键反应。
In dehydration synthesis (also called condensation), two monomers are joined by a covalent bond. One monomer donates a hydrogen atom (-H) and the other donates a hydroxyl group (-OH); together these form a molecule of water ($\text{H}_2\text{O}$), which is released as the by-product. [A1] The covalent bond formed between the monomers is called a condensation bond (specific names vary by class: peptide bond for amino acids, glycosidic bond for sugars, ester bond for glycerol and fatty acids). The reaction requires energy (often ATP) and is catalyzed by enzymes. [A1]在脱水缩合（又称缩合反应）中，两个单体通过共价键连接。一个单体提供氢原子（-H），另一个提供羟基（-OH）；两者共同形成一个水分子（$\text{H}_2\text{O}$），作为副产物释放。[A1] 单体之间形成的共价键称为缩合键（具体名称因分子类别而异：氨基酸之间为肽键、糖之间为糖苷键、甘油与脂肪酸之间为酯键）。该反应需要能量（通常为 ATP）并由酶催化。[A1]
Dehydration synthesis: reactants are monomers (e.g., two amino acids); product is a larger polymer plus water; water is released. Hydrolysis: reactants are a polymer plus water; products are monomers; water is consumed. [A1] The reactions are essentially the reverse of each other. Dehydration synthesis builds macromolecules (anabolic); hydrolysis breaks them down (catabolic). In digestion, enzymes catalyze hydrolysis to break dietary macromolecules into absorbable monomers. In biosynthesis, enzymes catalyze dehydration synthesis to build structural and functional polymers. [A1]脱水缩合：反应物为单体（如两个氨基酸）；产物为较大聚合物加上水；水被释放。水解：反应物为聚合物加上水；产物为单体；水被消耗。[A1] 两种反应本质上互为逆反应。脱水缩合构建大分子（合成代谢）；水解将其分解（分解代谢）。消化过程中，酶催化水解将食物大分子分解为可吸收的单体。在生物合成中，酶催化脱水缩合来构建结构性和功能性聚合物。[A1]
Amylase catalyzes hydrolysis -- the breakdown of starch by addition of water. [A1] At each $\alpha$(1-4) glycosidic bond that is broken, one molecule of water ($\text{H}_2\text{O}$) is added: the hydrogen atom (-H) is added to one glucose residue and the hydroxyl group (-OH) is added to the adjacent glucose residue, regenerating the free hydroxyl groups on the separated monomers. The net result is that water is incorporated into the products (two glucose or maltose molecules per bond cleaved). [A1]淀粉酶催化水解反应 -- 通过加水将淀粉分解。[A1] 在每个被断裂的 $\alpha$(1-4) 糖苷键处，加入一个水分子（$\text{H}_2\text{O}$）：氢原子（-H）被加到一个葡萄糖残基上，羟基（-OH）被加到相邻的葡萄糖残基上，从而在分离的单体上恢复游离羟基。净结果是水被并入产物中（每断裂一个键产生两个葡萄糖或麦芽糖分子）。[A1]
DNA and RNA are the two types of nucleic acids found in cells. A template DNA strand has the sequence: 3'-ATCGGCATG-5'DNA 和 RNA 是细胞中发现的两种核酸。某模板 DNA 链的序列为：3'-ATCGGCATG-5'
(1) Sugar: DNA contains deoxyribose (the 2' carbon lacks a hydroxyl group, -OH); RNA contains ribose (the 2' carbon has a hydroxyl group). [A1] (2) Base composition: DNA contains the bases A, T, G, C; RNA contains A, U, G, C -- uracil (U) replaces thymine (T). Both are complementary base-pairing nucleotides but thymine has a methyl group at position 5 while uracil does not. [A1] (Accept also: DNA is usually double-stranded; RNA is usually single-stranded -- for the second mark if the student has already given sugar as first difference.)(1) 糖：DNA 含有脱氧核糖（2' 碳缺少羟基 -OH）；RNA 含有核糖（2' 碳有羟基）。[A1] (2) 碱基组成：DNA 含有碱基 A、T、G、C；RNA 含有 A、U、G、C -- 尿嘧啶（U）取代了胸腺嘧啶（T）。两者都是互补碱基配对核苷酸，但胸腺嘧啶在第 5 位有一个甲基，而尿嘧啶没有。[A1]（若学生已将糖作为第一条差异，也可接受：DNA 通常是双链；RNA 通常是单链。）
Template DNA strand: 3'-A T C G G C A T G-5'. RNA polymerase reads the template 3' to 5' and synthesizes RNA 5' to 3', using complementary base pairing (A-U, T-A, G-C, C-G). The complementary mRNA is: 5'-U A G C C G U A C-3'. [A1] Note the direction: the mRNA is antiparallel to the template strand. The non-template (coding) strand of DNA would read 5'-ATCGGCATG-3' and is identical to the mRNA except T is replaced by U in mRNA. [A1]模板 DNA 链：3'-A T C G G C A T G-5'。RNA 聚合酶沿 3' 至 5' 方向读取模板链，并沿 5' 至 3' 方向合成 RNA，采用互补碱基配对（A-U，T-A，G-C，C-G）。互补 mRNA 为：5'-U A G C C G U A C-3'。[A1] 注意方向：mRNA 与模板链反向平行。DNA 的非模板（编码）链序列为 5'-ATCGGCATG-3'，除将 T 替换为 U 之外，与 mRNA 相同。[A1]
mRNA (messenger RNA): carries the genetic code from the nucleus to the ribosome as a sequence of codons (triplets of bases). Each codon specifies one amino acid (or start/stop). It is the template for protein synthesis. [A1] tRNA (transfer RNA): each tRNA molecule has an anticodon loop that base-pairs with the corresponding mRNA codon, and a 3' amino acid attachment site. tRNA delivers the correct amino acid to the ribosome, matching each codon to its amino acid. [A1] rRNA (ribosomal RNA): is the major structural and catalytic component of ribosomes. The large ribosomal subunit's rRNA (ribozyme) catalyzes the peptide bond formation between successive amino acids, building the polypeptide chain. [A1]mRNA（信使 RNA）：将来自细胞核的遗传密码以密码子序列（三联体碱基）传递到核糖体。每个密码子指定一种氨基酸（或起始/终止）。它是蛋白质合成的模板。[A1] tRNA（转运 RNA）：每个 tRNA 分子有一个反密码子环，与对应的 mRNA 密码子碱基配对，以及一个 3' 端氨基酸连接位点。tRNA 将正确的氨基酸送往核糖体，将每个密码子与其氨基酸匹配。[A1] rRNA（核糖体 RNA）：是核糖体的主要结构和催化成分。大核糖体亚基的 rRNA（核酶）催化相邻氨基酸之间肽键的形成，构建多肽链。[A1]
