Typical Example Questions Explained in IB Edexcel Biology | IB Edexcel 生物：典型例题详解

📚 Typical Example Questions Explained in IB Edexcel Biology | IB Edexcel 生物：典型例题详解

Biology exams in both the IB Diploma Programme and the Edexcel International A Level challenge students not only with factual recall but also with data analysis, application to unfamiliar contexts, and evaluation of experimental design. This article walks you through ten representative question types drawn from past papers and specimen assessments across these syllabi. For each example, you will find a clear breakdown of what the examiner is looking for, key command terms, common pitfalls, and model points that would score full marks. Whether you are tackling a structured question on membrane transport, a genetics probability table, or a data-based question on photosynthesis, the paired strategies below will help you decode the marking scheme and sharpen your exam technique.

无论是 IB 文凭课程还是 Edexcel 国际 A Level 的生物考试，都不仅仅考查学生对事实的记忆，更要求他们具备数据分析、在陌生情境中应用知识以及评价实验设计的能力。本文精选了这两种课程历年真题和样卷中的十种典型题型，逐一进行详解。每一个例题都包含了考官想看到什么、关键指令词、常见失分点，以及能够拿到满分的答题要点。不论你面对的是关于膜运输的结构题、遗传概率表格，还是光合作用的数据分析题，下文提供的双语法则都将帮助你拆解评分方案，优化答题技巧。

1. Membrane Structure and Transport | 细胞膜结构与运输题

Question type: ‘Explain how the structure of the cell membrane enables it to carry out its function.’ This appears frequently in both IB Paper 1 Section B and Edexcel Unit 2 ‘Cells and Transport’. The examiner expects you to move beyond stating the fluid mosaic model and to explicitly link each component to a specific role. You must show understanding of how the arrangement of phospholipids, proteins, glycoproteins, and cholesterol contributes to selective permeability, signalling, and compartmentalisation.

典型例题：“阐释细胞膜的结构如何使其能够完成功能。” 这道题常见于 IB 试卷一 B 部分和 Edexcel 单元二的“细胞与运输”中。考官期望你不仅仅说出流动镶嵌模型，而且要明确地将每一个组分与其特定作用联系起来。你需要展示对磷脂、蛋白质、糖蛋白和胆固醇的排列如何实现选择透过性、信号传导和区室化的理解。

The bilayer is formed by phospholipids: hydrophilic heads face outwards and inwards to aqueous environments, hydrophobic tails face inward — this creates a barrier to large or charged molecules (English).
双分子层由磷脂构成：亲水头部朝外朝内朝向水环境，疏水尾部朝内——这为大型分子或带电分子形成了一道屏障（中文）。
Integral and peripheral proteins act as channels, carriers, pumps (active transport), and enzymes; for example, aquaporins facilitate rapid water movement, and Na⁺/K⁺ ATPase establishes electrochemical gradients (English).
整合蛋白和外周蛋白起通道、载体、泵（主动运输）和酶的作用；例如，水通道蛋白促进水的快速运动，Na⁺/K⁺-ATP 酶建立起电化学梯度（中文）。
Cholesterol intercalates between phospholipid tails, reducing excessive fluidity and maintaining stability across a range of temperatures. Glycoproteins and glycolipids function in cell recognition and adhesion (English).
胆固醇嵌入磷脂尾部之间，降低过度流动性并在一定温度范围内维持稳定性。糖蛋白和糖脂在细胞识别和黏附中起作用（中文）。

Common pitfall: listing structures without linking to function, or giving vague statements like ‘proteins control what enters and leaves’. Always name a specific protein type and its transport mechanism.

常见失分点：罗列结构却不联系功能，或者给出“蛋白质控制物质进出”这样模糊的陈述。务必指明具体的蛋白质类型及其运输机制。

2. Enzyme Kinetics and Inhibition Graphs | 酶动力学与抑制曲线题

Both IB and Edexcel papers regularly present Lineweaver–Burk plots or rate–substrate concentration curves, asking you to distinguish competitive from non‑competitive inhibition. The key is to use the changes in Vmax and Km to deduce the inhibitor type, then explain the molecular reason.

