📚 Mastering Biology Exam Questions: A Step-by-Step Guide | 生物典型例题详解：步步为营攻克考题
Exam success in biology comes not only from knowing the facts, but from being able to apply them under pressure. This guide walks you through carefully chosen worked examples, covering the most common question types in IB and CCEA specifications. Each section models the thinking process you need to turn textbook knowledge into full-mark answers.
生物考试的成功不仅来自对知识的记忆，更来自在压力下灵活运用知识的能力。本指南精选IB和CCEA考试中最常见的题型，逐一详细解析。每个小节都会展示从课本知识到满分答案所需的完整思维过程。
1. Cell Structure & Function | 细胞结构与功能
A typical question asks you to compare a prokaryotic cell with a eukaryotic cell. Begin by listing the organelles that are unique to each, then state the shared features.
典型题目要求比较原核细胞和真核细胞。首先分别列出各自特有的细胞器，再说明共同特征。
Question: Outline the structural differences between a typical bacterium and a liver cell. (4 marks)
题目： 概述一个典型细菌与一个肝细胞在结构上的差异。（4分）
Model answer: The bacterium is prokaryotic, so it lacks a membrane-bound nucleus; its DNA lies free in the cytoplasm as a circular chromosome. It has 70S ribosomes, a cell wall made of peptidoglycan, and may have a flagellum. The liver cell is eukaryotic – it contains a nucleus enclosed by a double membrane, 80S ribosomes, mitochondria, endoplasmic reticulum and other membrane-bound organelles. Both have a cell membrane, cytoplasm and ribosomes. (4)
参考答案： 细菌是原核生物，因此没有膜包被的细胞核；其DNA以环状染色体的形式游离在细胞质中。它具有70S核糖体、由肽聚糖构成的细胞壁，并可能有鞭毛。肝细胞是真核细胞——它含有由双层膜包围的细胞核、80S核糖体、线粒体、内质网和其他膜结合细胞器。两者都有细胞膜、细胞质和核糖体。（4分）
Tip: Always mention both unique and common structures. Use comparative language such as ‘whereas’ and ‘by contrast’.
提示： 一定要提到特有结构和共同结构。使用比较性语言，如“而”、“相比之下”。
2. Enzyme Action & Inhibition | 酶的作用与抑制
Interpretation of graphs showing reaction rate against substrate concentration, or against inhibitor concentration, is a classic skill. You must link the shape of the curve to the molecular events at the active site.
解读反应速率随底物浓度或抑制剂浓度变化的曲线图是一项经典技能。你必须将曲线形状与活性位点上的分子事件联系起来。
Question: Explain how a non-competitive inhibitor reduces the rate of an enzyme-catalysed reaction, using a labeled graph to support your answer. (6 marks)
题目： 解释非竞争性抑制剂如何降低酶促反应速率，并用有标记的曲线图辅助说明。（6分）
Model answer: Draw a graph with rate on the y-axis and substrate concentration on the x-axis. Plot control reaction (asymptote at Vmax) and inhibited reaction (lower asymptote). Annotation: ‘Non-competitive inhibitor binds to an allosteric site, which changes the shape of the active site. The substrate can no longer bind, so the number of functional enzyme molecules is reduced. Vmax decreases, but Km (substrate affinity) remains unchanged because the remaining uninhibited enzyme molecules still bind substrate with the same affinity. The two curves never converge even at infinite substrate concentration.’ (6)
参考答案： 绘制一张以反应速率为纵轴、底物浓度为横轴的图。画出对照组曲线（渐近线为Vmax）和受抑制曲线（渐近线较低）。标注：“非竞争性抑制剂结合在变构位点上，从而改变活性位点的形状。底物无法结合，因此有功能的酶分子数量减少。Vmax降低，但Km（底物亲和力）保持不变，因为剩余未被抑制的酶分子仍以相同的亲和力结合底物。即使在无穷大的底物浓度下，两条曲线也不会相交。”（6分）
Tip: Always define Vmax and Km when describing enzyme kinetics, and explain why they change or stay the same.
