Docstoc

Lab pellet

Document Sample
Lab pellet Powered By Docstoc
					Biochemistry I
Fall 2007

                                 Lab 21
                            Enzyme purification
                   Part A: Ammonium sulfate precipitation.

In today’s experiment, we will begin the purification of an enzyme: lactate
dehydrogenase (LDH). LDH catalyzes the reaction:


                    CH3                                 CH3

                    C     O                           HC       OH
                                                                              Eq. 3-1

                    COO           NADH       NAD        COO

                 Pyruvate                              Lactate


This enzyme is particularly important in tissues where oxygen can be depleted; it allows
fermentation to take place so glycolysis can continue. (NAD+ must be regenerated for
glycolysis to continue). We will isolate the protein from beef heart. The enzyme is
found in large quantities in the cytosol of these cells, so it is easy to isolate.

In general, the basic steps of protein purification for soluble2 proteins include:
       1) Homogenization: Breaking open the cells. Homogenization can be done
            using a variety of tools including a French press3, a sonicator, lysozyme4,
            detergents, flash freezing, or a blender.
       2) Fractionation: Typically the initial steps of the purification procedure, these
            steps typically take place while the protein is in a fairly large volume of
            solution. Techniques of fractionation include: centrifugation, salting out
            (ammonium sulfate precipitation) and sucrose gradients.
       3) Chromatography: There are many different types of chromography used to
            purify proteins: affinity, ion-exchange, hydrophobic interaction, and size
            exclusion. These will be discussed in detail next week.

Centrifugation
In today’s lab, you will do some fractionation steps. First, you will centrifuge (at 20000
x g) the homogenate of the beef heart (this has been homogenized in a blender).
Centrifugation will pellet the larger particles (unbroken cells, cell membranes,

1
  Adapted from Experiments in Biochemistry (A Hands-on Approach) by S.O. Farrell and
R. T. Ranallo (2000). Harcourt Brace and Co. Orlando, FL.
2
  For membrane proteins, solubilization in a detergent is necessary.
3
  No, not a coffee maker. This is a hydraulic press with a steel chamber for the bacterial
cells. As the cells exit the container through a small spout, there is a drastic change in
pressure, which causes the cells to lyse.
4
  An enzyme that breaks the oligosaccharides in bacterial cell walls.


                                                                                         16
 Biochemistry I
 Fall 2007
 mitochondria, etc.). The protein will stay in the solution, which is called the
 supernatant. Large particles will pellet at lower centrifugation speeds, while smaller
 particles can pellet at higher centrifugation speeds.

 Salting out
 You will also salt out the LDH in the experiment today. Proteins are kept in solution by
 their interaction with water molecules—specifically, by the formation of hydrogen bonds.
 By adding a polar compound—usually ammonium sulfate but other salts also work—the
 water will interact with the salt instead of the protein. If there are fewer water molecules
 to interact with the protein, the protein molecules will begin to interact with one another.
 When the particles get large enough, the protein will precipitate. To collect the protein
 that has precipitated, the solution is centrifuged. Since some proteins precipitate out of
 solution at low concentration of ammonium sulfate while others precipitate out at a
 higher concentration of ammonium sulfate, one can purify the protein by using different
 percentages of ammonium sulfate (different ammonium sulfate cuts). The protein can be
 resolubilized in buffer after it has precipitated. Salts are not the only way to precipitate
 proteins; some can be precipitated by change in pH, with heat, and (for a few that are
 stable to the conditions) organic solvent such as ethanol or acetone.