IB 和 Edexcel 的试卷常常给出 Lineweaver–Burk 图或者速率-底物浓度曲线，要求你区分竞争性抑制和非竞争性抑制。关键是利用 Vmax 和 Km 的变化推断抑制剂类型，然后从分子层面解释原因。

Competitive inhibitor: resembles the substrate, binds to the active site, can be overcome by high substrate concentration. Km increases, Vmax unchanged (English).
竞争性抑制剂：与底物结构相似，与活性位点结合，可通过高底物浓度克服。Km 增大，Vmax 不变（中文）。
Non‑competitive inhibitor: binds to an allosteric site, changes enzyme shape so substrate can no longer bind. Vmax decreases, Km unchanged (English).
非竞争性抑制剂：与别构位点结合，改变酶的形状使底物无法结合。Vmax 下降，Km 不变（中文）。

When describing graphs, always refer to both axes, state what happens to the line, and connect this to the active‑site or allosteric site wording. In Edexcel, you may also need to interpret Vmax and Km from a table and calculate the inhibitor constant.

描述图表时，始终要说明两个轴的变化，指出曲线如何移动，并将其与活性位点或别构位点的术语联系起来。在 Edexcel 中，你可能还需要从数据表中读取 Vmax 和 Km 并计算抑制常数。

3. DNA Replication Steps | DNA 复制步骤题

‘Outline the process of semi‑conservative DNA replication’ is a guaranteed high‑mark question in IB Topic 7 and Edexcel Topic 2. Full‑mark answers must contain precise enzyme names, directionality (5′ → 3′), and the concepts of leading and lagging strands.

“简述半保留 DNA 复制过程”是 IB 主题 7 和 Edexcel 主题 2 中必出的高分题。满分的答案必须包含准确的酶名称、方向性（5′ → 3′），以及前导链和后随链的概念。

Helicase unwinds the double helix by breaking hydrogen bonds between bases, forming a replication fork (English).
解旋酶通过打破碱基间的氢键解开双螺旋，形成复制叉（中文）。
Single‑strand binding proteins stabilise the separated strands; topoisomerase relieves supercoiling ahead of the fork (English).
单链结合蛋白稳定已分开的链；拓扑异构酶在复制叉前方解除超螺旋（中文）。
Primase synthesises a short RNA primer; DNA polymerase III adds nucleotides in the 5′ → 3′ direction, complementary to the template. Continuous synthesis on the leading strand, discontinuous synthesis (Okazaki fragments) on the lagging strand (English).
引物酶合成短 RNA 引物；DNA 聚合酶 III 以 5′ → 3′ 方向添加核苷酸，与模板互补。前导链上连续合成，后随链上不连续合成（冈崎片段）（中文）。
DNA polymerase I replaces RNA primers with DNA; DNA ligase seals nicks between Okazaki fragments, forming a sugar–phosphate backbone (English).
DNA 聚合酶 I 将 RNA 引物替换为 DNA；DNA 连接酶将冈崎片段间的切口连接，形成糖–磷酸骨架（中文）。

Examiners penalise any reversal of direction or omission of the role of primase. Always underscore ‘semi‑conservative’: each new molecule contains one original and one new strand.

考官会扣分的方向性错误或遗漏引物酶的角色。务必强调“半保留”：每个新分子含有一条旧链和一条新链。

4. Genetics Probability Question | 遗传概率题

A typical question: ‘A man who is heterozygous for tongue rolling (Rr) and a woman who cannot roll her tongue (rr) have two children. Calculate the probability that both children can roll their tongue.’ In both syllabi, you must show a Punnett square (or multiplicative rule) and state the assumption of independent assortment for unlinked genes.

典型题目：“一个舌卷曲杂合子（Rr）的男性和一个不能卷舌（rr）的女性育有两个孩子。计算两个孩子均能卷舌的概率。” 在两个体系中，都需要展示庞纳特方格（或乘法法则）并说明非连锁基因的自由组合假设。

Punnett square: Rr × rr → ½ Rr (rollers), ½ rr (non‑rollers)

Probability one child rolls = ½. For two independent children: P = ½ × ½ = ¼.

一个孩子能卷舌的概率 = ½。对于两个独立的孩子：P = ½ × ½ = ¼。

In Edexcel, you may also need to use a chi‑squared test to compare observed and expected ratios. In IB, such a calculation often appears alongside a pedigree analysis where you must first deduce genotypes. Always justify your probabilities in words.

