4. enzyme-ii
TRANSCRIPT
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Enzyme kinetics
-- Michaelis Menten kinetics
Two approaches:1. Rapid equilibrium approach
2. uasi steady state approach
!ssumptions: Total enzyme concentration remains constant
durin" the reaction
!mount o# enzyme is $ery small compared toamount o# substrate
The product concentration is so low that theproduct inhibition is ne"li"ible.
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where
k1% #orward rate constant #or #ormation o# E&comple'
k2% backward rate constant #or #ormation o#
E& comple'k(% rate constant #or #ormation o# product )
++
2
31
k
kk PEESSE
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some notations**..
e % concentration o# enzyme
s % concentration o# substrate
p % concentration o# product +es, % concentration o# enzyme substrate
comple'
t % time $ % reaction rate or $elocity
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Michaelis Menten kinetics
--Rapid equilibrium approach
t is assumed that E& comple' is established$ery rapidly+since this equilibrium step is onlythe #ormation o# weak interaction between E
&,
The product releasin" step +k(, is $ery
slow**.which determines the rate
The rate o# re$erse reaction o# the second stepis ne"li"ible
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The equilibrium constant
The rate o# product #ormation/ +mol0 l.s,
The total enzyme concentration
)(
1
2
es
se
k
kK ==
)(3 eskdtdpv ==
)(3esk
dtdpv ==
)(1
2
es
se
k
k
K ==
1
2
)(0 esee += (
v
++
2
31
k
kk
PEESSE
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et an e'pression #or +es, in known quantities**
&ub. eqn +1, in +(,
)()3(0
esee +=
)()( essKes +=
+= 1)(s
Kes
+=
1
)( 0
s
K
ees
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ow sub. the $alue o# +es, in eqn 2.
)()2( 3 eskv =
+
=s
K
ek
1
03
sKsek+=
03
sK
svv
M
+
= max
velocityMaximalekv =03max
(
)tan(1
2 tconsMichaelisKk
kK M ==
3Michaelis Menten Equation
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There#ore**..
is a #unction o# enzyme concentration only
! low $alue o# means that the enzyme has hi"h a##inity #or
the substrate
Three special cases**..
4ase +s%5M,
4ase +s665M,
4ase +s775M,
maxvM
K
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4ase +s%5M,
Eqn. +3, %6
&o when s%5M/ the rate o# reaction is one hal# o#
its ma'imal $alue.
i.e. at which 389 o# enzyme acti$e sites areoccupied by substrate
ss
svv
+=
max
2
maxv
v =
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4ase +s665M,
Eqn. +3, %6
+=
1
max
s
Ks
svv
M
maxvv = ---------ER; ;R
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4ase +s775M,
Eqn. +3, %6
sK
vv
M
max
= ---------=R&T ;R
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Michaelis Menten kinetics
--uasi steady state approach
This approach is assumed that the chan"e in
the intermediate +transition comple',
concentration with respect to time is ne"li"ible.
(pseudo steady state/quasi steady state)---Briggs-Haldane approach
i.e. 0
)(=
dt
esd
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++
2
31
k
kk
PEESSE
)(3esk
dt
dpv ==
)(21eskesk
dt
ds+=
)()()(
321eskeskesk
dt
esd=
1
1
1
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>y p.s.s assumption/
Eqn ?%%6
0)(=
dt
esd
0))(()(
321=+= kkesesk
dt
esd
))((321eskkesk +=
)(1
32
esskkke
+=
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@e know/ )(0
esee +=
++=
sk
kkes
1
321)(
sk
kkees
1
32
0
1
)(+
+
=
321
10)(kksk
skees++
=
d
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sub the $alue o#
+es, in +1,
)(3esk
dt
dpv ==
sK
svv
k
kks
sek
k
kksk
sekk
kksk
sekk
M+
=
++
=
+
+
=
++=
max
1
32
03
1
32
1
013
321
013
321
013
kksk
sekk
++=
++
=
1
32
1
013
kkksk
sekk
++
=
1
32
03
kkks
sek
sK
svv
M +
= max
1
32
kkkKM +=
03maxekv =
Michaelis Menten Equation
C t f E ilib i h
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Controversy of Equilibrium approach:
by Equilibrium approach/
Means that A&B is costant*..which is notcorrect
21
10)(ksk
skees
+
=
)(21 eskeskdtds +=
)(0
eseebut += )()]([201esksesek +=
)()(2101esksesksek ++=
])[(2101
kskessek ++=
][
][
21
21
01
01ksk
ksk
seksek +
+
+=
0=dt
ds
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This has been recti#ied by )&& approach*..
by )&& assumption
321
10)(
kksk
skees
++
=
)(21eskesk
dt
ds+=
)()]([201
esksesek
+=])[(
2101kskessek ++=
][21
321
01
01ksk
kksk
seksek +
++
+=
321
021
2
0
2
1
01
kksk
sekkseksek
+++
+=
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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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Ci#e o# a man**
>e#ore marria"e*. &)