Lecture 14

– Intro to enzymes– This Friday: Seminar speaker Howard

Salis (148 Baker) 3PM. Extra credit seminar

Thermodynamics

• Determines if the reaction is spontaneous (does it occur).• Does not tell us how fast a reaction will proceed.• Catalysts (enzymes) can lower the activation barrier to get

from products to reactants.

Thermodynamics of Enzyme Function

• Catalysts lower the energy barrier between reactants and products.

Free energy diagrams of simple chemical catalysis

Figure 2-21 Energetics of catalysis (cont.)In the presence of a catalyst (red curve), which in this case is acting by changing the pathway of the reaction and introducing additional smaller activation-energy barriers, intermediate I1, formed by crossing transition-state barrier TSc1, leads to transition-state barrier TSc2.

Its free energy (ΔGc), although the highest point in the reaction, is considerably lower than the free energy (ΔGu) of the uncatalyzed transition state, TSu.

After formation of a second intermediate, I2, a third transition state, TSc3, leads to product.

Because TSc2 is the highest transition state in the catalyzed reaction, the rate at which the reactants pass over this barrier determines the overall rate and thus it is said to be the rate-determining transition state of the catalyzed reaction.

The rate-determining step of this reaction is thus the conversion of I1 to I2.

Figure 2-21 Energetics of catalysis

Figure 2-21 Energetics of catalysis (cont.)

The transition state is the highest point in free energy on the reaction pathway from substrate to product.

It is the top of the activation-energy barrier (see TSu in Fig. 2-21).

Chemically, it is a species that exists for about the time required for a single atomic vibration to occur (about 10-15

s).

In the transition state, the making or breaking of chemical bonds in the reaction is not yet complete: the atoms are "in flight".

The stereochemistry and charge configuration of the transition state is thus likely to be quite different from that of either the substrate or the product, although it may resemble one more than it does the other. Figure 2-21

Energetics of catalysis

Figure U2-3.2 The rate-determining step

• The rate-determining step of a reaction is the step with the largest energy barrier.

• In this example, the height of the barrier between intermediates I1 and I2 is greater than between S and I1 or I2 and P, and the rate determining step is therefore I1 to I2.

U2-1 Enzyme Kinetics: General Principles• Enzymes function as biological catalysts to increase the rate

(speed) of chemical catalysis.• Reaction rates reflect key properties of enzymes and the

reactions they catalyze• Kinetics is the study of how fast chemical reactions occur.• The free energy change can tell us in which direction the reaction will

spontaneously occur.• The free energy change does not tell us how rapidly. • Some spontaneous reactions occur quickly, e.g. sec., others occur

almost imperceptibly over many years.• The rate of a chemical reaction or process, or the reaction rate, is the

change in the concentration of reacting species (or of the products of their reaction) as a function of time (Fig. U2-1.1).

Figure U2-1.1 Reaction rates measure how fast processes occur

(a) In this example, two reacting species A and B (red and blue) combine to form a product C (purple).

(b) The concentration of product molecules increases as the reaction proceeds, and within two seconds of the reacting species being mixed together, the concentration of C becomes 10 mM.

(c) The rate of the reaction during these two seconds is therefore 5 mM s-1, which is the slope of the line when we plot the concentration of the product against time.

Figure 6.6 A plot of initial reaction velocity versus the concentration of enzyme [E].

Note that velocity increases in a linear fashion with an increase in enzyme concentration.

Add more catalyst, get faster reaction rate.

-Not surprising!

Figure U2-1.2 Rate constants are measured from reaction rates at different reactant

concentrations • Follow the progress of the

reaction between molecules A and B, as shown in Figure U2-1.1:

• Rate at which product C is produced decreases as the reaction proceeds (Fig. U2-1.2a).

• The rate decreases because the concentration of reactants decreases.

• The possibility also exists that the reverse reaction (C → A + B) becomes significant as the concentration of product C increases.

Figure U2-1.2 Rate constants are measured from reaction rates at different reactant

concentrations • To avoid these complications, we

can measure the initial rate of the reaction (i.e, 5-10% of total reaction time):

The rate before the concentration of reactants decreases significantly and before the accumulation of product is able to interfere with the reaction (Fig. U2-1.2b).

