Abiotic / Trophic levels

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Topic 2 The Ecosystem

Notes Ch. 2 – Part 1

2.1 Structure pp. 14 - 23

1. Biotic - Abiotic / Trophic levels – Food Chains –

Food Webs

Definitions

– keep simple a) biotic b) abiotic c) producer d) consumer e) primary consumer f) secondary consumer g) trophic level h) decomposer

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Read abiotic and biotic definitions and answer the following questions .

IB1. Which list contains only abiotic components of an ecosystem?

A. air, water, rock

B. air, producers, climate

C. soil, consumers, water

D. carnivores, herbivores, decomposers

IB2. Which list includes three abiotic factors that affect aquatic plants?

A. number of decomposers, pH of water, day length

B. water temperature, oxygen concentration, light intensity

C. carbon dioxide concentration, number of primary consumers, wavelength of light

D. water temperature, number of secondary consumers, carbon dioxide concentration

Explain , in general, why the other answers are wrong:

2

Read Trophic Levels, food chains and food webs

1. Explain which way the arrows point in a food chain.

2. Explain the sequence of why you have primary, secondary etc. consumers and only 1 producer.

DIAGRAM

IB3. Most food chains seldom have more than four trophic levels because

A. in most ecosystems, competition for food is very great.

B. the total biodiversity in any ecosystem is limited.

C. energy is lost as it moves along a food chain and little remains at the level of the top carnivore.

D. in many parts of the world, many species have become extinct and complex ecosystems are rare.

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IB4.

The diagram below represents the pyramid of biomass for a certain ecosystem. trophic level 3 carnivores trophic level 2 herbivores trophic level 1 producers

The most likely explanation for the biomass of trophic level 2 being greater than that of trophic level 1 is that the

A. system shows seasonal fluctuations and the plants have died off before the herbivores.

B. number of plants is greater than the number of herbivores.

C. number of herbivores is greater than the number of plants.

D. productivity of the herbivores is greater than the productivity of the plants.

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2. Pyramids , biomass, productivity

.

Definitions

– keep simple a) Pyramids of numbers b) Pyramid of biomass c) Pyramids of productivity d) Biomass e) Productivity f) Bioaccumulation

1. a) In general, how much energy ( biomass) is passed from one level to the next in a pyramid? b) How does this relate to thermodynamics? Explain the difference between pyramid a and b p. 17 fig. 2.4 ( see above)

2. How is biomass measured?

3. Explain pyramids of productivity and how they are fundamentally different than pyramids of biomass. Explain units involved.

5

Pyramid structure and ecosystem functioning

Explain the risk to top carnivores using DDT as an example :

3. Species, populations, habitats and niches

6

Definitions

– keep simple a) Species – b) Population

– c) Habitat – d) Niche – e) Community

– f) Ecosystem

Write one sentence that clearly illustrates the words species, population and habitat.

Use a specific organism. :

Write one sentence that clearly illustrates the words niche, community and ecosystem.

Use a specific organisms and a specific ecosystem. :

IB6. An ecological niche is

A. where a species lives.

B. all the populations of an ecosystem.

C. what a species eats.

D. the role a species has in an ecosystem.

IB7. A community is composed of

A. several habitats.

B. several populations.

C. several abiotic factors.

D. biotic and abiotic factors

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Population Interactions

Briefly explain each type of population interaction

Competition

Predation

Parasitism –

Mutualism –

Make sure you understand fig. 2.7 and 2.8

Fig. 2.7

8

Fig. 2.8

IB8. Which interaction would benefit both organisms in a relationship?

A. Predation

B. Commensalism

C. Mutualism

D. Competition

IB9. Spiders mainly use a web and poisonous fangs to trap and eat insects such as flies. What type of interaction is this?

A. Parasitism

B. Competition

C. Mutualism

D. Predation

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2.2 Measuring abiotic components pp. 24

– 28

1. Measuring

A. List the basic type of ecosystems and 1 example of each

B. Compare the abiotic factors of marine ecosystems and terrestrial systems.

By looking at the lists, how can you tell by common sense that some of the terrestrial abiotic factors are not marine abiotic factors?

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IB10. Which list contains only abiotic factors in a marine environment?

A. algae, temperature and nitrogen level

B. salinity, turbidity and plankton

C. decomposers, carbon dioxide and calcium level

D. pH, dissolved oxygen and light intensity

In the question above , even if you didn’t know the abiotic list of a marine ecosystem, how could you still get the right answer?

Use this strategy to answer the following question:

IB11. Which of the following is a list of abiotic factors that affect the distribution of organisms?

