Year 13 AQA Biology Unit Test Mock Paper Walkthrough | AQA 生物单元测试模拟卷解析

📚 Year 13 AQA Biology Unit Test Mock Paper Walkthrough | AQA 生物单元测试模拟卷解析

This walkthrough takes you through a representative AQA Year 13 unit test paper, unpacking the key skills and knowledge required for top marks. Each question targets a core topic from the specification – from photosynthesis and respiration to genetics, gene expression and ecosystems – and the detailed commentary highlights common pitfalls, essential terminology and exam technique. Use this as a revision tool to sharpen your understanding and build confidence for the real exam.

本文将对一套典型的 AQA Year 13 生物单元测试模拟卷进行逐题解析，梳理获取高分所需的关键技能与知识。每道题均针对考纲核心主题——涵盖光合作用与呼吸作用、遗传学、基因表达和生态系统等内容；详细的点评着重指出常见错误、关键术语以及答题技巧。将此解析作为复习工具，可加深理解并为正式考试积累信心。

1. Question 1 – Rate of Photosynthesis in Isolated Chloroplasts | 第1题 – 分离叶绿体的光合速率

Question: Isolated chloroplasts were provided with an artificial electron acceptor (DCPIP) which changes from blue to colourless when reduced. The rate of DCPIP decolourisation was measured under different light intensities. Explain why the rate of decolourisation can be used as a measure of the light-dependent reactions.

题目：分离的叶绿体被提供一种人工电子受体（DCPIP），当其被还原时从蓝色变为无色。测量不同光照强度下 DCPIP 脱色的速率。解释为何脱色速率可作为光反应速率的一种测量指标。

The light-dependent reactions involve the flow of electrons from water, through photosystem II and photosystem I, to NADP⁺, forming reduced NADP. In the practical set-up, DCPIP acts as a substitute for the terminal electron acceptor NADP⁺. As light energy drives the photoionisation of chlorophyll, excited electrons pass along the electron transport chain and are ultimately accepted by DCPIP, reducing it to its colourless form. Therefore, the faster the light-dependent reactions proceed, the quicker DCPIP is reduced and decolourised. The rate of decolourisation is thus directly proportional to the rate of electron transport, providing a valid proxy for the activity of the light-dependent stage.

光反应涉及电子从水经过光系统 II 和光系统 I 流向 NADP⁺，生成还原型 NADP。在此实验装置中，DCPIP 替代了终端电子受体 NADP⁺。当光能驱动叶绿素的光电离时，激发电子沿电子传递链传递，最终被 DCPIP 接受，将其还原为无色形式。因此，光反应进行得越快，DCPIP 被还原并脱色的速度就越快。脱色速率与电子传递速率成正比，可作为光反应阶段活性的有效指标。

---

2. Question 2 – Chemiosmosis and ATP Yield | 第2题 – 化学渗透与 ATP 产量

Question: DNP (dinitrophenol) makes the inner mitochondrial membrane permeable to protons. Predict and explain the effect of DNP on ATP production during aerobic respiration.

题目：DNP（二硝基苯酚）使线粒体内膜对质子通透。预测并解释 DNP 对有氧呼吸中 ATP 生成的影响。

During oxidative phosphorylation, protons are pumped from the mitochondrial matrix into the intermembrane space, creating an electrochemical gradient. Protons flow back through ATP synthase, driving the synthesis of ATP. DNP provides an alternative route for protons to re-enter the matrix, bypassing ATP synthase. This dissipates the proton gradient. The energy released by the flow of protons is lost as heat rather than being coupled to ATP synthesis. Consequently, less ATP is produced, although the electron transport chain may continue to run. The cell compensates by oxidising more respiratory substrate, leading to increased oxygen consumption but reduced ATP yield.