Rate constants are measured from reaction rates at different reactant concentrations

• The initial reaction rate for this example depends upon The initial reaction rate for this example depends upon the concentrations of the reactants (denoted as [A] and the concentrations of the reactants (denoted as [A] and [B]) according to the equation:[B]) according to the equation:

Rate Rate ∝∝ [A] [B] [A] [B]• The reaction rate doubles if the concentration of A is The reaction rate doubles if the concentration of A is

doubled, as the reactants collide doubled, as the reactants collide twice as oftentwice as often..• A more useful measure is called the rate constant, k,

which tells us how the reaction rate varies with the concentrations of the reactants:

Rate = k [A] [B]• Note: Kinetic rates use small letter k, not to be confused

with equilibrium constant, Keq

Figure U2-1.2 Rate constants are measured from reaction rates at different reactant concentrations

• The units of k will depend on the number of reacting molecules.

• In this example (bimolecular reaction) the units would be M-1 s-1.

• For a unimolecular reaction, such as the conversion of molecule A to molecule B, the units of k would be s-1.

The rate constant is measured from the slope of the line when we plot the initial reaction rate at different concentrations of one of the reactants (Fig. U2-1.2c).

Reaction orders

• Rate equations show the frequency with which reacting molecules come together dependent upon their concentration.

• Rate = k[A]a[B]b…[Z]z

• The order of a reaction is defined as the sum of exponents in the rate equation.

R First order (unimolecular)

k1P

k is the rate constant

Rate of formation of P = k1[R]

Velocity (v) = -d[A] = d[P]

dtdt

Reaction orders

• For a reaction that is reversible

R First order (unimolecular)

k1

k2

P

Rate of formation of P = k1[R] - k2[P]

Reaction orders

A + B Second order (bimolecular)

k1

k2

P

Rate of formation of P = k1[A][B] - k2[P]

If A = B: 2Ak1

k2

P

2nd order forward and 1st order reverse

Rate of formation of P = k1[A]2 - k2[P]

1st and 2nd order reactions are common

Reaction orders

A + B + C Third order (termolecular)

k1

k2

P

Rate of formation of P = k1[A][B][C] - k2[P]

3rd order forward and 1st order reverse, nearly impossible and 4th order reactions are not known

Reaction orders

A + B

The steady state assumption: assumes that the reverse reactions are slow

k1

k2

[AB] + C

If k3 is slow, the rate of P formation = k3[C][AB]. However, [AB] is k1[A][B]Therefore, the net rate is k1[A][B][C] = 3rd order.

If k1 is slow, the rate will appear as k1[A][B] = 2nd order.

k3

k4

P

Figure U2-2.1 The binding of substrate to an enzyme is dependent on the substrate concentration

At low substrate concentrations, only a small fraction of enzyme molecules (blue) will have substrate (red) bound at any moment (top).

If the substrate concentration is high, If the substrate concentration is high, practically all the enzyme molecules practically all the enzyme molecules will contain bound substrate (bottom).will contain bound substrate (bottom).

At these high substrate concentrations, At these high substrate concentrations, if a substrate molecule dissociates from if a substrate molecule dissociates from the enzyme, another substrate the enzyme, another substrate molecule will almost immediately molecule will almost immediately collide with the enzyme to take its collide with the enzyme to take its place.place.

Catalytic activity maybe so fast that the reaction is not rate limiting but rather the binding of the substrate to the enzyme. Therefore, by studying the order of binding, you have some idea of the reaction mechanism.

Enzyme kinetics

Early observation:

• Unusual effect of substrate concentration on the rate of reaction (catalytic rate)

• v0 - Initial velocity (rate of reaction) increases upon adding more substrate (we are talking about molar ratios of Enzyme E and substrate S);

• But only up to a point!

• Net behavior observed is a hyperbola.

• Has asymptotic upper limit in the number of substrate molecules processes per unit time per mole of enzyme.

Enzyme kinetics

Figure 6.3 Experimental procedure to study the kinetics of an enzyme-catalyzed reaction.

An identical amount of enzyme is added to a set of tubes containing increasing amounts of a substrate.

The reaction rate or initial velocity is measured for each reaction mixture by determining the rate of product formation.