在 Edexcel 中，可能还需要使用卡方检验比较观察值与预期值的比例。在 IB 中，此类计算常与系谱分析结合，需先推导基因型。始终用文字说明概率的推导过程。

5. Natural Selection and Antibiotic Resistance | 自然选择与抗生素耐药题

Both syllabi love to assess the evolution of antibiotic resistance as a real‑world example of natural selection. The question often begins with a graph showing bacterial population size over time or a zone of inhibition on an agar plate. You must explain stepwise how resistance spreads through a population.

两个课程体系都喜欢将从抗生素耐药性作为自然选择的现实案例来考查。题目通常以一幅细菌数量随时间变化的曲线图或者琼脂平板上的抑菌圈为开端。你需要逐步解释耐药性如何在群体中扩散。

Variation: within a bacterial population, random mutations produce some individuals with reduced susceptibility to an antibiotic (English).
变异：在细菌群体中，随机突变产生了一些对抗生素敏感性降低的个体（中文）。
Selection pressure: when the antibiotic is applied, susceptible bacteria die, while resistant ones survive and reproduce (English).
选择压力：当使用抗生素时，敏感细菌死亡，而耐药菌存活并繁殖（中文）。
Heritability: the resistance alleles are passed to offspring, increasing their frequency in the population over generations (English).
遗传：耐药等位基因传递给子代，在世代间增加了它们在群体中的频率（中文）。
Result: the population becomes predominantly resistant, making the antibiotic ineffective. Use specific terminology: ‘selective advantage’, ‘differential survival’, ‘allele frequency change’ (English).
结果：群体主要由耐药菌组成，使抗生素失效。须使用特定术语：“选择优势”、“差异存活”、“等位基因频率改变”（中文）。

In data‑response tasks, connect the graph shape to these stages: lag, exponential decline of susceptibles, resurgence of resistants. Edexcel may ask you to outline methods to reduce resistance (finish the course, no unnecessary prescriptions).

在数据分析题中，要把曲线形状与这些步骤相联系：滞后期、敏感菌指数的下降、耐药菌的重新崛起。Edexcel 可能会让你概述减缓耐药性的方法（完成疗程、不滥用处方）。

6. Photosynthesis – Limiting Factors and Graphs | 光合作用 – 限制因素与曲线题

A common IB Section B and Edexcel Unit 4 question presents a graph of photosynthesis rate against light intensity, CO₂ concentration, or temperature, often with multiple curves. The task is to identify the limiting factor at different regions of the graph. A full‑answer requires naming the Calvin cycle or light‑dependent reaction explicitly and explaining how the factor exerts its effect at the molecular level.

IB B 部分和 Edexcel 单元 4 中常出的题目是，给出光合作用速率随光照强度、二氧化碳浓度或温度变化的曲线图，通常含有多条曲线。任务是指出曲线上不同区段的限制因素。满分回答需要明确提到卡尔文循环或光反应，并从分子层面解释该因素如何产生影响。

Plateau at high light intensity: another factor (e.g. CO₂) is limiting. Reason: insufficient CO₂ for RuBisCO, so Calvin cycle slows, NADPH and ATP accumulate, electron transport chain backs up (English).
高光强处的平台：另一个因素（如 CO₂）成为限制因素。原因：RuBisCO 缺少 CO₂，卡尔文循环减慢，NADPH 和 ATP 积累，电子传递链受阻（中文）。
Plateau at high CO₂: light intensity or temperature now limits. Reason: less ATP and NADPH produced in light‑dependent reactions, so conversion of GP to TP is limited (English).
高 CO₂ 处的平台：此时光照强度或温度成为限制因素。原因：光反应产生的 ATP 和 NADPH 减少，限制了 GP 转化为 TP（中文）。
Temperature: effects on enzyme‑catalysed steps in the Calvin cycle and on RuBisCO activity. At very high temperatures, photorespiration and stomatal closure become relevant (English).
温度：影响卡尔文循环中酶催化步骤和 RuBisCO 的活性。在极高温度下，光呼吸和气孔关闭也变得重要（中文）。

Remember to always quantify the change: ‘Between point A and B, the rate doubles as light intensity increases from 200 to 400 µmol m⁻² s⁻¹, indicating light is the main limiting factor.’

