📚 Year 12 CAIE Biology: Unit Test Mock Paper Analysis | Year 12 CAIE 生物：单元测试模拟卷解析
This article walks you through a representative Year 12 CAIE Biology unit test, dissecting each question with model answers, common pitfalls and exam-savvy commentary. The mock paper covers the core AS topics that frequently appear in structured and multiple-choice sections, helping you consolidate key concepts and sharpen your technique for the real exam.
本文带你逐题解析一份典型的 Year 12 CAIE 生物单元测试模拟卷，提供标准答案、常见错误和备考思路。模拟卷覆盖了 AS 阶段的核心主题，常出现在结构题和选择题中，帮助你巩固关键概念，打磨应试技巧。
1. Cell Membrane Structure and Transport | 细胞膜结构与运输
Question: An experiment is set up using dialysis tubing filled with a 5% glucose solution, immersed in a beaker of distilled water. After 30 minutes, the mass of the bag increases by 12%. Explain the result in terms of water potential and the properties of the membrane. (3 marks)
题目：用透析管装入 5% 葡萄糖溶液，浸入蒸馏水烧杯中。30 分钟后袋质量增加 12%。请从水势和膜特性的角度解释结果。（3分）
The dialysis tubing is partially permeable, allowing small water molecules to pass but restricting larger glucose molecules. Distilled water has a water potential of 0 kPa, while the 5% glucose solution has a lower (more negative) water potential. Water therefore moves down the water potential gradient from the beaker into the bag by osmosis, causing the bag to swell and increase in mass.
透析管是部分通透膜，允许小水分子通过但限制较大葡萄糖分子。蒸馏水水势为 0 kPa，而 5% 葡萄糖溶液水势更低（更负）。因此水沿水势梯度从烧杯通过渗透作用进入袋内，使袋膨胀、质量增加。
2. Enzyme Activity and Temperature | 酶活性与温度
Question: The graph shows the rate of an enzyme-catalysed reaction at various temperatures. At 50 °C the rate drops to zero. Explain why, using the concepts of tertiary structure and active site complementarity. (4 marks)
题目：图中显示酶促反应速率随温度变化的情况。在 50 °C 时反应速率降至零。请利用三级结构和活性位点互补性的概念加以解释。（4分）
At temperatures above the optimum, increased kinetic energy disrupts the hydrogen bonds, ionic bonds and hydrophobic interactions that maintain the enzyme's precise tertiary structure. This denaturation alters the shape of the active site so that the substrate can no longer fit. The lock-and-key or induced-fit model can no longer operate, and no enzyme-substrate complexes form, reducing the rate to zero.
当温度超过最适温度时，增加的功能会破坏维持酶精确三级结构的氢键、离子键和疏水相互作用。这种变性改变了活性位点的形状，使底物不再匹配。锁钥模型或诱导契合模型无法运作，不能形成酶-底物复合物，反应速率降至零。
3. Mitosis and the Cell Cycle | 有丝分裂与细胞周期
Question: A root tip squash is examined under the microscope. 60 cells are counted: 8 in prophase, 3 in metaphase, 4 in anaphase, 10 in telophase and 35 in interphase. Calculate the mitotic index, showing your working. Suggest why most cells are in interphase. (3 marks)
题目：镜下观察根尖压片，计数 60 个细胞：前期 8 个，中期 3 个，后期 4 个，末期 10 个，间期 35 个。计算有丝分裂指数，写出计算过程。并说明为何多数细胞处于间期。（3分）
phase + anaphase + telophase = 8 + 3 + 4 + 10 = 25. Mitotic index = 25/60 = 0.417 (or 41.7% l functions, replicates DNA and prepares for division. Mitosis itself is relatively brief.
有丝分裂指数 = 进行有丝分裂的细胞数 / 总细胞数。有丝分裂细胞数 = 前期 + 中期 + 后期 + 末期 = 8 + 3 + 4 + 10 = 25。指数 = 25/60 = 0.417（或 41.7%）。多数细胞处于间期是因为间期占据了细胞周期最长的时段；间期细胞执行正常功能、复制 DNA 并为分裂做准备。有丝分裂本身相对短暂。
4. Biological Molecules – Food Tests | 生物分子——食品检测
Question: An unknown biological sample tests positive with Biuret reagent and remains blue-black when iodine solution is added, but yields a brick‑red precipitate with Benedict's reagent after heating. Identify the food groups present. Deduce what the sample is likely to be, justifying your answer. (3 marks)
题目：某未知生物样本遇双缩脲试剂呈阳性，滴加碘液后保持蓝黑色，但与班氏试剂加热后产生砖红色沉淀。鉴别存在的食物成分。推断该样本可能是什么并说明理由。（3分）
Biuret reagent turns purple in the presence of peptide bonds, indicating protein. Iodine solution tests for starch; remaining blue-black means starch is absent. A brick‑red precipitate with Benedict's reagent reveals a reducing sugar. Therefore the sample contains protein and reducing sugar but no starch. A possible identification is milk, because milk contains the protein casein and the reducing sugar lactose; starch is not a component of milk.
