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![Looks complex but it’s not..!
D = 1 – [ ∑ ( n / N)2 ]
n = number of individuals
N = total number of individuals](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-8-320.jpg)
![Calculating Simpsons diversity index
(D)
Species observed Percentage cover
Field A (n) Field B (n)
Cocksfoot grass 57 38
Timothy grass 32 16
Buttercup 3 14
Clover 3 22
Thistle 1 5
Dandelion 4 5
Total (N) 100 100
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-9-320.jpg)
![Calculating Simpsons diversity index
(D)
Species observed Percentage cover
Field A (n) n/N
Cocksfoot grass 57 0.57
Timothy grass 32 0.32
Buttercup 3 0.03
Clover 3 0.03
Thistle 1 0.01
Dandelion 4 0.04
Total (N) 100
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-10-320.jpg)
![Calculating Simpsons diversity index
(D)
Species observed Percentage cover
Field A (n) n/N (n/N)2
Cocksfoot grass 57 0.57 0.349
Timothy grass 32 0.32 0.1024
Buttercup 3 0.03 0.0009
Clover 3 0.03 0.0009
Thistle 1 0.01 0.0001
Dandelion 4 0.04 0.0016
Total (N) 100 ∑ = 0.4308
D = 1 – [ ∑ ( n / N)2 ]
D = 1 – 0.4308
D = 0.5692](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-11-320.jpg)
![Now calculate (D) for Field B…
Species observed Percentage cover
Field A (n) Field B (n)
Cocksfoot grass 57 38
Timothy grass 32 16
Buttercup 3 14
Clover 3 22
Thistle 1 5
Dandelion 4 5
Total (N) 100 100
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-12-320.jpg)
![Field B (D)
Species observed Percentage cover
Field B (n) n/N (n/N)2
Cocksfoot grass 38 0.38 0.1444
Timothy grass 16 0.16 0.0256
Buttercup 14 0.14 0.0196
Clover 22 0.22 0.0484
Thistle 5 0.05 0.0025
Dandelion 5 0.05 0.0016
Total (N) 100 ∑ = 0.243
D = 1 – [ ∑ ( n / N)2 ]
D = 1 – 0.243
D = 0.757](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-13-320.jpg)
![Survey of animals in a freshwater
stream.
species Number (n) n / N (n / N)2
Gammarus pulex ( water shrimp) 150 0.289 0.084
Asellus aquaticus ( water louse) 32 0.062 0.004
Baetis rhodani ( mayfly nymph) 113 0.218 0.047
Lymnaea peregra ( snail) 2 0.004 0.000016
Rhyacophila ( caddis-fly nymph) 12 0.023 0.000529
Chironimidae ( midge larvae) 210 0.405 0.164
Total 519 ∑ = 0.299
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-20-320.jpg)
![Survey of animals in a freshwater
stream.
species Number (n) n / N (n / N)2
Gammarus pulex ( water shrimp) 150 0.289 0.084
Asellus aquaticus ( water louse) 32 0.062 0.004
Baetis rhodani ( mayfly nymph) 113 0.218 0.047
Lymnaea peregra ( snail) 2 0.004 0.000016
Rhyacophila ( caddis-fly nymph) 12 0.023 0.000529
Chironimidae ( midge larvae) 210 0.405 0.164
Total 519 ∑ = 0.299
D = 1 – [ ∑ ( n / N)2 ]
D = 1 – 0.299
= 0.7](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/85/Measuring-biodiversity-21-320.jpg)
![Looks complex but it’s not..!
D = 1 – [ ∑ ( n / N)2 ]
n = number of individuals
N = total number of individuals](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-8-2048.jpg)
![Calculating Simpsons diversity index
(D)
Species observed Percentage cover
Field A (n) Field B (n)
Cocksfoot grass 57 38
Timothy grass 32 16
Buttercup 3 14
Clover 3 22
Thistle 1 5
Dandelion 4 5
Total (N) 100 100
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-9-2048.jpg)
![Calculating Simpsons diversity index
(D)
Species observed Percentage cover
Field A (n) n/N
Cocksfoot grass 57 0.57
Timothy grass 32 0.32
Buttercup 3 0.03
Clover 3 0.03
Thistle 1 0.01
Dandelion 4 0.04
Total (N) 100
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-10-2048.jpg)
![Calculating Simpsons diversity index
(D)
Species observed Percentage cover
Field A (n) n/N (n/N)2
Cocksfoot grass 57 0.57 0.349
Timothy grass 32 0.32 0.1024
Buttercup 3 0.03 0.0009
Clover 3 0.03 0.0009
Thistle 1 0.01 0.0001
Dandelion 4 0.04 0.0016
Total (N) 100 ∑ = 0.4308
D = 1 – [ ∑ ( n / N)2 ]
D = 1 – 0.4308
D = 0.5692](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-11-2048.jpg)
![Now calculate (D) for Field B…
Species observed Percentage cover
Field A (n) Field B (n)
Cocksfoot grass 57 38
Timothy grass 32 16
Buttercup 3 14
Clover 3 22
Thistle 1 5
Dandelion 4 5
Total (N) 100 100
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-12-2048.jpg)
![Field B (D)
Species observed Percentage cover
Field B (n) n/N (n/N)2
Cocksfoot grass 38 0.38 0.1444
Timothy grass 16 0.16 0.0256
Buttercup 14 0.14 0.0196
Clover 22 0.22 0.0484
Thistle 5 0.05 0.0025
Dandelion 5 0.05 0.0016
Total (N) 100 ∑ = 0.243
D = 1 – [ ∑ ( n / N)2 ]
D = 1 – 0.243
D = 0.757](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-13-2048.jpg)
![Survey of animals in a freshwater
stream.