请记住，一定要量化变化：“在 A 点和 B 点之间，当光强度从 200 增加到 400 µmol m⁻² s⁻¹ 时，速率翻倍，表明光是主要限制因素。”

7. Cellular Respiration – Coenzymes and Yield | 细胞呼吸 – 辅酶与能量产量题

IB and Edexcel examiners often ask students to ‘explain the role of coenzymes in aerobic respiration’ or ‘calculate the total number of ATP molecules produced from one glucose molecule’. An exemplary answer distinguishes substrate‑level phosphorylation from oxidative phosphorylation and correctly accounts for the NADH shuttle systems.

IB 和 Edexcel 的考官经常要求学生“解释辅酶在有氧呼吸中的作用”或“计算一分子葡萄糖产生的 ATP 总数”。示范性答案要区分底物水平磷酸化和氧化磷酸化，并正确计算 NADH 穿梭系统的能量。

Glycolysis: glucose → 2 pyruvate; net 2 ATP (substrate level) + 2 NADH. In cytoplasm (English).
糖酵解：葡萄糖 → 2 丙酮酸；净产生 2 ATP（底物水平）+ 2 NADH。发生在细胞质（中文）。
Link reaction: pyruvate → acetyl‑CoA; 2 NADH, CO₂ released (English).
连接反应：丙酮酸 → 乙酰辅酶 A；2 NADH，释放 CO₂（中文）。
Krebs cycle: total per glucose: 2 ATP (or GTP), 6 NADH, 2 FADH₂ (English).
三羧酸循环：每分子葡萄糖总计：2 ATP（或 GTP），6 NADH，2 FADH₂（中文）。
Oxidative phosphorylation: NADH and FADH₂ donate electrons to ETC; proton gradient drives ATP synthase. 1 NADH ≈ 3 ATP (or 2.5 depending on shuttle), 1 FADH₂ ≈ 2 ATP (or 1.5). Use the standard IB value: 1 NADH = 3 ATP for Edexcel and IB (unless specified). Total: prokaryote ~38 ATP; eukaryote ~36 ATP due to cost of mitochondrial shuttle (English).
氧化磷酸化：NADH 和 FADH₂ 将电子提供给电子传递链；质子梯度驱动 ATP 合酶。1 NADH 约产生 3 ATP（或依穿梭系统不同为 2.5），1 FADH₂ 约产生 2 ATP（或 1.5）。除非特别说明，Edexcel 和 IB 通常使用 1 NADH = 3 ATP 的标准值。总计：原核生物约 38 ATP；真核生物因线粒体穿梭消耗约 36 ATP（中文）。

Coenzyme roles: NAD⁺ and FAD act as hydrogen/electron carriers, becoming reduced during glycolysis, link reaction, and Krebs cycle, and reoxidised at the ETC. Emphasise that they are oxidising agents.

辅酶的作用：NAD⁺ 和 FAD 作为氢 / 电子载体，在糖酵解、连接反应和 Krebs 循环中被还原，在电子传递链处被重新氧化。强调它们是氧化剂。

8. Immune Response – Primary vs Secondary | 免疫应答 – 初次与再次应答题

A data‑based question may present a graph of antibody concentration over time after first and second exposure to an antigen. The explanation must cover clonal selection, memory cells, and the faster, stronger secondary response. Use precise immunological terminology.

数据分析题可能给出初次和再次接触抗原后抗体浓度随时间变化的曲线。解释必须涵盖克隆选择、记忆细胞，以及更快更强的再次应答。使用精确的免疫学术语。

Primary response: first exposure, lag phase while specific B‑cells with complementary receptors are activated, clonal expansion, and differentiation into plasma cells and a few memory cells. IgM appears first, then IgG (English).
初次应答：首次接触，存在滞后期，此时带有互补受体的特异性 B 细胞被激活，进行克隆扩增，并分化为浆细胞和少数记忆细胞。先出现 IgM，后出现 IgG（中文）。
Secondary response: upon re‑exposure, memory cells (both B and T) rapidly proliferate and differentiate, producing a larger quantity of antibodies (mainly IgG) in a shorter time. The antibody concentration peaks higher and stays elevated longer (English).
再次应答：再次暴露时，记忆细胞（B 细胞和 T 细胞）迅速增殖并分化，在更短时间内产生大量抗体（主要为 IgG）。抗体浓度峰值更高，维持时间更长（中文）。
Explain the graph: label the lag phase for primary, shorter lag for secondary, higher peak for secondary. Connect to vaccination principles (English).
解释曲线：标出初次应答的滞后期、再次应答较短的滞后期、更高的峰值。联系疫苗接种原理（中文）。

Edexcel often asks students to compare active and passive immunity, using examples such as vaccination versus antivenom injection. Always state whether memory cells are produced.