双缩脲试剂遇肽键变紫色，表明含蛋白质。碘液检测淀粉；保持蓝黑色说明不含淀粉。班氏试剂加热后出现砖红色沉淀表明有还原糖。因此样本含蛋白质和还原糖，不含淀粉。牛奶是可能的答案，因为牛奶含有蛋白质酪蛋白和还原糖乳糖，且不含淀粉。
5. DNA Replication | DNA 复制
Question: Describe the role of DNA helicase and DNA polymerase in semi‑conservative replication. Explain why one new strand is synthesised continuously and the other in Okazaki fragments. (5 marks)
题目：描述 DNA 解旋酶和 DNA 聚合酶在半保留复制中的作用。解释为何一条新链是连续合成，另一条则形成冈崎片段。（5分）
DNA helicase unwinds the double helix by breaking hydrogen bonds between complementary base pairs, forming a replication fork. DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides, using the exposed parental strands as templates. It can only add nucleotides in the 5' to 3' direction. On the leading strand, synthesis moves towards the replication fork and is continuous. On the lagging strand, the polymerase must work away from the fork, so it synthesises short segments (Okazaki fragments) which are later joined by DNA ligase.
DNA 解旋酶通过打开互补碱基对间的氢键解开双螺旋，形成复制叉。DNA 聚合酶以暴露的亲本链为模板，催化相邻核苷酸之间形成磷酸二酯键。它只能沿 5' 至 3' 方向添加核苷酸。前导链的合成方向朝向复制叉，因此是连续的。后随链的聚合酶必须远离复制叉工作，因此合成短片段（冈崎片段），后续由 DNA 连接酶连接。
6. Plant Transport – Transpiration | 植物运输——蒸腾作用
Question: A potometer is used to measure the rate of water uptake by a leafy shoot. Explain how three environmental factors – light intensity, humidity and air movement – affect the transpiration stream. (6 marks)
题目：使用蒸腾计测量带叶枝条的吸水量。解释三种环境因子——光照强度、湿度和空气流动——如何影响蒸腾流。（6分）
High light intensity stimulates stomatal opening for photosynthesis, increasing the diffusion gradient for water vapour out of the leaf and thus accelerating transpiration. Low humidity means a steeper water potential gradient between the leaf's sub‑stomatal air spaces and the external atmosphere; this increases the rate of water loss. Greater air movement removes the boundary layer of humid air near the stomata, maintaining a steep diffusion gradient. All three factors therefore promote transpiration, which pulls water up the xylem in the transpiration stream through the cohesion‑tension mechanism.
高光照强度促使气孔开放进行光合作用，增大水蒸气向外的扩散梯度，从而加速蒸腾。低湿度意味着叶片气室与外界大气之间的水势梯度更陡，这加大了水分散失速率。较大的空气流动移除气孔附近潮湿空气的边界层，维持陡峭的扩散梯度。因此三因子均促进蒸腾，通过内聚力-张力机制牵引水分沿木质部上升形成蒸腾流。
7. Mammalian Circulation – Cardiac Output | 哺乳动物循环——心输出量
Question: An athlete has a resting heart rate of 55 beats min⁻¹ and a stroke volume of 80 cm³. Calculate resting cardiac output. After exercise, the heart rate rises to 160 beats min⁻¹ and cardiac output to 24 dm³ min⁻¹. Calculate the new stroke volume. Show all working and units. (3 marks)
题目：一名运动员静息心率为 55 次/分，每搏输出量为 80 cm³。计算静息心输出量。运动后心率升至 160 次/分，心输出量为 24 dm³/min。计算新的每搏输出量。显示所有计算过程及单位。（3分）
000 cm³ min⁻¹. New stroke volume = cardiac output / heart rate = 24 000 cm³ min⁻¹ / 160 min⁻¹ = 150 cm³. The stroke volume increased from 80 cm³ to 150 cm³, demonstrating the heart's ability to pump more blood per beat during strenuous activity.
