Grossmont Extraction & Purification of Photosystem I & II from Spinach Lab Report

D. Results

The length of this section will vary from experiment to experiment.  Present the results in the most clear and understandable fashion possible.  This may involve Figures, Tables, Charts, and/or Graphs, but must also contain typed explanatory text.  You cannot just present a collection of tables and graphs and call that your Results section; you must explain in words what the charts and graphs demonstrate.  Data shown in a graph also must be presented in a table.

E. Conclusion and discussion

Discuss the results of your experiment.  In your introduction there was a purpose statement.  Address this statement by explaining the results gathered, and if the experiment accomplished what was said in the purpose.  Finally, do include error analysis.  Describe what errors may have taken place and the impact that these had on the results.

Extraction & Purification of Photosystem I & II from
Spinach.
Part A: Introduction
Overview
One of the most important steps in characterizing chemical and biochemical systems is
separation of components. The reason this is an important step is because the analytical
instruments and tools available, such as UV-VIS, tend to measure a limited number of
parameters and those parameters can be affected by neighboring material. So if follows that
to study the particular processes involved in plant based photosynthesis, the parts of the
plants involved in photosynthesis must be separated from the rest of the plant material to
eliminate interactions and isolate and quantify those systems.
In plants, photosynthesis occurs in two photosystems, I and II. Photosystem I consists of a
series of at least 13 polypeptide chains, chlorophyll molecules (a and b) and carotenoids and
produces NADPH. NADPH (nicotinamide adenine dinucleotide phosphate) is involved in
electron transport within biochemical reactions. Photosystem II also contains chlorophyll a
and b, carotenoids, but contains different polypeptides and produces free oxygen through a
different mechanism. The two photosystems are essential to life.
Both photosystems are found in chloroplasts in plant cells. Chloroplasts are organelle
(cellular sub-structures) found within cells and bound by a double membrane. Within the
chloroplasts are flat “sacs” called thylakoids. Photosystems I and II along with chlorophyll a
and B are found within the double membrane and the thylakoids.
In addition, since chlorophyll a and chlorophyll b are both photosensitive chemicals, and since
they have different structures, they will both absorb light at slightly different frequencies.
Chlorophyll a absorbs 642nm red light while chlorophyll b absorbs 625 nm red light. By using a
spectrophotometer with those two frequencies, the concentration of chlorophyll a and b in a
given sample can readily be measured. In this this experiment, there are carotenoids present
which also absorb light at those two frequencies. To adjust for this, the frequencies are
slightly shifted off peak for chlorophyll to a wavelength that interference from carotenoids is
minimal. 663nm and 645nm.
In this experiment, chlorophyll a and b from chloroplast walls and chlorophyll a and b from
thylakoids was separated from plant material through a process called “differential
centrifugation”. That process involves centrifuging materials at different conditions and
different resulting forces, to separate by density and size. The concentrations were then
measured with a UV-VIS (ultraviolet – visible light) spectrometer.
Purpose
The purpose of this experiment is to demonstrate the technique of differential centrifugation
to isolate the chlorophyll a and chlorophyll b components contained in the cellar walls of
chloroplasts from chlorophyll a and b contained in the thylakoids within spinach leafs.
General Plan
In this experiment, approximately 50g of spinach leaves was prepared by washing with
copious amounts of deionized water followed by removal of midribs and petioles. Once dried,
the leaves where chilled in a refrigerator for 10-15minutes then added to a Waring blender
along with 120mL of cold STN (a solution of sucrose, Tris, and NaCl). The STN solution
provided a iso-osmotic media which prevented water transport (and cellular/organelle
destruction) through the cell walls. The mixture was homogenized by blending for 15 seconds
