Purification of Mitochondria by Differential Sedimentation and Monitoring of Fractions for Specific

Activity of Succinate Dehydrogenase

Abstract:

The main goal of this experiment was to find purified mitochondrial protein from liver. This was done by making a series of pellets and supernatants, and fractions were then made. After the fractions were made, they were treated with Bradford Reagent to indicate the amount of protein present. Then the fractions were then put through an assay procedure, where they were read in a spectrometer. The spectrometer gave the absorbance and %transmittance. From the absorbencies, a graph was made, along with an equation. When the absorbencies from another dilution were plugged into the equation four protein concentrations were found; homogenate = 14.749, nuclear = 2.609, mitochondrial = 12.896, and soluble = 6.041. From these the total proteins were found; Homo. = 265.5, nuclear = 13.045, mito. = 64.5, and soluble = 87.6. The next step was the making of standards. From the spectrometer readings from the standards calculations were made of the units of activity, units per ml, crude activity and specific activity. The fraction with the largest unit of activity (.3055) was the nuclear. The nuclear also had the largest units per ml (1.5275). The mitochondrial-10 had the largest crude activity (7.540). The nuclear had the largest specific activity (1.1709); however, there was a contamination of proteins in the nuclear fraction because the nuclear was not supposed to be the largest one. A way of improving this experiment was that it could have been done slower and more preparation could have gone into it. However, this experiment was very useful in describing where the largest amount of protein concentration was in a cell. Because where the highest concentration of protein was in a cell, that organelle was bound to be the ‘work house’ of the cell.

Introduction:

The main objective of this experiment was to isolate a purified sample of

mitochondrial proteins from liver. This was done by a procedure called differential

sedimentation. This was done by centrifuging the liquid, letting the more dense particles

settle to the bottom of the tube faster. When the procedure was done a pellet was formed,

along with a supernatant. Depending on what stage the experiment was at, the pellet will be

consist of crude homogenate, nuclear, soluble, or mitochondrial. The crude homogenate was

everything, including the nuclear, mitochondria and everything else in the cell. The nuclear

was everything inside the nucleus, including the nucleoplasm and nuclear envelope. The

mitochondrial was usually pure mitochondria. The soluble was everything except the nuclear

and mitochondria.

At the initial stages of the experiment sucrose buffer was used to rinse off the liver

during homogenizing. The sucrose buffer kept the environment at a pH of 7.0. Another

reagent used was Bradford reagent, its purpose was to stain the proteins, and this was how the

amount of protein was indicated. The bluer the more proteins were present, and vice versa.

Another reagent used during the course of this experiment was in the assay medium. The

assay medium included 10mM succinate, which acted as the substrate. Also 0.02mg/ml DPIP

(2, 6-dichlorophenol indophenol), when oxidized it absorbs 600nm strongly, thus becoming

colorless when it was reduced. This was why the best reading happens when there was very

little DCPIP present. Also 10mM phosphate buffer was included, this controlled the pH at

7.4. 2mM of KCN were in there to block the electron transport chain from happening. 10mM

of CaCl2 was there to increase the mitochondria’s permeability, so that the amount of SDH

could be known. Albumin was present in the amount of 0.5mg/ml, since the permeability was

increased; the albumin was there to keep the mitochondria from bursting.

The only new machine that was used during this experiment was the spectrometer. A

spectrometer was an instrument that was used for measuring the amount of light of different

wavelengths that are absorbed by a solution. The spectrometer was used to measure the

%transmittance and absorbance (OD) of each fraction in the experiment.

Another main objective of this experiment was to look for the succinate

dehydrogenase (SDH). As a coenzyme, SDH will catalyze the oxidation of the fumarate and

succinate. And since KCN was present the electron transport chain was stopped and the FAD

2

keeps the electrons and reroutes it to the DCPIP, thus changing its color. The SDH was found

bounded to the inner membrane of the mitochondria, here the Krebs’ cycle occurs, the whole

reaction takes place with pyruvate and in an aerobic environment.

