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Please see the attachment. This is a lab assignment. Forest Composition Lab.
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General Environmental Science I
ENV 121
Forest Composition Lab
Name: __________________________
Background
(Deleted for Plagiarism Purposes)
Day 2 – Back in the Lab
Data Sheet
species
# in plot
relative density
sum basal area
Chestnut oak
Northern Red
Oak
Black Oak
White Oak
3
0.30
111.95
relative
dominance
0.072
2
0.20
248.26
0.162
0.362
1
4
0.10
0.40
140.45
1,039.82
0.091
0.675
0.191
1.075
Total
10
1
1,540.48
1
2
Relative Density:
Chestnut Oak: Number of trees/Total Number of trees = 3/10 = 0.30
Northern Red Oak: 2/10 = 0.20
Black Oak: 1/10 = 0.10
White Oak: 4/10 = 0.40
Sum Basal Area:
1- Chestnut Oak ➔ Sum = 29.55 + 46.62 + 35.78 = 111.95 cm2
Tree 1: A = π * (C/2π)2 = 29.55
Tree 2: A = π * (C/2π)2 = 46.62
Tree 3: A = π * (C/2π)2 = 35.78
2- Northern Red Oak ➔ Sum = 122.34 + 125.92 = 248.26 cm2
Tree 1: A = π * (C/2π)2 = 122.34
Tree 2: A = π * (C/2π)2 = 125.92
3- Black Oak: Tree 1: A = π * (C/2π)2 = 140.45 cm2
4- White Oak ➔ Sum = 207.08 + 296.26 + 243.29 + 293.19 = 1,039.82 cm2
Tree 1: A = π * (C/2π)2 = 207.08
Tree 2: A = π * (C/2π)2 = 296.26
Tree 3: A = π * (C/2π)2 = 243.29
Tree 4: A = π * (C/2π)2 = 293.19
Relative Dominance:
Chestnut Oak: 111.95 / 1540.48 = 0.072
Northern Red Oak: 248.26 / 1540.48 = 0.162
Black Oak: 140.45 / 1540.48 = 0.091
White Oak: 1039.82 / 1540.48 = 0.675
Importance Value:
Chestnut Oak: 0.30 + 0.072 = 0.372
1
importance value
0.372
Northern Red Oak: 0.20 + 0.162 = 0.362
Black Oak: 0.10 + 0.091 = 0.191
White Oak: 0.40+0.675 = 1.075
D = Σ (n / N)2 = (0.2)2 + (0.1)2 + (0.4)2 + (0.3)2 = 0.3
➔ S = (?)
Discussion Questions
1- Species richness measures the number o...