Biodiversity Exam Practice | 生物多样性真题精练

📚 Biodiversity Exam Practice | 生物多样性真题精练

This revision article focuses on exam-style practice for Topic 4.2 Biodiversity. We will unpack key definitions, explore diversity indices, interpret data, and tackle typical command words found in A-level biology papers. Working through worked examples and model answers will help you refine your technique and avoid common pitfalls.

本复习文章专攻4.2 生物多样性的真题训练。我们将拆解关键定义，探讨多样性指数，解读数据，并解决A-level生物试卷中常见的指令词。通过细致演算和模型答案，你将打磨答题技巧，避开常见失分点。

1. Defining Biodiversity | 生物多样性的定义

Biodiversity is the variety of living organisms in a given area. It can be considered at three levels: species diversity (richness and evenness of species), genetic diversity (variation in alleles within a gene pool), and ecosystem diversity (range of different habitats and communities).

生物多样性指特定区域内生物的多样性。可从三个层面考量：物种多样性（物种的丰富度和均匀度）、遗传多样性（基因库中等位基因的变异）以及生态系统多样性（不同栖息地和群落的范围）。

In exam questions, you may be asked to distinguish these levels. For example, a woodland may have high species richness but low genetic diversity if populations are inbred.

在考题中，你可能被要求区分这些层面。例如，一片林地可能物种丰富度高，但如果种群近交则遗传多样性低。

2. Species Richness and Evenness | 物种丰富度与均匀度

Species richness is simply the number of different species present in a sample. It does not account for how many individuals of each species are found. Evenness describes how equal the abundances of the species are. A community with one dominant species and several rare ones has low evenness.

物种丰富度仅仅是样本中不同物种的数量，不考虑各物种有多少个体。均匀度描述各物种丰度是否均等。如果一个群落只有一个优势种和多个稀有种，其均匀度低。

Exam hint: a high species richness does not automatically mean high diversity if evenness is very low. Always link both concepts when analysing data.

应试提示：若均匀度很低，即使物种丰富度高也不一定代表多样性高。分析数据时务必将两个概念联系起来。

3. Simpson’s Index of Diversity | 辛普森多样性指数

The Simpson’s diversity index (D) is a widely used measure that combines richness and evenness. In A-level specifications, it is commonly calculated as:

D = 1 – Σ(n/N)²

辛普森多样性指数（D）是结合丰富度与均匀度的常用指标。在A-level大纲中，通常按下式计算：

D = 1 – Σ(n/N)²

where n = total number of organisms of a particular species, N = total number of organisms of all species. The value of D ranges from 0 (lowest diversity) to 1 (highest diversity). A high D means more diversity, often implying a stable, complex ecosystem.

其中 n = 某一物种的个体总数，N = 所有物种的个体总数。D 值的范围从 0（最低多样性）到 1（最高多样性）。D 值高意味着多样性更高，通常表示生态系统稳定而复杂。

You must be able to perform this calculation from tabulated data as well as interpret the result.

你必须能够根据表格数据进行计算并对结果做出解释。

4. Worked Example: Simpson’s Index Calculation | 计算例题：辛普森指数计算

A sweep net sample from a meadow gave the following counts:

Species	Number (n)
Grasshopper	15
Ladybird	8
Spider	5
Bee	10
Beetle	12

Step 1: Calculate N. N = 15 + 8 + 5 + 10 + 12 = 50.

第一步：计算N。N = 15 + 8 + 5 + 10 + 12 = 50。

Step 2: Calculate (n/N)² for each species:

第二步：计算每种生物的(n/N)²：

Grasshopper (15/50)² = 0.3² = 0.09

蚱蜢 (15/50)² = 0.3² = 0.09

Ladybird (8/50)² = 0.16² = 0.0256

瓢虫 (8/50)² = 0.16² = 0.0256

Spider (5/50)² = 0.1² = 0.01

蜘蛛 (5/50)² = 0.1² = 0.01

Bee (10/50)² = 0.2² = 0.04

蜜蜂 (10/50)² = 0.2² = 0.04

Beetle (12/50)² = 0.24² = 0.0576

甲虫 (12/50)² = 0.24² = 0.0576

Step 3: Sum these values: Σ = 0.09 + 0.0256 + 0.01 + 0.04 + 0.0576 = 0.2232

第三步：求和：Σ = 0.09 + 0.0256 + 0.01 + 0.04 + 0.0576 = 0.2232

Step 4: Simpson’s index D = 1 – 0.2232 = 0.7768 (approx. 0.78). This value indicates relatively high diversity.

第四步：辛普森指数 D = 1 – 0.2232 = 0.7768（约 0.78）。该值表明多样性相对较高。

Always square (n/N) accurately and keep enough decimal places during working to avoid rounding errors.

