enzyme kinetics - chapter 4
TRANSCRIPT
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Why study enzyme kinetics?
To quantitate enzyme characteristicsdefine substrate and inhibitor affinities
define maximum catalytic rates
Describe how reaction rates vary with reaction
conditions Provide an understanding of an enzymes role in a
metabolic pathway
Define the conditions under which the rate of thereaction is proportional to the amount of enzymepresentbiochemical and clinical enzyme assays
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Enzyme Kinetics
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First- and Second- order reactions
First-order Second order
PA
][][
Akdt
Adv
][
][ 00
]ln[A
A
t
dtkAd
ekt
oAA ][][
ktAA 0]ln[]ln[
PA2
2][][ AkdtAdv
tA
Adtk
A
Ad
02
][
][
][
][
0
ktAA o
][
1
][
1
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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/][lnthen 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[
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e kt
A
A
0][
][
ktAA
A
][1][
][
0
Decay curves for 1st and 2nd orderreactions of same half-life
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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
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[A][B]k
or at equilibrium:
U + PV
G=H- TS
for most biological systems:
U
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QPXBA k
K
'
][']][[][
XkBAkdt
Pd
]][[
][
BA
XK
GKRT ln
]][['][ / BAek
dt
Pd RTG
'k
h
Tkk B
'
RTGB e
h
Tkk /
Thermodynamics of the Transition State
h/TkB
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Plots of Initial rates and Reactant ConcentrationsDiffer for enzymatic and non-enzymatic reactions
Non-enzymatic enzymatic
[S]
RATE
[A]
RATE
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-ES complex is formed
-[E]
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where Km = (k-1+ k2)/k1
since [S] = [St] early in the reaction and [E] = [Et] - [ES] then
PEESSE k
k
k
2
1
1 0][][]][[][
211
ESESSEdt
ESdkkk
][]][[
21
1 ESSE
kk
k
][)
]][[
121(
ESSE
kkk
K
SEES
m
]][[][
K
SESEES
m
t ]])[[]([][
Michaelis-Menten ModelDerivations
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K
SESEES
m
t ]])[[]([][
K
SES
K
SEES
mm
t ]][[]][[][
][as)][1(
][][][
2 ESv
KS
KSEES k
m
mt
KSEKSES mtm ]][[)][1]([
][Km
][Vmax
][
]][[2
S
S
SKm
SEtv k
Michaelis-Menten ModelDerivations (contd)
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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
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Lineweaver_BurkDouble reciprocal plot
1/Vmax
slope=Km/Vmax
1/ [ S]
1/v
-1/Km
][Km
][Vmax
S
Sv
Vmax
1
][
1
Vmax
Km1
Svbmxy
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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
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Graphical Representations of Changes in
Km and Vmax (contd)
1/[S]
1/v
Vmax
Km same
Km
Vmax same
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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 k
k
k
2
1
1
kcat= turnover number = Vmax
[Et]
turnover number is the first order rateconstant describing the number ofmolecules of substrate converted toproduct per molecule of enzyme in
units ofsec-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) ofan enzyme is a second order rate
constant in units of M-1sec-1
For highly efficient enzymes whoserates are nearly diffusion limiting:
keffis near 108to 109M-1sec-1
carbonic anhydrase = 8 x 107M-1sec-1
acetylcholinesterase =1.5 x 108M-1sec-1
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Kinetic constants for some enzymes and
substrates
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Reversibility of the reaction adds complexity
to the rate equation
PEESSE
k
k
k
k
2
2
1
1
P
M
S
M
P
M
r
S
M
f
K
P
K
S
K
PV
K
SV
v][][
1
][][ maxmax
For the simple reversible reaction
The velocity of the reaction is:
WhereT
f EkV ][2max Tr EkV ][1max
1
21
k
kkK
S
M
2
21
k
kkKPM
When [P] =0, i.e. when v = v0then the more familiar
Michelis-Menten form is
evident:
][
][
][1
][max
max
SK
SV
K
S
KSV
vS
M
f
S
M
S
M
f
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3O
F
CH
3CH
3CH
3CH
OH SER-PROTEIN
B:
3 O
F
CH
3CH
3CH
3CH
O
SER-PROTEIN
B:H
-
HO
H
3 OCH
3CH
3CH
3
CH
O
SER-PROTEIN
O H
H
B:
FAST VERY
SLOW
C H - O - P - O - C H C H - O - P - O - C H C H - O - P - O - C H
Enzyme Inhibition:
Irreversible
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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
FAST
Un-competitive
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S
I
S
P
I I
S
P
I
+S
+ I+ I
+S
FAST
Non-competitive
Enzyme Inhibition:
Reversible (mixed)
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Enzyme Inhibition:
Reversible (competitive)
v
V m ax
[S ]
K m K m
-I
+ I
I
m
KI
SK
SVv
][1
][
][max
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1/ [S]
1/v
1 / V m a x
-1/Km
-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
][1where
1
][
11
maxmax
Enzyme Inhibition:
Reversible (competitive)
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Enzyme Inhibition:
Reversible (competitive)
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Enzyme Inhibition:Reversible (uncompetitive)
E E S E + P
E IS E I + P
I
S
[ES][ I ]
[E IS ]K i =
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 /
=(1 + [I]/Ki)
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Enzyme Inhibition:Reversible (uncompetitive)
maxmax ][
11
VSV
K
v
m
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Enzyme Inhibition:Reversible mixed (non-competitive)
E E S E + P
E I E IS E I + P
I I
S
S
[E + ES ] [I]
[E I + E IS ]K i =
V m a x [S ]
K m + [S ]( )V =
v
V m a x
V m a x
[S ]
K m
-I
+ I
=(1 + [I]/Ki)
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Enzyme Inhibition:Reversible mixed (non-competitive)
K m
[S ]V m a x
1
V m a x+
1
v=
1/ [S]
1/v
-I
+ I
V m a x
1 / V m a x-1/Km
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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
Enzyme Inhibition:Reversible mixed (non-competitive)
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maxmax
'][
11VSV
Kv
M
Enzyme Inhibition:Reversible mixed (non-competitive)
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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
ABSORB
ANCE
t
A
LACTATE + NAD+ PYRUVATE +
H+ + NADH
Use of Substrate or Product Absorbance to Measure
Rates of a Reaction