📚 Index of Diversity Exam Practice | 生物多样性指数真题精练
Mastering the index of diversity, particularly Simpson's Index, is essential for A-level Biology students tackling ecology and biodiversity topics. This article provides clear explanations, worked examples, and examination-style practice to build your confidence and ensure you can calculate and interpret diversity indices accurately under exam conditions.
掌握生物多样性指数，尤其是辛普森指数，对于学习生态学和生物多样性专题的 A-level 生物学生至关重要。本文通过清晰的解释、例题演练和考试风格的练习，帮助你建立信心，确保你在考试中也能准确计算和解读多样性指数。
1. What Is an Index of Diversity? | 什么是多样性指数？
An index of diversity is a quantitative measure that reflects how many different species there are in a community (species richness) and how evenly the individuals are distributed among those species (species evenness). A high diversity index indicates a stable, complex ecosystem, while a low index may signal environmental stress or dominance by a few species.
多样性指数是一种定量指标，反映一个群落中有多少不同物种（物种丰富度）以及个体在各个物种中分布的均匀程度（物种均匀度）。高多样性指数表明一个稳定、复杂的生态系统，而低指数可能意味着环境胁迫或少数物种占主导。
The most commonly used index in A-level Biology is Simpson's Index of Diversity (D). It gives a value between 0 and 1, with 1 representing infinite diversity and 0 representing no diversity. However, many exam boards use the reciprocal version or a formula that yields higher numbers for greater diversity. Always check the specific formula provided in your exam.
A-level 生物中最常用的指数是辛普森多样性指数 (D)。其值介于 0 到 1 之间，1 代表无限多样性，0 代表没有多样性。然而，许多考试局使用倒数版本或得出较高数字对应较高多样性的公式。请务必核对考试中给出的具体公式。
2. Simpson's Index Formula Breakdown | 辛普森指数公式拆解
The standard Simpson's Index is calculated using the formula:
标准辛普森指数的计算公式为：
D = 1 − ( Σ n(n − 1) / N(N − 1) )
Where: n = total number of organisms of a particular species; N ives a value greater than 1, with higher values indicating greater diversity. We will focus on the more common 0–1 version, but both are tested.
其中：n = 某一物种的个体总数；N = 所有物种的个体总数。总和 Σ 是对所有物种求和。有些课程大纲使用替代形式 D = N(N − 1) / Σ n(n − 1)，得到的值大于 1，值越高表示多样性越高。我们将重点讲解更常见的 0–1 版本，但两种都会考查。
It is crucial to understand that the expression n(n − 1) is calculated for each species and then summed. This sum is divided by N(N − 1), and the result is subtracted from 1. The subtraction reflects the probability that two randomly selected individuals belong to the same species; a low probability of same-species selection means high diversity.
关键是理解表达式 n(n − 1) 要对每个物种分别计算然后求和。这个总和除以 N(N − 1)，再从 1 中减去结果。减法体现的是随机抽取两个个体属于同一物种的概率；同种概率低则多样性高。
3. Understanding n and N With a Worked Table | 通过工作表理解 n 和 N
Consider a simple sample from a pond habitat:
考虑一份取自池塘生境的简单样本：
| Species | Number of individuals (n) | n(n − 1) |
|---|---|---|
| Water flea | 12 | 12 × 11 = 132 |
| Pond snail | 5 | 5 × 4 = 20 |
| Damselfly nymph | 8 | 8 × 7 = 56 |
| Leech | 3 | 3 × 2 = 6 |
| Total | N = 28 | Σ n(n − 1) = 214 |
y high index suggests a reasonably diverse community, though it could be compared with another habitat.
N(N − 1) = 28 × 27 = 756。那么 D = 1 − (214 / 756) = 1 − 0.283 = 0.717。这个相对较高的指数表明群落具有相当高的多样性，尽管它可以与另一生境进行比较。
4. Step-by-Step Calculation Guide | 逐步计算指南
To avoid errors in exams, always follow a systematic approach:
为避免考试中出错，请始终遵循系统步骤：
- List all species and their abundances (n). If only percentage cover or frequency is given, treat these as n values only if the method clearly indicates relative abundance. For point quadrat data, convert to frequency counts.
