📚 Answers to End-of-Chapter Questions | 章末问题答案解析
This article provides model answers to common end-of-chapter questions from the Cambridge International AS & A Level Biology course. Each section presents a typical question, followed by a concise marking-scheme style answer that focuses on key terms and concepts. Use these answers to check your understanding and to learn how to structure your own responses.
本文提供剑桥国际 AS 与 A Level 生物课程中常见章末问题的范例答案。每个小节呈现一个典型问题，并给出简洁的评分方案式答案，重点突出关键术语与概念。使用这些答案来检查你的理解，并学习如何组织自己的回答。
1. Cell Structure | 细胞结构
Question: Outline the key differences between prokaryotic and eukaryotic cells.
问题：概述原核细胞与真核细胞的主要区别。
Answer: Prokaryotic cells lack a true nucleus; their DNA is circular and lies free in the cytoplasm. They have smaller 70S ribosomes, no membrane-bound organelles (such as mitochondria or chloroplasts), and a cell wall made of peptidoglycan. Eukaryotic cells have a membrane-bound nucleus containing linear DNA, larger 80S ribosomes, and membrane-bound organelles. Plant and fungal cell walls, if present, are made of cellulose or chitin respectively, not peptidoglycan.
答案：原核细胞缺乏真正的细胞核；其 DNA 为环状，游离在细胞质中。它们具有较小的 70S 核糖体，没有膜包被的细胞器（如线粒体或叶绿体），细胞壁由肽聚糖构成。真核细胞具有由膜包围的细胞核，内含线性 DNA，较大的 80S 核糖体，以及膜包被的细胞器。植物和真菌细胞壁（如果存在）分别由纤维素或几丁质构成，而非肽聚糖。
2. Biological Molecules | 生物分子
Question: Describe the structure of a triglyceride and explain how its structure relates to its function as an energy store.
问题：描述甘油三酯的结构，并解释其结构如何与其作为能量储存物质的功能相关。
Answer: A triglyceride consists of one glycerol molecule bonded to three fatty acid molecules by ester bonds, formed through condensation reactions. The fatty acid chains are long hydrocarbon tails that are non-polar and hydrophobic. These chains contain many carbon-hydrogen bonds, which can be oxidised in respiration to release a large amount of energy. Because triglycerides are insoluble in water, they do not affect the water potential of cells and can be stored in adipose tissue without causing osmotic problems. Their compact, globular shape allows efficient packing and storage.
答案：甘油三酯由一个甘油分子通过缩合反应形成的酯键与三个脂肪酸分子结合而成。脂肪酸链是长碳氢链，非极性且疏水。这些链含有许多碳氢键，在呼吸作用中可被氧化，释放大量能量。由于甘油三酯不溶于水，不会影响细胞的水势，因此可以储存在脂肪组织中而不会引起渗透问题。其紧凑的球形结构便于高效堆积和储存。
3. Enzymes | 酶
Question: Explain the induced-fit model of enzyme action and how it differs from the lock-and-key model.
问题：解释酶作用的诱导契合模型，并说明它与锁钥模型的区别。
Answer: In the induced-fit model, the active site is not a rigid shape. When the substrate binds, the active site undergoes a conformational change, moulding itself around the substrate. This interaction forms an enzyme-substrate complex and places strain on the substrate bonds, lowering the activation energy. The lock-and-key model proposed that the active site is exactly complementary to the substrate and does not change shape. The induced-fit model is now preferred because it explains how enzymes can catalyse reactions with different but related substrates and how the transition state is stabilised.
答案：在诱导契合模型中，活性位点不是刚性的形状。当底物结合时，活性位点发生构象变化，围绕底物进行模塑。这种相互作用形成酶-底物复合物，并对底物键施加应力，降低活化能。锁钥模型提出活性位点与底物完全互补且形状不变。诱导契合模型目前更被接受，因为它解释了酶如何催化不同但相关的底物反应，以及过渡态如何被稳定。
4. Cell Membranes and Transport | 细胞膜与运输
Question: Explain how the fluid mosaic model accounts for the selective permeability of the cell membrane.
