GCSE Edexcel生物 生态学 样方 样带 食物链
生态学是 Edexcel GCSE 生物里少数既要求背准定义、又要求算对数据的章节，Paper 2 的数据题经常从这一章取材。按考试里实际出现的顺序，把定义、调查方法、计算和物质循环串起来看，比单背一份术语清单好用。
个体 种群 群落 生态系统
生态学的第一组定义必须背准，因为后面每一道题都压在这几个词的边界上。个体指单个生物体。种群是同一物种、同一区域、同一时间的全部个体，比如一片草地上所有的蒲公英。群落是同一区域内所有种群的集合，从植物到动物再到微生物都算在内。生态系统是群落加上它周围的无机环境，阳光、土壤、水、空气都是生态系统的组成部分，范围可以小到一个池塘，也可以大到整片森林。考试里最常见的失分是把「一片森林里所有的鸟」写成种群，鸟包含很多个物种，它只能算群落里的一部分。Edexcel 1BI0 的 Topic 9 标题是 Ecosystems and material cycles，2016 版大纲自 2018 年首次考试沿用至今，这一章的术语直接决定 Paper 2 的数据题能不能读懂。
These four words carry marks on their own in the Edexcel 1BI0 specification, Topic 9, Ecosystems and material cycles. A population is every individual of one species in one area at one time; a community adds all the other species living there; an ecosystem is that community together with its non-living surroundings, so light, water, soil and minerals belong to it, whether the scale is a pond or a whole woodland. Calling all the birds in a wood a population costs the easiest mark on the paper, because birds are many species and the term the mark scheme wants is community.
生物因素与非生物因素
描述物种分布时，先把影响因素分成两类再动笔。非生物因素是物理与化学条件，包括光照强度、温度、含水量、土壤 pH、矿物离子浓度，以及水中的溶氧量。生物因素来自其他生物，包括食物供应、捕食者、病原体，以及同种或异种之间的竞争。判断时问自己一句：这个因素是不是活的，活的归生物因素，其余归非生物因素。答案要落到具体名称上，只写「环境」是拿不到分的。很多学生以为光照只影响光合作用，其实它还决定地面的温度与湿度，树荫下与空地上的物种组成差别很大，这正是树荫样带能画出明显趋势的原因。
Sort the factors before writing. Abiotic factors are non-living conditions such as light intensity, temperature, moisture, soil pH, mineral ions and dissolved oxygen; biotic factors come from other organisms, whether that is food supply, a predator, a pathogen or competition between or within species. Naming the factor earns the mark, while writing the environment earns nothing. Shade is the influence students underestimate, since light intensity also sets ground temperature and moisture, which is why the species list under a canopy differs from open ground a few metres away.
样方取样 25cm 方框里的估算公式
Edexcel 的核心实验 CP8 要求用实地调查技术研究生物与环境的关系，2016 版核心实验指南里写明要使用样方与样带。随机取样的操作是先拉两条互相垂直的卷尺，再用随机数生成器给出两个数字作为坐标。把 25cm × 25cm 的样方框放到坐标点上，数出框内目标物种的个体数。压在上方与左侧边线上的植株算进来，压在下方与右侧的排除，这条规则让不同人计数同一块样方得到同一个数字。同一步骤换坐标重复十次，取十次结果的平均值。估算公式是总面积除以样方面积，再乘以样方内的平均个体数。代入一组具体数字更清楚：20m × 20m 的样地面积是 400 m²，样方面积是 0.0625 m²，十次平均每个样方 8 株，估计总数就是 400 ÷ 0.0625 × 8，约 51200 株。常见错误是走到草长得最密的地方才放下样方，这样得到的平均值偏高，估算结果也跟着偏大。
CP8 in the Edexcel core practical guide is the fieldwork practical, and quadrats plus belt transects are the techniques it names. Random sampling is the part that gets tested: coordinates from a random number generator, a 25 cm by 25 cm quadrat placed on those coordinates, counts recorded ten times and then averaged. Edge rules matter as well, since plants on the top and left edges are counted while those on the bottom and right are not, so two students counting the same quadrat reach the same number. The estimate is total area divided by quadrat area multiplied by the mean count, so a 400 m² field, 0.0625 m² quadrats and a mean of 8 plants give 400 ÷ 0.0625 × 8, about 51200 plants. Dropping the quadrat on the thickest patch of vegetation is the mistake behind most unrealistic estimates.
