Purification by Differential Centrifugation
This is a long and complicated lab. It involves learning two new techniques and then applying them to determine an experimental result. Since this experiment involves two different assays and many steps, it will be performed over three weeks. For the first two weeks you will practice the protein concentration and SDH activity assays so that by the third week the lab will run more smoothly. In week one you will do a Lowry assay and develop a standard curve for protein concentration. In week two you will perform the SDH test on a sample of mitochondria given to you. Once you have mastered these two techniques, you will perform them in conjunction with the differential centrifugation and determine the amount of activity in each fraction and the fold enrichment of mitochondria from the original homogenate to the final product.
Week 1 will start with the Background. For the quiz you should understand the basics of differential centrifugation. We will then do the Lowry Assay for protein determination.
Week 2 we will do the SDH Assay.
Week 3 we will do the whole experiment, starting with mycelia and measuring the activity of the isolated mitochondria.
Background:
The purpose of this lab is to show how organelles can be purified from a tissue homogenate by differential centrifugation. This is a technique that is very commonly used in cell biology to purify a specific target (i.e., organelle) from a lysate or homogenate of a whole organism or tissue.
The process of differential centrifugation is based on the fact that organelles have differences in size, shape and density. As a result, the effect of gravity on each is different. We can use this principle to separate an organelle from a homogenous solution of particles by artificially controlling the gravity of a solution. This is done by putting the solution in a variable speed centrifuge and rotating them at a high rate of speed. This creates a force that can be much greater than the force of gravity, and particles that would normally stay in solution will fall out and form a pellet at the bottom of the tube. The relative centrifugal force can be calculated by the following equation:
R.C.F. = 1.119 x 10 -5 (rpm2) r
where rpm is the revolutions per minute of the rotor and r is the distance (in cm) of the particle from the axis of rotation. The radius used is the distance from the center of the axis of rotation to the middle of the centrifuge tube. The forces created at low speeds are small (e.g. 600 X g) and only very large or dense particles will fall out of solution (nuclei, whole cells and large cellular debris). At high speeds, the force created can be quite great (e.g. as much as 300,000 X g). At these speeds, most particles will fall out of solution and only very small, highly soluble molecules will remain in solution.
Differential centrifugation schemes involve stepwise increases in the speed of centrifugation. At each step, more dense particles are separated from less dense particles, and the successive speed of centrifugation is increased until the target particle is pelleted out. The final supernatant is removed, the pellet is resuspended, and further study or purification can be done on it. The fractionation of rat liver is an example of how this process works:
Figure 1
An important thing to note is that there is cross contamination between the second and third pellets. Mitochondria show up in Pellet 3 and lysosomes show up in Pellet 2. This shows that the separations made by this technique aren't absolute purifications, but relative enrichments of organelles.
Marker Enzyme
In order to develop a differential centrifugation scheme to isolate a particular organelle, a marker must be used to follow its isolation. The marker can be the activity of an enzyme that is confined to that organelle. For example, enzymes of the electron transport chain are membrane bound and confined to the inner membrane of the mitochondria. Therefore, after a centrifugation to isolate mitochondria, both the pellet and supernatant can be analyzed to see which has more of the activity associated with these enzymes. The fraction with more of the activity has been "enriched" with mitochondria. Purification of the organelle is accomplished by following the enrichment through successive steps.
Another way to follow enrichment is by binding a radioactive label to protein on the organelle. For example, cell surface membranes can be isolated by first binding a radioactive drug that has its protein receptor in the cell membrane. The drug binds tightly to the receptor and remains there, so the fractions that contain the most radioactivity are enriched for cell surface membranes.
Enrichment
Enrichment of the organelle in question is determined with respect to its specific activity. Specific activity is expressed in units of enzyme per milligram protein (un/mg). The reason for this is that during the differential centrifugation, proteins are being separated. Some are going into the pellet and some stay in the supernatant. Therefore, if your enzyme stays in the supernatant and many other proteins pellet out, the enzyme will represent a higher percentage of the protein in the supernatant than before it was centrifuged. However, if you measured enzyme activity in units per ml of supernatant (unit/ml) the activity would seem the same, since the volume of the supernatant hasn't changed. Similarly, if the enzyme ended up in the pellet, its activity expressed in units/ml would be dependent on what volume you resuspended the pellet in before doing the assay, and not on any enrichment created by the centrifugation. For this reason we measure both the enzyme activity in a fraction, and its protein concentration, and then calculate the specific activity of enzyme in units/mg protein.
