why study enzyme kinetics? to quantitate enzyme characteristics define substrate and inhibitor...

Why study enzyme kinetics?

To quantitate enzyme characteristics define substrate and inhibitor affinities define maximum catalytic rates

Describe how reaction rates vary with reaction conditions

Provide an understanding of an enzyme’s role in a metabolic pathway

Define the conditions under which the rate of the reaction is proportional to the amount of enzyme present

biochemical and clinical enzyme assays

Enzyme Kinetics

First- and Second- order reactionsFirst-order Second order

PA

][][

Akdt

Adv

][

][ 00

]ln[A

A

tdtkAd

ekt

oAA ][][

ktAA 0]ln[]ln[

PA2

2][][

Akdt

Adv

tA

Adtk

A

Ad02

][

][ ][

][0

ktAA o

][

1

][

1

Half-life of first-order reactions

When [A] = 1/2 the initial concentration [A]0

ekt

oAA ][][

ekt

oAA 2/1][2

][ 0

2/1][

2/][ln then kt

A

A

o

o

kkt

693.02ln and 2/1

(For a second-order reaction:t 1/2 = 1/k [A]o)

ktAA 0]ln[]ln[

ekt

A

A 0][

][

ktAA

A][1

][

][

0

Decay curves for 1st and 2nd order reactions of same half-life

First- and Second- order reactions

First-order Second order

Ln[A]

Time Time

1/[A]

ln[A] = ln[A]0 -kt 1/A = 1/A0 + ktln[A]0

Slope = -k

1/A0

Slope = k

[A][B]k1 = k-1[C][D]

U = Ub - Ua = Q - W

A + B C + Dk1

k-1

=Keq=[C][D][A][B]

k1

k-1

S

P

S

REACTION COORDINATE

EN

ER

GY

Go'

G

S Pk1

k-1

0'G =-RT ln KeqG = G + RT ln

0' [C][D]

[A][B]or at equilibrium:

U + PV

G =H - TS

for most biological systems:U

QPXBA kK

'

][']][[][ XkBAk

dt

Pd

]][[

][

BA

XK

GKRT ln

]][['][ / BAek

dt

Pd RTG

'k

h

Tkk B

'

RTGB eh

Tkk /

Thermodynamics of the Transition State

h/

TkB

Plots of Initial rates and Reactant Concentrations Differ for enzymatic and non-enzymatic reactions

Non-enzymatic enzymatic

[S]R

AT

E[A]

RA

TE

-ES complex is formed

-[E]<<[S]

-initial rate of [P] formation negligible

-EP----> E + P very fast

-Briggs-Haldane modification assumes [ES] is constant at steady-state.

PEESSE kk

k

2

1

1

Michaelis-Menten ModelAssumptions

where Km = (k-1 + k2)/k1

since [S] = [St] early in the reaction and [E] = [Et] - [ES] then

PEESSE kk

k

2

1

1 0][][]][[][

211

ESESSEdt

ESdkkk

][]][[

21

1 ESSE

kkk

][)

]][[

121(

ESSE

kkk

K

SEES

m

]][[][

K

SESEESm

t ]])[[]([][

Michaelis-Menten ModelDerivations

K

SESEESm

t ]])[[]([][

K

SES

K

SEESmm

t ]][[]][[][

][ as )][1(

][][][

2ESv

KSKSE

ES km

mt

KSEKSES mtm ]][[)][1]([

][Km

][Vmax

][

]][[2

S

S

SKm

SEtv k

Michaelis-Menten ModelDerivations (cont’d)

Km and VmaxDerivation and significance

When [S] >> Km then v = Vmax [S]/[S] = Vmax

When v = Vmax/2 x then Vmax = 2 Vmax [S] Km + [S]

or Km + [S] = 2 [S] and Km = [S]

since Km = (K -1+ K2)/K1 thus when K2 << K -1

then Km = K -1/K1 = Kd

A high Km = weak binding

A low Km = tight binding of the substrate to the enzyme

v = Vmax [S] Km + [S]

[S]v

Vmax

Vmax/2

Km

PEESSE kk

k

2

1

1

Lineweaver_BurkDouble reciprocal plot

1/Vmax

slope=Km/Vmax

1/ [ S]

1/v

-1/Km

][Km

][Vmax

S

Sv

Vmax

1

][

1

Vmax

Km1 Sv

bmxy

Graphical Representations of Changes in Km and Vmax

v

V m a x

V m a x

[S ]

2

1

K m

v

Vmax

[S]Km Km1 2

Graphical Representations of Changes in Km and Vmax (cont’d)

1/[S]

1/v

VmaxKm same

KmVmax same

Use of “turnover number” and “catalytic efficiency” as measures of enzyme behavior

For the above reaction at saturatingsubstrate concentrations Vmax = k2 [Et]or Vmax = kcat [Et]

v = Vmax [S] Km + [S]

