M.Prasad NaiduMSc Medical Biochemistry,Ph.D.Research Scholar

Enzymes follow zero order kinetics when substrate

concentrations are high. Zero order means there is no increase

in the rate of the reaction when more substrate is added.

Given the following breakdown of sucrose to glucose and

fructose

Sucrose + H20 Glucose + Fructose

O

H

HO

H

HO

H

OH

OHHH

OH

OH

HOH

H OH

O

H

H

HO

H

H

H

OH

P EES S E 2

1

1-

kk

k

E = Enzyme S = Substrate P = Product

ES = Enzyme-Substrate complex

k1 rate constant for the forward reaction

k-1 = rate constant for the breakdown of the ES to

substrate

k2 = rate constant for the formation of the

products

When the substrate concentration becomes large

enough to force the equilibrium to form completely

all ES the second step in the reaction becomes rate

limiting because no more ES can be made and the

enzyme-substrate complex is at its maximum value.

ESP

2kdt

dv

[ES] is the difference between the

rates of ES formation minus the

rates of its disappearance.

ESESSEES

211 kkkdt

d

1

Assumption of equilibrium

k-1>>k2 the formation of product is so

much slower than the formation of the ES

complex. That we can assume:

ES

SE

1

1

k

kK s

Ks is the dissociation constant for the ES complex.

Assumption of steady state

Transient phase where in the course of a reaction the

concentration of ES does not change

0ES

dt

d

2

ES E E T 3

Combining 1 + 2 + 3

ESk k SES-Ek 21-T1

SEk Sk k kES T1121-

S K

SE ES T

M

1

21-

k

k k KM

rearranging

Divide by k1 and solve for [ES] Where

SK

SEES

P T22

0 Mt

o

kk

dt

dv

vo is the initial velocity when the reaction is just starting out.

And is the maximum velocityT2max Ek V

SK

SVmax

M

ovThe Michaelis - Menten

equation

The Km is the substrate concentration where vo equals

one-half Vmax

The KM

can be expressed as:

1

2

1

2

1

1 KKk

k

k

k

k

ksM

As Ks decreases, the affinity for the substrate

increases. The KM can be a measure for substrate

affinity if k2<k-1

There are a wide range of KM, Vmax , and efficiency

seen in enzymes

But how do we analyze kinetic data?

maxmax

M

V

1

S

1

V

K1

ov

Lineweaver-Burk plot: slope = KM/Vmax,

1/vo intercept is equal to 1/Vmax

the extrapolated x intercept is equal to -1/KM

For small errors in at low [S] leads to large errors in 1/vo

T

max

E

Vcatk

kcat is how many reactions an

enzyme can catalyze per second

The turnover number

For Michaelis -Menton kinetics k2= kcat

When [S] << KM very little ES is formed and [E] = [E]T

and

SEK

kSE

K

k

M

catT

M

2ov

Kcat/KM is a measure of catalytic efficiency

When k2>>k-1 or the ratio21

21

kk

kkis maximum

Then1

MKk

kcatOr when every substrate that hits

the enzyme causes a reaction to

take place. This is catalytic

perfection.

Diffusion-controlled limit- diffusion rate of a substrate

is in the range of 108 to 109 M-1s-1. An enzyme lowers

the transition state so there is no activation energy

and the catalyzed rate is as fast as molecules collide.

All substrates must combine with enzyme before reaction can occur

Bisubstrate reactions

Group transfer reactions One or more products released before all

substrates added

Although a phenomenological description can be

obtained the nature of the reaction intermediates

remain indeterminate and other independent

measurements are needed.

There are three types of inhibition kinetics competitive,

mixed and uncompetitive.

•Competitive- Where the inhibitor competes with the

substrate.

Competitive Inhibition

SK

SV

M

maxov

IK

I1

EI

IEK I

Competitive Inhibition: Lineweaver-Burke Plot

Uncompetitive Inhibition

Uncompetitive Inhibition: Lineweaver-Burke Plot

Mixed inhibition is when the inhibitor binds to the

enzyme at a location distinct from the substrate

binding site. The binding of the inhibitor will either

alter the KM or Vmax or both.

EI

IEK I ESI

IESK I

SK

SV

M

maxov

IK

I1