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Exercise 7. Enzyme Kinetics Objectives This experiment is designed to familiarize the student with basic enzyme kinetics and analysis of initial rate data. The enzyme chosen for study is -galactosidase, a large enzyme that is fairly stable and easily assayed. The substrate is chosen such that a colored product results upon hydrolysis. This product can then be measured spectrophotometrically. An understanding of the Michaelis-Menten equation is important. Pre-lab 1) Prior to lab, come ready with a table listing all 18 reactions, and the amount of enzyme solution (0.25 g/ml), Z buffer, and ONPG solution (0.5 mg/ml) that each cuvette will contain. 2) In the procedure text, the Michaelis Menten equation is derived assuming a rapid equilibrium step. Show the derivation for an alternative form, instead assuming a pseudosteady state. Experimentally, how could one test verify the accuracy of this assumption if direct observation of the intermediate is not available? 3) What order is the MM equation with respect to enzyme? How might this change if the enzyme were immobilized on a solid support instead of in solution and why? 4) Consider an enzymatic reaction product that is an inhibitor of the free enzyme. Where would you expect to find such a system and why? Background Consider the following simple mechanism for an enzyme (E)-catalyzed reaction in which a substrate (S) is converted to a product (P): K k s 2 E S E P E S Eq (1) In the first step, the enzyme and substrate reversibly bind to form an enzyme-substrate complex. In the second step, the substrate is converted to a product and released from the enzyme. (The kinetics of product release are neglected here.) If we assume that the first step is in equilibrium, the following expression can be obtained: Eq (2) Ks E S E S 40 The constant Ks is the ratio of the reverse (k-1) and forward (k1) rate constants for step one, also known as the dissociation constant. In a closed system, the rate of product formation is then derived from the second step to give the following expression: k E S dP 2 dt Eq (3) The ES complex is not easily measured and is often a very short-lived intermediate. Therefore, we need to obtain an expression for [ES] in terms of other, measurable variables. Eq (2) expresses the complex concentration in terms of [S] and [E], the free enzyme concentration. However, the free enzyme concentration is also difficult to determine. Remembering that the enzyme only participates in the reaction as a catalyst, a mass balance around the enzyme yields: E E E S 0 Eq (4) where [E]0 is the initial, or total enzyme concentration. Now, Eqs (2) and (4) can be combined to give an expression for [ES] in terms of the substrate concentration and the total enzyme concentration. This expression can be inserted into Eq (3) to give expression for the rate of production formation, or the velocity of the reaction: v k E S V S dt K S K S dP 2 max 0 s m Eq (5) where Vmax is the maximum reaction velocity and Km is the Michaelis constant. Note that Km is equal to the dissociation constant, Ks, for the simple mechanism of Eq (1) (but this is not always the case for enzyme kinetics). The final expression given in Eq (5) is known as the MichaelisMenten equation for enzyme kinetics. The parameters Vmax and Km give important insight into the behavior of the enzyme for a particular set of experimental conditions. These parameters can be determined graphically, given the reaction velocity as a function of the substrate concentration and the initial enzyme concentration. Vmax is further defined as: Vmax kcat E 0 Eq (6) where kcat is the catalytic rate constant (or turnover number). For this simple mechanism, kcat is equal to k2, the first order rate constant for the second step of Eq (1). More complicated reaction mechanisms may also yield a rate expression in the Michaelis-Menten form; kcat may then be a 41 function of several rate constants, each representing a different step in the reaction sequence. The Michaelis constant, Km, is also a more complex function of rate constants in such cases. In this exercise, we will investigate the kinetics of -galactosidase, a large enzyme (monomer molecular weight 116,000) that is fairly stable and easily assayed. The enzyme galactosidase hydrolyzes -D-galactosides, such as lactose. It can be easily assayed with chromogenic substrates, which contain a reactant that releases a colored product when hydrolyzed. One such material is ONPG (o-nitrophenyl--D-galactopyranoside). This colorless compound is converted to galactose and o-nitrophenol in the presence of -galactosidase. Onitrophenol is yellow and can be measured by its absorption at 420 nm. The reaction is halted by the addition of concentrated sodium carbonate, which shifts the pH to 11, inactivating the galactosidase. The amount of o-nitrophenol produced is proportional to the amount of enzyme present under conditions of excess ONPG, permitting an assay of the amount of enzyme present in a sample. In this exercise however, we will determine the kinetics of hydrolysis by varying the ONPG concentration and measuring the initial rate of formation of o-nitrophenol. PROCEDURE 1. A 0.5 g/ml solution of -galactosidase in Z buffer will be given to you. Stock ONPG solution (0.5 mg/ml) will also be provided for use as the substrate. Z buffer contains a small amount of 2-mercaptoethanol, a chemical which is toxic and can be absorbed through the skin. It also smells terrible! Wear gloves at all times during this exercise and be careful not to spill reagents. 