Thomas Caudle
Practical Report 2: Photosynthesis (Set A)
A1. Experiments on photosynthetic electron transport and
photophosphorylation using isolated chloroplasts
In a preliminary stage, chloroplasts were isolated from pea leaves, to be used in all of the
following experiments. The procedure can be found in the modules handbook, and no
changes were made to this procedure.
In this chloroplast assay, the photosynthesis reaction will be carried out in cuvettes, outside
the leaves. As a result, an artificial electron acceptor needs to be used. 2,6dichlorophenolindophenol (DCPIP) will act as the electron acceptor, and changes from a
blue colour to colourless when it accepts electrons, and is therefore reduced. This reaction
is often called the “Hill Reaction”.
i ) The reduction of DCPIP by chloroplasts
The procedure for i) can be found in the module handbook. In this particular experiment,
there were no changes made. However, this procedure is used as the basis for the
procedures of other assays in this practical, but with slight adjustments, which i will make
clear as I get to them.
Raw Data: Absorbance Values
Time (secs)
Absorbance
0
20
40
60
80
100
120
140
160
180
200
220
240
1.069
0.902
0.811
0.705
0.562
0.467
0.313
0.191
0.089
0.047
0.012
0.019
0.002
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Thomas Caudle
Absorbance
Time vs. Absorbance of DCPIP with
Chloroplasts
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Absorbance without uncoupler
0
20 40 60 80 100 120 140 160 180 200 220 240 260
Time (secs)
My data and graph clearly show that as time progresses, the absorbance of the reaction
mixture decreases. Therefore, the reaction mixture is becoming more colourless as the
DCPIP accepts electrons and is reduced.
There was one anomalous result which appears at 220 seconds, when the absorbance
increases slightly after it has fallen to a low level. I have no explanation as to why this
happened, other than that there may have been a smudge on the side of the cuvette which
would have raised the absorbance. There is no way that the increase could be due to the
DCPIP being oxidised and turning blue again, as the cuvette was in the light the whole time.
ii ) The effects of an uncoupler of photophosphorylation on photosynthetic
electron transport measured as DCPIP reduction
For this assay, an uncoupler of photophosphorylation will be added to the reaction mixture.
The mixture is set up exactly as in the procedure of i) only with the addition of 10µl of 3 M
ammonium chloride to act as the uncoupler.
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Thomas Caudle
Raw Data: Absorbance values
Time (secs)
Absorbance
0
20
40
60
80
100
120
140
1.059
0.976
0.871
0.689
0.360
0.145
0.011
0.001
Absorbance
Time vs. Absorbance with Uncoupler
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Absorbance
0
20
40
60
80
100 120 140 160 180 200 220 240 260
Time (secs)
From looking at the raw data and graph, it is clear to see that as time progresses,
absorbance decreases, much like i). However, it is a much more rapid decrease in
absorbance, which can only be due to the uncoupler.
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Thomas Caudle
Question: Ammonium ions uncouple photosynthetic electron transport from ATP synthesis.
How do they do this? Account for any differences you observe in the rate of electron
transport in this experiment compared with the rate observed in i) above.
The ammonium ion acts as a proton carrier and moves protons from either side of the
thylakoid membrane. The extra H+ on the ammonia molecule can dissociate and move
elsewhere. This means that a proton gradient can’t be established and the energy from this
gradient that would have driven the ATP synthase protein isn’t there, so ATP isn’t
synthesised.
Absorbance
Time vs. Absorbance of DCPIP with
Chloroplasts
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Absorbance without uncoupler
Absorbance with uncoupler
0
20 40 60 80 100 120 140 160 180 200 220 240 260
Time (secs)
A difference in the rate of electron transport can clearly be seen when the uncoupler is
present. This could be the chloroplasts photosynthesising more to try and create the proton
gradient, resulting in more electrons passing down the ETC and reducing the DCPIP more
quickly.
Iii ) The effects of the presence or absence of the envelope membranes on the
rate of electron transport measured as DCPIP reduction
In this experiment, the 0.35ml of water used in the basic procedure of i) is replaced with
0.35ml of 1.0M sorbitol. The addition of sorbitol should stop the chloroplasts bursting due
to osmotic pressure, therefore leaving the envelope membranes intact rather than them
being absent in the other assays.
