Year 12 Edexcel Biology Unit Test Mock Paper Analysis | Year 12 Edexcel 生物：单元测试模拟卷解析

📚 Year 12 Edexcel Biology Unit Test Mock Paper Analysis | Year 12 Edexcel 生物：单元测试模拟卷解析

Welcome to this comprehensive walkthrough of a Unit Test mock paper designed for Year 12 Edexcel Biology. This article provides detailed solutions and examiner-style commentary for a 60-mark assessment covering Topics 1 to 4: Biological Molecules, Cells, Exchange and Transport, and Genetics. Each question is deconstructed to highlight the key knowledge points, common pitfalls, and the precise command words that Edexcel examiners expect you to address. Use this analysis to identify gaps in your understanding, refine your exam technique, and build confidence for your end-of-year assessments.

欢迎来到这份为 Year 12 Edexcel 生物设计的单元测试模拟卷全面解析。本文提供了一份涵盖主题1至4（生物分子、细胞、交换与运输、遗传学）共60分的评估试卷的详细解答和考官式点评。每道题目都被逐一拆解，突出关键知识点、常见失分点以及 Edexcel 考官期望你把握的精确指令词。利用这份解析来识别你理解中的薄弱环节，完善你的考试技巧，并为年终评估建立信心。

1. Multiple Choice: Identifying Biological Molecules | 选择题：识别生物分子

Question: A student analysed a sample of an unknown biological molecule. The sample was found to contain carbon, hydrogen, and oxygen atoms in a ratio of approximately 1:2:1. It dissolved readily in water but did not react with Biuret reagent. Which molecule is most likely present in the sample? A) Protein B) Starch C) Sucrose D) Triglyceride

题目：一名学生分析了一份未知生物分子样本。该样本被发现含有碳、氢和氧原子，比例约为1:2:1。它易溶于水，但不与双缩脲试剂反应。样本中最可能存在的分子是哪一种？A) 蛋白质 B) 淀粉 C) 蔗糖 D) 甘油三酯

The correct answer is C — sucrose. The atomic ratio of approximately 1:2:1 corresponds to the general formula (CH₂O)ₙ, which is characteristic of carbohydrates. The molecule dissolves readily in water, confirming it is a small, polar carbohydrate rather than a large polysaccharide such as starch (which is insoluble). The negative Biuret test rules out protein entirely, as Biuret reagent detects peptide bonds. Triglycerides contain a much higher proportion of hydrogen relative to oxygen and are non-polar and insoluble, so D is incorrect. Starch has the correct atom ratio but is a large polysaccharide that does not dissolve readily in water, making B less suitable than C.

正确答案是C——蔗糖。约1:2:1的原子比对应于通式(CH₂O)ₙ，这是碳水化合物的特征。该分子易溶于水，证实它是一种小型极性碳水化合物，而非像淀粉这样的大型多糖（淀粉不溶于水）。双缩脲试验呈阴性结果完全排除了蛋白质的可能性，因为双缩脲试剂检测的是肽键。甘油三酯含有相对于氧而言比例高得多的氢，且为非极性、不溶性分子，因此D不正确。淀粉具有正确的原子比例，但它是大型多糖，不易溶于水，因此B不如C合适。

Examiner tip: Always link molecular structure to observable properties. The general formula (CH₂O)ₙ indicates a carbohydrate, and solubility distinguishes monosaccharides and disaccharides from polysaccharides. Edexcel frequently tests the Biuret, Benedict’s, and emulsion tests — know which bonds or structures each reagent detects.

考官提示：始终将分子结构与可观测性质联系起来。通式(CH₂O)ₙ表明它是碳水化合物，而溶解性则可以区分单糖/二糖与多糖。Edexcel 经常考查双缩脲试验、本尼迪克特试验和乳化试验——务必掌握每种试剂检测的是哪种键或结构。

---

2. Multiple Choice: Cell Organelle Functions | 选择题：细胞器功能

Question: Which row correctly matches the organelle to its primary function? A) Ribosome — synthesis of ATP B) Golgi apparatus — modification and packaging of proteins C) Smooth endoplasmic reticulum — synthesis of ribosomal RNA D) Lysosome — synthesis of lipids

题目：哪一行正确匹配了细胞器与其主要功能？A) 核糖体——合成ATP B) 高尔基体——修饰和包装蛋白质 C) 光面内质网——合成核糖体RNA D) 溶酶体——合成脂质

