PRIMARY PRODUCTIVITY
•Productivity is the rate of biomass formation
•Primary productivity (photosynthesis) of phytoplankton
can be measured directly by O2 production
•Photosynthesis can be summarized as
6CO2 + 12 H20 ----> C6H12O6 + 6O2 + 6H2O
•Respiration is the reverse reaction
The most common techniques for PP measurement are:
•the light-dark bottles technique for O2 production
•14C uptake measurements of CO2 assimilation
•harvest method
•Whole water column O2 or CO2 flux
Primary Producers in Aquatic Ecosystems
Cyanobacteria, Protista (microalgae, macroalgae), Aquatic plants
These groups are all found in both standing and flowing water
In standing water the two main primary producer communities are:
a.) the phytoplankton
(suspended cells or colonies of microalgae or cyanobacteria)
b.) the littoral community
(benthic micro- or macroalgae or cyanobacteria,
epiphytic microalgae, and
macophytes (macroalgae or plants)-emergent/submerged
In running water the two main primary producer communities are:
a.) benthic algae attached to rocks (generally in fast current)
b.) submerged macrophytes & associated epiphytes—in areas of slow current
Nightime
mg/L O2
Afternoon
mg/L O2
2
4
6
5
8
10
2
4
6
8
5
Depth
m
Depth
m
10
10
Average [O2]
Epilimnion =
9.8 mg/L
Average [O2]
Epilimnion =
6.9 mg/L
10
Van Dorn sampler obtains samples of water from a desired depth
A Yellow Springs Instrument DO meter being calibrated
Changes in Dissolved Oxygen
•Water samples from various depths are enclosed in light
(transparent) and dark (completely opaque) bottles,
•For each depth initial readings of dissolved O2 are taken (IB) and
the light (LB) and dark (DB) samples are incubated for a period
long enough to produce measurable changes in O2.
•During the incubation, we expect that the initial DO concentration
(IB) at a given depth will decrease to a lower concentration in the
dark bottles (DB) due to respiration of phytoplankton.
•Conversely, we expect light bottle (LB) should increase from their
initial values (IB).
IB  DB
 respiratio n rate/unit volume
t
LB  IB
 net photosynth etic rate/unit volume
t
LB  DB
 gross photosynth etic rate/unit volume
t
Productivity measurements are usually expressed in terms of
carbon not oxygen
Therefore we need to convert the changes in oxygen
concentration to corresponding changes in carbon.
For this we use the photosynthetic quotient (PQ)
and the respiratory quotient (RQ), which are ratios describing
the relative amounts of oxygen and carbon involved in
photosynthesis and respiration.
:
PQ = (molecules of oxygen liberated during photosynthesis) /
(molecules of CO2 assimilated)= usually around 1.2
RQ = (molecules of CO2 liberated during respiration) /
(molecules of oxygen consumed) = usually very close to 1.0
Calculations
•Gross photosynthesis = [(LB - DB) * 1000 * 0.375] / (PQ * t)
•Net photosynthesis = [(LB - IB) * 1000 * 0.375] / (PQ *  t)
•Respiration = [(IB - DB) * RQ * 1000 * 0.375] /  t
•Gross and net photosynthesis and respiration are expressed
as mg C/m3/t
•LB, DB, and IB are dissolved oxygen concentrations in mg/L
•
 t is the incubation period eg. hours
•1000 converts L to m3 (1 L = 1000 cm3)
•0.375 converts mass of oxygen to mass of carbon and is a
ratio of moles of carbon to moles of oxygen (12 mg C/32 mg
O2 = 0.375)
Because we are using bottles incubated at various depths in the photic zone to
measure primary productivity in situ, and the measurements vary with depth,
we need to obtain a profile of primary productivity with depth.
% of surface irradiance
Primary productivity
(mgC/m3/d)
Depth (z)
Let us say that we wish to incubate samples at the 75% , 50%, 25%, 10% and 1%
light levels. How do we decide at what depths to incubate samples at?
The Secchi disk—a simple way to estimate light extinction
1.7
k
Secchi disk trans parency (m)
Light extinction --Light enters from above and its intensity (I) is sharply
attenuated with depth (z)—absorption by water or solute molecules or
scattered by particles
Section 10.6
Iz
z 50%
z 10%
At what depth z 1% while the light intensity be 1% of I 0
Iz
 e  kz so 0.01  e  kz and  ln 0.01  kz...
Photic I 0
zone  z  4.6 ,
k
for the depth at which light intensity is 10% of I 0
2.3
,
k
for the 50% light level
z
z 1%
z
z
Page 144 in text
0.69
k
For a lake with Secchi disk transparency of 3m k=0.57 (1.7/3)
%light
z=-ln
(%light/100)/k
(k=0.57)
ln(%light/100)
0.75
-0.29
0.50 m
0.5
-0.69
1.22 m
0.25
-1.39
2.43 m
0.1
-2.30
4.04 m
0.01
-4.61
8.08 m
If we place dark and light bottles at each of these depths, and calculate
the Gross Primary productivity at each
GPPz gC/m3/d
1.5
0.5 1.0
z 75%
z 50%
x
x
x
z 10%
x
m
z 1%
z
x
Photic
zone
Find the area under the GPP vertical profile
Its units will be gC/m2/d, Why?
If we place dark and light bottles at each of these depths, and calculate
the Gross Primary productivity at each
Find the area under the GPP vertical profile
Its units will be gC/m2/d, Why?
