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

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

This article provides a detailed breakdown of a mock unit test for Year 13 OCR Biology, covering core topics from Module 5 (Communication, homeostasis and energy) and Module 6 (Genetics, evolution and ecosystems). Each question is analysed to reinforce key concepts and exam technique.

本文详细解析一份面向 Year 13 OCR 生物的单元测试模拟卷，涵盖模块5（交流、稳态与能量）和模块6（遗传、进化与生态系统）的核心主题。每道题均配有考点梳理和答题技巧分析。

---

1. Multiple-Choice: Refractory Period | 选择题：不应期

Question: Which of the following correctly describes the refractory period? A. It is the time when potassium ions enter the axon. B. It ensures that action potentials can only travel in one direction. C. It occurs during depolarisation. D. It is caused by the opening of voltage-gated sodium channels. Correct answer: B.

题目：下列哪项正确描述了不应期？A. 是钾离子进入轴突的时间；B. 确保动作电位只能单向传播；C. 发生在去极化期间；D. 由电压门控钠通道打开引起。正确答案：B。

The refractory period is the interval after an action potential during which the axon is less responsive to stimulation. In the absolute refractory period, voltage-gated Na⁺ channels are inactivated, so no new action potential can be generated. In the relative refractory period, some Na⁺ channels have recovered but K⁺ channels are still open, making the membrane hyperpolarised; a stronger-than-normal stimulus is required. This ensures that action potentials travel away from the cell body and do not propagate backwards, maintaining unidirectional signalling.

不应期是动作电位后轴突对刺激反应降低的一段时间。在绝对不应期中，电压门控Na⁺通道失活，无法产生新动作电位。在相对不应期中，部分Na⁺通道恢复但K⁺通道仍开放，膜电位超极化，需要比正常更强的刺激。这确保了动作电位从胞体向远处单向传播，不会反向传递，维持了单向信号传导。

---

2. Structured Question: Muscle Contraction | 结构化问题：肌肉收缩

Question: Outline the roles of calcium ions (Ca²⁺) and ATP in the sliding filament model of muscle contraction. (4 marks)

题目：概述钙离子(Ca²⁺)和ATP在肌肉收缩滑丝模型中的作用。(4分)

Model answer: Ca²⁺ ions bind to troponin, causing a conformational change that moves tropomyosin away from the myosin-binding sites on actin filaments. This allows myosin heads to bind to actin. ATP binds to the myosin head, breaking the cross-bridge and enabling the head to detach from actin. ATP is then hydrolysed to ADP and inorganic phosphate (Pi), which cocks the myosin head into a high-energy position. When the head reattaches to a new actin-binding site, the power stroke occurs as ADP and Pi are released, pulling the actin filament inward.

参考答案：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，让肌球蛋白头部能结合肌动蛋白。ATP与肌球蛋白头部结合，打开横桥使头部与肌动蛋白分离。随后ATP水解为ADP和磷酸(Pi)，将肌球蛋白头部重新翘起至高能状态。当头部附着于新的肌动蛋白结合位点，释放ADP和Pi时发生动力冲程，拉动肌动蛋白丝向内。

The sequential binding and hydrolysis of ATP drives the cross-bridge cycle, while Ca²⁺ acts as the regulatory switch that initiates contraction by exposing the binding sites. Without ATP, rigor mortis occurs because the cross-bridges cannot detach.

ATP的连续结合与水解驱动横桥循环，而Ca²⁺作为调控开关，通过暴露结合位点启动收缩。没有ATP时，横桥无法分离，就会发生尸僵。

---

3. Structured Question: Loop of Henlé | 结构化问题：亨勒袢

Explain how the loop of Henlé acts as a countercurrent multiplier to produce concentrated urine. (6 marks)

解释亨勒袢如何作为逆流倍增器产生浓缩尿液。(6分)

The descending limb is permeable to water but impermeable to Na⁺ and Cl⁻. As the filtrate moves down, water leaves by osmosis because the surrounding medulla has a higher solute concentration. The ascending limb actively transports Na⁺ and Cl⁻ out into the medulla, contributing to the hypertonic interstitial fluid. This ascending limb is impermeable to water. Because of the countercurrent flow, the filtrate in the descending limb continuously encounters a region of increasing osmolarity, causing more water to leave. The active transport in the ascending limb multiplies the gradient. The vasa recta maintain the countercurrent exchange by removing reabsorbed water. Consequently, the fluid arriving at the distal tubule is dilute, and the collecting duct passes through the high medullary osmolarity, allowing a large amount of water to be reabsorbed under the influence of ADH, producing concentrated urine.