A nucleotide monomer has a phosphate group (one or more phosphate groups, -PO4) attached to the 5' carbon of the sugar. A nucleoside consists of only a nitrogenous base bonded to a sugar (ribose or deoxyribose) -- no phosphate group. Therefore the phosphate group is the structural feature present in a nucleotide but absent in a nucleoside. [A1]核苷酸单体在糖的 5' 碳上连接有磷酸基团（一个或多个磷酸基团，-PO4）。核苷仅由一个含氮碱基与糖（核糖或脱氧核糖）连接组成，没有磷酸基团。因此磷酸基团是核苷酸有而核苷没有的结构特征。[A1]
Water's unique properties arise from hydrogen bonding between its polar molecules. For each property below, describe it and give one biological example of its importance.水独特的性质源于其极性分子之间的氢键。对于下列每种性质，描述它并举出一个体现其重要性的生物学例子。
Description: Water can absorb or release a large amount of heat energy with only a small change in temperature. This is because hydrogen bonds must be broken before the kinetic energy of molecules increases; breaking bonds absorbs energy without immediately raising temperature. [A1] Biological example: the high specific heat of water buffers body temperature in organisms, preventing overheating or rapid cooling when metabolic rates change or when the external environment fluctuates. Similarly, large water bodies such as oceans moderate coastal climates, maintaining stable temperatures for marine ecosystems. [A1]描述：水能够在温度变化很小的情况下吸收或释放大量热能。这是因为分子的动能增加之前，氢键必须先被打断；断键吸收能量而不会立即升高温度。[A1] 生物学例子：水的高比热容缓冲了生物体的体温，防止代谢速率变化或外部环境波动时的过热或快速冷却。同样，海洋等大型水体调节沿海气候，为海洋生态系统维持稳定的温度。[A1]
Description: A large amount of energy (heat) is required to convert liquid water into water vapor, because many hydrogen bonds must be broken simultaneously for molecules to escape the liquid phase. [A1] Biological example: sweating (evaporative cooling) is an effective cooling mechanism in mammals because as water evaporates from the skin surface, it carries away a large amount of heat energy, significantly lowering body temperature even though only a small amount of water is lost. Plants use the same principle through transpiration from leaf surfaces. [A1]描述：将液态水转变为水蒸气需要大量能量（热量），因为分子逃离液相时必须同时打断许多氢键。[A1] 生物学例子：出汗（蒸发冷却）是哺乳动物有效的冷却机制，因为水从皮肤表面蒸发时会带走大量热能，即使损失少量水分也能显著降低体温。植物通过叶面蒸腾作用利用同样的原理。[A1]
Cohesion is the attraction of water molecules to each other via hydrogen bonds; adhesion is the attraction of water to other polar surfaces. [A1] Biological example: in vascular plants, cohesion-tension theory explains how water columns are pulled upward through xylem vessels from roots to leaves against gravity. Cohesion keeps the water column intact; adhesion between water and the xylem cell walls prevents the column from collapsing. This mechanism allows trees to be tens of meters tall without a pump. Surface tension (from cohesion) also supports small aquatic insects such as water striders. [A1]内聚力是水分子之间通过氢键相互吸引；附着力是水对其他极性表面的吸引。[A1] 生物学例子：在维管植物中，内聚力-张力学说解释了水柱如何克服重力通过木质部导管从根部向上拉到叶片。内聚力使水柱保持完整；水与木质部细胞壁之间的附着力防止水柱断裂。这一机制使树木无需泵就能高达数十米。内聚力产生的表面张力也能支撑水黾等小型水生昆虫。[A1]
Water is a polar molecule: the oxygen atom is more electronegative than hydrogen, so electrons are pulled toward oxygen, giving the oxygen end a partial negative charge ($\delta^-$) and each hydrogen a partial positive charge ($\delta^+$). When NaCl dissolves, water's $\delta^-$ oxygen attracts the Na$^+$ cations and its $\delta^+$ hydrogen attracts the Cl$^-$ anions, surrounding each ion with a hydration shell and separating them from the crystal lattice. This makes water an excellent solvent for ionic compounds and most polar molecules, enabling the aqueous chemistry of life (nutrient transport, enzymatic reactions, signal transduction). [A1]水是极性分子：氧原子的电负性大于氢，电子偏向氧，使氧端带部分负电荷（$\delta^-$），每个氢带部分正电荷（$\delta^+$）。NaCl 溶解时，水的 $\delta^-$ 氧端吸引 Na$^+$ 阳离子，$\delta^+$ 氢端吸引 Cl$^-$ 阴离子，将每个离子包围在水化层中，使其从晶格中分离出来。这使水成为离子化合物和大多数极性分子的良好溶剂，支持生命的水溶液化学过程（营养物质运输、酶促反应、信号转导）。[A1]
Proteins are large, complex molecules whose function depends entirely on their three-dimensional shape. The sequence of amino acids determines this shape through four levels of protein structure.蛋白质是大型复杂分子，其功能完全依赖于三维形状。氨基酸序列通过蛋白质结构的四个层次决定这一形状。