样带 沿光照梯度记录物种分布
要比较同一物种在阴处与阳处的分布差异，样方不够用，得换成样带。从树基部向开阔地拉一条卷尺，把样方框的一角对齐 0m 处开始计数。随后每 5m 放一次，一直到样带末端，同时用光强计在贴地高度读数，注意别让自己的影子挡住探头。把距离作横轴、个体数作纵轴作图，得到的曲线反映的正是分布沿环境梯度的变化。控制变量要写清样方大小、计数方式和每个位置的重复次数，否则数据没法比较。结论只能写成分布沿梯度存在差异，不能写成光照造成了这种分布，样带给出的始终是相关性。样方回答有多少的问题，样带回答分布怎么变的问题，两者在答案里不能互换。
A transect answers a different question from a quadrat: not how many, but how the distribution shifts along a gradient. Run a tape from the base of a tree into open ground, put one corner of the quadrat on the 0 m mark, count, then repeat every 5 m to the end while reading light intensity at ground level with your own shadow kept off the sensor. Plot distance against number of individuals and the trend appears, provided quadrat size, counting method and the number of repeats at each point stay the same. The relationship a transect shows is a correlation along the gradient rather than proof of cause, and the conclusion has to be worded that way. Swapping the two methods in a written answer loses marks even when the fieldwork itself was done well.
标记重捕法 数不清的动物怎么估
数目多、活动范围大的动物没法用样方数，要用标记重捕法。第一轮捕到 n1 只，做好标记后放回原处。等标记在种群中混匀，第二轮捕到 n2 只，数出其中带标记的 m2 只。种群大小的估计值是 n1 乘以 n2 再除以 m2。举一组数字：第一次捕到并标记 40 只蜗牛，第二次捕到 50 只，其中 5 只带标记，估计种群约为 40 × 50 ÷ 5，也就是 400 只。这个估算成立有前提，标记不能影响动物存活与被捕的概率，标记不能脱落，两次捕获之间不能有明显的出生、死亡和迁入迁出。重捕到的带标记个体太少时误差会迅速放大，m2 小于 5 的估算结果只能当参考。阅卷时经常见到学生把两个捕获数一乘就交卷，忘了除以重捕到的标记个体数，公式错一步整道题的分都没了。
Mobile animals cannot be counted in a quadrat, so mark and recapture estimates them instead: catch a sample, mark and release, then catch again later and record how many carry a mark. The estimate is n1 × n2 ÷ m2, so marking 40 snails, catching 50 the second time with 5 of them marked, gives about 400. The calculation only holds if marking does not change an animal's chance of surviving or being caught again, marks stay on, and the population does not gain or lose members between visits. Fewer than about five recaptured marks makes sampling error dominate the estimate. In practice, examiners see answers that multiply the two catches and stop, which throws the estimate away.
营养级与能量传递效率
食物链上每一级叫一个营养级。能量在传递时大部分通过呼吸作用以热量形式散失，余下的一部分留在未被消化的物质和排泄物里，真正传到下一级的通常只有大约十分之一，所以食物链很少超过四到五级。生物量金字塔按每一级生物的总质量绘制，多数情况下下宽上尖。海洋是常被问到的例外，浮游动物的总质量可以超过浮游植物，因为浮游植物繁殖快、被摄食也快，这种倒置只出现在生物量金字塔，能量金字塔不会倒。数量金字塔按个体数绘制，一棵树上的昆虫就有上万只，塔形照样会翻过来，所以它的适用范围比前两种窄得多。计算题常给出两级的具体数值让你求传递效率，例如生产者固定 20000 kJ、初级消费者同化 2000 kJ，效率就是 10%，单位要统一，kg 与 kJ 不能混用。
Each level of a food chain is a trophic level. Most energy leaves the chain as heat from respiration and some stays locked in undigested material and faeces, so roughly a tenth passes to the next level and chains rarely run past four or five links. A pyramid of biomass shows the mass at each level and is usually widest at the base; the ocean is the exception examiners like, since zooplankton can outweigh the phytoplankton they feed on because the producers reproduce and get eaten so quickly, and no such inversion appears in a pyramid of energy. A pyramid of numbers counts individuals instead, and one tree carrying thousands of insects inverts the shape just as easily, which is why it is used far less than the other two. Calculation questions usually hand you two levels and ask for the transfer efficiency, which for 20000 kJ fixed by producers and 2000 kJ assimilated by primary consumers works out at 10%, with mixed units the usual source of error.
