Enzyme Kinetics overheadsjsharp/enzyme.pdfEnzymes with kcat/KM values between 108 and 109 M‐1...
Transcript of Enzyme Kinetics overheadsjsharp/enzyme.pdfEnzymes with kcat/KM values between 108 and 109 M‐1...
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Dr. Joshua SharppRoom 1088
Complex CarbohydrateComplex Carbohydrate Research Center
Phone: 706‐542‐3712Phone: 706‐542‐3712e‐mail:
jsharp@ccrc uga [email protected]
E Ki tiEnzyme Kinetics
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Catalyst does not change energy of substrate or products. The equilibrium constant remains unchangedconstant remains unchanged.
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To increase rate of reaction 10‐foldTo increase rate of reaction 10 fold, energy of activation must be decreasedby 5.71 kJ/mol.
Since the equilibrium constant is the same, catalysts increase the rate of both the forward and the reverseboth the forward and the reverse reactions.
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When [sucrose] >>>When [sucrose] >>> [invertase], kinetics are independent of sucrose concentration.
Why? All free enzyme has been converted into an enzyme‐sucrose complex.
k1 k2E + S ES E + P
1
k‐1
2
k‐2
k1 k2
SES
P
k‐1 k‐2E E
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E + S ES E + Pk1
k
k2
kk‐1 k‐2
So, the initial velocity of product formation can be expressed as:formation can be expressed as:
]ES[]P[ 20 k
dtdv ==dt
[ES])[ES]([S][E][ES]211 kkkd
+−= − [ES]) [ES]([S] [E] 211 kkkdt
+
Formation of ES l
Dissociation f ES l
Can we integrate this equation?
ES complex of ES complex
[ES]dIf we assume that: 0[ES]=
dtd
That is, the rate of formation of the [ES] complex equals the rate of dissociation of the [ES] complex
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[ES] and [E] are not readily measurable, but the initial concentration of enzyme [E]0 can be determined. [ES]dIf then: [ES][E][E]0 += 0[ES]
=dt
d
[ES])([S][ES])([E] 2101 kkk +=, or:
[ES])([S] [ES])([E] 2101 kkk +=− −
[S] [E] [S])( [ES] 01121 kkkk =++−
If you solve for [ES], then:
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S][
[S] E][[ES] )( 21
0
++
=− kk 1
21
kkkKM
+=
−
S][)(1
+k
1k
[S][E][P] 02)( kd[S] [S] [E][ES][P] 02
200 )(+
=== =M
tK
kkdt
dv
The fastest an enzymatic reaction can go is when theThe fastest an enzymatic reaction can go is when the enzyme is saturated with substrate; thus [E]0 = [ES]. This maximal velocity is termed Vmax.
Combining the Vmax , KM, and the equation for
02max [E] kV =Combining the Vmax , KM, and the equation for determining initial velocity, we get:
[S]maxV[S]
[S]max 0
+=
MKVv
This is the Michaelis‐Menten equation, and it is the basic equation of enzyme kinetics
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Variations of the Michaelis‐Menten Equation
[S][S]max
0+
=MK
Vv[ ]
max0
Vv =[S]1
0MKv
+=
0
max 1 S][V
VKM−
=0V
])[(0 SKV M +[S]
])[(0max
SKVV M +=
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When [S] = Km, the velocity of the reaction v = Vmax/2
k1 k2E + S ES E + P
1
k‐1
2
k‐2
1
21
kkkKM
+=
− or:1
2
1
1
kk
kkKM +=
−
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[S][S]max
0 =KVv
[S]+MK
When the reverse reaction is negligible:When the reverse reaction is negligible:
[S]S][ max − VdS][
[ ]][+
==MKdt
v
[S] ])S[( S][ maxVdtKd M =+−
K )( dtVKd M max)( 1
[S] S][ −=+
tVKM )[S]]S([S][ln 0)( =+ tVKM )[S]]S([S][
ln max00)( −=−+
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1max
0MK
Vv =[S]1 MK+
)(maxmax0
1[S]
1 1 )( VVK
vM +=
Lineweaver Burk plotLineweaver‐Burk plot
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Hanes–Woolf plot
01
[S]11 )( VV
Kv
M +=maxmax0 [S])( VVv
1[S] Kmaxmax0
S][1[S]V
KVv
M+=
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Eadie‐Hofstee plot
01
[S]11 )( VV
Kv
M +=maxmax0 [S])( VVv
00 )( VVKv M += max00 )( [S] VVKv M +−=
Vo/[S]
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E + S ES E + Pk1
k
k2
kk‐1 k‐2
For standard one‐intermediate Michaelis‐Menten kinetics, kcat = k2. For models where each enzyme has multiple active sites, n is the number of active sites per enzyme.
0
maxcat
[E]nVk =
0[E]n
[S] [S] [E] 0cat
0+
=MK
kv
When [S] << KM, the steady‐state [ES] is very small, so [E]0 ≈ [E].
k [S] [E]cat0
MKkv =
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kcat/KM = specificity constantkcat/KM specificity constant
E + A EA E + A’E + A EA E + AE + B EB E + B’
For two reactions competing for the same p genzyme:
[A] cat )( AMK
kvA=
[B] cat )( BMK
kvB
=
So kcat/KM is a measurement of which reaction is more efficient for the enzyme, independent of the substrate concentrationthe substrate concentration.
