Solution:
"I pledge on my honor that I have not given or received any
unauthorized assistance on this examination (or assignment)."
Solution:
Being able to grow an organ for transplantation is one of the
goals that hardworking biochemical engineers (including myself)
strive for. I dream a day when I can supply a liver affordably
to whoever needs it. We can culture a skin cell into a sheet to
help heal, say, badly burnt skin. A skin cell can reproduce on
its own; thus, it is alive. Yes, you are indeed killing a
million lives each time you rub your skin. However, since cells
constantly regenerate and die in a natural renewal process and
since these cells have no self-awareness, my conscience is clear
regardless of whether technology exists to grow a single cell
into an whole organism. And, yes, even lowly bacteria are living
organisms; you are killing a million lives each time you
disinfect a surface or autoclave glassware before culturing
cells. If you answer no, you should not be a biochemical
engineer and you shoud not be in this class. Furthermore, if you
object to altering nature, you should not be in engineering,
because that is what an engineer does: altering nature for the
betterment of humanity.
This question also tests you on your comprehension of life, which
is the ability to self-reproduce. One does not have to be a
whole human to be considered alive. Each cell in our body (with
a few exceptions) is capable of reproduction, thus is alive. As
far as reproduction of human cells are concerned, for each sexual
reproduction, there are gazillions of instances of asexual
reproduction. In my lab, we culture, in addition to an
assortment of cells derived from other mammals and insects, a
variety of human cells: human embryonic kidney (HEK) cells to
produce antibodies, human umbilical vein endothelial cells
(HUVEC) to study angiogenesis, human fetal retinal pigment
epithelial cells (hfRPE) to study eye diseases, macrophage cells
to study inflammatory responses, human cartilage to address joint
deterioration, etc.
Solution:
Life: Having the ability to reproduce on its own.
Organism: a living entity, i.e., an entity that has the ability
to reproduce on its own.
Solution:
Chemical engineering: economic manufacture of chemicals.
Biomolecular engineering: modification and/or studies of
biomolecules (DNA, RNA, proteins, sugars, etc). (Remember,
biomolecules are a subset of chemicals, and everything is a
chemical. Thus, chemical engineering covers just about
everything.)
Solution:
Species: A group of organisms that can interbreed with other members of
the same species. A liger is not a distinct species, because it is sterile
and cannot breed with another liger to propagate.
Solution:
Steady-State: a condition where time variant terms (d*/dt terms or accumulation terms) are zero, i.e., time invariant.
Equilibrium: a condition where the forward reaction equals the
backward reaction, resulting in no net observable change.
Solution:
These are the fundamental tenets in mathematical/quantitative analysis in chemical engineering.
accumulation of mass within a volume boundary = in - out + generation - consumption
accumulation of energy within a volume boundary = in - out + generation - consumption
ka
Einactive + C <------> E
k-a
k1 k2
E + S <------> ES ------> E + P
k-1
Solution:
rate: v = dp/dt = k2*ES
conservation of enzyme: E0 = E + ES + Einactive
equilibrium assumption: ka*C*Einactive = k-a*E
k1*S*E = k-1*ES
Solve for 4 unknowns: E, ES, Einactive, and v
Solution:
rate: v = dp/dt = k2*ES
conservation of enzyme: E0 = E + ES + Einactive
quasi-steady state assumption: d(E)/dt = 0 = ka*C*Einactive - k-a*E - k1*S*E + k-1*ES + k2*ES
d(ES)/dt = 0 = k1*S*E - k-1*ES - k2*ES
Solve for 4 unknowns: E, ES, Einactive, and v
Solution:
The resulting rate expression from either Part a) and Part b)
should be a function of S and C: v=v(s,c). Usually s changes with
time during reaction, but c does not; thus, v=v(s) The kinetic
equation governing a batch bioreactor with a given c is:
ds/dt = -v(s) I.C.: s(0)=s0 ... given
Integrate the above equation to find changes in s with time: s=s(t).
Then, find conversion from p(t)=s0-s(t).