species Number (n) n / N (n / N)2
Gammarus pulex ( water shrimp) 150 0.289 0.084
Asellus aquaticus ( water louse) 32 0.062 0.004
Baetis rhodani ( mayfly nymph) 113 0.218 0.047
Lymnaea peregra ( snail) 2 0.004 0.000016
Rhyacophila ( caddis-fly nymph) 12 0.023 0.000529
Chironimidae ( midge larvae) 210 0.405 0.164
Total 519 ∑ = 0.299
D = 1 – [ ∑ ( n / N)2 ]](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-20-2048.jpg)
![Survey of animals in a freshwater
stream.
species Number (n) n / N (n / N)2
Gammarus pulex ( water shrimp) 150 0.289 0.084
Asellus aquaticus ( water louse) 32 0.062 0.004
Baetis rhodani ( mayfly nymph) 113 0.218 0.047
Lymnaea peregra ( snail) 2 0.004 0.000016
Rhyacophila ( caddis-fly nymph) 12 0.023 0.000529
Chironimidae ( midge larvae) 210 0.405 0.164
Total 519 ∑ = 0.299
D = 1 – [ ∑ ( n / N)2 ]
D = 1 – 0.299
= 0.7](https://image.slidesharecdn.com/measuringbiodiversity-190502082806/75/Measuring-biodiversity-21-2048.jpg)
Uploaded byGuerillateacher
Measuring biodiversity
The document discusses methods for measuring biodiversity, including species richness, species evenness, and Simpson's Index of Diversity. Species richness refers to the number of different species, while evenness refers to the abundance of individuals within each species. Simpson's Index takes into account both richness and evenness, with a higher index value indicating more diverse habitats that can withstand environmental impacts. The document provides examples of calculating Simpson's Index for various habitats and explains that an index value of 0.7 for a freshwater stream survey indicates good species richness and evenness, making the stream more resilient to changes.
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Measuring biodiversity
- 1. Measuring Biodiversity • KeyConcepts: • Species richness • Species evenness • Simpson’s Index of Diversity (D)
- 2. Species richness • Thisis a qualitative description… • Eg “how many different species can I see?” • More species does not always mean more biodiversity…why not? …because there may not be many individuals of each species (evenness)
- 3. Species evenness • Thisis a quantitative measurement • It is a measure of the abundance of individuals in each species.
- 4. Abundance can bemeasured… • Pecentage cover – the proportion of each quadrat occupied by the species. • Population density – the number of individuals per quadrat • Species frequency – the proportion of quadrats with the species in it.
- 5. When both speciesrichness and species evenness increase, there is by definition an increase in BIODIVERSITY.
- 6. Which field showsthe most biodiversity? Species observed Percentage cover Field A Field B Cocksfoot grass 57 38 Timothy grass 32 16 Buttercup 3 14 Clover 3 22 Thistle 1 5 Dandelion 4 5 Total 100 100 Both have the same ‘richness’ ( 6 species), but Field B has greater ‘evenness’; so Field B is more diverse.
- 7. Simpson’s diversity index(D) is a measure of biodiversity that takes into account richness and evenness. • A high value for D is ‘good’ and means the habitat is diverse, species rich, and able to withstand some environmental impact. • A low value for D is ‘poor’ and means the habitat is low in species, so a small change to the environment ( eg pollution) would have a serious impact.