在氧化磷酸化过程中，质子从线粒体基质泵入膜间隙，形成电化学梯度。质子通过 ATP 合酶回流，驱动 ATP 合成。DNP 为质子提供了一条不经过 ATP 合酶的返回基质的备用通道，从而耗散质子梯度。质子回流释放的能量以热的形式散失，无法与 ATP 合成偶联。因此，ATP 产量减少，尽管电子传递链可能继续运行。细胞通过氧化更多呼吸底物来补偿，导致耗氧量增加但 ATP 得率降低。

---

3. Question 3 – Action Potential Propagation | 第3题 – 动作电位的传播

Question: Explain why an action potential is conducted in one direction along a myelinated axon, using your knowledge of ion channels and refractory periods.

题目：运用离子通道和不应期的知识，解释为何动作电位在有髓轴突上单向传导。

An action potential is generated when voltage-gated Na⁺ channels open, causing rapid depolarisation. Immediately after opening, these channels become inactivated during the absolute refractory period. Behind the active region, the membrane is in its refractory state and cannot generate a new action potential. Ahead of the active zone, the membrane is at the resting potential with voltage-gated Na⁺ channels in the closed but activatable state. Local circuits depolarise the adjacent membrane, triggering the opening of Na⁺ channels and propagation of the action potential forward. In myelinated axons, saltatory conduction causes the action potential to jump between nodes of Ranvier, reinforcing unidirectional transmission because the preceding node remains refractory.

动作电位产生时，电压门控 Na⁺ 通道开放，引发快速去极化。通道开放后立即进入绝对不应期的失活状态。在兴奋区域后方，膜处于不应状态，无法引发新的动作电位。兴奋区域前方，膜处于静息电位，电压门控 Na⁺ 通道处于关闭但可激活的状态。局部电流使相邻膜去极化，触发 Na⁺ 通道开放，动作电位向前传播。在有髓轴突中，跳跃传导使动作电位在郎飞结之间跳跃式前进，进一步强化了单向传播，因为前一个结区仍处于不应期。

---

4. Question 4 – Hardy–Weinberg Principle Application | 第4题 – 哈代–温伯格定律的应用

Question: In a population of 2000 fruit flies, the frequency of the recessive allele for vestigial wings is 0.3. Assuming Hardy–Weinberg equilibrium, calculate the expected number of heterozygous individuals.

题目：在 2000 只果蝇的种群中，残翅隐性等位基因的频率为 0.3。假设符合哈代–温伯格平衡，计算预期杂合子个体数。

gotes is 2pq = 2 × 0.7 × 0.3 = 0.42. Expected number of heterozygotes = 0.42 × 2000 = 840.

设 p = 显性等位基因频率，q = 隐性等位基因频率。已知 q = 0.3，则 p = 1 − q = 0.7。杂合子频率为 2pq = 2 × 0.7 × 0.3 = 0.42。预期杂合子个体数 = 0.42 × 2000 = 840。

---

5. Question 5 – Epigenetic Modification of Gene Expression | 第5题 – 表观遗传修饰与基因表达

Question: Outline how increased methylation of CpG islands in a promoter region can lead to reduced transcription of a tumour suppressor gene.

题目：概述启动子区域 CpG 岛甲基化程度增加如何导致抑癌基因转录减少。

Methylation involves the addition of methyl groups to cytosine bases within CpG dinucleotides in the promoter region. This modification alters the chromatin structure, making it more condensed. The condensed chromatin prevents the binding of transcription factors and RNA polymerase to the promoter, thereby reducing the rate of transcription initiation. In the case of a tumour suppressor gene, reduced expression removes its normal brake on cell division, increasing the risk of uncontrolled cell proliferation and cancer development.

甲基化是指在启动子区域的 CpG 二核苷酸中的胞嘧啶碱基上添加甲基基团。该修饰改变染色质结构，使其更加凝缩。凝缩的染色质阻止转录因子和 RNA 聚合酶与启动子结合，从而降低转录起始速率。对于抑癌基因而言，表达降低解除了其对细胞分裂的正常抑制，增加了细胞失控增殖和癌症发生的风险。

---

6. Question 6 – Energy Flow in Ecosystems | 第6题 – 生态系统中的能量流动

Question: In a woodland food chain, the gross primary production (GPP) is 28 000 kJ m⁻² yr⁻¹. Plants use 35% of GPP in respiration. Calculate the net primary production (NPP) and explain why only a small percentage of NPP is transferred to the primary consumer trophic level.

