Enzyme kinetics -- Michaelis Menten kinetics • Two approaches: 1. Rapid equilibrium approach 2. 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.
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Enzyme kinetics -- Michaelis Menten kinetics Two approaches: 1.Rapid equilibrium approach 2.Quasi steady state approach Assumptions: –Total enzyme concentration.
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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,