The correct answer is B — the Golgi apparatus receives proteins from the rough endoplasmic reticulum, modifies them (for example, by adding carbohydrate groups to form glycoproteins), and packages them into vesicles for transport within the cell or for secretion via exocytosis. Ribosomes are the site of protein synthesis, not ATP synthesis; ATP is primarily produced in mitochondria during oxidative phosphorylation. The smooth endoplasmic reticulum is involved in lipid synthesis and detoxification, not in ribosomal RNA synthesis (which occurs in the nucleolus). Lysosomes contain hydrolytic enzymes for intracellular digestion; they do not synthesise lipids.

正确答案是B——高尔基体从粗面内质网接收蛋白质，对其进行修饰（例如通过添加碳水化合物基团形成糖蛋白），并将其包装入囊泡以便在细胞内运输或通过胞吐作用分泌。核糖体是蛋白质合成的场所，而非ATP合成的场所；ATP主要在线粒体中通过氧化磷酸化产生。光面内质网参与脂质合成和解毒作用，不参与核糖体RNA的合成（后者发生在核仁中）。溶酶体含有用于细胞内消化的水解酶，它们不合成脂质。

Examiner tip: Edexcel requires precise language when describing organelle functions. Avoid vague terms like “processes” or “deals with.” Instead, use specific verbs: synthesises, modifies, packages, hydrolyses, and detoxifies. Create a revision table with two columns — organelle and specific function — and test yourself regularly.

考官提示：Edexcel 要求描述细胞器功能时使用精确语言。避免使用”处理”或”应对”等模糊术语。应使用具体动词：合成、修饰、包装、水解、解毒。制作一个两列的复习表格——细胞器和具体功能——并定期自测。

---

3. Multiple Choice: Enzyme Inhibition Kinetics | 选择题：酶抑制动力学

Question: The graph shows the effect of increasing substrate concentration on the rate of an enzyme-catalysed reaction in the presence and absence of an inhibitor. Curve X (without inhibitor) reaches a higher Vmax than Curve Y (with inhibitor), but both curves intersect the x-axis at the same point. What type of inhibition is demonstrated? A) Competitive B) Non-competitive C) Uncompetitive D) Mixed

题目：图中显示了在存在和不存在抑制剂的情况下，增加底物浓度对酶催化反应速率的影响。曲线X（无抑制剂）达到的Vmax高于曲线Y（有抑制剂），但两条曲线在x轴上的交点相同。这展示的是哪种抑制类型？A) 竞争性 B) 非竞争性 C) 反竞争性 D) 混合型

The correct answer is A — competitive inhibition. In competitive inhibition, the inhibitor resembles the substrate and binds reversibly to the active site of the enzyme. At sufficiently high substrate concentrations, the substrate can outcompete the inhibitor, meaning the same maximum rate (Vmax) can eventually be achieved — however, a higher substrate concentration is needed to reach half of Vmax, so the apparent Km increases. The question states that the Vmax values differ, which might initially suggest non-competitive inhibition, but careful reading reveals this is a tricky detail: if the inhibitor is present at a high enough concentration and substrate concentrations tested are not saturating, the apparent Vmax in the experimental range may appear lower. The key diagnostic feature here is the intersection pattern — competitive inhibitors show the characteristic rightward shift of the curve with the same origin on the x-axis when plotting rate against [S].

正确答案是A——竞争性抑制。在竞争性抑制中，抑制剂与底物结构相似，可逆地结合在酶的活性位点上。当底物浓度足够高时，底物可以胜过抑制剂，这意味着最终可以达到相同的最大速率（Vmax）——然而，需要更高的底物浓度才能达到Vmax的一半，因此表观Km增大。题目中提到Vmax值不同，这最初可能暗示非竞争性抑制，但仔细阅读会发现这是一个微妙的细节：如果抑制剂浓度足够高，且测试的底物浓度未达到饱和水平，那么在实验范围内的表观Vmax可能看起来较低。此处关键的诊断特征是交点模式——在绘制速率对[S]的曲线时，竞争性抑制剂显示曲线向右移动但在x轴上原点相同。

Examiner tip: Many candidates confuse competitive and non-competitive inhibition. Remember the mnemonic: Competitive = Compete for the active site, reversible by high substrate. Non-competitive = No competition, binds elsewhere, Vmax permanently lowered. In exam questions, check whether Vmax changes with and without the inhibitor at saturating substrate concentrations — this is the definitive test.

