See the following tables and accordingly please answer the question. P.S: please
ID: 300127 • Letter: S
Question
See the following tables and accordingly please answer the question. P.S: please show your work so I can know how to solve such questions.
* What are the species richness using each of the two methods
* Calculate Shannon Diversity index, eH and evenness using each survey method
* Calculate similarity index between the two methodologies
Line Transect method species LT1 LT2 LT3 LT4 LT5 LT6 LT7 LT8 LT9 LT10 LT11 LT12 LT13 LT14 LT15 LT16 LT17 LT18 LT19 LT20 LT21 Total Zygophylum qatarense 8 3 1 1 4 4 3 2 6 2 6 5 7 6 1 1 4 10 9 9 2 94 Heliotropium bacciferum 1 0 0 0 0 0 0 1 2 1 0 2 3 4 3 6 0 0 0 0 0 23 Aeluropus logopoides 1 8 4 4 10 13 10 15 2 3 16 2 1 1 1 22 8 0 2 7 9 139 Acacia tortilis 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 Tamarix aphla 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 1 1 5 Prosopis juliflora 0 0 0 0 0 0 0 0 0 0 3 5 0 0 0 0 0 0 0 0 0 8 Sporobolus arabicue 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 9 0 0 0 0 0 11 Phagonia sp. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 SUM 284Explanation / Answer
Species richness in the question is eight. Species richness is count of the species in an ecological environment.
To count Shannon diversity index we use following formula:
H = Pi x lnPi
Where H is Shannon diversity index, Pi is proportion of Species richness made up of ith number species.
Line transect
Total number of species in each Line transect
Total; number of individuals
LT1
3
10
LT2
2
11
LT3
2
5
LT4
2
5
LT5
3
14
LT6
2
17
LT7
4
17
LT8
4
19
In Line transects LT1 the species Zygophylum qatarense has 8 individuals in respect of total 10 species so it has 0.8p [(8/10) x 100 = 80%] as proportion in species richness.
So Shannon diversity index for Zygophylum qatarense is:
H = - Pi x lnPi = - 0.8 x ln 0.8
= - 0.8 x 2.303 x log0.8 [ln x =2.303 logx]
= -1.84 x (-0.097)
H = 0.1784
Shannon's equitability eH is calculated by H/ ln S
= 0.1784 / ln 3 = 0.1784/ (2.303 log 3)
= 0.1784 / (2.3 x 0.478)
= 0.1784/ 1.099 0.178 / 1.1
= 0.1618
Line transect
Total number of species in each Line transect
Total; number of individuals
LT1
3
10
LT2
2
11
LT3
2
5
LT4
2
5
LT5
3
14
LT6
2
17
LT7
4
17
LT8
4
19