提示： 描述酶动力学时务必定义Vmax和Km，并解释它们为何变化或不变。
3. Cellular Respiration | 细胞呼吸
Questions on respiration often require you to trace the fate of carbon atoms or the production of ATP. Use a systematic approach: name the stage, its location, the inputs, and the outputs.
关于呼吸作用的题目常常要求你追踪碳原子的去向或ATP的生成。采用系统的方法：说出阶段名称、发生位置、输入物和输出物。
Question: Describe how a molecule of glucose is completely oxidised to carbon dioxide and water in an aerobic muscle cell, stating where each stage occurs and how many ATP molecules are produced by substrate-level phosphorylation. (8 marks)
题目： 描述一个葡萄糖分子在有氧肌细胞中如何被完全氧化为二氧化碳和水，说明每个阶段发生的位置，以及通过底物水平磷酸化产生多少个ATP分子。（8分）
Model answer: Glycolysis occurs in the cytoplasm; glucose (6C) is split into two molecules of pyruvate (3C). 2 ATP are produced by substrate-level phosphorylation. Pyruvate enters the mitochondrial matrix, where the link reaction removes one carbon as CO₂ per pyruvate, producing acetyl-CoA (2C). Acetyl-CoA enters the Krebs cycle in the matrix, where each turn produces 2 CO₂, 1 ATP (by substrate-level phosphorylation), 3 NADH and 1 FADH₂. For one glucose, two turns occur, so 2 ATP are made. Overall, 4 ATP come from substrate-level phosphorylation (2 from glycolysis, 2 from Krebs). The NADH and FADH₂ pass electrons to the electron transport chain on the cristae, producing many ATP via oxidative phosphorylation. (8)
参考答案： 糖酵解发生在细胞质中；葡萄糖（6碳）被分解为两个丙酮酸分子（3碳）。通过底物水平磷酸化产生2个ATP。丙酮酸进入线粒体基质，在那里进行连接反应，每个丙酮酸脱去一个碳生成CO₂，形成乙酰辅酶A（2碳）。乙酰辅酶A进入基质中的克雷布斯循环，每轮循环生成2个CO₂、1个ATP（通过底物水平磷酸化）、3个NADH和1个FADH₂。对于一个葡萄糖，循环进行两轮，因此生成2个ATP。总体上，来自底物水平磷酸化的ATP共有4个（糖酵解2个，克雷布斯循环2个）。NADH和FADH₂将电子传递给嵴上的电子传递链，通过氧化磷酸化产生大量ATP。（8分）
Tip: Distinguish between substrate-level phosphorylation and oxidative phosphorylation. Emphasise that the majority of ATP is produced by the latter.
提示： 区分底物水平磷酸化和氧化磷酸化。强调绝大多数ATP是由后者生成的。
4. Photosynthesis | 光合作用
Questions often link the light-dependent reactions to the light-independent reactions. You must explain how the products of the first stage drive the second.
题目常将光反应与暗反应联系起来。你必须解释第一阶段的产物如何推动第二阶段。
Question: Explain how the light-dependent reactions of photosynthesis produce the substances needed for the Calvin cycle. (5 marks)
题目： 解释光合作用的光反应如何产生卡尔文循环所需的物质。（5分）
Model answer: Light energy is absorbed by chlorophyll in photosystem II, exciting electrons that travel along an electron transport chain. Photolysis of water replaces these electrons, releasing protons (H⁺) into the thylakoid space and O₂ as a by-product. The electron transport chain pumps H⁺ into the thylakoid space, creating a proton gradient; H⁺ diffuses back through ATP synthase, generating ATP. The electrons are passed to photosystem I, where light excites them again; they are transferred to NADP⁺, together with H⁺, forming reduced NADP (NADPH). Hence the light-dependent reactions produce ATP and NADPH, which supply energy and reducing power for the Calvin cycle to fix CO₂. (5)
参考答案： 光能被光系统II中的叶绿素吸收，激发电子，电子沿电子传递链移动。水的光解为这些电子提供替代，释放质子（H⁺）进入类囊体空间，同时产生副产物O₂。电子传递链将H⁺泵入类囊体空间，形成质子梯度；H⁺通过ATP合酶扩散回来，生成ATP。电子传递至光系统I，在那里光再次激发它们；电子与H⁺一起传递给NADP⁺，形成还原型NADP（NADPH）。因此，光反应产生ATP和NADPH，为卡尔文循环固定CO₂提供能量和还原力。（5分）
Tip: Use the terms ‘photophosphorylation’ for ATP synthesis and ‘photolysis’ for water splitting. Highlight that NADP is the final electron acceptor.
