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Chapter 3
Problem 3.1
( )
( )
3
15
24
k
kk
2
1
kk
52
S E ES (ES) P E
rate ESk
+ ⎯⎯→ +
=
a.) Equilibrium approach
* r values are the same as k values all are rate constants
( )( ) ( )
12
1
r S E r ES=
(1)
( ) ( )
34
12
r ES r ES=
(2)
( ) ( ) ( ) ( )
012
E E ES ES= + +
(3)
from (2)
( ) ( )
4
12
3
r
ES ES
r
=
(4)
from (1) + (4)
( ) ( )
( )
2 4 2
13
ES
rr
Er r S
=
(5)
Substitute (4) and (5) into (3):
( ) ( ) ( )
( ) ( )
( ) ( )( )
( ) ( )
( )
2 4 1 4 1 3
02
13
1 3 0
22 4 1 4 1 3
50 24
12
2 1 1 3
r r r r S r r S
E ES
r r S
r r E S
ES r r r r S r r S
r E S rr
rate where Km , Km
Km Km S S r r
++
=
=
++
= = =
++
b.) Quasi–steady–state approach
( ) ( )( ) ( ) ( ) ( )
11 2 4 3
1 2 1
d ES r E S r ES r ES r ES 0
dt = − + −
(6)
( ) ( ) ( ) ( )
23 4 5
1 2 2
d ES r ES r ES r ES 0
dt = − −
(7)
from (7)
( ) ( )
45
12
3
rr
ES ES
r
+
=
(8)
from (6) + (8):
( ) ( ) ( )
2 4 2 5 3 5
2
13
r r r r r r
E ES
r r S
++
=
(9)
substitute (8) and (9) into (3):
( ) ( ) ( ) ( )
( ) ( )
2 4 2 5 3 5 1 4 1 5 1 3
02
13
r r r r r r r r S r r S r r S
E ES
r r S
+ + + + +
=
( ) ( )( )
( ) ( ) ( )
1 3 0
22 4 2 5 3 5 1 4 1 5 1 3
r r E S
ES r r r r r r r r S r r S r r S
=
+ + + + +
( )( )
50
2 5 3 1 5 1
r E S
rate Km r r Km S r r S
= + + + +
where
2
1
1
r
Km r
=
and
4
2
3
r
Km r
=
Problem 3.2
* r values are the same as k values all are rate constants
( )
12
12
rr
rr
E S ES E P
−−
++
( ) ( )( ) ( )( ) ( ) ( )
1 2 1 2
d ES r E S r E P r ES r ES 0
dt −−
= + − −
(1)
( ) ( ) ( )( )
22
dP
rate r ES r E P
dt −
= = −
(2)
( ) ( ) ( ) ( ) ( ) ( )
00
E E ES , E E ES= + = −
(3)
from (1):
( ) ( ) ( ) ( )
12
12
rr
E ES
r S r P
−
−
+
=+
(4)
Combine (3) + (4):
( ) ( ) ( )
( ) ( ) ( )
1 2 1 2
0
12
r S r P r r
E ES
r S r P
−−
−
+ + +
=
+
( ) ( ) ( )
( )
( ) ( )
0 1 2
1 2 1 2
E r S r P
ES r S r P r r
−
−−
+
= + + +
(5)
Substitute (3) and (5) into (2):
( ) ( )
( ) ( ) ( )
( )
( )
1 2 0 2 2 0 0
1 2 1 2
r r E S r r E P
rate r 2 E ES P
r r r S r P
−
−−
+
= − − −
+ + +
( ) ( )
1 2 1 2
−−
Problem 3.3
5
12
m
1
kk
k
−
−+
b)
6
0
3
m
1
m
E 10 M
S 10 M
VS
VKS
−
−
=
=
+
=+
but
m 2 0
V r E=
41
V 9.58 10 Msec
−−
=
Problem 3.4.
At low substrate concentrations So< 150 mg/l substrate inhibition is negligible.
Problem 3.5.
Plot 1/V versus 1/S at different inhibitor concentrations
4
Problem 3.6 .
a. V = Vm (1 + A/KA) S /(Ks + S) = Vm (1+A/KA) / (1+Ks/S)
Problem 3.7.
Problem 3.8
E
2 2 3
H O CO HCO H
−+
++
Plot 1/V versus 1/S (Lineweaver – Burk plot)
11
1
−
51
41 m
41
m2
412
m
m
V 7.61 10 Msec
11.31 10 M sec
Vr 2.72 10 sec
1 1.31 10 M K 1.24 10 M
K 163.15
−−
−
−
−−
=
= → =
− −
= → =
Problem 3.9
a.) Plot 1/V versus 1/S:
11
9.525 0.60
VS
=+
1m,app
1 0.60 0.063M K 16.0gS/ L
−
−−
b.)
( )
( )
m,app m
I
I
m,app m
I
K K 1 K
I
K 1.37g I L
K K 1
=+
==
−
6
Problem 3.10
Plot 1/V versus 1/S
a.) For E0 = 1.6 g/L
b.)
m
1 mmol
V 3.31
0.302 ml.min
==
c.) The inhibitor is competitive.
