Factors that influence primary producers –biomass, productivity, diversity
Light
Temperature
Flow of water
Consumers
Scouring action of floods
Toxic substances
Nutrients—Chemical substances required for biomass to grow.—Any of the 13 or 14
essential elements could potentially be growth limiting.
All organic compounds require C
Proteins require a lot of N and some S
Energy stores (ATP) Phospholipids (membranes) and nucleic acids (DNA, RNA) require a
lot of phosphorus
Membranes also need a lot of K
Photosynthetic efficiency is low (light or CO2 taken up both not converted to biomass)
under severe nutrient limitation.
Redfield Ratio--C:N:P=106:16:1 for marine plankton, Diatoms, C:Si:N:P = 106:15:16:1.
Nutrient limitation of primary producers
•Productivity and biomass of most unenriched periphyton and
phytoplankton communities are usually limited by the availability of Nor P.
•In rare situations, Si, Fe, or some other trace element (eg Co) prove to
be rate limiting
•For aquatic macrophytes which obtain nutrients from sediment pore
water and not the water column, but generally obtain their C from DIC in
the water column, DIC can sometimes be the rate limiting factor.
•Although light is usually the main limiting factor, near the bottom of the
photic zone where water is deep, nutrient enrichment (eutrophication,
usually with N and P can make whole systems light limited.
Bioassays can be used to test the responses of algal communities to additions of
individual nutrients or mixtures.
How do we detect nutrient limitation?
Experimentally
a) Incubate bottles of water from lakes or rivers, or enclosures of rocks or artificial
substrates with different nutrients added, and measure primary productivity and/or
algal biomass. The factor that gives the largest growth response is the limiting
factor.
a) The response can depend on the scale of time and space, so experiments are
often done on a large-scale—whole system manipulations.
b) The response of primary productivity to nutrient addition is more noticeable in
measurements made volumetrically (/m3), than in areal measurements (/m2),
where increased light penetration can compensate at least partially for low
nutrient supply.
Comparative studies—Correlations
Measure primary productivity and primary producer biomass in different systems that
differ in their resource regimes (eg nutrients, light etc) and see which factor show
the best correlation to productivity and biomass.
Nutrient addition experiments can
be carried out on may different
spatial and temporal scales
Water column enclosures are
bags that extend downward—not
open at the bottom—usually to
the bottom of the photic zone, or
to the thermocline.
A large-scale experiment at the
Experimental Lakes Area (ELA) in
Canada
(Schindler et al. 1971)
The famous curtain experiment in
lake 226.
The lake was divided in half by a
curtain
the half on the right was fertilized
with a mixture of C, N and P.
The blue half on the left was
fertilized with a mixture of C and N.
The green colour is due to a
massive bloom of colonial
cyanobacteria (Anabaena).
Lake 227 in the foreground-26 years of Phosphorus addition
Lake 305 in the background—no fertilization
Mean annual biomass (as indicated by Chlorophyll a) in lakes is usually best
correlated with the phosphorus supply
Each point on the
graph represents
the average value
of P and Chl a for a
lake.
Lakes with poor P supply usually however have much greater benthic algal
biomass and productivity, so total biomass(per unit area) is usually much more
poorly correlated with nutrient supply in the water column
Lakes can be classified into trophic categories on the basis of productivity and
nutrient measures—however such boundaries are arbitrary.
Oligotrophic
P<10
Chla<3
Mesotrophic
10<P<40
3< Chla <25
Eutrophic
P>40
Chla > 25
For ultra-high P levels, chlorophyll a is no longer correlated with P
Most such systems have their primary productivity limited by light.
Nutrient limitation of stream algae??
Many people speculated that in flowing water
nutrient supply would be constantly renewed, so
that increasing its concentration would have little
effect on algal productivity or biomass.
Testing was difficult. It is difficult to make an
enclosure in flowing water, and it is hard to fertilize
a whole stream because a lot of water is flowing by.
Carnation Creek as it looks now—it used to have a densely forested watershed
The Stockner and Shortreed experiment on Vancouver Island.
The stream was partitioned into troughs and different nutrients were added to each.
Diatoms communities
In mountain streams sometimes
Respond to nutrient addition
… but not always
Why would the community
sometimes not respond?
Why is the response greater to a combination of N and P than to either alone?
In this nutrient addition experiment, no response
was detectable for the first 2 week but as the
biomass built up it was clear that the community
was responding.
Why do you think the algal biomass started to
diminish toward the end of the experiment while
nutrients were still being added?
Diatoms communities
In mountain streams sometimes
respond to nutrient addition
… but not always
Why would the community
sometimes not respond?
Limitation by light—Forest canopy
Grazers—sometimes biomass of
grazers responds but algae don’t
Flood Scouring—after floods there
might not be enough biomass
available to mount a response to
nutrient addition.
Managing releases of nutrients
to lakes and rivers
Awareness of the role of nutrients in eutrophication of waterways led to legislation
aimed at controlling nutrient release (eg improved sewage treatment, P-free
detergents). Such nutrient abatement measures were often highly successful in
restoring water quality
In Lake Washington, which formerly received the sewage from Seattle Washington,
nutrient diversion had a dramatic effect on the trophic status of the lake
Attempts like the Lake Washington experiment, where point source nutrient sources
were diverted from the lakes, failed more often than they succeed. Reasons?
1.)Large non-point source nutrient loading
2.) Lake not nutrient limited
3.) Internal loading from P trapped in sediments may have compensated for reduced
external loading.
Phosphorus models have played an important role in
eutrophication management
Since it requires a lot of investment of effort and funds
both public and private to improve sewage treatment or
build structures to divert point sources, it is best to try to
have some idea how well it will work.
Modelling incorporates what we know about the
important processes involved, and tells us how to expect
the lake to respond.
We can write an equation t hat describes this mass balance
Let P  total amount of phosphorus in the lake
P  It  Rt  hPt  ksedPt
I and R are the rates of external and internal loading,
both independen t of P, the phosphorus in the water column
and outflow  hP, where h is the fraction of the water column flowing
out per yr.
sedimentat ion  ksedP,
where ksed is the fraction t hat settles to the bottom per yr.
All of these changes are measured as kg P / yr
single compartment mass balance model
I+R
hP
P
(external+internal
loading all sources)
Downstream export
[P]=P/V
Sedimentation rate
ksedP
All processes (ie arrows) have the units (kg P/yr)
What are the units of h and ksed ?
The mass balance equation is
P  It  Rt  hPt  ksedPt
If we divide through by t and assume t 
 0 .
dP
 I  R  hP  ksedP
dt
How can we reconfigur e this equation so that it deals with
P concentrat ions (i.e [P] )rather th an the total amount P ?
Let P  P V , where V is the volume of the lake, which we assume to be
constant. Then
d P 
V
 I  R  hP V  ksed P V
dt
and
d P  I  R