General Procedure:
In this lab you will isolate mitochondria from cauliflower. This will be done by two centrifugations, resulting in two pellets and two supernatant fractions:
Figure 2
In order to determine the enrichment of mitochondria through the procedure, you will be following the activity of the enzyme succinate dehydrogenase (SDH). SDH is an inner mitochondrial membrane protein involved in the Krebs cycle. It converts succinate to fumarate with the production of one FADH2.
Succinate + FAD ______> fumarate + FADH2
In the SDH assay, a chemical (DCPIP) accepts the electrons from the FADH2 and becomes reduced. This reduction results in a color change that can be detected spectrophotometrically.
To determine enrichment, you will also need to measure the total amount of protein in each fraction (enzyme concentration is reported as units/mg protein). The procedure you will use to determine protein concentration is the Lowry method, which is explained in detail below.
Enrichment is calculated by taking a sample of the original homogenate and of each of the pellets and supernatants and measuring both enzyme activity and protein concentration in each. The enzyme concentration in units/ml is then divided by the protein concentration in mg/ml and the enzyme concentration in each sample is then reported as units/mg protein, for example:
Enzyme concentration / Protein concentration = Specific Activity
(10 units/ml) / (4.0 mg/ml) = 2.5 units/mg
The fold enrichment through a step of differential centrifugation is calculated by dividing the activity in the second sample by that in the first. For example, if the enzyme activity in the homogenate is 2.5 units/mg and the activity in the first supernatant is 10.0 units/mg, the fold enrichment is:
S1 / H = Enrichment
(10.0 units/mg) / (2.5 units/mg) = 4.0 fold
In addition to the calculation of fold enrichment, it is important to show that the total amount of protein and enzyme remains constant throughout the separation. That is to say, if you stared out with 80 mg of protein and 100 units of enzyme in the homogenate, then when you add up all the protein and enzyme in the resultant fractions it should add up to those figures. So, in this experiment, if you add up the total protein in the P1, P2 and S2 fractions, it should equal the total protein in the homogenate (and the same is true for total enzyme). This will never be exact due to experimental error. The total protein in the fractions may be sightly higher than that of the original (due to inaccurate measurements), or, more likely, somewhat lower than the original protein amount (due to sample lost in transfers). By doing this calculation you can determine where the largest experimental error entered into the calculations. For example, if your fold enrichment is much smaller than you would expect, this may be due to either overestimation of the protein or under estimation of enzyme activity. If the total protein in fractions P1, P2 and S2 add up to much more than that in the homogenate, then your experimental error is in the determination of protein concentration. Total protein or total enzyme in a fraction is calculated by multiplying the concentration by the volume of the fraction:
Protein: 4.0 mg/ml X 20 ml = 80 mg
Enzyme: 10 unit/ml X 20 ml = 20 units
Calculate this for H, P1, P2 and S2 and then:
P1 + P2 + S2 = H
for both enzyme and protein.
Week 1: Protein Determination (the Lowry Assay)
(This lab was downloaded by anonymous FTP from ns1.faseb.org and was originally written by Dr. Jason Wolfe [Department of Biology, Wesleyan University, Middletown, Connecticut 06457] and was modified to fit this experiment.)
Reference: Lowry, O.H. N.J. Rosebrough, A.L. Farr and R.J. Randall (1951) Protein measurement with the Folin phenol reagent. J. Biol. Chem. 193: 265-275.
Introduction:
The paper describing this assay procedure is one of the most widely referred to in the biological literature. It permits the determination of protein in amounts as dilute as 10 ug/ml by measuring the intensity of the blue color that develops after first reacting the protein solution with alkaline copper and then with Folin's phenol reagent. The Cu2+ ions form a pink-to-violet chelation complex with peptide bonds of proteins in alkaline solution. The phenol reagent is a phosphomolybdic-phosphotungstic acid reagent which is reduced to a dye, molybdenum blue, by aromatic amino acids, such as tryptophan and tyrosine. The action of both reagents leads to a stable and sensitive color whose maximum absorbance is at 600 nm.