Edexcel 经常要求学生比较主动免疫和被动免疫，举例疫苗接种与抗毒血清注射。应始终说明是否产生记忆细胞。

9. Neuronal Transmission and Synapses | 神经传导与突触题

Questions on the propagation of an action potential and synaptic transmission are staples of both IB Option A (Neurobiology) and Edexcel Topic 8 (Grey Matter). You may be asked to interpret a graph of membrane potential or to explain how a drug affects synaptic function.

动作电位的传播和突触传递是 IB 选修 A（神经生物学）和 Edexcel 主题 8（灰质）中的必考内容。题目可能要求解读膜电位曲线，或解释药物如何影响突触功能。

Resting potential: −70 mV maintained by Na⁺/K⁺ pump and differential permeability (English).
静息电位：−70 mV，由 Na⁺/K⁺ 泵和差异通透性维持（中文）。
Depolarisation: voltage‑gated Na⁺ channels open, Na⁺ influx, membrane potential rises to +30 mV (English).
去极化：电压门控 Na⁺ 通道打开，Na⁺ 内流，膜电位上升至 +30 mV（中文）。
Repolarisation: Na⁺ channels inactivate, voltage‑gated K⁺ channels open, K⁺ efflux. Hyperpolarisation then restores resting state (English).
复极化：Na⁺ 通道失活，电压门控 K⁺ 通道打开，K⁺ 外流。然后超极化过程恢复静息状态（中文）。
Synapse: arrival of action potential opens Ca²⁺ channels, vesicles fuse, neurotransmitter released, binds to postsynaptic receptors, causes either EPSP or IPSP (English).
突触：动作电位到达，开启 Ca²⁺ 通道，囊泡融合，神经递质释放，与突触后受体结合，引起 EPSP 或 IPSP（中文）。
Drugs: e.g. neonicotinoids mimic acetylcholine and irreversibly bind receptors, causing continuous stimulation. Mention enzyme acetylcholinesterase and re‑uptake (English).
药物：例如新烟碱类农药模拟乙酰胆碱，不可逆地与受体结合，引起持续刺激。须提及乙酰胆碱酯酶和重摄取机制（中文）。

Always refer to ions by name and charge, and specify the direction of movement. Avoid simply saying ‘more positive’; give approximate voltage values.

始终明确写出离子的名称和电荷，并指定运动方向。避免只说“电位变得更正”；给出近似的电压值。

10. Ecology Data Interpretation | 生态学数据解读题

Both syllabi feature questions with quadrat data, mark‑release‑recapture calculations, or Simpson’s diversity index. Typical instruction: ‘Calculate the index of diversity and suggest what it shows about the community.’ A model answer combines accurate arithmetic with biological interpretation.

两个体系都会出现关于样方数据、标记重捕法计算或者辛普森多样性指数的问题。典型指令：“计算多样性指数，并指出它说明了群落的什么特征。” 示范性回答要兼顾精确的算术与生物学解释。

Simpson’s Index: D = 1 − (Σ n(n−1) / N(N−1))

Worked example: species counts are given. First compute N (total organisms), then for each species compute n(n−1), sum them, divide by N(N−1), subtract from 1. Interpret: a high value (close to 1) indicates high diversity; community is more stable and resilient to environmental change. A low value may indicate dominance by one species and a stressed ecosystem.

计算示例：给出各物种数量。先求出 N（总个体数），然后对每个物种计算 n(n−1)，求和，除以 N(N−1)，用 1 减去该值。解读：数值高（接近 1）表明多样性高；群落更稳定，对环境变化有更强的恢复能力。数值低可能表明某一物种占优势，生态系统可能受到胁迫。

For mark‑release‑recapture, use Lincoln Index: N = (n₁ × n₂) / n₃, where n₁ = first capture (marked), n₂ = second capture total, n₃ = marked recaptures. Discuss assumptions: no immigration/emigration, marking does not affect survival, random mixing (English).
对于标记重捕法，使用林肯指数：N = (n₁ × n₂) / n₃，其中 n₁ 为第一次标记个体数，n₂ 为第二次捕获总数，n₃ 为第二次捕获中带标记的个数。讨论假设前提：没有迁徙，标记不影响存活，个体已随机混合（中文）。

Edexcel may also ask for a transect description: how species distribution changes from low to high tide zones, linking to abiotic factors like salinity and desiccation.