提示： 用术语“光合磷酸化”表示ATP合成，用“光解”表示水的分解。强调NADP是最终电子受体。
5. Genetics & Inheritance | 遗传与遗传规律
Monohybrid and dihybrid cross questions test your ability to use Punnett squares and to interpret phenotypic ratios. Learn to recognise the patterns of dominance, co-dominance, and sex linkage.
单因子杂交和双因子杂交题目考查你运用庞纳特方格及解读表型比例的能力。学会识别显隐性、共显性和伴性遗传的模式。
Question: In pea plants, tall (T) is dominant to dwarf (t), and round seed (R) is dominant to wrinkled seed (r). A plant heterozygous for both traits is crossed with a dwarf plant that has wrinkled seeds. Determine the phenotypic ratio of the offspring. (6 marks)
题目： 在豌豆中，高茎（T）对矮茎（t）为显性，圆粒种子（R）对皱粒种子（r）为显性。一个对这两种性状都是杂合的植株与一个矮茎且结皱粒种子的植株杂交。确定后代的表型比例。（6分）
Model answer: The heterozygous plant has genotype TtRr, producing gametes TR, Tr, tR, tr. The dwarf wrinkled plant is ttrr, producing only tr gametes. Construct a 4×1 Punnett square. Offspring genotypes: TtRr (tall, round), Ttrr (tall, wrinkled), ttRr (dwarf, round), ttrr (dwarf, wrinkled). All four combinations are equally likely, so the phenotypic ratio is 1 tall round : 1 tall wrinkled : 1 dwarf round : 1 dwarf wrinkled. (6)
参考答案： 杂合植株的基因型为TtRr，产生配子TR、Tr、tR、tr。矮茎皱粒植株是ttrr，只产生tr配子。构建一个4×1的庞纳特方格。后代的基因型为：TtRr（高茎圆粒）、Ttrr（高茎皱粒）、ttRr（矮茎圆粒）、ttrr（矮茎皱粒）。所有四种组合概率相等，因此表型比例为 1 高圆 : 1 高皱 : 1 矮圆 : 1 矮皱。（6分）
Tip: When a heterozygous individual is crossed with a homozygous recessive, the result is a 1:1:1:1 ratio if the genes are unlinked. Always state the ratio clearly.
提示： 杂合个体与纯合隐性个体杂交时，若基因不连锁，结果就是1:1:1:1的比例。始终清晰说明该比例。
6. Ecology & Energy Flow | 生态学与能量流动
Calculations of energy transfer efficiency and construction of pyramids of biomass are common. Precision in units and careful interpretation of given data are key.
能量传递效率的计算和生物量金字塔的绘制是常见考点。单位的精确性和对给定数据的细致解读是关键。
Question: The primary producers in a lake capture 2.8 × 10⁶ kJ m⁻² yr⁻¹ of solar energy. The primary consumers contain 3.4 × 10⁴ kJ m⁻² yr⁻¹. Calculate the percentage energy transfer from the first to the second trophic level, and suggest two reasons for the low efficiency. (5 marks)
题目： 某湖泊中的初级生产者捕获了 2.8 × 10⁶ kJ m⁻² yr⁻¹ 的太阳能。初级消费者含能量为 3.4 × 10⁴ kJ m⁻² yr⁻¹。计算从第一营养级到第二营养级的能量传递百分比，并提出效率低下的两点原因。（5分）
Model answer: Efficiency = (energy in primary consumers / energy in producers) × 100 = (3.4 × 10⁴ / 2.8 × 10⁶) × 100 = 1.21%. Two reasons: not all the light captured is used in photosynthesis (some is reflected or transmitted); a large proportion of the gross production is lost as heat in respiration by the producers. Also, not all of the primary production is consumed by herbivores – some parts are indigestible or die uneaten. (5)
参考答案： 效率 = (初级消费者能量 / 生产者能量) × 100 = (3.4 × 10⁴ / 2.8 × 10⁶) × 100 = 1.21%。两点原因：并非所有捕获的光能都用于光合作用（部分被反射或透射）；总初级生产量中有很大一部分通过生产者的呼吸作用以热能形式散失。此外，并非所有初级生产量都植食动物被取食——部分器官难以消化或未待取食就已死亡。（5分）
Tip: Always show the formula and the calculation step-by-step. When giving reasons, use ecological terms such as ‘respiratory heat loss’ and ‘indigestible material’.