 Dialysis
 Dialysis is another way to separate molecule on the basis of size. It is typically used to
 remove (or add) salt from a protein sample. (After resolubilization, the sample will still
 contain a lot of ammonium sulfate.) Dialysis tubing is a semi-permeable, thin
 membrane made of nitrocellulose or cellulose. Different types of dialysis membrane
 have different molecular weight cutoffs—molecules above the molecular weight cutoff
 cannot pass through the membrane, while smaller particles pass through the membrane.
 The protein sample is placed in a dialysis bag or “cassette” of the membrane material.
 The dialysis bag is placed in a large beaker of the equilibration buffer. Over time, the salt
 diffuses out of the pores and the ammonium sulfate will be diluted to a very low
 (negligible) concentration.

 Analysis of protein purification: Activity
         In any protein purification, the goal of each step is to isolate the desired protein
 away from other proteins. A high yield of protein in its active form is also desirable.
 For an enzyme, the best way to check the progress of our purification is to do enzymatic
 activity assays. Activity is a measure of the enzyme’s ability to catalyze a specific
 reaction and is defined as:

                               Amount of product
                  Activity =                                                   Eq. 3-2
                                    time

 Enzyme activity has the unit of “Units”, which has the standard definition of the mol of
 product produced per minute.
     
                                             1 mol
        1 Unit of Enzymatic Activity(U)=                                       Eq. 3-3
                                               min


                                                                                            17

  Biochemistry I
  Fall 2007



  Units of enzyme activity sometime have other units (nmol/min; mmol/sec). The
  enzymatic activity can also vary with reaction conditions like pH, temperature and ionic
  strength; therefore it is important to record these variables for any experiment.
          The concentration of the active enzyme (or the relative activity) is defined as the
  activity of an enzyme per milliliter.

                                                Units
                          Relative activity =                                   Eq. 3-4
                                                 mL

  To determine the relative activity of a sample, one must take into account the volume of
  the sample that has been analyzed. For example, if one obtains 50 Units of activity (50
               
  mol/ min of product produced) from the analysis of 0.50 ml of sample, the relative
  activity is:
                                     50mol/min         Units
                 Relative activity =               100
                                        0.50 ml           ml

   From the relative activity and the overall volume5 of the analyzed fraction (the total
  volume of the from each fractionation step, not the analyzed amount of the fraction), the
  total 
        amount of the enzyme in each fraction can be found.

                                            Units 
          Total activity = Relative activity       ml fraction              Eq. 3-5
                                             ml 

 From the total activity of the first step and the total activity of the subsequent steps, one
 can determine the % yield of each step.

                              total activity of fraction
                %Yield =                                                         Eq. 3-6
                            total activity of first fraction

  Percent yield of the enzyme should decrease in subsequent step; however, sometimes the
  activity of a later step will be larger because the purification procedure removes
       
  inhibitors of the enzyme.

  Next week, we will also check the protein concentration. From the combination of the
  two, we will be able to determine the specific activity of the protein sample. The specific
  activity is a measure of the purity of the enzyme.




  5
    Volume becomes a confusing term in these calculations. Make sure you understand the
  calculations so you know which volume to use. The total volume refers to the
  fractionation step. The volume assayed refers to the amount of sample used in an assay.


                                                                                             18
     Biochemistry I
     Fall 2007
                                                         Units 
                        Units       Relative activity      
                                                          mL 
     Specific activity            =                                           Eq. 3-7
                                                             
                       mg protein Protein concentration mg 
                                                              
                                                             mL 

     A larger number for specific activity signifies that more of the protein in the sample is the
     active enzyme.

     Determining the activity of LDH
              To quantify the amount of active enzyme present in a protein preparation, one
     must be able to observe how it catalyzes a reaction. There are many different strategies
     for this including:

            1) Observing the appearance or disappearance of a product or substrate by
               absorption spectroscopy
            2) Observing the appearance or disappearance of a product or substrate by
               fluorescence spectroscopy.
            3) If the product or reactant are hard to observe by the above techniques, the
               appearance of a product can be observe by coupling it to a reaction that
               produces a more easily observable product.

