The food web
Death and sedimentation
Primary producers
A1
A2
D
Detritus and associated
Microflora (bacteria/fungi)
inedible
Primary consumers
H1
H2
detritivore
herbivore
P
Productivity
Biomass
Secondary Productivity:
Primary production supports
a web of consumers—a simple example
Defining some productivity terms
A rate equation for biomass can be written as a mass balance
bB/t
Birth (production) term
B
Death (loss) term
dB
mB/t
 bB  mB
dt
that is the net growth rate of biomass is the difference between
the rate at which it is formed and lost.
Productivi ty is the rate at which new biomass is formed.
The birth term is called the productivi ty of the population ,
b is the specific production or growth rate and m is the specific death rate
The productivity is a combination of the birth of new organisms and the
growth of the organisms already present
Similarly, the death process is a combination of death of organisms and
weight loss by existing organisms.
If the productivity (birth term) exceeds the death term the biomass is
increasing, and if the death term is larger, the biomass is decreasing
Biomass
of a
consumer
dB
0
dt
B
t
Time (t)
dB
0
dt
dB
0
dt
dB
 bB  mB  b  m B
dt
 production minus losses
or
1 dB
bm
B dt
dR
0
Rotifer dt
Biomass
R
dR
0
dt
dR
0
dt
t
Time (t)
dB
and
dt
So if we measure
B at any point in time and we can
estimate the specific birth rate, we can then obtain the
specific death rate by subtraction.
 1 dB 
m b

B
dt


A tiny organism like a rotifer is born at full size, so productivity amounts to measuring the
rate at which new animals are born
Rotifers
carrying
eggs
•For a tiny consumer like a rotifer the birth rate is easy to estimate since the
adult females carry their eggs around until they hatch
•When they hatch they come out as full sized rotifers
.
•If we know the fraction of adults carrying egs and the average time it takes for
eggs to hatch, we can calculate the birth rate.
•Since the rotifers are born more or less full size, so there is no need to model
or measure the growth of individuals.
# eggs
E
,
# adults
T  avg time required for hatching (d)
1
 hatching rate
T
E
b
 d -1
T
E
1 dB
m

T
B dt
dR
0
Rotifer dt
Biomass
R
dR
0
dt
dR
0
dt
t
Time (t)
For a large organism like a fish, biomass production occurs mostly from individual
growth. New born fish are so tiny that birth of individuals makes a negligible
contribution to biomass production
W4
W3
Wt  1
SGRt  ln
Wt
Year classes
W2
W1
The Wt represent the Weights of each year class
•We can calculate the growth rate of biomass individual fish by weighing fish and
determining their age and then seeing how much weight they gain each year.
•The productivity of each age class is the Specific growth rate (SGR) of that age
class times the total biomass of that age class in the population.
This approach assumes that the size vs age relationship is relatively constant.
W4
W3
P    SGRt * Bt
i
t
SGRt  ln
Wt  1
Wt
W2
W1
The Wtrepresent the Weights of each year class
Production sum of [specific growth rate times biomass present]
for each of t age classes,and summed over all i species in the trophic category
Average SGR for a species over the whole life table would be
SGRavg   SGRt *
t
Bt
 Bt
By this method the average SGR for the whole population can be calculated as
the weighted average over all age classes
T
SGR   wt  SGRt
t 1
SGRt  ln
where wt 
Wt  1
Wt
Bt
, the fraction of Biomass in the t ' th age class
B
However there is a problem with this approach…???
However there is a problem with this approach.
By considering only the gain in weight across age classes, this method
ignores weight gained and lost within the same year, eg Gonad tissue
Adult fish usually convert a considerable portion of their body mass to gonads
and release it during spawning every year.
Thus it does not add to next year’s weight and would not be recorded as growth
We can quite easily correct for this by factoring in gonad production (add GSI)
SGRt  GSI
where GSI (Gonadosom atic index) 
mature gonad weight
total weight
We would of course only make this GSI correction on adult age classes.
How can we tell how old a fish is?
Scales of a chum salmon
Measure
distances from
scale center to each
annulus along a
chosen
2+
3+
axis
distance to annulus
LA

total scale length
LT
4+
This allows us to construct a growth curve based on length and age.
LA
Age yr
Convert the growth curve based on length to weights using
a length-weight plot for the species
Many types of bony structures are commonly used to determine age of fish
Scale
Otolith
Opercular bone
These three structures are all from the same 3+ year old 30 cm cutthroat trout
The specific death rates can also be estimated from the population structure. This time
we assume that the age structure of the population is constant, and that the numbers of
individuals of each age within a sample reflects the proportion of that age group in the
population.
Assume we have a sample of 215 pike from a population.
