📚 Answers to Self-Assessment Questions | 自评问题解答
Self-assessment questions are a cornerstone of active revision for Cambridge International AS and A Level Biology. They give you a chance to test your understanding of key concepts, practise structuring answers and identify gaps that need more work. The sample answers provided in this article focus on common questions drawn from across the syllabus and are written in the clear, precise style examiners expect.
自评问题是剑桥国际AS和A Level生物主动复习的基石。它们让你有机会检验自己对核心概念的理解，练习组织答案，并找出需要加强的薄弱环节。本文提供的样题答案聚焦大纲中常见的考点，并以考官所期待的那种清晰、精准的风格写成。
1. Why Self-Assessment Questions Matter | 自评问题为何重要
Self-assessment questions force you to recall information without the aid of notes, strengthening long-term memory. They also help you become familiar with command words such as 'describe', 'explain' and 'compare'. Regular practice with these questions builds the confidence needed to tackle exam-style papers under timed conditions.
自评题迫使你在没有笔记辅助的情况下回忆信息，从而强化长期记忆。它们还能帮助你熟悉“描述”、“解释”和“比较”之类的指令词。定期用这些问题进行训练，可以培养在计时条件下应对考试型试卷的信心。
2. How to Approach Self-Assessment Questions | 如何应对自评问题
Before writing, underline the command word and check the mark allocation. For a 3-mark 'describe' question, stick to stating facts without lengthy explanations. For an 'explain' question, always link cause and effect. Use precise biological terminology and, where appropriate, annotate diagrams to support your written answer.
动笔之前，划出指令词并查看分值。对于3分的“描述”题，只陈述事实，无需长篇解释。对于“解释”题，一定要将因果联系起来。使用准确的生物学术语，并在合适时用带注释的示意图来辅助书面回答。
3. Biological Molecules – Sample Questions and Answers | 生物分子 – 样题与答案
Q1: Compare the structures of α-glucose and β-glucose. (2 marks)
A1: Both are hexose monosaccharides with the formula C₆H₁₂O₆. The difference lies in the position of the –OH group on carbon 1: in α-glucose the –OH points downwards (below the ring), whereas in β-glucose it points upwards.
Q1：比较α-葡萄糖和β-葡萄糖的结构。(2分)
A1：两者都是分子式为C₆H₁₂O₆的己糖单糖。区别在于1号碳上–OH基团的位置：在α-葡萄糖中，–OH指向下方（环平面之下），而在β-葡萄糖中它指向上方。
Q2: Explain how the structure of cellulose relates to its function as a structural polysaccharide in plant cell walls. (4 marks)
A2: Cellulose is composed of β-glucose monomers linked by β-1,4 glycosidic bonds. Alternate glucose molecules are inverted, producing long, straight chains. Many chains run parallel and are cross-linked by hydrogen bonds between –OH groups, forming microfibrils. This arrangement gives the cell wall high tensile strength, preventing cells from bursting under turgor pressure.
Q2：解释纤维素的结构如何与其作为植物细胞壁结构多糖的功能相联系。(4分)
A2：纤维素由β-葡萄糖单体通过β-1,4糖苷键连接而成。相邻的葡萄糖分子交替翻转，产生长而直的链。许多链平行排列，并通过–OH基团之间的氢键交联，形成微纤维。这种排列赋予细胞壁很高的抗拉强度，防止细胞在膨压下破裂。
4. Cell Structure – Sample Questions and Answers | 细胞结构 – 样题与答案
Q: Outline the role of the rough endoplasmic reticulum and Golgi apparatus in the secretion of proteins from eukaryotic cells. (5 marks)
A: Ribosomes on the rough endoplasmic reticulum (RER) synthesise polypeptide chains, which enter the RER lumen for folding and modification. Transport vesicles bud off the RER and fuse with the cis face of the Golgi apparatus. Inside the Golgi, proteins undergo further modification, such as glycosylation, and are sorted and packaged into secretory vesicles. These vesicles move to the plasma membrane and release the proteins by exocytosis.
Q：概述粗面内质网和高尔基体在真核细胞分泌蛋白质过程中的作用。(5分)
A：粗面内质网上的核糖体合成多肽链，多肽进入RER腔进行折叠和修饰。运输小泡从RER出芽并与高尔基体的顺面融合。在高尔基体内部，蛋白质进一步修饰，例如糖基化，并被分选、包装成分泌小泡。这些小泡移动到质膜，通过胞吐作用释放蛋白质。
5. Enzymes – Sample Questions and Answers | 酶 – 样题与答案
Q: Explain the induced-fit model of enzyme action. (4 marks)
A: The induced-fit model states that the active site of an enzyme is not a rigid shape complementary to the substrate. Instead, when the substrate binds, the active site undergoes a conformational change, moulding itself more closely around the substrate. This puts a strain on bonds in the substrate and lowers the activation energy, allowing the reaction to proceed more rapidly. The enzyme returns to its original shape after product release.