提示： 务必逐步展示公式和计算过程。阐述原因时使用生态学术语，如“呼吸热损耗”和“不可消化物质”。
7. Homeostasis & Kidney Function | 稳态与肾脏功能
Explanation of ultrafiltration and selective reabsorption in the nephron is a recurring theme. You must be precise about the names of structures and the composition of fluid at different points.
描述肾单位中的超滤和选择性重吸收是一个常见考点。你必须准确说明各结构的名称以及不同位置液体的成分。
Question: Describe the process of ultrafiltration in the Bowman’s capsule. (4 marks)
题目： 描述鲍曼氏囊中的超滤过程。（4分）
Model answer: Blood enters the glomerulus via the afferent arteriole, which has a wider diameter than the efferent arteriole. This creates high hydrostatic pressure in the glomerular capillaries. The capillary wall, basement membrane, and podocytes form a filtration barrier. Water, glucose, amino acids, urea, and ions are forced through into the Bowman’s capsule to form glomerular filtrate. Blood cells and large proteins remain in the capillary because they are too large to pass through. (4)
参考答案： 血液通过入球小动脉进入肾小球，入球小动脉的直径大于出球小动脉，这使得肾小球毛细血管内产生高静水压。毛细血管壁、基膜和足细胞构成滤过屏障。水、葡萄糖、氨基酸、尿素和离子被迫滤出，进入鲍曼氏囊形成肾小球滤液。血细胞和大分子蛋白质因尺寸太大而无法通过，留在毛细血管内。（4分）
Tip: Use ‘hydrostatic pressure’, ‘filtration barrier’, and ‘podocytes’ for full marks. Mention the difference in arteriole diameters.
提示： 要获得满分需使用“静水压”、“滤过屏障”和“足细胞”等术语。提及入球和出球小动脉的直径差异。
8. Molecular Biology – DNA Replication | 分子生物学——DNA复制
Students often lose marks by omitting enzyme names or by not stating the directionality of synthesis. A flow diagram in words is the most reliable way to answer.
学生常因遗漏酶的名称或未说明合成的方向性而丢分。用文字流程图是回答此类问题的最可靠方法。
Question: Explain how the structure of DNA ensures semi-conservative replication. Refer to the roles of the main enzymes involved. (7 marks)
题目： 解释DNA的结构如何确保半保留复制。请提及涉及的主要酶的作用。（7分）
Model answer: DNA consists of two antiparallel strands held together by complementary base pairing (A-T, C-G) via hydrogen bonds. During replication, DNA helicase unwinds the double helix and breaks the hydrogen bonds, exposing the template strands. This allows semi-conservative replication, where each original strand serves as a template for a new strand. DNA polymerase adds free DNA nucleotides to a growing chain in the 5′ to 3′ direction, complementary to the template. On the lagging strand, synthesis is discontinuous, forming Okazaki fragments that are later joined by DNA ligase. Thus each daughter molecule contains one original strand and one newly synthesised strand. (7)
参考答案： DNA由两条通过氢键和互补碱基配对（A-T，C-G）结合的反平行链组成。复制时，DNA解旋酶解开双螺旋并断裂氢键，暴露出模板链。这使得半保留复制得以进行，即每条母链充当新链合成的模板。DNA聚合酶将游离的DNA核苷酸按5′ → 3’方向添加到延伸的链上，该链与模板互补。在后随链上，合成是不连续的，形成冈崎片段，随后由DNA连接酶连接。因此，每个子代DNA分子都含有一条母链和一条新合成的链。（7分）
Tip: Emphasise ‘antiparallel strands’ and ‘5’ to 3′ direction’. Name the enzymes in the order they act: helicase, polymerase, ligase.
