Lecture 6: Enzyme Kinetics

Isomerases

Catalyze structureal change within a single molecule -Because these reactions have only one substrate and one product, they are among the simplest enzymatic reactions

Ligases

Catalyze the ligationor joining of two substrates -These reactions require input of chemical potential energy in the form of a nucleoside triphosphate such as ATP -Ligases usually are referred to as synthetases

Michaelis-Menten equation

Describes the relationship between the initial velocity of a reaction and the substrate concentration. Vo= k2[E]total[S]/ Km + [S] -This equation describes the kinetics at both low and high [S]. -L et's say [S] is very low, << Km. __ __ Km + [S] ~ Km. __ __Equation simplifies to v0 = k2[E]total[S]/ Km or v0 = (K2/ Km) ([E]total[S]) __ __ k2/Km is the apparent rate constant for Step 1. __ __Note: In your textbook on page 145, "[E]" should be "[E]total" __ __Let's say [S] is very high, >> Km. __ __ Km + [S] ~ [S]. k2[E]total [S] __ __Equation simplifies to v0 = —————— [S] __ __ . . . or k2[E]total -So now we have one equation that accounts for observed initial rates at all S concentrations. __ __But there's another consequence . . . __ __The curve is a hyperbola: y = ax / b + x __ __a is the asymptote of the hyperbola, ymax __ __b is the value of x when y = ½ ymax __ __First observed about a hundred years ago by Michaelis and Menten ^Let's define its variables. __ __Obviously, y is v0 and x is [S] __ __a corresponds to k2[E]total. __ __Since a is ymax and y is v0, a is Vmax. __ __That is, the theoretical fastest rate you'd get at some insanely high [S] __ __Therefore Vmax = k2[E]total. __ __Now our equation looks like this: v0 = Vmax[S] / Km + [S] _b is Km. __ __Since b is the value of x when y = ½ ymax . . . __ __ . . . then b is the [S] when v0 = ½ Vmax

Rate equation

Expresses the velocity of an experiment in relation to the concentration of each reactant -reflects the fact that the velocity of a reaction depends on the concentration of the substrate -K is the rate constant

Consider the data in the table below: enzyme Km kcat chymotrypsin 1.5 × 10-2 M 0.14 s-1 pepsin 3.0 × 10-4 M 0.50 s-1 tyrosyl-tRNA-synthetase 9.0 × 10-4 M 7.6 s-1 ribonuclease 7.9 × 10-3 M 790 s-1 carbonic anhydrase 2.6 × 10-2 M 40,000 s-1 fumarase 5.0 × 10-6 M 800 s-1 Which enzyme binds most strongly to its substrate? Please explain your choice

Fumarase would bind most strongly to its substrate, as it has the lowest value for Km.

. Explain why Km isn't always the same thing as the equilibrium constant for the reverse reaction of the binding of a substrate to an enzyme.

In a simple two-step enzymatically-catalyzed reaction, Km is equal to the equilibrium constant for the reverse reaction of the binding of a substrate to an enzyme, but in a more complex reaction involving more steps, Km may be a function of several equilibrium constants of several reaction steps.

Explain why kcat and k2 aren't always the same things

In a simple two-step enzymatically-catalyzed reaction, kcat is equal to k2, but in a more complex reaction involving more steps, kcat may be a function of several rate constants of several reaction steps.

Turnover number

It indicates the maximum number of substrate molecules converted to product each second of the active site

Chemical Kinetics

Kinetic experiments examine the relationship between the amount of product formed in a unit of time and the experimental conditions under which the reaction takes place

Making sense of Km

Km =K2+K-1/K1 -We often see kcat instead of k2 in these equations: __ __In our simple two-step mechanism, k2 is the rate constant of the second step. __ __It is the overall rate constant when [S] >> Km. __ __v0 = k2[E]total __ __It's part of the apparent rate constant when [S] << Km. __ __We call it the catalytic rate constant, kcat. __ __In the simple two-step reaction, k2 = kcat. __ __But many enzymatic reactions have more steps. __ __In such cases, kcat is a more complex function of several rate constants for multiple steps. __ __So it's more generally applicable to write the Michaelis-Menten equation as: __ __v0 = kcat[E]total[S] / Km + [S] __ __At high [S], v0 = kcat[E]total __ __At low [S], v0 = (kcat/Km)[E]total[S] __ __ kcat/Km is the apparent rate constant. __ __We call it the "catalytic efficiency" - products per substrate binding incident.

