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UNIVERSITY OF TORONTO
FACULTY OF APPLIED SCIENCES AND ENGINEERING
Final examination
BME 205S: Engineering Biology
Examiner: M.V.Sefton
April 19, 2002
Answer all questions
All questions are of equal value
Clearly state your assumptions.
If you do not understand a question, make an assumption, clearly state it and
proceed on the basis of your assumption.
1. Bernstein, Yang and Langer of MIT1 devised an immobilised heparinase
bioreactor to remove heparin from blood. Heparin is an anticoagulant and it
prevents blood from clotting (i.e. turning into a solid gel) while blood is taken
out of the body and passed through an oxygenator circuit or any other blood
contacting device. An oxygenator is used for patients undergoing open heart
surgery and it is desirable to remove the heparin before the blood is returned
to a patient. Heparinase is an enzyme (a 43kDa single chain protein) that
degrades heparin into an inactive substance
Unlike the hollow fibre bioreactor we discussed
in a tutorial, Bernstein, Yang and Langer chose
to use a different geometry (Figure 1). They
immobilised the enzyme to a porous agarose2
gel bead (diameter: 50 m) and then suspended
many of these beads in a stirred tank reactor.
The ability of the immobilised enzyme beads to
degrade heparin is given in the table following.
The diffusivity of heparin in the beads was such
that the Thiele modulus was 0.1.
Figure 1
Agarose
beads
containing
immobilised
heparinase
1
Langer reference
Agarose is a 'plastic' derived from seaweed. It is a polysaccharide. Because of the ease of
chemical modification it is often used as a base for enzyme immobilisation.
2
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(a) Estimate the volume of beads (per mL of reactor) needed to reduce the
heparin concentration from 1 mg/mL to 0.05 mg/mL in one hour in a batch
reactor. Batch operation would be used to test the device prior to actual
use with a patient.
Heparin concentration, mg/mL
0.05
0.1
0.25
0.5
1
2
v, mg/hr/mL of bead
31.6
46.3
57.1
75.8
88
89
(b) Heparin is a relatively large molecule (approx. 15 - 25 kDa), a linear
polysaccharide, which is highly negatively charged (due to many sulphate
groups). Speculate as to the structure of the heparinase active site (i.e.,
likely amino acid composition, see attached page) and its location on the
enzyme.
2.
The BME205 exam is unlike any exam you have taken before, plus you
had a short night of sleep, you had car problems yesterday, you had too
much to drink last night, and the phone kept ringing early this morning.
The result - you have a terrible headache, unlike any headache you have
ever had before. You reach for your bottle of a new headache medicine
that claims maximum relief within 30 minutes and contains a well-known
pain reliever. An amazing coincidence is that the exam includes a
problem related to the pharmacokinetics of this pain reliever. You are
given the following information:
Elimination rate constant = 0.277 hr-1
Apparent distribution volume = 28 L
Minimum effective level = 10 μg/mL
Maximum tolerable level = 20 g/mL
(a) You plan to take this pain reliever. What dose in milligrams should you
take?
(b) Outline how you would estimate the time you could think about taking
another dose? [There is no need to actually calculate the time]
3. (a) Calculate the power output (in watts) of the heart in pumping blood. The
cardiac output is normally about 6 L/min. Blood enters the right side of the
heart at a pressure of about 0 mm Hg and flows via the pulmonary arteries
to the lungs at a mean pressure of 11 mm Hg. Blood returns to the left
side of the heart through the pulmonary veins at a mean pressure of 8 mm
Hg. The blood is then ejected from the heart through the aorta at a mean
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pressure of 90 mm Hg. (1 watt = 1 Joule/sec). Clearly state your
assumptions.
(b) Hollow fibre membrane systems are used as kidney dialysis units (to
remove urea and other toxins from blood) or oxygenators (to oxygenate
blood). These units containing many membrane tubes in parallel, 200 m
inside diameter. For example for kidney dialysis, blood is drawn from the
arm of a patient (and returned to the patient) at 200 mL/min. It is pumped
through the membrane unit, which has a total membrane area of 2.5 m 2. If
the membrane tube has a length of 20 cm, calculate the pressure drop in
mm Hg. (1 atmosphere pressure = 760 mm Hg = 1.01 x105 Pa)
Blood viscosity is 0.003 N-s/m2, density is 1.05 g/mL.