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21
12 catkk
kkK
kM +=
− 21 kk +If k2 >> k‐1, then kcat/KM approaches k1. Therefore, the ultimate limit on the value of k /K is the rate of formation of the enzyme‐kcat/KM is the rate of formation of the enzyme‐substrate complex.
If the enzyme‐substrate complex forms in every instance that the enzyme encounters the substrate, k1 will still be limited by the limit of diffusion. Depending on conditions, the limit of diffusion is generally around 108‐109 M‐1 s‐1.
Enzymes with kcat/KM values between 108 and 109 M 1 1 f id h h d ki i109 M‐1 s‐1 are often said to have reached kinetic perfection—that is, the enzymatic efficiency is at the diffusion limit.
Some enzymes have partially overcome the diffusion limit by working together, passing off the product from one enzyme to be thethe product from one enzyme to be the substrate for the next and preventing the need for diffusion.
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[S][S] [E] 02
0+
=MK
kv[S]+MK
E + S ES EP E + Pk1 k2 k3
S Sk‐1 k‐2 k‐3
If we assume that k‐2 and k‐3 are negligible, then:2 3
[S]'[E] [S] 0
cat+
=MK
kv[ ]
23cat
kkk =32
catkk
k+
123' kkkK −+×
132 kkkK M ×
+=
If k3 >> k2 (that is, catalysis is the rate‐limitingIf k3 >> k2 (that is, catalysis is the rate limiting step), then we get the original equation.
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Enzyme Inhibition
SES
Pk1
k‐1
k2
k‐2E E
I
k k
I
k ' k ’
ESI
k3 k‐3k3 k‐3
P
EI
S
k1'
k‐1’EI
k2'
k‐2’S 1
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Competitive Inhibition
SES
Pk1
k‐1
k2
k‐2E E
I
k k
I
k kXESI
k3 k‐3k3 k‐3
PX
EI
S
k1
k‐1
EIXX3kKI −=
S 1
33k
kKI =
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E + S ES E + Pk1
k
k2
kk‐1 k‐2+I
k k
EI
k3 k‐3
1
21
kkkKM
+=
−
33k
kKI −=
The apparent KM of the enzyme is dependent on the equilibrium constant of the enzyme‐inhibitor complex and the concentration of inhibitor:complex and the concentration of inhibitor:
)( I][1' MM KKK += )(IK
[S]maxV[S] '
[S]max 0
+=
MKVv
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Pharmacological Example of Competitive InhibitionCompetitive Inhibition
Disulfiram
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Non‐competitive Inhibition
SES
Pk1
k‐1
k2
k‐2E E
I
k k
I
k k
ESI
k3 k‐3k3 k‐3
P
EI
S
k1
k‐1
EIXS 1
02max [E]kV = 02max [E] kV =
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E + S ES E + Pk1
k
k2
kk‐1 k‐2+I
k k
+I
k k
EI + S EIS
k3 k‐3
33k
kKI −=k1
k 1
k3 k‐3
The apparent Vmax of the enzyme is dependent on the apparent amount of available enzyme,
k‐1
on the apparent amount of available enzyme, which in turn is dependent on the affinity of the inhibitor for the enzyme and the concentration of the inhibitor.
IKVV
]I[1' max
max+
=IK]I[1+
[S]'maxV[S] [ ]
0+
=MK
v
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Uncompetitive Inhibition
SES
Pk1
k‐1
k2
k‐2E E
I
k k
I
k kXESI
k3 k‐3k3 k‐3
PX
EI
S
k1
k‐1
EIXXS 1
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E + S ES E + Pk1
k
k2
kk‐1 k‐2+I
k k
EIS 33k
kKI −=k3 k‐3
KVV
]I[1' max
max+
=IK]I[1+
K I][1
'I
MM
KKK
+=
[S]'maxV[S] '[ ]
0+
=MK
v
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Substrate Inhibition
SES
Pk1
k‐1
k2
k‐2E E
k k k kXSS
ESS
k3 k‐3k3 k‐3
PX
EI k1
k‐1
ES*XX1
S
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E + S ES E + Pk1
k 1
k2
k 2+k‐1 ‐2
k
+S
k3 k‐3
ESS 33k
kKI −=3 3
v 0
[S]
1/v 0
V [S]1/[S]
IM KKVv 2
max 0
[S][S] [S]++
=
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Some Other Modes of Inhibition
• Mixed Inhibition—inhibitor binds to both free enzyme and enzyme substrate complex, but with different affinities. This causes an increase in the apparent KM, and a decrease in the apparent Vmax.
• Partially competitive inhibition a variation• Partially competitive inhibition—a variation of non‐competitive inhibition where the enzyme‐substrate‐inhibitor complex still has some catalytic activity, but reduced from the y y,enzyme‐substrate complex
• Product inhibition—inhibition where the d i hibi P dproduct can act as an inhibitor. Product
inhibition is a relatively common form of negative feedback regulation in biochemical pathwayspathways
• Irreversible inhibition—inhibition where the rate of release of the inhibitor is essentially zero. Often involves covalent modification of the enzyme by the inhibitor
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Problems to go over next class
bl• Problem 8.27
• Problem 8.28
• Problem 8.36
• Problem 8.37
• BMB 4110 Exam on Enzyme Kinetics from Sept. 2008