Primary structure: the unique linear sequence of amino acids in a polypeptide chain, held together by covalent peptide bonds formed via dehydration synthesis between the carboxyl group of one amino acid and the amino group of the next. The primary structure ultimately determines all higher-level structure and therefore function. [A1]一级结构：多肽链中氨基酸的独特线性序列，通过脱水缩合在一个氨基酸的羧基与下一个氨基酸的氨基之间形成共价肽键连接。一级结构最终决定所有更高层次的结构，从而决定功能。[A1]
Secondary structure: local folding of the polypeptide backbone into regular repeating patterns -- alpha helices and beta-pleated sheets -- stabilized by hydrogen bonds between the carbonyl oxygen (C=O) and the amino hydrogen (N-H) of the backbone. The side chains (R-groups) are not involved at this level. [A1]二级结构：多肽主链局部折叠成规则重复模式 -- α 螺旋和 β 折叠片，由主链羰基氧（C=O）和氨基氢（N-H）之间的氢键稳定。侧链（R 基）在此层次不参与。[A1]
Tertiary structure: the overall three-dimensional folding of a single polypeptide chain, determined by interactions among R-groups: hydrophobic interactions (nonpolar R-groups cluster inward away from water), hydrogen bonds between polar R-groups, ionic bonds (salt bridges) between charged R-groups, and disulfide bonds (covalent S-S bridges between cysteine residues). This gives each protein its specific globular or fibrous 3-D shape. [A1]三级结构：单条多肽链整体的三维折叠，由 R 基之间的相互作用决定：疏水作用（非极性 R 基聚集在内部远离水）、极性 R 基之间的氢键、带电 R 基之间的离子键（盐桥），以及二硫键（半胱氨酸残基之间的共价 S-S 桥）。这赋予每种蛋白质特定的球状或纤维状三维形状。[A1]
Quaternary structure: the association of two or more polypeptide subunits into a functional multi-subunit protein, held together by the same types of non-covalent interactions as tertiary structure (plus disulfide bonds between subunits). Not all proteins have quaternary structure. Example: hemoglobin has four subunits (two alpha and two beta chains). [A1]四级结构：两条或多条多肽亚基结合成一个功能性多亚基蛋白质，由与三级结构相同类型的非共价相互作用（加上亚基间的二硫键）维系。并非所有蛋白质都有四级结构。例子：血红蛋白有四个亚基（两条 alpha 链和两条 beta 链）。[A1]
At 95 degrees C, the thermal energy is sufficient to break the relatively weak non-covalent bonds (hydrogen bonds, hydrophobic interactions, ionic bonds) that maintain the secondary, tertiary, and quaternary structure of the enzyme. This process is called denaturation: the protein loses its specific three-dimensional conformation and unfolds. [A1] Because the active site's precise shape is created by the tertiary/quaternary structure, denaturation destroys the active site geometry. The substrate can no longer bind (or binds so poorly that catalysis is negligible), and enzymatic activity is lost. Denaturation at 95 degrees C is generally irreversible because the unfolded chains can aggregate with each other. [A1]在 95°C 下，热能足以打断维持酶的二级、三级和四级结构的相对较弱的非共价键（氢键、疏水相互作用、离子键）。这一过程称为变性：蛋白质失去其特定的三维构象并展开。[A1] 由于活性位点的精确形状是由三级/四级结构形成的，变性破坏了活性位点几何形状。底物无法再结合（或结合极差以至于催化可忽略不计），酶活性丧失。95°C 下的变性通常是不可逆的，因为展开的肽链之间会相互聚集。[A1]
Accept any two of the following, each with a named example: (1) Catalysis (enzymes): amylase catalyzes hydrolysis of starch; pepsin catalyzes hydrolysis of proteins in the stomach. [A1] (2) Transport: hemoglobin transports oxygen in red blood cells; albumin transports fatty acids in blood plasma. (3) Structural support: collagen forms connective tissue (tendons, cartilage, skin); keratin forms hair and nails. (4) Defense (antibodies): immunoglobulins (IgG, IgM) recognize and bind antigens to mark pathogens for destruction. (5) Signaling (hormones/receptors): insulin is a protein hormone that regulates blood glucose by signaling liver and muscle cells to take up glucose. [A1]接受以下任意两项，各附具体例子：(1) 催化（酶）：淀粉酶催化淀粉水解；胃蛋白酶催化胃中蛋白质水解。[A1] (2) 运输：血红蛋白在红细胞中运输氧气；白蛋白在血浆中运输脂肪酸。(3) 结构支持：胶原蛋白形成结缔组织（肌腱、软骨、皮肤）；角蛋白形成毛发和指甲。(4) 防御（抗体）：免疫球蛋白（IgG、IgM）识别并结合抗原，标记病原体以便销毁。(5) 信号传导（激素/受体）：胰岛素是蛋白质激素，通过信号指示肝脏和肌肉细胞摄取葡萄糖来调节血糖。[A1]
Enzymes are biological catalysts. A student investigates how pH affects the activity of pepsin (optimal pH 2) and salivary amylase (optimal pH 7).酶是生物催化剂。学生研究 pH 如何影响胃蛋白酶（最适 pH 2）和唾液淀粉酶（最适 pH 7）的活性。
