UNIVERSITY OF TORONTO
FACULTY OF APPLIED SCIENCES AND ENGINEERING
Mid-term examination
BME 205S: Engineering Biology
Examiner: M.V.Sefton
Answer all questions
All questions are of equal value
February 14, 2002
1. It has been experimentally observed that the presence of hemoglobin augments the transport of oxygen through aqueous solution. Available data suggests that this is a case of facilitated transport by a mobile carrier
(hemoglobin) wherein oxygen reacts to form the oxyhemoglobin complex.
Diffusion of the oxyhemoglobin results in a greater flux of oxygen than that attributable to purely physical diffusion of oxygen alone.
Consider a stagnant aqueous hemoglobin solution (0.5 g/mL) that is in the form of a flat 'film', 0.2 cm thick. Calculate the maximum relative amount by which the hemoglobin can augment oxygen diffusion if the upstream face is maintained at 100 mm Hg oxygen partial pressure (97% of maximal hemoglobin saturation) and the downstream face at 10 mm Hg oxygen partial pressure (10% of maximal hemoglobin saturation).
Data:
Diffusivity of oxygen in hemoglobin solution = 1 x 10 -5 cm 2 /sec
Diffusivity of hemoglobin in hemoglobin solution = 5 x 10 -7 cm 2 /sec
Each gram of hemoglobin can bind with 1.34 ml (STP) of oxygen, maximum
Solubility coefficient for oxygen in hemoglobin solution (as free oxygen) is the same as that in water = 3 x 10 -5 mL O
2
(STP)/cm 3 mmHg
Assume that the diffusivity of oxyhemogobin is equal to that of hemoglobin.
2.
(a) A cell has a maximum specific growth rate, µ max
of 0.3 hr -1 when grown aerobically on glucose. The Monod constant on glucose, K s
is 2 g/L and the cell yield on glucose (Y x/s
) is 0.35 g cell/g glucose. Assume steady state operation and that all nutrients are in excess except glucose.
Assume 10 9 cells weigh 1g.
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We grow this cell in a 10 L continuous reactor using 1 L/hr. of a feed stream containing 10 g/L glucose. What is the glucose concentration exiting from the CSTR?
(b) Consider again the stem cell production problem of PBL 2 and exercise 4. The probability of self renewal (p) is dependent on the ratio of differentiated cells to stem cells, as it is the mix of growth and differentiation factors (and inhibitors) produced by the two cell types that determine the value of p. A starting point for this kind of analysis would be an estimate of the ratio of stem cells to differentiated cells (S/X), where S
= # of stem cells/unit volume and X = # of differentiated cells/unit volume, as a function of p. In addition to stem cell growth (self renewal) and differentiation, you need to account for differentiated cell growth which takes place with the same "doubling time" as the stem cells: i.e.,

is the same for both cell types. The doubling time is 24 hours.
Stream 2
Stem cell recycle
Stream 1
Differentiated cell effluent
However, the standard continuous well-stirred reactor situation must be modified in order for the differentiated cells to not outgrow and essentially displace the stem cells from the reactor (at steady state). In one approach to this problem a separation device is inserted into the effluent (Figure 1) to separate stem cells from differentiated cells (based on differences in protein receptors on the surfaces of the two cells). What is the ratio of stem cells to differentiated cells in this reactor (at steady state) for a separator that is 100% efficient for differentiated cells (no differentiated cells in stream 2), 95% efficient for stem cells (the concentration of stem cells in stream 1 is 5% of that leaving the reactor) and when stream 2 has a volumetric flowrate that is 5 x that of stream 1.
As before: rate of self renewal
 rate of differenti
 pS ation
 
( 1
 p ) S
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3. In order for birds to fly south for the winter, the birds store fuel for the long flight by depositing fat - as much as doubling their weight. Unfortunately this makes them heavier so that they expend more metabolic energy during flight.
Because their wing span does not change, heavier birds have to fly faster to stay airborne.
(a) According to aerodynamic theories of bird flight, flight speed scales with mass, M as M 1/6 . If output power (power associated with flying) scales with the product of weight and speed, what is the scaling relationship for output power. [this is a gift]
(b) Kvist et al recently (Nature, 413, 730-732, October 18, 2001) measured the metabolic power input for a particular type of bird (the red knot) flying at 15 m/s in a wind tunnel. The data is shown in figure 2. They found that power input scales with bird mass as M 0.35
.
Derive a scaling relationship for flight muscle efficiency,

:
 
P out
P in
1 .
1 

BMR where P out
= output power, P in
= input power and BMR is the power used in basal metabolism. BMR can be estimated from the following graph, even though a bird is not a mammal. Assume 1 mole of ATP/GTP can produce 7.3 kcal of energy. 1 cal = 4.19 J
Figure 2 Total metabolic power input ( P in
) in relation to average body mass ( m b
) in red knots flying at 15 m s -1 . Different symbols denote different individuals. An analysis of covariance (ANCOVA) with individual as factor and log
10 m b
as covariate revealed significant effects of log
10 m b
( F
1,23
= 7.00, P = 0.01) and individual ( F
3,23
= 3.11, P = 0.046) on log
10
P in
. The interaction term was not significant (that is, the slope relating log
10
P in
to log
10 m b
did not differ between individuals) and was excluded from the model. The common slope relating log
10
P in
to log
10 m b
, correcting for individual differences in intercept, was 0.35 (95% CI 0.08–0.62). Individual intercepts were 0.41, 0.39, 0.41 and
0.35
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Figure 3
4. What do you need to know/understand to solve the following problem. Do
NOT actually solve it. Explain or justify your answer. Restatement of the problem is not an adequate answer. No additional information is required.
_____________
Professor Yu-Ling Cheng worked for the ALZA Corporation in California before joining the University of Toronto. In those days, ALZA designed drug delivery systems especially for controlling the delivery of drugs across the skin
(transdermal delivery). The difficulty was that the skin was a variable barrier
(resistance) to the drugs of interest (eg anti-nausea drugs for motion sickness).
For example the flux of drug A through human skin under maximal concentration driving force was found to range from 2
 g/cm 2 hr to 10
 g/cm 2 hr, with an average of 6
 g/cm 2 hr. This was too great a variability to be useful, since the required therapeutic dose of A was 100
 g/hour and the range of variability in dosage was no more than

20
 g/hr. Professor Cheng/ALZA's ingenious solution was to add a rate controlling membrane in series with the skin to reduce the impact of the skin variability, as shown below. The drug reservoir contains a supersaturated solution of the drug and thereby generates a constant and maximal driving force. Specify the required membrane area and calculate the maximum and minimum flux of the drug through the rate controlling membrane/adhesive composite under the maximal concentration difference.
Backing membrane (impermeable)
Drug reservoir
Rate controlling membrane
Adhesive (very permeable) skin
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