1). In a competitive inhibition experiment a
structural analog was used along with the
substrate and the following kinetics was
observed:
At 10µM substrate the velocity was 25 µM/min.
With 2mM of the analog the velocity dropped
to 50%. Calculate the Ki of the inhibitor.
Given that the substrate concentration used
gives half-maximal velocity. Calculate how
much inhibitor should be used for increasing
the Km to 10 times the uninhibited value?
a). s  10M .......velocity  25
M
min
( without
i  2mM  2000M ......velocity, v  12.5
M
min
inhibition)
( with inhibition   50%)
v
M
 v  50
2
min
only when s  K 
G.T ...@ s  10M ......velocity 
m
m
M
 K  10M
b). if K M*  10 K
M
K  ???
I
Competitive Inhibition.....
v s
v

i 
K 1    s
 K 
m
M
I
 K  1000M
I
M
 i ?


i
  10 K
K 1 
M K 
M

I
 i  9000M
2).(Shuler3.7) An enzyme ATPase has a molecular weight of 5x104 Daltons, a
Km value of 10-4 M and a k3(rate constant for product formation) value of
104 molecules ATP/min molecule enzyme at 37°C . The reaction catalysed
is the following:
ATPase
ATP
ADP + Pi
which can be also be represented as
E+S
ES
E+P
where S is ATP. The enzyme at this temperature is unstable. The enzyme
inactivation kinetics are first order:
E =Eo EXP (-kdt)
where Eo is the initial enzyme concentration and kd = 0.1min-1.
In an experiment with a partially pure enzyme preparation , 10µg of
total crude protein (containing enzyme) is added to a 1 ml reaction mixture
containing 0.02M ATP and incubated at 37°C. After 12 hours the reaction
ends (i.e., t  ∞) and the inorganic phosphate (Pi) concentration is found to
be 0.002M, which was initially zero. What fraction of the crude protein
preparation was the enzyme?
Hint: Since [S] >> Km, the reaction rate can be represented by
d [ P]
 k3[E]
dt
dp
 k [ E ]  k E exp( k t )
dt
3
3
0.002
0
d
720
 dp  k E  exp( k t ) dt
3
0
0.002
0
d
0
720
 dp  k E  exp(  k t ) dt
3
0
0
d
0
10
1
0.002 
E
min 0.1 min
4
0
1
exp( 0.1x720)  1  E
mol
g
10 lt 5 x10
lt
mol
 1x10 g  1g
 2 x10
8
3
4
6
Fraction of enzyme in crude protein 
1g
 0.1  10%
10g
 2 x10 M
8
0
3.
The following data were obtained from enzymatic
oxidation of phenol oxidase at different phenol
concentrations.
S(mg/l)
10
20
30
50
60
80
90
110
130
140
150
Rate, V (mg/l.
h)
5
7.5
10
12.5
13.7
15
15
12.5
9.57
7.5
5.7
i). What type of inhibition is this?
ii). Determine the constants Vm, Km, and Ksi.
iii). Determine the oxidation rate at [S] = 70 mg/l.
• Plot ‘v’ vs ‘s’-----Substrate Inhibition
• Plot ‘1/v’ vs ‘1/s’…..find KM and Vm….since at
low substrate concentration no substrate
inhibition
we know s  K K
opt
M
I
 K  ??
I
• sub ' s '  in v 
vs
m
s
K s
K
2
M
s
 v  ??
4. What is the turn-over number of 10-6M
solution of Carbonic anhydrase catalyses the
formation of 0.6M H2CO3 per second when it is
fully saturated with substrate?
v
Turn over Number k   ??
e
m
cat
0
5. The following data were obtained for an enzymecatalyzed reaction. Determine Vm and Km by inspection.
Plot the data using the Eadie-Hofstee method and
determine these constants graphically. Explain the
discrepancy in your two determinations. The initial rate
data for the enzyme-catalyzed reaction are as follows:
[S]mol / l
5 x 10-4
2 x 10-4
6 x 10-5
4 x 10-5
3 x 10-5
2 x 10-
1.6 x 10-5
1.0 x 10-5
8 x 10-6
V(µmol /
min)
125
125
121
111
96.5
62.5
42.7
13.9
7.5
5
Do these data fit into Michaelis-Menten kinetics? If
not, what kind of rate expression would you suggest?
Use graphical methods.
Eadie Hofstee Plot
140
120
V/S
100
80
60
40
20
0
0
0.5
1
1.5
2
2.5
3
V
By inspection Vm =125 micromole/min
Km = 20 micromole / l
3.5
Eadie Hoftsee plot
140
120
100
y = -8.2507x + 130.36
80
Slope = Km=8.25micromole / l
Vm= 130.36 micromole / min
V
60
40
20
0
0
1
2
V/S
3
4
6. Lipase is being investigated as an additive to
laundry detergent for removal of stains from
fabric. The general reaction is:
Fats
Fatty acids + Glycerol
The Michaelis constant for pancreatic lipase is 5
mM. At 60ºC, lipase is subject to deactivation
with a half life of 8 min. Fat hydrolysis is carried
out in a well-mixed batch reactor which
simulates a top loading washing machine. The
initial fat concentration is 45 gmol /m3. At the
beginning of the reaction the rate of hydrolysis
is 0.07 m mol/ l s. How long does it take for the
enzyme to hydrolyze 80% of the fat present?
Enzyme subjected to Deactivation......
K  5mM
M
ln 2
t  8 min  480 sec 
k
1/ 2
d
 k  1.44 x10 sec
3
1
d
we know.....
mol
s
1  k 

