PRIMARY PRODUCTIVITY Productivity is the rate of biomass formation Primary productivity (photosynthesis) of phytoplankton can be measured directly by O 2 production Photosynthesis can be summarized as 6CO H > C 6 H 12 O 6 + 6O 2 + 6H 2 O Respiration is the reverse reaction The most common techniques for PP measurement are: the light-dark bottles technique for O 2 production 14 C uptake measurements of CO 2 assimilation harvest method Whole water column O 2 or CO 2 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

Afternoon Nightime Depth m mg/L O Depth m mg/L O Average [O 2 ] Epilimnion = 9.8 mg/L Average [O 2 ] Epilimnion = 6.9 mg/L

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 O 2 are taken (IB) and the light (LB) and dark (DB) samples are incubated for a period long enough to produce measurable changes in O 2. 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).

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 CO 2 assimilated)= usually around 1.2 RQ = (molecules of CO 2 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/m 3 /t LB, DB, and IB are dissolved oxygen concentrations in mg/L  t is the incubation period eg. hours 1000 converts L to m 3 (1 L = 1000 cm 3 ) 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 O 2 = 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 Depth (z) Primary productivity (mgC/m 3 /d) 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

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 IzIz z Photic zone z 50% z 1% z 10% Section 10.6 Page 144 in text

%light ln(%light/100) z=-ln (%light/100)/k (k=0.57) m m m m m For a lake with Secchi disk transparency of 3m k=0.57 (1.7/3)

If we place dark and light bottles at each of these depths, and calculate the Gross Primary productivity at each GPP z z 10% z Photic zone z 50% z 1% x x x x z 75% gC/m 3 /d m x Find the area under the GPP vertical profile Its units will be gC/m 2 /d, Why?

If we place dark and light bottles at each of these depths, and calculate the Gross Primary productivity at each z Photic zone GPP z x x x x gC/m 3 /d m x sq in 0.45 sq in A B Area B represents 0.5 gC/m 3 /d * 2 m = 1.0 gC/m 2 /d Area A represents [4.10/0.45] *1.0 gC/m 2 /d=9.1 gC/m 2 /d Average productivity per unit volume 9.1gC/m2/d / 9.1 m = 1.0 gC/m3/d If you average all the GPP estimates 1.2,1.65,1.70, 1.1,0.22, =1.174 gC/m 3 /d x 9.1 m = 10.7 gC/m 2 /d Averaging the readings leads to a nearly 20% overestimate, Why? Find the area under the GPP vertical profile Its units will be gC/m 2 /d, Why?

Primary producers differ in their photosynthetic responses to light intensity

GPP/t per unit of biomass A B C Light intensity 10% 25% 50% Comparing the Photosynthethis/irradiance curves among species

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 GPP/t per unit of biomass A B C Light intensity 10% 25% 50%

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 GPP/t per unit of biomass A B C Light intensity 10% 25% 50% Could C ever be expected to dominate?

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

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. Achnanthes longipes: alternation between a motile, solitary phase and a stalked sessile phase Why is this life cycle be advantageous for life in streams?

Daily fluctuations in O 2 provides and integrated measure of community metabolism over a reach.

O 2 mg/L A B C 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

GPP gC/m 3 / d Light availability (I z /I 0 ) Photosynthesis Irradiance curve constructed from water column GPP profile P max 

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 GPP/t per unit of biomass A B C Light intensity 10% 25% 50%

Diatom in valve view, What does the word diatom mean? Central area raphe stria Terminal nodule Central nodule Striae are composed of punctae (pores), what do you think their function is?