📚 6.5 Ecosystems Exam Practice | 6.5 生态系统真题精练
Welcome to the ultimate exam practice guide for 6.5 Ecosystems. This resource walks you through key terminology, energy flow, productivity calculations, nutrient cycles and succession, all framed by typical A‑level questions and model responses. Work through each section carefully to sharpen your data analysis and longer‑answer technique.
欢迎来到 6.5 生态系统终极真题精练指南。本资源带你梳理关键术语、能量流动、生产力计算、营养循环和演替，全部以典型 A‑level 试题和标准答案呈现。认真完成每个部分，提升你的数据分析能力和长答题技巧。
1. Key Ecosystem Terminology | 生态系统关键术语
An ecosystem is a dynamic system made up of a community of living organisms (biotic factors) interacting with their non‑living environment (abiotic factors such as temperature, pH and light intensity). Within any ecosystem you must be able to distinguish between a population – all the individuals of a single species in a particular habitat – and a community – all the populations of different species that live and interact in the same area.
生态系统是一个动态系统，由生物群落（生物因素）与其非生物环境（非生物因素，如温度、pH 和光照强度）相互作用构成。在任何生态系统中，你必须能够区分种群——某一特定栖息地中同一物种的所有个体——和群落——生活在同一区域并相互作用的所有不同物种的种群。
The niche of an organism encompasses its habitat, its feeding relationships, its pattern of activity and its interactions with other species. Examiners often stress that a niche is not simply where an organism lives; it is its 'role' – how it fits into the ecosystem, including where it feeds, what it eats and when it is active. Two species cannot occupy exactly the same niche for long in the same habitat because competitive exclusion will result.
生物的生态位包括其栖息地、食谱关系、活动规律以及与其他物种的相互作用。考官常强调生态位不仅仅是一个生物居住的地方；它是它的“角色”——它如何融入生态系统，包括在哪里取食、吃什么和何时活动。在同一个栖息地内，两个物种无法长期占据完全相同的生态位，因为会发生竞争排斥。
2. Energy Flow and Food Webs | 能量流与食物网
All ecosystems rely on a supply of energy, almost always from sunlight. Producers (autotrophs) trap light energy during photosynthesis and convert it to chemical energy in organic molecules. This energy is then transferred when organisms feed on one another, forming food chains and complex food webs. It is vital to remember that arrows in a food chain represent energy flow, not simply 'who eats whom'.
所有生态系统都依赖能量供应，能量几乎总是来自阳光。生产者（自养生物）在光合作用中捕获光能，并将其转化为有机物中的化学能。当生物以其他生物为食时，这种能量便沿着食物链和复杂的食物网传递。至关重要的是要记住，食物链中的箭头代表能量流动，而不仅仅是“谁吃谁”。
lost, so food chains rarely exceed four or five trophic levels. Being able to construct and interpret a food web from given organisms is a common exam task.
一个典型的陆地食物链可能是：草 → 兔 → 狐狸。在海洋环境中，你常常见到浮游植物 → 浮游动物 → 小鱼 → 大鱼 → 海豹。每一步都有能量损失，因此食物链很少超过四或五个营养级。能否根据给定的生物构建并解读食物网是一项常见的考试任务。
3. Trophic Levels and Energy Loss | 营养级与能量损失
When energy passes from one trophic level to the next, only about 10% is transferred to new biomass; the rest is lost. The energy is lost through respiration (released as heat), through undigested material in faeces, and because not all of an organism is consumed (e.g. bones, roots). This low transfer efficiency explains why biomass and numbers of organisms decrease at higher trophic levels.
当能量从一个营养级传递到下一个营养级时，只有大约10% 转化为新的生物量；其余的能量都损失了。能量通过呼吸作用（以热的形式释放）、粪便中未消化的物质以及部分未被消费的个体（如骨头、根）而损失。这种低传递效率解释了为什么生物量和生物个体数在较高营养级会减少。
The standard formula for ecological efficiency is: efficiency (% ate efficiencies from data presented in tables or pyramids. A classic question gives the energy content of producers and primary consumers and asks for the transfer efficiency.
生态传递效率的标准公式为：效率 (%) = (第 n 营养级生物量中的能量 / 第 n−1 营养级生物量中的能量) × 100。你必须能够根据表格或金字塔中的数据计算效率。一道经典题目是给出生产者和初级消费者的能量含量，然后要求计算传递效率。
contains 2 340 kJ m⁻² yr⁻¹. Calculate the efficiency. Answer: (2340 / 18000) × 100 = 13.0%.
例题：在草地中，生产者营养级含有 18 000 kJ m⁻² yr⁻¹；初级消费者营养级含有 2 340 kJ m⁻² yr⁻¹。计算其效率。答案：(2340 / 18000) × 100 = 13.0%。
4. Pyramids of Numbers, Biomass and Energy | 数量、生物量和能量金字塔
Ecologists use three types of pyramids to represent feeding relationships: pyramid of numbers (counts of individuals), pyramid of biomass (dry mass per unit area, usually g m⁻² or kg m⁻²), and pyramid of energy (energy flow per unit area per unit time, kJ m⁻² yr⁻¹). The pyramid of energy is always upright because energy is lost at every transfer. The pyramid of biomass is almost always upright, but can be inverted if producers have very rapid turnover (e.g. phytoplankton). The pyramid of numbers can be upright, inverted or spindle‑shaped.