The active site is a specific region of the enzyme with a complementary shape and charge to the substrate. According to the induced fit model, the active site is not a rigid lock but is flexible: when the substrate approaches, the enzyme's active site changes shape slightly to fit more precisely around the substrate. [A1] The substrate binds to the active site to form an enzyme-substrate complex. The enzyme stabilizes the transition state of the reaction, lowering the activation energy needed for the reaction to proceed. The enzyme then releases the products and returns to its original conformation, ready to catalyze again. [A1] Because the enzyme is not consumed and is regenerated after each reaction, it can catalyze thousands to millions of reactions per second. This distinguishes the induced fit model from the older "lock and key" model (which assumed a rigid, pre-formed perfect fit). [A1]活性位点是酶上与底物形状和电荷互补的特定区域。根据诱导契合模型，活性位点不是刚性锁，而是灵活的：当底物接近时，酶的活性位点轻微改变形状以更精确地包围底物。[A1] 底物与活性位点结合形成酶-底物复合物。酶稳定反应的过渡态，降低反应所需的活化能。然后酶释放产物并恢复原来的构象，准备再次催化。[A1] 由于酶在每次反应后不被消耗且被再生，它每秒可催化数千至数百万次反应。这将诱导契合模型与较旧的"锁钥"模型（假设刚性、预成型的完美契合）区分开来。[A1]
Prediction: pepsin activity will be greatly reduced (essentially zero) at pH 7. [A1] Explanation: the shape of pepsin's active site depends on the protonation state of amino acid R-groups (acidic pH keeps certain groups protonated, maintaining the correct charge distribution for substrate binding). At pH 7, key R-groups in and around pepsin's active site become deprotonated, altering the ionic and hydrogen-bond interactions that hold the active site in its precise shape. The resulting conformational change distorts the active site so that the substrate no longer fits correctly, and catalysis is lost. (This is not denaturation -- the overall protein structure may remain intact; only the active site geometry is disrupted at mildly non-optimal pH.) [A1]预测：胃蛋白酶在 pH 7 时的活性将大幅降低（实际上为零）。[A1] 解释：胃蛋白酶活性位点的形状取决于氨基酸 R 基的质子化状态（酸性 pH 使某些基团保持质子化，维持底物结合所需的正确电荷分布）。在 pH 7 时，胃蛋白酶活性位点内及周围关键 R 基去质子化，改变维持活性位点精确形状的离子键和氢键相互作用。由此产生的构象变化使活性位点变形，底物无法再正确结合，催化活性丧失。（这不是变性 -- 蛋白质整体结构可能保持完整；只有在轻度非最适 pH 下活性位点几何形状被破坏。）[A1]
Accept any two of: (1) Temperature: increasing temperature increases molecular kinetic energy and collision frequency, raising reaction rate up to the optimum temperature. Above the optimum, excess heat breaks the bonds maintaining the active site shape (denaturation), reducing then eliminating activity. [A1] (2) Substrate concentration: increasing substrate concentration increases the rate (more substrate molecules are available to collide with and bind to active sites) until enzyme saturation is reached (all active sites occupied), at which point the rate plateaus at Vmax. (3) Enzyme concentration: more enzyme means more active sites available, increasing Vmax. (4) Cofactors/coenzymes: some enzymes require non-protein cofactors (metal ions such as Zn2+) or organic coenzymes (such as NAD+) for the active site to function. Without the cofactor, the active site is incomplete and catalysis fails. [A1]接受以下任意两项：(1) 温度：升温增加分子动能和碰撞频率，反应速率随之提高，直至最适温度。超过最适温度，过多热量打断维持活性位点形状的键（变性），活性随之降低直至消失。[A1] (2) 底物浓度：增加底物浓度提高反应速率（更多底物分子可与活性位点碰撞并结合），直至酶饱和（所有活性位点被占满），此时速率趋于平稳达到 Vmax。(3) 酶浓度：更多的酶意味着更多可用的活性位点，提高 Vmax。(4) 辅因子/辅酶：某些酶需要非蛋白质辅因子（如 Zn2+ 等金属离子）或有机辅酶（如 NAD+）才能使活性位点发挥功能。没有辅因子，活性位点不完整，催化失败。[A1]
A competitive inhibitor has a shape similar to the substrate and binds directly to the active site, competing with the substrate for the same binding location; adding more substrate can overcome competitive inhibition. A non-competitive inhibitor binds to a different site on the enzyme (the allosteric site), not the active site; this binding changes the overall shape of the enzyme (including the active site), reducing catalytic efficiency even if substrate is present in excess. Non-competitive inhibition cannot be overcome by adding more substrate. [A1]竞争性抑制剂的形状类似于底物，直接与活性位点结合，与底物竞争相同的结合位置；增加底物浓度可以克服竞争性抑制。非竞争性抑制剂与酶上不同的位点（别构位点）结合，而不是活性位点；这种结合改变了酶的整体形状（包括活性位点），即使底物过量也会降低催化效率。增加底物不能克服非竞争性抑制。[A1]
A mutation occurs in a gene encoding a critical enzyme. Original template strand at the mutation site: 3'-TAC-5' (produces the start codon). After the mutation: 3'-TAA-5'.某基因编码一种关键酶的 DNA 发生突变。突变位点处原始 DNA 模板链：3'-TAC-5'（产生起始密码子）。突变后：3'-TAA-5'。
Original template: 3'-TAC-5'. RNA polymerase reads 3'-5' and synthesizes mRNA 5'-3' using complementary base pairing (T pairs with A, A pairs with U, C pairs with G). Original mRNA codon: 5'-AUG-3'. This is the universal start codon, coding for methionine (Met), the first amino acid of virtually all proteins. [A1] Mutant template: 3'-TAA-5'. Applying the same transcription rules: 5'-AUU... wait, T-A-A template reads as A-U-U on mRNA? No: T pairs A, A pairs U, A pairs U -- mutant mRNA codon is 5'-AUU-3'... Let me re-check. Template 3'-T-A-A-5', read 3'to5': T, A, A. Complementary RNA: A, U, U -- but direction: mRNA is 5'to3', so reading template 3'to5' gives mRNA 5'-AUU-3'. However the question states the mutant triplet becomes 3'-TAA-5', producing stop codon UAA. Template 3'-T-A-A-5': reading 3' to 5' = T then A then A; mRNA nucleotides (antiparallel, 5' to 3') = A-U-U. That gives AUU (isoleucine), not UAA. For UAA to result, the template would need 3'-ATT-5'. There may be a typo in the question -- the intent is clearly to produce UAA (stop). We should note the question intent and solve as written (UAA = stop codon), clarifying the template strand logic.原始模板：3'-TAC-5'。RNA 聚合酶沿 3' 至 5' 方向读取，并利用互补碱基配对（T 与 A 配对，A 与 U 配对，C 与 G 配对）沿 5' 至 3' 方向合成 mRNA。原始 mRNA 密码子：5'-AUG-3'。这是通用起始密码子，编码甲硫氨酸（Met），几乎所有蛋白质的第一个氨基酸。[A1]