     For LDH, we can easily observe the appearance of a product: NADH. NADH absorbs
     light at 340 nm; NAD+ does not. Therefore, if we supply lactate and NAD+ to the
     enzyme, it will catalyze the reverse reaction of Eq. 3-1 and produce pyruvate and NADH.
     As more NADH is produced (more mol are in the reaction tube), the absorbance at 340
     nm of the reaction tube will increase. As both the NAD+ and lactate are supplied in
     saturating concentrations, the change in absorption is directly related to the amount of
     enzyme in the sample.

     Calculating the initial velocity, mol product, and Units
              In enzymatic reaction, we measure the initial velocity of the reaction. The initial
     velocity is measured because the velocity of the reaction is dependent upon the
     concentration of the substrates and we estimate that for a short amount of time the
     concentration of the substrate does not change. After the reaction proceeds for a minute
     or two, this estimate will not be valid and the reaction will slow. Additionally, as the
     amount of product increases, the back reaction becomes a factor.
              The initial velocity can be determined by observing the change in absorbance
     (A) over time. Figure 3-1 illustrates the data points (in this example) that are in the
     initial velocity range and that are outside the initial velocity region. Notice that the
     absorbance changes linearly with time in the initial velocity region, while outside that
     region the A per unit time becomes smaller.




                                                                                               19
Biochemistry I
Fall 2007

                                         Determining the initial velocity of an enzyme catalyzed reaction
                             0.5


                            0.45


                             0.4


                            0.35


                             0.3

               Absorbance
                                         Region used to determine
                            0.25              initial velocity


                             0.2                                                        Not in initial velocity range

                            0.15


                             0.1


                            0.05


                              0
                                   0        20             40        60                80           100            120   140
                                                                          Time (sec)



                                                                                                                               
                            Figure 3-1: The initial velocity region in enzyme kinetics

It is not necessary to graph the data to find the region of initial velocity; one can just look
at the data points to see if the A between the time points becomes smaller.

Time (s)     Absorbance
    0            0
   15          0.07
   30          0.14                                         A ~ 0.07/ 15 sec
   45          0.21
   60          0.29
   75          0.34
   90          0.39                                         A < 0.07/ 15 sec
  105          0.43
  120          0.45

From the example above the initial A/min is 0.28. If one uses all the data points (up to
120 sec or 2 min.) the A/min is 0.23. This is an almost 20 % difference!
       To find the mol/min, you need to convert A to mol. In the first step, the
concentration change of the product (per unit time) can be found using Beer’s Law.


                                                    A lc                                                               Eq. 3-8

                                       where A is the absorbance,
                                              is the extinction coefficient of the absorbing molecule
                                             l is the pathlength of the cuvette (usually 1 cm)
                                             c is the concentration of the molecule.

           
                                                                                                                                   20
Biochemistry I
Fall 2007
The extinction coefficient is a property of the molecule and the wavelength of the light.
It has the units of cm-1 M-1. Notice that this is molarity not moles.
         Using the example above, and assuming the extinction coefficient of the
absorbing molecule to be 4000 cm-1 M-1, one can calculate the concentration of the
product in solution:

                 0.28  (4000 cm1 M 1 )(1cm)(c)
                          0.28                       mol
                 c           1
                                   70 x10 6 M  70
                      (4000 M )                       L

This is the change in molarity per minute. To find the moles, multiply this value by the
reaction volume:
     
                 Moles=Molarity (mole/L) x volume (L)                   Eq. 3-9

If the reaction volume is 3 ml, the mol produced in the reaction tube is:

                              70 mol/L x 0.003 L=0.21 mol

So 0.21 mol/min of product are produced in the reaction tube, meaning there are 0.21
Units of enzyme in the reaction tube.

However, the reaction is only a part of the original sample. Let’s say that 100 l of
enzyme was used in the 3 ml of the reaction. The relative Units (Eq.3-4) of the original
sample can be calculated by dividing the units from the reaction by the volume assayed.

