Year 9 Cambridge Biology: Mock Test Walkthrough | 剑桥Year 9生物：单元测试模拟卷解析

📚 Year 9 Cambridge Biology: Mock Test Walkthrough | 剑桥Year 9生物：单元测试模拟卷解析

Welcome to the walkthrough of a Year 9 Cambridge Biology unit test mock paper. This analysis will guide you through typical questions, highlight key concepts, and help you refine your exam technique. Each item is broken down so you can see exactly what the examiner expects.

欢迎阅读剑桥Year 9生物单元测试模拟卷解析。本文将带你逐一分析典型题目，强调关键概念，并帮助你优化答题技巧。每一项都被拆解开来，让你清晰了解考官的期望。

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1. Characteristics of Life: Sensitivity vs Movement | 生命的特征：敏感性与运动

Question: A woodlouse moves away from a torch beam. Which characteristic of a living organism is demonstrated? A Movement B Sensitivity C Growth D Reproduction

问题： 一只鼠妇在手电筒光束下移开。这体现了生物的哪一特征？A 运动 B 感性 C 生长 D 生殖

The correct answer is B Sensitivity. While the woodlouse does move, the key is that it detects the light stimulus and responds accordingly. Movement in living organisms is often a response to a stimulus; it is the ability to sense and react that defines sensitivity. Simply being able to move does not make a living thing sensitive – robots move without sensation. Therefore, option A is too broad, and options C and D are not shown here.

正确答案是 B 感性。虽然鼠妇确实移动了，但关键在于它侦测到光刺激并作出相应反应。生物的移动往往是对刺激的反应；定义感性的是感知并反应的能力。仅仅能移动并不能说明物体具有感性——机器人无感觉却可移动。因此选项A过于宽泛，选项C和D在此并未体现。

Remember the seven life processes using MRS GREN: Movement, Respiration, Sensitivity, Growth, Reproduction, Excretion and Nutrition. In an exam, always identify which specific process is being described. If a seedling grows taller, the characteristic is growth; if a plant roots grow towards water, it is sensitivity.

记住七个生命活动 MRS GREN：运动、呼吸、感性、生长、生殖、排泄和营养。考试中务必确认所描述的具体是哪一过程。若幼苗长高，特征是生长；若植物根向水生长，则是感性。

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2. Cell Structure: Plant vs Animal Cells | 细胞结构：植物细胞与动物细胞

Question: Which structure is present in a palisade leaf cell but not in a human cheek cell? A Nucleus B Cytoplasm C Cell membrane D Chloroplast

问题： 哪种结构存在于栅栏叶肉细胞中，但不存在于人口腔上皮细胞中？A 细胞核 B 细胞质 C 细胞膜 D 叶绿体

The answer is D Chloroplast. Palisade cells are plant cells specialised for photosynthesis; they contain many chloroplasts, which house the green pigment chlorophyll. Cheek cells are animal cells and never contain chloroplasts. All the other options – nucleus, cytoplasm and cell membrane – are found in both plant and animal cells.

答案是 D 叶绿体。栅栏细胞是特化用于光合作用的植物细胞；它们含有大量叶绿体，叶绿体内含绿色色素叶绿素。口腔上皮细胞是动物细胞，绝不含叶绿体。其余选项——细胞核、细胞质和细胞膜——在植物和动物细胞中均存在。

Feature / 特征	Palisade cell / 栅栏细胞	Cheek cell / 口腔上皮细胞
Cell wall / 细胞壁	Yes (cellulose) / 有（纤维素）	No / 无
Chloroplasts / 叶绿体	Yes / 有	No / 无
Large central vacuole / 大中央液泡	Yes / 有	No (small temporary vacuoles) / 无（小型暂时液泡）

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3. Classification and Dichotomous Keys | 分类与二分法检索表

Question: Use the key to identify organism X: 1a Has wings … go to 2; 1b No wings … go to 3; 2a Wings with coloured scales … butterfly; 2b Wings clear … dragonfly; 3a 6 legs … ant; 3b More than 6 legs … woodlouse. Organism X has wings that are transparent and membranous. What is X?

问题： 请使用检索表识别生物X：1a 有翅…至2；1b 无翅…至3；2a 翅具彩色鳞片…蝴蝶；2b 翅透明…蜻蜓；3a 6条腿…蚂蚁；3b 腿多于6条…鼠妇。生物X具有透明、膜状的翅。X是什么？

The organism is a dragonfly. Because it has wings, you follow 1a to step 2. At step 2, you encounter the choice between coloured scales and clear wings. Since X has transparent wings, 2b applies, leading directly to dragonfly. Dichotomous keys must be followed step by step, choosing one statement at each branch. Never skip a step.

该生物是 蜻蜓。因其有翅，故沿1a到步骤2。在步骤2，需在彩色鳞片和透明翅之间选择。由于X的翅透明，适用2b，直接得出蜻蜓。二分法检索表必须逐步进行，在每个分支处选择一条陈述，切勿跳步。

Keys are constructed by scientists to identify unknown organisms based on observable characteristics. If a specimen does not match any pathway perfectly, re-examine the feature carefully – you may have misread a measurement or colour.