The resulting mash was then placed into a centrifuge tube and centrifuged for 7min at 3300
RPM. This initial centrifugation separated the STN, water from the leaves, and material in
leaves smaller than chloroplasts from the chloroplasts and larger materials (such as dirt and
remaining stems / ribs). The supernatant was saved for later analysis.
The pellet from the centrifuge tube was then redispersed in 35mL of STN. The resulting
suspension was then centrifuged a second time at 1600 RPM for 1 minute to separate the
lighter chloroplasts from the heavier dirt and larger particles. The pellet was then discarded.
The supernatant from the second centrifugation was then treated with 10mL of DI water
which transported through the chloroplast outer membrane due to osmotic pressure. The
increased water content caused the chloroplasts to burst releasing whole thylakoids and
fractions of the chloroplast outer membranes. The resulting liquid was centrifuged a third
time at 12000 RPM and 20 minutes to separate out the larger thylakoids from the fractional
pieces of the chloroplast outer walls. The pellet, containing the thylakoids, was then
resuspended in buffer D. Buffer D is slightly buffered solution containing Mg2+ ions.
Magnesium ions deactivate proteins and prevent them from recombining with cell walls (forms
aggregates or agglomerates in solution).
Both solutions (the supernatant containing the chloroplast cellular walls and the resuspended
thylakoid) were then treated with 85% acetone to (1) solubilize the membranes, (2) denature
and precipitated proteins out of solution, (3) free chlorophyll a and b from proteins, and (4)
to act as a solvent for the chlorophyll (chlorophyll is a hydrophobic molecule, acetone is an
organic non-polar solvent).
The samples were then centrifuged to remove the proteins, and the resulting supernatants
were diluted again in acetone (80%) to create samples within the detection limits of a
laboratory UV-VIS spectrometer. Percent transmittance and absorbance were then measured
and the concentrations of chlorophyll a and b were then measured and calculated via the
Beer-Lambert law which for this system at the wavelengths given is
total chlorophyll concentration
μg
= 8.02 × abs663 nm + 20.2 × abs645 nm
[mL]
The values were then corrected for dilution factors and the percentage of chlorophyll
obtained from chloroplast walls to thylakoids was then determined.
Finally, the solution containing thylakoids was treated with a surfactant (Triton X 100) to
release photosystem I components from their respective proteins. Triton X accomplishes this
by solubilizing the ends of the proteins where the photosystem I components are peripherally
attached. Photosystem II components are integral to the proteins. The ratio of Triton X to
thylakoid must be tightly controlled because too much surfactant will solubilize the entire
protein releasing photosystem II components as well. In a 2.5:1 ratio, Photosystem II
remained bonded to their proteins. After addition of sucrose (to enhance separation by
density gradient) the solution was then centrifuged a 4th time at 12000 RPM for 15 minutes.
Photosystem one was floating on the top supernatant liquid, photosystem 2 was between the
supernatant and the sucrose layer.
Theoretical
Centrifugation follows two well-known physical laws. The first is
R PM =
F
r
→
Fcentrifuge = R PM 2 x r
For any particular centrifuge, r (radius of rotation) is essentially constant so that Fcentrifuge
(applied centrifugal force) is proportional to the RPM of the centrifuge. The second law is
“stokes law” which states that a body moving through a fluid has a particular drag force that
follows this equation.
Fdrag = 6πμRv
Stok e′s law
Where µ is the viscosity of the fluid, R is the radius of the particle moving through the fluid
and v is the velocity at the which the particle moves through a fluid. For a particle to move
in a centrifuge tube, Fcentrifuge > Fdrag. Therefore, a system of two particles of different
diameters can separated by carefully controlling the spin rate and operational time of a
centrifuge.
In the first step of the process, the cellular structures of leaves are destroyed in a blender to
release the chloroplast subcellular structures. Centrifugation is then used at an RPM and time
that overcomes Fdrag for the chloroplasts, dirt and other material. As those materials sink to
the bottom of the centrifuge tube, they displace the liquid above it. In the second step of the