Figure U2-2.2 Graph of rate against total substrate concentration for a typical enzyme-catalyzed reaction

The rate (or velocity, v) of an enzyme-catalyzed reaction increases as the concentration of substrate is increased.

The rate approaches a maximum value (Vmax) as the substrate concentration becomes so high that all of the enzyme molecules are occupied (saturated) with substrate.

The concentration of substrate at which the rate is 1/2 Vmax is denoted as Km (the Michaelis constant).Michaelis constant).

This curve can be fitted by the MichaelisThis curve can be fitted by the Michaelis--MMenten equation:enten equation:

v = v = VVmax max [S]/([S]/(KKm m + [S]). + [S]). (Direct fit to model requires non-linear optimization.)(Direct fit to model requires non-linear optimization.)

VVmax max depends only upon the enzyme concentration and the rate constant, depends only upon the enzyme concentration and the rate constant, kkcatcat..

(moles, molarity)

(mol

es f

orm

ed p

er s

econ

d)

Review of Enzyme Function

• Enzymes generally function in the following manner:

1. Recognize, bind specific chemical compounds (“substrate(s)”) in solution.

2. Convert bound substrate to product via lowering Gibbs free energy of intermediate transition state between initial substrate, final product.

3. Release weakly-bound product, prepare to repeat new catalytic reaction cycle.

Michaelis-Menten Steady-State Enzyme Kinetics

.

kkcatcat = “turnover number” [s = “turnover number” [s-1-1]]

KKmm = “Michaelis constant” [M] = “Michaelis constant” [M]

kkcatcat//KKmm = “catalytic efficiency” [s = “catalytic efficiency” [s-1-1 M M-1-1]]

VVmaxmax = maximum = maximum kkcatcat value possible when enzyme is saturated with value possible when enzyme is saturated with

substratesubstrate

• KKmm is simply defined as the substrate conc. at which rate (“velocity”) is simply defined as the substrate conc. at which rate (“velocity”) vv = =

1/2 1/2 VVmax. max. Note: Note: KKmm ≠ ≠ KKdd for substrate binding! for substrate binding!

Saturation kinetics with respect Saturation kinetics with respect to substrate concentration.to substrate concentration.

Where E is enzyme molecule

S is the substrate molecule

ES is the enzyme-substrate complex

P is the resulting product molecule• Assumption (mostly valid in initial stages of forward

reaction and in hydrolysis reactions):

• k4 = 0 (no back reaction of product!)

E + Sk1

k2

ESk3

k4

E + P

In 1913, based on experimental observations of enzyme kinetics, Michaelis and Menten proposed model:

Michaelis-Menten Equation

Michaelis-Menten Enzyme Kinetics

• ES is not the same as an activated complex– Activated complex is energy maximum going from R to P

– Enzyme-Substrate (ES) is amount of substrate bound to enzyme.

• Rate is of primary importance in [ES]. Maximum rate occurs when all of the enzyme is in [ES] form (∞ [S])– Assume steady state, therefore [S] = [S] initial (not - ES) and P

formation is irreversible.

E + S k1

k-1

[ES]k2

k-2

E + P

Michaelis-Menten Enzyme Kinetics

E + S k1

k-1

ESk2

k-2

E + P

Vform = k1[E][S]

Free enzymeFor ES

but E = Etotal - ES

so, Vform = k1[Etotal - ES][S]

Vdeg = k2[ES] + k-1[ES]

so, vform = vdeg

vdeg = k2[ES] + k-1[ES]

At the steady state, rate of formation = rate of degradation

vform = k1[Etotal - ES][S]

k1[Etotal - ES][S] = k2[ES] + k-1[ES]

E + S k1

k-1

ESk2

k-2

E + P

[Etotal - ES][S] = k2+ k-1

k1[ES]

This defines KM

E + S k1

k-1

ESk2

k-2

E + P

[Etotal - ES][S] = k2+ k-1

k1[ES]= KM

[Etotal][S] - [ES][S] = KM

[ES] [ES]

[Etotal][S] = KM + S

[ES]

E + S k1

k-1

ESk2

k-2

E + P

[Etotal][S] = KM + S

[ES]

[Etotal][S] = [ES]

KM + S

vform = k2[ES]

= k2[Etotal][S]