检索表是科学家基于可观察特征构建的，用以鉴别未知生物。若标本不能完美匹配任一路径，需仔细复查特征——你可能误读了某尺寸或颜色。

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4. Food Chains and Energy Transfer | 食物链与能量传递

Question: In the food chain grass → rabbit → fox, which statement is correct? reases at each trophic level. D The grass transfers energy from sunlight to the ecosystem.

问题： 在食物链“草→兔→狐狸”中，哪一陈述正确？A 兔是生产者。B 狐狸是初级消费者。C 能量在每一营养级增加。D 草将太阳光能传递入生态系统。

The answer is D. Grass, as a green plant, carries out photosynthesis and converts light energy into chemical energy stored in its tissues; it is the producer. Rabbits eat grass, making them primary consumers. Foxes are secondary consumers. Energy is lost (about 90%) at each trophic level, primarily as heat from respiration, so it does not increase – option C is incorrect.

答案是 D。草作为绿色植物进行光合作用，将光能转化为化学能储存在组织中；它是生产者。兔吃草，故为初级消费者。狐狸是次级消费者。能量在每一营养级均会损失（约90%），主要以呼吸产热形式散失，因而能量不会增加——选项C错误。

Grass (Producer) → Rabbit (Primary consumer) → Fox (Secondary consumer)

草（生产者）→ 兔（初级消费者）→ 狐狸（次级消费者）

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5. Plant Transport Systems: Xylem and Phloem | 植物运输系统：木质部和韧皮部

Question: Which tissue transports water and dissolved mineral ions from the roots to the leaves? A Phloem B Xylem C Epidermis D Stomata

问题： 哪种组织将水和溶解的矿质离子从根运输到叶？A 韧皮部 B 木质部 C 表皮 D 气孔

The correct choice is B Xylem. Xylem vessels are hollow, dead tubes strengthened with lignin; they form continuous pipelines from the root hairs upward. Water moves by transpiration pull. Mineral ions are actively taken up by root hair cells using energy, then carried in the water stream. Phloem transports sugars (sucrose) and amino acids in both directions, so A is not correct for water transport.

正确选项是 B 木质部。木质部导管是由死细胞构成的中空管，壁具木质素加固；它们从根毛向上形成持续的管道。水分因蒸腾拉力而移动。矿质离子由根毛细胞耗能主动吸收，随后随水流向上运输。韧皮部运输糖类（蔗糖）和氨基酸，且是双向的，因此A不负责水分运输。

Remember the adaptation of xylem: they have no end walls, which allows an uninterrupted flow. Root hair cells increase surface area for osmosis and active uptake. In an exam question on transpiration, link the effect of temperature or wind to the rate of water loss.

记住木质部的适应特征：无端壁，使液流无阻碍。根毛细胞增大表面积，利于渗透和主动吸收。在涉及蒸腾作用的考题中，要将温度或风的影响与失水速率联系起来。

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6. Human Digestive System: Bile and Emulsification | 人体消化系统：胆汁与乳化

Question: Bile is produced by the liver and stored in the gall bladder. What is the role of bile in digestion? A It breaks starch into glucose. B It emulsifies fats. C It kills bacteria in food. D It absorbs water.

问题： 胆汁由肝生成，储存在胆囊中。胆汁在消化中的作用是什么？A 将淀粉分解为葡萄糖。B 乳化脂肪。C 杀灭食物中的细菌。D 吸收水分。

Bile emulsifies fats (answer B). It breaks large fat globules into smaller droplets, greatly increasing the surface area for lipase enzymes to act on. This is physical digestion, not chemical – bile contains no enzymes. Lipase then chemically digests fats into fatty acids and glycerol. Without bile, fat digestion would be extremely slow.

胆汁可 乳化脂肪（答案B）。它将大脂肪球分解成小液滴，极大增加脂肪酶作用的表面积。这是物理消化，而非化学消化——胆汁不含酶。随后脂肪酶将脂肪化学消化为脂肪酸和甘油。若无胆汁，脂肪消化将极其缓慢。

In the mock test you might also be asked about the site of absorption. The majority of digested nutrients, including glucose and amino acids, are absorbed in the small intestine through villi. The large intestine mainly absorbs water.

模拟考中你也可能被问及吸收部位。绝大多数已消化养分，包括葡萄糖和氨基酸，通过绒毛在小肠被吸收。大肠主要吸收水分。

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7. Respiration: Anaerobic in Animals | 呼吸作用：动物无氧呼吸

Question: During vigorous exercise, muscle cells may no longer receive enough oxygen. Which substance is produced by anaerobic respiration in muscle cells? A Ethanol B Lactic acid C Carbon dioxide D Water

问题： 剧烈运动时，肌肉细胞可能无法获得足够氧气。肌肉细胞无氧呼吸会产生哪种物质？A 乙醇 B 乳酸 C 二氧化碳 D 水

The product is lactic acid (B). Animal cells respire anaerobically to release a small amount of energy quickly, but lactic acid builds up and causes muscle fatigue and cramps. Anaerobic respiration in yeast produces ethanol and carbon dioxide, not lactic acid. Note that anaerobic respiration in animals does not release carbon dioxide.