N 4  20
N 3  30
Nt  1
mt   ln
Nt
N 2  55
N 1  110
 N  215
20
m3   ln
 0.41
30
30
m 2   ln
 0.61
55
55
m1   ln
 0.69
110
t
t
The Nt /  Nt represent the proportion of the population
in each year class
150
The survivorship curve assuming stable age
structure for the population looks like this
*
100
110
Nt  1
 mt  ln
Nt
m1 =0.69
#
55
*
50
m2 =0.61
30
*
20
m3 =0.41
0
0
1
3
*
5
Age
Summary
The rate of change of biomass for a species is usually modelled as
the difference between its growth and loss processes
dB
 B  SGR  GSI   B  m  GSI   B  SGR  m
dt
Productivity term
loss term
If productivi ty exceeds losses, the population is growing in biomass
otherwise the population is decreasing in biomass
For a stable population the two terms are equal.
Ecological efficiency (n) for a consumer at the nth trophic level
Pn
n 
Pn  1
Ecological
efficiency of
zooplankton is
usually around
10% of NPP in
lakes
Variability??
Fecal pellets
Zooplankton such as Daphnia filter-feed using currents generated by
their thoracic appendages. Fecal pellets sediment rapidly to the bottom
80-95% of energy is lost at each trophic step, much of it as feces
Assimilation efficiency
Herbivores depends on diet
≈100% for sugary nectar
≈40-80% for small phytoplankton
and filamentous algae
<20% for mud and detritus
Carnivores
60-70% for aquatic insects
70-90% for meat
Fecal pellets
The undigested material in the zooplankton fecal pellets was not assimilated.
Assimilation efficiency depends on the digestibility of the diet
Cellulose, chitin, lignin or other undigestible material makes AE low
Ingested energy ─ egested energy = assimilated energy
Assimilation efficiency (AE, %)= assimilated energy/ingested energy x 100
Exploitation efficiency or Consumption Efficiency (EE)
In
EEn 
 100
Pn  1
Exploitation efficiency is the consumption rate at a given trophic level divided by
the productivity of the trophic level it feeds on.
Zooplankton will have low EE (CE) when phytoplankton are sedimenting rapidly
to the bottom before they are being eaten.
If EE(CE) is high then most of the sedimentation will be in the form of fecal
pellets, which sink more rapidly than individual cells.
Zooplankton fecal pellets are good food for benthic invertebrates
If EE for herbivorous zooplankton is low then dead (sedimenting) phytoplankton
will be readily available for detritivores (zoobenthos)
Activity is energetically expensive and high Metabolic rate means
low Production efficiency
Gross PE
Endotherms
≈5% or less
≈1% some birds
Ectotherms
≈10-30% for fish
≈ 5-15% insects
Otter swim about rapidly and spend large amounts of energy looking for fish to eat
Assimilated energy ─ respiration ─ excretion = production (growth)
Net Production efficiency (NPE, %)= growth/assimilation x 100
Gross PE (%)=[assim/ingest x growth/assim] x100=growth/ingest x 100
Pelagic fish like kokanee salmon
expend a huge amount of energy
actively searching for prey--they
have high basal metabolic rates low
conversion efficiencies
The deepwater sculpin sits
on the bottom and ambushes
unsuspecting prey. They
have very low basal
metabolic rates and high
conversion efficiencies
If these two species were fed the same amount of food, the sculpin
would grow more than twice as fast as the salmon
Copepod dominated communites have lower trophic efficiency than cladoceran
dominated communities—possible reasons?
Filter-feeding by a calanoid copepod
•Copepods are rapid swmmers and generate feeding currents as they swim
•Copepods filter-feed by generating currents with their 1st antennae and their thoracic
appendages..
•Water from small eddy currents around the mouthparts is drawn over the fine setae of the
maxillae, where small algae are collected and moved to the mouth.
http://www.ucmp.berkeley.edu/arthropoda/crustacea/images/copepoda03.jpg
Energy budget for herbivorous zooplantkon
NPP = rate of formation of
phytoplankton biomass
S = rate of production of
uneaten algae, mostly inedible
species (sedimentation)
F= rate of production of
fecal pellets (sedimentation)
 The Bioenergetic budget for an organism
C= G + M*A+ SDA+F+U
Metabolic costs include basal metabolism, activity costs
and specific dynamic action (costs of digestion etc)
Zooplankton production is the rate at which biomass (energy)
becomes available for consumption by zooplanktivores
Energetic losses in the food chain
Less than 1% of the incident light
energy is captured by
photosynthesis
as NPP.
Productivity declines by
about 10-fold for each
trophic level in the food
chain.
Most of the losses are
are in the form of waste
heat.