务必准确平方(n/N)，计算过程中保留足够小数位，避免四舍五入误差。

5. Interpreting Diversity Index Values | 多样性指数值的解读

When comparing two habitats, the one with the larger D is more diverse. A low D (e.g., 0.2) often indicates a stressed environment dominated by few species, such as polluted water or monoculture farmland. A high D (e.g., 0.85) suggests many species with balanced populations, typical of ancient woodlands or coral reefs.

比较两个栖息地时，D值较大的那个更有多样性。低D（如0.2）通常表示环境受胁迫，少数物种占主导，如污染水体或单一种植农田。高D（如0.85）表明物种丰富且种群均衡，常见于古老林地或珊瑚礁。

Students often confuse high diversity with high productivity. Diversity indices measure variety and evenness, not biomass. In an exam, link high D to ecosystem stability, resilience to change, and resource partitioning.

学生常把高多样性与高生产力混淆。多样性指数衡量的是种类与均衡度，并非生物量。考试中应将高D与生态系统稳定性、抗干扰能力以及资源划分联系起来。

6. Genetic Diversity and Polymorphism | 遗传多样性与多态性

Genetic diversity refers to the total number of different alleles in a population’s gene pool. It can be measured by the proportion of polymorphic gene loci or by heterozygosity index. For example, if a locus has two alleles present in a population, it is polymorphic.

遗传多样性指种群基因库中不同等位基因的总数。可以通过多态基因座比例或杂合度指数来测量。例如，若某基因座在种群中存在两个等位基因，则为多态性。

Heterozygosity (H) is often used as an index of genetic diversity: H = number of heterozygotes at a locus / total number of individuals. A higher heterozygosity suggests greater genetic variation and a greater ability to adapt to environmental changes.

杂合度(H)常被用作遗传多样性指标：H = 某基因座的杂合个体数 / 个体总数。杂合度越高，说明遗传变异越大，适应环境变化的能力越强。

Exam command words: ‘Explain how a genetic bottleneck reduces genetic diversity’ requires reference to a loss of alleles and reduction in heterozygosity due to a drastic population decline.

试题指令词：‘解释遗传瓶颈如何降低遗传多样性’需要提及由于种群数量剧减导致等位基因丢失与杂合度下降。

7. Sampling Methods to Assess Biodiversity | 评估生物多样性的取样方法

For animals, techniques include sweep netting, pitfall traps, kick sampling in streams, and mark-release-recapture for motile species. For plants, quadrats (frame or point) along a transect are standard. Sampling must be random to avoid bias, or systematic to study zonation.

动物的取样技术包括扫网、陷阱采集、溪流的踢样法，以及用于活动物种的标记-释放-重捕法。植物的取样标准是沿样带放置样方（方形框或点样方）。取样必须随机以避免偏差，或采用系统取样研究带状分布。

A common exam question asks to describe how to estimate species richness in a grassland using a quadrat. You should mention random number generation, placing a 0.5m² quadrat, recording each species present, repeating many times, and calculating mean species richness.

常见考题要求描述如何用样方估算草地的物种丰富度。应答时提及随机数字生成、放置0.5m²样方、记录出现的每种植物的物种、多次重复，并计算平均物种丰富度。

For Simpson’s index, you will often be given raw count data from random quadrat sampling and asked to calculate D.

辛普森指数的题目常给出随机样方调查的原始计数数据，要求计算D。

8. Threats to Biodiversity | 对生物多样性的威胁

Human activities reduce biodiversity through habitat destruction, overexploitation, introduction of invasive species, pollution, and climate change. Deforestation in tropical regions causes a sharp drop in both species richness and genetic diversity. Coral bleaching, driven by rising sea temperatures, exemplifies a climatic threat.

人类活动通过栖息地破坏、过度开发、外来物种入侵、污染和气候变化降低生物多样性。热带地区的森林砍伐导致物种丰富度和遗传多样性急剧下降。由海温升高引发的珊瑚白化是气候威胁的典例。

In exam essays, you need to explain the specific mechanism: e.g., monoculture replaces complex food webs with a single crop, eliminating niches and reducing Simpson’s index. Link loss of keystone species to cascading effects on entire ecosystems.

在考试论述中需解释具体机制：例如，单一栽培以单一作物取代复杂的食物网，消除了生态位，降低了辛普森指数。将关键种丧失与整个生态系统的级联效应联系起来。

9. Conservation Approaches | 保护方法

In situ conservation protects species in their natural habitats (nature reserves, national parks, marine protected areas). Ex situ conservation involves zoos, seed banks, and botanic gardens. A-level questions often ask to evaluate the advantages of seed banks: they store genetic diversity, require less space, and can protect against extinction.

就地保护是在自然栖息地中保护物种（自然保护区、国家公园、海洋保护区）。迁地保护涉及动物园、种子库和植物园。A-level问题常要求评估种子库的优势：它们保存遗传多样性，所需空间小，能预防物种灭绝。

International agreements like CITES regulate trade in endangered species. Habitat corridors can reconnect fragmented landscapes, increasing gene flow and reducing inbreeding.

