1) Given the species list in Table 5.1, calculate the following community metric
ID: 115250 • Letter: 1
Question
1) Given the species list in Table 5.1, calculate the following community metrics: species richness (S) Shannon diversity index (H), maximum diversity (Hmax) and evenness index (J) for each of the 100 m-2 forest plots. See above for the formulae. a. b. c. 2) What calculations would you do to estimate the community metrics for the overall forest regeneration on the Eastern Shore of NS? Do the calculations you are proposing. Forest inventory (number of individuals per 100m2) of five forest plots that were harvested Table 5.1 using clearcutting in the same year on the Eastern Shore of Nova Scotia Plot 1 15 10 Plot 2 12 14 2 Plot 3 11 17 Plot 4 21 Plot 5 18 12 Species Abies balsamea Picea rubens Pinus strobus Betula alleghaniensis Acer rubrum 6 43 6 38 Total Number of Trees 30 41 36Explanation / Answer
1)
a) species richness (s)
for Plot 1 = 4
for plot 2 = 6
for plot 3 = 4
for plot 4 = 6
for plot 5 = 5
b) shannon diversity index = - Summation[P(i) * lnP(i)]
for plot 1
for abies balsamea = P(i) = 15/30 = 0.5
InP(i) = -0.69
= 0.5*(-0.69)
= -0.345
for picea rubens = P(i) = 10/30
= 0.33
InP(i) = -1.10
= 0.33*-1.10
= -0.363
betula alleghaniensis = P(i) = 2/30
= 0.06
InP(i) = -2.81
= 0.66*-2.81
= -1.85
for acer saccharum = P(i) = 3/30
= 0.1
InP(i) = -2.30
= 0.1*-2.30
= -0.23
so shannon diversity index for plot 1 = 2.78
for plot 2
for abies balsamea = P(i) = 12/41 = 0.29
InP(i) = -1.22
= 0.29*(-1.22)
= -0.353
for picea rubens = P(i) = 14/41
= 0.34
InP(i) = -1.07
= 0.34*-1.07
= -0.363
for pinus strobus = p(i) = 2/41
= 0.04
InP(i) = -3.02
= 0.04*-3.02
= -0.120
betula alleghaniensis = P(i) = 5/41
= 0.12
InP(i) = -2.10
= 0.12*-2.10
= -0.252
for acer rubrum = P(i) = 1/71
= 0.01
InP(i) = -4.26
= 0.01*-4.26
= -0.0426
for acer saccharum = P(i) = 7/41
= 0.17
InP(i) = -1.76
= 0.17*-1.76
= -0.29
so shannon diversity index for plot 2 = 1.42
for plot 3
for abies balsamea = P(i) = 11/36 = 0.30
InP(i) = -1.18
= 0.30*(-1.18)
= -0.354
for picea rubens = P(i) = 17/36
= 0.47
InP(i) = -0.750
= 0.47*-0.750
= -0.352
for acer rubrum = P(i) = 5/36
= 0.13
InP(i) = -1.97
= 0.13*-1.97
= -0.256
for acer saccharum = P(i) = 3/36
= 0.08
InP(i) = -2.48
= 0.08*-2.48
= -0.19
so shannon diversity index for plot 3 = 1.15
for plot 4
for abies balsamea = P(i) = 21/43= 0.48
InP(i) = -0.71
= 0.48*(-0.71)
= -0.34
for picea rubens = P(i) = 7/43
= 0.16
InP(i) = -1.81
= 0.16*-1.81
= -0.29
for pinus strobus = p(i) = 5/43
= 0.11
InP(i) = -2.15
= 0.11*-2.15
= -0.23
betula alleghaniensis = P(i) = 2/43
= 0.04
InP(i) = -3.06
= 0..04*-3.06
= -0.12
for acer rubrum = P(i) = 2/43
= 0.04
InP(i) = -3.06
= 0.04*-3.06
= -0.12
for acer saccharum = P(i) = 6/43
= 0.13
InP(i) = -1.96
= 0.13*-1.96
= -0.25
so shannon diversity index for plot 4 = 1.35
for plot 5
for abies balsamea = P(i) = 18/38= 0.47
InP(i) = -0.74
= 0.47*(-0.74)
= -0.35
for picea rubens = P(i) = 12/38
= 0.31
InP(i) = -1.15
= 0.31*-1.15
= -0.35
for pinus strobus = p(i) = 1/38
= 0.02
InP(i) = -3.63
= 0.02*-3.63
= -0.07
betula alleghaniensis = P(i) = 1/38
= 0.02
InP(i) = -3.63
= 0..02*-3.63
= -0.07
for acer saccharum = P(i) = 6/38
= 0.15
InP(i) = -1.84
= 0.15*-1.84
= -0.27
so shannon diversity index for plot 5 = 1.11
maximum diversity(Hmax) is for plot 1 = 2.78
c) evenness index (J) for plot 1 = Shannon’s diversity index H/In(S)
= 2.78/1.38
= 2.01
evenness index (J) for plot 2 = Shannon’s diversity index H/In(S)
= 1.42/1.79
= 0.79
evenness index (J) for plot 3 = Shannon’s diversity index H/In(S)
= 1.15/1.38
= 0.83
evenness index (J) for plot 4 = Shannon’s diversity index H/In(S)
= 1.35/1.79
= 0.75
evenness index (J) for plot 5 = Shannon’s diversity index H/In(S)
= 1.11/1.60
= 0.69
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