PEESSE kk

k

2

1

1

kcat = turnover number = Vmax [Et]

turnover number is the first order rateconstant describing the number of molecules of substrate converted to product per molecule of enzyme in

units of sec-1

At low [S], v approaches the value: v= Vmax[S] or kcat [Et] [S] Km Km

v = kcat [Et] [S] = keff [Et] [S] Km

Thus, the catalytic efficiency (keff) of an enzyme is a second order rate constant in units of M-1sec-1

For highly efficient enzymes whoserates are nearly diffusion limiting: keff is near 108 to 109 M-1sec-1

carbonic anhydrase = 8 x 107 M-1sec-1

acetylcholinesterase =1.5 x 108 M-1sec-1

Kinetic constants for some enzymes and substrates

Reversibility of the reaction adds complexity to the rate equation

PEESSE

k

k

k

k

2

2

1

1

PM

SM

PM

r

SM

f

KP

KS

KPV

KSV

v][][

1

][][ maxmax

For the simple reversible reaction

The velocity of the reaction is:

Where Tf EkV ][2max T

r EkV ][1max

1

21

k

kkK SM

2

21

k

kkK PM

When [P] =0, i.e. when v = v0

then the more familiar Michelis-Menten form is evident:

][

][][

1

][

max

max

SK

SV

KS

KSV

vSM

f

SM

SM

f

3O

F

CH

3CH

3CH

3CH

OH SER-PROTEIN

B:

3 O

F

CH

3CH

3CH

3CH

OSER-PROTEIN

B:H

-

HO

H

3 OCH

3CH

3CH

3CH

OSER-PROTEIN

O H

H

B:

FAST VERY

SLOWC H - O - P - O - C H C H - O - P - O - C H C H - O - P - O - C H

Enzyme Inhibition:Irreversible

Enzyme Inhibition:Reversible

ENZ ENZ ENZ

S

P

+

FAST

ENZ

I

I

+I

+S

S

Competitive

S

S

P

I

S

P

I

+ S

+ I

F A S T

Un-competitive

S

I

S

P

I I

S

P

I

+ S

+ I+ I

+ S

F A S T

Non-competitive

Enzyme Inhibition:Reversible (mixed)

Enzyme Inhibition:Reversible (competitive)

v

V m ax

[S ]

K m K m

-I

+ I

I

m

KI

SK

SVv

][1

][

][max

1 /[S ]

1/v

1 /V m ax

-1 /K m-1

S L O P E = K m V m a x

S L O P E = K m /V m a x

+ I

-I

K m

I

m

K

I

VSV

K

v

][1 where

1

][

11

maxmax


Enzyme Inhibition:Reversible (uncompetitive)

E E S E + P

E IS E I + P

I

S

[E S ][I]

[E IS ]K i =

V m a x [S ]

K m + [S ]V =

v

V m ax

V m ax

[S ]K m

-I

+ I

K m /

=(1 + [I]/Ki)

Enzyme Inhibition:Reversible (uncompetitive)

maxmax ][

11

VSV

K

vm

Enzyme Inhibition:Reversible mixed (non-competitive)

E E S E + P

E I E IS E I + P

I I

S

S

[E + E S ][I]

[E I + E IS ]K i =

V m a x [S ]K m + [S ]( )V =

v

V m ax

V m ax

[S ]

K m

-I

+ I

=(1 + [I]/Ki)


K m

[S ]V m a x1

V m a x+

1

v=

1 /[S ]

1/v

- I

+ I

V m a x

1 /V m ax-1 /K m

E E S E + P

E I E IS E I + P

I I

S

S

V m a x [S ]

K m + '[S ]V =

v

V m a x

V m a x

[S ]

K m / '

-I

+ I'

K m

][

]][[

][

]][[

'

ESI

IESK

EI

IEK

I

I

IK

I ][1

'

][1'

IK

I


maxmax

'

][

11

VSV

K

vM


FOR A= 0.10; t = 1 MIN.; PATH = 1 CM

VOLUME = 1 ML

RATE = (0.16 X 10 M/MIN)(0.001 LITERS)= 16 NMOLES/MIN

-4

-4

L=

t

I o I

CUVETTE

DETECTOR

ABSORBANCE = A = LOG (I/ I o)

BEER'S LAW A = CL

L

C = CONCENTRATION (MOLAR)L = PATH LENGTH (CM)

= MOLAR EXTINCTION COEFFICIENT

(M CM )-1 -1

= 6.23 X 10 M CM

FOR NADH AT 340 nm

3 -1 -1

A/t C = 0.16 X 10 M/MIN

TIME

AB

SO

RB

AN

CE

t

A

LACTATE + NAD+ PYRUVATE + H+ + NADH

Use of Substrate or Product Absorbance to Measure Rates of a Reaction

why study enzyme kinetics? to quantitate enzyme characteristics define substrate and inhibitor...

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