Unformatted text preview: CHE 339/BME 339/BIO 335
Exam I
Thursday, October 2, 2007 Your name ________________________________ Part 1: Short Questions: ANSWER IN THE SPACE PROVIDED
1. Name 4 ways for measuring cell growth (4 points).
1)
2)
3)
4) Direct counting (microcope count, particle counting, single cell forward light scattering
using flow cytometry, colony forming units) – 1 Point
Turbidity measurements – 1 Point
Dry cell weight – 1 Point
Indirect measurement of cell mass (measure metabolite concentration that is proportional
to cell mass or perform a mass balance) – 1 Point 2. In ideal chemostats we assume that the Yield coefficient is constant and independent of the
dilution rate but in reality we have to take into account the maintenance energy requirement of the
cells. Derive an expression for obtaining the maintenance energy requirement from a set of values
of the observed yield coefficient, Yx/s vs the dilution rate D (10 points).
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! 3. Name two key advantages of cell recycle in continuous fermenters? (6 points).
Any 2 of the following 3 – 3 Point each
1) Better substrate utilization
2) Higher X and better productivity
3) More stable to perturbations and washout 4. Continuous reactors display a higher volumetric productivity than batch reactors, yet they are
not as widely used in industry. Name the two major drawbacks of continuous reactors
with respect to biotechnology applications (6 points).
Higher risk of contamination
Higher probability of selecting a mutant with lower productivity 5. Oxygen transfer [16 points total]
A. What is the critical dissolved oxygen concentration? (4 points).
Minimum O2 concentration to avoid adverse effects on cell metabolism
B. How does the Oxygen Transfer Rate (OTR) depend on the interfacial area and the
driving force for transfer (i.e. write the equation and explain the meaning of the
terms (4 points).
OTR = kL a (CO2*  CO2)
1) a = interfacial area: larger area results in higher OTR
2) kL = driving force: mass transfer coefficient.
3) (CO2*  CO2) = driving force: lower CO2 results in higher OTR
C. Explain using the OTR equation why increasing the gas pressure increases the rate of
oxygen transfer (4 points).
Increasing the gas pressure increases CO2* since it’s proportional to the partial
pressure of O2 using Henry’s Law
D. At steady state OTR=OUR. To maintain the steady state and thus a sufficiently
high oxygen concentration the supply of the limiting nutrient is reduced as the cell
concentration increases. Why? (4 points)
Lower nutrient concentrations results in a slower the growth rate. A slower
growth rate decreases the O2 demand 5. A. Briefly describe the five key stages in the development of new drugs? (5 points – only one
of the descriptive terms is necessary for each phase to get credit).
B. Which one is the most expensive and why? (2 point) [7 points total]
A) Discovery/Preclinical Testing  A) 6.5 Years
B) Lab and animal studies to assess safety, biological
activity and formulations
C) 5,000 compounds enter
Phase 1 – A) 1.5 Years
B) 20100 healthy volunteers
C) Study how the drug works and is it safe
D) 5 compounds enter
Phase 2 – A) 2 Years
B) 100500 patents volunteers
C) Drug tested for effectiveness and side effects
Phase 3 – A) 3.5 Years
B) 10005000 patents volunteers
C) Drug retested for effectiveness and side effects from long term use
Phase 4 – A) Additional postmarketing testing required by FDA
B) Phase 3 because the number of human patients used requires a lot people to be used
to administer the study over a longer period of time
6. Animal cell cultivation [15 points total]:
A. What is apoptosis and what are the characteristics of transformed cells? (4 points)
Apoptosis – programmed cell death  2 Points
Transformed cells are immortal  2 Points
B. What is the role of a 5% CO2 atmosphere in animal cell cultivation? (4 points).
Buffering
C. Animal cells require glucose and glutamine as the carbon and nitrogen source,
respectively. However, as a result of growth they produce three main inhibitory
products, which if left to accumulate in the culture medium, inhibit cell growth.
