📚 Answers to Self-Assessment Questions | 自我评估题答案解析
In Cambridge International AS & A Level Biology, self-assessment questions are a powerful way to test your understanding and application of core concepts. This article provides model answers to a selection of typical self-assessment questions across the syllabus, with clear explanations in both English and Chinese.
在剑桥国际 AS 与 A Level 生物课程中，自我评估题是检验你对核心概念理解与应用的强大工具。本文为一系列典型自评题提供标准答案解析，并用中英双语给出清晰说明。
1. Cell Structure: Calculating Magnification | 细胞结构：计算放大倍数
A student measures the length of a chloroplast in an electron micrograph as 42 mm. The actual length of the chloroplast is 7 µm. Calculate the magnification.
一名学生测得电镜照片中叶绿体的长度为 42 mm。该叶绿体实际长度为 7 µm。计算放大倍数。
Model answer: First convert both measurements to the same unit. Since 1 mm = 1000 µm, 42 mm = 42000 µm. Magnification = image size ÷ actual size = 42000 µm ÷ 7 µm = 6000×. Therefore the electron micrograph has a magnification of 6000 times. Remember to state magnification as a number followed by ×, because magnification is dimensionless and has no unit.
标准答案：先将两者换算为同一单位。因为 1 mm = 1000 µm，所以 42 mm = 42000 µm。放大倍数 = 图像尺寸 ÷ 实际尺寸 = 42000 µm ÷ 7 µm = 6000×。因此该电镜照片的放大倍数为 6000 倍。请记住放大倍数以数字加 × 表示，不能加单位，因为放大倍数是无量纲的。
Magnification = Image size ÷ Actual size = 6000× | 放大倍数 = 图像尺寸 ÷ 实际尺寸 = 6000×
2. Biological Molecules: Glycosidic Bonds in Starch and Cellulose | 生物分子：淀粉与纤维素中的糖苷键
Explain why humans can digest starch but not cellulose, despite both being polymers of glucose.
解释为什么人类能消化淀粉却不能消化纤维素，尽管两者都是葡萄糖的聚合物。
Model answer: Starch is made of α-glucose units joined by α-1,4 glycosidic bonds, with occasional α-1,6 bonds at branch points. Human amylase enzymes can recognise and hydrolyse these α-glycosidic bonds. Cellulose is made of β-glucose units joined by β-1,4 glycosidic bonds. These form straight, unbranched chains with hydrogen bonds between adjacent chains. Human digestive enzymes do not have active sites complementary to β-glycosidic bonds, so cellulose passes through the gut undigested.
标准答案：淀粉由 α-葡萄糖通过 α-1,4 糖苷键连接而成，在分支点偶尔出现 α-1,6 糖苷键。人类淀粉酶能够识别并水解这些 α-糖苷键。纤维素由 β-葡萄糖通过 β-1,4 糖苷键连接而成，形成直链，且相邻链之间存在氢键。人类消化酶的活性位点与 β-糖苷键不互补，因此纤维素无法被消化而直接通过肠道。
3. Enzymes: Effect of a Competitive Inhibitor | 酶：竞争性抑制剂的影响
Describe and explain the effect of a competitive inhibitor on the rate of an enzyme-catalysed reaction as substrate concentration increases.
描述并解释随着底物浓度增加，竞争性抑制剂对酶促反应速率的影响。
Model answer: A competitive inhibitor has a shape similar to the substrate and binds reversibly to the active site of the enzyme. At low substrate concentration, the inhibitor competes effectively with substrate, so the rate of reaction is reduced. As substrate concentration increases, the substrate molecules outnumber the inhibitor molecules and are more likely to bind to active sites. Therefore the maximum rate (Vmax) can still be reached, but a higher substrate concentration is needed. This means the Michaelis constant (Km) increases while Vmax remains unchanged.
标准答案：竞争性抑制剂的形状与底物相似，可逆地与酶的活性位点结合。在底物浓度较低时，抑制剂与底物有效竞争，因此反应速率降低。随着底物浓度增加，底物分子数量超过抑制剂分子，底物更有可能与活性位点结合。因此仍可达到最大反应速率（Vmax），但需要更高的底物浓度。这意味着米氏常数（Km）增大，而 Vmax 保持不变。
4. Cell Membranes: Water Potential and Animal Cells | 细胞膜：水势与动物细胞
A red blood cell is placed in pure water. The water potential of pure water is 0 kPa and the water potential of the cytoplasm is -300 kPa. Describe and explain what happens to the cell.
将红细胞放入纯水中。纯水的水势为 0 kPa，细胞质的水势为 -300 kPa。描述并解释细胞会发生什么变化。
Model answer: Water moves by osmosis from a region of higher water potential (0 kPa) to a region of lower water potential (-300 kPa) across the partially permeable cell membrane. Water therefore enters the red blood cell. The cell swells and may burst, a process called lysis. Animal cells do not have a cell wall, so there is no rigid structure to prevent excessive water uptake.