s  45
 45mM
t   ln 1   K ln  ( s  s)
k
s
m

 v 
@ the beginning of the rxn.
mmol
mM
 t  1642.83 sec  27.38 min
v  0.07
 0.07
lt. sec
sec
d
0
3
d
m0
80% conversion
i.e. s  (1  0.8)45  9mM
t  ????
0
M
m0
0
7.
Amyloglucosidase from Endomycopsis bispora is
immobilized in polyacrylamide gel. Activities of immobilized
and soluble enzyme are compared at 80ºC. Initial rate data
measured at a fixed substrate concentration are listed
below: What is the half life for each enzyme?
Time
(min)
Enzyme Activity
(µmol / ml min)
Soluble enzyme
Immobilized enzyme
0
0.86
0.45
3
0.79
0.44
6
0.70
0.43
9
0.65
0.43
15
0.58
0.41
20
0.46
0.40
25
0.41
0.39
30
-------
0.38
40
-------
0.37
• We know ln v = ln v0 - kd t
• Plot ln ‘v’ vs ‘t’
EZNYME DEACTIVATION
0
-0.2 0
5
10
15
20
25
30
35
40
-0.4
Soluble Enzyme
ln v
-0.6
-0.8
-1
y = -0.0051x - 0.8078
Immobilized Enzyme
-1.2
y = -0.0295x - 0.1546
-1.4
-1.6
tim e (m in)
• Therefore for soluble enzyme….th = ln 2 /
kd==>ln2/0.029=23.79 min
• For immobilized enzyme….
th =ln2/0.0051=135.91min stability is
enhanced
45
8).Shuler 3.2 consider the reversible product formation in an
enzyme catalyzed reaction:
ES 
 ES 
 E  P
k1
k3




k2
k4
Develop a rate expression for product formation using the
quasi-steady-state approximation and show that
dp v K s  v K  p
v

s
p
dt
1

K
K
s
M
p
M
p
p
where v  k e , v  k e
s
3
0
p
2
k k
k k
K 
,K 
k
k
2
3
M
2
p
1
4
0
3
ES 
 ES 
 E  P
k1
k3




k2
k4
dp
v
 k (es)  k ep
dt
3
4
d (es)
 k es  k (es)  k (es)  k ep
dt
@ pss  k es  k (es)  k (es)  k ep  0
1
2
1
3
2
4
3
 k k 
 e
(es)

k s  k p 
2
3
1
4
e  e  (es)
3
1
4
0
 k k 

(es)  (es)

k s  k p 
1
k  k  k s  k p 
 e  (es) 

k
s

k
p


2
0
2
4
3
4
1
4
 v  k (es)  k ep
3
4
 k k 
 k (es)  k p 
(es)

k s  k p 


 k k 
k sk p
ke 
 k p
e


k s  k p 
k  k  k s  k p 
2
3
3
4
1
4
1
3
2
4
0
3
4
2
3
1
4
1
4


k sk p
k  k  k s  k p 


1
0
2
4
3
1
4
 k sk e  k k e p  k k e p  k k e p 


k k k sk p


1
3
0
3
2
4
0
2
3
1
4
0
3
4
4
0
take v  k e , v  k e
s
3
0
p
2
0
devide both sides by (k  k )
2
3
k k
k k
and take K 
,K 
k
k
2
3
2
M
3
p
1
4
k
k
v
sv
p
k k
k k
v
k
k
1
s
p
k k
k k
1
4
s
p
2
3
2
1
2
3
4
3
 v 
2
3
v K s  v K  p
s
M
p
p
s
p
1

K
K
M
p
S.No.
E0
(g/l)
T (0C)
I (mmol/ml)
S (mmol/ml)
V
(mmol/
ml-min)
1
1.6 30
0
0.1
2.63
2
1.6 30
0
0.033
1.92
3
1.6 30
0
0.02
1.47
4
1.6 30
0
0.01
0.96
5
1.6 30
0
0.005
0.56
6
1.6 49.6
0
0.1
5.13
7
1.6 49.6
0
0.033
3.70
8
1.6 49.6
0
0.01
1.89
9
1.6 49.6
0
0.0067
1.43
10
1.6 49.6
0
0.005
1.11
11
0.92 30
0
0.1
1.64
12
0.92 30
0
0.02
0.90
13
0.92 30
0
0.01
0.58
14
0.92 30
0.6
0.1
1.33
15
0.92 30
0.6
0.033
0.80
16
0.92 30
0.6
0.02
0.57
• Determine the MM constant for the
reaction with no inhibitor present at 30 0C
and at 49.6 0C and an enzyme
concentration of 1.6 g/l
• Det. The maximum velocity of the
uninhibited reaction at 30 0C and an
enzyme concentration of 1.6 g/l
• Det. The Ki for the inhibitor at 30 0C and
decide what type of inhibitor is being used.
13.2 Batch production of aspartic
acid using cell bound enzyme
• Aspartase enzyme is used industrially for manufacture of aspartic
acid, a component of low calorie sweetener.
• Under investigation is a process using aspartase in intact
B.cadaveris cells. In the substrate range of interest, the conversion
can be described using Michaelis-Menten kinetics with Km 4 g/l. The
substrate solution contains 15% (w/v) fumaric acid; enzyme is added
in the form of lyophilized cells and the reaction stopped when 85%
of the substrate is converted. At 32ºC, vmax for the enzyme is 5.9g/l.h
and its half life is 10.5days. At 37ºC, vmax increases to 8.5 g/l.h but
the half life is reduced to 2.3 days.
(a). Which operating temperature would you recommend?
(b). The average downtime between batch reaction is 28h. At the
temperature chosen in (a), calculate the reactor volume required to
produce 5000tons of aspartic acid per year