2. You will be running 18 reactions simultaneously. Three will be controls, and the rest will be your enzymatic reactions. In this experiment, you will be varying the substrate concentration by changing the proportion of substrate solution added to the reaction mixtures. There will be a total of 13 different substrate concentrations. There will also be two additional reactions run at high substrate concentration in which the enzyme concentration is varied. 3. In one cuvette you will have 1.0 ml of Z buffer equilibrated to the assay temperature (all assays will be run at room temperature), to which is added 0.2 ml of ONPG solution. Later, 100 L of the enzyme solution will be added to start the reaction, for a total reaction volume of 1.3 ml. This is your 1X substrate concentration. 4. Into subsequent cuvettes, add appropriate amounts of Z buffer and ONPG solution, such that you arrive at the following substrate concentrations: 0.1X (ten-fold dilution), 0.2X, 0.3X, 0.4X, 0.5X, 0.6X, 0.8X, 1.2X, 1.4X, 1.6X, 2X, and 3X. (Keep in mind that the amount of enzyme added to these reactions will be the same--100 L.) The total volume for each of these will be 1.3mL 5. Make three control cuvettes at three substrate concentrations: 1X, 0.2X, and 2X. These cuvettes should contain 100 L of Z buffer in place of enzyme. For example, the 1X control should contain 1.1 ml Z buffer and 0.2 ml ONPG solution. 42 6. The final two cuvettes should contain 0.2X and 2X of the enzyme solution and 3X of the ONPG solution. Add the Z buffer and substrate first. All cuvettes should now contain the correct proportion of Z buffer and ONPG solution. 8. As quickly as possible, add the appropriate amounts of enzyme solution to the cuvettes. Stir each well with your pipette tip and start the timer after the addition of the first. 9. You will be adding 150uL of Na2CO3 to each cuvette to halt the reaction. 0.1X thru 0.8X OPNG, and also the reactions from step 6. should be stopped after 2 minutes. 1X thru 3X OPNG should be stopped after 10 minutes. Mix the Na2CO3 thoroughly with your pipette tip 10. Blank the spectrophotometer with a cuvette containing 1.3 ml Z buffer plus 0.15 ml Na2CO3 solution. Read the OD at 420 nm for all 18 cuvettes. One student should take measurements while the other records the data. Guideline for Analysis & Conclusions Section The extinction coefficient of o-nitrophenol is 4500 M-1cm-1; the molecular weight of ONPG is 301.3. Use concentration units of M (mol/L) and time units of seconds in your calculations. Explain the mechanism of this enzyme for hydrolysis of ONPG. Be specific with respect to the important amino acids on the enzyme, the role of crucial salts and water. Why would one want to stop those reactions with lower OPNG earlier than those with higher OPNG? Prepare Lineweaver-Burke and Eadie-Hofstee plots for the data obtained during the lab. Use these plots to determine values of Vmax and Km. Also perform a non-linear regression using mathematical software (Mathcad, KaleidaGraph, etc.) to determine another set of values for vmax and Km. Create a velocity vs. concentration plot showing the data you obtained and the three curves generated by incorporating your values for vmax and Km into the Michaelis-Menten equation. Remember, data is plotted as x’s or dots on the graph, and equations are represented by lines. Comment on the determined value of KM. What values (range) would you expect for an enzyme such as -galactosidase for this substrate relative to (e.g.) lactose? Determine the value of kcat from the concentration of the enzyme in your solutions and the value of vmax. What assumptions did you make concerning the integrity of the enzyme preparation? 43 EQUIPMENT AND REAGENTS Z buffer: 16.1 gm Na2HPO47H2O 5.5 g NaH2PO4.H2O 0.75 g KCl 0.246 g MgSO47H2O 2.7 ml 2-mercaptoethanol Adjust to pH 7.0 and bring to 1.0 liter with distilled water Do not autoclave. -galactosidase, 0.5 g/ml in Z buffer o-nitrophenyl--D-galactosopyranoside (ONPG), 0.5 mg/ml 1.0 M Na2CO3 solution Spectrophotometer (OD at 420 nm) Spectrophotometer cuvettes, 1 cm width (path length) REFERENCES Bailey, J.E., and Ollis, D.F. (1986) Biochemical Engineering Fundamentals, 2nd Ed., pp. 89108, McGraw Hill, Inc., New York. Blanch, H.W., and Clark, D.S. (1996) Biochemical Engineering, pp. 1-1-18, Marcel Dekker, Inc., New York. Fogler, F.S. (1992) Elements of Chemical Reaction Engineering, 2nd Ed., pp. 359-364, Prentice Hall, Englewood Cliffs, NJ. 44