[SOLVED] BIO2101 Exercise 6 Cell Fractionation by Differential Centrifugation

BIO2101

Exercise 6:    Cell Fractionation by Differential Centrifugation

Purpose: To integrate microscopic observation and quantitative analysis with cell fractionation. Purification of mitochondria by   differential sedimentation and monitoring of fractions for specific activity of succinate dehydrogenase.

Part A: Preparation of cell fractions Introduction:

Eukaryotic cells contain many kinds of subcellular organelles.    In order to determine  the  physical  and  chemical  properties  of  these  organelles,  it  is necessary to study them in isolation from other cellular components.   The most common  way  to  perform.  this  cell  fractionation  is  to  disrupt  the  cells  and separate their components by use of centrifugation.    Particles subjected to the acceleration  of  a  centrifugal  field  will   move,   or  sediment,   at  a  velocity determined by various properties of the particle, including its mass, its density compared to that of the suspending fluid, and its frictional properties.

If the suspending fluid forms a gradient of density around the density of the particle itself, the particle will form a band at a position in the centrifuge tube where its density is matched by the density of the fluid.    The forces impelling it further down the tube are countered by the increasing resistance of the ever denser fluid.   The tendency of diffusion to disperse the particles is countered by  the  centrifugal  field.      Cell  homogenates  treated   in  this  way  can   be separated in one step into multiple bands of species differing in density.    This technique is known as equilibrium density gradient centrifugation.   The best separation, however, occurs only after centrifugation for a long time, when equilibrium  has  become  established  between  the  forces  concentrating  the particles and those leading to their dispersion.

A less complete, but more rapid, separation can be achieved by a technique called  differential  sedimentation.     The  fluid  in  which  the  particles  are suspended is less dense than any of the particles.   Thus instead of forming bands,  they  pass  down  the  centrifuge  tube  at  a  rate  determined  by  the properties described above.   When they  reach the  bottom of the tube, they form a pellet.    If particles of several different kinds are present initially, the pellet’s composition will change over the period of the centrifugation, initially consisting of the most rapidly sedimenting particles, with the slower ones being added later.   Thus a separation can be effected by stopping the centrifugation periodically, removing the pellet, and continuing to centrifuge the supernatant.

This procedure has been standardized into a scheme for fractionating cells, in which the various pellets have been identified according to the predominant organelle  they  contain:  the  nuclear  fraction,  the  mitochondrial  (or  plastid) fraction, the microsomal fraction, and the soluble fraction.   It is important to remember, though, that each fraction is contaminated by elements of the other fractions, unless further purified.   Also the soluble fraction contains material, e.g. ribosomes, which could be pelleted if the centrifugal forces were further increased.    Thus,  in  a  rigorous  application,  it  is  necessary  to  confirm  by independent means the purity of a fraction.    This can be done by microscopic analysis or by assaying for an enzyme activity characteristically associated with a particular organelle.

The rate at which a particle moves is also a property of the centrifugal field, usually expressed as a multiple of g, the acceleration due to gravity.    Thus the relative centrifugal force,  R.C.F. =  1.118 x  10-5  (rpm)2r,  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.   A convenient way to determine the R.C.F. for a given rotor at a given speed is by use of a nomogram.   The radius used is usually the mean between the radius to the top of the centrifuge tube and the radius to the bottom.

Since distance = rate x time, a given particle may be sedimented the length of the centrifuge tube (i.e., pelleted), by subjecting it to high speeds for short times or to lower speeds for longer times.    Other factors influence the choice of conditions, such as diffusion properties of the particle of interest compared to other particles in the homogenate, or other time-dependent factors.

Various kinds of centrifuge are available to carry out this kind of work.  Clinical centrifuges, both refrigerated and not, seldom run faster than 5,000 rpm; they can readily sediment cells, nuclei, or chloroplasts, but smaller particles  require impractically long times.    High-speed centrifuges go as fast as 18,000  rpm; ultracentrifuges can manage 60,000 – 75,000 rpm.   Both these types of  instrument are usually refrigerated and may also be operated in a vacuum, to  reduce heat generated by friction with air molecules.    The  rotors, too, are  designed of titanium and strong alloys to withstand the strain of the centrifugal  forces and to conserve angular momentum.

Procedure:

The exercise today will  use  differential  sedimentation to  prepare  three fractions from a homogenate of rat liver: a “nuclear” fraction, a “mitochondrial” fraction, and a “soluble” fraction.

Note: Soluble fraction would still include cell components such as microsomes and ribosomes, which could be sedimented in stronger centrifugal fields.

1.    Add about 2 g fresh rat liver into a 50 ml centrifuge tube containing 5 ml ice-cold buffered sucrose.   Put the tube on ice.    Finely mince the tissue with scissors.

*Buffered sucrose: 0.25 M sucrose in 10 mM phosphate buffer, pH 7.0.

2.    Transfer  the tissue  pieces with the ice-cold sucrose to a homogenizer vessel.   Add additional 5 ml ice-cold buffered sucrose to rinse the tube and pour out all remaining tissues to the vessel again.