考官提示：许多考生混淆竞争性抑制和非竞争性抑制。记住这个记忆口诀：竞争性=争夺活性位点，高浓度底物可逆转。非竞争性=不争夺活性位点，结合在别处，Vmax永久性降低。在考试题目中，检查在饱和底物浓度下有抑制剂和无抑制剂的Vmax是否发生变化——这是决定性的检验标准。

---

4. Short Answer: The Fluid Mosaic Model | 简答题：流动镶嵌模型

Question (4 marks): Describe the arrangement of phospholipids and proteins in the cell surface membrane according to the fluid mosaic model.

题目（4分）：根据流动镶嵌模型，描述细胞表面膜中磷脂和蛋白质的排列方式。

A model answer would state: phospholipids form a bilayer, with their hydrophilic phosphate heads facing outwards towards the aqueous environment on both the extracellular and cytoplasmic sides, and their hydrophobic fatty acid tails facing inwards, away from water, forming the core of the membrane. The term “fluid” refers to the fact that individual phospholipids can move laterally within their own monolayer, and the membrane remains flexible. The term “mosaic” describes the pattern of protein molecules that are scattered throughout the phospholipid bilayer. Intrinsic (integral) proteins span the entire bilayer, while extrinsic (peripheral) proteins are confined to one surface of the membrane. Some proteins act as channels or carriers for transport, while others function as receptors or enzymes.

一份标准答案会这样表述：磷脂形成双分子层，其亲水性磷酸头部分别朝外，面向细胞外和细胞质两侧的水性环境，而疏水性脂肪酸尾部朝内，远离水，形成膜的核心。”流动”一词指的是单个磷脂分子可以在其自身的单层内横向移动，膜保持柔韧性。”镶嵌”一词描述了散布在整个磷脂双分子层中的蛋白质分子的图案。内在蛋白（整合蛋白）横跨整个双分子层，而外在蛋白（外周蛋白）仅限于膜的一个表面。一些蛋白质充当运输的通道或载体，而另一些则充当受体或酶。

Mark scheme allocation: 1 mark for bilayer arrangement; 1 mark for hydrophilic heads out / hydrophobic tails in; 1 mark for protein distribution described as scattered / mosaic; 1 mark for distinction between intrinsic and extrinsic proteins OR for a named function. Edexcel expects you to use the terms “fluid” and “mosaic” explicitly and link them to the structural features they describe. Vague answers like “proteins float in a sea of phospholipids” will not gain full marks without additional structural detail.

评分方案分配：1分给双分子层排列；1分给亲水性头部朝外/疏水性尾部朝内；1分给蛋白质分布描述为散布/镶嵌；1分给区分内在蛋白和外在蛋白，或给一个命名功能。Edexcel 期望你明确使用”流动”和”镶嵌”这两个术语，并将其与它们所描述的结构特征联系起来。像”蛋白质漂浮在磷脂海洋中”这样模糊的回答，如果没有额外的结构细节，将无法获得满分。

---

5. Short Answer: Semi-Conservative DNA Replication | 简答题：DNA半保留复制

Question (5 marks): Outline the key events in semi-conservative DNA replication, naming the enzymes involved at each stage.

题目（5分）：概述半保留DNA复制的关键事件，并说出每个阶段涉及到的酶的名称。

The process begins when the enzyme DNA helicase unwinds the double helix by breaking the hydrogen bonds between complementary base pairs, forming a replication fork. Single-stranded binding proteins stabilise the separated strands, preventing them from re-annealing. The enzyme primase synthesises a short RNA primer, providing a free 3′-OH group for DNA polymerase to extend from. DNA polymerase III then catalyses the addition of free deoxyribonucleoside triphosphates to the growing strand in the 5′ to 3′ direction, using complementary base pairing rules (adenine pairs with thymine, cytosine pairs with guanine). On the leading strand, synthesis is continuous; on the lagging strand, synthesis occurs discontinuously in short fragments called Okazaki fragments. Finally, DNA ligase seals the gaps between adjacent Okazaki fragments, forming a continuous sugar-phosphate backbone. The result is two identical DNA molecules, each containing one original (parental) strand and one newly synthesised strand — hence the term semi-conservative.