- 列出所有物种及其丰度 (n)。如果只给出百分比盖度或频度，仅在方法明确表示相对丰度时将其视作 n 值。对于点样方数据，转换为频次计数。
- Calculate n(n − 1) for each species. Watch for values of n = 1; then 1 × 0 = 0, which is important for accuracy.
- 计算每个物种的 n(n − 1)。注意 n = 1 的情况；此时 1 × 0 = 0，这对准确性很重要。
- Sum all n(n − 1) to get Σ n(n − 1).
- 将所有 n(n − 1) 求和得到 Σ n(n − 1)。
- Calculate N by summing all n values, then compute N(N − 1).
- 通过将所有 n 值求和算出 N，然后计算 N(N − 1)。
- Divide Σ n(n − 1) by N(N − 1) and subtract from 1. Keep intermediate values as fractions to maintain precision, or round only at the final step.
- 用 Σ n(n − 1) 除以 N(N − 1) 再从 1 中减去。保留中间值为分数以保持精度，或仅在最后一步四舍五入。
Always quote your final answer to the same number of decimal places as the data or as required by the question (often 2 or 3 d.p.).
最终答案始终保留与数据相同或题目要求的小数位数（通常 2 或 3 位小数）。
5. Interpreting the Index: What Does 0.6 Actually Mean? | 解读指数：0.6 究竟意味着什么？
A Simpson' all equally abundant might have D ≈ 0.8, while another with 5 species but one making up 90% D values between sites allows you to comment on which habitat is healthier or more stable.
辛普森指数 0.6 并不能直接告诉你物种数量，它整合了丰富度和均匀度。例如，一个拥有 5 个物种且个体数均匀的群落，D 值可能约为 0.8，而另一个同样有 5 个物种但其中一个物种占个体数的 90% 的群落，D 值可能为 0.2。因此，比较不同地点的 D 值，你可以评述哪个栖息地更健康或更稳定。
In exam answers, always link diversity values to ecological factors: high diversity often correlates with greater resource availability, more niches, or less disturbance. Low diversity can be due to pollution, extreme conditions, or pioneer stages of succession.
在考试答案中，始终将多样性值与生态因素联系起来：高多样性通常与更丰富的资源、更多的生态位或较少的干扰相关。低多样性可能是由于污染、极端条件或演替的先锋阶段。
6. Exam Question 1: Calculating D for a Farmland Ecosystem | 真题示例 1：计算农田生态系统的 D 值
Question: conventionally managed field. The data for the organic field are shown. Calculate Simpson' using the formula D = 1 − [ Σ n(n − 1) / N(N − 1) ]. Give your answer to 2 decimal places.
题目：一名学生对两块田地中的无脊椎动物进行取样：一块有机田和一块常规管理田。有机田的数据如下表所示。利用公式 D = 1 − [ Σ n(n − 1) / N(N − 1) ] 计算辛普森多样性指数 (D)。答案保留两位小数。
| Species | Individuals (n) |
|---|---|
| Earthworm | 15 |
| Beetle | 9 |
| Spider | 6 |
| Woodlouse | 10 |
| Centipede | 4 |
Worked solution: +30+90+12 = 414. Total N = 15+9+6+10+4 = 44. N(N − 1) = 44×43 = 1892. D = 1 − (414/1892) = 1 − 0.2188... = 0.78 (to 2 d.p.).
解题步骤：首先，计算每个 n(n − 1)：蚯蚓：15×14=210；甲虫：9×8=72；蜘蛛：6×5=30；潮虫：10×9=90；蜈蚣：4×3=12。Σ n(n − 1) = 210+72+30+90+12 = 414。总个体数 N = 15+9+6+10+4 = 44。N(N − 1) = 44×43 = 1892。D = 1 − (414/1892) = 1 − 0.2188... = 0.78（保留两位小数）。
s greater species diversity, likely due to no pesticide use and more varied microhabitats.
