Pre-U Edexcel Biology: Unit Test Mock Paper Walkthrough | Pre-U Edexcel 生物：单元测试模拟卷解析

📚 Pre-U Edexcel Biology: Unit Test Mock Paper Walkthrough | Pre-U Edexcel 生物：单元测试模拟卷解析

Welcome to this comprehensive walkthrough of a mock unit test tailored to the Edexcel Pre-U Biology specification. By exploring a set of representative questions, you will sharpen your understanding of core concepts and learn how to craft high-scoring responses. We will unpick each question type, linking knowledge to the mark scheme and revealing common pitfalls.

欢迎阅读针对 Edexcel Pre-U 生物课程精心设计的单元模拟卷解析。通过逐一剖析典型试题，你将加深对核心概念的理解，并学会构建高分答案。我们将拆解各种题型，将知识点与评分标准挂钩，同时揭示常见失分点。

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1. Multiple Choice on Membrane Components | 细胞膜成分选择题

The multiple choice question states: “Which of the following is NOT a component of the plasma membrane? A. Phospholipids, B. Glycoproteins, C. Ribosomes, D. Cholesterol.” The correct answer is C.

这道选择题的问题是：“以下哪一项不是细胞膜的组成成分？A. 磷脂，B. 糖蛋白，C. 核糖体，D. 胆固醇。”正确答案是 C。

Ribosomes are not embedded in the plasma membrane; they are cytoplasmic organelles that synthesise proteins, either free in the cytosol or attached to the rough endoplasmic reticulum. All other options are genuine membrane constituents.

核糖体并不镶嵌在细胞膜中，它们是蛋白质合成的细胞器，游离于胞质溶胶或附着在粗面内质网上。其余选项都是真正的膜结构成分。

Phospholipids arrange themselves into a bilayer with hydrophilic heads facing the aqueous surroundings and hydrophobic tails pointing inward. Glycoproteins, which are proteins with attached carbohydrate chains, are integral to cell–cell recognition and adhesion. Cholesterol molecules fit between the phospholipid tails, modulating membrane fluidity and stability.

磷脂排列成双分子层，亲水头部朝向水相环境，疏水尾部向内。糖蛋白是连接了糖链的蛋白质，对细胞识别和粘附至关重要。胆固醇分子嵌入磷脂尾部之间，调节膜的流动性和稳定性。

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2. Tabular Comparison: Prokaryotes vs Eukaryotes | 表格比较：原核细胞与真核细胞

A table-completion question asks you to tick features belonging to prokaryotes, eukaryotes, or both. Let’s examine key differences that frequently appear in such tables.

一道表格填写题要求你勾选属于原核细胞、真核细胞或两者共有的特征。我们来审视这类表格中常出现的关键差异。

Feature	Prokaryotes	Eukaryotes
Nucleus with double membrane	✘	✔
70S ribosomes	✔	✘ (80S in cytoplasm)
Membrane-bound organelles	✘	✔
Circular DNA	✔	✘ (linear chromosomes)
Peptidoglycan cell wall	✔ (most)	✘ (if present, chitin or cellulose)

The table above highlights that prokaryotes, such as bacteria, lack a true nucleus and membrane-bounded organelles. Their DNA is a single circular chromosome in the nucleoid, and their smaller 70S ribosomes differ from the 80S ribosomes found in eukaryotic cytoplasm, although mitochondria and chloroplasts contain 70S ribosomes – an evolutionary clue.

上表清晰地表明，原核生物（如细菌）没有真正的细胞核和膜包被的细胞器。它们的 DNA 是一个裸露的环状染色体，位于拟核中，其 70S 核糖体比真核细胞质中的 80S 核糖体小，但真核生物的线粒体和叶绿体含有 70S 核糖体——这是一个进化线索。

When tackling such tabular questions, always recall the exceptions: red blood cells lack nuclei, sieve tube elements lose most organelles, and yeast cells possess a wall made of chitin, not cellulose.

解答这类表格题时，务必记住例外情况：红细胞无核、筛管失去大多数细胞器、酵母菌细胞壁由几丁质而非纤维素组成。

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3. Enzyme Specificity and Inhibition | 酶的专一性与抑制作用

A short-answer question asks: “Outline the induced-fit model of enzyme action and explain how a competitive inhibitor reduces the rate of an enzyme-catalysed reaction.” This tests both core theory and application.