降支对水通透，对Na⁺和Cl⁻不通透。滤液向下流动时，由于髓质组织液溶质浓度更高，水通过渗透流出。升支主动将Na⁺和Cl⁻转运至髓质，维持高渗组织液；升支对水不通透。由于逆流流动，降支中的滤液持续遇到浓度渐增的髓质，使更多水渗出。升支的主动转运放大了梯度。直小血管通过移除重吸收的水维持逆流交换。因此，到达远曲小管的液体是稀释的，而集合管穿过高渗髓质，在ADH作用下大量水被重吸收，产生浓缩尿液。

---

4. Data Analysis: Photosynthetic Spectra | 数据分析：光合作用光谱

A graph shows the absorption spectrum of chlorophyll a and the action spectrum of photosynthesis. Explain why the action spectrum does not perfectly match the absorption spectrum. (3 marks)

图显示了叶绿素a的吸收光谱和光合作用的作用光谱。解释为什么作用光谱不能完全匹配吸收光谱。(3分)

Chlorophyll a alone does not absorb light efficiently across the entire visible spectrum; its absorption peaks are mainly in the red and blue regions. Accessory pigments such as chlorophyll b, carotenoids, and xanthophylls absorb light in other wavelengths (e.g., green and orange) and transfer the energy to chlorophyll a. This broadens the range of light wavelengths that can drive photosynthesis. As a result, the action spectrum shows higher photosynthetic activity at wavelengths that are not strongly absorbed by chlorophyll a itself, such as blue-green and orange-red light.

叶绿素a本身无法在整个可见光谱高效吸收光，其吸收峰主要在红光和蓝光区域。辅助色素如叶绿素b、类胡萝卜素和叶黄素吸收其他波长（如绿光和橙光）的光，并将能量传递给叶绿素a，拓宽了可驱动光合作用的光波长范围。因此，作用光谱在叶绿素a本身吸收较弱的蓝绿和橙红区域显示出较高的光合活性。

---

5. Calculation: Respiratory Quotient | 计算：呼吸商

A respirometer is used to measure the volume of oxygen consumed and carbon dioxide produced by germinating seeds. Results: O₂ consumed = 1.5 cm³, CO₂ produced = 1.5 cm³. Calculate the respiratory quotient (RQ) and suggest the likely respiratory substrate. (2 marks)

用呼吸计测量萌发种子的O₂消耗量和CO₂产生量。结果：O₂消耗量 = 1.5 cm³，CO₂产生量 = 1.5 cm³。计算呼吸商(RQ)并推测可能的呼吸底物。(2分)

RQ = CO₂ produced / O₂ consumed = 1.5 / 1.5 = 1.0. An RQ of 1.0 indicates that carbohydrates are the main respiratory substrate, because the balanced equation for aerobic respiration of glucose is C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, giving a CO₂:O₂ ratio of 1.0. Substrates such as lipids would yield an RQ below 1.0, and proteins typically give an RQ around 0.9.

RQ = 产生的CO₂ / 消耗的O₂ = 1.5/1.5 = 1.0。RQ为1.0表明主要呼吸底物是碳水化合物，因为葡萄糖有氧呼吸的配平方程式为C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O，CO₂与O₂的比值为1.0。脂质的RQ低于1.0，蛋白质的RQ一般约为0.9。

---

6. Genetics: Epistasis in Labrador Retrievers | 遗传学：拉布拉多犬的上位效应

In Labrador retrievers, coat colour is controlled by two genes: B/b for pigment colour (B = black, b = brown) and E/e for pigment deposition (E = deposition occurs, e = no deposition). State the expected phenotypic ratio from a dihybrid cross BbEe × BbEe, and explain the type of epistasis shown. (4 marks)

在拉布拉多犬中，毛色由两对基因控制：B/b决定色素颜色（B=黑，b=棕），E/e决定色素沉积（E=沉积，e=不沉积）。请写出双杂交BbEe × BbEe预期的表型比，并解释显示的上位类型。(4分)

The expected phenotypic ratio is 9 black : 3 brown : 4 yellow. Genotypes: black (B_E_), brown (bbE_), yellow ( _ _ ee). This is an example of recessive epistasis, where the homozygous recessive genotype at the E locus (ee) masks the effect of the B locus. Regardless of whether the dog has B or b alleles, if it is ee, no pigment is deposited in the fur, resulting in yellow coat colour. The epistatic gene is the extension gene (E), and its

Published by TutorHao | Year 13 Biology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）