OCR生物 生物多样性 Simpson指数 采样 保护
OCR A-Level Biology A 把生物多样性放在 Module 4 的第一个小节 4.2.1，现行考纲在这一节下列了 (a) 到 (i) 九个学习目标，从三个层次的多样性一路排到国际公约。这一节的题面往往很短，给一张物种个体数表格或一句 D 值，真正要写的是解释。下面把它拆成可执行的部分：考纲要求、计算口径、常见失分和一份六周复习安排。
The biodiversity section of OCR A-Level Biology A (4.2.1) carries nine learning outcomes, from the three levels of biodiversity through to named international agreements. Exam questions on it are usually short, but most of the marks come from explanation rather than recall, so the calculations have to be paired with interpretation.
4.2.1 在 Paper 02 里的位置
OCR Biology A（H420）有三门笔试：Biological processes（100 分、2 小时 15 分、占 37%）、Biological diversity（100 分、2 小时 15 分、占 37%）、Unified biology（70 分、1 小时 30 分、占 26%），三门口径都写在 OCR 官网的考核说明里。生物多样性是第 2 门的卷名来源，4.2.1 的内容正面考在这门卷子里，第 3 门会拿它当综合题材料。很多学生以为 4.2.1 只是背定义，翻一翻 H420/02 的真题会发现计算与解释各占一半。考纲还写明笔试里至少 10% 的分值给到 Level 2 及以上的数学，采样数据的整理和指数计算正好落在这条线上。复习时把这两件事一起排，效率比先背书再刷题高。
OCR Biology A has three written papers: Biological processes (100 marks, 2 hours 15 minutes, 37%), Biological diversity (100 marks, 2 hours 15 minutes, 37%) and Unified biology (70 marks, 1 hour 30 minutes, 26%), as listed on the OCR website. Section 4.2.1 is examined directly in the second of these and used synoptically in the third. At least 10% of the written marks are for Level 2 mathematics, which is exactly where sampling tables and the index calculation sit, so the arithmetic deserves its own revision slot.
三个层次的多样性
考纲 4.2.1(a) 把多样性分成三层：栖息地多样性举的例子是沙丘、林地、草甸和溪流，物种多样性要求同时看物种丰富度与物种均匀度，遗传多样性指同一物种内的不同品种。很多学生以为多样性就是数物种个数，题目给一张 30 个物种、其中一个种占九成个体的表，算出来的 D 值只有 0.2 上下。均匀度和丰富度必须一起看，OCR 的评分说明里这两点常常分开给分。三层之间还会互相影响，一块沙丘被道路切成两半，栖息地层面先碎掉，物种层面的下降随后才出现。答题时先判断题干给的是哪一层证据，再决定用哪个词。
The specification separates biodiversity into three levels that candidates must keep apart. Habitat biodiversity is illustrated with sand dunes, woodland, meadows and streams, species biodiversity is defined through species richness plus species evenness, and genetic biodiversity covers the different breeds within one species. A table with 30 species where one species supplies most of the individuals returns a low D value, simply because evenness is poor. Splitting a dune with a road damages the habitat level first and species richness later, so the evidence in a question usually points to one level only.
采样为什么重要 随机与非随机
4.2.1(b) 要求知道采样为什么重要，还要在野外真的做随机与非随机取样。随机取样用随机数生成坐标，样方落点不能由人挑，否则草多的角落会被多采几次。非随机取样分三种：机会取样是看到什么记什么，分层取样按生境先划片再在每片内取样，系统取样沿一条线每固定间隔放一个样方。工具不止样方，扫网抓草丛昆虫，陷阱罐收地面爬虫，吸虫器吸小型节肢动物，这一组技能在 OCR 的实践清单里归到 PAG 3 sampling techniques。常见的错误是只在好走的地方放样方，算出的 D 值比真实值高，讨论题里这一条要写出来。考纲还要求采到栖息地里各类生物，只数植物会被扣分。
Sampling matters because a habitat cannot be counted in full, so a small number of samples has to represent the whole. Random sampling places quadrats at coordinates from a random number generator, while opportunistic, stratified and systematic sampling are the non-random options named in the specification. Quadrats, sweep nets, pitfall traps and pooters all appear under PAG 3 sampling techniques, and the full range of organisms has to be sampled rather than plants alone. Quadrats placed only where the ground is easy to walk give an inflated D value, which is a standard discussion point.
测物种丰富度与均匀度
4.2.1(c) 要求会测量物种丰富度与均匀度。做法是先数出样方里每个物种的个体数，列成物种与个体数两列的表，再核对物种数与个体总数。丰富度是一个样方里出现了多少种，均匀度看各物种的个体数相差多少，两个量都会直接进 Simpson 的计算。这里最容易出错的是把同一物种的不同叶形当成两个种，比如不同叶形的车前草。样方边上一半在内一半在外的植株，要先定一条规则再数，否则两次记录会互相矛盾。表格整理干净，后面两步计算才不容易崩。
Measuring richness and evenness starts with a two-column table of species and individuals per quadrat. Richness counts how many species are present, while evenness describes how the individuals are spread across those species, and both values go straight into the index. Records are lost when one plant is entered twice under two names, which happens with species that vary in leaf shape. Edge rules for half-covered individuals have to be fixed before counting begins.
