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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 Page 1 of 7 (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 Page 2 of 7 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. Page 3 of 7 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 Page 4 of 7 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) Page 5 of 7 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 8L 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 Page 6 of 7 Page 7 of 7