生态学家使用三种金字塔来呈现摄食关系：数量金字塔（个体数量）、生物量金字塔（单位面积干重，通常为 g m⁻² 或 kg m⁻²）以及能量金字塔（单位面积单位时间的能量流，kJ m⁻² yr⁻¹）。能量金字塔总是正的（上窄下宽），因为每次传递都损失能量。生物量金字塔几乎总是正的，但如果生产者的周转极快（如浮游植物），也可能倒置。数量金字塔可以是正的、倒置的或纺锤形。
Exam question: 'Explain why the pyramid of numbers for an oak tree may be inverted.' An oak tree is a single large producer, yet it can support thousands of insect consumers. Thus one large producer supports many consumers, producing an inverted pyramid for numbers.
考题：“解释为什么橡树的数量金字塔可能是倒置的。” 一棵橡树是单个大型生产者，但它可以支持成千上万的昆虫消费者。因此，一个大型生产者支持大量消费者，从而产生倒置的数量金字塔。
The table below summarises the three pyramids and their typical shape.
下表总结了三种金字塔及其典型形状。
| Pyramid type | What it shows | Shape | Notes |
|---|---|---|---|
| Numbers | Number of organisms at each trophic level | Variable | May be inverted when producers are few but large |
| Biomass | Dry mass of living material per unit area | Usually upright | Can invert if producer biomass is small but turnover is high |
| Energy | Energy content per unit area per unit time | Always upright | Energy is lost at each trophic level |
5. Calculating Primary and Secondary Productivity | 计算初级生产力和次级生产力
ionship is NPP = GPP − respiration. NPP is the energy available to the next trophic level.
总初级生产力 (GPP) 是单位面积单位时间内光合作用将光能转化为化学能的总量。植物将部分能量用于自身的呼吸作用；剩余部分为净初级生产力 (NPP)。其关系为 NPP = GPP − 呼吸消耗。NPP 是可供下一营养级利用的能量。
ata may ask you to compare efficiencies of different livestock or contrasting food chains.
次级生产力指消费者产生生物量的速率。次级生产效率的公式为：效率 = (并入消费者新生物量中的能量 / 摄入或同化的能量) × 100。考试数据可能要求你比较不同牲畜或不同食物链的效率。
ulate the efficiency based on ingested energy. Answer: (8.0 × 10³ / 2.0 × 10⁵) × 100 = 4.0%.
典型计算：一头奶牛每年摄入草中的能量 2.0 × 10⁵ kJ。它同化 1.2 × 10⁵ kJ，并将 8.0 × 10³ kJ 转化为新的组织。计算基于摄入能量的效率。答案：(8.0 × 10³ / 2.0 × 10⁵) × 100 = 4.0%。
6. The Carbon Cycle | 碳循环
Carbon moves between the atmosphere, oceans, living organisms and fossil deposits. The key pools are atmospheric CO₂, dissolved CO₂ in water, organic carbon in living things, and carbonate rocks. Photosynthesis removes CO₂ from the atmosphere; respiration, decomposition and combustion return it. Understanding the role of decomposers (bacteria and fungi) is critical: they break down dead organic matter extracellularly and release CO₂ through respiration, while also returning mineral ions to the soil.
碳在大气、海洋、生物体和化石贮库之间移动。关键碳库为大气中的 CO₂、水中溶解的 CO₂、生物体内的有机碳以及碳酸盐岩石。光合作用从大气中移除 CO₂；呼吸作用、分解作用和燃烧将 CO₂ 送回大气。理解分解者（细菌和真菌）的作用至关重要：它们在细胞外分解死亡有机质，并通过呼吸作用释放 CO₂，同时将矿质离子归还土壤。
A standard six‑mark question might read: 'Describe the processes that contribute to the cycling of carbon in an oak woodland ecosystem.' A full answer would include photosynthesis by leaves, respiration by all organisms, decomposition of leaf litter, soil respiration by microbes, feeding relationships, and, where applicable, combustion if a fire occurs.
一道标准的六分题可能是：“描述橡树林地生态系统中参与碳循环的过程。” 一份完整的答案需要包括树叶的光合作用、所有生物的呼吸作用、落叶的分解、土壤微生物的呼吸、摄食关系，以及如适用火灾时的燃烧过程。
7. The Nitrogen Cycle | 氮循环
Nitrogen is required for the synthesis of proteins and nucleic acids, yet most organisms cannot use atmospheric N₂ directly. The nitrogen cycle includes four main processes: nitrogen fixation (conversion of N₂ to ammonia NH₃, then ammonium NH₄⁺, by free‑living bacteria such as Azotobacter or mutualistic Rhizobium in root nodules); nitrification (oxidation of NH₄⁺ to nitrite NO₂⁻ and then to nitrate NO₃⁻ by nitrifying bacteria); assimilation (uptake of NO₃⁻ by plants, then through food chains); and denitrification (reduction of NO₃⁻ to N₂ gas by denitrifying bacteria under anaerobic conditions).
氮是合成蛋白质和核酸所必需的，但大多数生物不能直接利用大气中的 N₂。氮循环包括四个主要过程：固氮作用（由自由生活的细菌如固氮菌，或与豆科植物共生的根瘤菌，将 N₂ 转化为氨 NH₃，再转化为铵 NH₄⁺）；硝化作用（硝化细菌将 NH₄⁺ 氧化为亚硝酸盐 NO₂⁻ 再氧化为硝酸盐 NO₃⁻）；同化作用（植物吸收 NO₃⁻，然后进入食物链）；以及反硝化作用（反硝化细菌在厌氧条件下将 NO₃⁻ 还原为 N₂ 气）。
Ammonification also occurs when decomposers break down proteins and urea from dead organisms and excreta, releasing NH₄⁺. A frequent exam trap is confusing nitrification with denitrification: nitrification produces NO₃⁻, denitrification removes it and returns N₂ to the air. Remember that nitrogen‑fixing bacteria contain the enzyme nitrogenase, which is inhibited by oxygen, hence the root nodule structure protects the enzyme.
当分解者分解来自死亡生物
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