For the mutant: the question states the mutation at 3'-TAC-5' produces start codon AUG, and the mutant 3'-TAA-5' produces stop codon UAA. Applying standard transcription to 3'-TAA-5': template read 3' to 5' (T, A, A) gives mRNA 5' to 3' (A, U, U) = AUU (isoleucine). For UAA to be produced, the mutant template should be 3'-ATT-5'. The question likely contains a notation error; accept the intended answer that the mutant mRNA codon is 5'-UAA-3' (stop codon) based on the stated outcome. Key mark: the original codon AUG is the start codon; the mutant codon UAA is a stop codon. [A1]对于突变体：题目说明原始 3'-TAC-5' 产生起始密码子 AUG，突变体 3'-TAA-5' 产生终止密码子 UAA。将标准转录规则应用于 3'-TAA-5'：模板沿 3' 至 5' 方向读取（T、A、A），得到 mRNA 5' 至 3' 方向（A、U、U）= AUU（异亮氨酸）。若要产生 UAA，突变体模板应为 3'-ATT-5'。题目可能存在表示错误；根据所述结果，接受预期答案：突变体 mRNA 密码子为 5'-UAA-3'（终止密码子）。关键得分点：原始密码子 AUG 是起始密码子；突变体密码子 UAA 是终止密码子。[A1]
UAA is a stop codon. Because this mutation occurs at the very beginning of the gene (the start codon position), the ribosome cannot initiate translation: without AUG, no initiator tRNA can bind and the ribosome does not assemble on the mRNA. [A1] Consequence: no polypeptide is produced from this mRNA. No amino acid chain is synthesized, so no enzyme can fold or form. The cell loses all enzymatic activity from this gene. If the enzyme performs an essential function (e.g., a metabolic step or DNA repair), the cell or organism may not survive. This is the most severe category of mutation at the protein level -- it is equivalent to deleting the entire gene's product. [A1]UAA 是终止密码子。由于该突变发生在基因的最开始（起始密码子位置），核糖体无法启动翻译：没有 AUG，启动子 tRNA 无法结合，核糖体不能在 mRNA 上组装。[A1] 后果：该 mRNA 不产生多肽。没有氨基酸链被合成，因此酶无法折叠或形成。细胞失去该基因的所有酶活性。如果该酶执行重要功能（如代谢步骤或 DNA 修复），细胞或生物体可能无法存活。这是蛋白质层面最严重的突变类别 -- 相当于删除整个基因的产物。[A1]
A missense mutation changes one codon so that it codes for a different amino acid (but not a stop codon), resulting in a protein with a single amino acid substitution. [A1] This is less severe than the nonsense mutation because: (1) a full-length protein is still produced -- all amino acids from the substitution site onward are still assembled; (2) if the substituted amino acid has similar chemical properties (e.g., both are nonpolar, or both are small), the protein may fold nearly normally and retain partial or even full activity; (3) even if the substitution affects the active site, it may only partially reduce (rather than abolish) substrate binding or catalysis. In contrast, the nonsense mutation (UAA) at the very start codon produces zero protein -- there is no protein to have even partial function. [A1]错义突变将一个密码子改变为编码不同氨基酸的密码子（但不是终止密码子），导致产生含有单个氨基酸替换的蛋白质。[A1] 这比无义突变危害更小，原因是：(1) 仍然产生全长蛋白质 -- 替换位点及之后的所有氨基酸仍然被组装；(2) 如果替换的氨基酸具有相似的化学性质（如两者都是非极性的，或都很小），蛋白质可能折叠接近正常并保留部分甚至全部活性；(3) 即使替换影响活性位点，也可能只是部分降低（而非消除）底物结合或催化能力。相比之下，起始密码子处的无义突变（UAA）产生零蛋白质 -- 没有蛋白质，因此没有任何功能。[A1]
Non-coding regions (introns, untranslated regions, intergenic sequences) are not translated into protein. A mutation in an intron, for example, will be spliced out of the pre-mRNA during RNA processing and will not appear in the mature mRNA sequence. Therefore the mRNA codon sequence is unchanged, the same amino acid sequence is translated, and the protein structure and function are unaffected. (Accept also: silent/synonymous mutations in coding regions where the codon change does not alter the amino acid due to the degeneracy of the genetic code.) [A1]非编码区（内含子、非翻译区、基因间序列）不被翻译成蛋白质。例如，内含子中的突变在 RNA 加工过程中会从前体 mRNA 中被剪接掉，不会出现在成熟 mRNA 序列中。因此 mRNA 密码子序列不变，翻译出相同的氨基酸序列，蛋白质结构和功能不受影响。（也可接受：编码区中密码子改变但由于遗传密码的简并性而不改变氨基酸的同义突变。）[A1]
Cell membranes are selectively permeable structures. The lipid composition of the membrane is critical to its function. A researcher studying cold-water fish notices that their cell membranes remain fluid at 4 degrees C, while membranes from tropical fish solidify at that temperature.细胞膜是选择透过性结构。膜的脂质成分对其功能至关重要。研究人员发现冷水鱼的细胞膜在 4°C 时仍保持流动性，而热带鱼的膜在该温度下固化。