Procedure:
1. Preparation of samples: * For the first week, when you only do the Lowry procedure, dilute the sample that you're given 1/5 and 1/10.
During the actual experiment you will dilute your samples in water as follows:
Dilute your samples in water as follows:
homogenate 1/6 (1/6 means 1 part sample to 5 parts water)
S1 1/6
P1 1/6
S2 1/2
P2 1/2
For each diluted sample put 0.1 ml of the sample into each of 3 clean 13 x 100 tubes. Add 0.9 ml distilled water (Important! This is another 1/10 dilution, which will be necessary for your calculations).
2. Prepare a standard curve and blank. To duplicate tubes add
0.1 ml BSA + 0.9 ml H2O
0.3 ml BSA + 0.7 ml H2O
0.5 ml BSA + 0.5 ml H2O
1.0 ml BSA
1.0 ml H2O
BSA refers to bovine serum albumin, a protein used as a reference standard and provided to you at a concentration of 200 ug/ml.
3. Prepare reagent C: In a beaker, mix 200 ml reagent A (2% Na2CO3, 0.1 M NaOH, 0.02 % Na, K tartrate) with 4 ml reagent B (0.5 % CuSO4.5 H2O)
4. Prepare reagent D:
6 ml phenol reagent + 12 ml distilled water
5. You are now ready to begin. Add 5 ml reagent C to each tube. Mix well. Wait ten minutes. Add 0.5 ml reagent D to each tube. Mix. Wait 10 minutes while color develops.
6. Use the spectrometer to read the absorbance of each sample at 600 nm.
7. Calculate the protein concentrations of the standards and plot absorbance versus concentration for the samples of the standard curve. Use this graph to determine the average concentration, in ug/ml, of your diluted samples. If the absorbencies of one of your samples is less than that of your most dilute standard, repeat the sample at a lower dilution. Using the dilution factors, calculate the protein concentrations in the original samples. Duplicate determinations should agree closely, as should different dilutions of the same original sample. (In your final report, show your calculations, data, and graph. Multiply by the volume in each fraction to get the protein content per fraction. Does the sum of the protein in all of the fractions equal the amount in the original homogenate? If not, give an explanation of where the error may have occurred.)
8. When finished, rinse all tubes and glassware thoroughly in running tap water. (Before discarding your fractions, you will want to be sure the assay worked properly. Otherwise you may need to repeat it!). Wash the spectrometer cuvettes carefully and leave to drain at your bench.
Week 2: Succinate Dehydrogenase (SDH) Assay
(This lab was also downloaded from ns1.faseb.org, originally written by Dr. Jason Wolfe [Department of Biology, Wesleyan University, Middletown, Connecticut 06457], and modified to fit this experiment.)
Introduction:
This enzyme catalyzes the oxidation of succinate to fumarate in the Krebs cycle. It is a good choice as a marker enzyme for mitochondria because not only is it readily assayed, but also it is the only Krebs cycle enzyme to remain bound to the inner mitochondrial membrane. Thus even if mitochondria lose some of their soluble contents through damage during isolation they remain able to exhibit SDH activity.
The chemical reaction is shown in the introduction above. Flavine adenine dinucleotide (FAD) is a coenzyme covalently bound to the SDH enzyme. For the enzyme to complete its catalytic cycle the electrons it receives from succinate are ordinarily passed on down the electron transport chain to oxygen, and the reduced FADH2 becomes reoxidized to FAD, ready to encounter another succinate. This procedure can be altered, however, by blocking the electron transport chain and providing the assay system with artificial electron acceptor to draw the electrons from reduced FADH2. If these acceptors are dyes with characteristic absorbance in the oxidized and reduced form, the progress of the reaction may be assayed by the change in the amount of color. One very useful artificial electron acceptor is 2,6-dichlorophenol indophenol (DCPIP). It absorbs strongly at 600 nm when oxidized, but becomes colorless in its reduced form.