Enzyme kinetics

Only specific substrates can fit to a given enzyme -Enzyme kinetics are different from chemical kinetics because the rates of the enzyme-catalyzed reactions depend on the concentration of enzyme and the enzyme is neither a product nor a substrate of the reaction -The rate also differ because substrate has to bind to enzyme before it can be converted to product -The initial velocities are obtained from progress curves, just as they are in chemical reactions ^The progress of the curve for an enzyme catalyzed reaction is dependent on the the concentration of the product which increases as the reaction proceeds. ^The initial velocities of the reaction is the slope of the initial linear portion of the curve *Note that the rate of the reaction doubles when twice as much enzyme is added to the identical reaction mixture

The enzyme lactase acts on the substrate lactose. Looking at the structures of lactose and maltose below, do you think the Km for the binding of maltose to lactase would be higher or lower than the Km value for the binding of lactose to lactase?

Since maltose is shaped differently than lactose, one would not expect maltose to fit into the active site of lactose very well. Therefore one would not expect maltose to bind as strongly to lactase as lactose does. For this reason, one would expect the Km for the binding of maltose to lactase to be higher than the Km value for the binding of lactose to lactase.

Suppose you're trying to design a drug that will block an enzyme at the active site where the substrate would normally bind. Would you want your drug to have a higher Km than the substrate, or a lower Km than the substrate?

The Km of the drug should be lower than the Km of the substrate because the drug has to bind more strongly to the enzyme than the substrate does in order to inhibit the enzyme more effectively.

Vmax

The asymptotic value (a) -The maximum velocity of the reaction at infinitely large substrate concentrations -Vmax value is often shown as initial velocity rather than the concentration of substrate plots *When the asymptotic curve is flattened out at moderate substrate concentrations at a lebel that seems far less than the Vmax value -The true Vmax is not determined by trying to estimate the position of the asymptote from the shape of the curve ^instead, it is precisely and correctly determined by fitting the data to the general equation for rectangular hyperbola

velocity

The basis of most kinetic measurements is the observation of the rate or velocity of a reaction, which varies directly with the concentration of each reactant

3. Carbonic anhydrase (CA) has a 25,000- fold higher activity (Kcat=10^6s^-1) than orotidine monophosphate decarboxylase (OMPD) (Kcat= 40 s^-1). However, OMPD provides more than a 10^10 higher rate acceleration than CA. Explain how this is possible

The catalytic constant (Kcat) is first order rate constant for the conversion of ES to E+P under saturating substrate converting substrate to product than does OMPD. However, the efficiency of an enzyme can also be measured by the rate acceleration provided by the enzyme over the corresponding uncatalyzed reaction. The reaction of the substrate for OMPD in the absence of enzyme is very slow than the CA reaction in terms of Kcat, OMPD is one of the most efficient enzymes known and provides a much higher rate acceleration than does CA when the reactions of each enzyme are compared to the corresponding uncatalyzed reactions

Kcat

The constant that is the number of moles of substrate converted to product per second per mole of enzyme AKA catalytic constant -represents the number of moles of substrate converted to product per second per mole of enzyme (or mole of active site for a multisubunit enzyme) under saturatinf conditions ^It indicates the maximum number of substrate molecules converted to product each second of the active site -The catalytic constant measure how quickly an enzyme can catalyze specific reaction ^describes the effectiveness of the enzyme

1. Initial velocities have been measured for the reaction of alpha-chymotrypsin with tyrosine benzyl ester [S] at six different substrate concentrations.

The initial velocities are approaching a constant value at the higher substrate concentrations so we can estimate the Vmax as 70 mM/min. -Since Km equals the concentration of substrat [S] required to reach half the max velocity, we can estimate the Km to be 0.01 M since that's the concentration of substrate that yields a rate of 35 mM/min (=Vmax>2)

Michaelis constant

The term b in the general equation for the rectangular hyperbola AKA the Michaelis Constant ^b is the point on the x-axis corresponding to a value of a/2 -The Michaelis Constant is defined as the concentration of the substrate when instant velocity is equal to one-half Vmaz The ratio of rate constants, (k2 + k −1) ÷ k1, is called the Michaelis-Menten constant, or Km. ([E]total − [ES])[S] Km = ————————— [ES] Multiply both sides by [ES]: [ES] ([E]total − [ES])[S] Km[ES] = ——————————— [ES] or Km[ES] = ([E]total − [ES])[S] or Km[ES] = [S][E]total − [S][ES] Add [ES][S] to both sides: Km[ES] + [S][ES] = [S][E]total or [ES](Km + [S]) = [S][E]total Divide both sides by Km + [S]: [ES]( Km + [S] ) [S][E]total —————————— = ——————— Km + [S] Km + [S] or [S][E]total [ES] = ——————— Km + [S]

What is the initial rate of an enzyme catalyzed reaction, relative to Vmax, when [S] = Km?