静息心输出量 = 心率 × 每搏输出量 = 55 min⁻¹ × 80 cm³ = 4400 cm³/min = 4.4 dm³/min。运动后：心输出量 = 24 dm³/min = 24 000 cm³/min。新每搏输出量 = 心输出量 / 心率 = 24 000 cm³/min / 160 min⁻¹ = 150 cm³。每搏输出量从 80 cm³ 增至 150 cm³，表明心脏在剧烈活动期间每搏泵血能力增强。
8. Gas Exchange and the Oxygen Dissociation Curve | 气体交换与氧解离曲线
Question: The oxygen dissociation curve for foetal haemoglobin lies to the left of that for adult haemoglobin. Explain the significance of this difference. Name and explain the effect that shifts the adult curve to the right during exercise. (4 marks)
题目：胎儿血红蛋白的氧解离曲线位于成人血红蛋白的左侧。解释这一差异的意义。命名并解释运动期间使成人曲线右移的效应。（4分）
A left‑shifted curve means foetal haemoglobin has a higher affinity for oxygen at the same partial pressure of oxygen, enabling it to take up oxygen from the mother's blood across the placenta. The Bohr effect shifts the adult curve to the right during exercise: increased CO₂ production lowers pH, which reduces haemoglobin's affinity for oxygen, promoting oxygen unloading in respiring tissues. This ensures active muscles receive more oxygen when demand is high.
曲线左移意味着在相同氧分压下胎儿血红蛋白对氧的亲和力更高，这使其能经由胎盘从母体血液中摄取氧气。运动期间波尔效应使成人曲线右移：增加的 CO₂ 生成降低 pH，减弱血红蛋白对氧的亲和力，促进氧在呼吸组织中释放。这确保活跃的肌肉在需求高时获得更多氧。
9. Immunity – Antibodies and Vaccination | 免疫——抗体与疫苗接种
Question: A child receives a booster dose of the MMR vaccine. Explain the cellular events that result in a rapid, high‑level production of anti‑measles antibodies, referring to memory cells and plasma cells. (5 marks)
题目：一名儿童接种 MMR 加强针。解释导致快速、高水平产生抗麻疹抗体的细胞事件，涉及记忆细胞和浆细胞。（5分）
On first exposure, the primary immune response generates memory B cells specific to measles antigens. Upon re‑exposure via the booster, these memory cells rapidly divide by mitosis to form a large clone of plasma cells. Plasma cells are effector B cells that synthesise and secrete large quantities of complementary antibodies. The response is faster, stronger and longer‑lasting than the primary response because of the expanded population of memory cells that act as immunological memory.
初次暴露时，初次免疫应答产生针对麻疹抗原的特异性记忆 B 细胞。通过加强针再次暴露时，这些记忆细胞通过有丝分裂迅速增殖，形成大量浆细胞克隆。浆细胞是效应 B 细胞，合成并分泌大量互补抗体。由于记忆细胞群体的扩增充当免疫记忆，应答比初次更快、更强且更持久。
10. Infectious Disease – Tuberculosis | 传染病——结核病
Question: Discuss how the global increase in antibiotic resistance in Mycobacterium tuberculosis makes treatment difficult. In your answer, refer to natural selection and suggest one non‑antibiotic control measure. (4 marks)
题目：讨论结核分枝杆菌对抗生素耐药性的全球上升如何使治疗变得困难。回答中需提及自然选择并提出一种非抗生素控制措施。（4分）
Within a bacterial population, random mutations may confer resistance to a specific antibiotic. When antibiotics are used, susceptible bacteria are killed while resistant ones survive and reproduce, passing on the resistance alleles. Over time, the frequency of resistant strains increases through natural selection, making standard antibiotic regimens ineffective. This necessitates longer, more expensive multi‑drug therapy. A non‑antibiotic control measure is the BCG vaccination, which stimulates the immune system to recognise and fight M. tuberculosis, reducing the incidence of active TB.
在细菌群体中，随机突变可能赋予对特定抗生素的抗性。使用抗生素时，敏感菌被杀灭，而耐药菌存活并繁殖，将抗性等位基因传递给后代。随着时间推移，耐药菌株的频率通过自然选择升高，使标准抗生素方案失效。这需要更长、更昂贵的多药疗法。非抗生素控制措施如卡介苗接种，能刺激免疫系统识别并抗击结核分枝杆菌，降低活动性结核病的发病率。
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