提示： 强调“反平行链”和“5’到3’方向”。按作用顺序列出酶的名称：解旋酶、聚合酶、连接酶。
9. Practical Skills – Osmosis Experiment | 实验技能——渗透作用实验
Controlled assessment and IB internal assessment require you to identify variables, justify method choices, and evaluate reliability. A typical question will present a scenario and ask for a critique.
实验评估和IB内部评估要求你识别变量、论证方法选择并评估可靠性。典型题目会给出一个情境并要求作出评价。
Question: A student investigates the effect of sucrose concentration on the mass of potato cylinders. The cylinders are blotted, weighed, placed in solutions for 30 minutes, then reweighed. State the independent and dependent variables, two control variables, and explain why the cylinders were blotted before the final weighing. (4 marks)
题目： 某学生研究了蔗糖浓度对马铃薯圆柱体质量的影响。将圆柱体吸干、称重，浸入溶液中30分钟后再称重。说出自变量、因变量和两个控制变量，并解释为什么最后称重前要将圆柱体吸干。（4分）
Model answer: Independent variable: sucrose concentration. Dependent variable: change in mass (or percentage change in mass). Controlled variables: temperature (all solutions at same temperature), volume of solution, incubation time, potato variety, surface area of cylinders (any two). Blotting removes excess surface water, which would otherwise add to the measured mass and lead to an overestimate of water uptake (or underestimate of water loss). (4)
参考答案： 自变量：蔗糖浓度。因变量：质量变化（或质量变化百分比）。控制变量：温度（所有溶液温度相同）、溶液体积、浸泡时间、马铃薯品种、圆柱体表面积（任选两个）。吸干是为了去除表面多余的水分，否则这些水分会增加到测得的质量中，导致对水分吸收的高估（或导致对水分丢失的低估）。（4分）
Tip: Always say ‘percentage change in mass’ when analysing osmolarity, as it allows comparison between samples with different starting masses.
提示： 分析渗透浓度时总是使用“质量变化百分比”，因为这样可以比较不同起始质量的样本。
10. Data Analysis & Evaluation | 数据分析与评价
Interpreting data tables, graphs, and arriving at a justified conclusion is a transferable skill. Always quote figures, describe trends, and discuss whether the data fully support the hypothesis.
解读数据表、图表并得出有依据的结论是一项可迁移的技能。要引用数据、描述趋势，并讨论数据是否完全支持假设。
Question: A drug trial measures the mean blood glucose concentration (mmol L⁻¹) of 10 patients before and after treatment. Pre-treatment: 12.4 ± 2.1; post-treatment: 7.9 ± 1.8. The t-test yields p = 0.03. State what the p-value indicates and whether the difference is statistically significant. (3 marks)
题目： 一项药物试验测量了10名患者治疗前后的平均血糖浓度（mmol L⁻¹）。治疗前：12.4 ± 2.1；治疗后：7.9 ± 1.8。t检验给出p = 0.03。说明p值表示什么，以及该差异是否具有统计学显著性。（3分）
Model answer: The p-value indicates the probability that the observed difference (or a more extreme one) could have occurred purely by chance, assuming the null hypothesis is true. Here, p = 0.03, which is less than the conventional significance level of 0.05. Therefore, there is less than 5% probability that the difference is due to chance; the difference is statistically significant. The null hypothesis is rejected. (3)
参考答案： p值表示假设零假设成立时，观察到的差异（或更极端的差异）纯粹由偶然因素造成的概率。此处p = 0.03，小于常规显著性水平0.05。因此，该差异由偶然因素造成的概率小于5%；该差异具有统计学显著性。拒绝零假设。（3分）
Tip: Define the null hypothesis explicitly and always compare the p-value to 0.05. Remember, ‘statistically significant’ does not automatically mean ‘biologically important’.
提示： 明确定义零假设，并始终将p值与0.05比较。记住，“统计学上显著”并不自动等同于“生物学上重要”。
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