KM + S

Solve in terms of products

E + S k1

k-1

ESk2

k-2

E + P

v = k2[ES]

= k2[Etotal][S]

KM + S

vmax= k2[Etotal]

v [Etotal] = v(KM + S)

k2 [S]

Rearrange the terms

v(KM + S)

k2 [S]=

k2

vmax= v(KM + S)

[S]

[Etotal][S] = [ES]

KM + S

substitute

E + S k1

k-1

ESk2

k-2

E + P

Can be rearranged in terms of v

vmax= v(KM + S)

[S]

v0 = Vmax[S]

KM + [S]

Where v0 = initial velocity concentration at substrate concentration

[S].

v0 [Substrate]/t [Product]/t

Vmax = maximum velocity, rate of reaction

{moles per second }

[S] is the initial substrate concentration

KM = Michaelis constant

Basic Michaelis-Menten Equation:

Michaelis-Menten Equation

€

KM =k2 + k3

k1

v0 = Vmax[S]

KM + [S]

Enzyme-catalyzed reactions must involve formation of an enzyme-substrate complex, followed by one or more chemical

steps

• Vmax and Km are two key measurable properties of enzymes.

Vmax:

• kkcatcat: : the rate constant for the catalytic step carried out by the rate constant for the catalytic step carried out by

the enzyme.the enzyme.

• VVmaxmax: : the rate at which a given amount of enzyme catalyzes the rate at which a given amount of enzyme catalyzes

a reaction at saturating concentrations of substrate.a reaction at saturating concentrations of substrate.

kcat [Etotal] = Vmax

• kkcat cat is interpreted as a measure of the rate of chemical is interpreted as a measure of the rate of chemical

conversion of substrate to product.conversion of substrate to product.

Enzyme-catalyzed reactions must involve formation of an enzyme-substrate complex,

followed by one or more chemical steps• Vmax and Km are two key measurable properties of

enzymes - Km

Km:

• KKm m is the ratio of the rate constants for the individual steps:is the ratio of the rate constants for the individual steps:

Km = (k-1 + kcat)/k1.

• If If kkcat cat is small compared to is small compared to kk-1-1, then , then KKm m = = KKss, the dissociation , the dissociation

constant for the enzymeconstant for the enzyme--ssubstrate complex. ubstrate complex.

• Thus for many enzymes, Thus for many enzymes, KKm m can be interpreted as a can be interpreted as a

measure of the affinity of the enzyme for its substrate. measure of the affinity of the enzyme for its substrate.

• KKm m also = [S] at which v = 1/2 also = [S] at which v = 1/2 VVmax max

Some Rules on Interpreting Michaelis-Menten Enzyme Kinetic Parameters (from A. Fersht

text)

• kcat is a 1st order rate constant that refers to the properties and reactions of the enzyme-substrate, enzyme-intermediate (‡) and enzyme product complexes.

• Km is an apparent dissociation constant that may be treated as the overall dissociation constant of all enzyme bound species.

• kcat/Km is an apparent second order rate constant that refers to properties of the free enzyme and the free substrate.

Michaelis-Menten Enzyme Kinetics

E + S k1

k-1

[ES]k2

k-2

E + P

v0 = Vmax[S]

KM + [S]

KM + [S]

Vmax[S]1/v0 =

1/v0 =KM

Vmax[S]

1+

1

Vmax

Basic Michaelis-Menten Equation: If we take the inverse of MM eq:

Use the equation for a straight line: y = ax + b

Lineweaver-Burk Inversion

1/V0 =KM

Vmax[S]

1+

1

Vmax

slope y-intercept

Plot 1/v (= y) versus 1/[S] (= x)

1/V

1/[S]

y-intercept = 1/Vmax

slope = KM/Vmax

x-intercept = -1/KM

Figure U2-2.3 Lineweaver-Burk or reciprocal kinetic plot of 1/v against 1/[S]

1. Original raw kinetic data 2. Transformed (inverse) data

Note: this L-B method weights data incorrectly, puts emphasis on data near minimum rate (Vmax) : (e.g., 1/ 0.05 = 20, 1/0.1 = 10)

You should perform non-linear fit of actual raw data to equation with You should perform non-linear fit of actual raw data to equation with computer software (KaleidaGraph, Sigmaplot, Orig, GraphPad computer software (KaleidaGraph, Sigmaplot, Orig, GraphPad

Prism, kinetic fitting packages, etc.)Prism, kinetic fitting packages, etc.)

