SME
Principles of Biology II
Ecology
Use this for “Producers store 25,000 kJ/(m²·yr) and 10% passes up each level. How much reaches the tertiary consumers?”
Diagram
10% of 25,000 kJ/(
m
2
·yr) = 2,500 kJ/(
m
2
·yr)
10% of 2,500 kJ/(
m
2
·yr) = 250 kJ/(
m
2
·yr)
10% of 250 kJ/(
m
2
·yr) = 25 kJ/(
m
2
·yr)
This is an example. Type your own values to start a new problem.
E₁
Producers
example · kJ/(m²·yr)
e
Transfer efficiency
example · %
E₂
Primary consumers
calculated · kJ/(m²·yr)
E₃
Secondary consumers
calculated · kJ/(m²·yr)
E₄
Tertiary consumers
calculated · kJ/(m²·yr)
Formulas
E₂
=
E₁
×
e
100
2,500
=
25,000
×
10
100
E₃
=
E₂
×
e
100
250
=
2,500
×
10
100
E₄
=
E₃
×
e
100
25
=
250
×
10
100
Assumptions
•
Each energy is per square meter of ground per year, in kJ/(m²·yr).
•
The transfer efficiency e is the share of one level’s energy stored in the level above, about 5–20%.
•
The rest is lost as heat in respiration or left in waste and dead matter, so energy flows one way.
•
One efficiency holds for every step up the chain.
Step-by-step
WE KNOW
E₁ = 25,000 kJ/(m²·yr), e = 10%
FIND
primary consumers (E₂), secondary consumers (E₃), tertiary consumers (E₄)
Step 1 · Find primary consumers (E₂)
Use E₂ = E₁ × e ÷ 100
Only
e
%
of
the
energy
in
the
producers
is
stored
in
the
level
above.
E₂
=
25,000
×
10
100
E₂
=
250,000
100
E₂
=
2,500
kJ/(
m
2
·yr)
Step 2 · Find secondary consumers (E₃)
Use E₃ = E₂ × e ÷ 100
Only
e
%
of
the
energy
in
the
primary
consumers
is
stored
in
the
level
above.
E₃
=
2,500
×
10
100
E₃
=
25,000
100
E₃
=
250
kJ/(
m
2
·yr)
Step 3 · Find tertiary consumers (E₄)
Use E₄ = E₃ × e ÷ 100
Only
e
%
of
the
energy
in
the
secondary
consumers
is
stored
in
the
level
above.
E₄
=
250
×
10
100
E₄
=
2,500
100
E₄
=
25
kJ/(
m
2
·yr)
Check
Put the numbers back in. Both sides should match.
2,500
=
25,000
×
10
100
✓
250
=
2,500
×
10
100
✓
25
=
250
×
10
100
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.