When [S] = Km, v0 = ½ Vmax.

Explain why [ES] remains relatively constant during most of the duration of an enzyme catalyzed reaction when [S] is high.

When [S] >> Km, the enzyme becomes saturated, and [ES] can't increase unless some [ES] is consumed by the step 2 of the reaction. At very high [S]s step 1 is faster than step 2, so as soon as any ES is consumed by reacting to produce E + P in step 2, step 1 will produce ES quickly enough to replace any that is consumed by step 2. For this reason, [ES] remains fairly constant for most of the duration of an enzyme-catalyzed reaction when [S] is high.

4. An enzyme That follows Michaelis-Menten kinetics has a Km of 1µM. The initial velocity is 0.1 µM min⁻¹ at a substance concentration of 100µM. What is the initial velocity when [S] is equal to a. 1mM b. 1µM c. 2µM

When [S]=100µM, [S] >>> Km, so Vo=Vmax =0.1µM min⁻¹ a. For any substrate concentration greater than 100µM , Vo=Vmax= 0.1min⁻¹ b. When [S]= Km, vo=Vmax/2 or 0.50 µM min⁻¹ c. Since Km and Vmax are known, the Michaelis-Menten equation can be used to calculate Vo at any substrate concentration. For [S]=2µM

Lineweaver-Burk Plots

You can roughly estimate Vmax from the asymptote of a v0 vs. [S] plot. __ __A better way it to plot 1/v0 vs. 1/[S]. __ __The derivation is in the book if you care to look it up 1/Vo=Km/Vmax * 1/[S] + 1/Vmax -(1/ [S]) is the slope (m) __ __This plot is linear. __ __ y-intercept is 1 / Vmax __ __ x-intercept is - (1 / Km) __ __Not really used to calculate Vmax or Km anymore. __ __You won't have to draw one! __ __But the plot is useful when looking at enzy

Estimating [ES] at Steady-State Conditions

_Suppose an enzyme-catalyzed reaction takes place in two steps, as shown below: E + S k1/k2^⇌ ES K2^---> E + P where E is the enzyme, S is the substrate, ES is the bound enzyme-substrate complex, and P is the product. The rate of ES formation is: v = k1[E][S] Since at any given moment, [E] = [E]total − [ES] v = k1([E]total − [ES])[S] The rate of ES consumption is v = k2[ES] + k −1[ES] or v = (k2 + k −1)[ES] Since at the steady-state, rate of ES formation = rate of ES consumption, (k2 + k −1)[ES] = k1([E]total − [ES])[S] Now we can rearrange everything to solve for [ES]. First, we'll divide both sides by k1: (k2 + k −1)[ES] k1 ([E]total − [ES])[S] ——————— = —————————— k1 k1- or (k2 + k −1)[ES] ——————— = ([E]total − [ES])[S] k1 Then divide both sides by [ES]: (k2 + k −1) [ES] ([E]total − [ES])[S] ———————— = ————————— k1 [ES] [ES] or k2 + k −1 ([E]total − [ES])[S] ————— = ————————— k1 [ES]

Again consider the enzymes in the table from question 3. Suppose each was observed catalyzing its usual reaction, and all reactions were observed at the same enzyme concentrations and the same substrate concentrations. a) Which enzyme would result in the highest initial reaction rate, provided, if [S] << Km? Please explain your choice. b. Which enzyme would result in the highest initial reaction rate, provided, if [S] >> Km? Please explain your choice

a. If the [S] << Km, then fumarase would result in the highest reaction rate, as fumarase has the highest value for kcat/Km: kcat 800 s-1 —— = —————— = 1.6 × 108 s-1M- Km 5.0 × 10-6 M b. If [S] >> Km, then carbonic anhydrase would result in the highest reaction rate, as it has the highest value for kcat.