国际协定如CITES管制濒危物种贸易。栖息地廊道可重新连接破碎化的景观，增加基因流动，减少近交。

Be prepared to discuss the roles of local communities and sustainable development in conservation success.

准备讨论地方社区和可持续发展在保护成功中的作用。

10. Common Exam Pitfalls | 常见失分点

One major error is forgetting to subtract from 1 in the Simpson’s index formula, giving a value that represents the probability of picking two individuals of the same species rather than diversity. Always write D = 1 – Σ(n/N)².

一个主要错误是忘记辛普森指数公式中的‘1 -’，导致得出的值代表抽到同种两个个体的概率，而非多样性。务必写出 D = 1 – Σ(n/N)²。

Another pitfall is confusing species richness with diversity. A habitat with 10 species, each represented by one individual, has high richness but low index value due to a different reason? Actually, if each species has one individual, evenness is perfect, so D would be high. But be careful: richness alone doesn’t guarantee high D, especially if one species dominates.

另一个易错点是混淆物种丰富度和多样性。虽然一个拥有10个物种、每种各一个个体的栖息地丰富度和D值都高，但丰富度本身并不保证高D，特别是当某一物种占优势时。

In describing sampling, failing to mention randomisation or adequate replication loses marks. In data interpretation, not referring to the numerical values in the table when making comparisons is a common flaw.

描述取样时未提及随机化或足够的重复会丢分。在数据解读中做比较时不引用表格中的具体数值也是常见缺陷。

11. Exam-Style Question and Model Answer | 真题风格题与模型答案

Question: Two woodland sites were sampled for invertebrates using pitfall traps. Site A had 5 species with total individuals: 40, 10, 5, 3, 2. Site B had 4 species with totals: 20, 20, 18, 2. Calculate the Simpson’s diversity index for each site and comment on the difference. (4 marks)

问题：使用陷阱法对两个林地样点的无脊椎动物进行取样。样点A有5个物种，个体总数分别为：40, 10, 5, 3, 2。样点B有4个物种，个体总数分别为：20, 20, 18, 2。计算各样点的辛普森多样性指数，并评论差异。(4分)

Model Answer: For Site A, N = 60. Σ(n/N)² = (40/60)² + (10/60)² + (5/60)² + (3/60)² + (2/60)² = 0.4444 + 0.0278 + 0.0069 + 0.0025 + 0.0011 = 0.4827. D = 1 – 0.4827 = 0.5173 (≈0.52).

模型答案：样点A：N = 60。Σ(n/N)² = (40/60)² + (10/60)² + (5/60)² + (3/60)² + (2/60)² = 0.4444 + 0.0278 + 0.0069 + 0.0025 + 0.0011 = 0.4827。D = 1 – 0.4827 = 0.5173 (≈0.52)。

For Site B, N = 60. Σ(n/N)² = (20/60)² × 2 + (18/60)² + (2/60)² = 0.1111×2 + 0.09 + 0.0011 = 0.2222 + 0.09 + 0.0011 = 0.3133. D = 1 – 0.3133 = 0.6867 (≈0.69).

样点B：N = 60。Σ(n/N)² = (20/60)² × 2 + (18/60)² + (2/60)² = 0.1111×2 + 0.09 + 0.0011 = 0.3133。D = 1 – 0.3133 = 0.6867 (≈0.69)。

Site B has a higher Simpson’s index (0.69 vs 0.52), indicating greater species diversity. Although Site A has higher species richness (5 vs 4), Site B’s evenness is much higher, resulting in a larger D. This shows that both richness and evenness determine diversity.

样点B的辛普森指数更高（0.69 vs 0.52），表明物种多样性更高。尽管样点A物种丰富度更高（5 vs 4），但样点B的均匀度远高于A，导致D值更大。这说明多样性的高低同时取决于丰富度和均匀度。

Marking notes: Correct calculation of both D values (2 marks), correct comparison and reference to evenness/richness (2 marks).

评分注释：正确计算两个D值(2分)，正确比较并提及均匀度/丰富度(2分)。

12. Final Revision Summary | 最后复习总结

Remember to define biodiversity at all three levels; practise Simpson’s index calculations until they are second nature; use numerical evidence when comparing sites; and link genetic diversity to adaptability and conservation. Review sampling techniques and their limitations.

牢记在三个层面定义生物多样性；反复练习辛普森指数计算直到成为本能；比较样点时使用数字证据；将遗传多样性与适应性和保护联系起来。复习取样技术及其局限性。

With these skills, you will confidently handle any 4.2 Biodiversity exam question.

掌握这些技能，你将自信应对任何4.2生物多样性考题。

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Biodiversity Exam Practice | 生物多样性真题精练