Q：解释酶作用的诱导契合模型。(4分)
A：诱导契合模型指出，酶的活性位点并不是与底物刚性互补的形状。当底物结合时，活性位点发生构象变化，更紧密地包裹底物。这使底物内部的化学键产生张力，降低活化能，从而使反应更快进行。产物释放后，酶恢复原来的构象。
6. Cell Membranes and Transport – Sample Questions and Answers | 细胞膜与运输 – 样题与答案
Q: Describe the fluid mosaic model of membrane structure. (5 marks)
A: The membrane is composed of a phospholipid bilayer, with hydrophilic phosphate heads facing outwards towards aqueous environments and hydrophobic fatty acid tails pointing inwards. Proteins are scattered throughout the bilayer in a mosaic pattern. Some are integral proteins spanning the membrane, while others are peripheral. Cholesterol molecules sit between phospholipids, regulating membrane fluidity. The components can move laterally, giving the membrane a fluid nature.
Q：描述膜结构的流动镶嵌模型。(5分)
A：膜由磷脂双分子层构成，亲水的磷酸头朝向外侧的水环境，疏水的脂肪酸尾指向内部。蛋白质以镶嵌图案散布于双分子层中，一些是贯穿膜的整合蛋白，另一些是外周蛋白。胆固醇分子插在磷脂之间，调节膜的流动性。各组分可以横向移动，赋予膜流动性。
Q: Distinguish between diffusion and active transport. (2 marks)
A: Diffusion is the passive movement of particles down a concentration gradient and does not require metabolic energy. Active transport moves substances against a concentration gradient, requires carrier proteins and uses energy from ATP.
Q：区分扩散和主动运输。(2分)
A：扩散是粒子顺浓度梯度的被动运动，不需要代谢能。主动运输逆浓度梯度进行物质转运，需要载体蛋白，并利用ATP提供的能量。
7. The Immune System – Sample Questions and Answers | 免疫系统 – 样题与答案
Q: Distinguish between specific and non-specific immune responses. (4 marks)
A: Non-specific defences, such as skin, phagocytosis and inflammation, respond in the same way to all pathogens and act immediately. Specific immune responses involve lymphocytes (B cells and T cells) and are slower to activate upon first exposure. They are highly specific to a particular pathogen, produce memory cells and result in long-lasting immunity.
Q：区分特异性免疫应答与非特异性免疫应答。(4分)
A：非特异性防御，如皮肤、吞噬作用和炎症，对所有病原体以相同方式反应并立即起作用。特异性免疫应答涉及淋巴细胞（B细胞和T细胞），首次接触时激活较慢。它们对特定病原体具有高度特异性，产生记忆细胞，并带来持久的免疫力。
8. Mitosis and the Cell Cycle – Sample Questions and Answers | 有丝分裂与细胞周期 – 样题与答案
Q: Outline the events of prophase in mitosis. (3 marks)
A: Chromatin condenses into visible chromosomes, each consisting of two sister chromatids joined at the centromere. The nuclear envelope begins to break down. Centrioles migrate to opposite poles of the cell and spindle fibres begin to form.
Q：概括有丝分裂前期的事件。(3分)
A：染色质凝集成为可见的染色体，每条染色体由两条在着丝粒处相连的姐妹染色单体组成。核膜开始解体。中心粒移向细胞两极，纺锤丝开始形成。
9. DNA Replication and Protein Synthesis – Sample Questions and Answers | DNA复制与蛋白质合成 – 样题与答案
Q: Describe the role of DNA helicase in DNA replication. (3 marks)
A: DNA helicase unwinds the double helix by breaking hydrogen bonds between complementary base pairs. This separates the two strands and creates a replication fork, exposing the template bases for free nucleotides to pair with.
Q：描述DNA解旋酶在DNA复制中的作用。(3分)
A：DNA解旋酶通过断开互补碱基对之间的氢键来解开双螺旋。这使两条链分开，形成复制叉，将模板碱基暴露出来以便游离核苷酸进行配对。
10. Ecology and Energy Flow – Sample Questions and Answers | 生态与能量流 – 样题与答案
Q: Explain why energy transfer between trophic levels is usually less than 10% efficient. (4 marks)
A: Not all of the organism at one trophic level is consumed by the next level; parts such as bones and roots are left uneaten. Of the material that is eaten, a large proportion is lost in faeces as undigested matter. Much of the assimilated energy is released as heat through respiration for movement and metabolism. Only a small fraction is converted into new biomass available to the next trophic level.
Q：解释为什么营养级之间的能量传递效率通常低于10%。(4分)
A：某一营养级的生物并非全部被下一营养级取食；骨骼和根等部分未被食用。在被吃掉的物质中，很大一部分以未消化的残渣形式随粪便排出。被同化的能量大部分通过呼吸作用以热的形式释放，用于运动和新陈代谢。只有一小部分转化为可供下一营养级利用的新生物量。
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