What are these (3 points)
1) Lactate – 1 Point
2) Ammonium – 1 Point
3) CO2 – 1 Point D. Why does bubbling air can result in cell damage and what is done to reduce such
damage (4 points).
Bubble bursting results in cell lysis – 2 Points
Antifoam/smaller bubbles – 2 Points
7. Sterilization:
A. Explain graphically (preferably, but not necessarily by referring to the relevant
equations) why sterilization at high temperatures results in lower nutrient destruction relative to
sterilization at lower temperatures. [4 points total].
Equation: ln (kx) = (Ex/R)T + ln (A)
Graph ln (kx) vs. 1/T. Graphically show ln (kx) increases with increasing T since the slope
is negative and determined by Ex As the temperature increases, the reaction rate, k, increases more rapidly for the reaction with the higher activation energy. Considering the difference between the two activation energies of spore destruction versus substrate destruction, an increase in temperature would accelerate spore destruction more than nutrient destruction. B. Name two methods for sterilization, other than heat (2 points)
Any 2 of the following 3 – 1 Point each
1) Physical removal (i.e. filtering)
2) Chemical
3) Radiation PROBLEM 1
1. A pharmaceutical company is having problems with the production of its new drug
“Remembrex”. Over the last several weeks there has been consistent contamination in the batch
tank.
a) Calculate the time required to obtain a concentration of “Remembrex” of 10mg/mL in
batch fermentation. Assume the product is growth associated (P is proportional to the growth
rate) and the cells obey Minod kinetics. µ = 0.6hr1; YP/X = 0.001 g “Remembrex” per g of
cell. Initial cell concentration is 0.05 g cell/mL. (12 points)
b) Now derive an expression for P(t) assuming the kinetics of product formation depend on
both the growth rate and biomass concentration. Otherwise use the same assumptions as in
part a. (8 points).
A) Biomass Material Balance: dX
= µX
dt
X = X o exp(µt )
Product Material Balance: dP
= YX / P rx = YX / P µX
dt
dP YX / P µX o exp(µt )
=
dt
P = YX / P X o exp(µt ) + C
At time zero there is no product:
P = 0 = YX / P X o + C
C = YX/PX0
Therefore:
P(t) = YX/PX0 [exp(µ t) – 1]
The timer required to obtain 10mg/mL of product is then:
ȹ 1 ȹ P
ȹ
1 ȹ P
1 ȹ 0.01
ȹ
lnȹ + 1ȹ
t = lnȹ + 1ȹ = lnȹ + 1ȹ t =
ȹ X Y
ȹ µ ȹ X Y
ȹ
0.6 ȹ 0.5 × 0.001 Ⱥ
µ ȹ o X / P Ⱥ
ȹ o P / X
Ⱥ
t = 8.84 hours B) dX
= µX
dt
X = X o exp(µt )
dP
dX
= YX / P
+ βX = YX / P µX + βX
dt
dt
dP
= (YX / P µX 0 + βX o )exp(µt )
dt
1
P = (YX / P X 0 + βX o ) exp(µt ) + C µ βX o ȹ
ȹ ȹ exp(µt ) + C
P = ȹ YX / P X 0 +
ȹ µ ȹ
ȹ Ⱥ
At time zero there is no product: βX o ȹ
ȹ ȹ + C
P = 0 = ȹ YX / P X 0 +
ȹ µ ȹ
ȹ Ⱥ
βX o ȹ
ȹ ȹ
C = −ȹ YX / P X 0 +
ȹ µ ȹ
ȹ Ⱥ
βX o ȹ
ȹ ȹ[exp(µt ) − 1]
P(t ) = ȹ YX / P X 0 +
ȹ µ ȹ
ȹ Ⱥ ...
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 Spring '11
 Georgiou
 Chemical Engineering, cell biology, Apoptosis, Bacteria

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