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Thomas Caudle
Raw Data: Absorbance Values
Time (secs)
Absorbance
0
20
40
60
80
100
120
140
160
180
200
220
240
1.103
1.063
0.996
0.891
0.770
0.644
0.516
0.395
0.273
0.158
0.062
0.008
0.001
Time vs. Absorbance with Sorbitol
1.2
1.1
1
0.9
Absorbance
0.8
0.7
0.6
0.5
Absorbance
0.4
0.3
0.2
0.1
0
0
20
40
60
80
100 120 140 160 180 200 220 240 260
Time (secs)
The data and graph shown above for the experiment with sorbitol added are similar to the
first set of data obtained in i), except these data show a much more consistent decrease in
absorbance over the time frame.
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Thomas Caudle
Question: Compare the rates of dye reduction in the presence of sorbitol with the rates in the
absence of sorbitol. How does sorbitol maintain the intactness of the envelope? Account for
any differences you observe in the rates of electron transport in the plus- and minus-sorbitol
experiments.
Absorbance
Time vs. Absorbance of DCPIP with and
without Sorbitol
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Absorbance without sorbitol
Absorbance with sorbitol
0
20 40 60 80 100 120 140 160 180 200 220 240 260
Time (secs)
When the reaction mixture contains sorbitol, the reaction is slightly slower than when
sorbitol is absent. Sorbitol maintains the intactness of the envelope by lowering the water
potential outside the chloroplasts so there is no net movement of water by osmosis into the
chloroplasts, and therefore they don’t burst, keeping the envelope intact.
The decrease in the rate of DCPIP reduction when sorbitol is present could be attributed to
the fact that all the thylakoids are retained inside individual chloroplasts, so essential
elements to photosynthetic electron transport ie. Water, H+ ions, cannot be shared with
other thylakoids that are lacking them in other chloroplasts.
iv ) Control Experiments : Design and conduct control experiments, perfomed
essentially as in i) above, to demonstrate that:a ) Light is required for DCPIP reduction
For this control experiment, ill will use exactly the same procedure as in i) with the same
reaction mixture, but instead of putting the cuvette in front of a lamp between absorbance
readings, I will keep the cuvette containing the reaction mixture in the spectrophotometer
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Thomas Caudle
so it gets no light. The reason I am keeping the cuvette in the spectrophotometer instead of
any other method of blocking out light, is because spectrophotometers are designed so that
no ambient light can enter them, otherwise absorbance readings would be
misrepresentative. Therefore, I can ensure that my chloroplasts will get no light energy. I
will take readings at the same 20 second intervals as before.
Here are my results:
Time (secs)
Absorbance
0
20
40
60
80
100
120
140
160
180
1.078
1.061
1.055
1.055
1.052
1.050
1.049
1.049
1.048
1.047
Absorbance
Time vs. Absorbance in the Dark
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80 100 120 140 160 180 200 220 240 260
Time (secs)
My results clearly show that when no light is present, DCPIP reduction doesn’t occur, and
that light is critical for DCPIP reduction. The small decrease in the first minute can be
attributed to DCPIP being reduced when light hit the chloroplasts in the brief moment
before the cuvette entered the spectrophotometer.
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Thomas Caudle
b ) Chloroplasts are required for DCPIP reduction
The procedure for this experiment is exactly as described in i), but without the addition of
the 0.05ml of chloroplast suspension. The absence of chloroplasts will prove wether or not
they are needed for the reduction reaction. The cuvette will still be placed in front of the
lamp and absorbance readings taken every 20 seconds. The absence of chloroplasts will
give a definitive answer as to whether or not they are needed for the reduction reaction.
Here are my results:
Time (secs)
Absorbance
0
20
40
60
80
100
1.368
1.370
1.366
1.369
1.369
1.368
It is easy to tell just by looking at the raw data that there was no change in absorbance of
the solution, and therefore no reduction took place, proving that chloroplasts are necessary
for reduction of DCPIP
c ) The observed light-induced absorption changes are not due to changes in
the chloroplasts themselves
In this experiment, I will not be adding any DCPIP solution to the reaction mixture stated in
i). Taking absorption readings of the mixture with just the chloroplast solution will rule out
any interference they may be having with the absorption of the DCPIP solution. The
procedure is the same as in i), but with the change stated above. I will still be placing the
mixture in front of the lamp between readings.
Here are my results:
Time (secs)
Absorbance
0
20
40
60
80
100
0.799
0.800
0.798
0.797
0.798
0.795
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Thomas Caudle
The data tells me that the chloroplasts are not affecting the changes in absorption of the
DCPIP solution, as the absorption of the reaction mixture without DCPIP didn’t change from
its original level, despite still being placed in front of the lamp.
