enzyme kinetics -- michaelis menten kinetics two approaches: 1.rapid equilibrium approach 2.quasi...

Enzyme kinetics -- Michaelis Menten kinetics

• Two approaches:1. Rapid equilibrium approach2. Quasi steady state approach

• Assumptions:– Total enzyme concentration remains constant

during the reaction– Amount of enzyme is very small compared to

amount of substrate– The product concentration is so low that the

product inhibition is negligible.

where

k1= forward rate constant for formation of ES complex

k2= backward rate constant for formation of ES complex

k3= rate constant for formation of product P

2

31

k

kk PEESSE

some notations……..

• e = concentration of enzyme

• s = concentration of substrate

• p = concentration of product

• (es) = concentration of enzyme substrate complex

• t = time

• v = reaction rate or velocity

Michaelis Menten kinetics --Rapid equilibrium

approach• It is assumed that ES complex is established

very rapidly (since this equilibrium step is only the formation of weak interaction between E & S)

• The product releasing step (k3) is very slow…….which determines the rate

• The rate of reverse reaction of the second step is negligible

The equilibrium constant

The rate of product formation, (mol/ l.s)

The total enzyme concentration

)(3esk

dt

dpv

)(1

2esse

k

kK 1

2

)(0 esee 3

v

2

31

k

kk PEESSE

Get an expression for (es) in known quantities……

Sub. eqn (1) in (3)

)()3(0

esee

)()( ess

Kes

1)(s

Kes

1

)( 0

sKe

es

• Now sub. the value of (es) in eqn 2.

)()2(3eskv

sKek

1

03

sK

sek

03

sK

svv

M

max

velocityMaximalekv 03max

(

)tan(1

2 tconsMichaelisKk

kK

M

5Michaelis Menten Equation

• Therefore……..

is a function of enzyme concentration only

A low value of means that the enzyme has high affinity for the substrate

• Three special cases……..Case I (s=KM)

Case II (s>>KM)

Case III (s<<KM)

maxv

MK

Case I (s=KM)

Eqn. (5) =>

So when s=KM, the rate of reaction is one half of its maximal value.

i.e. at which 50% of enzyme active sites are occupied by substrate

ss

svv

max

2maxv

v

Case II (s>>KM)

Eqn. (5) =>

1

max

sK

s

svv

M

maxvv --------- ZERO ORDER

Case III (s<<KM)

Eqn. (5) =>

sK

vv

M

max ---------FIRST ORDER

1

max

M

M Ks

K

svv

Michaelis Menten kinetics --Quasi steady state

approach• This approach is assumed that the change in

the intermediate (transition complex) concentration with respect to time is negligible. (pseudo steady state/quasi steady state)---

Briggs-Haldane approach

i.e. 0)(

dt

esd

2

31

k

kk PEESSE

)(3esk

dt

dpv

)(21eskesk

dt

ds

)()()(

321eskeskesk

dt

esd

1

1

1

By p.s.s assumption,

Eqn 4==>

0)(

dt

esd

0))(()(

321 kkesesk

dt

esd

))((321eskkesk

)(1

32 essk

kke

We know, )(0

esee

sk

kkes

1

321)(

sk

kke

es

1

32

0

1)(

321

10)(kksk

skees

• sub the value of (es) in (1)

)(3esk

dt

dpv

321

013

kksk

sekk

1

32

1

013

k

kksk

sekk

1

32

03

k

kks

sek

sK

svv

M

max

1

32

k

kkK

M

03maxekv

Michaelis Menten Equation

• Controversy of Equilibrium approach:by Equilibrium approach,

Means that ‘S’ is costant…..which is not correct

21

10)(ksk

skees

)(21eskesk

dt

ds

)(0

eseebut )()]([201esksesek

)()(2101esksesksek

])[(2101kskessek

][][ 21

21

01

01ksk

ksk

seksek

0dt

ds

• This has been rectified by PSS approach…..

by PSS assumption321

10)(kksk

skees

)(21eskesk

dt

ds

)()]([201esksesek

])[(2101kskessek

][21

321

01

01ksk

kksk

seksek

321

021

2

0

2

1

01 kksk

sekkseksek

321

021

2

0

2

1031021

2

0

2

1

kksk

sekkseksekksekksek

1

32

1

031

k

kksk

sekk

dt

dp

sK

sv

dt

ds

M

m

321

031

kksk

sekk

goodholdsandcorrectiswhich

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