SOLUTIONS Problem 3.12 1. Is Harryβs reasoning right? Do you agree with him? Harryβs 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 h1 . 100 β΄ π(Dp = 0.5cm) = = 0.5 200 π(Dp = 0.7cm) =
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 40 30 20 10 0
DX 0.10402 0.104027 0.10404 0.104082 0.1
Productivity of Biomass, DX
DX vs D 0.105 0.1045 0.104 0.1035 0.103 0.1025 0.102 0.1015 0.101 0.1005 0.1 0.0995 0
5
10
12.5
15
20
25
30
35
Dilution Rate, D
Optimum dilution rate maximizing productivity of biomass, Dopt = 12.5
40
45
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 0 20 40 60 80
DP 0.2 0.208081 0.20804 0.208027 0.20802
Productivity of Product, DP
DP vs D 0.21 0.209 0.208 0.207 0.206 0.205 0.204 0.203 0.202 0.201 0.2 0.199 0
10
20
25
30
40
50
60
Dilution Rate, D
Optimum dilution rate maximizing productivity of product, Dopt = 25