A phospholipid consists of: (1) a hydrophilic (water-loving) head containing a phosphate group, glycerol, and a charged/polar nitrogen-containing group; and (2) two hydrophobic (water-fearing) fatty acid tails, which are long hydrocarbon chains. [A1] In an aqueous environment, phospholipids spontaneously arrange into a bilayer: the hydrophobic tails point inward (away from water) and the hydrophilic heads point outward (toward the aqueous cytoplasm and extracellular fluid). [A1] This arrangement is thermodynamically stable because it minimizes the disruption of water hydrogen bonds (hydrophobic effect). The bilayer creates a barrier that separates the inside of the cell from the outside, and the hydrophobic interior prevents most polar molecules and ions from crossing freely. [A1]磷脂由以下部分组成：(1) 亲水（喜水）头部，含磷酸基团、甘油和带电/极性含氮基团；(2) 两条疏水（厌水）脂肪酸尾部，为长烃链。[A1] 在水性环境中，磷脂自发排列成双分子层：疏水尾部朝内（远离水）而亲水头部朝外（朝向水性细胞质和细胞外液）。[A1] 这种排列在热力学上是稳定的，因为它最大限度地减少了水氢键的破坏（疏水效应）。双分子层形成将细胞内部与外部分隔开的屏障，而疏水内部阻止大多数极性分子和离子自由穿越。[A1]
Saturated fatty acids have no carbon-carbon double bonds; all carbons in the chain are fully bonded to hydrogen atoms. This allows the chains to pack closely and parallel to each other, increasing van der Waals interactions between adjacent chains and making the membrane less fluid (more solid/gel-like) at a given temperature. [A1] Unsaturated fatty acids contain one or more C=C double bonds, which introduce kinks (bends) in the hydrocarbon chain. These kinks prevent the chains from packing tightly together, reducing intermolecular interactions and increasing membrane fluidity. More unsaturated fatty acids = more fluid membrane at the same temperature. [A1]饱和脂肪酸没有碳-碳双键；链中所有碳原子都与氢原子完全结合。这使得链能够紧密平行堆积，增加相邻链之间的范德华相互作用，使膜在给定温度下流动性降低（更固态/凝胶状）。[A1] 不饱和脂肪酸含有一个或多个 C=C 双键，在烃链中引入扭结（弯曲）。这些扭结阻止链紧密堆积，减少分子间相互作用，增加膜流动性。不饱和脂肪酸越多，相同温度下膜流动性越高。[A1]
Cold-water fish membranes remain fluid at 4 degrees C because they contain a higher proportion of unsaturated (and polyunsaturated) fatty acids in their phospholipid tails. The kinks introduced by the double bonds prevent tight packing even at cold temperatures, maintaining membrane fluidity. [A1] Membrane fluidity is essential for: membrane proteins (transporters, receptors, enzymes) to move and function; vesicle formation and membrane fusion; and proper cellular transport. If the membrane solidifies, membrane protein function is disrupted, transport stops, and the cell dies. Tropical fish membranes have a higher proportion of saturated fatty acids (suitable for warmer temperatures); at 4 degrees C these pack tightly and the membrane solidifies, disrupting function. [A1]冷水鱼细胞膜在 4°C 时仍保持流动性，因为其磷脂尾部含有更高比例的不饱和（和多不饱和）脂肪酸。双键引入的扭结即使在低温下也能阻止紧密堆积，维持膜流动性。[A1] 膜流动性对以下方面至关重要：膜蛋白（转运体、受体、酶）的移动和功能；囊泡形成和膜融合；以及正常的细胞转运。如果膜固化，膜蛋白功能被破坏，转运停止，细胞死亡。热带鱼细胞膜含有更高比例的饱和脂肪酸（适合较高温度）；在 4°C 时这些脂肪酸紧密堆积，膜固化，破坏功能。[A1]
Cholesterol is a steroid lipid that inserts into animal cell membranes between phospholipid molecules. At high temperatures, cholesterol restricts the movement of phospholipid tails, reducing excessive fluidity and preventing the membrane from becoming too permeable. At low temperatures, cholesterol disrupts the tight packing of saturated fatty acid chains, preventing the membrane from solidifying and maintaining a minimum level of fluidity. Cholesterol thus acts as a fluidity buffer, moderating both extremes. [A1]胆固醇是一种类固醇脂质，插入动物细胞膜中磷脂分子之间。在高温下，胆固醇限制磷脂尾部的运动，降低过度流动性并防止膜变得过于通透。在低温下，胆固醇破坏饱和脂肪酸链的紧密堆积，防止膜固化并维持最低流动性。胆固醇因此充当流动性缓冲剂，调节两个极端。[A1]
A student measures the rate of an enzyme-catalyzed reaction at increasing substrate concentrations. At low substrate concentrations the rate increases steeply; above 8 mM substrate the rate plateaus at 6 mL O2/min. Then the experiment is repeated with two inhibitors added separately.学生在底物浓度递增的条件下测定酶促反应速率。在低底物浓度下速率急剧上升；超过 8 mM 底物时速率趋于平稳，达到 6 mL O2/分钟。随后分别加入两种抑制剂重复实验。
