Biochem 507 - Exam 2 - Enzyme Kinetics (Lec. 11)

Product concentration [P] at any given time t

1. Commonly measured by UV-visible spectroscopy, fluorescence spectroscopy, or by separating substrate and product (chromatography, electrophoresis) 2. The experimenter chooses when to measure and analyze [P], and choosing wisely enables further simplification

Random

1. Either substrate can bind to the enzyme first, and then the other binds to the [ES] 2. Ex: In some enzymes, the order of substrate binding and the order of product release is random -a. Cleland notation for the creatine kinase reaction, a random bi bi reaction

[S] << KM

1. Reaction is first-order in [S] 2. Linear relationship between v0 and [S] 3.V0=Vmax[S]/Km -a. Remove the [S] from the denominator because it won't make a difference since it is so much smaller than Km

[S] >> KM

1. Reaction is zero-order in [S] 2. v0 is independent of [S] 3. [ES] [Et] (saturation) 4.V0=Vmax[S]/[S]=Vmax -a. Can neglect Km since it is so small, which allows for the [S] to cancel out

Ordered

1. Second substrate binds to the first [ES] complex 2. Ex: Reactions that use NADH and NAD+ co-substrates usually exhibit ordered sequential mechanisms -a. NADH binds first, then pyruvate binds to [ES] --i. Lactate is then released first followed by NAD+ --ii. Has to be in this order -b. Cleland notation is a convenient way of illustrating the order of binding in multisubstrate reactions: -c. Lactate dehydrogenase catalyzes an ordered bi bi reaction --i. First bi tells us two substrates, second bi tells us two products

Total substrate concentration [St]

1. [St] can be measured using UV/visible spectroscopy if it absorbs light, or by weight if necessary 2. The substrate concentration [St] is equal to the sum of the concentration of free substrate [S] and all species in the reaction that involve a complex with substrate (i.e., the ES complex): -a. [St] = [S] + [ES] (substrate conservation equation) 3. In general, experiments are carried out with [St] >> [Et], so [ES] is negligible compared to [St] and we can assume that [St] = [S] -a. Keep substrate concentration very high to assume the [ES] is so small that you don't need to worry about it

[S] = KM

1.V0=Vmax/2 -a. Substitute [S] for Km since they are the same

catalytic perfection

Their rates are limited by a physical step (collision in solution) rather than a chemical one

Total enzyme concentration [Et]

Typically measured by absorbance at 280 nm (A280) -a. Tryptophan residues absorb in this wavelength 2. Recall that A280 is related to protein concentration by Beer's law: A=Elc 3. The total enzyme concentration is equal to the sum of the concentration of free enzyme [E] and all species in the reaction that involve a complex with enzyme (i.e., the ES complex): -a. [Et] = [E] + [ES] (enzyme conservation equation)

The Michaelis-Menten equation

V0=Vmax[S]/[S]+Km

kcat/KM is a measure of catalytic efficiency

a. At [S] << KM -i. V0=kcat[Et][S]/Km b. kcat/KM is a pseudo-second order rate constant that takes into account both turnover number and the affinity of an enzyme for its substrate -i. Units is per Molar, per sec. c. Under physiological conditions, [S] is usually subsaturating, so kcat/KM is a relevant parameter d. kcat/KM is known as the catalytic efficiency of an enzyme e. The magnitude of kcat/KM is limited by diffusion, i.e., the rate at which enzyme and substrate encounter one another in solution and cannot be higher than ~108—109 M-1 s-1 f. Enzymes with kcat/KM approaching this limit are said to have achieved 'catalytic perfection'. Their rates are limited by a physical step (collision in solution) rather than a chemical one

Bisubstrate reactions involving a ternary complex

a. Binding of substrates may be ordered or random in bisubstrate reactions

Lineweaver-Burk plot caveats

a. Double-reciprocal plots are useful for recognizing patterns of inhibition and for distinguishing ordered vs. random substrate binding in bisubstrate reactions b. However, there are many caveats to using them: -i. Compound errors in the data -ii. Remove the data one more step from reality and make things seem obscure and complicated -iii. Hide experimental problems that would be immediately obvious from direct plots

Bisubstrate reactions that do not involve a ternary complex

a. In some bisubstrate reactions, the first substrate is converted to product before the second substrate binds, so no ternary complex is formed -i. An example of this is the ping-pong, or double-displacement, mechanism b. Cleland notation for the aspartate aminotransferase reaction, a ping-pong bi bi reaction: -i. A substrate binds and then a product is released before the other substrate binds

