1
6. GOT#3
3723 F02
Biochemistry 3723Lecture 6c Protein Purification 3Sept. 16, 2002
I.GOT Assays
A.Reaction:
B.Assay
1.Force the reaction in forward direction by high [L-asp] and [-KG]. Products not easily quantified. Therefore use coupled reaction.
2.Continually remove product in coupled reaction.
a.
b.Follow reaction: NADH: EM,340= 6.22103 L/mol-cm
NAD+:≈ 0
So can follow reaction by loss of absorbance at 340 nm
3.Problem: In crude extract there is another enzyme that uses NADH in the presence of -KG; Glutamate dehydrogenase (GDH)
a.NADH + -KG + NH4+ NAD+ + H+ + Glutamate
b.Control: by measure activity in absence of L-aspartate. Any reaction under these conditions due to GDH not GOT.
c.GDH is a mitochondrial enzyme. Should be removed in first purification step. Is not heat stable.
C.Calculation: Reaction Rates = Activity measurement
1.Define vi = ∆ C/∆time: measure
(– sign to correct for fact that A)
2.Convert ∆A/∆time to ∆Concentration/∆time;
a.Beer's Law: A = E C l . Thus: C = A/(E l)
b.∆C/∆time = (–∆A/∆time)/(E l); E = 6.22103 L/mol-cm
3.And vi expressed by:
4.Sample calculation:
a.3.00 ml reaction mix with 50.0 l enzyme, ∆A = –0.250/min
b.
c.Correct for dilution
D.Other Calculations for Purification Table
Fract / Volume(ml)
** / [Protein]
(mg/ml)
** / Activity
[µmol/min]
ml enzyme
** / Total
Protein
(mg) / Total
Activity
(µmol/min) / Specific
Activity
[µmol /min]
mg protein / Purif-ication
(fold) / Yield
(%)
Fr. I / 1 / 100
Fr. II
55% P
75% P
75% S
1.Total Protein = [Protein (mg/ml)] [Volume (ml)] = mg protein
a.Expect to decrease with ensuing steps
b.Remove unwanted proteins
2.Total Activity= [(µmol product/min)/ml sample] [Volume (ml)]
= µmol/min
a.Goal--to have total activity remain near constant
b.Remove other proteins but retain protein of interest
3.Specific Activity: activity per mg protein
a.A measure of purity of enzyme; increases as enzyme purified
b.= [(µmol product/min)/ml] / [mg/ml] = µmol/min/mg protein (IU/mg)
4.Purification
a.How has the purity of the enzyme changed with each step--hopefully increased. Goal to maximize purification
b.purification= (specific activity of this step)/(specific activity at start)
= Fold purification (unitless #)
c.Best steps give high purification over previous step.
5.Yield: How much enzyme is left?
a.[total activity at this step]/[total activity at start] 100 = % yield
b.Usually yield goes down. Sometimes goes up at early steps if inhibitor is removed.
c.Goal--to have yield go down the least possible
d.Realize that 55%P, 75%P and 75%S fractions together make up one step in the sense that the yield from the three fractions together should be Fr. II
C.Goal: Maximizespecific activity (purity) and yield at minimum cost (time and money)
II.Assaying the Activity and Protein concentration--practical
A.Specs:-- use the diode arrays--
1.Mix all but GOT in cuvette. Start "progress curve" program. Check that no reaction (~5 sec). Also that start at A340 ~0.75-1.00.
2.Add GOT as demonstrated in lab
3.Follow change in absorbance long enough to get a good rate (15-30 sec)
B.Prepare next reaction
1.Clean cuvettes
2.Careful pipetting
3.Adjust amt of GOT added to get good rate
a.should be between 5-50 µl
b.if need to go to less than 5 µl, dilute aliquot 1/10 just before use, add 5-10 µl (which is same as 0.5 -1.0 of undiluted)
c.WRITE DOWN the reaction # (from the computer), and amount of enzyme used in lab book!
C.Once both partners understand the enzyme assays, one partner MUST begin Bradford assays for protein concentration on samples or you will be here all night. Procedure for Bradford, exactly as in Exp. 5. Be careful to make the dilutions correctly!
III. Dialysis of samples.
A.Theory of dialysis--
1.Membrane (cellulose nitrate) of defined pore size allows small molecules in solution to pass through while retaining large particles
2.Place protein solution, inside bag or cassette, into large volume of the solution you want the protein dissolved in at end
3.Equilibrate several hours with gentle stirring(≈ 6 hrs to equilibrium)
a.small molecules at equal conc. in and out of bag at equilibrium
b.large molecules retained within membrane
4.Different membranes have different “cut-off” sizes. Usually ≈ 10 kDa
5.Desalting/buffer change: change dialysis buffer 2 or 3 times to completely remove undesirable salt
6.Other methodologies--ultrafiltration, centrifugal concentration--in text.