产物是 乳酸（B）。动物细胞进行无氧呼吸以快速释放少量能量，但乳酸堆积会引起肌肉疲劳和痉挛。酵母的无氧呼吸产生乙醇和二氧化碳，而不是乳酸。请注意，动物的无氧呼吸不释放二氧化碳。

Word equation for anaerobic respiration in animals: Glucose → Lactic acid (+ some energy)

动物无氧呼吸的文字方程式：葡萄糖 → 乳酸（+ 少量能量）

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8. Sexual Reproduction in Flowering Plants | 开花植物的有性生殖

Question: The diagram shows a flower. After pollination and fertilisation, which labelled part develops into the seed? (Parts: A Stigma, B Style, C Ovary, D Ovule)

问题： 图示一朵花。传粉和受精后，哪一标注部分发育为种子？（A 柱头，B 花柱，C 子房，D 胚珠）

The ovule, D, becomes the seed. The ovary (C) swells to become the fruit. The fertilised egg cell inside the ovule divides to form the embryo plant, and the surrounding ovule tissues harden into the seed coat. The stigma and style have no direct role after fertilisation.

胚珠 D 发育为种子。子房（C）膨大成为果实。胚珠内受精卵细胞分裂形成植物胚，胚珠外围组织变硬形成种皮。柱头和花柱在受精后无直接作用。

Understand the difference between pollination (transfer of pollen to stigma) and fertilisation (fusion of male and female gametes). Wind-pollinated flowers have small, dull petals and large, feathery stigmas; insect-pollinated flowers are brightly coloured and produce nectar.

要理解传粉（花粉传递到柱头）与受精（雌雄配子融合）的区别。风媒花花瓣小而暗淡，柱头大而呈羽毛状；虫媒花颜色鲜艳且产花蜜。

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9. Ecological Sampling: Quadrats | 生态采样：样方

Question: A student places a 0.5 m² quadrat in ten random locations in a field and counts the number of dandelions in each. What is the main purpose of using random quadrats? A To record the exact number of dandelions in the field B To estimate the population size and distribution C To attract more dandelions D To remove weeds

问题： 一学生在田野中随机布设十个0.5平方米的样方，并计数每个样方内的蒲公英数量。使用随机样方的主要目的是什么？A 记录田野中蒲公英的确切数量 B 估计种群大小和分布 C 吸引更多蒲公英 D 除去杂草

The purpose is B a to be scaled up (e.g., mean per quadrat × number of quadrats that fit in the field). Students should know how to use random numbers to place quadrats and how to calculate mean, median and mode from their counts.

目的是 B，即估计种群大小和分布。计数每一株蒲公英通常不切实际。随机布设可避免偏差，并允许数据放大（如每样方平均数 × 田野可容纳的样方数）。学生应掌握如何使用随机数布设样方，以及如何根据计数计算平均数、中位数和众数。

If a question asks why the results might still be unreliable, recall factors such as: too few quadrats, non-random placement, or uneven distribution of the species.

若题目问及为何结果仍可能不可靠，要想到：样方数量过少、非随机布设、或物种分布不均等因素。

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10. Data Analysis: Heart Rate Investigation | 数据分析：心率调查

Question: A student recorded her heart rate (beats per minute) before, during and after running. Results: rest 72 bpm; immediately after running 132 bpm; 5 minutes after running 85 bpm. Identify the independent variable and explain the trend.

问题： 一学生记录了自己跑步前、跑步中和跑步后的心率（次/分）。结果：静息72 bpm；跑步后立即 132 bpm；跑步后5分钟 85 bpm。请确定自变量，并解释趋势。

The independent variable is time or activity level (what you change). The dependent variable is heart rate (what you measure). The trend shows that heart rate increases during exercise to supply muscle cells with more oxygen and glucose for respiration, and to remove the lactic acid and carbon dioxide produced. After exercise, the heart rate gradually returns to resting level as the oxygen debt is repaid.

自变量是 时间或活动水平（你所改变的变量）。因变量是心率（你所测量的变量）。趋势表明：运动时心率增高，以便向肌细胞供给更多氧气和葡萄糖用于呼吸作用，并移除产生的乳酸和二氧化碳。运动后随着氧债被偿还，心率逐渐恢复至静息水平。

When answering such graph questions, always quote data from the table or graph. For example: 'Heart rate rose from 72 to 132 bpm immediately after running.'

回答此类图表题时，务必引用表格或图中的数据。例如：“心率从跑步前的72 bpm上升至跑步后立即的132 bpm。”

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