14. Use the Michaelis Mentent equation to demonstrate the following a. Vo becomes independent of [S] when [S]>>>Km b. The reaction is first order with respect to S when [S]<<<Km c. [S]>>>Km when Vo is one half Vmax

a. When [S]>>> Km, then Km+ [S]~~ [S]. Substrate concentration has no effect on velocity, and Vo =Vmax b.When [S] <<<Km, Km + [S]~~Km, and the Michaelis-Menten equation simplifies -Velocity is related to [S] by a constant value, and the reaction is first order with respect to S c. When Vo=Vmax/2, Km=[S]

1. Oxidoreductase

catalyze oxidation-reduction reactions ex./ lactate dehydrogenase

Catalytic Efficiency

is different from catalytic proficiency. __ __Catalytic proficiency is (kcat/Km) ÷ kn. __ __kn is the rate constant without the enzyme. __ __Enzymes with multiple active sites—remember hemoglobin. __ __Multiply [E]total by number of active sites. __ __All this ignores cooperative binding. __ __We've been treating strength of enzyme-substrate binding as a constant. __ __It isn't constant when cooperative binding is involved.

2. Why is the kcat/Km value used to measure the catalytic proficiency of an enzyme? a. What are the upper limits for Kcat/Km values for enzymes? b. Enzymes with Kcat/Km values approaching these upper limits are said to have reached "catalytic perfection" Explain

*confusion about 2a* _We're measuring the catalytic proficiency of one enzyme against another when they're competing for the same substrate. __ __The question doesn't say __ __The rate laws for each reaction will be identical except for kcat/Km. __ __The reaction with the higher kcat/Km will be the faster reaction. __ __It's enzyme has the higher catalytic efficiency __ __3 is a two-part question: 1) when [S] << Km, 2) when [S] >> Km. __ __There's a typo in 14c: [S] = Km, not [S] >>Km. a. The ratio Kcat/Km, or specificity constant, is a measure of the preference of an enzyme for different substrates -When two substrates at the same concentration complete for the active site of an enzyme, the ratio of their rates of conversion to product is equal to the ratio of the Kcat/Km values since Vo=(kcat/km)[E][S] for each substrate and [E] and [S] are the same b. The upper limit of Kcat/Km approaches 10^8 to 10^9s⁻¹, the fastest rate at which two uncharged molecules can approach each other by diffusion at physiological temperatures

What is the rate limiting factor of S + E → ES → P

*depends on the concentration of the substrate 1. At low [S], step 1 will be slow, as v0 ~ [S] a. So, the reaction rate depends on the rate law of step 1. -v0 = k1[E][S] ^This is a simplification, because step 1 is usually reversible. >k−1 is the rate constant for the reverse reaction. >>The rate constant is going to be more complex 2. At high [S], step 1 will be fast, as v0 ~ [S] a. So, the reaction rate depends on the rate law of step 2. -v0 = k2[ES] b. An enzyme might encounter a wide range of [S] in the body, but they need to be able to work well when [S] >> [E]. -Remember, proteins cost a lot of energy to make, so they need to be efficient! ^We really want one rate law, not two c. At high [S], the enzyme gets saturated. -If step 1 were irreversible, that would mean all enzyme molecules are bound. ^But step 1 is usually reversible, so it means E and ES have reached equilibrium. ^Either way, if [S] is high enough, step 1 makes new ES as quickly as step 2 can convert ES into E + P . . . . . . or step 1 reverse can convert ES into E + S. -So at high [S], [ES] doesn't change over time. ^This is called the steady-state approximation

Plots of initial velocity vs substrate concentration for an enzyme catalyzed reaction

1. Each experimental point is obtained from a substrate progress curve using the same concentrations of enzyme. 2. The shape of the curve is hyperbolic. 3. At low substrate concentrations, the curve aproximates a straight line that rises steeply. 4. In this region of the curve, the reaction is highly dependent on the concentrations of substrate 5. At high concentrations of substrate, the enzyme is almost saturated, and the initial rate of the reaction does not change much when substrate concentration is further increased 6. The concentration of substrate that corresponds to half-maximum velocity is called the The Michaelis Constant (Km) -The enzyme is half- saturated when S=Km

