Question

Hello, I just need help answering these 2 questions. I am confused as to what I am supposed to do since they seem pretty straightforward but I do not want to get them wrong. Thank you.

Exercise 3: Diversity and Spatial Scale Species diversity typically increases with greater area sampled or the scale of perception. For instance, you will tend to find the lowest number of species at local sites and the maximum number when a whole region is considered. This is because a region contains a pool of species that reflects the history of the region and are available to colonize local sites. However, even when comparing among different local sites, you may find that the membership of communities will differ because of differences in individual species niches and differences in (micro)habitats and their abiotic factors. You will likely find that sites closer together are more similar in terms of community membership and that similarity decreases with increasing distance. This difference among local sites across a region is referred to as species turnover as individual species begin to replace each other over space. Interestingly, there is a way to consider species diversity across spatial scale. If we standardize our sampling of local sites across a region, diversity and scale are related to each other as... y=aß Or in other words: Regional diversity (Gamma) = Local or within habitat diversity (Alpha) X Turnover or between habitat diversity (Beta). Regional diversity is all of the species in the region

Answer 1

3.1)

In order to find out the number of species in the regional pool , the regional diversity has to be found out.This is done by simply sustituting the values for that is given by :

aand 8

  

7= a. 8

= 20 * 1.5 ( since 20 is the number of average species)

= 30  

So,the number of species in the regional pool is 30.

3.2)

Inorder to find the pond turnover the following equation can be used

Turnover =$\gamma/ \alpha$

where: $\gamma =$ number of species in an entire study area and

$\alpha =$ the number of species per plot within the study area

Therefore ,

Substituting in the equation,

Turnover = 12/4.5

=2.66666