What does KM tell us about an enzyme?

a. KM approximates the amount of substrate needed for significant catalysis to occur -i. Units is concentration(Molar) b. KM is often close to physiological substrate concentration -i. Allows significant catalysis to occur in a cellular context -ii. Enables response to changes in cellular concentration of substrate c. KM is independent of the amount of enzyme used in the experiment d. KM includes contributions from the rate constants of chemical steps and is therefore not strictly an equilibrium constant -i. Km does not always equal Kd -ii. In the case where k-1 >> k2, KM = k-1/k1, the equilibrium dissociation (KD) constant of the ES complex -iii. In the case where k2 >> k-1, KM = k2/k-1 -iv. The only universal definition of KM is --1. Km=[S] where V0=Vmax/2

Bisubstrate reactions

a. Many enzymes catalyze reactions that utilize two or more substrates and make two or more products b. Substrates may be bound to the enzyme concurrently (to form a ternary complex) or sequentially c. Substrates may bind in a random sequence or a specific order d. Bisubstrate reactions can also be analyzed with a Michaelis-Menten approach by analyzing one substrate at a time -i. Michaelis-Menten kinetics provides limited information about how many steps or intermediates occur in the reaction -ii. Michaelis-Menten kinetics can distinguish between pathways involving a ternary complex and those that do not

Michaelis-Menten kinetics experiments to distinguish reactions that involve a ternary complex from those that do not

a. Michaelis-Menten kinetics experiments can be performed on enzymes that catalyze bisubstrate reactions by evaluating one substrate at a time -i. One substrate held constant, the other varied -ii. Repeat at several concentrations b. Overlay Lineweaver-Burk plots -i. If the lines intersect, a ternary complex is formed -ii. If the lines are parallel, no ternary complex is formed

How do we measure enzyme kinetics?

a. Monitor enzymatic reaction rate under controlled conditions b. Apply a kinetic model the provides a framework for understanding the results c. Goals: -i. Relate reaction rate to measurable and controllable parameters -ii. Simplify by defining assumptions and ensuring that they are met under experimental conditions -iii. Define a basic set of parameters that describe the catalytic properties of the enzyme of interest

The steady-state assumption

a. Phases of an enzymatic reaction: -i. Pre-steady-state phase --1. Initial burst of ES complex formation --2. microsec to millisec timescale—faster than hand manipulations -ii. Steady-state phase --1. [ES] and [E] remain constant over time --2. s to min timescale -iii. Simplifying assumption: 0 over the course of the measurement (the steady-state assumption)

The ES complex is the key to understanding the kinetic behavior of enzymes

a. Proposal: the overall reaction is composed of two steps a. At high [S], 100% of enzyme active sites are bound to S, so increasing [S] has no effect on the rate. This is called saturation kinetics. b. The velocity or rate of this reaction is given by -i. V=d[P]/dt=k(2)[ES]-k(-2)[E][P] c. [E] and [ES] are difficult to measure d. Challenge: Relate the rate equation to parameters that are measurable and controllable

Enzyme kinetics: an experimental approach to understanding how enzymes work

a. Put enzymes to work or turn them off b. Cancer therapeutics with enzyme inhibitors c. Solve problems in chemical synthesis -i. Makes designer materials d. Use them for chemical synthesis

Measuring initial rates simplifies the rate equation for enzymatic reactions

a. Rates are measured in the initial rate regime: -i. The initial rate regime is the time during which only a few percent of substrate has been converted to product. In this regime, the measured rate is the initial velocity, v0. --1. This ensures that the concentration of substrate does not change appreciably over time, i.e. [S] = [St] AND that approximately no product has built up, i.e. [P] = 0. -2. These approximations enable us to apply a simplifying assumption to our rate equation: -3. Simplifying assumption: [P]~0 over the course of the measurement, so = 0 and a.v=k2[ES] b.v0=k2[ES]

