📚 Year 10 OCR Biology: Unit Test Mock Paper Walkthrough | Year 10 OCR 生物：单元测试模拟卷解析
This article provides a comprehensive walkthrough of a typical Year 10 OCR Biology unit test mock paper. By working through exam-style questions on cell biology, enzymes, transport, and organ systems, you will strengthen key concepts and improve your exam technique. Each section breaks down the question, explains the scientific reasoning, and highlights common marking points.
本文全面解析了一份典型的 Year 10 OCR 生物单元测试模拟卷。通过练习细胞生物学、酶、运输和器官系统等考试题型，你将巩固关键概念并提高应试技巧。每个部分分解题目、解释科学原理并突出常见得分点。
1. Question 1: Labelling an Animal Cell | 问题一：标注动物细胞结构
The diagram shows a typical animal cell. You are asked to label the nucleus, cytoplasm, cell membrane, mitochondria and ribosomes. For full marks, spell 'mitochondria' and 'ribosomes' correctly.
示意图展示了一个典型的动物细胞。要求标注细胞核、细胞质、细胞膜、线粒体和核糖体。要想得满分，请正确拼写 'mitochondria' 和 'ribosomes'。
The nucleus is usually the largest, round organelle; it controls the cell and stores DNA. In the diagram, structure X is the nucleus.
细胞核通常是最大且呈圆形的细胞器；它控制细胞并储存DNA。图中，结构X是细胞核。
Mitochondria are oval-shaped and are the site of aerobic respiration, releasing energy. Ribosomes are tiny dots, either free in the cytoplasm or attached to the rough endoplasmic reticulum.
线粒体呈椭圆形，是有氧呼吸的场所，释放能量。核糖体是微小的颗粒，或游离于细胞质中，或附着在粗面内质网上。
The cell membrane forms the boundary and controls what enters and leaves the cell. The cytoplasm is the jelly-like substance where chemical reactions occur.
细胞膜形成边界，控制物质进出细胞。细胞质是果冻状物质，化学反应在其中进行。
2. Question 2: Calculating Magnification | 问题二：计算放大倍数
In the mock, an image of a red blood cell measures 30 mm, while its actual size is 7 µm. Always convert units first—this step often earns a mark.
在模拟卷中，一个红细胞的图像大小为30毫米，而实际大小为7微米。务必先转换单位——这一步通常能得分。
Convert 30 mm to micrometres: 30 mm = 30 × 1000 = 30 000 µm. Magnification = Image size ÷ Actual size = 30 000 ÷ 7 ≈ 4286.
将30毫米转换为微米：30 mm = 30 × 1000 = 30 000 µm。放大倍数 = 图像大小 ÷ 实际大小 = 30 000 ÷ 7 ≈ 4286倍。
Write the answer as '× 4300' (to two significant figures) and do not include units, because magnification is a ratio.
答案写作‘× 4300’（保留两位有效数字），不带单位，因为放大倍数是比值。
3. Question 3: Enzyme Action and Denaturation | 问题三：酶的作用与变性
A graph of enzyme activity against temperature shows that the rate increases up to 37 °C, then drops sharply. Enzymes are biological catalysts with an active site complementary to a specific substrate.
酶活性随温度变化的曲线图显示，反应速率在37 °C之前上升，随后急剧下降。酶是生物催化剂，具有与特定底物互补的活性位点。
At the optimum temperature, collisions are maximised and the enzyme-substrate complex forms efficiently. Beyond the optimum, bonds in the enzyme break, the active site changes shape and the enzyme is denatured.
在最适温度下，碰撞达到最大，酶-底物复合物高效形成。超过最适温度，酶内部化学键断裂，活性位点形状改变，酶发生变性。
Denaturation is permanent; the enzyme can no longer bind the substrate, so activity falls to zero. A similar pattern occurs with pH extremes.
变性是不可逆的；酶不能再与底物结合，因此活性降至零。极端的pH值也会产生类似的影响。
4. Question 4: Osmosis in Potato Cylinders | 问题四：土豆圆柱体渗透实验
The experiment placed potato cylinders in different sucrose solutions and measured the percentage change in mass. Use the table below to interpret the results.
该实验将土豆圆柱体置于不同蔗糖溶液中，并测量了质量变化的百分比。利用下表解读结果。
| Concentration (mol/dm³) | Initial mass (g) | Final mass (g) | Change in mass (%) |
|---|---|---|---|
| 0.0 | 3.8 | 4.2 | +10.5 |
| 0.4 | 3.9 | 3.9 | 0.0 |
| 0.8 | 4.0 | 3.6 | -10.0 |
If the potato gains mass, water has entered by osmosis from a hypotonic (high water concentration) solution. A loss of mass indicates water has left the cells into a hypertonic (low water concentration) solution.
若土豆质量增加，说明水通过渗透从低渗（水浓度高）溶液进入细胞。质量减少表明水离开细胞进入了高渗（水浓度低）溶液。
No net change (0.0 %) means the solution is isotonic to the potato cells. In pure water, plant cells become turgid; in strong sugar solution they become flaccid and the tissue turns limp.
无净变化（0.0 %）表示该溶液与土豆细胞等渗。在纯水中，植物细胞变得硬挺；在浓糖水中，细胞失水变软，组织萎蔫。
5. Question 5: Mitosis and the Cell Cycle | 问题五：有丝分裂与细胞周期
The image shows onion root
Published by TutorHao | Year 10 Biology Revision Series | aleveler.com
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