该过程始于DNA解旋酶通过断裂互补碱基对之间的氢键来解开双螺旋，形成复制叉。单链结合蛋白稳定分离后的链，防止它们重新退火结合。引物酶合成一段短的RNA引物，提供游离的3′-OH基团供DNA聚合酶延伸。随后，DNA聚合酶III催化将游离的脱氧核糖核苷三磷酸添加到生长链的3’端，方向为5’至3’，遵循互补碱基配对规则（腺嘌呤与胸腺嘧啶配对，胞嘧啶与鸟嘌呤配对）。在领导链上，合成是连续的；在滞后链上，合成是不连续的，形成称为冈崎片段的短片段。最后，DNA连接酶密封相邻冈崎片段之间的缺口，形成连续的糖-磷酸骨架。结果是两个相同的DNA分子，每个都含有一条原始（亲本）链和一条新合成的链——因此称为半保留。

Common errors: Students often forget to mention the directionality (5′ to 3′) of synthesis or confuse the roles of primase and DNA polymerase. Another frequent mistake is stating that DNA polymerase unwinds the helix — this is the role of helicase. In Edexcel mark schemes, naming the enzyme along with its precise catalytic action is essential for full marks.

常见错误：学生经常忘记提及合成方向（5’至3’），或混淆引物酶和DNA聚合酶的作用。另一个常见错误是说DNA聚合酶解开螺旋——这是解旋酶的职责。在Edexcel的评分方案中，要获得满分，必须说出酶的名称及其精确的催化作用。

---

6. Data Analysis: Enzyme Activity Under Varying pH | 数据分析：不同pH下的酶活性

Question (6 marks): A student investigated the effect of pH on the activity of amylase. The table shows the time taken for starch to be completely digested at each pH. (a) Calculate the rate of reaction at pH 7.0, expressing your answer in arbitrary units of min⁻¹. (b) Explain why the rate decreases sharply at pH 2.0 and pH 11.0. (c) Suggest one limitation of this method and how it could be improved.

题目（6分）：一名学生研究了pH对淀粉酶活性的影响。表格显示了在每个pH下淀粉被完全消化所需的时间。(a) 计算pH 7.0时的反应速率，答案以任意单位min⁻¹表示。(b) 解释为什么在pH 2.0和pH 11.0时速率急剧下降。(c) 提出该方法的一个局限性及改进方法。

For part (a), rate is the reciprocal of time: if digestion took 2.0 minutes, the rate = 1 ÷ 2.0 = 0.50 min⁻¹. For part (b), enzymes have an optimum pH at which the three-dimensional shape of the active site is complementary to the substrate. Amylase functions in the mouth and small intestine, with an optimum pH near neutral (around pH 7.0). At extreme pH values (pH 2.0 is highly acidic; pH 11.0 is highly alkaline), the excess hydrogen ions (H⁺) or hydroxide ions (OH⁻) disrupt the ionic bonds and hydrogen bonds that maintain the tertiary structure of the enzyme. This causes the active site to change shape irreversibly — the enzyme is denatured — so the substrate can no longer bind, and the rate drops to near zero. For part (c), a common limitation is that using the disappearance of starch (tested with iodine) as the endpoint is subjective — different students may judge the endpoint differently. Improvement: use a colorimeter to measure absorbance at regular time intervals for a more objective and quantitative measurement of reaction progress.

在(a)部分，速率是时间的倒数：如果消化耗时2.0分钟，速率 = 1 ÷ 2.0 = 0.50 min⁻¹。在(b)部分，酶具有最适pH，在此pH下活性位点的三维形状与底物互补。淀粉酶在口腔和小肠中发挥作用，最适pH接近中性（约pH 7.0）。在极端pH值下（pH 2.0为高酸性；pH 11.0为高碱性），过量的氢离子（H⁺）或氢氧根离子（OH⁻）破坏维持酶三级结构的离子键和氢键。这导致活性位点不可逆地改变形状——酶已变性——因此底物无法再结合，速率降至接近零。在(c)部分，一个常见的局限是使用淀粉消失（用碘液检测）作为终点是主观的——不同学生可能对终点的判断不同。改进方法：使用比色计定期测量吸光度，以更客观和定量地测量反应进程。

Edexcel CPAC link: This question directly tests skills from Core Practical 1: Investigating the Effect of pH on Enzyme Activity. You must be able to identify independent variables (pH), dependent variables (rate/time), and control variables (temperature, enzyme concentration, substrate concentration). Edexcel also expects you to evaluate methodology and suggest valid improvements — a skill assessed across all core practicals.

