Lectures 5 and 6
Inhibition Systems and Derivation of their Rate Equations
Vm[S]
S +Km =
Vmax
v
1)
v=
2)
Lineweaver Burk - 1
v
vs
v vs. [S] - Hyperbolic
1
[S]
Linear Plot or Double Reciprocal Plot
Basic Equation
V = Vm[S]
[S] +Km
1
v
=
1
Vmax
y
=
b
(TAKE RECIPROCAL)
Km
+ Vm
+
intercept
1
v
1
[S]
m
slope
Km
Vm
1
Vmax
∞,
@ [S]
1
Km
X
1
[S]
1
[S]
O,
1
∴ Y intercept = Vmax
@1
v
=O
1
=
Vm
-
Km
Vm
Cross Multiply
1
1
Km = S
1
S
Km 1
Vm x [S] = O
v
EADIE-HOFSTEE PLOTS - v vs [S]
V=
1
v
(Gives Km & Vmax directly)
Vm[S]
[S]+Km
Km
1
= Vm + Vm
1
Vmax
Multiply by Vmax • v
Slope = - Km
v
Vm = v + Km s
v
Vmax
Km
Or rearranging
v
v = Vm - (Km) x [S]
Y= b
m
x
v
[S]
Rules for deriving rate laws for simple systems
1.
Write reactions involved in forming P from S
2.
Write the conservation equation for expressing the total enzyme
concentration [E]total among the various species
3.
Write the velocity dependence equation, summing all the
catalytic rates constants multiplied by the concentration of the
respective produt - forming species.
4.
Divide the velocity dependence equation by the conservation
equation.
5.
Express the concentration of each enzymic species in terms
of free enzyme concentration & substitute
6.
Algebra
For a simple 1 step reaction, no inhibitor
(1)
k1[S]
E+S
k3
ES
E+P
k2
(2)
[E]T = [E] = [ES]
(3)
v = k3[ES]
k3[ES]
v
[E]T = [E] + [ES]
(4)
(5)
[ES] = k1[S] • [E] - k2[ES] - k3[ES]
[ES] = k1[S] • [E] - (k2+ k3)[ES]
@ steady state
d[ES]
dt
= O, ∴ rate of formation = rate of breakdown
k1 [S] • [E] = (k2 + k3)[ES]
[ES] =
(6)
k1[S]•[E]
,
k 2 + k3
Define Km = k2 + k3
k1
Substitute into step (4)
[S]
vv
K
k33 • Km •[E]
[ET]
[E
]
T
=
[S]
E + Km • [E]
=
[S]•[E]
Km
[S]
v
÷ Top and bottom = k3 Km = [E ]
T
By[E]
1+ S
Km
Multiply by Km
v
[ET]
k3 [S]
=
;
Km + [S]
= [ES]
Define
Vmax [S]
Vmax = k3 [ET] ; v = Km + [S]
EQUATION FOR COMPETITIVE INNIBITION • MUTUALLY EXCLUSIVE
BINDING OF S AND I
k1[S]
k3
E +S
ES
E+P
k2
k on [I]
k off
- Can drive all E to ES
By increasing [S]
- Since [I] & [S] are mutually
exclusive binders, [I] apparently
decreases affinity for E, e.g. Km
EI
[E]T = [E] + [ES] = [EI]
v = k3 [ES]
v
[E]T
=
k3[ES]
[E]+[ES]+[EI]
[ES] = k1 • [S] • [E]
(k2 + k3) =
[EI] =
1
Km
[S]
= Km • [E]
k on [E] • [I] = k off [EI]
@ steady-state d[EI]
dt
dissociation
= O;
[I] • [E]
KI
[S]
v =
k3 Km • [E]
[E]T
[S]
E + Km • [E] + [I] • [E]
k on
k off
1
KI=
i.e., receprocal
constant
[EI]
k3 [S]
Km + [S] + Km [I]
KI
Define Vmax &
Vmax [S]
Double
1
Collect Terms
Reciprocal
=
[I]
v =
Km 1+ KI + [S] ; Equation
[I]
1
Km 1
Vmax + Vm [S] 1+ KI
b+
x
m
DETERMINING KI FROM SLOPE REPLOT
-Measure
1
v
vs
1
S
@ several [I]
increasing [I]
X intercept @ 1 = O
v
∴
1
Vmax
= Km
Vm
1+
I
KI
I
S
1
v
[I] = O
As [I] , Slope increases
1
v
Cross multiply
1
- Km
I
1
- Km 1+ K
I
Slope of 1 vs 1 = Km (1+ [I]
s
v
Vm
KI
=
1
Km
1+
[I]
Km
1
[S]
1
[S]
PLOT SLOPE vs [I]
SLOPE
Km
Vm
Km
Vmax
[I]
Km
Vmax KI