Thus the basic procedure is to mix samples of fractions to be tested (enzyme) with an assay mixture containing, among other things, substrate, cyanide and DCPIP. Cyanide blocks the electron transport chain, so that SDH needs another electron acceptor for its reduced FADH2. The dark blue DCPIP serves as the acceptor, and as it is reduced it becomes colorless and the SDH then goes after another molecule of substrate. Thus the rate of the disappearance of the blue color is proportional to the concentration of enzyme. The change in absorbance of the mixture is measured as a function of time and the enzyme concentration is determined from these data.
The assay medium contains:
This medium is prewarmed to 37oC (the temperature of the reaction) and the enzyme sample is added. The loss of the blue color is measured as a function of time. It is sometimes difficult to obtain accurate rate measurements of this enzyme for two reasons: (1) The enzyme undergoes an activation in the initial phases of the incubation which interferes with observation of the reaction itself. (2) The most accurate readings of absorbance are obtained when DCPIP is present in only small amounts. It is reduced so rapidly, however, that rapid-response instrumentation is needed for the best readings. Once reduced, it may be reoxidized (albeit slowly) by atmospheric oxygen.
We will try to avoid these problems by taking readings both as we first mix the reaction and after 3 minutes (to allow for activation); we will also use a higher concentration of DCPIP than might be otherwise desirable, in order to prolong the reaction to the point that our manual methods give reliable data.
Procedure:
1. Pipette 2.8 ml of assay medium into a Spectronic tube, put parafilm on the top, and place in a 37oC water bath or heating block. Keep a tube of buffered sucrose on hand as a blank to set the zero on the spectrophotometer.
2. At t = 0 add 0.2 ml of your diluted enzyme preparation (see below), and mix well. Immediately read and record the zero-time absorbance at 600 nm, and return the tube to the water bath. Avoid shaking the tube from this point on; the DCPIP reoxidizes readily.
3. At t = 3 minutes, again read and record the absorbance at 600 nm and return the tube to the water bath.
4. At t = 15 minutes take a final reading and record the absorbance.
You will repeat the procedure (steps 1-4) on the following samples:
The first week, dilute the sample 1/2 and 1/4
Carry out each of the 6 assays in triplicate. You will thus have data for 18 assays. They can be done in groups of 3 or even 6 at a time, but be sure to stagger the starting times so that there is time to make the readings (it takes about 30 s to take each reading). It is also a good idea to mix the order of the assays so that the three assays of a single dilution are made at different times in the course of the lab period.
Data analysis:
The t = 0 readings are important, as they include a component of turbidity (caused by the suspension of the enzyme micelles in the solution) that varies from sample to sample. That component is determined by subtracting the zero-time readings of the "blank" assays from that of the sample assays. The difference is the turbidity for that sample. This value should now be subtracted from the t = 3 and t = 15 readings, yielding the component of the absorbance reading due to DCPIP. The rate of the reaction (v) is defined as the rate of disappearance of the colored substrate, and is proportional to the concentration of that substrate (s) over time (t):
v = -ds/dt = Ks
Rearranging
-d/s = Kdt.
Integrating from 0 to t
ln (so/st) = Kt
This is the equation of a straight line whose slope (K, in units of time-1) represents the activity of the enzyme. Thus the next step is to divide the reading at t = 3, (so) by that at t = 15 (st) and then to take the natural logarithm of the quotient. This value is divided by t = 12 min to give K in units of min-1.
We define a value of 0.1 min-1 to be 1 unit of enzyme activity. Calculate the number of units present in each sample by dividing by this. Correct this value for amount of sample you added (0.2 ml.) and the dilution factor for that sample to give the units of activity per ml of fraction:
K x 1 un/0.1 min-1 = units
(units/0.2 ml) x dilution factor = units/ml
Now divide the activity per ml of fraction by the mg protein per ml of fraction determined last time in the Lowry assay (as shown in the introduction).
(units/ml) / (mg/ml) = units/mg
This value is the specific activity of the enzyme in the fraction: units of enzyme per mg protein.
