📚 Year 7 Cambridge Biology: Mock Exam Analysis | Year 7 剑桥生物：单元测试模拟卷解析
This article guides you through a full mock exam for Year 7 Cambridge Biology. Each question includes a detailed explanation to help you understand key concepts and common mistakes. Working through these exam-style questions will build your confidence and improve your performance in end-of-unit tests.
本文带你完整解析一套 Year 7 剑桥生物单元测试模拟卷。每道题目配有详细讲解，帮助你掌握核心概念，避开常见错误。通过练习这些考试题型，你将增强信心，在单元测试中发挥得更好。
1. Cell Structures | 细胞结构
Question: The diagram shows an animal cell. Identify the parts labelled A, B and C. State one function of part B.
题目： 图中显示一个动物细胞。写出标号 A、B 和 C 的名称，并说出部分 B 的一项功能。
Answer: A – cytoplasm, B – nucleus, C – cell membrane. The nucleus controls all cell activities and contains genetic material (DNA).
答案： A – 细胞质，B – 细胞核，C – 细胞膜。细胞核控制细胞的一切活动并含有遗传物质（DNA）。
In an animal cell, the cytoplasm is the jelly-like substance where most chemical reactions occur. The nucleus is surrounded by its own membrane and acts as the control centre. The cell membrane is the outer boundary that controls what enters and leaves the cell. Many students confuse the nucleus with the vacuole, which is permanent and large in plant cells but small and temporary in animal cells.
在动物细胞中，细胞质是胶状物质，大多数化学反应在这里进行。细胞核由自身膜包裹，起着控制中心的作用。细胞膜是细胞的外边界，控制物质的进出。许多学生容易把细胞核和液泡混淆——植物细胞中的液泡大且永久存在，而动物细胞中的液泡小而临时。
If the question asks you to label a diagram of a plant cell, you must also identify the cell wall, vacuole and chloroplasts. Remember that chloroplasts contain chlorophyll for photosynthesis and are absent in animal cells. Always relate the structure to the function, e.g. the cell wall provides support and is made of cellulose.
如果题目要求标注植物细胞，还必须辨认细胞壁、液泡和叶绿体。记住叶绿体含有叶绿素，用于光合作用，动物细胞中没有。始终将结构与功能联系起来，例如细胞壁由纤维素构成，提供支撑。
2. Levels of Organisation | 结构层次
Question: Place these terms in the correct order from smallest to largest: organ, cell, tissue, organ system, organism. Give one example of an organ system in the human body and list two organs within it.
题目： 将下列术语按由小到大的顺序排列：器官、细胞、组织、器官系统、生物体。举出人体的一个器官系统并列出其中的两个器官。
Answer: cell → tissue → organ → organ system → organism. Example: digestive system – stomach and small intestine (also accept: circulatory system – heart and blood vessels; respiratory system – lungs and trachea).
答案： 细胞 → 组织 → 器官 → 器官系统 → 生物体。示例：消化系统——胃和小肠（也可接受：循环系统——心脏和血管；呼吸系统——肺和气管）。
The key is to remember that similar cells form a tissue, different tissues work together to form an organ, several organs make up an organ system, and all systems combine to form the organism. Students often mix up the middle steps, so using a mnemonic like ‘Cells That Open Organised Systems’ can help. Making a labelled diagram with arrows often earns extra marks in written questions.
关键在于记住：相似的细胞构成组织，不同的组织共同形成器官，多个器官组成器官系统，所有系统结合构成生物体。学生常混淆中间步骤，使用助记口诀如“细胞组织器官系统生物体”会很有帮助。在书面题中画出带箭头的标注图往往能获得额外分数。
3. Photosynthesis and Gas Exchange | 光合作用与气体交换
Question: A plant is kept in bright light. Explain what happens to the concentration of oxygen and carbon dioxide inside the leaf during the day. Use the word equation for photosynthesis to support your answer.
题目： 一株植物放在强光下。解释白天叶片内部氧气和二氧化碳浓度的变化。用光合作用的文字方程式来支持你的答案。
Answer: During the day, the rate of photosynthesis is high because light is available. The plant uses carbon dioxide and water to produce glucose and oxygen: carbon dioxide + water → glucose + oxygen. Therefore, carbon dioxide concentration decreases and oxygen concentration increases inside the leaf. Some oxygen is used in respiration, but photosynthesis produces more oxygen than respiration consumes.