Km
Km[I]
SLOPE = Vmax + Vmax•K
I
y
=
b
+
m
x
@ SLOPE O (i.e. x-intercept)
-Ki
Km
-Km
=
• I
Vmax Vmax KI
Cross Multiply
-KI = [I] @ slope = O
NON COMPETITIVE INHIBITION
k1[S]
E+S
k3
ES
E+P
k2
k on I
k off
k1[S]
EI + S
ESI
k2
[E]T = [E] + [ES] + [EI] + [ESI]
v = k3 [ES]
v
[E]T =
k3[ES]
[E] + [ES] + [EI] + [ESI]
[S]
[ES] = Km
I
• [E] ; [EI] = KI
• [E]
[ESI] = Cannot easily calculate [ESI] by steady state hypothesis
∴ Most assume equilibrium thus is valid since EI + ESI are in equilibrium
(i.e. k3 = O for ESI ∴ k2 > > > k3)
THUS:
k on [ES] • I
k off
[ESI] =
v
[E]T
v
[E]T
;
S
•
[I]
Km
=
[E]
KI
S
Km •[E]
S
E+
• E + [I] • E + [I]
Km
KI
KI
k3 [S]
Km + [S] + Km [I] + [S]
KI
k3
=
•
Vmax[S]
v =
1+
[I] S +
KI
1+
I
Km
KI
[S] • E
Km
1
1
=
Vmapp
Vm
I
KI
1+
1
v
Slope = Km
Vm
1+
[I]
KI
Increasing [I]
[I] =O
1+
1
=Yint
Vmax
=
1
Vmaxapp
I
KI
1
Vmax //
-Km
I
1
1
@Xint,
∴
1+
Vmax
K
S
I
V=O
1
S
Cross Multiply
1
1
=
[S]
Km
Replots to determine KI
Slope of
1 1
v, S
Slope =
y
Slope
Slope =
Km
Vm••KI
Km
Vmax
[I]
I
1
=Vmax 1+ KI
Km
Plot = Vm
1+
[I]
KI
Km + KM[I]
Vm
VmKI
b
+
m x
x int. of slope vs [I]
@ Slope = O
∴-
Km[I]
Km
= VmK [I]
I
Vm
Cross Multiply
-KI
-KI = [I] @ Slope =O
yCan also do intercept replot as well
1
1
[I]
Y int = Vmaxapp = Vmax
+ Vmax • K
I
x
m
Uncompetitive Inhibition
E+S
k1[S]
k2
Like non-competitive,
[S] @ ∞ cannot drive
enzyme to ES form
k3
ES
E+P
There is an obligate
order of binding
First S Then I
ESI
∴I should decrease
Km by driving
reaction E+S
ES
towards ES formation
V = k3 [ES]
[E]t = [E]+[ES]+[ESI]
[ES] =
k1[S] • [E]
k2+k3 = 1
Km
ESI=
Kon[I] • [ES]
Koff = 1
KI
V
=
[ET]
V
=
[ET]
V
=
=
[S] • [E]
Km
I • [ES] [S] • [I]
KI
Km • KI
=
k3 [ES]
[E]+[ES]+[ESI]
S
k3 Km • [E]
E+ S [E] + [S] • [I]
Km Km••kI
Vm [S]
Km+S+ [S][I]
Km+KI
Factor
Taking Reciprocal
1
V
=
I binds only to ES
Km 1
1
Vm[S]+ Vm
1+[I]
kI
Vm [S]
Km+S 1+ [I]
KI
∴Slope unaffected
Yint decreased by [I]
Increasing [I]
1
v
[I]=O
1
X int =Y = O;
1
V max
1
1
- Km
[S]
1
1+[I]/KI
- Kmapp = Km
1
Y int = Vmax
1
To obtain KI, Replot of V vs [I]
[I]
1
1
@ [S] = ∞ 1 = Vmax = Y intercept = Vmax 1+ kI
v
app
[I] - X
1
int = Vmax + Vmax••kI - m
b
1
Vmaxapp
1
@ X int = - Vmax app = O
1
KIVm
1
Vmax
[I]
- KI
∴
1
[I]
[]
Vm = - Vmax∞
∞KI
- KI = [I]
(I)
1+ KI
[I]
Km 1 1
Vm[S] = Vm 1+ KI
I
∴1 • 1
S V = O - 1+ KI
Km
=
COMPLEX INHIBITION
E+S
k1
k2
ES
k3
E+P
+
+
I
I
αKs
KI
αKs
EI + S
βk3
ESI
EI + P
α = O or ∞
- Competitive
α = 1, β = O
- Non-Competitive
α = 1, O < β < 1 - Partial Non-Competitive
1 < α < ∞, β = 1 - Partial Competitive
1 < α < ∞, β = O - Mixed Inhibition (Type 1)
1 < α < ∞, O < β <1 - Mixed Inhibition (Type 2)
In partial inhibition, the EI or ESI complexes are not
dead-end complexes as they are in simply inhibition
schemes
Partial Non Competitive Inhibition
k1
E+S
+
I
ES
k3
E+P
k2
KI
KI
k1
EI + S
ESI
βk3
E+P
k2
All forms of E (E + EI)
combine equally well with
S, ∴ Km does not change.