                                              0.21 Units       Units
                           Relative units =               2.1
                                                0.1 ml          ml

If the total fraction volume is 20 ml, then the total Units of this protein fraction is:

                                            Units
                          Total units = 2.1         x 20 ml= 42 Units
                                               ml

If the protein concentration is known, the specific activity can be calculated (Eq. 3-7). If
the protein concentration is 0.5 mg/ml, then the specific activity is:
               
                                               Units
                                           2.1
                       Specific activity=       ml  4.2 Units
                                                mg          mg
                                            0.5
                                                ml


Make an effort to understand the equations above! They will show up again (in class and
in the lab.) 


                                                                                            21
Biochemistry I
Fall 2007
Using a recovery table: Keeping track of purification.
         You will keep track of the progress of the purification of LDH using a recovery
table. It will look something like this:
Fractions                  2        3        4        5      6        7         8          9
              A/min      Volume    Units   Units/ml   Total   %        Protein   Specific   Fold
                          assayed                      units   Recove   (mg/ml)   activity   purificaiton
                                                               ry
Crude                                                          -----    -------   ------     --------
homogenate
20000 x g                                                      100                           1
supernatant
40 %
(NH4)2SO4
supernatant
40 %
(NH4)2SO4
pellet
resuspended
65%
(NH4)2SO4
supernatant
65 %
(NH4)2SO4
pellet
resuspended

Column 1 is the experimentally measured change in the absorbance (the initial velocity)
that occurs in the reaction.
Column 2 is the volume of the enzyme assayed. If you had to dilute the enzyme to obtain
a reasonable change in the absorbance, report this in this column too.
Column 3 reports the units of activity in the reaction tube. You find this value by taking
the data from column 1 and converting it to units using Eq. 3-8 and 3-9 or the conversion
factor from the prelab. The extinction coefficient ( of NADH is 6220 M-1 cm-1, the
pathlength of the cuvette is 1 cm, and the volume you will use for your reaction is 3.0 ml
Column 4 is the relative activity of the sample. This value is column 3 divided by
column 2 (converted to the correct units) unless you diluted the sample. If you made a
dilution, take this into account for the calculation.
Column 5 is the total units of the particular fractionation step. This value is calculated by
multiplying column 4 by the volume of the fractionation step.
Column 6 is the % recovery of each step. We will assume the supernatant step has 100 %
recovery because it is extremely difficult to obtain an accurate measurement of activity
from the crude homogenate.
Column 7 is the protein concentration in mg/ml. You will learn how to measure this in a
later lab.
Column 8 is the specific activity, as calculated from Eq. 3-7. You will fill in this column
next week.
Column 9 is the fold purification. To calculate this value, you take the specific activity of
the fraction and divide it by the original specific activity (from the 20000 x g
supernatant.)



                                                                                                 22
Biochemistry I
Fall 2007
Procedure
Materials
       For purification
       Beef heart homogenate in 0.05 M sodium phosphate, pH 7.0
       0.05 M sodium phosphate buffer pH 7.0
       Ammonium sulfate buffer
       Dialysis buffer: 0.03 M bicine
       40 ml centrifuge tubes

       For enzyme analysis
       6mM NAD+
       150mM lactic acid (pH ~10.0)
       0.15M CAPS pH 10.0