问题：解释流动镶嵌模型如何解释细胞膜的选择透过性。
Answer: The cell membrane is described as a fluid mosaic because it consists of a phospholipid bilayer in which proteins and other molecules are embedded and can move laterally. The hydrophobic fatty acid tails face inward, forming a barrier to ions and large polar molecules. Small non-polar molecules such as oxygen and carbon dioxide can diffuse through directly. Integral proteins include channel proteins and carrier proteins that allow specific ions and polar molecules to cross. Cholesterol maintains membrane fluidity. Glycoproteins and glycolipids function in cell recognition. The selective permeability therefore arises from the hydrophobic core and the specificity of transport proteins.
答案：细胞膜被描述为流动镶嵌模型，因为它由磷脂双分子层组成，蛋白质和其他分子嵌入其中并可侧向移动。疏水的脂肪酸尾部朝内，形成对离子和大极性分子的屏障。小的非极性分子如氧气和二氧化碳可以直接扩散通过。整合蛋白包括通道蛋白和载体蛋白，允许特定的离子和极性分子通过。胆固醇维持膜的流动性。糖蛋白和糖脂参与细胞识别。因此，选择透过性源于疏水核心和转运蛋白的特异性。
5. Cell Division | 细胞分裂
Question: Describe the stages of mitosis and explain its biological significance.
问题：描述有丝分裂的各阶段，并解释其生物学意义。
Answer: Mitosis consists of four stages. In prophase, chromatin condenses into visible chromosomes, and the nuclear envelope breaks down. In metaphase, chromosomes align at the equator of the cell, attached to spindle fibres. In anaphase, sister chromatids separate and are pulled to opposite poles. In telophase, nuclear envelopes reform around each set of chromosomes, which decondense. Mitosis produces two genetically identical daughter cells with the same chromosome number as the parent cell. It is essential for growth, replacement of damaged cells, and asexual reproduction in some organisms.
答案：有丝分裂包括四个阶段。在前期，染色质凝缩成可见的染色体，核膜解体。在中期，染色体排列在细胞赤道板上，与纺锤丝相连。在后期，姐妹染色单体分离并被拉向相反的两极。在末期，核膜围绕每组染色体重新形成，染色体解凝缩。有丝分裂产生两个与亲代细胞染色体数目相同的遗传上相同的子细胞。它对生长、损伤细胞的替换以及某些生物的无性生殖至关重要。
6. DNA and Protein Synthesis | DNA 与蛋白质合成
Question: Compare the structure and functions of DNA and RNA.
问题：比较 DNA 和 RNA 的结构与功能。
Answer: DNA is a double-stranded helix composed of deoxyribonucleotides. Its sugar is deoxyribose and its bases are adenine, thymine, cytosine, and guanine. It is very stable and stores genetic information. RNA is usually single-stranded, contains ribose sugar, and has uracil instead of thymine. There are three main types: messenger RNA (mRNA) carries the genetic code from DNA to ribosomes; transfer RNA (tRNA) brings specific amino acids during translation; ribosomal RNA (rRNA) is a structural component of ribosomes. DNA remains in the nucleus (in eukaryotes), while RNA is found in both nucleus and cytoplasm.
答案：DNA 是由脱氧核糖核苷酸组成的双链螺旋。其糖为脱氧核糖，碱基为腺嘌呤、胸腺嘧啶、胞嘧啶和鸟嘌呤。它非常稳定，储存遗传信息。RNA 通常为单链，含有核糖，并且以尿嘧啶代替胸腺嘧啶。主要有三种类型：信使 RNA（mRNA）将遗传密码从 DNA 携带到核糖体；转运 RNA（tRNA）在翻译过程中携带特定氨基酸；核糖体 RNA（rRNA）是核糖体的结构成分。DNA 停留在细胞核中（真核生物），而 RNA 存在于细胞核和细胞质中。
7. Genetics and Inheritance | 遗传与遗传
Question: In a monohybrid cross between two heterozygous tall pea plants (Tt), calculate the expected phenotypic ratio of offspring and state the assumptions.