题目：在林地食物链中，总初级生产量（GPP）为 28 000 kJ m⁻² yr⁻¹。植物将 GPP 的 35% 用于呼吸。计算净初级生产量（NPP）并解释为何只有小部分 NPP 传递到初级消费者营养级。

ory losses = 28 000 − (0.35 × 28 000) = 28 000 − 9 800 = 18 200 kJ m⁻² yr⁻¹. Only about 10% e parts are indigestible and egested as faeces; much of the consumed energy is used in respiration for movement and heat production; and energy is lost through excretion of nitrogenous waste. Therefore, energy transfer between trophic levels is highly inefficient.

NPP = GPP − 呼吸损失 = 28 000 − (0.35 × 28 000) = 28 000 − 9 800 = 18 200 kJ m⁻² yr⁻¹。植物生物量中储存的能量（NPP）只有约 10% 传递到初级消费者，原因在于能量损失：并非所有植物体都被取食；部分被食入的物质无法消化而以粪便形式排出；所消耗的能量大部分用于呼吸作用以支持运动和产热；含氮废物的排泄也带走能量。因此，营养级之间的能量传递效率很低。

---

7. Question 7 – Sliding Filament Model and Muscle Contraction | 第7题 – 滑动丝模型与肌肉收缩

Question: Describe the roles of calcium ions, troponin and tropomyosin in the contraction of skeletal muscle.

题目：描述钙离子、肌钙蛋白和原肌球蛋白在骨骼肌收缩中的作用。

When a muscle fibre is stimulated, an action potential travels along the sarcolemma and into the T-tubules, causing the sarcoplasmic reticulum to release Ca²⁺ ions into the sarcoplasm. Ca²⁺ binds to troponin, causing a conformational change that moves tropomyosin away from the myosin-binding sites on actin filaments. This exposes the binding sites, allowing myosin heads to attach to actin and form cross-bridges. The myosin heads then pivot, pulling the actin filaments towards the centre of the sarcomere, thereby shortening the muscle fibre. When stimulation ceases, Ca²⁺ is actively transported back into the sarcoplasmic reticulum, troponin releases Ca²⁺, and tropomyosin re-covers the binding sites, causing relaxation.

骨骼肌纤维受到刺激时，动作电位沿肌膜传播并进入 T 管，引发肌质网释放 Ca²⁺ 离子进入肌浆。Ca²⁺ 与肌钙蛋白结合，引起构象变化，使原肌球蛋白从肌动蛋白丝上的肌球蛋白结合位点移开。结合位点暴露后，肌球蛋白头部与肌动蛋白结合，形成横桥。肌球蛋白头部随后摆动，将肌动蛋白丝拉向肌节中央，使肌纤维缩短。刺激停止后，Ca²⁺ 被主动转运回肌质网，肌钙蛋白释放 Ca²⁺，原肌球蛋白重新覆盖结合位点，肌肉舒张。

---

广告 / Sponsored

8. Question 8 – ELISA Test for Pathogen Detection | 第8题 – 用于病原体检测的 ELISA 实验

Question: A patient is suspected of having a viral infection. Describe how a direct ELISA could be used to confirm the presence of viral antigens in a blood sample.

题目：怀疑某患者受到病毒感染。描述如何通过直接 ELISA 确认血液样本中是否存在病毒抗原。

A direct ELISA involves the following steps. The blood sample (containing possible viral antigens) is added to a well of a microtiter plate, allowing antigens to bind to the plastic surface. After washing, a blocking buffer is added to prevent non-specific binding. A primary antibody specific to a viral antigen is then added; this antibody is conjugated to an enzyme such as horseradish peroxidase. Following incubation and washing to remove unbound antibodies, the enzyme’s substrate is added. If the viral antigen was present, the enzyme converts the substrate into a coloured product. The intensity of colour, measured by spectrophotometry, is proportional to the amount of antigen present. A positive result confirms the presence of the virus.