Purification When possible, do each of these steps on ice.
   1) Obtain 40 ml of beef heart homogenate. Save 0.5 ml of this for the enzyme and
       protein assays. Put the rest in a ~40 ml centrifuge tube.
   2) Put the tube in the centrifuge. Make sure it is balanced with another tube.
       Centrifuge at 20000 x g for 20 minutes at 4C. You will need to check the chart
       on the centrifuge to determine what rpm (revolutions per minute) the centrifuge
       should be set to.
   3) When the centrifugation is over, save the supernatant and discard the pellet.
       Measure and record the volume of supernatant. Save 0.5 ml of the supernatant for
       the enzyme and protein assays.
   4) Add 0.230 g of ground ammonium sulfate for every milliliter of supernatant. Add
       this slowly with constant stirring. The ammonium sulfate should be ground (with
       a mortar and pestle) because it will go into solution more easily and it will be less
       likely that there will be a locally high concentration of ammonium sulfate. This
       will make a solution that is 40% ammonium sulfate, so this is the 40 %
       ammonium sulfate cut. Let the solution stand on ice for 10-15 minutes so that the
       protein precipitates.
   5) Centrifuge the solution at 15000 x g for 15 minutes.
   6) Keep the supernatant, measure and record the volume. Save 0.5 ml of the
       supernatant for the enzyme and protein assays. Slowly add 0.166 g/(ml of
       solution) of ground ammonium sulfate to the solution with constant stirring. This
       is the 65% cut.
   7) Let the solution stand on ice for 10-15 min. While you are waiting, resolubilize
       the 40% cut pellet with 2 ml of phosphate buffer. Save this for analysis.
   8) Centrifuge the 65% cut at 15000 x g for 15 minutes.
   9) Save the supernatant (this should have little LDH) and resuspend the pellet (this
       should have most of the LDH) in 5-7 ml of buffer. Carefully measure and record
       the volumes of the supernatant and the resuspended pellet. Save 0.5 ml of the
       supernatant and 0.1 ml of the resuspended pellet for later assays.
   10) You will analyze the activity in the resuspended pellet and the other saved
       fractions. After you have completed the analysis on the resuspended pellet and
       confirmed the presence of LDH, put this sample in the Slide-a-Lyzer cassette and



                                                                                         23
Biochemistry I
Fall 2007
       dialyze it in 0.03 M bicine. Store the samples of the other fractionation steps in
       the refrigerator.

Analysis of Fractions Note: Some of the fractions can be analyzed during the “down
time” of the purification procedure.

       Analyze each of your fractions; you should have 6 samples total. If it appears that
your assays are not working, there is some LDH from Sigma that you can use to test the
procedure.
       In each of the assays you will need to make a reaction “cocktail.” The reaction
cocktail contains lactate, NAD+, and a buffer at the appropriate pH (lactate
dehydrogenase works best around pH 10). Each assay will be 3.0 ml and contains:

                 1.9 ml of CAPS buffer, 0.14 M pH 10
                 0.5 ml of NAD+,6 mM
                 0.5 ml of lactate, 0.15M (close to pH 10)

The volumes add to 2.9 ml; the remaining 100 l will be a combination of water and the
enzyme sample. The enzyme sample is added only when you are ready to begin the
measurement. For most samples, 10 l of enzyme and 90 l of water is sufficient, but
you may need to alter this to obtain a reasonable A/min.

        Since you will be doing multiple assays, it is recommended that you make enough
reaction cocktail for all of you experiments. For example, if you are doing 5 reactions,
you should mix together:

                 1.9 ml x 5=9.5 ml CAPS
                 0.5 ml x 5= 2.5 ml NAD+,6 mM
                 0.5 ml x 5= 2.5 ml lactate, 0.15M

It is advisable to make a bit more reaction cocktail than you expect to need, in case you
need to repeat an experiment.

   1) Set the Spec 20 to 340 nm and zero it with a cuvette of water. (Make sure the
      filters are set correctly.)
   2) Pipet 2.9 ml of the reaction cocktail into the Spec 20 cuvette. Add the water and
      then add the enzyme sample last. Invert the cuvette 3 times (put parafilm on the
      top); do not vortex! Put the cuvette into the Spec 20.
   3) Record the change in absorbance every 10 to 15 seconds. To obtain reasonable
      data, the A/min should be between 0.1 and 0.6.
   4) Evaluate the data and find the initial velocity. After you have measured all the
      fractions, make a recovery table in your notebook.




                                                                                            24

				
DOCUMENT INFO