3.    Homogenize the minced liver tissue for about 12 seconds (until no lumps remain). Then add additional 5 ml ice-cold buffered sucrose to rinse the homogenizer vessel. Keep cold throughout.

(Step 1-3 will be done by TAs.)

4.    Decant  the  homogenate  through  a  cell  strainer  into  an  ice-cold  50-ml centrifuge tube.   Rinse the homogenizer vessel with 5 ml ice-cold buffered sucrose and pour it through the cell strainer into the tube too.   This is a 10% (w/v) crude homogenate (why 10%?).    Record the volume.

5.    Save 3*1 ml of crude homogenate in three chilled 1.5 ml microfuge tubes, for assay,  microscopy,  and  SDS-page  (next  experiment),  respectively. Label  the  tube  with  group  number  and  indicate  which  fraction  it  is. Centrifuge the rest of homogenate fraction in the centrifuge tube for 10 min at 600 g at 4oC.

6.    After centrifugation of Step 5:

(i)    Carefully transfer the supernatant to another high speed centrifuge tube (do not disturb the pellet).

(ii)   For  the  pellet  (which  is  the  nuclei),  resuspend  in  10  ml  buffered sucrose.   Centrifuge again for 10 min at 600 g to wash the nuclei. Discard the  supernatant.    Resuspend  the  pellet  in  5  ml  buffered sucrose, this  is  the  nuclear  fraction  (so  its  total  volume  is  5  ml). Save it on ice.

7.    Centrifuge again the supernatant from Step 6(i) for 20 min at 20,000 g at

4oC.

8.    After centrifugation of Step 7:

(i)    Carefully transfer he supernatant to another chilled test tube (do not disturb the pellet).   Record the volume.   This is the soluble fraction. Save it on ice.

(ii)   For  the   pellet,  resuspend  in  5  ml  ice-cold  sucrose,  this   is  the mitochondrial fraction (so its total volume is 5 ml).    Save it on ice.

9.    In  addition  to  crude  homogenate  (saved  from  Step  5),  you  now  have another three samples: nuclear fraction (Step 6 ii), mitochondrial fraction (Step 8 ii) and soluble fraction (Step 8 i).

Save 3*1 ml of crude homogenate in three chilled 1.5 ml microfuge tubes, for assay,  microscopy,  and  SDS-page  (next  experiment),  respectively. Label the tubes with group number and indicate which fraction it is.

Note: Submit the two of the saved 1 ml of the four fractions to instructor. They will be saved in refrigerator for the protein determination and the enzyme assay to be done in next week, and SDS-page in next experiment.

10.  For the microscopy observation of samples, prepare wet mounts of each and examine by microscopy as follows:

•     Add 10 ul of the fraction on a clean slide.

•     Spread the drop with the edge of a second slide to make a smear.

•     Allow the slide to air dry.

•     Add 10 ul of stain (methyl green pyronin).    Let sit 3 minutes.

•     Invert the slide and immerse slide in a rack in a pan of SLOW running tap water for about 30 seconds.   Avoid flushing  the  sample smear directly with water!

•     Add 10 ul glycerin and a coverslip.

* Please pipette the glycerin gently since glycerin is sticky

•     Observe in bright field.    Nuclei should stain purple, cytoplasm red or pink,  and  mitochondria  can  be  seen  as  small  dots  (under  400x magnification).

Note: Record the presence of cytoplasm and nuclei in crude homogenate and all three fractions, and describe their relative amount by the number

of “+” sign.

Part B, C and D should be done in next week

Part B: Sample Dilution Procedure:

1.    Thaw all four saved samples from last week.

2.    Dilute your fractions in buffered sucrose to 1 ml (=  1000 μl), with each diluted sample in microfuge tube.

1/25 dilution	40 µl sample + 960 µl sucrose buffer
1/50 dilution	20 µl sample + 980 µl sucrose buffer

Part C: Protein Assay Procedure:

1.    Prepare  the  following  blank  and  bovine  serum  sample  (BSA)  in  96- microplate wells for protein standard plot construction.

We can use Bradford reagent to determine the protein content of bovine serum albumin (BSA) in order to construct a standard plot as reference. Bradford reagent contains a dye, Coomassie blue, which binds to protein. The  dye/protein  complex  produces  a  blue  color  whose  absorbance  is directly proportional to the protein concentration.

1.    Bovine serum albumin (BSA) has been prepared in PBS at a concentration of 1.2 mg/ml.

2.    Using  an aliquot of the  1.2  mg/mL protein standard  prepare a range of standards  solutions.  Add  the  following  volumes  of  1.2  mg/mL  protein standard and H2O (Table 1) to the appropriate wells to construct a range of standards solutions between 0 and 0.6 mg/mL (e.g. 0, 0.0375, 0.075, 0.15, 0.3, 0.6 and 1.2 mg/mL). Mix the content well by pipetting up & down. A total volume of 50 μL of each standard is sufficient for all the required assays.

Table 1 Construction of Protein Standard Plot

(BSA is given at concentration of 1.2 mg/ml.)