Edexcel 还可能要求描述样线：物种分布如何从潮间带低处到高处变化，并联系到盐度和干燥等非生物因素。

11. Gene Technology and Ethical Evaluation | 基因技术与伦理评价题

A distinctive feature of both IB and Edexcel is the requirement to evaluate biotechnological methods. A question might present a case study on CRISPR‑Cas9 and ask for ‘discuss the ethical implications’. The answer must balance scientific explanation with arguments for and against.

IB 和 Edexcel 的一个显著特点是要求评价生物技术方法。题目可能给出一个关于 CRISPR‑Cas9 的案例，要求“讨论其伦理问题”。回答必须在科学解释的基础上，均衡地提出支持与反对的论点。

Scientific explanation: CRISPR‑Cas9 uses a guide RNA to direct Cas9 endonuclease to a specific DNA sequence, creating a double‑strand break. Cellular repair through non‑homologous end joining (NHEJ) or homology‑directed repair (HDR) can knock out or edit a gene (English).
科学解释：CRISPR‑Cas9 使用向导 RNA 将 Cas9 核酸内切酶引导至特定 DNA 序列，造成双链断裂。通过非同源末端连接（NHEJ）或同源定向修复（HDR）进行细胞修复，从而敲除或编辑基因（中文）。
Arguments for: potential to cure genetic disorders like sickle cell anaemia; could eradicate vector‑borne diseases by gene drive; advances in crop improvement (English).
支持论点：有潜力治愈镰刀型细胞贫血症等遗传病；可借助基因驱动消除媒介传播疾病；推动作物改良（中文）。
Arguments against: off‑target mutations; germline editing raises consent and ‘designer baby’ concerns; ecological risks of gene drives; equity in access to therapy (English).
反对论点：脱靶突变；生殖系编辑引发知情同意和“设计婴儿”的担忧；基因驱动的生态风险；治疗的可及性与公平性（中文）。

A top‑band answer makes a reasoned conclusion, weighing the evidence and suggesting regulation. Avoid one‑sided arguments.

高分段答案需要有基于证据的推理结论，并建议如何进行监管。避免单方面的论证。

12. Drawing and Annotating a Diagram | 绘图与标注题

In IB Paper 2 and Edexcel practical‑based questions, you may be asked to ‘draw a labelled diagram of the ultrastructure of a mitochondrion’ or ‘sketch the oxygen dissociation curve for adult and foetal haemoglobin’. Clear, pencil‑drawn lines with accurate proportions and labels touching the structure earn the marks.

在 IB 试卷二和 Edexcel 基于实验的题目中，可能会要求你“画出并标注线粒体的超微结构图”或“画出成人和胎儿血红蛋白的氧解离曲线”。用铅笔画出清晰、比例正确的线条，并让标注线接触到结构，才能拿到分数。

Mitochondrion: must show outer membrane, inner membrane with cristae, matrix, ribosomes (70S), circular DNA. No shading; labels on the right side (English).
线粒体：必须显示外膜、带嵴的内膜、基质、核糖体（70S）、环状 DNA。不要涂阴影；标注线在右侧（中文）。
Oxygen dissociation curve: sigmoid shape; normal adult haemoglobin curve, foetal haemoglobin curve shifted to the left. Label axes: partial pressure of O₂ (x‑axis) and % saturation of haemoglobin (y‑axis). Explain that left shift indicates higher affinity for O₂, facilitating transfer across placenta (English).
氧解离曲线：S 形曲线；正常成人血红蛋白曲线，胎儿血红蛋白曲线左移。坐标轴标注：氧分压（x 轴）和血红蛋白饱和度（%）（y 轴）。解释左移代表对 O₂ 的亲和力更高，有利于通过胎盘传递氧气（中文）。

In a diagram question always include a title and ensure that your labels do not cross each other. Underline labels for IB.

回答绘图题时，务必写上标题，并确保标注线不相互交叉。在 IB 中，标注文字要加下划线。

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Typical Example Questions Explained in IB Edexcel Biology | IB Edexcel 生物：典型例题详解