ENZYME ACTIVITY AS A FUNCTION of ph

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1 ENZYME ACTIVITY AS A FUNCTION of ph Paul Krause Department of Chemistry University of Central Arkansas Conway, Arkansas Copyright Paul F. Krause All rights reserved. You are welcome to use this document in your own classes but commercial use is not allowed without the permission of the author. Creation Date: August 1997-March 1998 Modification Date: INTRODUCTION: A simple model of ph regulation of enzyme activity is to consider that an enzyme can exist in three degrees of protonation, In addition, it is assumed in this exercise that the substrate is in great enough excess such that the equilibrium constant for the protonation of the free enzyme is the same as for the enzyme-substrate complex. Only one form of the enzyme is capable of binding substrate and catalyzing the reaction, and in this case it is EH +. The equilibria described below illustrate this model. MODEL: K 1 K 2 EH 2 +2 = EH + + H + = E + 2 H + GOAL of the Exercise: The exercise should help you understand that optimal activity of an enzyme is greatly influenced by the acidity of the medium. You will see that the values of the equilibrium constants are significant in determination of the optimal ph for reaction rate. Additionally, various aspects of Mathcad will be used in assisting you in looking at the chemistry but also in becoming more proficient in the use of the software. You will be required to use both symbolic and numerical processing as well as presenting results in graphical formats. NOTES to users 1. You should pay particular attention to text in red. As part of the completing the tutorial, these statements/questions are to be answered and included in your report. There is space in the tutorial to answer the requests

2 2. On a number of occasions throughout the document you will notice incomplete (e.g. "missing operand") statements. These are designed to assist you in deciding an approach and are to be completed in sequential fashion. 3. A goal is for you to turn in a completed document either on disc or in "hard copy". Therefore you should be clear in what you are doing by imbedding text so a reader can follow your logic. The key stroke CTRL F9 allows for adding a line at the cursor while CTRL F10 removes a line Beginning the Exercise: 1. Write the equilibrium constant expressions K 1 and K 2. Note the use of the symbolic equals sign. (It is convenient to use EH2, EH1, and E as the three enzyme forms). K 1 K 2 2. Generate a set of statements such that you input ph values (~1000 values) over a 0-15 ph range: The following MATHCAD statements allow for input of ph values: N i ph i 3. Because of the form of the Keq expressions we need the corresponding hydrogen ion concentrations H i

3 The fraction of an enzyme in a specific enzyme state can be derived from the equilibrium constant expressions. They will be derived below. 4. Do the derivations symbolically i.e. solve for EH2, EH1, and E. These represent the fraction of enzyme in the di-protonated, mono-protonated and free states. a. Copy and paste K1 and K2 b. Solve for EH2 and EH c. Solve for the fractions symbolically i.e. f(eh2), f(eh1), and f(e) first as a function of the various enzyme fractions, then substitute for EH2 and E, giving the fractions in terms of H and the equilibrium constants K1 and K2. (Set up the general expressions, do some substituting by cutting and pasting, and then let the symbolic processor simplify the expressions.) f( EH2) upon substituting yields: which simplifies to: f( EH1) upon substituting yields: which simplifies to:

4 f( E) upon substituting yields: which simplifies to: We need to generate data for the various enzyme fractions as a function of hydrogen ion concentration. This will involve using the variable equals rather than the symbolic equal you have been using. 5. Generate variable expressions for the fractions as a function of H i, K1 and K2 and even though Mathcad doesn't know what K1 and K2 are as of yet, we'll take care of that later with a universal "equals" so we can look at generated curves as a function of ph and K. Copy and paste the appropriate portion of the fractions above and adjust them to variable fractions (e.g. EH1 i ) EH2 i EH1 i E i

5 It is instructive to view the above enzyme concentrations as a function of ph for a variety of values of K1 and K2. As this is accomplished below pay attention to the shift in the ph for maximum effectiveness of EH1 and also its relative peak height as K1 and K2 are adjusted. 6. Prepare XY plots of the fraction of the three enzyme concentrations as a function of ph for various values of K1 and K2. Generate hard copy plots for at least three different combinations of K1 and K2. (Remember these are weak acid equilibria). Make sure you are printing the page you want. Also, K1 and K2 are used later in the document. Make sure those equations are "toggled-off" (see Math Menu). Normally K1 is several orders of magnitude greater than K2, but it is interesting to note what happens to the curves when K2~K1 and K2>K1. Also note the use of the universal equals sign. Why is it employed? K1 K2 Before continuing, did you title the plot and label the various traces. You might also want to adjust the presentation to make it more eye-appealing.