答案： 白天因为有光，光合作用速率很高。植物利用二氧化碳和水生成葡萄糖和氧气：二氧化碳 + 水 → 葡萄糖 + 氧气。因此，叶片内二氧化碳浓度下降，氧气浓度上升。部分氧气用于呼吸作用，但光合作用产生的氧气多于呼吸作用消耗的氧气。
A common error is to forget that plants respire all the time. At night, photosynthesis stops, so oxygen decreases and carbon dioxide increases due to respiration. Many exam questions test the balance between the two processes. Remember that the exchange of gases happens through tiny pores called stomata, mainly on the underside of the leaf.
常见的错误是忘记植物时刻都在进行呼吸作用。夜晚光合作用停止，由于呼吸作用，氧气浓度下降而二氧化碳浓度上升。许多考题考查这两个过程的平衡。记住气体交换通过称为气孔的小孔进行，气孔主要分布在叶片背面。
4. Food Chains and Food Webs | 食物链与食物网
Question: A garden food chain is: grass → snail → thrush → sparrowhawk. (a) Name the producer. (b) What would happen to the number of thrushes if all the snails died from a disease? (c) Add labels to show the trophic levels using the terms ‘primary consumer’ and ‘tertiary consumer’.
题目： 花园中的一条食物链为：草 → 蜗牛 → 画眉鸟 → 雀鹰。(a) 说出生产者的名称。(b) 如果所有的蜗牛因病死亡，画眉鸟的数量会发生什么变化？(c) 使用“初级消费者”和“三级消费者”标注营养级。
Answer: (a) Grass is the producer because it makes its own food by photosynthesis. (b) The number of thrushes would decrease because their food source (snails) is no longer available. (c) Snail – primary consumer; thrush – secondary consumer; sparrowhawk – tertiary consumer.
答案： (a) 草是生产者，因为它通过光合作用自己制造食物。(b) 画眉鸟的数量会减少，因为它们的食物（蜗牛）不复存在。(c) 蜗牛——初级消费者；画眉鸟——次级消费者；雀鹰——三级消费者。
Always read food chains in the direction of energy flow, from producer to top predator. If you are asked about the effect of removing one organism, think about what it eats and what eats it. Using arrows that point from the eaten to the eater is essential; reversed arrows lose marks. The sun is the ultimate source of energy for nearly all food chains.
阅读食物链时始终顺着能量流动方向，从生产者到顶级捕食者。如果被问及去除某一种生物的影响，思考它吃什么以及什么吃它。箭头从被食者指向捕食者至关重要，箭头反向会失分。太阳是几乎所有食物链的最终能量来源。
5. Digestive System and Absorption | 消化系统与吸收
Question: Describe the journey of a piece of bread through the digestive system, naming the organs in order. Explain the role of enzymes and villi in digestion and absorption.
题目： 描述一片面包在消化系统中的旅程，按顺序说出所经过的器官。解释酶和绒毛在消化和吸收中的作用。
Answer: Mouth (chewing and saliva) → oesophagus (food pipe) → stomach (churning and enzyme action) → small intestine (digestion by enzymes and absorption through villi) → large intestine (water absorption) → rectum (storage of faeces) → anus (egestion). Enzymes break down large food molecules into smaller, soluble ones. Villi are finger-like projections in the small intestine that increase the surface area for absorption of digested food into the blood.
答案： 口腔（咀嚼和唾液）→ 食道（食管）→ 胃（搅动和酶作用）→ 小肠（酶消化和通过绒毛吸收）→ 大肠（吸收水分）→ 直肠（储存粪便）→ 肛门（排遗）。酶将大的食物分子分解为较小的可溶性分子。绒毛是小肠中手指状的突起，增大表面积，促进消化后的食物吸收进入血液。
Students often confuse egestion (removal of undigested waste) with excretion (removal of metabolic waste like carbon dioxide). Bread contains mainly starch, which is broken down by amylase in the mouth and small intestine. Villi contain blood capillaries that carry absorbed nutrients away. A model or diagram of villi will help you remember how surface area aids efficient absorption.