直接 ELISA 步骤如下。将可能含病毒抗原的血液样本加入微孔板的孔中，使抗原吸附在塑料表面。洗涤后加入封闭缓冲液以防止非特异性结合。随后加入针对病毒抗原的特异性一抗，该抗体已与辣根过氧化物酶等酶结合。孵育并洗涤去除未结合抗体后，加入酶底物。若存在病毒抗原，酶将底物转化为有色产物。用分光光度计测定颜色强度，其与抗原量成正比。阳性结果证实病毒的存在。

---

9. Question 9 – Role of IAA in Phototropism | 第9题 – IAA 在向光性中的作用

Question: Explain the mechanism by which unilateral light causes a shoot to bend towards the light source, referring to the distribution and action of indoleacetic acid (IAA).

题目：解释单侧光照导致茎向光源弯曲的机制，涉及吲哚乙酸（IAA）的分布与作用。

Unilateral light causes IAA to redistribute laterally from the illuminated side of the shoot tip to the shaded side. The resulting higher concentration of IAA on the shaded side stimulates cell elongation more than the lower concentration on the illuminated side, via the acid growth hypothesis: IAA promotes the active transport of protons into the cell wall, lowering the pH and activating expansins, which break bonds between cellulose microfibrils, allowing cell wall loosening and turgor-driven extension. The asymmetric elongation causes the shoot to bend towards the light. Once equilibrium is reached, the curvature stops as IAA distribution evens out.

单侧光照导致 IAA 从茎尖的向光侧横向再分布至背光侧。背光侧 IAA 浓度升高，通过酸性生长假说更强烈地刺激细胞伸长：IAA 促进质子主动运输进入细胞壁，降低 pH 并激活扩展蛋白，后者破坏纤维素微纤丝之间的键，使细胞壁松弛，并依靠膨压进行伸展。两侧不对称的伸长导致茎向光弯曲。达到平衡后，IAA 分布趋于均匀，弯曲停止。

---

10. Question 10 – Statistical Analysis of Genetic Crosses | 第10题 – 遗传杂交的统计分析

Question: ed green. Use the chi-squared (χ²) test to determine whether the observed ratios are consistent with the expected 9:3:3:1 Mendelian ratio. Show your calculations and interpret the result at the 5% significance level (critical value = 7.82 for 3 degrees of freedom).

题目：在一项双因子杂交实验中，得到以下表型比：90 圆黄，30 圆绿，28 皱黄，12 皱绿。运用卡方 (χ²) 检验判断观察比是否符合预期的 9:3:3:1 孟德尔比例。展示计算过程，并在 5% 显著性水平（自由度 3 时临界值 = 7.82）下解释结果。

ow = 3/16 × 160 = 30; wrinkled green = 1/16 × 160 = 10. χ² = Σ (O−E)²/E = (90−90)²/90 + (30−30)²/30 + (28−30)²/30 + (12−10)²/10 = 0 + 0 + (4/30) + (4/10) = 0.133 + 0.4 = 0.533. The calculated χ² (0.533) is much less than the critical value of 7.82 at p = 0.05. Therefore, we fail to reject the null hypothesis; the observed ratios do not differ significantly from the expected 9:3:3:1 ratio. The variation is due to chance alone.

观察总数 = 90 + 30 + 28 + 12 = 160。基于 9:3:3:1 的预期数：圆黄 = 9/16 × 160 = 90；圆绿 = 3/16 × 160 = 30；皱黄 = 3/16 × 160 = 30；皱绿 = 1/16 × 160 = 10。 + (30−30)²/30 + (28−30)²/30 + (12−10)²/10 = 0 + 0 + (4/30) + (4/10) = 0.133 + 0.4 = 0.533。计算所得 χ² (0.533) 远小于 p = 0.05 时的临界值 7.82。因此，不能拒绝原假设；观察比与预期的 9:3:3:1 没有显著差异，偏差纯粹由偶然因素引起。

---

11. Question 11 – Recombinant DNA and Genetic Fingerprinting | 第11题 – 重组 DNA 与基因指纹识别

Question: Explain how restriction endonucleases and gel electrophoresis can be used to produce a genetic fingerprint for a DNA sample.

