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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
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BK10110302-Shuler Problems.pdf

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  • 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