Question 3. Ecologists have noted that individuals and populations of some speci

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Question 3. Ecologists have noted that individuals and populations of some species influence energy flow between trophic levels more than others. Using data from the table to the right, provide a rough estimate of how much energy would make it to the second, third, and fourth trophic levels in the following simplified food chains. Start with 100 units of energy in the autotrophic base of each (i.e., plants or algae). NOTE: the PE values in the table are listed as %! To calculate trophic transfer from, for example, algae to fish: PE = 9.8% = 0.098 = Pfish/Palgae. So Pfish = Palgae * 0.098... This table is what was given

a) plants ® non-insect invertebrate herbivores ® small mammals ® large mammals

b) algae ® aquatic non-insect invertebrate herbivores ® insect predators ® fish

c) plants ® large mammal herbivores ® large mammal predators ® large mammal predators

d) plants ® insect herbivores ® insect predators ® insect predators

e) Remembering that transfer of energy between trophic levels can influence the number of trophic levels an ecosystem can sustain, and that greater energy transfer usually enhances the establishment of higher trophic levels, which of the hypothetical food chains above would be MOST likely, and which LEAST likely, to sustain a fifth trophic level?

f) What key ecological difference between taxa is driving the predicted differences in energy flow between the two food chains you identified above? Explain.

Explanation / Answer

a. PE( non insect herbivorous)=20.9=0.209

P(non insectivorous herbivorous)=P plants*0.209=1*0.209=0.209

PE(small mammals)=1.5%=0.015

P(small mammals)=0.015*P(non insectivorous herbivorous)=0.015*0.209=3.135*10e-3

PE(Large mammals)=3.1%=0.031

P(large mammals)=0.031*Psmall mammals=0.031*3.135*10e-3=9.7*10e-5

b.PE(aquatic non insectivorous herbrivorus)=20.9%=0.209

P(aquatic non insectivorous herbivorous)=Palgae*0.209=1*0.209=0.209

PE(insect predators)=27.6%=0.276

P(insect predators* P( insect herbivorous)=0.276*0.209=0.057

PE(fish)=9.8%=0.98

Pfish=0.98*0.057=0.055

c.PE(large mammal herbivorous)=3.1%=0.031

P(large mammal herbivorous)=0.031*P plants=0.031*1=0.031

P(large mammal predators=0.031*0.031=9.6*10e-4

d.PE(insect herbivorous)=38.8=0.388

P(insect herbivorous)=Pplants*0.388=1*0.388=0.388

PE(insect predators)=55.6=0.556

P(insect predators)=0.556*0.388=0.215

P(insect predators)=0.215*0.556

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