B.Practical for today
1.Decide which fraction has highest GOT SA --Ask Judy or me.
2.Place 50 µl of GOT-1 into microfuge tube, add 950 µl H2O. Label and store frozen for SDS-PAGE. Record data as instructed on p 65.
3.Measure volume of remaining GOT-1 and then dialyze according to instructions in Exp. 8 (p 79)
a.Dialyze against 50 mM Tris.
b.To remove AmSO4 which would interfere with ion exchange chromatography of sample next week.
IV.Sample Problems: probably no time to do today, but include anyway
A.Assays
1.Design an assay for the enzyme -galactosidase, which catalyzes the hydrolysis of nitrophenyl galactose to nitrophenol + galactose. Nitrophenol is yellow at pH above 9 but the pH optimum for the reaction is about 7.2.
Solution:Since the product of the reaction is colored, an easy assay would involve the specrophotometer max for nitrophenol is 405 nm). However, the pH that leads to color formation is NOT the pH at which the enzyme is active. To get the best assay, you would need to do a fixed time assay, running the enzymatic reaction at pH 7.2 for a fixed time, stopping the reaction by adding OH– to raise the pH and then reading the absorbance at 405 nm. It would be necessary to establish that the reaction is linear for the time period it is run. Taking all this into account, a usable assay would be as follows: To a standard amount of nitrophenyl galactose in pH 7.2 buffer, add b-galactosidase, incubate for standard time at controlled temperature (37°C is optimum for bacterial enzyme), stop reaction by adding standard amount of NaOH. Read absorbance at 405 nm on spectrophotometer.
2.An assay for GOT using 25 µl of enzyme in a 3.00 ml reaction mixture showed a change of absorbance at 340 nm of – 0.375/min. What was the activity of this enzyme preparation? Remember E340 for NADH is 6.22x 103 L/mol-cm
Solution:– (– 0.375 /min)/[(6.22 ml/µmole-cm)(1.00 cm)]• 3.00 ml rxn /0.025 ml enz
= 7.23 (µmole/min)/ml enz
note: (6.22x103 L/mole-cm) X (103 ml/L) X (mole/106 µmole) = 6.22 ml/µmole-cm
B.Purification
1.Describe an experiment to determine the optimum ammonium sulfate concentration to use to salt out -galactosidase from a bacterial cell extract.
Solution:Prepare a series of samples (on a small scale, say 10 ml of extract for each tube) at, say, 20, 30, 40, 50, 60, 70, 80% saturated in AmSO4. Centrifuge each sample. Keep the supernatant and resuspend the pellet of each sample. Assay all samples for
-galactosidase activity. Determine which % AmSO4 cut still has the bulk of the enzyme activity in the supernatant and which % has the bulk in the pellet. Then make your cuts at these values. You may want to repeat to fine tune the precipitation. For example if 30% AmSO4 still has 95% activity in supernatant, 40% AmSO4 has 50% in supernatant and 50% AmSO4 has 5% in the supernatant, with the rest of the activity in the resuspended pellet, you might want to try cuts at 35% and 45% AmSO4. Ultimately, you want to choose two levels of AmSO4; one just below the % that begins to precipitate the enzyme and one high enough to precipitate the bulk of the enzyme. Add the first AmSO4, centrifuge and continue with the supernatant. Increase the AmSO4, centrifuge and resuspend and continue with the pellet.
2.Purification table: Fill in the purification table below:
Fraction / Volume (ml) / Protein (mg/ml) / Activity(µmol/min)/ml / SA / Purif / Yield
Start / 35 / 18 / 2.5
End / 2.5 / 0.53 / 23
Solution:SA (specific Activity) is µmol/min/mg so is Activity/Protein
Start: 2.5 /18 = 0.14 µmol/min/mg enz
End: 23/0.53 = 43 µol/min/mg enz
Purif is SA at step x /SA at step one
Start: 0.14/0.14 = 1 fold purified
End:43/0.24 =307 fold (really 310 to the appropriate sfs)
Yield is total activity at step x/total activity at step 1 x 100%
Start = 100%
End: [23µmol/min/ml x 2.5 ml] / [2.5 µmol/min/ml x 35 ml] x 100%
= 66%