Background: Phosphatases are enzymes that catalyze the hydrolysis of phosphate esters. Some phosphatases are very specific for their preferred substrates while others are nonspecific and hydrolyse phosphate from any phosphate ester. We will prepare a crude preparation of a phosphatase enzyme from wheat bran that is nonspecific. As a consequence, we can use it to hydrolyze a phosphorylated organic substrate in a reaction that produces a product, p-nitrophenol, which absorbs light at 410 nm.
para-nitrophenylphosphate (PNPP) para-nitrophenol
Since the Beer-Lambert Law (Abs. = ε · c · l) shows that absorbance of light at a specific wavelength is proportional to the concentration of the colored species in solution, we can monitor changes in product concentration under various conditions. We will perform several studies: a determination of the proper enzyme dilution for effective assay, the effect of substrate concentration on rate of reaction, and the effect of a known inhibitor on the rate of reaction.
Objectives: After this experiment, you will be able to:
- Follow a protocol for a crude enzyme preparation.
- Determine the most suitable dilution of the enzyme to give a useful assay result.
- Prepare a series of enzyme-substrate mixtures at various concentrations and measure the absorbance of the product using spectrophotometry.
- Convert corrected absorbance to product concentration and determine reaction velocity.
- Prepare Michaelis-Menten hyperbolic graphs and Lineweaver-Burk plots; use them to determine Vmax, KM and the effect of a known inhibitor of the reaction.
- Identify the type of inhibition observed for this enzyme-inhibitor pair.
See each section
This is an individual assignment, complete all parts on your own.
Part 1: Preparation of a crude phosphatase from wheat bran
Fresh wheat bran, 15 g per person
Ammonium sulfate, pulverized solid
Dialysis tubing, 12-inch lengths of ¾ inch tubing
Centrifuge capable of 5000 rpm and centrifuge tubes or 250-mL bottles
Instructions: Chill a beaker containing 150 mL of deionized water in an ice bath. Weigh out approximately 15 g of fresh wheat bran and gently stir it into the water. Keep the mixture cold and stir gently but frequently for 30 minutes or longer.
Filter the mixture through several layers of cheesecloth, then centrifuge at maximum speed for ~5 minutes. Discard the precipitate. Measure the volume of the clarified solution and cool it in an ice bath to 0-10°C.
Weigh out an amount of solid ammonium sulfate calculated from the following formula:
g (NH4)2SO4 = 0.53 x volume of solution in mL
With gentle continuous stirring, add the ammonium sulfate to the solution. Add slowly so that the rate of addition equals the rate of dissolving. Continue stirring until the solid has completely dissolved. The final solution will have a concentration of approximately 3.2 M or about 80% of a saturated solution. At this concentration, most of the proteins will precipitate out of solution, a procedure called “salting out”. The solution will become opaque. Allow the mixture to stand in the ice bath for 10–15 minutes to obtain complete precipitation.
Centrifuge to settle the precipitate. Decant and discard the supernatant. Suspend the precipitate in 25 mL of cold distilled water.
Prepare a section of dialysis tubing by soaking it in distilled water until it is pliable. Open the tubing by rubbing it gently between your fingers. Tie two overhand knots near one end. Check for leaks. Then pour the suspended enzyme solution into the tubing (use a funnel to avoid spilling) and tie off the open end. Place the tubing in a 600-mL beaker filled with chilled distilled water, label the beaker and place it in the refrigerator. The dialysis water must be changed several times to aid in the removal of the ammonium sulfate and the redissolving of the proteins.
Remove the enzyme preparation from the dialysis tubing and drain it into a small flask. Carefully heat the concentrated enzyme solution to 60°C (NO HIGHER!!!) and hold the solution at this temperature for 10 minutes. Chill rapidly in an ice bath. Centrifuge the cold enzyme preparation to remove precipitated unwanted denatured proteins while leaving the relatively heat-stable phosphatase in solution. The resulting clear solution will be used for several days; store any UNUSED and UNDILUTED enzyme in the refrigerator.
