PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 25 marksAP 风格选择题 + 安/卑省考短答 · 共 25 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter and write one sentence justifying your choice. For short-answer items, answer in complete sentences. No calculator required in Part I.本节包含选择题与短答题。选择题请圈出字母答案，并用一句话说明理由。短答题请用完整句子作答。第一部分无需计算器。
Q1EASY易🇺🇸 US美AP-style MCQAP 风格选择题§1 Water Properties水的性质 · HS-LS1-6[3 marks][3 分]
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？
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？
(A)Cellulose contains fructose monomers; starch contains only glucose纤维素含有果糖单体；淀粉仅含葡萄糖
(B)Cellulose has beta ($\beta$) glycosidic linkages; starch has alpha ($\alpha$) glycosidic linkages纤维素具有 $\beta$ 糖苷键；淀粉具有 $\alpha$ 糖苷键
(C)Cellulose is a lipid; starch is a carbohydrate纤维素是脂质；淀粉是碳水化合物
(D)Starch is a monosaccharide; cellulose is a polysaccharide淀粉是单糖；纤维素是多糖
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)State the structural feature that distinguishes a saturated fatty acid from an unsaturated fatty acid.说明区分饱和脂肪酸与不饱和脂肪酸的结构特征。[2]
(b)Explain why butter is solid at room temperature while most vegetable oils are liquid. Refer to molecular packing in your answer.解释为何黄油在室温下是固体，而大多数植物油是液体。在回答中请涉及分子堆积方式。[2]
(c)State one biological role of phospholipids that is distinct from the role of triglycerides.写出磷脂区别于甘油三酯的一种生物学功能。[1]
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。
(a)Describe what happens during a dehydration synthesis reaction. State what molecule is produced as a by-product.描述脱水缩合反应中发生的过程，并说明产生的副产物分子。[2]
(b)Compare dehydration synthesis and hydrolysis: state how they differ in terms of reactants, products, and whether water is consumed or released.比较脱水缩合与水解：说明两者在反应物、产物以及水的消耗或释放方面的区别。[2]
(c)Amylase is a digestive enzyme that breaks down starch to maltose. Identify the type of reaction amylase catalyzes and state what is added at each bond broken.淀粉酶是一种将淀粉分解为麦芽糖的消化酶。指出淀粉酶催化的反应类型，并说明每个断裂键处加入了什么。[2]
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'
(a)State two structural differences between DNA and RNA.写出 DNA 与 RNA 之间的两个结构差异。[2]
(b)Write the complementary mRNA sequence produced during transcription from the template strand given above. Use standard base notation (5' to 3').写出从上述模板链转录产生的互补 mRNA 序列。使用标准碱基符号（5' 至 3'）。[2]
(c)Explain the role of each of the three types of RNA (mRNA, tRNA, rRNA) in translation.解释三种 RNA（mRNA、tRNA、rRNA）在翻译过程中各自的作用。[3]
(d)State one way in which a nucleotide monomer differs structurally from a nucleoside.写出核苷酸单体与核苷在结构上的一种区别。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Support each claim with biological evidence or logic. Write in complete sentences. Conclude each sub-part with a clear, direct answer statement.每一步推理都要写出。用生物学证据或逻辑支撑每个论点。用完整句子作答。每个子问题以清晰直接的答案陈述收尾。
Q6EASY易🇺🇸 US美AP-feeder FRQAP 衔接简答题§1 Water Properties水的性质 · HS-LS1-6[7 marks][7 分]
Water's unique properties arise from hydrogen bonding between its polar molecules. These properties are essential for life. 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。这些性质对生命至关重要。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。
(a)High specific heat capacity高比热容[2]
(b)High heat of vaporization高汽化热[2]
(c)Cohesion and adhesion内聚力与附着力[2]
(d)State whether water is a polar or nonpolar molecule and explain why it effectively dissolves ionic compounds such as $\text{NaCl}$.判断水是极性还是非极性分子，并解释为何水能有效溶解离子化合物，如 $\text{NaCl}$。[1]
Q7MEDIUM中🇨🇦 ON安ON Provincial-style安大略省考风格§4 Proteins and Their Structure蛋白质及其结构 · SBI3U[8 marks][8 分]
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。
(a)Describe each of the four levels of protein structure (primary, secondary, tertiary, quaternary) and identify the type of bond or interaction that stabilizes each level.描述蛋白质结构的四个层次（一级、二级、三级、四级），并指出稳定各层次的键或相互作用类型。[4]