Procedure:

One watch glass was obtained, along with a new razor blade and a graduated cylinder.

The watch glass was used to zero out the scale, and it was also used because it served as a

hard surface to cut on. Next a piece of liver was obtained and 2 grams were weighed out. The

liver was then cut up on the watch glass. Next the liver was put into a homogenizer vessel

along with 5ml of sucrose buffer. The sucrose buffer contained: .25M sucrose in 10mM

phosphate buffer, which had a pH of 7.0, this helped keep the solution neutral through out the

experiment. Then the tube was put into the homogenizer, and homogenized until there were

no liver pieces left, about 15 times. The homogenized liver was then strained through a

monolayer of cheesecloth into a test tube. After that 15ml of the sucrose buffer was put in the

test tube, and stirred. This made a total of 20ml in the tube. Next 2ml were taken out of the

tube, 1ml was put in one test tube and 1ml placed in another tube, leaving 18ml in the original

test tube. This tube was then spun at 600g forces for a total of 10 minutes. This produced a

pellet (#1) and a supernatant (#1).

First the supernatant (#1) was poured out into a new test tube. Then the supernatant

was spun at 10,000g forces for 20 minutes. This produced a pellet (#2) and a supernatant

(#2). The supernatant (#2) was named Soluble Fraction, and it was poured out and divided

into two equal parts, 7.25ml went into one tube and 7.25ml went into another test tube. Then

the tubes were placed into the freezer. Next the pellet (#2) was resuspended with 5ml of

3

sucrose buffer and inverted to mix the pellet. This was named the Mitochondrial Fraction. It

was divided into two equal parts, 2.5ml were placed in a tube and 2.5ml were placed in

another tube. Then they were put into the freezer, until needed.

Then pellet #1 was resuspended with 20ml of sucrose buffer and inverted. It was then

spun at 600g forces for 10 minutes. The supernatant was poured out and discarded. The

pellet (#3) was resuspended with 5ml of sucrose buffer and inverted. This was labeled the

Nuclear Fraction. Then it was divided into two equal parts, 3ml in one tube and 3ml in

another tube. Then they were placed in the freezer, until needed.

One sample from each fraction was taken out of the freezer and placed on a test tube

rack, and allowed to thaw. Eight 1.5ml test tubes were gathered and labeled 1 through 7.

Then dilutions were made in each tube, and they were called the standards. The reagents that

were added to each tube included Bovine Serum Albumin (BSA), which was in a

concentration of 200g/ml, distilled water, and the assay. In tube #1, it had 1ml of BSA and

no water added to it, the dilution factor was zero, so the concentration was

1000micrograms/ml, and the amount in assay was 100 micrograms. In tube #2, 0.8ml of BSA

and 0.2ml of water was added to the tube, the dilution factor was 4:5, with a concentration of

800microgram/ml, and an assay of 80micrograms. In tube #3, there was 0.6ml of BSA and

0.4ml of water added to the tube; the dilution factor was 3:5, with a concentration of

600microgram/ml and an assay of 60micrograms. In tube #4, there was 0.4ml of BSA and

0.6ml of water added to the tube, the dilution factor was 2:5, with a concentration of

400microgram/ml, and the assay was 40microgram. In tube #5, 0.2ml of BSA and 0.8ml of

water was added to the tube, the concentration was 200 micrograms/ml, the assay was

20micrograms, with a dilution factor of 1:5. In tube #6, 0.1ml of BSA and 0.9ml of water was

4

added to the tube, with a dilution factor of 1:10, and a concentration of 100microgram/ml, the

assay was 10 micrograms. In tube #7, just 1.0ml of water was added to the tube, this was the

control (blank).

