Investigating Diversity Flashcards
A student investigated the distribution of plants in a heathland.
The table below shows the number of plants he found in a sample area of 1 m^2.
Species of plant
1 ) Common heather
2 ) Red fescue
3 ) Vetch
4 ) White clover
Number counted in 1 m^2
1 ) 2
2 ) 14
3 ) 2
4 ) 8
What is the species richness of this sample?
- 4
Calculate the index of diversity of this sample. Show your working.
Use the following formula to calculate the index of diversity.
where N is the total number of organisms of all species and n is the total number of organisms of each species
- sum of n( n - 1 ) = 242 ( Add a column to the table, labelled n( n - 1 )
- Sum of N = 26
- Index of diversity = ( 26 ( 26 - 1 ) / 242 ( N ( N - 1 ) / sum of n( n - 1 ) )
- = 2.67
Suggest how this student would obtain data to give a more precise value for the index of diversity of this habitat.
- Use a larger number of samples
- to identify and remove anomalies and then calculate a mean value
Species richness and an index of diversity can be used to measure biodiversity within a
community.
What is the difference between these two measures of biodiversity?
- Species richness only measures the number of different species
Scientists investigated the biodiversity of butterflies in a rainforest.
Their investigation lasted several months.
The scientists set one canopy trap and one understorey trap at five sites.
- The canopy traps were set among the leaves of the trees 16–27 m above ground level.
- The understorey traps were set under trees at 1.0–1.5 m above ground level.
The scientists recorded the number of each species of butterfly caught in the traps.
The table below summarises their results.
Species of butterfly
1 ) Prepona laertes
2 ) Archaeoprepona demophon
3 ) Zaretis itys
4 ) Memphis arachne
5 ) Memphis offa
6 ) Memphis xenocles
Mean number of butterflies
In canopy:
1 ) 15
2 ) 14
3 ) 25
4 ) 89
5 ) 21
6 ) 32
In understorey:
1 ) 0
2 ) 37
3 ) 11
4 ) 23
5 ) 3
6 ) 8
P value
1 ) < 0.001
2 ) < 0.001
3 ) > 0.05
4 ) <0.001
5 ) < 0.001
6 ) < 0.001
The traps in the canopy were set at 16–27 m above ground level.
Suggest why there was such great variation in the height of the traps.
- Trees vary in height
By how many times is the species diversity in the canopy greater than in the
understorey? Show your working.
Use the following formula to calculate species diversity.
d = ( N ( N - 1 ) ) / E n ( n - 1 )
where N is the total number of organisms of all species and n is the total number of
organisms of each species.
Canopy:
N = 196 ( Total number of organisms )
E n ( n - 1 ) = 10236
Understory:
N = 82
E n ( n - 1 ) = 2010
Index of diversity for Canopy:
- 196 ( 196 - 1 ) / 10236
- = 3.733880422
Index of diversity for Understory:
- 82 ( 82 - 1 ) / 2010
- = 3.304477612
Multiplier:
- 3.73… / 3.30… = 1.129945746
- = 1.13
- So the index of diversity in the canopy is 1.13 times bigger than in the understory
The scientists carried out a statistical test to see if the difference in the distribution
of each species between the canopy and understorey was due to chance.
The P values obtained are shown in the table.
1 ) < 0.001
2 ) < 0.001
3 ) > 0.05
4 ) <0.001
5 ) < 0.001
6 ) < 0.001
Explain what the results of these statistical tests show.
- For Zaretis itys ( 3 ), difference in distributions probably due to by chance
- For other species, difference in distribution is unlikely to be due to chance
Table 1 shows how a bird called the bluethroat ( Luscinia svecica ) is classified by biologists.
Taxon
1 ) Domain
2 ) ( Blank )
3 ) ( Blank )
4 ) ( Blank )
5 ) ( Blank )
6 ) ( Blank )
7 ) Genus
8 ) Species
Name of Taxon:
1 ) Eukaryota
2 ) Animalia
3 ) Chordata
4 ) Aves
5 ) Passeriformes
6 ) Muscicapidae
7 ) ( Blank )
8 ) ( Blank )
Complete Table 1 by filling the seven blank spaces with the correct terms.
Taxon:
2 ) Kingdom
3 ) Phylum
4 ) Class
5 ) Oder
6 ) Familiy
Name of taxon:
7 ) Luscina
8 ) Svecica
A group of scientists investigated genetic diversity in different species of bird.
For each species, the scientists:
• collected feathers from a large number of birds
• extracted DNA from cells attached to each feather
• analysed the samples of DNA to find genetic diversity.
Table 2 summarises their results.
Species of bird:
1 ) Willow flycatcher
2 ) House finch
3 ) Bluethroat
Number of genes examined:
1 ) 708
2 ) 269
3 ) 232
Number of genes examined that showed genetic diversity:
1 ) 197
2 ) 80
3 ) 81
In this investigation, what is meant by genetic diversity?
- Number of different alleles of each gene
The scientists concluded that the bluethroat showed greater genetic diversity than the willow flycatcher.
Explain why they reached this conclusion.
Use calculations to support your answer.
Species of bird:
1 ) Willow flycatcher
2 ) House finch
3 ) Bluethroat
Number of genes examined:
1 ) 708
2 ) 269
3 ) 232
Number of genes examined that showed genetic diversity:
1 ) 197
2 ) 80
3 ) 81
( The proportion of “ number of genes examined “ compared to “ number of genes examined that showed genetic diversity “ are closer in results with bluethroat than with the willow flycatcher )
- Bluethroat have a greater proportion of genes
- Percentage is 35% compared to 28%
( Percentage was worked out by doing the “ Number of genes examined that showed genetic diversity “ / “ Number of genes examined “ X 100 )
Ecologists investigated the size of an insect population on a small island.
They used a mark-release-recapture method. To mark the insects they used a fluorescent powder.
This powder glows bright red when exposed to ultraviolet (UV) light.
The ecologists captured insects from a number of sites on the island.
Suggest how they decided where to take their samples.
- Draw a grid over the area
- Select coordinates at random
Give two assumptions made when using the mark-release-recapture method.
- No losses to predation
- Marking doesn’t effect their survival
Suggest the advantage of using the fluorescent powder in this experiment.
( To mark the insects they used a fluorescent powder.
This powder glows bright red when exposed to ultraviolet (UV) light. )
- Only glows with UV light so doesn’t make insect more visible
- Doesn’t increase predation
The ecologists did not release any of the insects they captured 1–5 days after
release of the marked insects.
The table below shows the ecologists’ results.
Days after release
1
2
3
4
5
Number of marked insects remaining in population
1 ) 1508
2 ) 1430
3 ) 1400
4 ) 1382
5 ) 1380
Number of insects captured
1 ) 524
2 ) 421
3 ) 418
4 ) 284
5 ) 232
Number of captured insects that were marked
1 ) 78
2 ) 30
3 ) 18
4 ) 2
5 ) 9
Calculate the number of insects on this island 1 day after release of the marked insects.
Show your working.
- N = Number of marked insects X Number of insects captured / the number that were captured and marked
- 1508 X 524 / 78
- =10130.66667
- = 10131
The ecologists expected to obtain the same result from their calculations of the
number of insects on this island on each day during the period 1–5 days after release.
In fact, their estimated number increased after day 1.
During the same period, the number of insects they caught decreased.
The method used by the ecologists might have caused these changes.
Use the information provided to suggest one way in which the method used by the
ecologists might have caused the increase in their estimates of the size of the insect
population.
- Scientist removed larger numbers of insects from the same area
- Affecting ratio of marked and unmarked