题目：解释限制性内切酶和凝胶电泳如何用于产生 DNA 样品的基因指纹。

Restriction endonucleases cut DNA at specific recognition sites, producing fragments of varying lengths. Because the number and location of these recognition sites differ between individuals, the resulting fragment lengths are unique (except for identical twins). The DNA fragments are placed into wells in an agarose gel and subjected to an electric field. Because DNA is negatively charged, fragments migrate towards the positive electrode; smaller fragments move faster through the gel matrix. After electrophoresis, the fragments are transferred to a nylon membrane (Southern blotting) and probed with labelled DNA probes that bind to specific sequences. The pattern of bands visualised by autoradiography or fluorescence is the genetic fingerprint, which can be used for identification or paternity testing.

限制性内切酶在特定的识别位点切割 DNA，产生不同长度的片段。由于个体之间这些识别位点的数量和位置不同，产生的片段长度是独特的（同卵双胞胎除外）。将 DNA 片段加入琼脂糖凝胶的加样孔，并置于电场中。DNA 带负电，因此片段向阳极迁移；小片段在凝胶基质中移动更快。电泳后，将片段转移到尼龙膜上（Southern 印迹），用标记的 DNA 探针进行杂交，探针与特定序列结合。通过放射自显影或荧光显现的条带模式即为基因指纹，可应用于身份鉴定或亲子鉴定。

---

12. Question 12 – Essay-style: Control of Blood Glucose Concentration | 第12题 – 论述题：血糖浓度的调控

Question: Give an account of the mechanisms involved in the regulation of blood glucose concentration, including the roles of insulin, glucagon, and the second messenger model.

题目：阐述调节血糖浓度的机制，包括胰岛素、胰高血糖素以及第二信使模型的作用。

Blood glucose concentration is maintained within a narrow range by a negative feedback system involving the pancreas and liver. When blood glucose rises (e.g., after a meal), pancreatic β-cells detect the increase and secrete insulin. Insulin binds to specific receptors on the surface of target cells (hepatocytes, muscle cells), activating tyrosine kinase. This triggers a signaling cascade that leads to the translocation of GLUT4 glucose transporter vesicles to the cell membrane, increasing glucose uptake. Insulin also activates enzymes for glycogenesis, converting glucose to glycogen for storage. Glucagon acts antagonistically: when blood glucose falls, α-cells of the pancreas secrete glucagon. Glucagon binds to receptors on liver cells, activating adenylate cyclase via a G-protein; adenylate cyclase converts ATP to cyclic AMP (cAMP), a second messenger. cAMP activates protein kinase A, which phosphorylates and activates enzymes for glycogenolysis and gluconeogenesis, releasing glucose into the blood. Adrenaline also raises blood glucose by a similar second messenger mechanism, binding to liver receptors and promoting glycogen breakdown. These coordinated responses ensure a stable supply of glucose for cellular respiration, especially crucial for brain function.

血糖浓度通过涉及胰腺和肝脏的负反馈系统维持在较窄范围内。当血糖升高（如餐后），胰腺 β 细胞感知变化并分泌胰岛素。胰岛素与靶细胞（肝细胞、肌肉细胞）表面的特异性受体结合，激活酪氨酸激酶。这触发信号级联反应，导致 GLUT4 葡萄糖转运蛋白囊泡向细胞膜转位，增加葡萄糖摄取。胰岛素还激活糖原合成酶，将葡萄糖转化为糖原储存。胰高血糖素起拮抗作用：血糖下降时，胰腺 α 细胞分泌胰高血糖素。胰高血糖素与肝细胞受体结合，通过 G 蛋白激活腺苷酸环化酶；腺苷酸环化酶将 ATP 转化为环磷酸腺苷（cAMP），一种第二信使。cAMP 激活蛋白激酶 A，后者磷酸化并激活糖原分解和糖异生相关酶，将葡萄糖释放入血。肾上腺素也通过类似的第二信使机制升高血糖，与肝受体结合促进糖原分解。这些协同反应确保了细胞呼吸所需葡萄糖的稳定供应，对脑功能尤为关键。

Published by TutorHao | Biology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）