-
V.Prasarnth Raaj Shuler Problems BK10110302
SOLUTIONS
Problem 3.12
1. Is Harrys reasoning right? Do you agree with him?
Harrys reasoning is right. Immobilization often prolongs the
life of the protein. Thus I
agree with Harry that, immobilization can prolong the active
lifespan of enzymes (although
it can also kill enzyme with certain linkages).
2. Why is that so?
Aggregation is often a problem with proteins in solution, the
higher the concentration of
enzyme, the quicker the aggregation and it can lead the enzymes
to die faster. This can
be further increased if redox sites are involved, at least in
part due to cysteine reactivity
and divalent bonds forming between enzymes leading to inactive
sludge Additionally,
enzymes which undergo conformational changes during their
catalysis also can become
more prone to denature in a purified state denatured proteins
also tend to glom up more
readily, rendering dead enzyme quite quickly. Certain enzymes
(those designed to chew
up other molecules) also will exhibit some activity against
themselves (even if low, this
adds up quickly in the high concentration, low other-substrate
type environment of
storage). Immobilization solves several of these problems -
enzymes are at a relatively
low concentration for aggregation and inter-enzyme reactions
with each other, while they
can still be at a high relative concentration of reaction with
substrate flowed through the
beads. From the description the type of beads is Poros-type
beads
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 3.14
a. Because the reaction rate is almost the same for the 0.1 and
0.2 cm particle diameter, we
can assume that the rate of reaction without immobilizing
uricase enzyme is 200 mg l-1 h-
1.
(Dp = 0.5cm) =100
200= 0.5
(Dp = 0.7cm) =50
200= 0.25
b. Applying Lineweaver-Burk plot,
1
=
1
+
1
[]
0(mg UA 1) 1/0 (mg UA
1 h1) 1/
10 0.1 10 0.1
25 0.04 20 0.05
50 0.02 30 0.033333
100 0.01 40 0.025
200 0.005 45 0.022222
250 0.004 46 0.021739
=1
0.017= 58.82 mg UA 1 h1
= 0.821 58.82 = 48.29 mg UA 1
y = 0.8217x + 0.0175
0
0.02
0.04
0.06
0.08
0.1
0.12
-0.03 -0.01 0.01 0.03 0.05 0.07 0.09 0.11
1/v
1/S
Lineweaver-Burk Plot
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 3.15
a.
d=2mm ; r=1mm
[Sb]=0.5mM ; neglect liquid film resistance, therefore [Sb]=
[Ss]
v=10mM h-1 = 2.78x10-3mM s-1
De=1.5x10-5cm2/sec
Km' = 0.2 mM
=4
33
= 4.18 1033
, =2.78 1031
4.18 1033
= 0.665 31
=[]
+ []
0.665 31 =()(0.5)
0.2 + 0.5
= 0.931 31
=
= 0.1
0.931 31 0.2
1.5 10521
= 55.7
=3
=3
55.7
= 0.0538
b.
d=4mm ; r=2mm ; r=0.2cm
=4
33
= 0.0343
-
V.Prasarnth Raaj Shuler Problems BK10110302
, =2.78 1031
0.0343
= 0.082 31
=[]
+ []
0.082 31 =()(0.5)
0.2 + 0.5
= 0.1148 31
=
= 0.2
0.1148 31 0.2
1.5 10521
= 39.12
=3
=3
39.12
= 0.0767
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 3.17
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 3.18
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 6.15 (a)
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 6.15 (b)
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 6.17
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 6.17
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 6.19 (a)
-
V.Prasarnth Raaj Shuler Problems BK10110302
Problem 6.19 (b)
Two graph need to be plotted in order to find the optimum
dilution rate
Plot 1
DX vs D
Plot the table below using this equation
DX = 0.1-((0.004*D)/(0.2-D))
D DX
40 0.10402
30 0.104027
20 0.10404
10 0.104082
0 0.1
Optimum dilution rate maximizing productivity of biomass, Dopt =
12.5
0.0995
0.1
0.1005
0.101
0.1015
0.102
0.1025
0.103
0.1035
0.104
0.1045
0.105
0 5 10 15 20 25 30 35 40 45
Pro
du
ctiv
ity
of
Bio
mas
s, D
X
Dilution Rate, D
DX vs D
12.5
-
V.Prasarnth Raaj Shuler Problems BK10110302
Plot 2
DP vs D
Plot the table below using this equation
DP = 0.2-((0.008*D)/(0.2-D))
D DP
0 0.2
20 0.208081
40 0.20804
60 0.208027
80 0.20802
Optimum dilution rate maximizing productivity of product, Dopt =
25
0.199
0.2
0.201
0.202
0.203
0.204
0.205
0.206
0.207
0.208
0.209
0.21
0 10 20 30 40 50 60 70 80 90
Pro
du
ctiv
ity
of
Pro
du
ct, D
P
Dilution Rate, D
DP vs D
25