Suggestions for further study:
The biochemistry of this exercise can be further developed by using the most active fraction (usually the mitochondrial) to perform measurements at different (lower) substrate concentrations, while the enzyme concentration is held constant. From the resulting rate measurements Km and Vmax may be calculated.
Other sources of mitochondria can also be used. I have heard of success using Cauliflower florets or Dictyostelium amebae. My impression is that any reasonably metabolically active tissue with which you are familiar and which is abundant will do nicely.
COMMON PROBLEMS AND THEIR LIKELY CAUSE
The chief problem with this procedure is that it is sufficiently complex that students often forget to make a critical volume measurement or to save a critical fraction for assay. Vigilance on the part of the students is important. The cell fractionation parameters are reasonably flexible, so that enrichment of the mitochondrial fraction for mitochondria and for enzyme activity is usually seen. The best separations are gained if 1.) care is taken when removing the first supernatant not to include any of the pellet (even if it means leaving behind some of the supernatant); and 2.) care is taken when removing the second supernatant to take as much of it as possible (even if it means losing a little of the pellet). The Lowry procedure is reasonably foolproof. The major mistake students make is to use their sample dilutions straight, forgetting to use only 0.1 ml in the assay. They must begin again with fresh dilutions, so it is good if they notice the problem promptly. It is obvious, because once they add the phenol reagent, the tubes all turn almost black!
The enzyme assay also works quite well. If insufficient dilution is made for the samples, a large part of the reaction can take place during the initial activation period. In such cases there is apt to be a misleadingly small drop in absorbance between 3 and 15 minutes, complicated further by the tendency of the DCPIP to reoxidize, especially at the interface between the liquid and air in the tube. The instructor can monitor the course of students reactions and recommend additional dilution of samples that appear too concentrated. This will also help by reducing the turbidity.
Week 3: Purification of Mitochondria by Differential Centrifugation
(This lab was obtained from Dr. Bonnie Seidel-Rogol of the Dept. of Biological Science, SUNY Plattsburgh)
Introduction: (see above)
Procedure:
Remember to take a 2.0 ml samples of the H and S1 samples during the procedure.
1. Obtain a two florettes from a head of cauliflower. Cut the soft ends off the stalks and collect them in a beaker. Weigh out approximately 6 grams.
Do all the following procedures in the cold, keeping all solutions on ice:
2. Using a mortar and pestle, grind the cauliflower with an equal weight of acid-washed sea sand (about one scoop) until a uniform paste is obtained (no clumps of cauliflower). Add 15 ml of buffer and continue grinding to suspend the mixture. Pour off into a 50 ml centrifuge tube. Rinse the mortar with 5 ml of buffer and combine it in the centrifuge tube with the original mixture. Let the sand settle to the bottom for 2 min on ice, then take a 2.0 ml sample and label it H.
3. Balance the tubes using sand to adjust the weights. Centrifuge at 3,000 rpm for 10 min (Sorvall HS-4 rotor). This step removes the sand and cellular debris and most of the nuclei.
4. Decant the supernatant into a 30 ml capped centrifuge tube. Be careful not to disturb the pellet. Once the supernatant is removed, add 10 ml of buffer to the sand/pellet mixture and resuspend it by shaking vigorously. Allow the sand to settleand use this as P1. Take a 2.0 ml sample of the supernatant and label it S1.
5. Centrifuge the supernatant at 14,000 rpm (Sorvall SS-34 rotor, ) for 30 min. Remove 2.0 ml of supernatant and label it S2. Decant carefully to remove the supernatant. The pellet tends to be a little loose at this step, so care must be taken to avoid loss of the pellet. Resuspend the pellet in 4.0 ml of buffer and label it P2.
6. Determine the amount of protein and the SDH activity in each of the five samples. Calculate the specific activty in each fraction and calculate the fold enrichment of mitochondria in each fraction.
Materials: Buffer: 15% Sucrose; 10 mM Tris-HCl pH 8.0; 0.2 mM EDTA
Procedure adapted from: Lambowitz,A.M. (1979) Methods in Enzymology 59:421-433.
If you have questions please send me a message at my Email address
adapted from: Lambowitz,A.M. (1979) Methods in Enzymology 59:421-433.
If you have questions please send me a message at my Email address