( ) ( )
m,app m m,app
I
I 0.6 mmol mL
I
K K 1 K 0.052 mmol mL
K
=
=+
=
Problem 3.11
44
m
4
2
MW ATPase 5 10 g mol, K 1 10 mol L
S 0.02 mol L, r 1 10 min molecule enzyme
−
= =
= =
12
1
rr
ATPase
r
ATP ADP Pi or E S ES E P
−
⎯⎯⎯→ + + ⎯⎯→ +
Enzyme inactivation kinetics: E = E0e-rt, r = 0.1 min-1
Michaelis-Menten Kinetics:
m
m
VS
VKS
=+
But
m m 2
S K , V V r E =
rt
20
dp r E e
dt
−
=
0.002 rt
20
00
dp r E e dt
−
=
( )
8
20 0
rE
0.002 M 1 , E 2.0 10 M
r
−
−
= − =
( )
8 3 4 6
mol g g
2.0 10 1 10 L 5 10 1 10 1.0 g
L mol g
−−
=
Fraction of pure enzyme
1.0 g 0.1
10 g
==
10% of the crude protein was ATPase.
Problem 3.12
At steady – state: reaction rate = mass transfer rate
Problem 3.13
Some of the error may be attributed to the method since both axes contain v. However, there
are two possible explanations for the deviations from Michaelis – Menten Kinetics:
Problem 3.14
a.) Plot [P] versus t data for both cases.
9
Obtain rates by taking tangents at specific time points.
dt
Obtain the amount of substrate present for both cases by a mass balance.
Using initial time data, a plot of 1/V versus 1/S can be made.
10
Soluble:
Immobilized:
4
m
m
m
1 1 mg 1 mg
0.12, V 1.80 10
V 0.12 mL min 46 400 units mL min unit of enzyme
1
K 575 mg mL
0.00174
−
= = =
==
b.) Plot In Vm versus 1/T for both cases.
Ea RT
m 2 0 0
a
m0
V r E E Ae
E
Ea
ln V ln E ln A slope
RT R
−
==
= + − =
a
mol K mol
Problem 3.15
In vitro batch reactors represent a closed system of constant volume, thus Michaelis-Menten
Problem 3.16
Problem 3.17
(a) Plot of V us. S indicates that this is substrate inhibition
12
(b) Vm andK'm are determined from low substrate concentration where inhibition is not
ineffect:
'
m2
Kr
SI
2
m
2
'
m
SI
E S ES E P
S
K
ES
VS
uS
KS
K
+ ⎯⎯→ +
+
=
++
SI
m
K1
S
(c) Rate of reaction at S = 70 mg/l?
Since inhibition effect is observed at this substrate, we must use (eg) 3.34
( )( )
( ) ( ) ( )
m
22
'
m
SI
VS 20.3 70
u70
S31.1 70
KS 260.5
K
u 11.9 mg l hr
==
++
++
=
Problem 3.18
reaction rate with diffusion limitation
Effectiveness factor = reaction rate with no diffusion limitation
(a) Effectiveness factor at
P
P
100 mg l hr
D 0.5cm 200mg l hr
50
50 mg l hr
D 0.7cm 200mg l hr
0.25
= → =
=
= → =
=
14
b. Assume negligible film resistance, so Sbulk = Ssurface Determine rate u/o diffusional
limitation from effectiveness factor:
mm
1/ V is at x 0: y 0.0084626 V 118.2 mg uA / l h= = → =
m
K 49.0mg uA l
=
Problem 3.19 a
m0
S
S0
VS
rKS
= +
(1)
m
Se
V
RKD
= +
(2)
3 =
(3)
15
( )
mm
V 0.5
23.5
3 0.595 5.04
= =
b.
( )
( )
( )
m
m5
m m m
V
similarly, V 14; 0.2 0.2 1.5 10 3600
V 26.95 V 1.925 V
−
= =
= =
Problem 3.20
(a) Plot 1/V vs 1/S for all cases to determine the type of inhibition
(b) Determine Vm, Km, KI
Vm, Km can be determined from the I = 0 case.@ I = 0 (1/V) = 0.15782 + 0.19404 (1/S)
m
m
0.19404 K
==
KI must be determined from the I > 0 case
'
m, app
I 1.26 mM determine K
I 1.95 mM
=
=
( ) ( )
( )
m,app
m,app
m,app
I 1.26mM : 1 V 0.15917 0.3098 1 S
0 0.15917 0.3098 1 K
0.15917 1 K 1.95mM
0.3098 K
= = +
= + −
= → =
( ) ( )
( )
m,app
m,app
m,app
I 1.95mM : 1 V 0.15343 0.40383 1 S
0 0.15343 0.40383 1 K
0.15343 1 K 2.63mM
0.40383 K
= = +
= + −
= → =
m, app m
I
I
K K 1 K
=+
I
I
I
I
1.26mM
I 1.26mM : 1.95mM 1.23mM 1 K
1.26
1.58 1 K
1.26
0.58 K 2.16mM
K
= = +
=+
= → =
Problem 3.21
3
L
3
b
K 2.10 m sec 0.2cm sec
S 1000mg L 1mg cm
−
==
==
Then
SS
3
S
S 0.62mg cm 620 mg L
==
b.
( )
( )
( ) ( )
mS L b S
SS
2
VS
V K S S
KS
0.1 0.62
V 0.2 1 0.62 0.076 mg cm sec.
0.2 0.62
= = −
+
= = − =
+
Problem 3.22
a) Use the graphical technique:
where A is a Function of surface concentration and is determined through rxn Kinetics.
1
b)
( )
2
S L 2 3
3 3 6 2
mm
m2 6 2 5 3 3
2
m2
3
2
.005mg
rate of supply rate of diffusion away V K S surface cm
K 5e mg cm ; V 4e mgkm s
V S surface 3.2 10 mg cm s 4e cm s S surface 5e mg cm
K S surface
S surface .013mg cm 13mg L
dP 2.89
dt
−−
− − −
= − = − −
= =
= − −
+
==
=
62
10 mg cm s
−
19
2152
1.0 10 mg cm S
−
d) Sbulk = 370 mg/L