一道简答题要求：“概述酶的诱导契合模型，并解释竞争性抑制剂如何降低酶促反应的速率。”此题同时考查核心理论与应用。

According to the induced-fit model, the active site is not a rigid structure. When a substrate approaches, the active site undergoes a conformational change, moulding itself around the substrate to form an enzyme–substrate (ES) complex. This precise fit puts strain on the substrate, lowering the activation energy needed to convert it into products.

根据诱导契合模型，活性部位并非刚性结构。当底物靠近时，活性部位发生构象变化，围绕底物成形，形成酶-底物（ES）复合物。这种精确契合对底物施加压力，降低了将其转化为产物所需的活化能。

A competitive inhibitor has a shape similar to the substrate. It fits into the active site, blocking the genuine substrate from binding. As a result, fewer functional ES complexes form, and the reaction rate decreases. This inhibition is often reversible: increasing substrate concentration can outcompete the inhibitor, allowing Vₘₐₓ to remain unchanged even though Kₘ increases.

竞争性抑制剂具有与底物相似的形状，能占据活性部位，阻止真正的底物结合。因此，功能性 ES 复合物减少，反应速率下降。这种抑制通常可逆：提高底物浓度可以排挤抑制剂，使得 Vₘₐₓ 不变，但 Kₘ 增大。

When writing your answer, include the key terms ‘conformational change’, ‘ES complex’, ‘activation energy’, and ‘blocking of active site’. A clear diagram in your mind helps, even if you aren’t drawing it.

书写答案时，要包含“构象变化”“ES 复合物”“活化能”和“活性部位被阻断”等关键术语。即使不画图，脑海中也应有一幅清晰的图示。

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4. Active Transport and the Sodium-Potassium Pump | 主动运输与钠钾泵

A structured question describes: “The Na⁺/K⁺-ATPase pump is essential for maintaining the resting potential in neurones. Explain how it works and why it is described as an example of primary active transport.”

一道结构化题目描述：“Na⁺/K⁺-ATPase 泵对维持神经元静息电位至关重要。解释其工作原理，以及为何它被归为初级主动运输的实例。”

The sodium-potassium pump is a transmembrane carrier protein that uses energy from ATP hydrolysis to move ions against their concentration gradients. For every ATP molecule split, three sodium ions (Na⁺) are pumped out of the cell and two potassium ions (K⁺) are pumped into the cell.

钠钾泵是一种跨膜载体蛋白，利用 ATP 水解提供的能量逆浓度梯度运输离子。每消耗一个 ATP 分子，三个钠离子（Na⁺）被泵出细胞，两个钾离子（K⁺）被泵入细胞。

This unequal movement creates an electrochemical gradient: the inside of the cell becomes slightly negative relative to the outside because more positive ions leave than enter. The pump is thus electrogenic. It is called primary active transport because the energy is derived directly from ATP hydrolysis, not from the pre-existing gradient of another solute.

这种不对等的转运建立了电化学梯度：由于运出的正离子多于运入，细胞内相对于细胞外略呈负电性，因此该泵具生电性。它被称为初级主动运输，因为能量直接源自 ATP 的水解，而非来自另一溶质的预先浓度梯度。

In the context of a neurone, the Na⁺/K⁺ pump continuously works to restore the resting potential after an action potential has passed, ensuring the axon is ready to transmit another impulse.

在神经元中，动作电位通过后，钠钾泵持续工作以恢复静息电位，确保轴突准备好传递下一个冲动。

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5. Data Analysis: Osmotic Effects on Red Blood Cells | 数据分析：渗透压对红细胞的影响

A graph in the mock paper shows the percentage of haemolysed red blood cells when incubated in NaCl solutions of varying concentrations. Explain the trend.

模拟卷中的一张图显示，将红细胞置于不同浓度的氯化钠溶液中时，溶血百分比的曲线图。试解释该趋势。

In an isotonic solution (about 0.9% NaCl), the water potential inside and outside the cells is equal; there is no net movement of water, so cells remain intact. In a hypotonic solution (lower NaCl concentration, e.g. 0.3%), the external water potential is higher than that inside the cell. Water enters by osmosis, causing the cells to swell and eventually burst – a process called haemolysis.

在等渗溶液（约 0.9% NaCl）中，细胞内外的水势相等，没有净水流动，细胞保持完整。在低渗溶液（较低氯化钠浓度，如 0.3%）中，细胞外的水势高于细胞内，水分通过渗透作用进入细胞，导致细胞膨胀并最终破裂——这一过程称为溶血。

In a hypertonic solution (higher NaCl concentration, e.g. 2%), water leaves the cells, making them shrink and develop a crenated appearance, but they do not typically lyse. The graph would therefore show a peak of haemolysis at very low salt concentrations.