碳循环 水循环 分解者与答题写法
碳在生物与大气之间来回流动。光合作用把二氧化碳固定成有机物，呼吸作用把它送回大气，分解者把遗体与排泄物中的有机物拆解，同时释放二氧化碳，燃烧化石燃料是另一条出口。分解速度取决于温度、含水量和氧气，堆肥堆内部变热正是微生物呼吸放热的结果。水循环里蒸腾作用把水从植物送入大气，氮循环则要靠固氮细菌把空气中的氮气转成植物可以利用的含氮化合物。物质循环把元素送回土壤和水体，植物再吸收硝酸盐这类矿物离子，整条通路才算闭合。森林在这个循环里既是碳汇也是碳库，砍伐会让两个作用同时减弱。Paper 2 覆盖 Topic 6 到 Topic 9 并带 Topic 1 到 Topic 5 的综合题，两卷各 1 小时 45 分钟、各 100 分、各占最终成绩的 50%。描述题的分数给在过程和名词上，写「分解者把有机物分解成二氧化碳和水」比写「物质循环」拿到的分多。
Carbon moves in two directions: photosynthesis fixes carbon dioxide into organic compounds, respiration returns it to the air, decomposers break down dead material and waste and release carbon dioxide as they do so, and burning fossil fuels adds another route out. Temperature, moisture and oxygen set how fast decomposition runs, which is why the inside of a compost heap warms up as microbes respire. Transpiration moves water from plants into the atmosphere, while nitrogen fixation by bacteria turns atmospheric nitrogen into compounds plants can absorb, and only then does the element return to soil and water through the cycle. Forests act as both a carbon store and a carbon sink, so felling them weakens the cycle twice over. Paper 2 covers Topics 6 to 9 with synoptic questions drawn from Topics 1 to 5, and each paper is 1 hour 45 minutes, 100 marks and 50% of the qualification. Marks in written answers sit on processes and named terms: decomposers break down organic matter into carbon dioxide and water scores, while a general reference to material cycles does not.
生物多样性 食物安全与保护措施
Topic 9 的后半段把生态学落在人的需求上。生物多样性包含物种、基因与生态系统三个层面的差异，维持它最直接的理由是药物来源、作物的抗病基因和授粉昆虫带来的产量，也包括生态系统本身的稳定。重新造林能固定碳、保住表土、给物种提供栖息地，但单一树种的人工林恢复效果远不如保留原有林地。食物安全指所有人能稳定获得足够的食物，气温上升带来的减产、干旱与洪水、耕地被占用，以及把作物拿去制造生物燃料，都会削弱它。捕捞速度超过鱼类繁殖速度，渔业资源就会崩塌，配额、限制网眼尺寸和人工养殖是常见的应对手段。这类题通常要求给出一条措施并说明理由，措施与理由必须对应，只写「保护环境」不得分。
Topic 9 in the Edexcel 1BI0 specification ends with the human side of ecology. Biodiversity covers variety at the level of species, genes and ecosystems, and the case for keeping it rests on medicines, disease-resistance genes for crops, pollination and the stability of whole systems, while reforestation stores carbon, holds topsoil and gives species somewhere to live, though a single-species plantation restores far less than standing woodland. Food security means everyone can reliably obtain enough food, and rising temperatures, drought, flooding, land taken for other uses and crops diverted into fuel all work against it; where fishing removes fish faster than they reproduce, stocks collapse, so quotas, mesh-size limits and aquaculture are the usual answers. A question asking for one measure and a reason marks the pair together, and protect the environment on its own earns nothing.
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