这个数值表明有机田中多样性较高。常规田（数据未显示）得出的 D=0.42，因此有机田物种多样性更大，可能是因为未使用农药且拥有更多样的微生境。
7. Exam Question 2: Using the Reciprocal Simpson's Index | 真题示例 2：使用倒数辛普森指数
n(n − 1). Here, larger values mean higher diversity. A sample from a rocky shore gave the following: limpet 34, barnacle 22, periwinkle 18, dog whelk 6. Calculate D using this reciprocal index.
有些考试局，如 AQA，可能使用倒数公式：D = N(N − 1) / Σ n(n − 1)。此时数值越大表示多样性越高。一份来自岩岸的样本数据如下：帽贝 34，藤壶 22，滨螺 18，荔枝螺 6。使用此倒数指数计算 D。
= 80. N(N − 1)=80×79=6320. D = 6320 / 1920 = 3.29. This is a moderate diversity; compare with an adjacent polluted site which gave D=2.1, indicating lower diversity due to contamination.
首先计算 n(n − 1)：帽贝：34×33=1122；藤壶：22×21=462；滨螺：18×17=306；荔枝螺：6×5=30。总和 = 1122+462+306+30 = 1920。N = 34+22+18+6 = 80。N(N − 1)=80×79=6320。D = 6320 / 1920 = 3.29。这是中等多样性；与邻近受污染站点得出的 D=2.1 相比，污染导致多样性较低。
Always read the question carefully to identify which formula is required. Mixing up the two can cost significant marks.
务必仔细审题以确认要求使用哪个公式。混淆两者会丢失大量分数。
8. Dealing with Transect and Quadrat Data | 处理样带和样方数据
Exam questions often present diversity data in the context of a transect across a sand dune or a series of quadrats along an environmental gradient. For Simpson's Index, you must pool the data from all sampling points or calculate an index for each sampling point and then average, depending on the question. However, pooling individual counts is the standard method: treat all organisms from the combined samples as one community.
考题常常在沙丘样带或沿环境梯度放置一系列样方的背景下给出多样性数据。对于辛普森指数，你必须将所有采样点的数据合并计算，或根据问题要求为每个采样点计算指数然后取平均值。但合并个体计数是标准方法：将组合样本中的所有生物视作一个群落。
If you are given percentage cover data from quadrats, you cannot directly use percentage values as n because an n value of, say, 65% would give a huge n(n − 1) that is meaningless. Instead, you either use an abundance scale (e.g., Braun-Blanquet) converted to numerical values, or the question will suggest an alternative index like the Shannon-Wiener Index. For Simpson's index, you typically need actual counts.
如果给你的是样方的百分比盖度数据，你不能直接将百分比值用作 n，因为比如 65% 的 n 值将得出毫无意义的巨大 n(n − 1)。相反，你要么使用转化为数值的丰度等级（如 Braun-Blanquet 等级），要么题目会建议使用其他指数如香农-威纳指数。对于辛普森指数，你通常需要实际计数。
9. Common Pitfalls and How to Avoid Them | 常见错误及避免方法
Forgetting to subtract 1: Many students calculate n² instead of n(n − 1), leading to a wildly incorrect sum. Practise writing the full multiplication each time until it becomes automatic.
忘记减 1：许多学生计算的是 n² 而不是 n(n − 1)，导致总和完全错误。每次练习时写出完整的乘法算式，直到成为习惯。
Misidentifying N: N is the total number of all individuals, not the number of species. Always double-check your total by summing the n column.
误认 N：N 是所有个体的总数，而不是物种数。始终通过加总 n 列表来反复检查你的总数。
Incorrect rounding: If the question specifies 2 decimal places, keep all working values unrounded and only round the final answer. Rounding early can distort the result.
舍入错误：如果题目要求保留两位小数，则所有中间值都不应舍入，仅对最后答案进行四舍五入。过早舍入会扭曲结果。
Interpreting D backwards: 0 indicates very low diversity. With the reciprocal, a large number means high diversity. Make a note next to your answer to avoid losing interpretation marks.