在高渗溶液（较高氯化钠浓度，如 2%）中，水分从细胞中逸出，使细胞皱缩呈锯齿状，但通常不会破裂。因此，曲线图会在极低盐浓度处显示溶血峰。

When analysing such data, always refer to water potential gradients rather than simply ‘solute concentration’, and use the term ‘net water movement’ to be precise.

分析此类数据时，务必使用“水势梯度”而非单纯的“溶质浓度”，并精准使用“净水移动”这一术语。

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6. Meselson-Stahl Experiment and DNA Replication | Meselson-Stahl 实验与 DNA 复制

The question states: “Bacteria were cultured in a medium containing ¹⁵N for many generations and then transferred to ¹⁴N medium. DNA was sampled after one and two rounds of replication and subjected to density-gradient centrifugation.” Describe the expected results and explain what they demonstrate.

题目：“细菌在含 ¹⁵N 的培养基中培养多代，然后转移至含 ¹⁴N 的培养基。在经过一轮和两轮复制后提取 DNA 并进行密度梯度离心。”描述预期的结果，并解释其证明了什么。

After many generations in ¹⁵N, all DNA is heavy (¹⁵N/¹⁵N). After one replication round in ¹⁴N, a single band of intermediate density (¹⁵N/¹⁴N hybrid) appears. This rules out the conservative model, which would predict one heavy and one light band.

在 ¹⁵N 中增殖多代后，所有 DNA 为重型（¹⁵N/¹⁵N）。在 ¹⁴N 中复制一轮后，出现单一的中间密度条带（¹⁵N/¹⁴N 杂交链）。这排除了全保留模型，因为该模型预测会出现一条重链和一条轻链。

After two replication rounds in ¹⁴N, two bands are seen: one intermediate density (¹⁵N/¹⁴N) and one light density (¹⁴N/¹⁴N). This pattern is consistent with the semiconservative model, whereby each new double helix consists of one old (template) strand and one newly synthesised strand.

在 ¹⁴N 中复制两轮后，出现两条带：一条中间密度（¹⁵N/¹⁴N）和一条轻型（¹⁴N/¹⁴N）。该图式与半保留复制模型一致，即每个新双螺旋由一条旧链（模板链）和一条新合成的链组成。

The experiment provided pivotal evidence for the semiconservative nature of DNA replication, a concept you must be able to link to the role of DNA polymerase and complementary base pairing.

该实验为 DNA 半保留复制提供了关键证据，你需要能将此概念与 DNA 聚合酶的作用及互补碱基配对联系起来。

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7. Transcription and Translation | 转录与翻译

A question provides a template DNA sequence: 3′-TAC GGA TCG ACT-5′. Write the corresponding mRNA sequence, the anticodons of the relevant tRNA molecules, and determine the amino acid sequence using the genetic code table provided.

一道题给出 DNA 模板链序列：3′-TAC GGA TCG ACT-5’。写出对应的 mRNA 序列、相关 tRNA 分子的反密码子，并根据提供的遗传密码表确定氨基酸序列。

mRNA is synthesised complementary to the template strand, with uracil (U) replacing thymine (T). Thus, the mRNA sequence is 5′-AUG CCU AGC UGA-3′. Remember that RNA polymerase reads the template strand in the 3′ to 5′ direction, building mRNA in the 5′ to 3′ direction.

mRNA 以模板链为互补模板合成，同时尿嘧啶（U）替换胸腺嘧啶（T）。因此，mRNA 序列为 5′-AUG CCU AGC UGA-3’。请记住，RNA 聚合酶从 3′ 到 5′ 方向读取模板链，沿 5′ 到 3′ 方向合成 mRNA。

tRNA anticodons are complementary to the mRNA codons: for AUG the anticodon is 3′-UAC-5′, for CCU it is 3′-GGA-5′, for AGC it is 3′-UCG-5′. The codon UGA is a stop codon and does not have a corresponding tRNA with an amino acid.

tRNA 反密码子与 mRNA 密码子互补：AUG 的反密码子是 3′-UAC-5’，CCU 的反密码子是 3′-GGA-5’，AGC 的反密码子是 3′-UCG-5’。密码子 UGA 是终止密码子，没有携带氨基酸的对应 tRNA

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