(b)A student heats an enzyme solution to 95°C for 10 minutes. Explain what happens to the protein's structure and why enzymatic activity is lost. Use the term "denaturation" in your answer.学生将酶溶液加热至 95°C 持续 10 分钟。解释蛋白质结构发生了什么变化，以及为何酶活性丧失。在回答中使用"变性"一词。[2]
(c)Give two functional roles of proteins in the human body, using specific named examples.举出蛋白质在人体中的两种功能性作用，并提供具体命名的实例。[2]
Q8MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§5 Enzymes: Function and Factors酶：功能与影响因素 · Biology 12[8 marks][8 分]
Enzymes are biological catalysts. A student investigates how pH affects the activity of pepsin (a stomach enzyme with optimal pH of 2) and salivary amylase (a salivary enzyme with optimal pH of 7).酶是生物催化剂。学生研究 pH 如何影响胃蛋白酶（最适 pH 为 2 的胃部酶）和唾液淀粉酶（最适 pH 为 7 的唾液酶）的活性。
(a)Explain, using the induced fit model, how an enzyme catalyzes a reaction. Include the terms active site, substrate, and enzyme-substrate complex in your answer.用诱导契合模型解释酶如何催化反应。在回答中包含活性位点、底物和酶-底物复合物等术语。[3]
(b)Predict and explain the activity of pepsin at pH 7. Refer to the effect on the active site shape.预测并解释胃蛋白酶在 pH 7 时的活性。涉及对活性位点形状的影响。[2]
(c)Other than pH, state two additional factors that affect enzyme reaction rate and briefly explain how each affects the active site or substrate availability.除 pH 外，写出另外两个影响酶反应速率的因素，并简要解释各因素如何影响活性位点或底物可用性。[2]
(d)State one difference between a competitive inhibitor and a non-competitive inhibitor.写出竞争性抑制剂与非竞争性抑制剂的一种区别。[1]
A mutation occurs in the DNA of a gene encoding a critical enzyme. The original DNA template strand at the mutation site reads 3'-TAC-5' (produces the start codon). After the mutation, this triplet becomes 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 模板链序列为 3'-TAC-5'（产生起始密码子）。突变后，该三联体变为 3'-TAA-5'。
(a)Write the original mRNA codon and the mutant mRNA codon produced from each DNA triplet. Use the standard 5' to 3' mRNA convention.写出由每个 DNA 三联体产生的原始 mRNA 密码子和突变 mRNA 密码子。使用标准 5' 至 3' 的 mRNA 方向。[2]
(b)The mutant codon UAA is a stop codon. Explain what consequence this mutation would have on the protein produced and on the enzyme's function.突变密码子 UAA 是一个终止密码子。解释这一突变对所产生蛋白质及酶功能的影响。[2]
(c)This type of mutation, where a single nucleotide change creates a premature stop codon, is called a nonsense mutation. Explain why a missense mutation (amino acid change) at the same position might have a less severe effect on enzyme function than this nonsense mutation.这种因单个核苷酸改变而产生过早终止密码子的突变称为无义突变。解释为何同一位置的错义突变（氨基酸改变）对酶功能的影响可能比这种无义突变更轻微。[2]
(d)State one reason why a mutation in a non-coding region of DNA might have no effect on the protein produced.写出 DNA 非编码区突变可能对所产生蛋白质没有影响的一个原因。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 27 marks阿省毕业考 + 通用题型 · 共 27 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Support interpretations with biological reasoning. Conclude each question with a sentence in context. Calculator permitted throughout Part III where indicated.用生物学推理支撑解读。每题以结合情境的完整句子作答。第三部分在有指示的情况下可使用计算器。
Cell membranes are composed primarily of a phospholipid bilayer, along with cholesterol, proteins, and carbohydrates. A researcher studying cell membrane fluidity finds that membranes with more unsaturated fatty acid tails remain fluid at lower temperatures than membranes with saturated fatty acid tails.细胞膜主要由磷脂双分子层构成，还含有胆固醇、蛋白质和碳水化合物。研究人员研究细胞膜流动性时发现，含有更多不饱和脂肪酸尾的膜在较低温度下仍能保持流动性，而含饱和脂肪酸尾的膜则不能。
(a)Describe the structure of a phospholipid and explain why phospholipids spontaneously form a bilayer in an aqueous environment. Refer to their hydrophilic and hydrophobic regions.描述磷脂的结构，并解释为何磷脂在水溶液环境中自发形成双分子层。涉及其亲水区和疏水区。[3]
(b)Explain, at the molecular level, why membranes with unsaturated fatty acid tails are more fluid than membranes with saturated tails at the same temperature.在分子水平上解释，为何含有不饱和脂肪酸尾的膜在相同温度下比含饱和脂肪酸尾的膜流动性更高。[2]