反向解读 D 值：对于 D = 1 − [...] 公式，接近 0 的值表示极低的多样性。对于倒数公式，数值大表示高多样性。在你的答案旁边做注释，以避免丢掉解读分数。
10. Practice Questions for Self-Assessment | 自测练习题
Try these yourself before checking the answers.
在核对答案前先自己尝试一下。
Q1: − [ Σ n(n − 1) / N(N − 1) ] to calculate diversity. (Ans: N=27, Σ=254, N(N−1)=702, D=0.64)
第 1 题：一份林地地表样本包含：蛞蝓 7、蜗牛 3、潮虫 12、甲虫 5。使用 D = 1 − [ Σ n(n − 1) / N(N − 1) ] 计算多样性。（答案：N=27，Σ=254，N(N−1)=702，D=0.64）
Q2: Students compare two streams. Stream A has mayfly nymph 20, stonefly nymph 15, caddisfly larva 10, flatworm 5. Stream B has midge larva 40, leech 4, water hog 6. Calculate D for each (using 1 − formula) and comment on water quality. (Hint: more sensitive species in A; B likely polluted.)
第 2 题：学生对比两条溪流。溪流 A 有蜉蝣稚虫 20、石蝇稚虫 15、石蛾幼虫 10、涡虫 5。溪流 B 有摇蚊幼虫 40、水蛭 4、水鼠妇 6。计算各自的 D 值（使用 1 − 公式）并评述水质。（提示：A 中敏感物种较多；B 可能受污染。）
Q3 (reciprocal): ) to find which meadow is more diverse. (Ans: A: D=3.75; B: D=1.45; so A is more diverse.)
第 3 题（倒数公式）：草甸 A：毛茛 30、三叶草 25、黑麦草 45、车前草 15。草甸 B：黑麦草 80、三叶草 20、蓟 5。使用 D = N(N − 1) / Σ n(n − 1) 找出哪片草甸更具多样性。（答案：A：D=3.75；B：D=1.45；所以 A 更具多样性。）
11. Linking Index to Fieldwork and Statistical Tests | 将指数与野外调查和统计检验相联系
Once you have calculated diversity indices for multiple sites, you are often required to apply a statistical test, such as Spearman's rank correlation, to see if diversity correlates with an abiotic factor like soil pH or light intensity. The index of diversity serves as the dependent variable. You may also be asked to critique the sampling method: random sampling with quadrats is essential to avoid bias.
一旦你计算出多个地点的多样性指数，通常会被要求应用统计检验，如斯皮尔曼等级相关系数，来检验多样性是否与土壤 pH 或光照强度等非生物因素相关。多样性指数作为因变量。你还可能被要求评价采样方法：使用样方进行随机采样对于避免偏差至关重要。
Remember that comparisons between indices are only valid if the sampling effort (area, number of quadrats, season) is standardised. In exam evaluations, always state that the same size quadrat and sampling technique were used.
请记住，只有在取样努力（面积、样方数量、季节）标准化的前提下，指数间的比较才有效。在考试评估中，务必说明使用了相同大小的样方和相同的取样技术。
12. Final Exam Tips and Summary | 最终考试技巧与总结
When you encounter an index of diversity question, allocate your time wisely: the calculation is often worth 3–4 marks, while the interpretation and comparison can earn another 2–3 marks. Show all stages of working; even if your final answer is incorrect, you can gain method marks for the correct n(n − 1) expressions and correct sums.
当你遇到多样性指数题目时，要明智分配时间：计算通常值 3–4 分，而解读与比较又能获得 2–3 分。展示所有工作步骤；即使最终答案错误，正确的 n(n − 1) 表达式和正确求和也能让你得到方法分。
In summary, practise using both forms of Simpson's Index, double-check your arithmetic, and always relate the numerical result to biological concepts such as ecosystem stability, succession, or human impact. With the examples and practice provided here, you are well-prepared to tackle any index-related exam question with precision.
总之，练习辛普森指数的两种形式，反复检查算术，并始终将数值结果与生态系统稳定性、演替或人类影响等生物学概念联系起来。通过本文提供的示例与练习，你已做好充分准备，能够精准应对任何与指数相关的考题。
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