Bio360-Lab11
MEASURING BIODIVERSITY
Fall 2006
INTRODUCTION In this exercise we will examine several concepts related to biodiversity (otherwise known as diversity). The term biodiversity describes the number and abundance of species inhabiting a community or a habitat. The concept of biodiversity is fairly important to community ecologists, since this serves as a starting point in attempts to explain patterns of community structure and dynamics. In this lab we will examine biodiversity of plants in two area on campus. Measuring Biodiversity The most basic measure of biodiversity is a count of the total number of species present in a community, termed species richness and symbolized by S. However, simply determining S may not always give a complete picture of the community. For instance, consider the two hypothetical communities below: # of Individuals in: Community X Community Z 96 33 3 33 1 34
Species 1 Species 2 Species 3
If we were basing our assessment of the communities only on S, we would say they were the same, since S = 3 in both. However, this clearly is not the case. Community X is strongly dominated by species 1, while species in community Z are essentially equally abundant. To ecologists, community Z looks more “diverse” than community X, since community X consists, for practical purposes, of just a single species. Thus, to take abundance into account we often calculate a diversity index, the two most common of which are the Shannon index (a.k.a. Shannon-Weaver or Shannon-Weiner index) and the Simpson index. Both the Shannon and Simpson indices use the variable pi, which is simply the proportion of species i in the population (the number of individuals of species i divided by the total number of individuals in the population). The formulas for each index are as follows: Shannon index: H = - (pi *ln pi) Simpson index: D = 1 / (pi2), where ln is the natural log function. For example, in community X we have p 1 = 0.96 (=96/100), p2 = 0.03 (=3/100), and p3 = 0.01 (=1/100). Using these values, we can calculate: H = - [(0.96)*(ln 0.96) + (0.03)*(ln 0.03) + (0.01)*(ln 0.01)] = 0.19 D = 1 / [(0.96)2 + (0.03)2 + (0.01)2] = 1 / (0.9226) = 1.08. Using the same formulas, you can calculate values of H = 1.10 and D = 2.99 for community Z (try it and see!). As you can see, for a given community H and D do not give the same value. However, H and D are both larger in community Z. Thus, communities with more equal abundances among the different species will have a higher diversity, assuming they have equal richness. Finally, there is a measure called species evenness, which attempts to measure the equality of the abundances for each species. The formulas for evenness are given below:
Bio360-Lab11
Shannon evenness index: J = H / ln(S) Simpson evenness index: E = D / S.
Fall 2006
Using our values calculated above, we get J = 0.17 and E = 0.36 for community X, and J = 0.9999 and E = 0.9998 for community Z. Both E or J have a maximum value of 1, which occurs when all species have equal abundances. Thus, we can conclude that community Z has greater evenness than community X. The Species Accumulation Curve and the Rank-Abundance Curve Typically, our estimate of S is based on a series of samples taken from the habitat, with S increasing every time we find a new species. We can summarize the results of this process with a species accumulation curve (also called a species-area curve):
You should note two things about the graph. First, the y-axis counts cumulative number of species; that is, for each new sample you add only new species to the already existing total. Second, the curve flattens out as you increase the number of samples taken. For example, in the curve above you only have to look at about 22 organisms to reach 10 distinct species. However, to add another 10 distinct species you have to look at an additional 80 or so individuals (that is, you don’t get to 20 total species until you’ve gone through about 105 individuals). This indicates that, initially, you find a lot of new species, but as you go along the chances of finding something new gradually decreases, since in later samples most of the organisms you collect have previously been collected. A second way to illustrate our results is to construct a graph of proportional abundance versus rank (with rank = 1 being the most abundant, rank =2 being second most abundant, and so on), which we call a rank-abundance curve:
