UNIVERSITY OF TORONTO
FACULTY OF APPLIED SCIENCES AND ENGINEERING
Mid-term examination
BME 205S: Engineering Biology
Examiner: M.V.Sefton
February 26, 2004
Answer all questions
All questions are of equal value
1. Estimate the changes in the sodium conductance of an axon 0.2 ms after it is
depolarised from -60 mV to +60 mV.
Assume (in units of mS/cm2):
g Na  120
g K  36
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2. Following the suggestion of Ilia AuerbachZiogas in your class, an enzyme, alcohol
dehydrogenase is to be encapsulated (i.e.,
formed into capsules) as a means of
immobilization. The enzyme can then be
used to treat ethanol overdoses (and excess
Capsule
partying by Engineering Science students).
membrane
Thousands of these tiny capsules will be
made in the form of a pill, that can be swallowed.
Enzyme
solution
Alcohol dehydrogenase oxidizes ethanol and converts it into a non-toxic product
(with the aid of additional components that are also encapsulated). The capsule is a
hollow sphere, 500 m in diameter, containing the enzyme in solution. It is so small
that the substrate (ethanol) concentration can be considered uniform throughout the
capsule. The capsule membrane or wall (which is very thin) needs to be permeable to
alcohol so that the ethanol can diffuse into the capsule and react with the enzyme,
while the membrane needs to be robust to withstand the forces associated with
digestion and the pH of the stomach and intestines. The membrane is needed to
protect the enzyme from the low pH of the stomach and to protect the individual from
the enzyme which is derived from yeast.
What should the permeability of the capsule be, such that it poses a negligible effect
on the rate of enzyme action? The Km of the enzyme is 0.013 M and Vmax is 2.3 x 103
moles/L/s. Use 0.2 M as a typical ethanol concentration in the stomach, outside the
capsules.
[FYI: Do a Google search on RU-21, the “KGB pill” for a real (?) variant on this
idea. ]
3. One of the first steps in the development of a foetus (about a month after fertilisation
of the egg) is the creation of the internal blood vessel network, a process called
‘vasculogenesis’. The critical ligand is VEGF (vascular endothelial growth factor),
which binds to two receptors on endothelial cells (the cells that form the blood vessel
network), VEGF R1 and VEGF R2 [VEGF receptors 1 and 2]. Binding of VEGF to
VEGF R2 leads to signal transduction and endothelial cell proliferation. Binding to
VEGF R1 has no signalling effect (the receptor is “silent”) and so acts much like an
inhibitor reducing the effect of VEGF on cell proliferation.
Professor Zandstra and his PhD student Stephen Dang have been studying this system
and hypothesise that VEGF R1 enables “fine tuning” of the effect of VEGF R2. The
expression of VEGF R2 is fairly constant while that of VEGF R1 is sensitive to
external influences such as oxygen concentrations. For example, when oxygen is low,
VEGF R1 is low leading to less reduction in the effect of VEGF on cell proliferation
(i.e., more proliferation). Modulation of VEGF R2 signal through VEGF R1 binding,
furthermore enables the endothelial cell proliferation to be organised in the form of a
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branching network, rather than in a disorganised form with cells growing everywhere
without any pattern.
Using the information provided, estimate the effect of VEGF R1 binding on VEGF
R2 signalling; i.e, calculate what happens without VEGF R1 and then with VEGF R1.
Use an initial concentration of VEGF = 0.02 pM (i.e., 2 x 10-14 M) at a cell density of
107 cells/L. Unfortunately, at these conditions you cannot assume that ligand
depletion is negligible. On the other hand, these conditions will allow for a different
helpful simplification; clearly state the simplification. Avogadro’s number is 6 x 1023.
VEGF R1
VEGF R2
KD (M)
1.6 x 10-11
7.6 x 10-10
# of receptors/cell
50,000
150,000
[Note at this ligand concentration the effect of R1 binding is small, but measureable.
At higher, more realistic, concentrations, e.g., 20 pM, the effect is much larger and
more significant]
4. What do you need to know in order to perform the requested calculations. Explain
or justify your answer by outlining how you would do the calculations or through
some other means.
DO NOT SOLVE THE PROBLEM. Restating the problem is not an adequate answer.
________________________________________________________________
As a recent recruit to God’s engineering team, you are assigned to the nerve conduction
team. Your boss, a fairly senior angel, has decided that myelination is the way to go for
high conduction velocity. Unfortunately the prototypes do not seem to be working.
You suggest trying out saltatory conduction. You explain that rather than myelinating the
entire nerve, you expose the cell membrane at regular short intervals (see Figure), to
enable the action potential to “jump” down the axon.
Your boss does not like to be shown up and so expects you to do the relevant calculations
showing the length of myelinated and unmyelinated portions.
To perform these calculations you are given this information: axon diameter is 6 m
(without a myelin sheath) and 16 m with the sheath. The permittivity of myelin is
similar to that of cell membrane, 40. The resistivity of the cell membrane is 5.6 x 109
ohm-cm and the resistivity of myelin is approximately the same. The resistivity of the
axon cytoplasm is 125 ohm-cm. Without myelin the capacitance across the membrane is
1 F/cm2 and the membrane thickness is 75 Angstroms.
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Length of unmyelinated portion
Myelinated portion
Length of
myelinated
portion
axon
Unmyelinated portion
Node of Ranvier
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Formulas for mid term
V  4 r 3
3
A  4r 2
KD 
ji   D
[ R][ L]
[C ]
u eq 
[C ] 
c
l
[ RT ][ L0 ]
K D  [ L0 ]
[C ] 
[ RT ][ L0 ]
[I ]
[ L0 ]  K D (1  0
[ L0 ][1  u eq ]
[C ]
[ L]


[ RT ] K D  [ L] K D  [ L0 ][1  u eq ]
KI
)
= n[RT]/Navo[L]0
Ei 
RT Cio
ln(
)
Cii
zF
Vm 
z i FJi  Ii  g i (Vm - Ei )
RT [ K  ]o  [ Na  ]o 
ln 

zF  [ K  ]i  [ Na  ]i 
I  (Vm  EK ) g K  (Vm  ECl ) gCl  (Vm  ENa ) g Na  Cm
g Na  g Na m3 (V , t )h(V , t )

1
Cm
  RmCm
v
d
4 Ri Rm

Vm [ S ]
K m  [S ]
g K (V , t )  g K n 4 (V , t )

1
at
 0
2i  m
Rm d

4 Ri d 2
dVm
dt
e.g., m(t, Vm )  m (, Vm ) (1  e
Cm 
 0
t
R
t
L
A
 m dt
4i
Km
1
1


v Vmax [ S ]Vm
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
)