Simpson 指数怎么算
4.2.1(d) 只要求会用与解释，公式在试卷上会给，不需要背。D 等于 1 减去 Σ(n/N)²，n 是某个物种的个体数，N 是所有物种个体数之和。计算时逐种求 n/N，平方，把各项相加，再用 1 减去这个和。D 越接近 1 多样性越高，说明物种多而且个体分布均匀；越接近 0 说明少数物种占了绝大多数个体。学生最常把 N 写成物种数，一算出五花八门的数就慌。2022 年 6 月 H420/02 第 7 题只给 D=0.85 让判断，正确选项是多样性高，根本不用再算。
Simpson's Index of Diversity is D = 1 minus the sum of (n/N) squared, where n is the number of individuals of one species and N is the total across all species. OCR states that the formula is provided in assessments and does not need to be recalled. Work through the species one at a time, square each proportion, add the squares, subtract from 1, and read the result against the two ends of the scale. A value close to 1 means many species with even abundance, a value close to 0 means one or two species dominate, and the usual slip is treating N as the number of species.
拿一组具体数字走一遍：样方里 A 种 40 只、B 种 30 只、C 种 20 只、D 种 10 只，总数 100，四个比例是 0.4、0.3、0.2、0.1，平方和 0.16+0.09+0.04+0.01=0.30，D 等于 0.70。把数量改成 88、4、4、4，平方和升到 0.78，D 掉到 0.22，物种数没变，均匀度垮了。这类对比在真题里反复出现，算完顺手写一句均匀度下降，比单写数字稳。
A worked case makes the arithmetic concrete: 40, 30, 20 and 10 individuals give proportions of 0.4, 0.3, 0.2 and 0.1, a sum of squares of 0.30 and D = 0.70. Change the counts to 88, 4, 4 and 4 and the sum of squares rises to 0.78, so D falls to 0.22 even though four species are still present. Contrasts of this shape appear repeatedly in past papers.
遗传多样性的计算口径
遗传多样性在考纲里是要算的，口径是多态基因位点的比例，等于多态位点数除以位点总数。位点是否多态，要看有没有两个以上的等位基因，通常靠测序或电泳判断。考纲建议的样本是动物园圈养种群、稀有品种和纯种动物，这些群体数量小、亲缘近，比例低得很明显。很多学生以为遗传多样性只能定性描述，其实题目会给一张位点表让学生自己数。结果按比例或百分数写，只写一个光秃秃的小数容易被判表达不清。位点总数变了，结论会跟着变，答题前先确认分母。
Genetic biodiversity is assessed by calculation, using the proportion of polymorphic gene loci, which is the number of polymorphic loci divided by the total number of loci. A locus counts as polymorphic only when more than one allele is present, normally established by sequencing or electrophoresis. Zoos running captive breeding programmes, rare breeds and pedigree animals are the populations the specification suggests, because small closed groups lose variation quickly. Answers should be reported as a proportion or a percentage, with the denominator stated.
影响生物多样性的三个因素
4.2.1(f) 列了三个因素：人口增长、农业单一种植、气候变化。人口增长抬高对土地和食物的需求，栖息地被道路和农田切成小块。单一种植把复杂的群落换成一种作物，昆虫和土壤生物的食源与栖身处一起消失。气候变化改变物候，鸟类育雏与毛虫高峰错开就会缺食。这一节常与 Module 6 的种群与可持续一起出综合题，答题时把原因和机制分开写，只写人多了会丢分。三个因素的机制不同，混着写容易读成同一句话。
The specification names three factors that affect biodiversity: human population growth, monoculture agriculture and climate change. Population growth raises demand for land and food, so habitats are fragmented into smaller patches. Monoculture replaces a mixed community with a single crop and removes food and shelter for insects and soil organisms. Climate change shifts phenology, so nestling birds may miss the caterpillar peak, and this section is often examined together with populations and sustainability in module 6.
维持多样性的三类理由
4.2.1(g) 把理由分成生态、经济、美学三类。生态理由写关键种与遗传资源，关键种一旦消失，依赖它的物种会接连受影响。经济理由写连作导致的土壤耗竭，单一种植几年后产量下滑，恢复成本很高。美学理由写景观保护，这一条最常被学生漏掉。三类理由的写法不一样，生态讲相互作用，经济讲长期产出，美学落在人的体验上。评卷时能看到不少答案把三类揉成一类，只写一句保护环境很重要。写理由的时候顺手补一句机制，分数才拿得稳。
Reasons for maintaining biodiversity are grouped as ecological, economic and aesthetic. Ecological reasons include protecting keystone species and maintaining genetic resources, since losing a keystone species pulls apart the species that depend on it. Economic reasons centre on soil depletion caused by continuous monoculture, and aesthetic reasons concern protecting landscapes. Marks are usually awarded per reason with a mechanism attached, so a sentence about the importance of protecting nature earns nothing.