- 8. Looks complex butit’s not..! D = 1 – [ ∑ ( n / N)2 ] n = number of individuals N = total number of individuals
- 9. Calculating Simpsons diversityindex (D) Species observed Percentage cover Field A (n) Field B (n) Cocksfoot grass 57 38 Timothy grass 32 16 Buttercup 3 14 Clover 3 22 Thistle 1 5 Dandelion 4 5 Total (N) 100 100 D = 1 – [ ∑ ( n / N)2 ]
- 10. Calculating Simpsons diversityindex (D) Species observed Percentage cover Field A (n) n/N Cocksfoot grass 57 0.57 Timothy grass 32 0.32 Buttercup 3 0.03 Clover 3 0.03 Thistle 1 0.01 Dandelion 4 0.04 Total (N) 100 D = 1 – [ ∑ ( n / N)2 ]
- 11. Calculating Simpsons diversityindex (D) Species observed Percentage cover Field A (n) n/N (n/N)2 Cocksfoot grass 57 0.57 0.349 Timothy grass 32 0.32 0.1024 Buttercup 3 0.03 0.0009 Clover 3 0.03 0.0009 Thistle 1 0.01 0.0001 Dandelion 4 0.04 0.0016 Total (N) 100 ∑ = 0.4308 D = 1 – [ ∑ ( n / N)2 ] D = 1 – 0.4308 D = 0.5692
- 12. Now calculate (D)for Field B… Species observed Percentage cover Field A (n) Field B (n) Cocksfoot grass 57 38 Timothy grass 32 16 Buttercup 3 14 Clover 3 22 Thistle 1 5 Dandelion 4 5 Total (N) 100 100 D = 1 – [ ∑ ( n / N)2 ]
- 13. Field B (D) Speciesobserved Percentage cover Field B (n) n/N (n/N)2 Cocksfoot grass 38 0.38 0.1444 Timothy grass 16 0.16 0.0256 Buttercup 14 0.14 0.0196 Clover 22 0.22 0.0484 Thistle 5 0.05 0.0025 Dandelion 5 0.05 0.0016 Total (N) 100 ∑ = 0.243 D = 1 – [ ∑ ( n / N)2 ] D = 1 – 0.243 D = 0.757
- 14. Conclusion: • D forField A = 0.5692 • D for Field B = 0.757 • “Field B has the higher diversity index, so has more species richness AND evenness. It would be more resistant to any environmental damage or change.”
- 16. Survey of animalsin a freshwater stream. species Number (n) n / N (n / N)2 Gammarus pulex ( water shrimp) 150 Asellus aquaticus ( water louse) 32 Baetis rhodani ( mayfly nymph) 113 Lymnaea peregra ( snail) 2 Rhyacophila ( caddis-fly nymph) 12 Chironimidae ( midge larvae) 210 Total Calculate Simpsons diversity index D
- 17. Survey of animalsin a freshwater stream. species Number (n) n / N (n / N)2 Gammarus pulex ( water shrimp) 150 Asellus aquaticus ( water louse) 32 Baetis rhodani ( mayfly nymph) 113 Lymnaea peregra ( snail) 2 Rhyacophila ( caddis-fly nymph) 12 Chironimidae ( midge larvae) 210 Total 519
- 18. Survey of animalsin a freshwater stream. species Number (n) n / N (n / N)2 Gammarus pulex ( water shrimp) 150 0.289 Asellus aquaticus ( water louse) 32 0.062 Baetis rhodani ( mayfly nymph) 113 0.218 Lymnaea peregra ( snail) 2 0.004 Rhyacophila ( caddis-fly nymph) 12 0.023 Chironimidae ( midge larvae) 210 0.405 Total 519
- 19. Survey of animalsin a freshwater stream. species Number (n) n / N (n / N)2 Gammarus pulex ( water shrimp) 150 0.289 0.084 Asellus aquaticus ( water louse) 32 0.062 0.004 Baetis rhodani ( mayfly nymph) 113 0.218 0.047 Lymnaea peregra ( snail) 2 0.004 0.000016 Rhyacophila ( caddis-fly nymph) 12 0.023 0.000529 Chironimidae ( midge larvae) 210 0.405 0.164 Total 519 ∑ = 0.299
- 20. Survey of animalsin a freshwater stream. species Number (n) n / N (n / N)2 Gammarus pulex ( water shrimp) 150 0.289 0.084 Asellus aquaticus ( water louse) 32 0.062 0.004 Baetis rhodani ( mayfly nymph) 113 0.218 0.047 Lymnaea peregra ( snail) 2 0.004 0.000016 Rhyacophila ( caddis-fly nymph) 12 0.023 0.000529 Chironimidae ( midge larvae) 210 0.405 0.164 Total 519 ∑ = 0.299 D = 1 – [ ∑ ( n / N)2 ]
- 21. Survey of animalsin a freshwater stream. species Number (n) n / N (n / N)2 Gammarus pulex ( water shrimp) 150 0.289 0.084 Asellus aquaticus ( water louse) 32 0.062 0.004 Baetis rhodani ( mayfly nymph) 113 0.218 0.047 Lymnaea peregra ( snail) 2 0.004 0.000016 Rhyacophila ( caddis-fly nymph) 12 0.023 0.000529 Chironimidae ( midge larvae) 210 0.405 0.164 Total 519 ∑ = 0.299 D = 1 – [ ∑ ( n / N)2 ] D = 1 – 0.299 = 0.7
- 22. “Explain this result”(3) • An index value of 0.7 means there is a high probability that any two individuals taken from this stream will be from different species. • The stream shows good species richness and evenness. • The stream is more likely to withstand changes such as pollution.
- 23. Measuring Biodiversity –self check • Do you know? • Species richness • Species evenness • Simpson’s Index of Diversity (D)