GPPz gC/m3/d
1.5
0.5 1.0
0.45 sq in
x
1.0
B
x
x
3.0
Photic
zone
A
Area A represents
[4.10/0.45] *1.0 gC/m2/d=9.1 gC/m2/d
Average productivity per unit volume
9.1gC/m2/d / 9.1 m = 1.0 gC/m3/d
x
5.0
7.0
x
zm
Area B represents
0.5 gC/m3/d * 2 m = 1.0 gC/m2/d
4.10 sq in
If you average all the GPP estimates
1.2,1.65,1.70, 1.1,0.22,
=1.174 gC/m3/d x 9.1 m = 10.7 gC/m2/d
Averaging the readings leads to a nearly 20% overestimate, Why?
Primary producers
differ in their
photosynthetic
responses to light
intensity
Comparing the Photosynthethis/irradiance curves among species
A
GPP/t
per unit of
biomass
B
C
10%
25%
50%
Light intensity
Assuming that A, B and C are similar in all respects other than their P/I curves, which
of these species would you expect to perform best in the well mixed water column of a
deep lake (25% light level at 10 m, max depth 100 m)
a) A
b) B
c) C
d) A and B would do equally well
e) A and C would do equally well
A
GPP/t
per unit of
biomass
B
C
10%
25%
50%
Light
intensity
Assuming that A and B are equal in all respects other than their P/I curves, under what
conditions would you expect B to outperform A?
a) in the epilimnion of a clear stratified lake, (Assume 25% light level at the thermocline)
b) in the well-mixed water column of a deep lake (Assume 25% light level at 10m, max
depth 100 m)
c) in the hypolimnion of a clear stratified lake (Assume 25% light level at the thermocline)
d) growing on the substrate near shore in the littoral zone of a clear lake (assume 25 %
light level at the outer boundary of the littoral)
e) both a and d
f) both b and c
A
GPP/t
per unit of
biomass
B
Could C ever be
expected to
dominate?
C
10%
25%
50%
Light
intensity
The seasonal dyanamics of the phytoplankton in lakes
•Temperature adaptations of different algal groups
•Thermal stratification, sinking rates and nutrient dynamics
•Food-web interaction—effects of grazing zooplankton
mid-summer low biomass
community shifts to inedible forms
Early spring—diatoms dominate--under cold
temperatures and low light conditions
plenty of nutrients in the well mixed water column
Summer—lake warms up, thermocline forms
diatoms fall out of the mixing layer—low viscosity
and low mixing depth
Asterionella the only diatom that can still hang in.
Mid-summer—nutrients lost from mixed layer
(sedimentation of algae), warm temperatures favour
green algae, and zooplankton herbivory is high
favouring fast growing small species eg Chlorella
Late summer—herbivores eliminate edible
species, large colonial cyanobacteria dominate
eg. Microcystis
Fall—water cooling, thermocline breaks up,
mixing depth increases, nutrients increase,
diatoms dominate
Winter—low light and cold temp low biomass
Adaptations of benthic diatoms to life in streams
They can live either as solitary cells, usually raphed—capable of slow gliding movement on rocks
or in chains joined at the valvular surface, attached to substrate either by mucilage pad or by a stalk
http://craticula.ncl.ac.uk/EADiatomKey/html/Achnanthes.html
Achnanthes longipes: alternation between a motile, solitary phase and a stalked sessile phase
When inoculated into fresh media, the cells are at first motile (I), then become sessile and produce a stalk
that anchors them to the substratum (II). This stalk continues to be synthesized as the cell is pushed away
from the substratum (III), and eventually mitosis occurs producing a row of cells stacked one upon the
other (IV). These cells eventually detach from one another (I) and begin the cycle again.
Why is this life cycle be advantageous for life in streams?
http://www.bio.mtu.edu/the_wall/integrated_microscopy/a_longipes_general.html
Daily fluctuations in O2 provides and integrated measure of
community metabolism over a reach.
Daily cycle in dissolved oxygen (flow weighted average) at Pavan study site
Pavan
12.50
12.00
11.50
11.00
Dissolved O2
mg / l
10.50
100% saturation
10.00
9.50
9.00
8.50
8.00
0:00
8AM
2:24
4:48
2PM7:12
9:36
8PM
12:00
Tim e
14:24
2AM19:12
16:48
21:36
8AM
0:00
A
B
C
O2 mg/L
Day
Night
Which one of these curves is from a treated sewage effluent and
which is from an untreated sewage effluent?
The photosynthesis versus irradiance (light intensity) curve—fig. 21-23
Photosynthesis Irradiance curve constructed from water column GPP profile
Pmax
GPP
gC/m3/
d
1.8
1.6
1.4
1.2
a
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
Light availability (Iz/I0)
0.6
0.8
You find that C outperforms both A and B during the summer months but not in the
early spring. Assuming that all three species have similar temperature optima and
nutrient uptake affinities, which of the following explanations is most plausible?
a) C is the least palatable species to herbivorous zooplankton
b) B does best at low light intensity, and A does best at high light intensity, but C does
best under fluctuating light intensities
c) C has the most eccentric shape
d) a and b are both plausible
e) a and c are both plausible
A
GPP/t
per unit of
biomass
B
C
10%
25%
50%
Light
intensity
Diatom in valve view, What does the word diatom mean?
raphe
Central area
stria
Terminal
nodule
Central nodule
Striae are composed of punctae (pores), what do you think their function is?