📚 IB Biology: Worked Examples of Typical Exam Questions | IB 生物：典型例题详解
IB Biology assessments go beyond simple recall; they require you to interpret data, apply concepts to novel situations, and articulate explanations with precision. This article presents ten carefully chosen worked examples that mirror the style of IB exam questions. Each example is broken down step by step, highlighting the reasoning and key terminology examiners expect. Use these walkthroughs to sharpen your analytical skills and deepen your conceptual understanding.
IB 生物考核不仅考查记忆力，更要求你解读数据、将概念应用到新情境中，并精准地组织解释。本文精选了十道贴近 IB 考试风格的典型例题，逐层拆解每一步推理过程和考官期望的关键术语。通过这些解析，你可以磨练分析技巧、加深概念理解，为考试做好充分准备。
1. Enzyme Activity and Temperature | 酶活性与温度
Question: A student measured the rate of an enzyme-catalysed reaction at different temperatures. The results are summarised in Table 1. Describe the trend and explain why the rate falls dramatically above 40 °C.
题目：一名学生测定了不同温度下某种酶促反应的速率，结果见表 1。请描述变化趋势，并解释为何速率在 40 °C 以上急剧下降。
Table 1: Temperature (°C): 0, 10, 20, 30, 40, 50, 60. Reaction rate (arbitrary units): 2, 5, 10, 15, 17, 5, 0.
表 1：温度 (°C)：0、10、20、30、40、50、60。反应速率（任意单位）：2、5、10、15、17、5、0。
Step 1 – Identify the overall pattern. As temperature rises from 0 °C to 40 °C, the rate of reaction increases steadily, reaching a maximum at 40 °C.
步骤 1 – 确定总体规律。温度从 0 °C 升至 40 °C 时，反应速率持续增大，在 40 °C 达到最大值。
Step 2 – Explain the increase. Higher temperatures raise the kinetic energy of both enzyme and substrate molecules. This leads to more frequent collisions and a greater proportion of collisions exceeding the activation energy, so the rate rises.
步骤 2 – 解释升速原因。温度升高使酶和底物分子的动能增大，碰撞更频繁，且更多碰撞的能量超过活化能，因此速率上升。
Step 3 – Explain the sudden decline above 40 °C. Beyond the optimum temperature, the enzyme’s tertiary structure is disrupted. Hydrogen bonds and ionic interactions that maintain the precise shape of the active site break, causing denaturation. The active site is no longer complementary to the substrate, so the enzyme can no longer catalyse the reaction.
步骤 3 – 解释 40 °C 以上速率骤降。超过最适温度后，酶的三级结构被破坏。维持活性位点精确形状的氢键和离子键断裂，酶发生变性。活性位点不再与底物互补，酶无法继续催化反应。
Step 4 – Note reversibility. The decrease in rate at low temperatures (e.g. 0 °C) is due to reduced kinetic energy, not denaturation; the enzyme remains functional and activity can be restored upon warming.
步骤 4 – 注意可逆性。低温（如 0 °C）下速率降低是因为动能减小，并非变性；酶依然具有功能，升温后活性可恢复。
2. Osmosis and Water Potential Calculation | 渗透作用与水势计算
Question: Potato strips were immersed in a series of sucrose solutions until no further change in mass occurred. The sucrose concentration that gave zero net change in mass was 0.30 mol dm⁻³. At 25 °C, pure water has a water potential of 0 kPa. Calculate the water potential of the potato tissue. (Assume sucrose does not ionise, so i = 1; R = 8.31 J mol⁻¹ K⁻¹; T = 298 K.)
题目：将马铃薯条浸泡在一系列蔗糖溶液中，直至质量不再变化。引起净质量变化为零的蔗糖浓度为 0.30 mol dm⁻³。25 °C 时纯水的水势为 0 kPa。计算马铃薯组织的水势。（设蔗糖不解离，i = 1；R = 8.31 J mol⁻¹ K⁻¹；T = 298 K。）
Step 1 – State the key relationship. At equilibrium, the water potential of the potato tissue equals the water potential of the external solution. Since the solution contains sucrose, its water potential is determined solely by solute potential (ψₛ), because pressure potential (ψₚ) in an open container is 0.
步骤 1 – 列出关键关系。平衡时，马铃薯组织的水势等于外界溶液的水势。由于溶液中含蔗糖，且开口容器中压力势 (ψₚ) 为 0，水势仅由渗透势 (ψₛ) 决定。
Step 2 – Convert concentration units. The solute potential equation ψₛ = –iCRT requires concentration in mol m⁻³. 0.30 mol dm⁻³ = 0.30 mol per 0.001 m³ = 300 mol m⁻³.
步骤 2 – 转换浓度单位。渗透势公式 ψₛ = –iCRT 要求浓度单位为 mol m⁻³。0.30 mol dm⁻³ = 每 0.001 m³ 0.30 mol = 300 mol m⁻³。
Step 3 – Substitute values into the formula.