标准答案：水分通过部分透性的细胞膜，以渗透作用从水势较高的区域（0 kPa）向水势较低的区域（-300 kPa）移动。因此水进入红细胞。细胞吸水膨胀，可能破裂，这一过程称为溶血。动物细胞没有细胞壁，因此没有刚性结构来阻止水分的过度进入。
5. Mitosis: Calculating Mitotic Index | 有丝分裂：计算有丝分裂指数
A student observes 250 cells in an onion root tip squash. Of these, 50 cells are in visible stages of mitosis. Calculate the mitotic index and suggest what a high mitotic index indicates.
一名学生在洋葱根尖压片标本中观察了 250 个细胞。其中 50 个细胞处于可见的有丝分裂阶段。计算有丝分裂指数，并说明高有丝分裂指数代表什么。
Model answer: Mitotic index = number of cells in mitosis ÷ total number of cells = 50 ÷ 250 = 0.2. This can be stated as 0.2 or 20%. A high mitotic index indicates that a large proportion of cells are dividing, which is common in regions of rapid growth, such as root tips and shoot tips. Mitotic index is also used in cancer diagnosis because tumours show abnormally high rates of cell division.
标准答案：有丝分裂指数 = 处于有丝分裂的细胞数 ÷ 总细胞数 = 50 ÷ 250 = 0.2。可以表示为 0.2 或 20%。高有丝分裂指数表示很大比例的细胞正在分裂，这常见于快速生长的区域，如根尖和茎尖。有丝分裂指数也用于癌症诊断，因为肿瘤表现出异常高的细胞分裂速率。
Mitotic index = Cells in mitosis ÷ Total cells = 0.2 = 20% | 有丝分裂指数 = 有丝分裂细胞数 ÷ 总细胞数 = 0.2 = 20%
6. Nucleic Acids: Semi-Conservative Replication | 核酸：半保留复制
Explain the results of the Meselson and Stahl experiment after one generation of growth in ¹⁴N medium, assuming DNA replication is semi-conservative.
假设 DNA 复制是半保留的，解释梅塞尔森和斯塔尔实验在 ¹⁴N 培养基中培养一代后的结果。
Model answer: Bacteria were first grown in a medium containing heavy nitrogen, ¹⁵N, so all DNA was labelled heavy. They were then transferred to a medium containing light nitrogen, ¹⁴N, and allowed to divide once. In semi-conservative replication, each new DNA molecule consists of one original heavy strand and one newly synthesised light strand. After centrifugation, all DNA molecules settle at a single intermediate density band, between the heavy and light bands. This result supported semi-conservative replication and ruled out the conservative model, which would have produced two separate bands.
标准答案：细菌首先在含有重氮 ¹⁵N 的培养基中生长，因此所有 DNA 都被重氮标记。然后将细菌转移到含有轻氮 ¹⁴N 的培养基中，让其分裂一次。在半保留复制中，每个新 DNA 分子由一条原来的重链和一条新合成的轻链组成。离心后，所有 DNA 分子都位于一条密度介于重带和轻带之间的中间带。这一结果支持半保留复制，并排除了保守复制模型，因为保守复制会产生两条分开的带。
7. Transport in Plants: Cohesion-Tension Theory | 植物运输：内聚力-张力理论
Explain how water moves up a tall tree against gravity, according to the cohesion-tension theory.
根据内聚力-张力理论，解释水分如何克服重力在高达树木中向上运输。
Model answer: Transpiration from leaves lowers the water potential in the leaf air spaces. Water evaporates from the surfaces of mesophyll cells and diffuses out through stomata. This creates tension, or negative pressure, in the leaf cell walls. Because water molecules are polar, they form hydrogen bonds with each other, giving them strong cohesion. The tension is transmitted down continuous columns of water in xylem vessels, pulling the entire column upward. Adhesion of water molecules to the lignified xylem walls also helps support the column against gravity.
标准答案：叶片蒸腾作用降低了叶内空气间隙的水势。水分从叶肉细胞表面蒸发并通过气孔扩散出去。这在叶细胞壁中产生张力，即负压。由于水分子具有极性，它们之间形成氢键，产生很强的内聚力。张力沿木质部导管中连续的水柱向下传递，将整个水柱向上拉动。水分子与木质化导管壁之间的附着力也有助于支撑水柱对抗重力。
8. Gas Exchange: Oxygen Dissociation Curve of Fetal Haemoglobin | 气体交换：胎儿血红蛋白的氧解离曲线
The oxygen dissociation curve of fetal haemoglobin lies to the left of the adult haemoglobin curve. Explain the advantage of this difference.
胎儿血红蛋白的氧解离曲线位于成体血红蛋白曲线的左侧。解释这种差异的优势。
Model answer: A curve shifted to the left means fetal haemoglobin has a higher affinity for oxygen at any given partial pressure of oxygen. In the placenta, the partial pressure of oxygen is relatively low. Fetal haemoglobin can therefore load oxygen from maternal blood at this low pO₂, because adult haemoglobin releases oxygen more readily under these conditions. This allows efficient transfer of oxygen from the mother to the fetus across the placenta.