 hP   ksed P 
dt
V
d P  I  R

 hP   ksed P 
dt
V
d P 
At equilibriu m
 0; i.e. inputs and losses balance
dt
IR
 h  ksed P *,
V
where P * is the equilibriu m P concentrat ion.
IR
IR
P* 

V h  ksed  Qo  Vksed
IR
IR
P* 

V h  ksed  Qo  Vksed
What prediction s does this equation make?
What factors will cause [P]* to increase?
What factors will cause it to decrease?
What causes the [P]*to reach equilibriu m when I and R change?
What factors might cause ksed to change?
Example Problem
Pigeon lake is an AB lake whose total external P load is 5600 kg/yr, 900 of which is
estimated to be the result of cabins and other residential activities around the
shoreline, and 1800 of which is estimated to be the result of agricultural activities in
the watershed. The present P concentration in the lake water averages 32 mg/m3.
How much would this drop if the cabins and houses were removed and all
agricultural activities in the watershed were to cease?
The mean depth of the lake is 6.2 m, the Area is 96.7 km2. The internal load from
the basin in this lake has been estimated to be 3000kg/yr. Mean annual outflow of
water is 4.5 x 106 m3/yr.
What do we need to assume in order to apply the previous model to this question?
What will be the new P concentration in the lake water after the lake reaches
equilibrium with the new loading regime?
We need to assume
(1) that the lake is in equilibriu m, with respect to P and hydrology
(2) we need to assume that the sedimentat ion constant
(the fraction of water column P that settles out in a year)
won' t change appreciabl y
V  6.2m  96.7 km2  600 x 106 m 3
At equilibriu m
IR
IR

V h  ksed  Qo  Vksed
Since, we know the present P *, I ,V , Qo & R and can calculate h ,
we can solve for ksed
P* 
P* Qo  Vksed   I  R

I  R   P * Qo (8600 x 106 mg  yr 1 )  (32mg  m -3 )(4.5 x 106 m 3  yr -1 )
ksed 

P*V
(32mg  m -3 )(600 x 106 m 3 )
 ksed  0.44  yr 1
The new equilibriu m P concentrat ion would then be
IR
(5900 x 106 mg  yr 1 )
P* new 

Qo  Vksed (4.5 x 106 m 3  yr -1 )  (600 x 106 m 3 )(0.44 / yr )
 22 mg  m 3
a 31% decrease from the initial value of 32 mg  m 3 .
Lakes can be classified into trophic categories on the basis of productivity and
nutrient measures—however such boundaries are arbitrary.
Oligotrophic
P<10
Chla<3
Mesotrophic
10<P<40
3< Chla <25
Eutrophic
P>40
Chla > 25
A review of some terminology concerning Nutrient inputs
External loading (I)
-point source (domestic, agricultural or industrial)
-non-point source (stream or seepage flows)
-atmospheric (wet or dry depostion)
Internal loading (R)
-return from the sediment (diffusion)
Forms of Nutrient
-Available for direct uptake by primary producers
dissolved free ions e.g. PO4-3, NO3-, NH4+
-Non-available
particulate—organic or inorganic (e.g. clays, minerals)
dissolved organic matter,
e.g. sugar phosphates, ATP, amino acids etc
We generally model total nutrient pools, because over the residence time scale of
nutrients in a lake most unavailable forms will be become available through
decomposition or desorption
Nutrient Retention by lakes
Most lakes tend to retain nutrients--i.e. output flow contains less than
inflowing water, with much of the input retained in sediment.
Lakes with large internal loadings exhibit “negative retention”.
Although we made rapid progress in curtailing point source nutrient loadings
such eg (domestic sewage, feedlots etc) it took longer to realize that nutrient
export from the landscape (non-point source loadings) also contributed greatly
to eutrophication of lakes and rivers
Studies on the nutrient concentrations in stream flow, lead to the concept of
watershed nutrient export
For a stream watershed
N export coef. (g  km  yr
2
1