Part 2: Determination of the Effective Enzyme Concentration
For use, the enzyme will probably have to be diluted between 5- and 20-fold. Since the enzyme concentration and effectiveness in each sample will vary, the proper dilution factor will have to be determined before beginning any studies of its kinetics. The most useful enzyme concentration will produce at 45°C an amount of p-nitrophenol (the colored product) that has an absorbance of 0.5 – 0.8 at 410 nm in our main study. We will incubate substrate with enzyme at various dilutions to determine the optimal concentration for use in the remaining studies.
Fresh wheat bran phosphatase enzyme preparation
0.5 M acetate buffer with 0.0010 M MgCl2
0.046 M PNPP (substrate) solution
0.5 M KOH
Water bath set at 45°C
Colorimeters set to measure absorbance at 410 nm
Instructions: Store your stock enzyme preparation on ice while preparing the following diluted mixtures. Store the tubes containing diluted enzyme on ice also until you are ready to use them.
Tube label mL conc. enz. mL H2O “Relative E”
1 E 1.0 — 1.0
½ E 0.5 0.5 0.5
1/5 E 0.2 0.8 0.2
1/10 E 0.1 0.9 0.1
1/20E dilute some of the ½ sample tenfold or dilute some of the 1/10 sample 1:1
A 0.2 mL sample of each diluted enzyme solution will be mixed with constant amounts of substrate and buffer to a total volume of 5.0 mL. The mixture is then incubated at 45°C for 10 minutes. To “quench” the reaction, 2.5 mL of KOH will be added then the absorbance at 410 nm will be read.
To assist in organizing the study, prepare 6 test tubes. Label #1 “blank”, #2 “E”, #3 “½ E”, #4 “1/5 E”, #5 “1/10 E” and #6 “1/20 E”. To the blank, mix 1.00 mL buffer, 0.25 mL PNPP solution and 3.75 mL water (this one gets NO enzyme), incubate 10 minutes then add 2.5 mL of KOH; set aside. This is a control tube to allow us to correct for any spontaneous hydrolysis of the substrate into the colored product. Any color measured here will be subtracted as a correction factor from the Part 2 reaction tubes that contain enzyme.
For the reaction tubes, mix the buffer, substrate and water then place in a 45°C water bath. Space-out additions of 0.2 mL of the appropriately diluted enzyme then “quench” the reaction as follows.
|Tube||mL buffer||mL 0.046M PNPP||mL H2O||add enzyme at __min||add KOH at __min|
After the addition of KOH, mix thoroughly and read the absorbance of the blank and of each reaction mixture. Chose the dilution that gives A410 (corrected) = ~0.5 in this preliminary assay; use this dilution factor to prepare enzyme solution for all further studies.
Part 3: Determination of the Effect of Substrate Concentration and Inhibitor on Reaction Velocity
In this experiment we will examine changes in reaction velocity, v, with varying substrate concentrations in the absence of, and then in the presence of, a known phosphatase inhibitor, inorganic phosphate ion.
Diluted wheat bran phosphatase enzyme preparation
0.5 M acetate buffer with 0.0010 M MgCl2
0.0125 M PNPP (substrate) solution
0.5 M KOH
0.0050 M “inorganic phosphate” solution
1.00-mL and 10.0-mL graduated pipettes
Colorimeters set to measure Absorbance at 410 nm
Part 3A: v for various E + S mixtures
Instructions: First, prepare a “blank” by mixing 1.00 mL buffer, 0.50 mL of 0.0125 M PNPP and 3.50 mL water. Set this aside for 15 minutes in a 60°C water bath then add 2.5 mL KOH. Read the absorbance at 410 nm and use this as a correction factor for parts 3A and 3B.
Prepare 6 test tubes with buffer, substrate and water (if needed) according to the chart below. Each tube will contain 4.8 mL of solution before the addition of enzyme. Incubate all tubes at 45°C for 10 minutes before starting the sequence of adding 0.20 mL of enzyme solution. As before, space out the addition of enzyme according to the timing chart, incubate each reaction mixture 15 minutes at 45°C then add 2.5 mL KOH to quench the reaction. Finally, read A410 (uncorrected) for each tube.