ψₛ = –iCRT = –1 × 300 mol m⁻³ × 8.31 J mol⁻¹ K⁻¹ × 298 K
步骤 3 – 将数值代入公式。
ψₛ = –iCRT = –1 × 300 mol m⁻³ × 8.31 J mol⁻¹ K⁻¹ × 298 K
Step 4 – Calculate. ψₛ = –(1 × 300 × 8.31 × 298) = –743, 214 Pa. Since 1 kPa = 1000 Pa, ψₛ ≈ –743 kPa. Therefore, the water potential of the potato tissue is –743 kPa.
步骤 4 – 计算。ψₛ = –(1 × 300 × 8.31 × 298) = –743,214 Pa。因为 1 kPa = 1000 Pa，ψₛ ≈ –743 kPa。因此马铃薯组织的水势为 –743 kPa。
3. Genetic Cross: ABO Blood Group Inheritance | 遗传杂交：ABO 血型的遗传
Question: A man with blood type A (genotype IAi) and a woman with blood type B (genotype IBi) are expecting a child. Determine the probability that the child will have type O blood.
题目：一名血型为 A 型（基因型 IAi）的男性与一名血型为 B 型（基因型 IBi）的女性将要生育。求他们的孩子为 O 型血的概率。
Step 1 – Identify possible gametes. The man produces gametes carrying either IA or i. The woman produces gametes carrying either IB or i.
步骤 1 – 确定可能的配子。男方产生的配子携带 IA 或 i；女方产生的配子携带 IB 或 i。
Step 2 – Construct a Punnett square. Combine gametes to show all possible offspring genotypes: IAIB (type AB), IAi (type A), IBi (type B), and ii (type O).
步骤 2 – 构建庞纳特方格。配子组合产生所有可能子代基因型：IAIB（AB 型）、IAi（A 型）、IBi（B 型）和 ii（O 型）。
Step 3 – Determine the probability. Each genotype is equally likely, occurring with a probability of 1/4. Only the ii genotype gives type O blood, so the probability is 1/4 or 25%.
步骤 3 – 计算概率。每种基因型等可能出现，概率各为 1/4。只有 ii 基因型表现为 O 型血，故概率为 1/4 即 25%。
4. Action Spectrum and Absorption Spectrum | 作用光谱与吸收光谱
Question: The absorption spectrum of chlorophyll a and the action spectrum of photosynthesis are plotted against wavelength. Explain why the action spectrum shows a trough in the green region of the visible spectrum and why plants appear green.
题目：叶绿素 a 的吸收光谱与光合作用的作用光谱随波长绘制。解释为何作用光谱在可见光绿色区域呈现谷值，以及植物为何呈现绿色。
Step 1 – Recall what each spectrum represents. The absorption spectrum shows wavelengths of light absorbed by a pigment; the action spectrum shows the rate of photosynthesis at each wavelength.
步骤 1 – 回顾两种光谱的含义。吸收光谱显示色素所吸收的光波长；作用光谱显示各波长下的光合作用速率。
Step 2 – Link the trough in the action spectrum to absorption. Chlorophyll a absorbs mainly red (around 680 nm) and blue (around 430 nm) light. It absorbs very little green light (around 550 nm). Because photosynthesis depends on light absorption by chlorophyll, the rate is low in the green region, creating the trough in the action spectrum.
步骤 2 – 将光谱谷值与吸收联系起来。叶绿素 a 主要吸收红光（约 680 nm）和蓝光（约 430 nm），几乎不吸收绿光（约 550 nm）。由于光合作用依赖叶绿素对光的吸收，在绿光区反应速率低，因而作用光谱出现谷值。
Step 3 – Explain leaf colour. Green light is not absorbed; it is reflected and transmitted. Our eyes detect this reflected green light, so healthy leaves appear green.
步骤 3 – 解释叶片颜色。绿光未被吸收，而被反射和透射。人眼捕获到这些反射的绿光，因此健康叶片呈现绿色。
5. DNA Replication Fidelity and Proofreading | DNA 复制的准确性与校对
Question: DNA polymerase III possesses 3′ → 5′ exonuclease activity. Explain the importance of this proofreading function in maintaining genome stability.
题目：DNA 聚合酶 III 具有 3′ → 5′ 外切核酸酶活性。解释该校对功能对维持基因组稳定的重要性。
Step 1 – Describe normal nucleotide insertion. DNA polymerase III adds nucleotides to the growing DNA strand, but the error rate for initial incorporation is about 1 in 10⁵ bases.
步骤 1 – 描述正常的核苷酸插入。DNA 聚合酶 III 将核苷酸添加到生长链上，但初始插入的错配率约为每 10⁵ 个碱基出现一次。
Step 2 – Explain proofreading. The 3′ → 5′ exonuclease site acts as a ‘proofreader’. When a mismatched nucleotide is detected, the polymerase stalls, the exonuclease removes the incorrect nucleotide from the 3′ end, and the polymerase resumes synthesis with the correct base.
步骤 2 – 解释校对机制。3′ → 5′ 外切核酸酶位点充当“校对员”。一旦检测到错配核苷酸，聚合酶暂停，外切酶从 3′ 端切除错误核苷酸，随后聚合酶插入正确碱基继续合成。
Step 3 – State the biological significance. This proofreading reduces the overall error rate to about 1 in 10⁷–10⁸ bases, preventing the accumulation of harmful mutations. Defects in proofreading are linked to higher cancer risk and genetic diseases.