Solution:
Substitute the two equations arising from making assumption
with two additional equations dC/dt and dp/dt. Substitute enzyme
conservation equation with d(Einactive)/dt
v = dp/dt = k2*ES
d(E)/dt = ka*C*Einactive - k-a*E - k1*S*E + k-1*ES + k2*ES
d(ES)/dt = k1*S*E - k-1*ES - k2*ES
d(Einactive)/dt = -ka*C*Einactive + k-a*E
d(C)/dt = -ka*C*Einactive + k-a*E
d(S)/dt = - k1*S*E + k-1*ES
I.C.: p(0)=0, E(0)=0, ES(0)=0, Einactive(0)=E0, C(0)=C0, S(0)=S0
Integrate all ODEs simultaneously with given initial conditions to solve for: E, ES, Einactive, C, S, and p
Solution:
In case of either equilibrium assumption or quasi-steady state
assumption, we tag on a first-order deactivation term to the
eventual rate expression derived in Part a) and Part b).
ds/dt=-v(s)*exp(-kd*t) I.C.: s(0)=s0 ... given
p=s0-s
In case of no assumption, add a first-order deactivation term to
each of the enzyme dynamic equations.
d(E)/dt = terms from last part - kd.active*E
d(ES)/dt = terms from last part - kd.active*ES
d(Einactive)/dt = terms from last part - kd.inactive*Einactive
dp/dt = terms from last part
d(C)/dt = terms from last part
d(S)/dt = terms from last part
Integrate all ODEs simultaneously with given initial conditions.
Solution:
Solution:
For parameters in "Michaelis-Menten" type kinetic expression
(vm and Km): vary starch level & measure
initial reaction rate from either/both the decrease in the starch
level and/or increase in the sugar level. (This will likely
results in a constant value v=vm.) For thermo
deactivation constant kd and its dependence on
temperature: vary temperature and measure initial reaction rate.
Estimate parameters kd0 & Ed in
kd=kd0*exp(-Ed/R/T)
Solution:
First, find conversion with time.
Note that maximizing conversion alone is not really realistic,
because conversion increases with time and this does not allow
you to find the optimum batch length. Maximizing instantaneous
rate of reaction v makes no sense either.
rate: dp/dt = v = vm = k2*E
= k2*E0*exp(-kd*t)
Integrate dp/dt (assuming constant T) yields conversion with t,
p(t) = k2*E0/kd*(1-exp(-kd*t))
Let us define "optimal" in terms of maximizing profit.
profit = (value of product - value of substrate - fixed cost per run - cost proportion to to run time) / (run time)
Profit(T,t) = [k2/kd*(1-exp(-kd*t) - (1+γ)*α*β] / t
where k2 & kd depends on T: k2=k20*exp(-Ea/R/T) & kd=kd0*exp(-Ed/R/T)
α = value of the substrate (barley) relative to the product (sugars)
β = substrate to enzyme ratio
γ = fixed cost as a fraction of the substrate value
Varying T and t to maximize Profit, either numerically or ananlytically.
In the latter case, find a solution to d(Profit)/dT=0 and d(Profit)/dt=0.
Solution:
Apply the same profit objective as in Part b), except that
we integrate dp/dt in Part b without assuming a constant T.
Case a) When barley is added at room temperature, integrate
dp/dt in Part b) with an appropriate bioractor temperature
profile (say, one that increases linearly with t for constant
heating rate). Case b) When barley is added after the broth has
been heated, calculate profit with the formula in Part b), except
the run time in the denominator is the sum of two periods:
heat-up and reaction. Evaluate profit for each scenario. The
answer depends on model parameters and profit objective function,
e.g., the relative values of sugar product to barley starch
substrate and operting cost. When the substrate cost is high, we
want to allow reaction to proceed for a longer period to achieve
high conversion. This longer period also deactivates the enzyme
further, especially when the heat-up period is long. Change
parameters in the Mathcad file below and see. Intuitively,
adding barley at the beginning of heat-up period yields a better
performance because not only the enzyme acts on barley during
this heat-up period, but also deactivates slower especially
at the start of the heat-up period.
Solution:
At t=¥, we have from Part b):
p¥=s0=k2*E0/kd
where the rate "constants" depend on temperature in an Arrhenius fashion.
k2=k20*exp(-Ea/R/T)
kd=kd0*exp(-Ed/R/T)
Solve the above equation to find T¥.
T¥=(Ed-Ea)/R/ln(p¥/E0*kd0/k20)
Solution:
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