问题：在两株杂合高茎豌豆（Tt）之间的单因子杂交中，计算后代的预期表现型比例，并说明假设。
Answer: T is the dominant allele for tall; t is the recessive allele for dwarf. A Punnett square shows the possible genotypes: TT, Tt, Tt, tt. Therefore the expected phenotypic ratio is 3 tall : 1 dwarf. This assumes that the alleles segregate independently, fertilisation is random, the trait is controlled by a single gene with complete dominance, and no mutation or selection occurs.
答案：T 是高茎的显性等位基因；t 是矮茎的隐性等位基因。庞纳特方格显示可能的基因型为：TT、Tt、Tt、tt。因此预期表现型比例为 3 高茎 : 1 矮茎。这假设等位基因独立分离，受精是随机的，该性状由具有完全显性的单个基因控制，并且不发生突变或选择。
8. Natural Selection and Evolution | 自然选择与进化
Question: Explain how natural selection leads to the development of antibiotic resistance in bacterial populations.
问题：解释自然选择如何导致细菌种群产生抗生素耐药性。
Answer: Within a bacterial population, random mutation creates genetic variation. Some individuals possess alleles that confer resistance to a particular antibiotic. When the antibiotic is applied, susceptible bacteria are killed, but resistant bacteria survive. These resistant individuals reproduce, passing the resistance allele to their offspring. Over many generations, the frequency of the resistance allele increases, so the population becomes resistant. This is an example of natural selection acting on pre-existing variation.
答案：在细菌种群内，随机突变产生遗传变异。一些个体拥有赋予对特定抗生素耐药性的等位基因。当使用抗生素时，敏感细菌被杀死，而耐药细菌存活。这些耐药个体繁殖，将耐药等位基因传递给后代。经过许多代，耐药等位基因的频率增加，因此种群变得耐药。这是自然选择作用于预先存在的变异的例子。
9. Transport in Plants | 植物运输
Question: Explain the cohesion-tension theory for the movement of water through xylem vessels.
问题：解释水分通过木质部导管运动的蒸腾-内聚力-张力理论。
Answer: Water evaporates from the surfaces of mesophyll cells in leaves during transpiration. This creates tension (negative pressure) in the leaf air spaces and cell walls. Water molecules are cohesive due to hydrogen bonding, so the evaporation pulls a continuous column of water up the xylem. Adhesion of water molecules to xylem walls helps support the column. As water is pulled up the xylem, more water is drawn into the roots from the soil, maintaining the flow.
答案：蒸腾作用期间，水分从叶片叶肉细胞表面蒸发。这在叶片气腔和细胞壁中产生张力（负压）。水分子由于氢键而具有内聚力，因此蒸发会拉动木质部中连续的水柱向上移动。水分子与木质部壁的附着力有助于支撑水柱。当水分被拉上木质部时，更多的水从土壤中被吸入根部，维持水流。
10. Ecology and Energy Flow | 生态与能量流动
Question: Explain why the transfer of energy between trophic levels is inefficient.
问题：解释为什么营养级之间的能量传递效率低。
Answer: Only a fraction of the energy in one trophic level is transferred to the next. Some energy is lost because not all of the organism is eaten (e.g., bones, roots). Of the energy that is ingested, a proportion is lost in faeces because it cannot be digested. Of the assimilated energy, much is used for respiration, releasing energy as heat. Only a small part is converted into new biomass. Typically, only about 10% of energy is transferred from one trophic level to the next, which limits the length of food chains.
答案：从一个营养级传递到下一个营养级的能量只占一小部分。一些能量因生物体未被完全吃掉而损失（例如骨骼、根）。在摄入的能量中，一部分因无法消化而随粪便损失。在同化的能量中，大部分用于呼吸作用，以热能形式释放。只有一小部分转化为新的生物量。通常，从一个营养级传递到下一个营养级的能量仅约 10%，这限制了食物链的长度。
Published by TutorHao | Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导