标准答案：曲线左移意味着在任何给定的氧分压下，胎儿血红蛋白对氧的亲和力更高。胎盘中的氧分压相对较低。因此胎儿血红蛋白能够在低 pO₂ 条件下从母体血液中获取氧，因为成体血红蛋白在这种条件下更容易释放氧。这使得氧能够有效地经胎盘从母体转移到胎儿。
9. Immunity: Comparing Primary and Secondary Immune Responses | 免疫：初次与二次免疫应答比较
Explain why a secondary immune response is faster and produces more antibodies than the primary response.
解释为什么二次免疫应答比初次应答更快、产生更多抗体。
Model answer: During the primary response to an antigen, specific B and T lymphocytes must be selected, activated and undergo clonal expansion before they differentiate into plasma cells and memory cells. This process takes several days, so antibody levels rise slowly. During the primary response, long-lived memory cells are produced. On secondary exposure to the same antigen, memory cells recognise it quickly, divide rapidly and differentiate into plasma cells within a much shorter time. This produces a larger and faster antibody response, often before symptoms appear.
标准答案：在对抗原的初次应答期间，特异性 B 细胞和 T 细胞必须经过选择、激活和克隆扩增，然后才能分化为浆细胞和记忆细胞。这一过程需要数天，因此抗体水平上升缓慢。初次应答期间产生了长寿命的记忆细胞。当再次接触相同抗原时，记忆细胞能快速识别抗原、迅速分裂，并在很短的时间内分化为浆细胞。因此产生更快、更大量的抗体应答，通常在症状出现之前就已开始。
10. Ecology: Using the Lincoln Index | 生态学：使用林肯指数
A biologist captures 30 woodlice, marks them and releases them back into their habitat. A second sample of 35 woodlice contains 7 marked individuals. Estimate the population size and state two assumptions of this method.
一名生物学家捕获 30 只鼠妇，做标记后放回栖息地。第二次抽样 35 只鼠妇中有 7 只带有标记。估计种群大小，并说明该方法的两个假设。
Model answer: Use the Lincoln index, where n₁ is the number captured and marked first, n₂ is the number captured second, and n₃ is the number marked in the second sample. N = (n₁ × n₂) ÷ n₃ = (30 × 35) ÷ 7 = 1500 ÷ 7 ≈ 214 woodlice. Assumptions include: no migration into or out of the population, no births or deaths between samples, marked individuals mix randomly with the rest, and marking does not affect survival or recapture rate.
标准答案：使用林肯指数，其中 n₁ 为首次捕获并标记的数量，n₂ 为第二次捕获的数量，n₃ 为第二次捕获中带有标记的数量。N = (n₁ × n₂) ÷ n₃ = (30 × 35) ÷ 7 = 1500 ÷ 7 ≈ 214 只鼠妇。假设包括：种群没有迁入或迁出、两次抽样之间没有出生或死亡、标记个体与其余个体随机混合、标记不影响存活率或再捕获率。
N = (n₁ × n₂) ÷ n₃ = (30 × 35) ÷ 7 ≈ 214 | N = (n₁ × n₂) ÷ n₃ = (30 × 35) ÷ 7 ≈ 214
11. Respiration: Calculating Respiratory Quotient | 呼吸作用：计算呼吸商
During aerobic respiration of an unknown substrate, a respirometer shows that 0.60 cm³ of oxygen is consumed and 0.60 cm³ of carbon dioxide is produced. Calculate the respiratory quotient (RQ) and identify the likely type of substrate.
在未知底物的有氧呼吸中，呼吸计显示消耗了 0.60 cm³ 氧气，产生了 0.60 cm³ 二氧化碳。计算呼吸商（RQ），并判断可能的底物类型。
Model answer: RQ = volume of CO₂ produced ÷ volume of O₂ consumed = 0.60 ÷ 0.60 = 1.0. An RQ of 1.0 is typical for carbohydrates, because carbohydrate respiration produces equal volumes of carbon dioxide and oxygen. Fats have RQ values around 0.7 and proteins around 0.9. Therefore the substrate is likely to be a carbohydrate, such as glucose.
标准答案：RQ = 产生 CO₂ 的体积 ÷ 消耗 O₂ 的体积 = 0.60 ÷ 0.60 = 1.0。RQ 为 1.0 通常代表碳水化合物，因为碳水化合物呼吸产生的二氧化碳与消耗的氧气体积相等。脂肪的 RQ 约为 0.7，蛋白质约为 0.9。因此底物很可能是碳水化合物，如葡萄糖。
RQ = CO₂ produced ÷ O₂ consumed = 0.60 ÷ 0.60 = 1.0 | RQ = 产生的 CO₂ ÷ 消耗的 O₂ = 0.60 ÷ 0.60 = 1.0
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