Edexcel CPAC 链接：这道题目直接考查核心实验1的技能：研究pH对酶活性的影响。你必须能够识别自变量（pH）、因变量（速率/时间）和控制变量（温度、酶浓度、底物浓度）。Edexcel 还期望你评估方法并提出有效的改进建议——这是所有核心实验中都评估的技能。

---

7. Extended Response: Stages of Mitosis | 长答题：有丝分裂各阶段

Question (8 marks): Describe the behaviour of chromosomes during each stage of mitosis in an animal cell. Explain the significance of mitosis for growth and repair.

题目（8分）：描述动物细胞有丝分裂各阶段中染色体的行为。解释有丝分裂对生长和修复的意义。

Mitosis consists of four main stages: prophase, metaphase, anaphase, and telophase, followed by cytokinesis. In prophase, chromatin condenses into visible chromosomes, each consisting of two identical sister chromatids joined at the centromere. The nuclear envelope breaks down, and the centrioles migrate to opposite poles of the cell, forming spindle fibres composed of microtubules. In metaphase, the chromosomes align along the equatorial plate (metaphase plate) of the cell. The spindle fibres attach to the centromeres of each chromosome. In anaphase, the centromeres divide, and the sister chromatids are pulled apart to opposite poles of the cell as the spindle fibres shorten. The separated chromatids are now considered individual chromosomes. In telophase, the chromosomes decondense back into chromatin, a new nuclear envelope forms around each set of chromosomes, and the spindle fibres disassemble. Cytokinesis follows, where the cytoplasm divides, producing two genetically identical daughter cells. The significance of mitosis lies in its production of genetically identical nuclei: it enables growth by increasing the number of cells in an organism, allows for replacement of damaged or worn-out cells during repair, and is the basis for asexual reproduction in some organisms.

有丝分裂由四个主要阶段组成：前期、中期、后期和末期，随后是胞质分裂。在前期，染色质凝聚成可见的染色体，每条染色体由两条在着丝粒处连接的相同姐妹染色单体组成。核膜解体，中心粒迁移到细胞的两极，形成由微管组成的纺锤丝。在中期，染色体排列在细胞的赤道板（中期板）上。纺锤丝附着在每条染色体的着丝粒上。在后期，着丝粒分裂，姐妹染色单体随着纺锤丝的缩短被拉向细胞的两极。分离后的染色单体现在被视为独立的染色体。在末期，染色体解旋回到染色质状态，在每组染色体周围形成新的核膜，纺锤丝解体。随后是胞质分裂，细胞质分裂，产生两个遗传上相同的子细胞。有丝分裂的意义在于它产生遗传上相同的细胞核：它通过增加生物体内细胞数量实现生长，允许在修复过程中替换受损或老化的细胞，并且是某些生物无性繁殖的基础。

Common error — naming confusion: Many students confuse chromatids with chromosomes. After anaphase, what was one chromosome (with two chromatids) becomes two separate chromosomes, each with one chromatid. The chromosome number temporarily doubles. Also, avoid using the term “split” — chromatids separate, but chromosomes do not split in mitosis; the centromere divides.

常见错误——命名混淆：许多学生混淆染色单体和染色体。在后期之后，原来的一条染色体（带有两条染色单体）变成两条独立的染色体，每条带有一条染色单体。染色体数目暂时翻倍。此外，避免使用”分裂”一词——染色单体分离，但染色体在有丝分裂中不分裂；是着丝粒分裂。

---

8. Extended Response: Osmosis and Water Potential | 长答题：渗透作用与水势

Question (6 marks): A student placed pieces of potato tissue into sucrose solutions of different concentrations and measured the percentage change in mass after 24 hours. The results showed that the potato gained mass in 0.0 mol dm⁻³ sucrose solution and lost mass in 0.6 mol dm⁻³ sucrose solution. Explain these results in terms of water potential. Determine the approximate water potential of the potato tissue.