More complex enzymatic reactions

a. Some enzymatic reactions involve more than one intermediate. For example, the k3 step is kinetically significant in some cases -i. An addition of [EP] step b. The steady-state kinetics of more complex enzymatic reactions are still described by the Michaelis-Menten equation and kcat and KM are still useful parameters -i. Michaelis-Menten kinetics cannot tell us which step is rate-limiting, or how many intermediates there are in the reaction -ii. The equations that define kcat and KM may be more complicated for multi-step reactions. However, kcat is still the turnover number and KM is still the substrate concentration at which v0 = Vmax/2

If you want to kick it old-school...

a. The double-reciprocal or Lineweaver-Burk plot allows us to fit a straight line instead of using non-linear curve fitting b. Double-reciprocal plots were useful before computers with non-linear curve fitting software were widely available c. Take the reciprocal of MM equation d. Can determine KM, Vmax, and KM/Vmax from x- and y-intercepts and the slope of the line

Chemical kinetics is the study of reaction rates

a. The rate law or rate equation is the relationship between the reactant concentration(s) and the rate (v) of the reaction, which are related by a proportionality constant k, called the rate constant. -i. v=kA^mB^n -ii. How fast we will see product occur b. m is the order of the reaction in A and n is the order of the reaction in B -i. the exponent

The Michaelis-Menten equation relies on several assumptions and experimental considerations:

a. The substrate is in excess of the enzyme [St] >> [Et] -i. Allows us to assume that [ES] << [S] and [St] = [S] b. Rates are measured in an initial rate regime -i. Allows us to assume that [P] 0 during the course of the measurement -ii. Only a few percent of substrate is consumed, so we can assume [S] = [St] during the course of the rate measurement c. The reaction is carried out under steady-state conditions -i. Can assume that [ES] and [E] are constant during the course of the rate measurement d. The total amount of enzyme is constant during the reaction and is either free (E) or part of the enzyme-substrate complex (ES), i.e., [Et] = [E] + [ES]

How is a steady-state kinetics experiment performed and analyzed in the lab?

a. Vary substrate concentration [S] while keeping [Et] fixed and measure initial rates -i. Use different substrate concentrations from 10x below KM to 10x above KM (usually at least 7-8 concentrations) --1. You get a time course for [S] concentration for time and get a graph of [P] --2. Fit these plots to make a graph of [S] verses V0 ---a. Each dot on plot comes from [P] -ii. Lowest [S] should still be at least 5x greater than [Et]

How does uncatalyzed sucrose hydrolysis differ from catalyzed sucrose hydrolysis?

a. When [sucrose] >> [enzyme], the reaction rate is independent of sucrose concentration b. In uncatalyzed reaction, it is first order(linear line) c. In catalyzed reaction, its rate greatly increases, still at first order, then levels off to zero order in high concentration of [S]

A physiological example: acetaldehyde dehydrogenase

a. When we consume ethanol, it is converted to acetaldehyde by alcohol dehydrogenase in our livers b. A second enzyme, aldehyde dehydrogenase, converts acetaldehyde to acetate -i. Humans have two variants of aldehyde dehydrogenase (mitochondrial and cytosolic) whose kinetic parameters are listed below -ii. Mito: lower Kcat and Km

What does Vmax tell us about an enzyme?

a. is dependent on enzyme concentration -i. Unlike Km b. Measurement of enables us to calculate kcat -i. kcat is the rate constant for the rate-limiting step of the reaction(for slow step) 1. Units is per sec or per minute -ii. kcat is also called the turnover number and tells us how many substrate molecules an enzyme molecule converts to product per unit time c. Vmax=k2[Et], or more generally Vmax=kcat[Et]

Applying the steady-state assumption to solve for [ES] in terms of [Et] and [S]

a. steady-state assumption b. steady-state assumption, restated: rate of ES formation = rate of ES breakdown c. the enzyme conservation equation can be used to solve for [E] in terms of [Et] and [ES] d. substituting for [E] in steady-state assumption equation e. multiply through by [S] on left; consolidate rate constants on right f. add k1[ES][S] to both sides g. solve for ES h. simplify by multiplying by k1/k1 (=1) i. define KM, the Michaelis constant j. substitute KM into the equation for [ES] k.[ES]=[Et][S]/[S]+Km

I. What can we control and what can we measure in an enzymatic reaction?

i. Total enzyme concentration [Et] i. Total substrate concentration [St] i. Product concentration [P] at any given time t

catalytic efficiency

kcat/KM

rate law

the relationship between the reactant concentration(s) and the rate (v) of the reaction, which are related by a proportionality constant k, called the rate constant