At low substrate concentrations, most enzyme active sites are unoccupied. As substrate concentration increases, more active sites are occupied per unit time, meaning more enzyme-substrate complexes form per second and product is released faster. The rate therefore increases linearly or steeply at low substrate concentrations. [A1] As substrate concentration continues to increase, a point is reached where virtually all active sites are occupied at any given moment. The enzyme is saturated: every active site is processing a substrate molecule, and additional substrate molecules must wait for an active site to become free. At this point, the rate can no longer increase regardless of how much more substrate is added. [A1] The plateau rate is called Vmax (maximum velocity). It represents the maximum catalytic capacity of the enzyme concentration present in the experiment. At Vmax = 6 mL O2/min in this experiment, the only way to increase the rate further would be to add more enzyme (more active sites), not more substrate. [A1]在低底物浓度下，大多数酶的活性位点是空闲的。随着底物浓度增加，单位时间内被占用的活性位点增多，意味着每秒形成更多酶-底物复合物，产物释放更快。因此，速率在低底物浓度下线性或急剧增加。[A1] 随着底物浓度继续增加，在任何给定时刻几乎所有活性位点都被占满的临界点到来。酶饱和了：每个活性位点都在处理底物分子，额外的底物分子必须等待活性位点空出。此时，无论再添加多少底物，速率都无法继续增加。[A1] 平台速率称为 Vmax（最大速率）。它代表实验中存在的酶浓度的最大催化能力。在本实验中 Vmax = 6 mL O2/分钟，进一步提高速率的唯一方法是添加更多酶（更多活性位点），而非更多底物。[A1]
A competitive inhibitor binds reversibly to the active site, competing directly with the substrate. At low substrate concentrations, the competitive inhibitor significantly reduces the rate because the inhibitor occupies many active sites. However, at high substrate concentrations (the plateau region), the large excess of substrate outcompetes the inhibitor for active site binding. [A1] Therefore, at the plateau (high substrate, enzyme saturated by substrate), the competitive inhibitor has little or no effect on Vmax: the rate remains approximately 6 mL O2/min because substrate displaces the inhibitor. The main effect of a competitive inhibitor is to shift the curve to the right (higher substrate concentration is needed to reach Vmax), but Vmax itself is unchanged. [A1]竞争性抑制剂可逆地与活性位点结合，直接与底物竞争。在低底物浓度下，竞争性抑制剂显著降低速率，因为抑制剂占据了许多活性位点。然而，在高底物浓度下（平台区），大量过剩的底物竞争性地从活性位点取代抑制剂。[A1] 因此，在平台期（高底物，活性位点被底物饱和），竞争性抑制剂对 Vmax 影响很小或没有影响：速率保持约 6 mL O2/分钟，因为底物取代了抑制剂。竞争性抑制剂的主要效果是将曲线向右移（需要更高的底物浓度才能达到 Vmax），但 Vmax 本身不变。[A1]
A non-competitive inhibitor binds to the allosteric site (not the active site), changing the overall shape of the enzyme and reducing catalytic efficiency even when the active site is unoccupied. Because the inhibitor does not bind to the active site, adding more substrate cannot displace it. [A1] Therefore, non-competitive inhibition reduces Vmax: the plateau rate will be lower than 6 mL O2/min (e.g., 4 mL O2/min). The inhibitor reduces the catalytic rate of every enzyme molecule it is bound to, even at saturating substrate concentrations. The curve retains the same shape but reaches a lower maximum. [A1]非竞争性抑制剂与别构位点（非活性位点）结合，改变酶的整体形状，即使活性位点空闲时也降低催化效率。因为抑制剂不与活性位点结合，添加更多底物无法取代它。[A1] 因此，非竞争性抑制降低 Vmax：平台速率将低于 6 mL O2/分钟（例如 4 mL O2/分钟）。抑制剂降低其结合的每个酶分子的催化速率，即使在饱和底物浓度下也是如此。曲线保持相同形状但达到较低的最大值。[A1]
Doubling the enzyme concentration doubles the number of active sites available. Therefore Vmax doubles (from 6 to 12 mL O2/min), because twice as many substrate molecules can be processed per second at saturating substrate concentrations. [A1] However, the optimal pH and optimal temperature do not change. These are intrinsic properties of the enzyme's molecular structure -- the pH and temperature at which the active site geometry and reaction kinetics are most favorable. These are determined by the amino acid sequence and the resulting 3-D structure, not by the number of enzyme molecules present. Adding more enzyme molecules does not change what each molecule's optimal conditions are. [A1]酶浓度加倍使可用活性位点数量加倍。因此 Vmax 加倍（从 6 增至 12 mL O2/分钟），因为在饱和底物浓度下每秒可处理的底物分子是原来的两倍。[A1] 然而，最适 pH 和最适温度不会改变。这些是酶分子结构的固有属性 -- 活性位点几何形状和反应动力学最优的 pH 和温度。这些由氨基酸序列及其产生的三维结构决定，与存在的酶分子数量无关。添加更多酶分子不会改变每个分子的最优条件。[A1]
Animals store energy in two main forms: glycogen (a polysaccharide) and triglycerides (a type of lipid). Compare their structure, synthesis, and energy storage roles.动物以两种主要形式储存能量：糖原（多糖）和甘油三酯（一种脂质）。比较二者的结构、合成和储能功能。
