LECTURES 8‐12: Enzyme Kinetics
Problem 1
Determine the initial rate of product (P) formation from enzyme X and
substrate Y that form the complex XY. The reaction kinetics is given by
Solution:
Michaelis-Menten approach : The rate of reaction is given by
rp = k f2 C XY − k b2 C P C X
Si
Since
enzyme is
i preserved
d
C X0 = C X + C XY
(1.1)
(1.2)
S b tit ti equation
Substituting
ti (1.2)
(1 2) in
i (1.1)
(1 1) we gett
rp = (k f2 + k b2 C P )C XY − k b2 C P C X0
(1.3)
Assuming the first reversible reaction is in equilibrium gives
C XY
kf1
=
CXCY
k b1
(1.4)
Substituting equation (5) in equation (3) for CX and rearranging for CXY yields
CXY =
Substituting equation (6) into equation (2) gives
CX0 CS
k b1
CS
k f1
⎛
⎞
k b2 k b1
⎜
k f2 C X0 ⎜ C Y −
C P ⎟⎟
k f2 k f1
⎝
⎠
rp =
k b1
+ CS
k f1
Equation (7) is in Michaelis-Menten form of equation, where
rMax = k f2 C XO
and
KM
k b1
=
k f1
(1.5)
(1.6)
Problem 2
Determine
D
t
i the
th initial
i iti l rate
t off product
d t (P) formation
f
ti from
f
enzyme X andd
substrate Y that form the complex XY. The reaction kinetics is given by
Solution : We can write the mass balance for the intermediates and product as:
d[(XY )1 ]
= k b1 [X ][Y ] − (k f1 + k 2 )[(XY )1 ]
dt
d[(XY )2 ]
= k 2 [(XY )1 ] − (k f3 )[(XY )2 ]
dt
d[P ]
= (k f3 )[(XY )2 ]
dt
where
k2 =
(2.1)
(2.2)
(2.3)
k f2
k b2
Assuming [(XY)1] and [(XY)2] are in steady state, we can eliminate
[(XY)2] from equation (2.4)
k f1 [X ][Y ]
[(XY )1 ] =
k b1 + k 2
[(XY )2 ] =
k2
[(XY )1 ] = k 2 • k 1 [X][Y]
k f3
k f3 k b1 + k 2
(2.4)
( )
(2.5)
Substituting equation (2.6) into equation (2.4) we get
k 2 k f1 [X ][Y ]
d[P]
rp =
= k f3 •
•
dt
k f3 k b1 + k 2
Since the enzyme is preserved
(2 6)
(2.6)
[X 0 ] = [X] + [(XY )1 ] + [(XY )2 ]
(2.7)
Substituting for [X] in terms of [X0] in equation (2.8) we get
k 2 k f3
[X][Y]
(k 2 + k f3 )
rp =
k f3
k b1 + k 2
•
+ [Y ]
k 2 + k f3f
k f1f
(2.8)
Hence the KM and Vmax values are reduced by the ratio kf3/(k2+ kf3),
resulting from the presence of the second intermediate [(XY)2].
]
Problem3
The loading of O2 to Hb (hemoglobin) follows a cooperative binding
and is given by
Develop a graphical method for determination of the coefficient ‘n’
from measurements of Hb (O2)n
Solution:
Total hemoglobin concentration is given by
[Hb]T = [Hb] + [Hb(O 2 )n ] = Constant
[Hb(O 2 )n ] =
[Hb]T
[Hb(O 2 )n ]
[Hb] + [Hb(O 2 )n ]
(3.1)
Given reaction is
Hb + nO 2 ⎯
⎯→ Hb(O 2 )n
Assuming steady state, therefore
K 1 [Hb][O 2 ] = K 2 [Hb(O 2 )n ]
n
(3.2)
From equation (3.1) and equation (3.2) we get
K1
[Hb][O 2 ]n
[Hb(O 2 )n ] = K 2
[Hb]T
[Hb] + K 1 [Hb][O 2 ]n
K2
(3.3)
As concentration is directlyy proportional
p p
to ppressure,, we can write equation
q
(3.3) as
K1 n n
α PO 2
[Hb(O 2 )n ] = K 2
K
[Hb]T
1 + 1 α n POn 2
K2
(3.4)
After the algebraic manipulations
[Hb(O 2 )n ] =
[Hb]T
POn 2
(3.5)
K1
n
P
+
O
2
K 2α n
Now, binding of oxygen to hemoglobin in terms of partial pressure of
oxygen is given by Hill Equation.
Hill Equation:
SO2 =
POn2
P +P
n
50
n
O2
(3.6)
Therefore
SO2
[
Hb(O 2 )n ]
=
(3 7)
(3.7)
[Hb]T
From equation (3.6) and equation (3.7)
SO 2 =
POn 2
(3.8)
K1
n
P
+
O2
K 2α n
Rearranging we get
Taking log on both side, we get
⎧⎪ S O 2
⎨
⎪⎩1 − S O 2
⎧⎪ S O 2
ln ⎨
⎪⎩1 − S O 2
⎫⎪
K1
n
P
⎬=
O2
n
⎪⎭ K 2 α
⎫⎪
⎛ K1
⎜
ln
=
⎬
⎜ K αn
⎪⎭
⎝ 2
⎞
⎟⎟ + n ln PO 2
⎠
(3 9)
(3.9)
(3 10)
(3.10)
⎧⎪ S O 2
Thus a logarithmic plot of ⎨
⎪⎩1 − S O 2
slope ‘n’
⎫⎪
⎬ versus PO2 will give a straight line of
⎪⎭
Fig. 3.1 Binding of oxygen to hemoglobin versus Partial pressure of oxygen
Problem 4
An inhibitor Y is added to the enzymatic reaction at a level of 1.2 gm/L. Their data
were obtained for KM=10 gm/L.
Rate
1.8
1.3
0.98
0.8
0.67
0.6
0.45
Substrate
20
9.8
6.7
5.1
4.2
3.2
2.6
a. Identify
Id tif th
the ttype off iinhibition
hibiti (i
(i.e. C
Competitive,
titi N
Non-competitive
titi or
Uncompetitive)
b Find the Ki (Inhibitor kinetic constant)
b.
Solution: Lets us assume that the given enzymatic reaction is
inhibited by competitive inhibition.
inhibition For an instance of competitive
inhibition the Lineweaver-Burke equation is
K M 1 ⎡ [I] ⎤
1
1
=
⎢1 +
⎥+
V VMax [S] ⎣ K i ⎦ VMax
So we pplot
1
V
1
versus to gget the values of VMax and Ki
S
Fig. 4.1. 1/V versus 1/S plot
(4 1)
(4.1)
From fig.4.1 we get Vmax=1.8181. Slope = 2.33826, therefore
⎡ [I] ⎤
⎢1 +
⎥ = 2.3383
⎣ Ki ⎦
Hence inhibitor constant Ki = -2.087
1
VMax
For Competitive inhibition, Ki is the dissociation constant for the
enzyme –inhibitor complex. A lower value of Ki denotes a stronger
inhibition. So as the value of Ki (-2.087) is negative our assumption
was correct and it is a competitive inhibition.
It is important to note that competitive inhibition can be overcome by
adding additional substrate.