Some energy from each
trophic level winds up in
the detrital pool, and some
of this remains buried
as sediment (or soil) organic
matter (fossilized)
Pn
n 
Pn  1
In
An
Pn
Pn
EEn 
, AEn 
, PEn 
, GPEn 
Pn  1
In
An
In
Show
n  EEn  AEn  PEn
n  EEn  GPEn
Productivity at different levels in the food web
500 x (0.1)3 g/m /yr
2
500 x (0.1)2 g/m /yr
2
500 x 0.1
g/m2/yr
NPP around 500 g/m /yr
2
Zooplankton
Benthic & epiphytic
invertebrates
PhytoplanktonDiet shift
Benthic & epiphytic
algae plus detritus
Trophic link
Net productivity at level n = the rate of growth of biomass at that level
= [SGR +GSI] * Biomass
= NPP (TE) n-1
Yields of piscivorous fish are well correlated with primary productivity but are
several orders of magnitude lower than PP
•Intensive aquaculture can
produce yields that are
orders of magnitude
beyond natural
ecosystems
How to maximize energy flow to fish
Increased nutrient loading—fertilization + ammonia and anoxia tolerant species
Shortening the food chain—primary consumers (eg carps, tilapia or mullets)
Don’t rely on natural recruitment and managing the life cycle—stocking/hatcheries
Increasing consumption efficiency—small pens intensive feeding
Increased assimilation efficiency—feeding with easy to digest food pellets
Increased production efficiency—low activity species that don’t mind crowding,
, highly turbid water
Lepeophtheirus salmonis
Many aquaculture proponents
argue that aquaculture reduces
harvesting pressure on wild
fisheries.
Salmonid aquaculture not very trophically efficient,
food pellets made from by-catch of wild species
Major water quality issues—nutrientpollution from
cages, anti-fouling paint, antibiotics, habitat
destruction
Transmit diseases to wild salmonids—bacteria,
viruses, protozoans, fungi, “fish lice” –parasitic
copepods and other Crustacea
Genetic problems when domestic escapees
compete with or interbreed with wild fish
Argulus
Summarizing concepts on Secondary production
•The organic matter produced by primary producers (NPP) is used by
a web of consumers
•NPP is used directly by primary consumers (herbivores and detritivores), which are in
turn consumed by carnivores.
•Measurement of 2o Production is done by estimating the rate of growth of individuals
and multiplying by the number of individuals per unit area in the cohort (age or size group).
•The efficiency of secondary production ranges from 5-20% (Avg 10%)
at each trophic level.
•Efficiency depends on several factors--palatability, digestibility, energy requirements
for feeding (activity costs)(eg homeotherms vs poikilotherms , other limiting factors
eg water, and nutrient quality of food.
•Trophic efficiency can be represented as the product of CE*AE*PE, each of which
is dependent on one or more of the above factors.
•The yields of many important fisheries depends on a combination of NPP, the
length ofthe food chain leading to the fish being harvested, and the efficiency
of each step.
•Many of the species that we harvest or very high in the food chain, so a great deal
of NPP is required to support them.
Pn
n 
Pn  1
In
An
Pn
Pn
EEn 
, AEn 
, PEn 
, GPEn 
Pn  1
In
An
In
Show
n  EEn  AEn  PEn
n  EEn  GPEn
If the productivity of a phytoplankton population is 4000 k J (kilo Joules) /yr /
m2, If sedimentation rate of dead cells to the substrate constitutes 1600
kJ/m2yr, and the phytoplankton population is dB/dt=0. If the rain of
zooplankton fecal pellets to the bottom is 1400 kJ/m2/yr. What is the
assimilation efficiency of the zooplankton trophic level (assume that they are
all feeding on phytoplankton).
1.
2.
3.
4.
5.
0.42 or 42%
0.60 or 60%
0.35 or 35%
0.25 or 25%
None of these
What is the exploitation efficiency EE (or Consumption efficiency CE) of the
zooplankton trophic level
1.
2.
3.
4.
5.
0.42 or 42%
0.60 or 60%
0.35 or 35%
0.25 or 25%
None of these
If the net production efficiency of the zooplankton trophic level is
0.40 (40%) what is the ecological efficiency () of the trophic level
1. 0.15 or 15%
2. 0.05 or 5%
3. 0.10 or 10%
4. 0.25 or 25%
5. None of these
If the zooplanktivorous fish are consuming zooplankton at the rate
of 400 kJ/yr/m2, their EE (CE) is
1. 0.40 or 40%
2. 0.60 or 60%
3. 1.00 or 100%
4. 0.25 or 25%
5. None of these
If the zooplanktivorous fish have an assimilation efficiency of 0.70
(70%) and Net production efficiency (NPE) of 0.20 (20%), the
productivity at this trophic level is
1. 40 kJ/yr/m2
2. 56 kJ/yr/m2
3. 100 kJ/yr/m2
4. 280 kJ/yr/m2
5. None of these
If in another lake with similar zooplankton productivity the
planktivore fish productivity was 2 X higher, a possible explanation
for this would be
1. the AE of the fish in that lake was 2X as high
2. the NPE of the fish in that lake was 2X as high
3. the EE (CE) in that lake was 2X as high
4. the AE*NPE in that lake was 2X as high
5. both b and d are true
6. both b and c are true
Residence time and turnover of energy by trophic levels
The standing stock of energy in the
plankton is low but it is turned over
rapidly, because the organisms are
small, grow rapidly and don’t live
long
Planktonic
Herbivore
(50mg) life span
1 month
A
H1
P
Phytoplankton (0.01mg,
life span, few days
H2
Benthic Detritivore
(0.1 g) life span 1yr
Carnivorous fish
(100g) life span
5-10 yr
Turnover is slower at higher trophic levels, since larger organisms accumulate
energy over a longer life span—longer residence time and slower turnover