(c)Cold-climate fish often have a higher proportion of unsaturated fatty acids in their cell membranes than tropical fish. Explain the adaptive advantage of this difference.寒冷气候中的鱼类细胞膜中不饱和脂肪酸的比例通常高于热带鱼。解释这一差异的适应性优势。[2]
(d)State one role of cholesterol in animal cell membranes.写出胆固醇在动物细胞膜中的一种作用。[1]
A student investigates the effect of substrate concentration on the rate of catalase activity. Catalase breaks down hydrogen peroxide ($\text{H}_2\text{O}_2$) into water and oxygen gas. The student measures oxygen production (mL/min) at increasing substrate concentrations and finds the reaction rate increases rapidly at first, then levels off, reaching a maximum rate ($V_{max}$) of 12 mL/min.学生研究底物浓度对过氧化氢酶活性速率的影响。过氧化氢酶将过氧化氢（$\text{H}_2\text{O}_2$）分解为水和氧气。学生在递增底物浓度下测量氧气产生量（mL/min），发现反应速率起初迅速增加，然后趋于平稳，达到最大速率（$V_{max}$）12 mL/min。
(a)Explain why the reaction rate increases rapidly at low substrate concentrations but levels off at high substrate concentrations. Use the concept of enzyme saturation in your answer.解释为何反应速率在低底物浓度时迅速增加，但在高底物浓度时趋于平稳。在回答中使用酶饱和的概念。[3]
(b)At a substrate concentration that produces a rate of 6 mL/min, a competitive inhibitor is added without changing substrate concentration. Predict whether the rate will increase, decrease, or stay the same, and explain your reasoning.在产生 6 mL/min 速率的底物浓度下，加入竞争性抑制剂但不改变底物浓度。预测速率是增加、减小还是保持不变，并解释你的推理。[2]
(c)The student repeats the experiment with a non-competitive inhibitor. Describe how this inhibitor affects $V_{max}$ compared to the uninhibited reaction, and explain why.学生用非竞争性抑制剂重复实验。描述与无抑制剂反应相比，该抑制剂如何影响 $V_{max}$，并解释原因。[2]
(d)The student finds that doubling the enzyme concentration doubles the $V_{max}$. Explain why increasing enzyme concentration increases $V_{max}$ but does NOT change the enzyme's optimal pH or temperature.学生发现将酶浓度加倍可使 $V_{max}$ 加倍。解释为何增加酶浓度能提高 $V_{max}$，但不改变酶的最适 pH 或温度。[2]
After a meal, the body digests food macromolecules and either uses the resulting monomers for energy or stores them. A person eats a meal containing starch, triglycerides, and proteins. After digestion and absorption, excess glucose is converted to glycogen (short-term storage) or triglycerides (long-term storage).餐后，身体消化食物大分子，并将所得单体用于能量或储存。某人摄入含淀粉、甘油三酯和蛋白质的一餐。消化吸收后，多余的葡萄糖被转化为糖原（短期储存）或甘油三酯（长期储存）。
(a)For each of the three macromolecules in the meal (starch, triglycerides, proteins), state the monomers produced upon complete digestion and identify the type of reaction that breaks them down.对于这顿饭中的三种大分子（淀粉、甘油三酯、蛋白质），分别说明完全消化后产生的单体，并指出将其分解的反应类型。[3]
(b)Describe how glycogen is synthesized from glucose monomers, naming the type of bond formed and the reaction used.描述如何由葡萄糖单体合成糖原，命名所形成键的类型及所用反应。[2]
(c)A person on a low-carbohydrate diet begins breaking down stored triglycerides for energy. Identify the products of triglyceride hydrolysis and explain why fat stores more energy per gram than carbohydrate stores.某人进行低碳水化合物饮食，开始分解储存的甘油三酯以获取能量。指出甘油三酯水解的产物，并解释为何脂肪每克储存的能量多于碳水化合物。[3]
(d)Compare the energy storage roles of glycogen and triglycerides: state which stores more energy per unit mass and explain one advantage of having BOTH types of energy storage rather than relying solely on one.比较糖原和甘油三酯的能量储存角色：说明哪种每单位质量储存的能量更多，并解释同时拥有两种能量储存类型（而非仅依赖一种）的一个优点。[2]
🇨🇦 British Columbia不列颠哥伦比亚Biology 12: Biochemistry and cell biology生物 12：生物化学与细胞生物学
🇨🇦 Alberta阿尔伯塔Biology 30 Unit A · 30-A1.1k · 30-A1.2k · 30-A1.3k · 30-A2.1k
Full Syllabus Map lives in ../Study Guides/Unit_2_Biochemistry_Molecules_of_Life.html. Enzyme kinetics calculations (Q11) are Honors / AB Biology 30 only; all other regions treat enzyme rate factors qualitatively.完整大纲对照表见 ../Study Guides/Unit_2_Biochemistry_Molecules_of_Life.html。酶动力学计算（Q11）仅适用于荣誉级 / 阿尔伯塔 Biology 30；其他地区对该主题仅作定性考查。