bk10110302-shuler problems.pdf
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Shuler & Kargi SolutionTRANSCRIPT
V.Prasarnth Raaj Shuler Problems BK10110302
SOLUTIONS
Problem 3.12
1. Is Harry’s reasoning right? Do you agree with him?
Harry’s 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.
∴ 𝜂(Dp = 0.5cm) =100
200= 0.5
𝜂(Dp = 0.7cm) =50
200= 0.25
b. Applying Lineweaver-Burk plot,
1
𝑣=
1
𝑉𝑚+
𝐾𝑚
𝑉𝑚
1
[𝑆]
𝑆0(mg UA ∙ 𝑙−1) 1/𝑆0 𝑣 (mg UA ∙ 𝑙−1 ∙ h−1) 1/𝑣
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
𝑉𝑚 =1
0.017= 58.82 mg UA ∙ 𝑙−1 ∙ h−1
𝐾𝑚 = 0.821 × 58.82 = 48.29 mg UA ∙ 𝑙−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
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑝ℎ𝑒𝑟𝑒 =4
3𝜋𝑟3
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑝ℎ𝑒𝑟𝑒 = 4.18 × 10−3𝑐𝑚3
𝑟𝑎𝑡𝑒 𝑜𝑓𝑢𝑟𝑒𝑎 ℎ𝑦𝑑𝑟𝑜𝑙𝑦𝑠𝑖𝑠, 𝑣 =2.78 × 10−3𝑚𝑀𝑠−1
4.18 × 10−3𝑐𝑚3
𝑣 = 0.665𝑚𝑀 𝑐𝑚−3𝑠−1
𝑣 =𝑉𝑚[𝑆𝑠]
𝑘𝑚 + [𝑆𝑠]
0.665𝑚𝑀 𝑐𝑚−3𝑠−1 =(𝑉𝑚)(0.5𝑚𝑀)
0.2𝑚𝑀 + 0.5𝑚𝑀
𝑉𝑚 = 0.931𝑚𝑀 𝑐𝑚−3𝑠−1
𝜙 = 𝑅√𝑉𝑚 𝑘𝑚⁄
𝐷𝑒𝜙 = 0.1𝑐𝑚√
0.931𝑚𝑀 𝑐𝑚−3𝑠−1 0.2𝑚𝑀⁄
1.5 × 10−5𝑐𝑚2𝑠−1
𝜙 = 55.7
ɳ =3
𝜙
ɳ =3
55.7
ɳ = 0.0538
b.
d=4mm ; r=2mm ; r=0.2cm
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑝ℎ𝑒𝑟𝑒 =4
3𝜋𝑟3
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑝ℎ𝑒𝑟𝑒 = 0.034𝑐𝑚3
V.Prasarnth Raaj Shuler Problems BK10110302
𝑟𝑎𝑡𝑒 𝑜𝑓𝑢𝑟𝑒𝑎 ℎ𝑦𝑑𝑟𝑜𝑙𝑦𝑠𝑖𝑠, 𝑣 =2.78 × 10−3𝑚𝑀𝑠−1
0.034𝑐𝑚3
𝑣 = 0.082𝑚𝑀 𝑐𝑚−3𝑠−1
𝑣 =𝑉𝑚[𝑆𝑠]
𝑘𝑚 + [𝑆𝑠]
0.082𝑚𝑀 𝑐𝑚−3𝑠−1 =(𝑉𝑚)(0.5𝑚𝑀)
0.2𝑚𝑀 + 0.5𝑚𝑀
𝑉𝑚 = 0.1148𝑚𝑀 𝑐𝑚−3𝑠−1
𝜙 = 𝑅√𝑉𝑚 𝑘𝑚⁄
𝐷𝑒𝜙 = 0.2𝑐𝑚√
0.1148𝑚𝑀 𝑐𝑚−3𝑠−1 0.2𝑚𝑀⁄
1.5 × 10−5𝑐𝑚2𝑠−1
𝜙 = 39.12
ɳ =3
𝜙
ɳ =3
39.12
ɳ = 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