Tube mL buffer mL H2O mL PNPP sol. add 0.2 mL E at __min add KOH at __min
1 1.0 3.8 0.0 0 15
2 1.0 3.7 0.10 2 17
3 1.0 3.6 0.20 4 19
4 1.0 3.4 0.40 6 21
5 1.0 3.0 0.80 8 23
6 1.0 2.2 1.60 10 25
Part 3B: v for various E + S + I mixtures
Instructions: Prepare 6 test tubes with buffer, substrate, phosphate (inhibitor) solution and water (if needed) according to the chart below. Each tube will contain 4.8 mL of solution before the addition of enzyme. Incubate all tubes at 45°C for 10 minutes before starting the sequence of adding 0.20 mL of enzyme solution. As before, space out the addition of enzyme according to the timing chart, incubate each reaction mixture 15 minutes at 45°C then add 2.5 mL KOH to quench the reaction. Finally, read A410 (uncorrected) for each tube.
|Label||mL buffer||mL phosphate||mL H2O||mL PNPP sol.||add E at __min||add KOH at __min|
Presentation of Results:
Calculate the concentration of substrate in each final reaction mixture for Parts 2 and 3. Remember that the final volume of each reaction mixture is 5.0 mL and that the other components of the reaction mixture dilute the substrate sample that was added to the reaction mixture.
Convert all A410 readings to reaction velocities. In order to do this calculation, we must first correct the observed A410 values for any spontaneous nonenzymatic hydrolysis of the substrate. The Abs. of the “blank” tube should be subtracted from the Abs. of any tube that contains enzyme.
A410(corr) = A410(measured) – A410(blank)
The Beer-Lambert Law can be rearranged to c = Abs. /(ε · l) where c represents concentration of the colored product species, l represents the standard 1.00 cm path length of the solution across the cuvette and ε is the extinction coefficient, a constant for the system. ε = 1.88 x 104 M-1 cm-1 for this reaction mixture. The ratio of product concentration after reaction ([P]initial = 0; [P]final is determined as above so Δ[P] = [P]final – 0 = [P]final) over time (15 minutes) is reaction velocity, v. Remember also that the final addition of 2.5 mL of KOH dilutes the product formed during the reaction period. The product was more concentrated in the reaction mixture by a factor of 7.5 mL/5.0 mL. Thus:
v = A410(corr) (7.5 mL/5.0 mL) in units of M/min.
(15 min) (1.88 x 104 M-1cm-1) (1.00 cm)
Calculate reaction velocity for each experimental sample in parts 2, 3A and 3B.
Calculate the final concentration of the phosphate inhibitor in the reactions of part 3B.
Prepare the following graphs of your data. Show sample calculation(s), where necessary, on the reverse of the graph page.
- Effect of Enzyme Concentration on the Rate of Reaction (v vs. [E]). Plot reaction velocity as a function of “relative enzyme” from 0.1 to 1.0. Discuss the relationship that you observe.
- The Michaelis-Menton (hyperbolic) Graph (v vs. [S]). Show the data for both the uninhibited reaction and the reaction in the presence of inhibitor on the same axes. Clearly distinguish between the two curves and include a key that indicates which curve represents which experiment.
What would have been the effect on the graph of the uninhibited reaction if you had used a) half as much enzyme? b) …twice as much enzyme?
- The Lineweaver-Burk (linear double reciprocal) Graph (1/v vs. 1/[S]). Again, show the data for both the uninhibited reaction and the reaction in the presence of inhibitor on the same axes. Clearly distinguish between the two.
From the Lineweaver-Burk plot of the uninhibited reaction, determine Vmax and KM. Show the calculations clearly on the back of the graph.
Discuss the nature of the inorganic phosphate’s inhibition. Base your discussion on the relationship of the two Lineweaver-Burk plots and on the chemical nature of phosphate ion with respect to the substrate.