5.07 Lecture 6Enzyme Kinetics Unraveling metabolic pathways and their regulation

For the model reaction S P, the time course of product formation is shown below.

v = -d[S]/dt = d[P]/dt = k[S]

Initial rates. To simplify analysis, all our discussions of enzyme kinetics will be limited to initial rates (vo; also referred to as initial velocities), that is, up to ~10% of the substrate-to-product conversions. During this initial time period, [S] [S]o.

Rates (v) of enzymatic and non-enzymatic reactions as a function of time, for example, product concentration ([P]) versus time (t).

How do kinetics of enzymatic reactions differ from those of uncatalyzed chemical reactions?Initial rate of product formation (vo) vs. [S]o

For enzymatic reactions, the initial rate (vo) of the reaction is directly proportional to [S]o at low substrate concentrations, as it is for the non-catalyzed reaction. However, at high [S]o, the enzymatic reaction, in contrast to an uncatalyzed one, becomes zero order in [S]o; i.e., vo is independent of the [S]o. Since the active sites of the enzymes are fully saturated with S, further increasing its [S]o has no effect on the reaction rate.

How can we mathematically describe these experimental observations?

Michaelis-Menten Equation k-1 k2 E + S ES E + P k1 We need a description of the overall rate of the reaction, i.e., the concentration of the product formed per unit of time: v = d[P]/dt = k2[ES] In general, if a step is irreversible, then choosing this step to describe therate of product formation simplifies the algebra. The answer is the sameno matter what step we choose. The [ES] usually is not experimentallymeasurable. One, therefore, needs to define [ES] in terms ofexperimentally measurable parameters, namely [E]o and [S]o.

Since the enzyme-substrate complex ES is formed in one reaction and decomposed in two, then d[ES]/dt = k1[E][S]o k-1[ES] - k2[ES]

To integrate this differential expression with its three variables, additional simplifications need to be made.

Michaelis-Menten Equation E + S ES E + PES is a non-covalent enzyme-substrate complexConditions and assumptions:1. In a typical experiment, [S]o >> [E]o: the [S]o is mM to M and the [E]o is nM to M.The reaction is run under conditions were only a few % of substrate (under ~ 10%) is converted to product (initial velocity conditions). Therefore, v = vo and [S] = [S]o Substrate conservation equation: [S]o = [S] + [ES] [S]o The [S]o can be experimentally measured (UV/Vis absorbance spectroscopy if S has a chromophore or scintillation counting if the substrate is made radioactive). Enzyme conservation equation: [E]o = [E] + [ES] The [E]o is experimentally measurable. Use the A280; see website http://ca.expasy.org/ under Expasy proteomics tools, one can estimate the extinction coefficient of a protein whose sequence is known. k-1

k1

k2

Michaelis-Menten EquationSteady-state assumption (generally valid under the conditions outlinedabove and greatly facilitating solution to the problem): d[ES]/dt = 0Therefore, after a brief time period called the pre-steady state(milliseconds, see lag phase below), the rate of formation of ES is equal tothe rate of its disappearance: k1[E][S]o = k-1[ES] + k2[ES]

Michaelis-Menten EquationNow rearrange the expression for [ES] in terms of experimentally measurable parameters given that [E]o = [E] + [ES] : [ES] = k1[E]o[S]o/(k-1 + k2 + k1[S]o)

Using the rate expression v = k2[ES] gives vo = k1k2[E]o[S]o/(k-1 + k2 + k1[S]o) where vo is experimentally measurable, [E]o and [S]o are known, and the [S]o can be varied.

We thus have two known variables ([E]o and [S]o) and three unknown rate constants (k1, k-1, and k2). We cannot solve the equation for the individual rate constants and need to regroup them into new constants, Vmax and KM, as shown on the next two slides.

Michaelis-Menten Equation Vmax Vmax = k2[E]o = kcat [E]o In this case, k2 (or, in general, kcat for more complex enzymatic reactions) is the turnover number of the enzyme. It is the number of reaction processes that each enzyme active site catalyzes per unit time. kcats units are time-1, usually s-1. kcat is in general composed of a number of first order rate constants that cannot be evaluated individually in the steady-state analysis. Attaining Vmax implies that all the enzyme active sites are tied up with the substrate and, therefore, the reaction cannot go any faster under given conditions (e.g., of pH and temperature).

Michaelis-Menten Equation KM = (k-1 + k2)/k1

KM (Michaelis-Menten constant; alternatively denoted as Km) is the [S]o at which the vo reaches 1/2 Vmax. The units of KM are those of concentration, usually M or mM. If k-1 >> k2, then KM = k-1/k1 = Kd; Kd is the thermodynamic dissociation constant reflecting the affinity of the substrate to the enzyme. In most cases, however,the KM is not the measured thermodynamic affinity of the E for the S; rather, as with kcat, it is also composed of a number of rate constants. KM is unique for each given enzyme pair under a particular set of conditions (such as pH and temperature). The smaller the KM value, the lower is the substrate concentration at which the enzymatic reaction rate reaches the Vmax, that is, the maximal catalytic efficiency under given conditions (e.g., of pH and temperature).