学生常混淆排遗（排出未消化的废物）和排泄（排出代谢废物如二氧化碳）。面包主要含有淀粉，淀粉被口腔和小肠中的淀粉酶分解。绒毛内含有血液毛细血管，运走吸收的营养物质。绒毛的模型或示意图有助于记住表面积如何促进高效吸收。
6. Sexual and Asexual Reproduction | 有性生殖与无性生殖
Question: A gardener grows new rose bushes by taking cuttings from a parent plant. Explain one advantage and one disadvantage of this method compared to growing from seeds. Identify the type of reproduction involved.
题目： 一位园丁从亲本植株上剪取插条来培育新玫瑰。与种子繁殖相比，解释这种方法的一个优点和一个缺点。指出所涉及的生殖类型。
Answer: This is asexual reproduction because there is only one parent and no fusion of gametes. Advantage: the offspring are genetically identical to the parent, so all desirable characteristics (e.g. flower colour, disease resistance) are kept. Disadvantage: no genetic variation, so if a disease attacks, all plants could be wiped out. Seeds allow variation and adaptation, but offspring may not be exactly like the parent.
答案： 这是无性生殖，因为只有一个亲本，没有配子融合。优点：后代与亲本遗传上完全相同，因此所有优良性状（如花的颜色、抗病性）都得以保留。缺点：没有遗传变异，因此一旦病害侵袭，所有植株可能全部死亡。种子繁殖允许变异和适应，但后代可能不完全像亲本。
In sexual reproduction, the fusion of male and female gametes produces offspring with a mix of characteristics from both parents, leading to variation. Common examples of asexual reproduction in plants include runners (strawberries), tubers (potatoes) and bulbs (onions). You should be able to compare the two methods clearly with reference to the number of parents and genetic similarity.
在有性生殖中，雄性和雌性配子融合产生具有父母双方混合特征的后代，导致变异。植物无性生殖的常见例子包括匍匐茎（草莓）、块茎（马铃薯）和鳞茎（洋葱）。你应能清晰地比较这两种生殖方式，涉及亲本数量和遗传相似性。
7. Basic Inheritance | 基础遗传
Question: In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). Two heterozygous tall plants are crossed. Draw a Punnett square to show the possible offspring and state the probability of a short plant.
题目： 在豌豆中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。两株杂合高茎豌豆杂交。画出庞纳特方格，显示可能的后代并说出矮茎植株的概率。
Answer: Parent genotypes: Tt × Tt. Punnett square gives TT, Tt, Tt, tt. The probability of a short plant (tt) is 1/4 or 25%. The phenotypic ratio of tall to short is 3:1. The allele for shortness is recessive and only appears when an individual inherits two recessive alleles.
答案： 亲本基因型：Tt × Tt。庞纳特方格得出 TT, Tt, Tt, tt。矮茎植株 (tt) 的概率是 1/4 或 25%。高茎与矮茎的表现型比例是 3:1。矮茎等位基因是隐性的，仅当个体遗传到两个隐性等位基因时才表现出来。
Make sure you can differentiate between genotype (the pairs of alleles) and phenotype (the physical appearance). When drawing a Punnett square, always place the parental gametes along the top and side. Many students forget to write the probability as a fraction or percentage. Remember that dominant alleles mask recessive ones in a heterozygous individual.
务必区分基因型（等位基因的组合）和表现型（外在特征）。画庞纳特方格时，始终将亲本的配子写在顶部和侧边。许多学生忘记将概率写成分数或百分数。记住在杂合个体中，显性等位基因会掩盖隐性等位基因的表现。
8. Ecological Roles | 生态系统的角色
Question: A pond ecosystem contains algae, water fleas, small fish and a heron. (a) Classify each organism as producer, primary consumer, secondary consumer or tertiary consumer. (b) Name one decomposer that might live in the pond and explain its importance.
题目： 一个池塘生态系统含有藻类、水蚤、小鱼和苍鹭。(a) 将每种生物划分为生产者、初级消费者、次级消费者或三级消费者。(b) 说出可能生活在池塘中的一种分解者，并解释它的重要性。
Answer: (a) Algae – producer, water fleas – primary consumer, small fish – secondary consumer, heron – tertiary consumer. (b) Bacteria or fungi are decomposers. They break down dead organic matter and waste, returning mineral nutrients (such as nitrates) to the water for algae to use. Without decomposers, nutrients would remain locked in dead organisms and the ecosystem would collapse.