A. Sunlight, decomposers and soil pH

B. Temperature, sunlight and soil pH

C. Water, sunlight and competition

D. Disease, pH and dissolved oxygen

2. Evaluating abiotic factors

Make sure you know the basic aspects of each type of measurement

Light meter - sun light

Temperature – indicates DO pH

– indicates DO

Wind - direction

Particle size – indicates drainage and water – holding capacity ( sand vs. pebbles vs clay)

Slope – indicates surface runoff

Soil moisture- heat soils before and after to measure amount of moisture it can hold

Mineral content - N, S, K etc.

Flow velocity – water flow in rivers indicate rain fall , how much water is being released by dams for farming etc. Also affect DO by generating bubbles.

Salinity

– indicates DO .

DO – most important :

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as salinity, temp. increases

, DO decreases. Salt “ dehydrates “ water and removes DO. Higher temp. increases kinetic energy and causes O2 to move around more and may evaporate so overall conc. of DO goes down.

Atmospheric pressure above water affects DO. As press. Above water increases it “ pushes O2 into water and increases DO. At higher elevations there is less atm press. so less DO in general. pH – decreases or increase of pH away from neutral 7.0 generally lowers DO due to acid base reactions.

Photosynthesis by plants etc. increases DO and respiration by organisms in water decreases DO.

Eutrophication ( know this term)

– due to increase of N or P in water causes increase of nutrients in water especially for plants. This causes too much plant growth than normal causing algae blooms ( large masses of plant and alga growth). Although plants will produce O2 during the day via photosynthesis plants also “ RESPIRE” – absorb O2 at night via photorespiration. This will decrease overall DO. Too much plant growth on surface of lakes etc. will also block sunlight getting to bottom of lakes reducing photosynthesis.

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IB 12 : Below is a diagram of part of a food web. snakes frogs spiders insects snails green plants

Which of the following is an abiotic factor that might affect this food web?

A. Presence of parasites

B. Spread of diseases in frogs

C. Increase in temperature

D. Introduction of new plant species

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2.3 Measuring abiotic components pp. 29 – 35

1. Key for species identification – Dichotomous key : Make sure you know how it works.

14

15

16

2. Describe and evaluate methods for estimating abundance of organisms. Make sure you understand the following examples:

Lincoln Index – allows you to estimate the total population size of an animal in your study area. Called a “ capture – mark – release – recapture” technique.

Basically you capture some animals ,mark them and release them back into the population. Then on another day, you capture animals from the population again and count how many of the marked animals you have .

Formula :

N = n1 x n2

m

N = total population size of animals in the study site n1 = number of animals captured on the first day = number of animals marked n2 = number of animals captured on the second day m = number of marked animals recaptured on the second day a) All of the animals that were marked the first day are recaptured. This proves that this number is the total population size of animals in the study site. b) Only half the animals that were marked the first day are recaptured. This proves that the total population size is twice as big as the number of animals that were recaptured and marked the second day.

Only half the animals that were marked the first day are recaptured ( n

2

= ½ n

1

, n

2

= m). This proves that the total population size (N) is twice as big as the number of animals that were recaptured and marked the second day.

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“Only half the animals that were marked the first day are recaptured “ big statement and you have to deduce that :

n

2

= ½ n

1

AND n

2

= m remember : n1 = number of animals captured on the first day ; all animals are marked and released n2 = number of animals captured on the second day. States that only ½ are recaptured so n

2

= ½ n

1 m = number of marked animals recaptured on the second day have to assume that the number of animals recaptured were all marked so m = n

2

So , use any number for n

1

, for example 6 : n

2

= ½ n

1

= 3 m = n

2

= 3

N = n1 x n2 = 6 x 3 = 6

m 3

This proves that the total population size (N) is twice as big as the number of animals that were recaptured and marked the second day ( n

2

= m)

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3. Quadrats

Used for non mobile organisms ( plants, trees, fungi etc. ) and basically measure number of species per m2.

4. Simpson’s diversity index

Takes into account the total number of all species ( N) and the number of a particular species ( n)

D = N ( N-1)

Σ n ( n-1)

D = diversity index

N = total number of organisms of all species found n = number of individuals of a particular species

Σ = sum of

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SEE FOLLOWING EXAMPLE

IB 13 : The table below gives the number of individuals of four species of trees in two small patches of Australian forest.

Tree species Area A Area B

Allocasurina huegelina 4 1

Banksia grandis

Eucalyptus calophylla

Acacia saligna

5

7

4

8

9

2

(b) (i) Using the formula for Simpson’s diversity index

D =

N(N

1 )

 n(n

1 ) calculate which of area A or area B has the higher diversity index.

A has been done for you . Try to do B:

A = 5=4+7+4 = N = 20

N (N-1) = 20(20-1) = 380

Area A n ( n-1)

4 ( 4-1) = 12

5( 5-1) = 20

7(7-1) = 42

4( 4-1) = 12

____________________

Σ n ( n-1) = 86

D =

N(N

1 )

 n(n

1 )

=

380 = 4.4 FINAL ANSWER NOW TRY B

86

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