2.   Add  10  μL  of  standards  to  the  appropriate  wells. Then  add  200  μl  of Bradford reagent to each well. Mix the contents well. Wait for 3-minutes

3.    Repeat  Step  2  for  the  following  diluted  samples  (which  are  already prepared in Part B). Add 10 μL of samples to the appropriate wells

Homogenate          1:50

Nuclear                   1:25

Mitochondrial         1:25

Soluble                  1:25

4.    After   3-minutes,   measure   absorbance   at   595   nm   with   microplate spectrophotometer   (BioTek    Epoch    plate    reader).        Record    the absorbance value on the datasheet.

Part D:   Succinate Dehydrogenase (SDH) Assay 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 their soluble contents through damage during isolation, they remain able to exhibit SDH activity.

The chemical reaction is given in Figure 1.    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  FAD  becomes  reoxidized,  ready  to  encounter  another  succinate. This procedure can be altered, however, by providing the assay system with artificial electron acceptors to draw the electrons from reduced FAD.    If these acceptors are dyes with characteristic absorbance in the oxidized and reduced forms, the progress of the reaction may be assayed by the change in the amount of color.

Figure 1.    The chemical reaction catalyzed by the enzyme succinate

dehydrogenase (SDH)

One  very   useful   artificial   electron   acceptor   is   2,   6-dichlorophenol indophenol (DCPIP) (see  Figure 2).    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 (which supposed to contain different level of SDH) with an assay mixture containing, among other things, substrate and DCPIP.   The absorbance of the mixture is measured as a function of time, and thus the rate of the enzyme-catalyzed reaction is determined.

Figure 2.      The  structure  of  dichlorophenol-indophenol  (DCPIP)  in  its

oxidized and reduced forms

Procedure:

For each assay follow this procedure:

1.    Centrifuge homogenate and nuclear fraction at 600x g, 5min.

2.    Add 10 μL of the following samples to the appropriate wells. Then add 180 μl of pre-warm assay medium to each well. Mix the contents well.

Well #	Assay medium (μL)	Buffered sucrose or Diluted fractions
1	180	10 μL of buffered sucrose (blank)
2	180	10 μL of homogenate supernatant
3	180	10 μL of nuclear fraction supernatant
4	180	10 μL of mitochondrial fraction
5	180	10 μL of soluble fraction

3.    Then add 10 μlof Solution 5 to each well. Mix the contents well

4.    Measure  absorbance  at  0  min  and  2  min  (600  nm)  with  microplate spectrophotometer (BioTek Epoch plate reader). Record the absorbance value on the datasheet.

Data Processing of SDH Assay

1.  This calculation assumes that the components in the assay medium

remain constant over the assay, valid unless some component of the    mixture precipitates. Therefore, the change in absorbance at 600 nm is solely due to the activity of SDH which indirectly lead to the conversion of oxidized DCPIP (absorbs at 600 nm) into reduced DCPIP (invisible)

2.  Calculate the specific activity of succinate dehydrogenase in the

original homogenate and in each fraction by dividing the activity per ml  of that fraction by the protein content protein concentration (mg per ml)  of the same fraction. The result, expressed in units of activity per mg of  protein (U/mg), should be in some cases greater than the homogenate  and in some cases less. It represents the degree to which the particular fraction is enriched for succinate dehydrogenase (for example,

mitochondria) or the extent to which the enzyme has been removed from that fraction.

Experimental Datasheet of Exercise 6

Part C:

Protein Standard Plot

BSA Standard	Protein concentration (mg/mL)	A595
#1	0
#2	0.0375
#3	0.075
#4	0.15
#5	0.3
#6	0.6
#7	1.2

On your laboratory write-up attach your standard curve and calculate the followings: Correlation coefficient, slope, Y-intercept, linear equation of the plot

Protein content per fraction

Fraction	Dilution	A595	Protein conc. in diluted sample   (mg/ml)	Protein conc. in fraction     (mg/ml)	Volume of fraction     (ml)	Amount of protein in fraction   (mg)
Homogenate	1:50
Nuclear	1:25
Mitochondrial	1:25
Soluble	1:25

On laboratory write-up, show the calculation steps accordingly

Data Processing of Part C

1.    Record absorbance at 595 nm in the column labeled “A595” .

2.    Using the known concentrations of the standards, construct a STANDARD PLOT with absorbance (dependent variable, y-axis) versus concentration (independent variable, x-axis).

3.    For each diluted sample, find the absorbance on the standard curve and record the corresponding concentration.

4.    Multiply this value by the dilution factor to get “protein concentration (mg/ml) of  fraction” .     Then  calculate  the  “averaged   protein  concentration  of  fraction” .

5.    Multiply the averaged protein concentration of fraction by the total original volume of the fraction to get the amount of protein (mg) of each fraction.

6.    Check the protein recovery: the sum of the protein in the three fractions (nuclear, mitochondrial, soluble) should approximate to the protein in the crude homogenate that was fractionated.   Express the recovery in each fraction as a percent of the total protein.