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Flashcards in Lab 5 Deck (10)
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photosytnesis primary step

light rxn hill reaction (photolyses water, traps ATP


photosystheis 2nd step

dark rxns, calvin cycle, CO2 fixed to form glucose


the higher the [ ] of oxygen in the solution...

...highter the rate of o2 difusion, greater the pen deflection on chart


what does the nirtogen do?

disapare the o2 gas and set it to zero oxygen


The procedure for Part D (page 34) was performed separately from the oxygen evolution measurements but is important in the context of this experiment. What information is obtained from this procedure and why do you need it?

The procedure for part D was important to the context of the experiment, because it allowed for the calculation of the total pigment concentration in the original cell suspension. This is what was used for the oxygen evolution experiment so we now know the total pigment concentration for the first part of the experiment.


I asked you to measure the rate of respiration of the cells in the dark. Were you able to observe a measurable rate of oxygen consumption at the three temperatures you tried? Given that some oxygen is consumed by algal respiration in your experiment, how do you think this compromises any measurement of oxygen evolution? Explain your answer

Test 1: 72.5 - 71 (1.5 chart units)
0.153 μg/min of oxygen consumed
Test 2: 61.5 - 60 (1.5 chart units)
0.153 μg/min of oxygen consumed

Test 1: 66 - 63.5 (2.5 chart units)
0.232 μg/min of oxygen consumed
Test 2: 65 - 63-5 (1.5 chart units)
0.140 μg/min of oxygen consumed

Test 1: 54 - 52 (2 chart units)
0.172 μg/min of oxygen consumed
Test 2: 52 dramatic drop/jump - 60

A measurable rate of oxygen consumption was observed for each experiment of a loss of around 1.5-2.5 chart units in the two minutes time. This was a steady value that didn’t seem to charge between the three temperatures. The oxygen is consumed by algal respiration in the experiment will cause the rate of oxygen evolved to go down and skew our results as the molecular oxygen is turning into water in respiration.


Under the bright light conditions used in this experiment, an increase in temperature usually results in an increase in the rate of oxygen evolution. If this experiment were performed under dim light conditions, would you expect the effect of temperature to be the same? Explain.

If this experiment were performed under dim conditions, the affect of temperature on the rate of oxygen evolution would still exist. Having a higher temperature would still lead to a higher level of oxygen evolution. Since it would be dim conditions, these numbers would be less then the bright light conditions, but the effect of temperature would still exist due to the different levels of oxygen concentration in the solution being measured. (the solubility of oxygen in solution goes up as temperature goes up)


Explain how you would expect temperature to affect the rate of oxygen evolution with reference to both the light and the dark reactions in photosynthesis (be specific). Do your results support your expectations? If not, suggest why your results differ from the expected results.

light rxns are temp insensitive,
dark rxns are temp sensitive
dark+light rxns are linked (products of light runs are used by dark rxns and vice versa).
Increase dark rxns=increase ATP+NADP for light rxns-->increase O2 prod


Given the fact that the herbicide you used inhibits the light reactions of photosynthesis, explain the observed effect on oxygen evolution (be specific about the actual mechanisms in the light reactions that are being inhibited). If we were to use a herbicide that inhibited the Calvin cycle, how would this affect the rate of oxygen evolution? (be specific)

Common herbicides bind to a core protein of PSII, herbicide can bind to D1 and occupy the empty PQb site after the release of PQH2. This inhibits the electron transport chain. Since the light reaction is inhibited, the amount of oxygen goes up since water is slip into molecular oxygen, but the rest of the photosystem is inhibited. If we inhibited the calvin cycle, no more ADP+NADP for light runs to use and CO2 would therefore increase, and the rate of oxygen evolution would go down.


Calculate the rate of measured oxygen evolution (expressed as micrograms oxygen evolved/minute/microgram pigment) at

Total Chart units used for total amount of oxygen present in chamber= 90
From table, 18 degress=9.18 μg/mL
2mL were used of algal suspension so total oxygen solubility in chamber was:
2ml x 9.18 μg/ml=18.36mg
18.36 μg/90chart units= 0.204 μg/chart unit of oxygen solubility

Total pigment concentration= 460.15 μgchla/L (as calculated from equation below)
Pigment concentration for 2ml= 0.002 x 460.15= 0.9203 μgchla

Test 1 for 18 degrees:
Light Reaction (5 min):
start: 56 End: 72 (difference of 16 chart units)
16x0.204= 3.264 μg/5min
3.264/5= 0.6528 μg/min of oxygen evolved
(0.6528 μg/min) / (0.9203 μgchla)= 0.709 μg/min/μgchla

Test 2 for 18 degrees:
Light Reaction (5 min):
start: 47.5 End: 61 (difference of 13.5 chart units)
13.5 x 0.204=2.754 μg/5min
0.551 μg/min
0.551 / 0.9203 = 0.599 μg/min/μgchla

Average for 18 degrees:
(0.709+0.599) / 2= 0.654 μg/min/μgchla

Total pigment concentration:
A665-A750= 0.388-0.006=0.382
A650-A750= 0.211-0.006=0.285
#μgchla = [1.65x OD665 - 0.83xOD650] x10 / 0.01
[(1.65x0.382)-(0.83x0.285)] x10 / 0.01
=460.15 μgchla/L