Another inversion method: multiply MM eq by V0Vmax

Eadie-Hofstee Plot

Plot v (= y) versus v/[S] (= x)

v

v/[S]

Vmax

slope = -KM

Vmax/KM

V0 = - KM [S]+ Vmax

V0v0 = Vmax[S]

KM + [S]

Hanes-Wilkinson Plot

S/V0 =KM

Vmax

[S] +1

Vmax

slope y-intercept

Plot 1/v (= y) versus 1/[S] (= x)

[S]/V

[S]

Slope =1/Vmax

KM/Vmax

-KM

v0 = Vmax[S]

KM + [S]

Another inversion method: multiply MM eq by [S]

Enzyme Kinetics: General Principles• Remember, Vmax = maximal rate of an enzyme (Etotal = ES)• KM = (k-1 + k2)/k1

• KM is the [S] at which V = 1/2 (Vmax)

v0 = Vmax[S]

KM + [S]

Basic Michaelis-Menten Equation:

If KM = [S]:

v0 = Vmax[S]

[S] + [S]

v0 = Vmax[S]

2 [S]

= Vmax

2

Enzyme Kinetics: General Principles

• If k-1 >> k2

Dissociation constant of the enzyme and substrate

k2+ k-1

k1

for KM = k-1

k1

= = KS

E + S k1

k-1

ESk2

k-2

E + P

• If k2 >> k-1

E + S k1

k-1

ESk2

k-2

E + P

Enzyme reacts every time it interacts with substrate. (see p. 480)

[S] moles/L 1/[S] v (µmol/min) 1/v2.00E-01 5.00 60 0.016666672.00E-02 50.00 60 0.016666672.00E-03 500.00 60 0.016666672.00E-04 5000.00 48 0.020833331.50E-04 6666.67 45 0.022222221.30E-05 76923.08 12 0.08333333

Example of Michaelis-Menten Enzyme KineticsGiven this data, what is Vmax? What is KM?

First, graph [S] vs. v to make sure it obeys MM kinetics

0

10

20

30

40

50

60

70

1.00E-05 1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00

v (µ

mol

/min

)

[S]

Vmax is 60 by inspection

[S] moles/L 1/[S] v (µmol/min) 1/v2.00E-01 5.00 60 0.016666672.00E-02 50.00 60 0.016666672.00E-03 500.00 60 0.016666672.00E-04 5000.00 48 0.020833331.50E-04 6666.67 45 0.022222221.30E-05 76923.08 12 0.08333333

Example of Michaelis-Menten Enzyme KineticsGiven this data, what is Vmax? What is KM?

Since Vmax = 60 we can solve for KM, plug this into MM eq.

v0 = Vmax[S]

KM + [S]

KM = [S] Vmax

v0

-1

If v = 48, [S]= 2 X 10-4, KM = 5.0 X 10-5

If v = 12, [S]= 1.3 X 10-5, KM = 5.2 X 10-5

These should agree with one another!

[S] moles/L 1/[S] v (µmol/min) 1/v2.00E-01 5.00 60 0.016666672.00E-02 50.00 60 0.016666672.00E-03 500.00 60 0.016666672.00E-04 5000.00 48 0.020833331.50E-04 6666.67 45 0.022222221.30E-05 76923.08 12 0.08333333

Example of Michaelis-Menten Enzyme KineticsGiven this data, what is Vmax? What is KM?

We can also check by Lineweaver-Burke plot

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

-40000 -20000 0 20000 40000 60000 80000 100000

-1/KM

1/Vmax 1/V0 =KM

Vmax[S]

1+

1

Vmax

Scale is important

1/v

1/[S]

Graph of rate against total substrate concentration for a typical enzyme-catalyzed

reaction • In this example, saturating

concentrations of substrate B are converted to product more slowly than saturating concentrations of substrate A

• (Vmax is lower for substrate B than for substrate A), but substrate B binds more tightly to the enzyme (reflected in its lower Km).

• The value of kcat/Km is higher for substrate B than A, indicating that the enzyme is more specific for substrate B.