题目（6分）：一名学生将土豆组织块放入不同浓度的蔗糖溶液中，24小时后测量质量百分比变化。结果显示，土豆在0.0 mol dm⁻³蔗糖溶液中质量增加，在0.6 mol dm⁻³蔗糖溶液中质量减少。用水势解释这些结果。确定土豆组织的近似水势。

Water moves by osmosis from a region of higher water potential to a region of lower water potential across a partially permeable membrane. In the 0.0 mol dm⁻³ sucrose solution (pure water), the water potential of the solution is 0 kPa — the highest possible value. The potato cells contain dissolved solutes, so their water potential is lower (more negative). Water therefore enters the cells by osmosis, causing the tissue to gain mass. In the 0.6 mol dm⁻³ sucrose solution, the solution has a very low (highly negative) water potential due to the high solute concentration. This is lower than the water potential of the potato cells, so water leaves the cells by osmosis, causing the tissue to lose mass. The approximate water potential of the potato tissue can be determined by finding the sucrose concentration at which there is no net change in mass — this is where the water potential of the solution equals the water potential of the potato cells. If the graph of percentage change in mass against sucrose concentration crosses the x-axis at 0.28 mol dm⁻³, then the water potential of the potato tissue is approximately equal to that of a 0.28 mol dm⁻³ sucrose solution.

水通过渗透作用，经部分透性膜从水势较高的区域向水势较低的区域移动。在0.0 mol dm⁻³蔗糖溶液（纯水）中，溶液的水势为0 kPa——这是可能的最高值。土豆细胞含有溶解的溶质，因此它们的水势较低（更负值）。因此，水通过渗透进入细胞，导致组织质量增加。在0.6 mol dm⁻³蔗糖溶液中，由于溶质浓度高，溶液的水势非常低（高度负值）。该值低于土豆细胞的水势，因此水通过渗透离开细胞，导致组织质量减少。土豆组织的近似水势可以通过找到没有净质量变化时的蔗糖浓度来确定——此时溶液的水势等于土豆细胞的水势。如果质量百分比变化对蔗糖浓度的曲线在0.28 mol dm⁻³处穿过x轴，那么土豆组织的水势大约等于0.28 mol dm⁻³蔗糖溶液的水势。

Examiner note: Edexcel mark schemes frequently penalise students who say water moves “up” or “down” a concentration gradient, rather than using the correct term “water potential gradient.” Water does not move in direct response to solute concentration differences; it moves in response to differences in water potential, which is influenced by solute concentration and pressure. Always frame your answer around water potential.

考官说明：Edexcel 的评分方案经常惩罚那些说水沿浓度梯度”向上”或”向下”移动的学生，要求使用正确的术语”水势梯度”。水并非直接响应溶质浓度差异而移动；它响应水势差异而移动，水势受溶质浓度和压力的影响。始终围绕水势构建你的答案。

---

9. Practical Skills: Colorimetry and Calibration Curves | 实验技能：比色法与校准曲线

Question (7 marks): A biochemist is using a colorimeter to determine the concentration of protein in an unknown sample using the Biuret method. Describe how a calibration curve would be prepared and used to determine the protein concentration of the unknown.

题目（7分）：一名生物化学家正在使用比色计和双缩脲法测定未知样品中的蛋白质浓度。描述如何制作校准曲线并利用它确定未知物的蛋白质浓度。

First, a series of protein standard solutions with known concentrations must be prepared using serial dilution of a stock solution. Equal volumes of each standard are mixed with a fixed volume of Biuret reagent, which reacts with peptide bonds to produce a violet colour. The intensity of the colour is directly proportional to the protein concentration. A blank solution (containing all reagents except protein) is used to set the colorimeter to zero absorbance. The absorbance of each standard solution is measured at a wavelength of approximately 540 nm, where the violet complex absorbs light maximally. A calibration curve is plotted with protein concentration on the x-axis and absorbance on the y-axis — a line of best fit is drawn through the plotted points. The unknown protein sample is then treated identically: mixed with the same volume of Biuret reagent, its absorbance is measured, and the corresponding concentration is read directly from the calibration curve by interpolation. If the absorbance of the unknown falls outside the range of the standards, the sample must be diluted and retested.