6 Because we are assuming only the mono-protonated form is capable of binding substrate, the reaction rate is simply the maximum reaction rate multiplied by the fraction of the enzyme in the singularly protonated state. 7. Assign a maximum relative reaction rate (v max ) and then generate an equation which calculates the reaction rate (v) as a function of [H + ]. Make XY plots and produce hard copies of at least three combinations of K1 and K2. Toggle the K1 and K2 values used in part #6 "off" and activate the ones on this page. v max v i The reaction rate as a function of ph. Vary K1 and K2 and observe the change in the peak, K1 K2

7 The optimum ph for enzyme activity is obviously the maximum in the of the rate versus ph curve. It is not so obvious that value can be found based on the following equation. H opt. K1 K2 8. Show that the H opt expression is correct. Hint: you already have an expression for EH1. You are interested in when it is a maximum. Use the symbolic processor to find the relative maximum in the curve. EH1( H) What is the optimum ph for the current values of K1 and K2?

8 A range of ph values can be determined where the EH + fraction is at least half of its maximum value. The upper and lower limit can be found by solving for the roots of a quadratic in [H] 9. Generate the quadratic described above. A few hints: (a) Generate an expression for the maximum fraction of EH1 and multiply it by 1/2 (b) Set the expression symbolically equal to the general expression for the fraction of EH1 (c) Simplify (b) (d) Cut and paste (c) until it is a polynomial equal to zero (e) Simplify (d) and use the "Collect on Subexpression" option of the Symbolic Menu to generate the quadratic

9 The quadratic above can be now be solved for the [H+] concentration range where the activity is at least half of the maximum value. 10. Copy the result from #9 and substitute the expression for [H] opt. Solve for the roots of the quadratic. 11. Use values of K1 and K2 again to determine the ph range. Deactivate any K1 and K2 from above (parts #6 and #7), and input values below. Again generate three sets of ranges for different Ks. F( H) which has solutions: F( H) = K1 K2

10 ph low ph high 12. A TIME of DEBRIEFING - Answer the following questions in your report: a. Recall the plots of the various enzyme forms as a function of ph. Interpret them in terms of the influence of ph on enzyme activity. b. The rate versus ph plot is a classic curve found in many texts. Interpret in the light of only one form of the enzyme capable of binding substrate. c. If the model is correct, what must be happening to the enzyme as the equilibrium shifts from left to right i.e. from EH 2+ to EH + to E. d. Implicit in this model is the assumption that mechanistically we are dealing with large substrate concentrations. How is this deduced from the derivations? e. Derive a more complex mechanistic scheme where the substrate (S) concentration is not large. In this case the equilibrium constant for the free enzyme protonation may not be the same as for the enzyme substrate complex although one can still assume only one form of the enzyme is biologically active. Use as a starting point k 1 k 2 S + EH + = EH + S ---> EH + + Products k -1 Although not part of this assignment, it is of importance to understand that the mechanism which you are to generate yields an expression for the reaction rate which is of the same form as the classic Michaelis-Menten Equation. The difference is that the Michaelis-Menten constant and v max are ph dependent. (This would be an excellent special project). f. Find an experimental plot of v max vs. ph for histidase acting on histidine. Compare it to the plot in #7 giving similarities and differences. What is the structure of histidine? Comment on the reaction being catalyzed, and as much as possible comment on structural characteristics of histidase. References: Cantor,C.R.;Schimmel, P.R. Biophysical Chenistry, Part III: The Behavior of Biological Macromolecules, Marshall, A.G. Biophysical Chemistry: Principles, Techniques, and Applications, Acknowledgment: PFK acknowledges the National Science Foundation for support of the 1997 NSF-UFE Workshop on "Numerical Methods in the Undergraduate Chemistry Curriculum Using the Mathcad Software" and its organizers Jeff Madura, Andrzej Wierzbicki and Sidney Young, Department of Chemistry, University of South Alabama, Mobile, AL

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