就地保护与迁地保护
4.2.1(h) 要求区分就地与迁地保护。就地保护在原生地做，考纲举的例子是海洋保护区和野生动物保护区，物种留在原来的生态关系里，成本较低但抗干扰能力差。迁地保护把物种移出原生地，种子库、植物园、动物园都属于这一类，适合抢救性保存，代价是样本量小、遗传漂变的压力大。两者不是二选一，多数保护项目同时用。学生常把动物园放进就地保护，这一步最容易丢分。答比较题时把标准先立起来，成本、遗传风险和生态关系三项一一对照。
In situ conservation keeps species in their original habitat, and the specification gives marine conservation zones and wildlife reserves as examples. Ex situ conservation moves species out of the habitat, covering seed banks, botanic gardens and zoos, which suits rescue work but carries the risk of small samples and genetic drift. The two are usually combined rather than chosen between, so an answer that treats them as alternatives loses the comparison marks. Zoos sit firmly in ex situ.
国际与地方协议
4.2.1(i) 要求知道国际与地方协议，考纲点名 CITES、Rio Convention on Biological Diversity 和英国的 Countryside Stewardship Scheme。CITES 于 1973 年 3 月 3 日在华盛顿签署，1975 年 7 月 1 日生效，管的是濒危物种的国际贸易，靠进出口许可证执行。Rio 生物多样性公约 1992 年 6 月 5 日在里约开放签署，1993 年 12 月 29 日生效，把就地与迁地保护写进条文。地方层面可以接英国 Environment Act 2021，英格兰的大型开发项目自 2024 年 2 月 12 日起必须交付至少 10% 的生物多样性净增益，这条来自 gov.uk 的公告。举一个能说出年份的例子，比列三个缩写更有说服力。
International and local agreements are a named learning outcome, with CITES, the Rio Convention on Biological Diversity and the Countryside Stewardship Scheme listed in the specification. CITES was signed in Washington on 3 March 1973 and entered into force on 1 July 1975, controlling trade in endangered species through import and export permits. The Rio convention opened for signature on 5 June 1992 and took effect on 29 December 1993, writing both in situ and ex situ conservation into its articles. In England, the Environment Act 2021 has required major developments to deliver at least 10% biodiversity net gain since 12 February 2024.
六周复习安排
六周复习按内容块切开，每周一块，周末留半天做整卷。第 1 周读 (a)(b)，把三层多样性和四种采样方式的区别写成一张对照表，顺手过一遍 PAG 3 的操作步骤。第 2 周做 (c)(d)，用三张真题的数据表手算 D 值，算完再读评分说明，看分子分母写错的答案扣在哪一步。第 3 周处理 (e)，把多态位点比例的计算题做透。第 4 周背 (f)(g)(h) 的三类因素、三类理由与两种保护方式，用关键词卡片自测。第 5 周专攻 (i) 与 6.3.2 的可持续管理，两节常常一起出题。第 6 周只做整卷和错题，把 H420/02 近三年的题按题型归一遍。
A six-week block works better than rereading notes, with one topic area per week and half a day at the weekend for a full paper. The first three weeks cover the levels of biodiversity, sampling under PAG 3, the index calculation and the polymorphic loci calculation, each checked against a past paper data table. Weeks four and five take the factors, the three reasons, in situ and ex situ methods, the named agreements and the sustainable management content in 6.3.2. The last week is past papers only, with every H420/02 question from the last three years sorted by question type.
答题语言与失分点
评分说明喜欢看到机制词，比如物种数、个体数、均匀度、栖息地完整性，写环境变好了这样的结论不给分。解释 D 值高低要同时提物种多少和个体分布是否均匀，只说物种多通常只拿一半分。讨论采样误差时把随机性、样方大小、样方数量三个变量分开说，混在一起写会被判含糊。阅卷时见过不少答案把结论写在开头、证据写在后面，读起来对不上。计算题写完数字再加一句解释，这半句常常就是第二个得分点。把这三条写进错题本，比多刷一套卷子实在。
Mark schemes reward mechanism words such as species number, individual number, evenness and habitat integrity, while a bare statement that the environment improved scores nothing. Explaining a D value needs both the number of species and how evenly the individuals are spread, because species number alone usually earns half the marks. Discussions of sampling error should separate randomness, quadrat size and quadrat number instead of bundling them together. Calculations should finish with a short interpretation, which is frequently the second marking point.
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