Bio360-Lab11
Fall 2006
You’ll note that all rank-abundance curves decline as the rank increases, but some do so more steeply than others. For example, the curve labeled “Bromilead, Rain” goes down much more rapidly than does, for instance, the curve “Cup, Rain.” This tells you that the “Bromilead, Rain” community is less even than the “Cup, Rain” community; in other words, the bromeliad community is dominated by a few common species, while the cup community has a number of equally common species. GENERAL METHODS We will examine plant communities from two distinct habitats on campus; details on these habitats will be given in class. From each habitat, you will gather data on numbers of individuals of species in a given area. Using a 1m plot you will estimate number of species presence from a random sampling of the environment. Identify species to order. STATISTICAL ANALYSIS 1. For each habitat type, you will calculate species richness (S), species diversity (H and D), and species evenness (J and E). 2. For each habitat type, you will draw a species accumulation curve and a rank-abundance curve. 3. We also want to compare the habitats to see if they differ, and we’ll use two different methods. a. First, use the class data to perform a t-test comparing the Shannon diversity index (H) between the two habitats. b. Second, we’ll use the Kolmogorov-Smirnov (K-S) test. This test compares the distribution of the two habitats, and again we’ll use the data from the entire class. The procedure works like this. To begin, rank the abundance data for each site from highest to lowest (do this separately for each site). Next, calculate pi for each species, and then calculate the cumulative percentages (that is, add each p i to the sum of all preceding pi’s). The K-S test is then performed by taking the absolute differences of the cumulative pi’s (called D), and then taking the maximum of these differences (Dmax). Dmax is then compared against critical values of D (Dcrit). To calculate this, we use the formula:
Dcrit 1.358 * (n1 n2 ) n1n2 ,
where n1 = the total number of organisms in habitat 1 and n2 = total number of organisms in habitat 2. If Dmax > Dcrit, then our two habitats are significantly different in overall biodiversity. I’ve given an example at the end of the handout.
Bio360-Lab11
LAB WRITE-UP The TA will inform you as to what s/he wants in terms of a lab writeup. AN EXAMPLE OF THE KOLMOGOV-SMIRNOV TEST Suppose we have the following set of data for two habitats, site 1 and site 2: Number of Individuals Species Site 1 Site 2 a 2 14 b 4 2 c 6 0 d 1 0 e 8 0 f 9 12 g 3 7 h 4 6 i 6 1 j 3 1 Total 47 44 If we rank the species by abundance and calculate the pi’s, we get: Site 1 Species No. pi Cumulative Proportion g 9 0.192 0.192 e 8 0.170 0.362 c 6 0.128 0.490 j 6 0.128 0.618 b 4 0.085 0.703 i 4 0.085 0.788 h 3 0.064 0.852 l 3 0.064 0.916 a 2 0.043 0.959 d 1 0.021 0.98 k 1 0.021 1.00
Fall 2006
Site 2 Species No. pi Cumulative Proportion a 14 0.318 0.318 g 12 0.273 0.591 h 7 0.159 0.750 i 6 0.136 0.886 b 2 0.046 0.932 j 1 0.023 0.955 k 1 0.023 0.978 l 1 0.023 1.000 c 0 0.000 1.000 d 0 0.000 1.000 e 0 0.000 1.000
Bio360-Lab11
Fall 2006
Note that species c, d, and e are still listed in site 2 (despite being absent from this site), because we are including all species found at either site. Next, we find D by subtracting the cumulative proportion from site 2 from the corresponding cumulative proportion from site 1 (and ignore any negative signs): Site 1 Cum. Proportion Site 2 Cum. Proportion D 0.192 0.318 0.126 0.362 0.591 0.229 0.490 0.750 0.260 0.618 0.886 0.268 0.703 0.932 0.229 0.788 0.955 0.167 0.852 0.978 0.126 0.916 1.000 0.084 0.959 1.000 0.041 0.980 1.000 0.020 1.000 1.000 0.000 The greatest value of D, or Dmax, is highlighted; it equals 0.268. We now want to determine Dcrit. Since n1 = 47 and n2 = 44, we have:
Dcrit 1.358 * (47 44) (47 * 44) 1.358 * 0.2098 0.285 .
Since Dmax (=0.268) is less than Dcrit, our two sites do not differ in species diversity.