1. Rules for deriving rate laws for simple systems 1.Write reactions involved in forming P from S 2. Write the conservation equation expressing the distribution of total enzyme concentration [E] total among the various species 3.Write the velocity dependence equation, summing all the catalytic rates constants multiplied by the concentration of the respective product-forming species. 4.Divide the velocity dependence equation by the conservation equation. 5.Express the concentration of each enzymic species in termsof free enzyme concentration & substitute 6.Substitute & algebra

2. 1. Write reactions involved in forming P from S E + S  ES  E + P k 1 k 3 k 2 For a simple 1 step reaction, no inhibitor

3. 2.Write the conservation equation expressing the distribution of total enzyme concentration [E] total among the various species [E] T = [E] + [ES]

4. 3.Write the velocity dependence equation, summing all the catalytic rates constants multiplied by the concentration of the respective product-forming species. v = k 3 [ES]

5. 4. Divide the velocity dependence equation by the enzyme conservation equation. [E] T [E] + [ES] v = k 3 [ES]

6. 5. Express the concentration of each enzymic species in terms of free enzyme concentration & substitute [ES] = k 1 [S]  [E] - k 2 [ES] - k 3 [ES] k 1 [S]  [E] = (k 2 + k 3 )[ES] d[ES] dt @ steady state = 0,  rate of formation = rate of breakdown [ES] = k 1 [S]  [E] - (k 2 + k 3 )[ES] Collecting terms: k 1 [S]  [E] k 2 + k 3 [ES] = Solve for [ES]; k 2 + k 3 k 1 Define K M = then [S]  [E] Km = [ES] Make steady state assumption:

7. 6. Substitute & Algebra v = k 3 [ES] [E] T = [E] + [ES] [S]  [E] Km v = k 3 [E] T = [E]+ [S]  [E] Km Divide by [E] [S] Km v = k 3 [E] T = 1+ [S] Km [S]  [E] Km = [ES]

8. Substitute & Algebra (cont’d) Multiply by Km [S] Km v = k 3 [E] T = 1+ [S] Km [S] v = k 3 [E] T = K M + [S] Define V max = k 3 [E T ] [S] v = V max K M + [S]

9. Expressing With Measurable Quantities The total concentration of enzyme [E] T can be determined, but it is very difficult to measure [E] and [ES]. Steady state rate equation: v = k 1 [E][S] = k -1 [ES] + k 2 [ES] By defining the Michaelis Constant (K M ) = (k -1 + k 2 )/k 1 then rearranging and substituting, [ES] can be expressed as: [ES] = [E] T [S]/(K M + [S]) The initial velocity (v o ) = k 2 [ES] = k 2 [E ] T [S]/(K M + [S]) The key point is that [E] T and [S] are measurable.

10. Importance of Initial Velocity In reality, v o is velocity measured before 10% of S is consumed. Before [ES] has built up- can assume reverse reaction is negligible. Working with v o minimizes complications  Reverse reactions  Inhibition of reaction by product

11. Implications of M-M Equation V o = V max [S] / (K M + [S]) 1. From the M-M equation, when [S] = K M, v o = V max /2.  This means that low values of K M imply the enzyme achieves maximal catalytic efficiency at low [S]. 2. The catalytic constant, k cat = V max / [E] T  k cat is also called the turnover number of the enzyme, i.e. the number of reaction processes (turn-overs) that each active site catalyzes per unit time.  Turnover numbers vary over many orders of magnitude for different enzymes in accord with need.

12. Implications of M-M Equation (cont.) 3. When [S] << K M, very little ES is formed. Under these conditions: v o ~ (k cat /K m )[E][S]  The k cat /K m term is a measure of the enzyme’s catalytic efficiency.  The upper limit to k cat /K m is k 1, I.e. the decomposition of ES to E + P can occur no more frequently than ES is formed.  The most efficient enzymes have k cat /K m values near the diffusion-controlled limit of 10 8 -10 9 M -1 s -1. They will catalyze a reaction almost every time a substrate is bound in the active site- catalytic perfection!!!

13. Practical Summary- Analysis of Kinetics  When [S] << K M, the reaction increases linearly with [S]; I.e. v o = (V max / K M ) [S]  When [S] = K M, v o = V max /2 (half maximal velocity); this is a definition of K M : the concentration of substrate which gives ½ of V max.  When [S] >> K m, v o = V max

14. Practical Summary - Practical Summary - V max and K m  K M gives an idea of the range of [S] at which a reaction will occur. The larger the K M, the WEAKER the binding affinity of enzyme for substrate.  V max gives an idea of how fast the reaction can occur under ideal circumstances.  K cat / K M gives a practical idea of the catalytic efficiency, i.e. how often a molecule of substrate that is bound reacts to give product.

15. Inhibition of Enzymes Enzyme Inhibitor: a molecule that reduces the effectiveness of an enzyme. Mechanisms for enzyme inhibition  Competitive inhibition- substrate analogs bind in the active site reducing availability of free enzyme  Uncompetitive inhibition- inhibitor binds to the substrate-enzyme complex and presumably distorts the active site making the enzyme less active  Mixed or non-competitive inhibition- combination of competitive and uncompetitive inhibition  Inactivator- irriversible reaction with enzyme

16. Enzyme Inhibitors in Medicine Many current pharmaceuticals are enzyme inhibitors (e.g. HIV protease inhibitors for treatment of AIDS- see pages 336-337). An example: Ethanol is used as a competitive inhibitor to treat methanol poisoning.  Methanol is metabolized by the enzyme alcohol dehydrogenase producing highly toxic formaldehyde.  Ethanol competes for the same enzyme.  Administration of ethanol occupies the enzyme thereby delaying methanol metabolism long enough for clearance through the kidneys.