Vmax is decreased because
a portion of ES is ESI and
ESI EI + P is slower
V = k3 [ES] + βk3 [ESI]
[E]t = [E] + [ES] + [EI] + [ESI]
[S]
[ES] =
[E]
Km
[EI] = [I] [E]
KI
[I] [S]
I
[ESI] = K [ES] = KI Km [E]
I
[S]
[I]
V = k3 km E + βk3 K
I
[F]t E +
(÷
÷) by E
(x) by Km
Define
Vm = k3 [E]t
V=
[S]
E
Km
[S] [I]
[S] + [I]
+
Km KI
KI
Km
β[I]
Vm 1 + KI)
[S]
_______________________
[I]
[I]
Km 1 + KI + [S] 1 + KI)
V=
Vm [S]
[I]
Km 1 + KI
+ [S] 1 +
βI
1 + KI
1+
[I]
KI
βI
KI
Take Reciprocal
1
V
1
Vmax;
=
[I]
Km 1 + KI
βI
Vm 1 + KI
[I]
1+ KI
1
+
βI
Vmax
1+ KI
1
[S]
1
v
[I]
= l+ KI
Increasing [I]
βI
Ymax 1+KI
[I]=0
@ [I] = ∞
Ks
Slope = βVm
1
Y int = βVm
1
Vmax
1
[S]
[I]
1+ KI
Y intercept =
βI
Vmax 1+ KI
Simplifies to
1
1
=
Y int = Ym
app Vmax
•
[I] + KI
β[I] + KI
Hyperbolic
1
Limit = βVmax
Y int =
1
Vmaxapp
To obtain βVmax
1
Plot Vmaxapp vs [I]
[I]
Vmapp = Vmax (β
β[I]+KI)
[I] + KI
Vmapp = βVmax + VmKI
[I]
multiply both
terms B
I+Kt
I
Vmapp = βVm[I] ± VmKI
[I]+KI
[I]+KI
Partial Competitive
k1
k2
E+S
+
k3
ES
k2k3
Km= k1
E+P
+
I
I
δKI
KI
αKm
EI + S
ESI
k3
EI
V = k3 [ES] + k3 [ESI]
[E]t E + EI + ES + ESI
[ES] = [S] • [E]
Km
[I]
[I]
ESI = αKI • [ES] = αKI
•E
[EI] = [I]
KI
The presence of I inhibits on
rate of S, ∴ Km is increased
(affinity lowered) by a factor
α (kα
α < 0). Since ESI, EI and
E are in equilibrium, and
since it makes intuitive sense
that S would interfere with I
binding, KI is increased by a
factor 1 < α < 00.
[S]
• Km
S
[I]
[S]
[E]
V = k3 Km • [E] + k3 αKI • Km
÷ by [E]
x Km
assume
Vm=k3 [E]t
[S]
[I]
[I] [S]
[E]
[E]t [E] + Km + KI • [E] + αKI Km
[I] [S]
Vm [S] + αKI
Km [I]
Km + [S] + KI
+
Factor
[I] [S]
αKI • Km
[I]
Vm 1 + αKI S
[I]
Km 1 + KI
[I]
+ S 1 + αKI
Vmax [S]
I
÷ by 1 + αKI
[I]
=
1 + KI
Km 1 + [I] + [S]
αKI
0r
Vmax[S]
Km App + [S]
Take Reciprocal
[I]
1 + KI
[I]
1 + αKI
Km
= Vmax
1
V
Slope =
[I]
Km 1+ KI
[I]
Vm 1+ αKI
1
[S]
Increasing
[I]
1
V
1
Vmax
Slope =
I=0
1
S
-1
Kmapp
1
- Km
Km
Vmax
+
1
Vmax
[I]
1 + KI
[I]
1 + αKI
= Km
+
Vmax
I + KI
I + αKI
=
I
Km KI
I
Vm αKI
Simplified to:
Slope =
αKm
Vm
Hyperbolic Plot
limit = αk3
Vmax
Slope
Makes sense in that in presence of
infinite [I] can still bind S, albeit weaker
than in absence of I
[I]