4.
A simplified model of the circulation
is shown in Figure 2. Normally the
arterial pressure (part) is the same as
that in the carotid artery sinus: part =
psinus. The carotid artery leads from
the aorta to the head and the
pressure at one particular point (the
"sinus") is used as a detector of
arterial pressure. When the pressure
is too high, the vagus nerve is
Figure 2
stimulated causing the arterial
pressure to be lowered (by changing
heart rate). In experiments on dogs whose vagus nerves were cut, the
carotid arteries were isolated and perfused by a separate pump. This
broke the feedback loop and allowed the curve shown on the graph below
(Figure 3) to be obtained. The empirical equation to describe the curve is
[Sher and Young, 1962]:
part = 90 +
120___________
[1 + exp [(psinus -165)/5]
(a) Draw a block diagram of the complete feedback system. Label the blocks
and indicate the proper cause and effect relationship
(b) Find the operating point of the feedback loop.
(c) Find the open loop gain.
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Figure 3
5. (a) For the following problem, identify the critical step in solving this problem.
Do not actually solve it. Clearly explain your answer.
Your answer should be focused and clear. It will likely be longer than a
few words, but less than a page. Rough work should be shown on the
unruled facing page as indicated in the examination book instructions.
Note that the problem statement may not contain all the information that is
needed. Hence, depending on your approach, it is likely that you will need
to clearly state and explain your assumptions.
_____________________________________________
While glucose monitors (using immobilised glucose oxidase) are of great
benefit to diabetics who need to control their blood glucose levels, they
have two disadvantages. They require a slightly painful pin-prick to obtain
a drop of blood and they cannot be used for continuous sensing of blood
glucose. Hence there is considerable interest in developing an implantable
glucose sensor based on the same immobilised enzyme.
Glucose oxidase catalyses the oxidation of glucose to gluconic acid and
the latter is detected via a pH electrode embedded within the implanted
sensor.
Glucose (C6H12O6) + O2
H2O
gluconic acid (C6H12O7) + H2O2
Unfortunately the hydrogen peroxide (H2O2) that is produced in this
reaction causes inactivation of the glucose oxidase. Hence it must be
removed using a second enzyme, catalase. Catalase reduces hydrogen
peroxide back to water:
H2O2
H2O
+
1/2 O2
Consider the glucose sensor design shown in Figure 4. The sensor
consists of two thin layers (each: 25 m thick x 1 mm wide x 5 mm long) of
immobilised enzyme on top of each other. One layer contains immobilised
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glucose oxidase and one contains immobilised catalase both in permeable
gels. Figure 4 shows the catalase layer adjacent to the electrode and the
glucose oxidase layer separated from the blood by a very thin
biocompatible membrane; for reasons beyond the scope of this problem,
this may not be the optimum configuration of the layers.
How much catalase should be immobilised?
The sensor should have a working range of 2 to 25 mM of glucose in
blood; normal glucose concentration is 5.5 mM. The very thin
biocompatible membrane is not a significant resistance to diffusion of
glucose (or oxygen), so you can assume that the glucose concentration at
the surface of the glucose oxidase layer is equal to that in blood.
Some useful data [this is not necessarily all the information that is
needed]:
Immobilised enzyme
Km (mM) Vmax (mmole/min/mg protein)
25 x 10-3
Glucose oxidase (estimated)
3.6
Catalase3
27.7
1.0
Figure 4
blood
Biocompatible
membrane
Immob. glucose oxidase
50 m
Immob. catalase
Electrode and electronics
5 mm
1 mm
__________________
(b)
3
What mark (letter grade: A, B, C, etc; use +/- if you wish) do you think you
deserve for this course. Justify your answer.
Enzym & Microbial Tech. 26(7) 497-501 (2000)
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Formulas for exam
ji   D
ln

 max Ci
K i  Ci
X
 
Xo
Vm [ S ]
K m  [S ]
v
v' 
c
l
Km
1
1


v Vmax [S ]Vm
V 'max [ S ]
K m,app  [ S ]
R
V "max
R 4 P
8L
4 Q
w 
R 3
K m,app  K m 
C2  C1o
x ' 
3 1
1
  
 
  tanh   
Km D
P
Q


V 'max K m
k L ([ S ]  K m )

V2
 gh  h f  constant
2
e u  e u
tanh u  u
e  e u
hf 
32 LV
D 2

  23 w
kaV1 e  k a t  e  k e t
V2 (ke  ka )
x
x
f
f


y  p
1  G1G2 1  OLG y
p
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