Next seven spectronic tubes were gathered and numbered 1 through 7. Then 0.1ml of

each standard that was in the test tubes was pipetted into the corresponding spec tube. Then

5.0ml of the Bradford reagent were added to each tube, and stirred with a piece of Parafilm on

top, so as not to get contaminated. Each tube was allowed to sit for five minutes, to allow the

color to develop. The Bradford reagent indicated the amount of protein present in the solution

in each tube. This was indicated by the color of blue that was shown in each tube. The more

blue, the more protein that was present. The less blue, the less protein. One tube was

prepared with just Bradford reagent, this became the blank. The blank was used to zero out

the spectrometer, so that the spectrometer would read 100 and 0. The spectrometer was set to

595nm. The spectrometer was used to measure the absorbance and % transmittance for each

of the fractions. Before each tube was placed in the spectrometer, a Kim wipe was used to

clean off any fingerprints off the tube that could foul up the readings. Also before a tube was

placed in the spectrometer the blank was used to zero out the machine. After the data was

taken for each tube it was plotted on a graph, with the BSA concentrations on the x-axis and

the absorbance on the y-axis. Then a straight line was plotted in between all of the points; this

became the standard curve. This graph was used later on in the experiment.

Next eight more 1.5ml tubes were gathered. And reagents were added to each tube,

the reagents included, the fractions previously made, and distilled water. Tube 1 had a dilution

factor of 1:10, with 0.5ml of the homogenate fraction with 4.5ml of distilled water added to

the tube. Tube 2 had a dilution factor of 1:20, with 0.5 homogenate fraction and 9.5ml of

5

distilled water added together. Tube 3 had a dilution factor of 1:5, with 0.5ml of nuclear

fraction and 2.0ml of distilled water. Tube 4 had a dilution factor of 1:10, with 0.5ml of

nuclear fraction and 4.5ml of distilled water. Tube 5 had a dilution factor of 1:5, with 0.5ml

of mitochondrial fraction and 2.0ml of water. Tube 6 had a dilution factor of 1:10, with 0.5ml

of mitochondrial fraction and 4.5ml of distilled water. Tube 7 had a dilution factor of 1:5,

with 0.5ml of soluble fraction and 2.0ml of distilled water. Tube 8 had a dilution factor of

1:10, with 0.5ml of soluble fraction and 4.5ml of distilled water. When the reagents were

added to each tube, pieces of Parafilm were placed on top of the tubes and inverted, this was

to protect the hands from unnecessary contamination from the chemicals that were used. Next

0.1ml of the diluted samples were put in the next set of spec tubes. Along with the samples,

5.0ml of the Bradford reagent were put in the tubes too. Then the tubes were allowed to sit

for five minutes, to let the color develop. The spectrometer was set to 595nm. The blank from

the previous section of the experiment was used to zero out the machine; also Kim wipes were

used. The data gathered from the spectrometer was used to compare with the graph that was

given from the previous section.

Next an assay was done on the fractions for the succinate dehydrogenase (SDH). First

the spectrometer was set to 600nm. Then 2.8ml of the assay medium was pipetted into a

spectrometer tube, and placed in a 37OC water bath. The assay medium included 10mM

succinate, which was the substrate, 0.02mg/ml DCPIP, which acted as the indicator dye that

changed from purple to clear when reduced, 10mM phosphate buffer, which kept the pH at

7.4, 2mM KCN, which helped block the electron transport chain, 10mM CaCl2, which

increased the mitochondrial permeability, and 0.5mg/ml albumin, which kept the osmotic

pressure from increasing in the mitochondria, to keep it from bursting. A blank, which

6

included buffered sucrose, was used to zero out the spectrometer. The reactions were ran in

sets of three. Group1 included crude homogenate, nuclear and soluble. Group 2 included the

mitochondrial fractions. At time 0 minutes, 200l of group1 was added to the tube and

mixed thoroughly. The tube was placed in the spectrometer and the absorbency was read

immediately. The tube was then placed in the water bath immediately afterwards. At time 3

minutes, the absorbency was read again, and the tube was placed in the water bath afterwards.