N ( g  m )Qm
)
3
3
 yr 1

DA(km2 )
For a lake watershed
Total Non - point source loading( g/yr )
N export coef. (g  km  yr ) 
Drainage area (km2 )
where N refers to any nutrient
2
1
By studying streams that drain different types of watersheds reflecting different
types of land use, it is possilble to determine how land use affects nutrient export
from the landscape.
Nutrient export rates from undisturbed watersheds are much lower than
those measured in disturbed landscapes (e.g. Agricultural and Urban)
Landscape
N export coef (kg/km2/yr) P export coef
N:P
Tundra
Boreal forest
Temperate forests
Agricultural land
27
97
300 (130-500)
600 (500-1000)
1.4
4.1
10 (5-12)
30 (5-50)
19
24
30
20
Urban
700 (500-880)
480 (30-1660)
<5
Aquatic macrophytes
don’t respond to N and P
additions in the water
column and only to
sediment fertilizers when
sediment is very nutrient
poor.
Eg sand or gravel
Mud always has plenty of
N and P.
Primary productivity by aquatic macrophytes is frequently limited by
availability of inorganic C, which the leaves take up from the water
Ocean acidification –the ongoing
decrease in the pH of the oceans, caused
by uptake of anthropogenic CO2 from the
atmosphere.
• Since the industrial revolution surface
ocean pH is estimated to have decreased
from approximately 8.179 to 8.104 (a
change of −0.075).
•~25% increase in [H+]
•NOAA scientists have found large portions of the ocean to be undersaturated in
carbonates.
•Organisms such as Foraminifera, red algae and coral reefs produce carbonate
skeletons and require water to saturated or super saturated.
•Such organisms help trap carbon and sediment it on the ocean floor, and
acidification will greatly impair this process—a strong positive feedback which will
gradually reduce the ocean’s capacity to trap carbon
Primary productivity by
aquatic macrophytes is
frequently limited by
availability of inorganic C
Potamogeton pectinatus is
switching to bicarbonate as
carbon dioxide becomes
depleted in the weed bed
Callitriche stagnalis does not
have this capability
The ability to use
bicarbonate depends on the
presence of the enzyme
carbonic anhydrase which
catalyses the hydration of
carbon dioxide
Hard water has more dissolved inorganic C (carbonates and bicarbonates)
than soft water. Production, biomass and species diversity of aquatic plants is
always higher in hard water lakes and ponds than softwater systems.
Carbon in the form of
CO2, HCO3- and CO3-2,
are oxidized forms of C,
and tend to be the only
forms present where O2
is plentiful.
In anoxic environments
methanogens
(Archaea) convert
organic C and CO2 into
methane (CH4).
Methane is a gas and
can bubble out of the
water or it can be
oxidized to CO2 by
Methanogens are not true bacteria, they belong to the Archaea
Most methanogens can grow on CO2 and H2 as their sole energy source:
Chemoautotrophs —chemical bond energy is their energy source
they utilize CO2 as their C source
http://faculty.plattsburgh.edu/jose.deondarza/images/Organisms/methanogen.jpg
C-transformations in aerobic and anaerobic environments
Oxidation
state
Where do we find methanogens?
-4
0
+4
•Under anaerobic conditions organic molecules break down to methane instead of CO2—This
process is facilitated by methanogens (Archaea), which are chemoautotrophic bacteria.
•They utilize the energy released from 2H2 + Organic C (CH2O)→CH4 + H20 to build their
biomass.
How to assign Oxidation numbers
We keep track of the e- transfer using Oxidation numbers (Ox#)
For each e- transferred the Ox# changes by 1
2H2 + O2
0 0
2H2O
+1 -2
Some rules for Oxidation numbers
1. In free elements Ox# =0
2. For ions with one atom Ox# = charge. eg H+ Ox# of H+ = 1
3. Ox# of O in most compounds is -2,
4. Ox# of H in most compounds is +1,
5. For a complex ion like SO4-2 , the net Ox# = charge (Thus S=+
Carbon and Oxygen play a major role in biological
energy transformations
Oxidation means giving up electrons, and reduction means taking on
electrons (─)
The oxidation state of C in CO2 is +4, in Carbohydrates is 0
How many electrons are taken up by each C during photosynthesis?