The Michaelis-Menton Model For non-allosteric enzymes, the most widely used kinetic model is based upon work done by Leonor Michaelis and Maud Menton For a given enzyme-catalyzed reaction: Substrate --> Product E+ S ESPE+ k1k1 k -1 k2k2 Note: There is no appreciable back reaction of Product to substrate!

E+ S ESPE+ k1k1 k -1 k2k2 k 1 = Rate constant for the Formation of the enzyme/substrate complex k -1 = Rate constant for the Dissociation of the enzyme/substrate complex k 2 = Rate constant for the formation of product and its release The enzyme is specifically (or explicitly) described in the mechanism This means it appears in the rate equation

When we carry out an experiment to characterize the kinetics of an enzyme, we do so at several concentrations of substrate We want to mix the solution and then immediately measure the initial velocity (V 0 ) Why? The [P] is low, so we do not have to be concerned with the back-conversion of product to substrate We plot V 0 versus the [S]

We can see two distinct portions in the curve: First Order Kinetics: Increasing [S] yields an increased V 0 Zero Order Kientics: The enzyme molecules are saturated with substrate and increasing [S] does not yield an increased V 0 Two useful terms can be extracted from this plot: V m or V max : The maximum velocity of the enzyme (where do we find it?) K m : The concentration of substrate that gives (1/2)V m

Deriving K m and the Steady State Assumption E+ S ESPE+ k1k1 k -1 k2k2 In the first step of the reaction, 2 competing processes are taking place. Rate of Formation of ES: Rate of Breakdown of ES: Note: The conversion of ES --> E+P pulls ES out of the pool, so we have to include it in the breakdown rate!

The Steady State Assumption 1.When we measure the velocity of a reaction, there is no product present in the solution during the initial time periods This means that E+P --> ES does not occur 2.We also know that enzymes are very efficient, so during the initial time periods, the lifetime of the ES complex is VERY short 3.Based upon these two facts, we can assume that within milliseconds (or microseconds or less), the reaction will reach a Steady State in which the rate of ES formation equals the rate of ES breakdown. This is the Steady State Assumption

K m and the Steady State Assumption What is [E]? (We know [S] from the experiment?) At any given moment in the reaction, the amount of free enzyme is: [E] = [E T ] - [ES] Plug this into the equation: From the steady state assumption

K m and the Steady State Assumption Multiply by [ES] Multiply the left side through Reorganize Factor out [ES] Isolate [ES] Now we have a term to describe [ES], we can use it to describe V 0 in terms of K m and V m

V 0, K m and V m 1.Part of our steady state assumption is that the E+P --> ES back reaction doesn’t occur 2.We also know that the rate limiting step will be the conversion of ES to E+P going through the transition state –Does this make sense to you? –If the rate limiting step (energetically most expensive) step was the formation of the ES complex, would the reaction even occur? 3.Based upon these observations, we can say that the observed velocity of the reaction is: V 0 =k 2 [ES]

Deriving the Michaelis-Menton Equation With the mathematical definition of [ES], V 0, V m and K m, we are ready to derive the Michaelis-Menton equation If we are in the zero order part of the V 0 vs. [S] curve, then V 0 =V M =k 2 [E T ] Substituting V m for k 2 [E T ], we get:

The Michaelis-Menton Equation With the Michaelis-Menton equation, we can relate the observed enzymatic velocity at any substrate concentration to the value of V m This allows us to analyze enzymes and: Identify differences in enzymatic function between species Compare the same enzyme from different tissues of the same organism Analyze the efficiency of newly discovered enzymes Identify the effect of mutations on enzymatic function

Working with / Thinking about the Michaelis- Menton Equation 1.If [S] >>> K m, what does V 0 equal? 2.If [S] = K m, what does V 0 equal?

Working with Real Data to Calculate K m and V m Using V 0 vs. [S] plots to calculate V m and K m can be difficult and inaccurate because of the asymptotic nature of the plot Two scientists, Hans Lineweaver and Dean Burk, realized that if you plotted the reciprocal of the Michaelis-Menton equation, you’d get a straight line: This has the form: y = mx + b

Lineweaver-Burk Plots This has the form of a linear equation: y = mx + b If we plot the data as 1/V 0 vs 1/[S]: The slope = ? The y-intercept = ? (This occurs when 1/[S] = ?) The x-intercept = ? (This occurs when 1/V 0 = ?) This is the common method to analyze non-allosteric Enzymological Data

Plot the data on p.156 of Campbell and Farrell on your own Also: Check the Useful Links page on the course website!

What do K m and V m Really mean? We know that V 0 =V m /2 when [S] = K m We can use this to say that when [S] = Km, 50% of the active sites of the population of enzymes are occupied If an enzyme has a low K m, the enzyme achieves maximum efficiency at low [S] K m is unique for each and every enzyme- substrate pair Some enzymes catalyze reactions that utilize 2 substrates: E + A EA + B EAB E + P + Q There is a unique K m value for A and one for B

Thinking about K m As long as k 2 << k -1, the Km can be viewed as a measure of the affinity of the enzyme for the substrate –If k -1 is high, what does that tell you? The Km value is also a function of pH and temperature. –Why?

Turnover number and k cat In order to better understand the catalytic efficiency of enzymes, we need to define another term: k cat The catalytic efficiency of an enzyme will be defined as:

k cat and Catalytic Efficiency k cat /K M results in the rate constant that measures catalytic efficiency. –This measure of efficiency is helpful in determining whether the rate is limited by the creation of product or the amount of substrate in the environment. If the rate of efficiency is: HIGH, k cat is much larger than K M, thus the enzyme complex turns more enzyme-substrate into product that the enzyme encounters substrate, which is defined as k 1. LOW, this means that the rate of product turnover is much lower than the substrate concentration, thus the enzyme and substrate have high affinity for each other.

k cat and Catalytic Efficiency k cat /K M is only useful when the substrate concentration is much less than the K M. Looking at the enzyme/substrate catalytic reaction equation: And knowing that k 2 could be considered to be equal to k cat, it is evident from that even if k cat is much greater than k -1, the equation will still be limited by k 1, which is the rate of ES formation. This tells us that k cat /K M has a limit placed on efficiency in that it cannot be faster than the diffusion controlled encounter of an enzyme and its substrate (k 1 ). E+ S ESPE+ k1k1 k -1 k2k2

k cat and Catalytic Efficiency Therefore, enzymes that have high k cat /K M ratios have essentially attained kinetic perfection because they have come very close to reaching complete efficiency only being limited by the rate at which they encounter the substrate in solution.