A2. Chromatographic Separation of Chloroplast Pigments
In this experiment, thin layer chromatography will be used to separate out the different
pigments in a chloroplast. Thin layer chromatography, in this experiment, involves
application of a mixture of the chloroplast solution and acetone onto a TLC plate coated
with cellulose, acting as the stationary phase. The end of the plate is then dipped into
diethyl ether which moves up the plate and acts as the mobile phase. The pigments are
carried up the plate at different rates, therefore separating them out and making them
identifiable.
The full procedure for this assay can be found in the module handbook.
Questions:
a) How can chlorophyll a and b be distinguished on your chromatogram? Identify
chlorophylls a and b, and note also the presence of yellow carotenoid pigments.
Annotate your chromatogram accordingly.
Chlorophyll a can be distinguished by its blue/green colour, compared to the
yellow/green colour of chlorophyll b.1 Furthermore, chlorophyll a has less affinity for
the stationary phase, and so moves further up the TLC plate than chlorophyll b does.
1
http://www.uni-r.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/D-TLC-e.htm, Feb 2009
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Thomas Caudle
b) Why do the pigments migrate at different rates?
The pigments migrate at different rates because they have different affinities for the
stationary and mobile phases. These affinities are determined by how well they dissolve in
the solvent, and how strongly they bind to the cellulose layer (stationary phase) on the TLC
plate. The more strongly a pigment adsorbs to the stationary phase, the shorter the
distance it travels, and the same goes for how easily it dissolves in the solvent.
c) Which of the chlorophyll pigments is most abundant? Comment on your conclusion.
Based on my chromatogram, I would say chlorophyll a is most abundant, because the band
is thicker and darker, suggesting that more pigment is situated on that part of the
chromatogram, and that there is more of it.
A3. Absorption Spectrum of Pea Leaf Extract
In this experiment, I will use a spectrophotometer to generate an absorption spectrum for
the pigments in some pea leaf extract. The procedure for extracting the pigments is simple,
and can be found in the handbook. I will also use the spectrophotometer in multi
wavelength mode to take the absorbance readings of the pigments at two set wavelengths
of 647nm and 664nm to compare.
Questions
a) Estimate wavelengths of the major absorption peaks in your spectrum. How do these
compare with textbook values?
From my generated spectrum, I would estimate that the two major peaks occur at 430nm
and 660nm.
After looking in textbooks, my estimates fall within the 400-440nm and 660-680nm ranges
for chlorophyll a absorption. Also, a small peak of chlorophyll b incorporated into the first
peak of chlorophyll a can be made out at around 450nm, which is also confirmed by
textbook values.
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Thomas Caudle
b) What information does the absorption spectrum of the pigment extract give about
photosynthesis in the leaf?
The absorption spectrum of the extract tells us that photosynthesis in the leaf is totally
dependent on blue and red light, as there is little or no absorption in the green and yellow
areas of the spectrum. Also, the spectrum shows that chlorophyll a is the main pigment in
pea leaves, because there is only a very small peak where chlorophyll b should be absorbing,
suggesting that there is less of it in the leaf.
Calculations
a) Calculate the concentrations of chlorophyll a and b, and total chlorophyll in your
extracts.
Wavelength (nm)
Absorbance
647
664
0.371
0.745
[Chlorophyll a] in cuvette = (13.10 x 0.745) – (2.57 x 0.371) = 8.80 µM
[Chlorophyll b] in cuvette = (22.10 x 0.371) – (5.26 x 0.745) = 4.28 µM
b) Based on the values obtained in a), calculate the chlorophyll a/b ratio and comment
on the value obtained. Discuss the value obtained in relation to the TLC result
obtained in A2
8.80 µM ÷ 4.28 µM = 2.05:1
There is roughly a 2:1 ratio of chlorophyll a to chlorophyll b in the extract. This is supported
by my TLC plate, as the chlorophyll a band is darker, and at roughly twice as thick as the
chlorophyll b band.
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Thomas Caudle
c) Calculate the total amount of chlorophyll as micromoles chlorophyll per gram fresh
weight of leaf tissue.
8.80 µM + 4.28 µM = 13.08 µM of chlorophyll
Leaf pigments dissolved in 5.0ml of acetone
Moles in solution = molarity / volume
Moles in solution = 13.08 µM / 5000 µl = 2.616 x10-3
2.616 x10-3 µmoles of chlorophyll in 0.05g of pea leaf.
2.616 x10-3 µmoles x 20 = 0.052 µmoles of chlorophyll per 1g of leaf tissue.
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