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BIOM5100 - Midterm Exam Winter 2007

You have 80 minutes to complete this exam. The exam has four questions; you are required to answer any three (3) of them. All questions are worth equal marks. This is a closed book exam; however, you are permitted to bring one (1) 8.5"×11" sheet of notes into the exam. You are permitted to use a non-programmable calculator. You may not communicate with anyone during the exam except the instructor.

You may use the following conditions and equations for your calculations:

  • E = mc², c= 3×108 m/s = 1.8 Terafurlongs per fortnight.
  • 1 e− = 1.6×10−19 Coul, electron mass = 9.11×10−31 kg
  • 1 mol = 6.02×1023 particles
  • F = ma, V = IR, Rc = 8ηL/π/r4, Lc = ρL/π/r2

1. Electrical signals from muscles

    1A. The EMG is a stochastic signal with a random amplitude and phase with respect to the underlying muscle signals, while the ECG shows a deterministic structure. However, both signals originate from contracting muscle fibres. Explain why the ECG and EMG differ in this way?

    1B. The EMG is measured using a platinum needle electrode. The Nernst equation for the half call potential is: E = E0 − (RT) / (nF) × ln( [C]n ). Using this equation, and the fact that the solubility of platinum in tissue is extremely low, discuss: i) if platinum electrodes are a good choice to measure the EMG? Why or why not? ii) Do platinum needle electrodes show a movement artefact, and, if so, what can be done to deal with the artefact?

2. ECG amplifiers and electrical safety

    2A. Sketch the electronic design of the ECG differential amplifier which uses a driven right leg circuit. Describe the function of each component. (note: it is not required to show it at the level of op-amps. A block diagram is adequate, but you must describe what each block does)

    2B. One practical clinical problem is that electrodes tend to become detached as the patient moves and changes posture. In order to detect this, some manufacturers apply a 10mA current through the electrodes. When the impedance to current flow increases greatly, the electrode is detected as disconnected.
    i) What type of electrical shock hazzard does this possibly represent and why? ii) Briefly discuss one electrical design issue that needs to taken into account when designing this circuit (ie. stimulation frequency, isolation)

3. Lung function measurement#1

    Two different instruments are designed to measure a baby's breathing. Design #1 follows:

    3A. A mask attached to a pneumotachometer is placed onto the face of a newborn. The mask pneumotachometer is composed of 30 closely packed channels, each with a length of 2cm and an inner diameter of 0.5mm. i) What is the flow resistance of this pneumotachometer? ii) If the baby expires at 20 ml/s, what additional pressure must the baby's lung muscles generate in order to breathe through the pneumotachometer?

    answer: for each channel Rc= 8ηL/π/r4
    Rc= 8*(18×10−6Pa-s)(.02m)/π/(.00025m)4 = 2.35×108Pa/(m³/s)
    for 30 channels R = Rc/30 = 7.82×106Pa/(m³/s)
    R = 7.82 Pa/(ml/s)
    20 ml/s = 20×10−6m³/s
    Pressure = Flow×R = ( 7.82×106Pa/(m³/s) ) ( 20×10−6m³/s ) = 156 Pa

    3B. i) sketch a diagram of this mask/pneumotachometer system. ii) a newborn will typically produce a lot of flegm and fluid while breathing. Briefly discuss problems this may cause to the instrument.

4. Lung function measurement#2

    Two different instruments are designed to measure a baby's breathing. Design #2 follows:

    4A. A newborn is placed into a 'head-out' body box, which encloses her body except the head. The box is is sealed at the neck, allowing the baby to freely breathe room air. The rigid box contains a pressure transducer to measure the pressure difference between the box pressure and the atmosphere (101.3 kPa).
    The pressure transducer is built using a flexible diaphragm with attached strain guages i) Sketch a diagram of such a pressure transducer. Strain guages of this type are strongly affected by the temperature. ii) Describe how such a strain guage pressure transucer is designed to reduce the contribution from temperature changes.

    4B. A baby has 100 ml of air in her lungs. She breathes in (inspires) a tidal volume of 50 ml of 100% humidity room air (at 27°C) and warms this air to body temperature (37°C). At the begining of expiration (just before airflow begins), the baby applies a 1 kPa pressure to the air in the lungs. The volume of air in the body box is 44 L and starts out at atmospheric pressure. Calculate the pressure change in the box between beginning of inspiration and beginning of expiration.

    answer: 50ml changes to 50*(273+37)/(273+27) = 51.7ml Total volume in lungs is 151.7ml
    Apply 1kPa: 151.7ml * (101.3)/(101.3+1) = 150.1 ml
    Change in volume is 150.1 ml - 100ml = 50.1ml
    1: If we assume isothermal box, then pressure change can be calculated by volume ratio (is Boyle's law):
    Change of 50.1ml in 44L box is .0501L/44*101.3kPa = 115Pa
    2: If we assume adiabatic box (α=1.4).
    P1*(V1/N)α = P2*(V2/N)α
    P2 = P1*(V1/V2)α
    101.3kpa + ΔP = 101.3kpa * (44L / (44L−.0501L) )α = 101.415kPa
    ΔP = 101.415kPa − 101.3kpa = 161Pa
    (Note that we could use this approach with α=1 to solve for the answer #1. Both approaches are accepted

Last Updated: $Date: 2007-11-28 23:52:32 -0500 (Wed, 28 Nov 2007) $