Six categories of enzymes

1. Oxidoreductase 2. transferases 3. hydrolases 4. lyases 5. isomerases 6. ligases

Meaning of Kcat and Kcat/Km

1. The catalytic constant (Kcat) is the rate constant for conversion of the ES complex to E + P. 2. It is measured most easily when the enzyme is saturated with substrate 3. The ratio Kcat/Km is the rate constant for the conversion of E + S to E + P at very low concentrations of substrate 4. The reactions measured by these rate constants are summarized below the graph

Reaction S->P

1. Vo is the initial rate a. Vo~ the concentration of enzyme -Vo vs the concentration of enzyme is a linear plot 2. The concentration of the substrate shows a hyperbolic curve 2. Rate seems to depend on the concentration substrate at low substrate BUT NOT high substrate -The rate depends on the concentration of enzyme at all concentration of either E or S So, v0 ~ [E] and [S] at low [S], but v0 ~ [E] only at high [S] __ __Apparently two separate rate laws: __ __at low [S]: v0 = k1[E][S] __ __at high [S]: v0 = k2[E] 3. This happens because we have a two-step reaction: S + E → ES → P

Derivation of the Michaelis-Menten ewuation

1. one common derivation is termed the Steady State Derivation: Postulates a period of time (called the steady state) during which the ES complex is formed at the same rate that it decomposes so that the concentration of ES is constant -The initial velocity is used in the steady state derivation because we assume that the concentration of product is negligible 2. Can be measured by the rate constant, the total enzyme concentration, and the substrate concentration *The concentration of substrate is greater than the concentration of total enzyme

Enzymes are catalysts

1. reinforce the definition of a catalyst 2. They exist to make chemical reactions go faster- a lot faster

Enzyme-substrate complex

A transient intermediate in an enzyme catalyzed reaction -only specific substrates can fit into a given enzyme -Early studies of enzyme kinetics confirmed that an enzyme binds to a substrate to form the enzyme-substrate complex -Are formed when ligands bind noncovalently in their proper places in the active site ^The substrate interacts transiently with the protein catalyst (and with other substrates in a multisubstrate reaction) on its way to forming the product of the reaction

Why is step 1 the rate limiting step of a simple two-step enzyme-catalyzed reaction at low substrate concentrations, but at high substrate concentrations step 2 is the rate-liming step?

At low [S], step 1 proceeds slowly, because the rate of step 1 is dependent on [S]. This makes step 1 the rate-limiting step at low [S]. As [S] increases, the rate of step 1 increases. When [S] high enough, the rate of step 1 becomes faster than step 2, and step 2 becomes the rate-limiting step.

Measurement of Km and Vmax

Can be measured in several ways: 1. Both values can be obtained by the analysis of initial velocities at a series of substrate concentrations and a fixed concentration of enzyme 2. In order to obtain reliable values of the kinetic constants the concentration of substrate points must be spread out both below and above Km to produce the hyperbola 3. It is difficult to determine either Km or Vmax directly from a graph of initial velocity vs concentration because the curve approaches Vmax asymptotically -However, accurate values can be determined by using a suitable computer program to fit the experimental results to the equation of the hyperbola 4. Values of Kcat can be obtained from measurements of Vmax only when the absolute concentration of the enzyme is known 5. Values of Km can be determined even when enzymes have not been purified provided that only one enzyme in the impure preparation can catalyze the observed reaction

Explain why determining the catalytic proficiency of an enzyme is not always possible.

Catalytic proficiency is a measure of how much an enzyme increases the rate of a reaction, compared to the rate of the reaction in the absence of a catalyst. In the cases of many enzyme catalyzed reactions, the non-catalyzed reaction takes place to slowly to measure its rate.

Enzymes

Catalyze Reactions -are named by adding the suffix -ase to the name of their substrates or to a descriptive term for the reactions they catalyze -many newly discovered enzymes are named after their genes or for some nondescriptive characteristic

Transferases

Catalyze group transfer reactions and many require the presence of coenzymes -In group transfer reactions a portion of the substrate molecule usually binds covalently to the enzyme or its coenzyme -This group included kinases, enzymes that catalyze the transfer of a phosphoryl group from ATP

Hydrolases

Catalyze hydrolysis -They are a special class of transferases with water serving as the acceptor of the group transferred ex./ Pyrophosphatase

Lyases

Catalyze lysis of a substrate generating a double bond in nonoxidative, elimination reactions -In the reverse direction, lysases catalyze the addition of one substrate to the double bond of a second substrate