At time 4 minutes Group2 was started the same way as Group1. At time 15 minutes, the

absorbency was measured again and the tube was placed in the water bath afterwards. This

procedure was repeated for the blank, homogenate, mitochondria, nuclear, and soluble

fraction. The dilutions were made with the fraction and sucrose buffer as follows: for the

crude homogenate, nuclear, and soluble there was a 1:2 dilution, 500l of fraction and 500l

of sucrose buffer. For the mitochondria, three dilutions were made: 1:5, 200l of fraction and

80l of sucrose buffer, 1:10, 100l of fraction and 900l of sucrose buffer, 1:25, 40l of

fraction and 960l of sucrose buffer. The time line for each group was as follows; Group1=

0-3-10-15-20: Group2 = 4-7-12-17-22. This data with the data from the above section was

used to find the protein concentration, total proteins present, crude activity of the proteins, and

specific activity for the proteins.

Results:

There was an error in the data gathered during the laboratory session, so data from

another laboratory session was used for the results below.

The first time that the tubes were put in the spectrometer the readings were:

7

Tubes % Transmittance Concentration Absorbance Colors

1 13 1000mg/ml .886 Turquoise

2 18 800mg/ml .745 “

3 29 600mg/ml .538 “

4 41 400mg/ml .387 Gray

5 64 200mg/ml .1939 “

6 80 100mg/ml .0969 “

The logarithmic graph made from the data above is at the back of this report. The

equation that the graph gave was; Y=0.0009x – 0.0057. When the absorbencies above were

plugged into the Y, then multiplied by the dilution factor, protein concentrations (mg/ml)

were given. They are as follows:

Fractions Protein Concentrations (mg/ml)

Homogenate 14.749

Nuclear 2.609

Mitochondrial 12.896

Soluble 6.041

From the protein concentrations the total proteins were found, they were as follows:

Fractions Total Proteins (mg)

Homogenate 265.480

Nuclear 13.045

Mitochondrial 64.483

Soluble 87.590

From the total protein information found the percent recovery was found by adding the

nuclear (13.045) by the mitochondrial (64.483) by the soluble (87.590), which equaled

8

165.118. Then by dividing this by the total proteins found in the homogenate fraction

(265.480), the total percent recovery was 62.2%.

The next set of results that were obtained from the standards when they were read in

the spectrometer were as follows:

Tubes % Transmittance Absorbance

1 12 0.920

2 22 0.658

3 34 0.469

4 59 0.229

5 1 2.000

6 7 1.155

7 10 1.000

8 29 0.538

The readings in the spectrometer after the fractions were timed were as follows:

9

O minutes 3 minutes 10minutes 15minutes 20minutes

Crude

Homogenate

Abs. .90 .88 .75 .70 .61

% Trans. 11 13 18 20 24.5

Nuclear

Abs. .85 .79 .70 .62 .47

% Trans. 14 17 20 24 34

Soluble Abs. .68 .66 .63 .63 .60

% Trans. 21 22 23.5 23.5 25

4 minutes 7 minutes 12minutes 17minutes 22minutes

Mito. 5

Abs. .88 .80 .75 .68 .58

% Trans. 13.5 16 18 21 25

Mito 10

Abs. .76 .79 .71 .67 .63

% Trans 17.5 18 19.9 21.5 23

Mito 25

Abs. .70 .70 .69 .66 .64

% Trans. 20 20 20.5 22 22

From the data above the units of activity (Z), units per ml (Y), crude activity (W), and specific

activity (V) were found. The calculated values were as follows:

Fractions Units of Activity

(Z)

Units per ml

(Y)

Crude Activity (activity/ml)

(W)

Specific Activity

(V)

Crude Homogenate 0.2155 1.0775 2.155 0.1461

Nuclear 0.3055 1.5275 3.055 1.1709

Soluble 0.0561 0.2805 0.561 0.0928

Mitochondrial 5 0.2144 1.0720 5.360 0.4156

Mitochondrial 10 0.1508 0.7540 7.540 0.5846

Mitochondrial 25 0.0597 0.2980 7.450 0.5777

10

From this data, bar graphs were made on Excel. The bar graphs that were made were protein

concentration, total proteins, crude activities, and specific activities. They are in the back of

this report.

