Biochemistry 412 Enzyme Kinetics I March 29 th, 2005

Reading: Mathews & van Holde, Biochemistry, Benjamin/Cummings Publishing Co., Redwood City, CA, pp in 1990 edition (or equivalent pages in a later edition) Other (optional) resources: >>> And special thanks for this lecture goes to Dr. Gabriel Fenteany, Department of Chemistry, University of Illinois at Chicago ( whose slides I liberally borrowed!

Enzymes Are Uniquely Powerful Catalysts Enzymes are proteins that can accelerate biochemical reactions often by factors of 10 6 to ! This is much higher than chemical catalysts. Enzymes can be extremely specific in terms of reaction substrates and products. Enzymes catalyze reactions under mild conditions (e.g. pH 7.4, 37ºC). The catalytic activities of many enzymes can be regulated by allosteric effectors.

Triose Phosphate Isomerase For example:

And…

Chemical Kinetics

Irreversible First-Order Reactions A  B v = d[B]/dt = -d[A]/dt = k[A] (k = first-order rate constant (s -1 )) Rearrange: d[A]/[A] = dln[A] = -kdt Integrate and express [A] as a function of time (t): [A]/[A] o = e –kt or [A]= [A] o e –kt ([A] o = initial concentration) k

Reversible First-Order Reactions A B v = -d[A]/dt = k 1 [A] - k -1 [B] At equilibrium: k 1 [A] eq - k -1 [B] eq = 0 [B] eq /[A] eq = k 1 /k -1 = K eq k1k1 k -1  

Second-Order Reactions 2A  P v = -d[A]/dt = k[A] 2 A + B  P v = -d[A]/dt = -d[B]/dt = k[A][B] (k = second-order rate constant (M -1 s -1 )) Change in [A] with time: 1/[A] = 1/[A] o + kt k k Note: third-order reactions rare, fourth- and higher-order reactions unknown.

Free Energy Diagrams K = e –∆Gº/RT For A  A ‡ [A] ‡ /[A] o = e –∆Gº‡/RT [A] ‡ = [A] o e –∆Gº‡/RT K = equilibrium constant ‡ = transition state [A] ‡ = concentration of molecules having the activation energy [A] o = total concentration –∆Gº ‡ = standard free energy change of activation (activation energy)

Relationship of Reaction Rate Constant to Activation Energy and Temperature: The Arrhenius Equation Reaction rate constant (k) determined by activation energy (∆Gº‡) and temperature (T) and proportional to frequency of forming product (Q = k B T/h, where k B = Boltzmann’s constant, h = Planck’s constant): k = Q e -  G°‡/RT = (k B T/h) e -  G°‡/RT (  G =  H - T  S) k = Q e  S°‡/R  e -  H°‡/RT k = Q´e -  H°‡/RT (Q´ = Q e -  S°‡/R ) ln k = ln Q´ -  H° ‡ /RT L-malate  fumarate + H 2 0 ln k

The Transition State Energy Barrier Opposes the Reaction in Both Directions K = k 1 /k -1 K = ( Q e -  G 1 °‡/RT )/( Q e -  G -1 °‡/RT ) K = e -(  G 1 °‡ -  G -1 °‡)/RT ∆Gº =  G 1 º‡ -  G -1 º‡ K = e –∆Gº/RT Equilibrium constant K says nothing about rate of reaction, only free energy difference between final and initial states.

Effect of a Catalyst on Activation Energy Catalysts do not affect G A (initial) or G B (final) and so do not affect overall free energy change (∆Gº = G B - G A ) or equilibrium constant K. Equilibrium concentrations of A and B still determined solely by overall free energy change. Catalysts only affect ∆Gº ‡, lowering the activation energy. They accelerate both the forward and reverse reaction (increase kinetic rate constants k 1 and k -1 ).

Intermediate States in Multi-step Reactions

Q: How do enzymes work? A: A number of ways: “propinquity”, catalytic groups at active site, catalytic metals at active site, etc. (see assigned reading). >>> however, primitive enzymes may have behaved like catalytic antibodies, which can accelerate reactions merely by binding to and increasing the relative concentration of the transition state ([A] ‡ = [A] o e –∆Gº‡/RT effect).

Enzyme Kinetics

Types of Enzymes 1.Oxidoreductases catalyze oxidation-reduction reactions. 2.Transferases catalyze transfer of functional groups from one molecule to another. 3.Hydrolases catalyze hydrolytic cleavage. 4.Lyases catalyze removal of a group from or addition of a group to a double bond, or other cleavages involving electron rearrangement. 5.Isomerases catalyze intramolecular rearrangement. 6.Ligases catalyze reactions in which two molecules are joined.

Two Models for Enzyme-Substrate Interaction

Induced Conformational Change in Hexokinase

Triose Phosphate Isomerase Mutational analysis to study enzyme mechanism: Site-directed mutagenesis - substitution mutation at specific position in sequence Deletion mutation

Free Energy Barrier to the Glyceraldehyde-3-Phosphate Dihydroxyacetone Phosphate Reaction Mutant in flexible loop that closes over active site.

OK, but now let’s talk about kinetics….

The Effect of Substrate Concentration on Reaction Velocity Q: for a fixed amount of enzyme, what happens if you keep adding more and more substrate?

The Steady State in Enzyme Kinetics

Michaelis-Menten Kinetics (1) v = k 2 [ES], if this is the rate-limiting step* [Enzyme] total = [E] t = [E] + [ES] How to solve for [ES]? 1. Assume equilibrium, if k -1 >> k 2 : K S = k -1 /k 1 = [E][S]/[ES] or 2. Assume steady state: d[ES]/dt = 0 [*Note: v is always measured as an initial rate!] (Michaelis and Menten, 1913) (Briggs and Haldane, 1925) E = free enzyme, S = substrate ES = enzyme-substrate complex P = product

Michaelis-Menten Kinetics Continued (2) Because of steady state assumption: d[ES]/dt = k -1 [ES] + k 2 [ES] - k 1 [E][S] = 0 So: k 1 [E][S] = k -1 [ES] + k 2 [ES] Rearranging: [ES] = (k 1 /(k -1 + k 2 ))[E][S] Substituting (the “M” constant* = K M = (k -1 + k 2 )/k 1 ): [ES] = ([E][S])/K M So: K M [ES] = [E][S] *Note: Briggs & Haldane came up with this, but they lost out when it came time to name things!

Michaelis-Menten Kinetics Continued (3) Substituting ([E] = [E] t - [ES]): K M [ES] = [E] t [S] - [ES][S] Rearranging: [ES](K M + [S]) = [E] t [S] So: [ES] = [E] t [S]/(K M + [S]) Now we can substitute for [ES] in the rate equation v = k 2 [ES], so…

The Michaelis-Menten Equation (4) v = k 2 [E] t [S]/(K M + [S]) or v = V max [S]/(K M + [S]) (since V max = k 2 [E] t )

At [S] « K m, V o is proportional to [S] At [S] » K m, V o = V max

A Lineweaver-Burk Plot

An Eadie-Hofstee Plot

Multi-step Reactions E + S ES  EP  E + P v = k cat [E] t [S]/(K M + [S]) k 2 k 3 k1k1 k -1   (k cat = general rate constant that incorporates k 2 and k 3 )

k cat, K M, and k cat /K m : Catalytic Efficiency => “Perfect enzyme” Diffusion-controlled limit: M -1 s -1 Substrate preferences for chymotrypsin