答案： (a) 藻类——生产者，水蚤——初级消费者，小鱼——次级消费者，苍鹭——三级消费者。(b) 细菌或真菌是分解者。它们分解死去的有机物和废物，将矿物养分（如硝酸盐）释放回水中供藻类利用。没有分解者，养分将锁定在死去的生物体内，生态系统将会崩溃。
Decomposers are often tested, and students frequently forget their vital role. They are not shown in food chain diagrams but are absolutely essential. Think of them as nature’s recyclers. When asked about the effect of removing a trophic level, always consider the knock-on effect on all other levels, including the producers.
分解者是常考的知识点，学生常常忘记它们的重要作用。它们不显示在食物链图中，但绝对不可或缺。把它们看作大自然的回收者。当被问及去除某一营养级的影响时，始终考虑对其他所有营养级（包括生产者）的连锁效应。
9. Plant Life Cycle and Pollination | 植物的生命周期与传粉
Question: A bee visits a flower. Explain how this helps the plant reproduce. Describe what happens inside the flower after pollination and before seed dispersal.
题目： 一只蜜蜂访问了一朵花。解释这如何帮助植物繁殖。描述传粉之后、种子传播之前花内部发生了什么。
Answer: The bee transfers pollen grains from the anther of one flower to the stigma of another flower of the same species. This is cross-pollination. After pollination, a pollen tube grows down the style, and the male gamete travels to the ovary to fuse with the female gamete (ovule). This is fertilisation. The fertilised ovule develops into a seed, and the ovary develops into a fruit.
答案： 蜜蜂将花粉粒从一朵花的雄蕊花药传递到同种另一朵花的雌蕊柱头上。这是异花传粉。传粉之后，花粉管沿花柱向下生长，雄配子到达子房与雌配子（胚珠）融合，这就是受精。受精后的胚珠发育成种子，子房发育成果实。
Many students mix up pollination and fertilisation. Pollination is simply the transfer of pollen; fertilisation is the fusion of gametes. You should be able to label the parts of an insect-pollinated flower: petals (brightly coloured to attract insects), anther (produces pollen), filament, stigma (sticky to catch pollen), style, ovary, ovule and nectary.
很多学生混淆传粉和受精。传粉仅仅是花粉的传递；受精是配子的融合。你应能标注虫媒花的各个部分：花瓣（颜色鲜艳以吸引昆虫）、花药（产生花粉）、花丝、柱头（有粘性以捕捉花粉）、花柱、子房、胚珠和蜜腺。
10. Adaptations for Survival | 生存适应
Question: A camel lives in a hot desert. Explain three ways the camel is adapted to survive with very little water. For each adaptation, state how it helps the camel conserve water or stay cool.
题目： 骆驼生活在炎热的沙漠。解释骆驼在缺水环境中生存的三种适应方式。对每项适应，说明它如何帮助骆驼节约水分或保持凉爽。
Answer: 1. Camels have long eyelashes and closable nostrils that keep out blowing sand and reduce water loss from the respiratory surfaces. 2. They produce very concentrated urine and dry faeces, which minimises water loss from the body. 3. Camels store fat in their humps, not water. When the fat is broken down during respiration, metabolic water is released. 4. They can tolerate large changes in body temperature, reducing the need for sweating. 5. Their thick fur insulates against the sun’s heat. (Any three with correct reasoning.)
答案： 1. 骆驼有长长的睫毛和可以关闭的鼻孔，阻挡风沙并减少呼吸面的水分蒸发。2. 它们产生高度浓缩的尿液和干燥的粪便，最大限度地减少水分从体内流失。3. 骆驼将脂肪储存在驼峰中而非水分。脂肪在呼吸作用分解时会释放代谢水。4. 它们能耐受较大的体温变化，减少出汗的需求。5. 它们厚实的皮毛能隔绝太阳的热量。（任写三项并给出正确理由皆可。）
The most frequent misunderstanding is that the hump stores water. It does not; it stores fat. When answering adaptation questions, always link the feature to the environmental challenge. Use the phrase ‘so that’ or ‘which allows’ to make the connection explicit. You might also be asked about adaptations of polar animals, such as thick blubber and white fur for insulation and camouflage.