Michaelis-Menten Equation

Substituting these new constants into the above equation vo = k1k2[E]o[S]o/(k-1 + k2 + k1[S]o) gives the Michaelis-Menten equation:

vo = Vmax[S]o/(KM + [S]o) or

kcat and kcat/KM are the two most important enzyme/substrate kinetic parameters. kcat: turnover number, tells us how good our enzyme is as a catalyst.

kcat/KM: proficiency constant or specificity constant, tells us how near to perfection our enzyme has evolved as a catalyst and how well it likes one substrate over another.

Why are these the important kinetic parameters?

Michaelis-Menten kinetics vo = Vmax[S]o/(KM + [S]o)

Let us look at the two extreme scenarios:1. If [S]o , then vo Vmax Reaction is zero-order in substrate. Tells us how good our enzyme is as a catalyst.2. If [S]o 0, then vo (k2/KM)[E]o[S]o. Under these conditions, the reaction is first order in both substrate and enzyme. k2/KM is a second-order rate constant called the specificity (or proficiency) constant. This constant is upper-limited by diffusion, i.e., when the enzyme and the substrate need to find each other in aqueous solution. The diffusion-controlled rate constant in enzymatic reactions is typically between 106 and 109 M-1s-1. The physical step is rate-limiting, and the enzyme has reached catalytic perfection.Initial rate of product formation (vo) Initial substrateconcentration ([S]o)

Accounts of Chemical Research (2001) article by Wolfendenkcat/KM values are relatively similar for enzymatic reactions as compared to uncatalyzed reactions

Evaluation of kinetic parameters is based on fits to Michaelis-Menten equation vo = Vmax[S]o/(KM + [S]o) Lineweaver-Burk plots provide a method, often used in enzyme kinetics, to linearize the vo vs. [S]o data and quickly assess the Vmax and KM values and visually present kinetic data:

1/vo = KM/Vmax(1/[S]o) + 1/Vmax [ y = mx + b ] Since Vmax= kcat[E]o, if [E]o is known,then kcat = Vmax/[E]o

Limitations of steady-state enzyme kineticsBy means of steady-state enzyme kinetics, the scheme E + S ES E + P cannot be distinguished from that where ES comprises ES1 ES2 ES3 ESn i.e., when there are several enzyme intermediates in equilibrium with each other. Therefore, the steady-state kinetic analysis of an enzyme-catalyzed reaction cannot unambiguously establish its mechanism (but can rule out a mechanism whose predictions contradict experimentally observed kinetic data).

Nor can the steady-state kinetic analysis of an enzyme-catalyzed reaction determine k1, k-1, or k2 individually. Pre-steady-state methods are required for that. For such more complex reactions, the expressions for kcat and KM become very complicated (denoted as kcat(app) and KM(app), respectively). Importantly, however, the specificity constant kcat(app)/KM(app) still equals kcat/KM.

Bisubstrate (or multisubstrate) enzyme reaction kinetics

Most known enzymatic reactions involve two (rather than just one) substrates and form two products. They are often group-transfer reactions or redox reactions: E A X + B A + B X

For example, the enzyme alcohol dehydrogenase (ADH) catalyzes oxidation of ethanol to acetaldehyde using NAD+ as an oxidizing agent (i.e., ethanols hydride is formally transferred from ethanol to NAD+): ADH CH3CH2OH + NAD+ CH3CH(=O) + NADH Such bi- (or multi-) substrate reactions typically occur via one of the following two kinetic mechanisms: (i) sequential (single-displacement) reactions, and (ii) ping-pong (double-displacement) reactions.

Bisubstrate (or multisubstrate) enzyme reaction kinetics

(i) Sequential (single-displacement) reactions All substrates must combine with the enzyme before the reaction can occur and products be released. For the enzyme-catalyzed reaction A X + B, the X group being transferred is directly passed from A to B. These sequential reactions may have an ordered mechanism, where there is a compulsory order of substrate addition to the enzyme, or a random mechanism, where there is no preference for the order of substrate addition to the enzyme.

In an ordered mechanism, the binding of the first substrate causes a conformational change in the enzyme molecule which forms the binding site for the second substrate. In a random mechanism, both binding sites pre-exist in the free enzyme. The just-considered reaction CH3CH2OH + NAD+ CH3C(=O)H + NADH follows an ordered bisubstrate mechanism, in which NAD+ is the leading substrate.

Bisubstrate (or multisubstrate) enzyme reaction kinetics

(ii) Ping-pong (double-displacement) reactions One or more products is/are released before all substrates have combined with the enzyme. In this case, for the enzyme-catalyzed reaction A X + B, the X group being transferred is not directly passed from A to B: rather, it typically covalently binds to the enzyme to form the enzyme intermediate E X, while the first product A is released (Ping). In the second, Pong stage, the X is transferred from the E X to the second substrate B and the enzyme E is regenerated.Bisubstrate enzyme-catalyzed reactions have two pairs of kcat(app) and KM(app) (one pair for each substrate) and complicated Michaelis-Menten equations. Importantly, however, the sequential and ping-pong mechanisms can be distinguished from each other by steady-state enzyme kinetic measurements.

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