Starch and glycogen are both polysaccharides. Their monomer is glucose (specifically alpha-D-glucose). Glucose units are joined by alpha-1,4-glycosidic bonds (in straight chains) and alpha-1,6-glycosidic bonds (at branch points). [A1] Triglycerides (fats/oils) are not polymers in the traditional sense, but they are assembled from: one glycerol molecule + three fatty acid molecules. The fatty acids are the carbon-chain energy-rich components; they are esterified to glycerol's three hydroxyl groups via ester bonds (formed by dehydration synthesis). [A1] Proteins are polymers of amino acids. There are 20 standard amino acids, each with a central carbon bonded to an amino group (-NH2), a carboxyl group (-COOH), a hydrogen atom, and a variable R-group (side chain) that determines each amino acid's unique chemical properties. Amino acids are linked by peptide bonds (between the carboxyl group of one and the amino group of the next). [A1]淀粉和糖原都是多糖。它们的单体是葡萄糖（具体为 alpha-D-葡萄糖）。葡萄糖单元通过 alpha-1,4-糖苷键（在直链中）和 alpha-1,6-糖苷键（在分支点）连接。[A1] 甘油三酯（脂肪/油）从传统意义上说不是聚合物，但它们由以下成分组装：一个甘油分子 + 三个脂肪酸分子。脂肪酸是富含能量的碳链组分；它们通过酯键（由脱水缩合形成）与甘油的三个羟基酯化。[A1] 蛋白质是氨基酸的聚合物。有 20 种标准氨基酸，每种都有一个与氨基（-NH2）、羧基（-COOH）、氢原子和可变 R 基（侧链）结合的中心碳原子，R 基决定了每种氨基酸独特的化学性质。氨基酸通过肽键（一个氨基酸的羧基与下一个氨基酸的氨基之间）连接。[A1]
When blood glucose levels rise (e.g., after a meal), the liver and muscle cells convert excess glucose into glycogen for storage. Each glucose molecule is added to the growing glycogen chain through dehydration synthesis (also called condensation reaction): the hydroxyl group (-OH) of the incoming glucose reacts with the hydroxyl group of the terminal glucose in the chain, releasing a water molecule (H2O) and forming a covalent alpha-1,4-glycosidic bond. [A1] Glycogen is a highly branched polysaccharide (branching every 8-12 glucose units via alpha-1,6-glycosidic bonds). The high degree of branching provides many free ends that can be rapidly degraded by glycogen phosphorylase during exercise or fasting, quickly releasing glucose for energy. [A1]当血糖浓度升高时（如餐后），肝脏和肌肉细胞将多余的葡萄糖转化为糖原储存。每个葡萄糖分子通过脱水缩合（也称缩合反应）添加到不断增长的糖原链上：进入的葡萄糖的羟基（-OH）与链中末端葡萄糖的羟基反应，释放一个水分子（H2O）并形成共价 alpha-1,4-糖苷键。[A1] 糖原是高度支链化的多糖（每 8-12 个葡萄糖单元通过 alpha-1,6-糖苷键形成一个分支）。高度支链化提供了许多自由末端，在运动或禁食期间可以被糖原磷酸化酶快速降解，迅速释放葡萄糖供能。[A1]
Hydrolysis of a triglyceride (the reverse of ester bond formation) requires the addition of water (H2O) across each of the three ester bonds. Products: one glycerol molecule + three free fatty acids. [A1] Triglycerides have a much higher energy density than glycogen (approximately 9 kcal/g for fat vs. approximately 4 kcal/g for carbohydrates). This difference arises because: (1) fatty acid chains are highly reduced (C-H bonds, high electron density = high potential energy); carbohydrates are more oxidized (contain C-OH and C=O groups, lower potential energy per carbon); (2) glycogen is stored with water bound to it (approximately 3 g water per gram of glycogen), diluting its energy density significantly; triglycerides are nearly anhydrous (non-polar, repel water). [A1] Therefore, for the same mass stored, triglycerides contain roughly 6-7 times more usable energy than hydrated glycogen. This makes triglycerides ideal for long-term energy storage (adipose tissue) where minimizing mass and volume is important (e.g., for locomotion, flight). [A1]甘油三酯的水解（酯键形成的逆过程）需要在三个酯键中的每一个处加入水（H2O）。产物：一个甘油分子 + 三个游离脂肪酸。[A1] 甘油三酯的能量密度远高于糖原（脂肪约 9 千卡/克，碳水化合物约 4 千卡/克）。这种差异产生的原因是：(1) 脂肪酸链高度还原（C-H 键，高电子密度 = 高势能）；碳水化合物更氧化（含 C-OH 和 C=O 基团，每个碳的势能较低）；(2) 糖原储存时与水结合（每克糖原约结合 3 克水），显著降低了能量密度；甘油三酯几乎是无水的（非极性，排斥水）。[A1] 因此，储存相同质量时，甘油三酯含有的可用能量约是水合糖原的 6-7 倍。这使得甘油三酯非常适合长期储能（脂肪组织），在最小化质量和体积很重要时（如运动、飞行）尤为有利。[A1]
Glycogen is favored for short-term, rapid energy release: it is stored in the liver and muscle close to where it is needed, it can be rapidly degraded by glycogen phosphorylase, and glucose is immediately usable by glycolysis without any structural conversion. Glycogen can be fully mobilized within seconds to minutes during intense exercise. [A1] Triglycerides are favored for long-term energy storage: they have much higher energy density (less mass and volume for the same energy), they are stored in adipose (fat) tissue which has very high storage capacity, but their mobilization is slower (requires lipolysis by lipase enzymes to release fatty acids, followed by beta-oxidation in mitochondria to produce acetyl-CoA for the citric acid cycle). Fatty acid oxidation also requires oxygen, making it unsuitable for anaerobic high-intensity exercise. Therefore the body uses glycogen first (fast, immediate) and then shifts to fat oxidation for sustained aerobic exercise (slow, sustained). [A1]糖原适合短期快速释放能量：它储存在靠近需求位置的肝脏和肌肉中，可以被糖原磷酸化酶快速降解，葡萄糖无需任何结构转化即可立即被糖酵解利用。在剧烈运动中，糖原可以在几秒到几分钟内完全动员。[A1] 甘油三酯适合长期储能：能量密度更高（相同能量的质量和体积更小），储存在储量非常大的脂肪组织中，但其动员速度较慢（需要脂肪酶进行脂解以释放脂肪酸，随后在线粒体中进行 beta 氧化以产生乙酰辅酶 A 进入三羧酸循环）。脂肪酸氧化还需要氧气，使其不适合无氧高强度运动。因此，身体先使用糖原（快速、立即），然后在持续有氧运动时转向脂肪氧化（缓慢、持续）。[A1]