步骤 3 – 阐述生物学意义。校对功能将总错误率降至约每 10⁷–10⁸ 个碱基一次，防止有害突变累积。校对功能缺陷会增加癌症风险和遗传病发生率。
6. Bacterial Growth Curve and Generation Time | 细菌生长曲线与代时
Question: A bacterial culture is in exponential growth phase. At the start (t = 0), the population is 1.0 × 10⁴ cells ml⁻¹. After 3 hours, the population reaches 6.4 × 10⁵ cells ml⁻¹. Calculate the generation time.
题目：某细菌培养物处于指数生长期。初始（t = 0）菌浓度为 1.0 × 10⁴ cells ml⁻¹，3 小时后达到 6.4 × 10⁵ cells ml⁻¹。计算代时。
Step 1 – Use the exponential growth model. N = N₀ × 2ⁿ, where N is the final cell count, N₀ the initial count, and n the number of generations.
步骤 1 – 使用指数生长模型。N = N₀ × 2ⁿ，其中 N 为最终细胞数，N₀ 为初始细胞数，n 为繁殖代数。
Step 2 – Calculate the number of generations. Divide both sides by N₀: N/N₀ = 2ⁿ. (6.4 × 10⁵) / (1.0 × 10⁴) = 64. So 2ⁿ = 64. Since 64 = 2⁶, n = 6 generations in 3 hours.
步骤 2 – 求代数。N/N₀ = 2ⁿ，代入得 (6.4 × 10⁵) / (1.0 × 10⁴) = 64。因此 2ⁿ = 64，64 = 2⁶，故 n = 6 代（3 小时内）。
Step 3 – Determine generation time. Generation time g = total time / number of generations. 3 hours = 180 minutes. g = 180 min / 6 = 30 minutes. Hence the bacteria double every 30 minutes.
步骤 3 – 求代时。代时 g = 总时间 / 繁殖代数。3 小时 = 180 分钟，g = 180 分钟 / 6 = 30 分钟。因此该菌每 30 分钟分裂一次。
7. Synaptic Transmission | 突触传递
Question: Describe the sequence of events that occurs when an action potential arrives at the presynaptic terminal of a cholinergic synapse.
题目：描述动作电位到达胆碱能突触前膜时发生的一系列事件。
Step 1 – Depolarisation opens Ca²⁺ channels. The action potential depolarises the presynaptic membrane, causing voltage-gated calcium ion channels to open.
步骤 1 – 去极化开启钙离子通道。动作电位使突触前膜去极化，导致电压门控钙离子通道开放。
Step 2 – Calcium influx triggers exocytosis. Calcium ions (Ca²⁺) diffuse into the presynaptic knob down their concentration gradient. The rise in intracellular Ca²⁺ causes synaptic vesicles containing acetylcholine (ACh) to fuse with the presynaptic membrane and release ACh into the synaptic cleft by exocytosis.
步骤 2 – 钙内流触发胞吐。钙离子沿浓度梯度扩散进入突触前膜。胞内 Ca²⁺ 浓度升高促使含有乙酰胆碱 (ACh) 的突触囊泡与突触前膜融合，以胞吐方式将 ACh 释放到突触间隙。
Step 3 – Neurotransmitter binds to receptors. ACh diffuses across the cleft and binds to specific receptor proteins on the postsynaptic membrane, which are ligand-gated sodium ion channels.
步骤 3 – 神经递质与受体结合。ACh 扩散穿越间隙，与突触后膜上特定的受体蛋白结合，这类受体是配体门控钠离子通道。
Step 4 – Postsynaptic potential is generated. Channel opening allows Na⁺ to enter the postsynaptic neuron, causing depolarisation. If the threshold is reached, an action potential is initiated in the postsynaptic cell.
步骤 4 – 产生突触后电位。通道开放允许 Na⁺ 进入突触后神经元，引发去极化。若达到阈值，则在突触后细胞引发动作电位。
Step 5 – Signal termination. ACh is rapidly broken down by acetylcholinesterase in the cleft, and the products are reabsorbed. This stops continuous stimulation and allows the synapse to reset.
步骤 5 – 信号终止。突触间隙中的乙酰胆碱酯酶迅速分解 ACh，分解产物被重吸收。这阻止了持续刺激，使突触得以恢复。
8. Direct ELISA for Detecting Antibodies | 用于抗体检测的直接 ELISA
Question: Outline the principle of a direct ELISA test used to screen a patient’s blood for HIV antibodies.
题目：简述使用直接 ELISA 检测筛查患者血液中 HIV 抗体的原理。
Step 1 – Coating the well. Purified HIV antigens are attached to the bottom of a well in a microtiter plate.
步骤 1 – 包被。将纯化的 HIV 抗原固定在微孔板底部的孔内。
Step 2 – Adding the patient’s serum. The patient’s blood serum is added. If HIV antibodies are present, they bind specifically to the immobilised antigens. Unbound antibodies are washed away.
步骤 2 – 加入患者
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