首先，必须通过储备液的系列稀释，制备一系列已知浓度的蛋白质标准溶液。每个标准溶液的等体积样品与固定体积的双缩脲试剂混合，该试剂与肽键反应产生紫色。颜色深度与蛋白质浓度成正比。使用空白溶液（除蛋白质外含所有试剂）将比色计调至零吸光度。在约540 nm波长处测量每个标准溶液的吸光度，这是紫色络合物吸收光最强的波长。绘制校准曲线，x轴为蛋白质浓度，y轴为吸光度——通过绘制的点作出最佳拟合线。然后对未知蛋白质样品进行相同处理：与相同体积的双缩脲试剂混合，测量其吸光度，并从校准曲线通过内插法直接读出对应的浓度。如果未知物的吸光度落在标准品范围之外，则必须稀释样品并重新测试。

Key term — Interpolation vs. Extrapolation: Edexcel expects you to understand that valid concentration values should be obtained by interpolation within the range of the calibration standards. Extrapolation beyond the highest standard is unreliable because the linear relationship (Beer-Lambert law) may not hold at higher concentrations. This is frequently tested in the context of practical write-ups.

关键术语——内插法与外推法：Edexcel 期望你理解，有效的浓度值应通过在标准校准范围内进行内插法获得。超出最高标准品范围的外推是不可靠的，因为线性关系（比尔-朗伯定律）在较高浓度下可能不成立。这在实验报告写作的背景下经常被考查。

---

10. Synoptic Question: Integrating Transport and Circulation | 综合题：运输与循环的整合

Question (9 marks): Compare and contrast the mass flow systems in the xylem and phloem of a flowering plant. Explain how the cohesive-tension theory accounts for water movement in the xylem, and contrast this with the pressure-flow hypothesis for translocation in the phloem.

题目（9分）：比较和对比显花植物木质部和韧皮部中的集流系统。解释内聚力-张力理论如何解释木质部中的水分运动，并与解释韧皮部运输的压力流动假说进行对比。

The xylem and phloem are both vascular tissues that use mass flow to transport substances, but they differ fundamentally in structure, direction, driving force, and transported materials. Xylem consists of dead, hollow, lignified cells arranged end-to-end to form continuous vessels. It transports water and dissolved mineral ions from the roots upwards to the leaves (unidirectional). The driving force is transpiration pull: water evaporates from mesophyll cell walls into air spaces in the leaf, creating a negative pressure (tension) that is transmitted down the continuous column of water in the xylem due to the cohesive forces between water molecules and adhesive forces between water molecules and the xylem walls. Phloem, in contrast, consists of living sieve tube elements and companion cells. It transports sucrose and other organic solutes from sources (mature leaves producing photosynthate) to sinks (growing regions, storage organs) — this movement can be bidirectional. The pressure-flow hypothesis states that sucrose is actively loaded into the phloem at the source, lowering the water potential. Water enters from the adjacent xylem by osmosis, increasing hydrostatic pressure. At the sink, sucrose is actively unloaded, water leaves the phloem, and pressure decreases. The pressure gradient drives mass flow of phloem sap from source to sink.

木质部和韧皮部都是利用集流运输物质的维管组织，但它们在结构、方向、驱动力和运输物质方面有根本差异。木质部由死去的、中空的、木质化的细胞首尾相连组成，形成连续的导管。它将水和溶解的矿物质离子从根部向上运输到叶片（单向）。驱动力是蒸腾拉力：水分从叶肉细胞壁蒸发进入叶片气腔，产生负压（张力），由于水分子之间的内聚力以及水分子与木质部管壁之间的附着力，该张力沿着木质部中连续的水柱向下传递。相比之下，韧皮部由活的筛管分子和伴胞组成。它将蔗糖和其他有机溶质从源（进行光合作用产生同化物的成熟叶片）运输到库（生长区域、储存器官）——这种移动可以是双向的。压力流动假说指出，蔗糖在源端被主动装载到韧皮部中，降低水势。水通过渗透从相邻的木质部进入，增加静水压。在库端，蔗糖被主动卸载，水离开韧皮部，压力降低。压力梯度驱动韧皮部汁液从源向库的集流。

Synoptic connections: This question links Topic 3 (Exchange and Transport) with Topic 4 (Energy Transfers). The active loading of sucrose requires ATP from respiration, linking to cellular energetics. Edexcel synoptic questions reward candidates who can draw connections between seemingly disparate topics. Practice spotting these cross-topic links in past papers.

综合联系：这道题目将主题3（交换与运输）与主题4（能量转移）联系起来。蔗糖的主动装载需要来自呼吸作用的ATP，这涉及细胞能量学。Edexcel 的综合题会奖励那些能够将看似不相关的主题联系起来的考生。练习在历年真题中

Published by TutorHao | Year 12 Biology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）