Discussion:

When the absorbencies were plugged into the equation given on the logarithmic graph,

the equation had to be equaled to X ; X= (Y+.0057)/(.0009). Only the absorbencies that were

closer to the standards were plugged into the equation, these were; for homogenate = 0.658,

for nuclear = .2291, for mitochondrial = 1.155, for soluble = .538. Then once these numbers

were taken, they were all multiplied by the dilution factors for each. For the homogenate

fraction, 0.658 was multiplied by 20. For the nuclear, mitochondrial, and soluble the

absorbencies were multiplied by 10. The final answers were the protein concentrations

(mg/ml). Then the protein concentrations were multiplied by the volume in each of the

fraction tubes; this gave the total proteins. These were 18 for the homogenate, 5 for the

nuclear, 5 for the mitochondrial, and 14.25 for the soluble. Since these were in milliliters, the

protein concentration, mg/ml, the milliliters canceled out and that made the total proteins were

in milligrams. When the total proteins were found the percent recovery was calculated. This

was done by finding the sum of the nuclear, mitochondrial, and soluble, which ended up being

165.12mg. This was then compared with the original total of the crude homogenate, which

were 265.48mg. The difference was almost 100mg. This huge difference could have been

because there was some of the fraction that was lost during the course of the experiment.

The units of activity (Z) were calculated by using the equation given in the laboratory

handout; K={ln(r3/r15)}/12 = X *10 . For the crude homogenate, nuclear, and soluble, the

11

times were r3 and r20 , so the time interval that they were divided by was 17 minutes for the

crude, nuclear, and soluble. For all three mitochondrial fractions the times were r7 and r22, so

the time interval that they were divided by was 15 minutes.

The units per milliliter (Y) were calculated by multiplying Z by 5, which was a

constant because the fractions were only assayed by 1/5 of a milliliter.

The crude activity (W) was calculated by multiplying Y by the dilution factor. For the

crude homogenate, nuclear, and soluble the dilution factor was 2. For all three mitochondria’s

the dilution factor was 5, 10, and 25. The units were enzyme / mg of protein.

The specific activity (V) for the proteins was found by dividing W by the protein

concentrations. The meaning of the specific activity is the amount of the SDH activity

compared to the total amount of protein in each sample of a fraction.

Looking at the graphs, starting with protein concentrations; the crude homogenate had

the most of all the fractions. This was because the crude homogenate included the entire cell,

and all proteins present within that cell. Since the mitochondria are the organelle with the

highest concentration of protein, that was the second highest column on the graph. The next

graph to look at was the total proteins. The highest numbers of proteins were found in the

crude homogenate, again, for the same reason as above. The soluble had the second largest

amount of total proteins because the soluble was everything except the nuclear and the

mitochondria. The next graph, the crude activity, showed the highest crude activity in the

mitochondria, and the lowest crude activity in the soluble. The last graph, specific activity,

shows the most activity in the nuclear. However, the most specific activity was supposed to

be found in the mitochondrial25. The error could have occurred because there was some

contamination by mitochondria in the nuclear fraction. Also there was a slight increase in

12

specific activity in the mitochondrial10 from the mitochondrial25. This might be because the

M10 got switched with the M25 accidentally. Another reason that the results were different

than they should have been was because when the liver was strained through the monolayer

cheesecloth, not all got into the tube, so the cheesecloth did not do its job very well.

One way that could improve this experiment, was to purify each sample more further.

Especially the mitochondrial fractions, then perhaps not as many contaminations would have

occurred. Also if this experiment was done in a lengthier manner, and in not such a hurry,

that could also improve the quality of the experiment, and its data.

13