experiment the pellet containing the chloroplasts and other material is resuspended and
centrifuged at an RPM and time that allows for the larger material to travel to the bottom of
the tube. The chloroplasts remain suspended. In the third step of the experiment, the
chloroplasts are destroyed by osmotic pressure releasing particles of cell membranes and
complete thylakoids. Those are again separated by adjusting the centrifugal force and time.
All three of these centrifugations are “differential” and do not rely on density gradients to
achieve separation.
In the second part of the experiment, sucrose is added to create a differential gradient zone.
This is called a “rate-zonal centrifugation”. This essentially changes the viscosity of the fluid
the particles are traveling through and the density of the fluid. The result is a change in the
amount of force required and time required to achieve Fcentrifuge > Fdrag for our particles of
interest. In addition, the less dense particles float on the surface of the liquid.
There is third type of centrifugation processed called Isopycnic centrifugation. Iso means
same and Pycno is greek for density. The process involves creating a density gradient that is
higher at the bottom of the tube and lower at the surface than the material of interest. The
result is the material of interest is found in the middle of the tube.
Part B: procedure
Flowchart
Sample preparation
(deribing, washing, drying, chilling)
Redisperse pellet in 35 mL STN
Centrifuge at 1600 RPM for 1 min
Blend with 50 mL STN 15 sec.
Dispose of
solids
Roughly filter with EDTA saturated
Muslin cloth.
Centrifuge at 3300 RPM for 7
minutes
Retain
supernate
For other
experiment
Centrifuge supernatant liquid at 3300
RPM for 7 minutes
Pellet
Redisperse pellet in 2mL STN
Measure total volume
Add 10mL DI H2O
Centrifuge at 12000 RPM for 20 min
Supernatant
liquid contains
chloroplast
membranes
Retain solids
(Chloroplasts)
Redisperse pellet with 2mL buffer D
Pellet contains thylakoids
Retain solids
(Chloroplasts)
Dispose of
supernate
Add 4.8mL 85%
acetone to 0.2mL
solution
Add 4.8mL 85%
acetone to 0.2mL
solution
Centrifuge at
12000 RPM for 20
min to remove
proteins
Centrifuge at
12000 RPM for 20
min to remove
proteins
Add 4.5mL
acetone to 500uL
supernate
Add 4.5mL
acetone to 500uL
supernate
UV-Vis
UV-Vis
Dilute thylakoids to 1mg/mL with
buffer 2
Add 2.5 mg triton X100 per 1mg
thylakoid
Add 15mL 15% sucrose as
centrifugation gradient
Centrifuge at 12000 RPM for 15
minutes
Top layer is photosystem 1
Middle layer is photosystem 2
Data Table: (calculations shown below)
Abs @
663nm
Abs @
645nm
Chlorophyll
concentratio
n stock
solution
µg / mL
Total
Volume
Stock
mL
Total
Total
chlorophyl chlorophyll
l
Stock
Stock
mg
µg
Chloroplasts
0.553
0.236
2290
3.12
7145
7.14
Thylakoids
0.663
0.160
2130
3.69
7860
7.86
Chlorophyll concentration in cuvette tubes
Concentration ofchlorophyll
Chloroplast chlorophyll
T hylak oid chlorophyll
μg
= 8.02*Abs663 + 20.2*Abs645
[mL]
μg
μg
= 8.02*0.553 + 20.2*0.236 = 9.160
[mL]
m L dilute
μg
μg
= 8.02*0663 + 20.2*0.160 = 8.520
[mL]
m L dilute
Chlorophyll concentration in stock solution
Cuvet te t ube concentration x dilution factor
dilution factor =
5.0 m L solution #1 5.0 m L dilute solution
0.2 m L stock
0.5m L solution #1
Chloroplast chlorophyll
T hylak oid chlorophyll
=
250 m L dilution solution
1 m L stock solution
μg
9.160 μg 250 m L dilute
μg
=
= 2290
[mL]
m L dilute
1 m L stock
m L stock
μg
8.520 μg 250 m L dilute
μg
=
= 2130
[mL]
m L dilute
1 m L stock
m L stock
Total Chlorophyll in stock solution (µg)
3.12 m L 2290 μg
= 7145 μg
1
m L dilute
Chloroplast chlorophyll[μg] =
thylak oid chlorophyll[μg] =
3.69 m L 2130 μg
= 7860 μg
1
m L dilute
Total Chlorophyll in stock solution (mg)
Chloroplast chlorophyll[mg] =
thylak oid chlorophyll[mg] =
3.12 m L 2290 μg
1 mg
= 7 . 145 mg
1
m L dilute 1000 μg
3.69 m L 2130 μg
1 mg
= 7 . 860 mg
1
m L dilute 1000 μg
Yield
% yield =
mg chlorophyll f rom thylak oids
7.860mg
× 100% =
× 100% = 110.%
mg chlorophyll f rom chloroplasts
7.145mg
Potential reasons why yield > 100%
1. Not all the chlorophyll was separated from the chloroplasts by the acetone
2. Errors in volume measurement of stock solution
3. It’s possible that a contaminate in the thylakoid stock solution caused elevated
absorbance in the UV-VIS analysis leading to higher than expected thylakoid
chlorophyll concentrations