最常见的误解是驼峰储水。实际上它储存的是脂肪。回答适应性问题时，始终将特征与环境挑战联系起来。使用“以便”或“这使得”等短语来明确揭示这种联系。你还可能被问到极地动物的适应，比如厚脂肪层和白毛皮用于保温和伪装。
11. Experimental Design and Variables | 实验设计与变量
Question: A student wants to test whether light intensity affects the rate of photosynthesis in pondweed. Describe a fair test, identifying the independent, dependent and control variables. Suggest an appropriate safety precaution.
题目： 一位学生想测试光照强度是否影响水草的光合作用速率。描述一个公平实验，明确自变量、因变量和控制变量。提出一项合适的安全措施。
Answer: Independent variable: light intensity (distance of lamp from pondweed). Dependent variable: rate of photosynthesis (measured by counting oxygen bubbles produced per minute or using a gas syringe). Control variables: same piece of pondweed, same temperature (use a water bath), same concentration of carbon dioxide (add sodium hydrogencarbonate), wait for the plant to equilibrate. Safety: be careful with water near electrical equipment, or handle hot lamp with care. Using LED lamps avoids heating the water, which would introduce a second variable.
答案： 自变量：光照强度（灯到水草的距离）。因变量：光合作用速率（通过每分钟计数产生的氧气气泡数或使用气体注射器测量）。控制变量：同一段水草，相同温度（使用恒温水浴），相同二氧化碳浓度（加入碳酸氢钠），等待植物稳定。安全：注意水电安全，或小心烫手。使用 LED 灯可避免加热水体，否则会引入第二个变量。
In any investigation question, you must clearly identify the variables and explain how you will keep control variables constant. Always refer to the table of results or the graph if provided. Students often describe the method without mentioning why a step is done; adding ‘so that’ shows you understand the purpose. A fair test means only the independent variable is changed.
在任何探究题中，你必须清晰识别变量并解释如何使控制变量保持恒定。如题目提供了结果表或图表，始终要加以引用。学生描述方法时常常不说为什么要做某一步；加上“以便”能显示你理解该步骤的目的。公平实验意味着只改变自变量。
12. Data Interpretation and Conclusions | 数据分析与结论
Question: The table shows the number of bubbles released by pondweed at different light intensities. Light intensity (lux): 200, 400, 600, 800, 1000; Bubbles per minute: 5, 12, 18, 18, 18. Describe the pattern and suggest a limiting factor at the plateau.
题目： 表格显示了不同光照强度下水草释放的气泡数。光照强度（勒克斯）：200, 400, 600, 800, 1000；每分钟气泡数：5, 12, 18, 18, 18。描述变化规律并提出在平台期起限制作用的一个因素。
Answer: As light intensity increases from 200 to 600 lux, the rate of photosynthesis increases. Beyond 600 lux, the rate remains constant at 18 bubbles per minute, forming a plateau. At this point, light intensity is no longer the limiting factor; something else, such as carbon dioxide concentration or temperature, is limiting the rate. A conclusion must state that photosynthesis is affected by light only up to a certain point.
答案： 当光照强度从 200 勒克斯增加到 600 勒克斯时，光合作用速率升高。超过 600 勒克斯后，速率保持在每分钟 18 个气泡，形成平台。此时，光照强度不再是限制因素；其他因素，如二氧化碳浓度或温度，限制了速率。结论必须指出光合作用受光的影响仅到某个临界点。
When interpreting graphs, use precise terms like ‘increase’, ‘decrease’, ‘remain constant’, ‘plateau’. Never say ‘it goes up and stays the same’ in a disjointed way. Always relate the data to biological knowledge. If asked about reliability, mention repeats and calculating a mean. A strong conclusion uses data (e.g. numbers from the table) to support the statement.
在解读图表时，使用精确的术语，如“增加”、“减少”、“保持不变”、“平台”。切莫以割裂的方式说“它上升然后保持不变”。始终把数据与生物学知识联系起来。如果被问到可靠性，要提及重复实验并计算平均值。强有力的结论会使用数据（例如表格中的数字）来支持陈述。
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