A new reality TV show includes a "contest" to survive solitary confinement. To win, the contestant must remain for 12 hours in an air-tight room that is 5ft X 5ft X 8 ft. One contestant realizes that the producers of this show are way too idiotic to have made a calculation of oxygen requirements! If his basal metabolic rate at the temperature of this room (25C) is 75 watts, should he be worried about running low on O2 (hypoxia)? To answer this question, assume the contestant is fasting (using fats for energy) that this contest is taking place at sea level and the room is filled with dry room air. A) What will be the percent O2 content in the room after 12 hours? B) What will be the alveolar PO2 at that time? C) Using the oxygen dissociation curve from the in class powerpoint, what would be the contestants percent oxygen saturation in arterial blood after 12 hours in this room? D) Should he be worried?
Potentially useful information: One foot = 0.3048m. 1.0 L = cube of 10cm sides.
PV = nRT, one mole of STP gas occupies 22.4L. Atmospheric air is 21% O2. Sea level barometric pressure (Pb) is 760 mmHg.
Alveolar PO2 if in "fresh" room air: PaO2 = 0.21 (Pb - 47) - PACO2/RER (a =alveolar, A= arterial)
Note RQ = RER = respiratory exchange ratio = CO2 produced / O2 consumed
|
Food
|
kJ/g
|
LO2/g
|
kJ/LO2
|
RER
|
|
Carbohydrates
|
17.6
|
0.84
|
20.9
|
1.0
|
|
Fat
|
39.3
|
2.0
|
19.7
|
0.71
|
|
Protein
|
18.0
|
0.96
|
18.8
|
0.81
|
2. As I mentioned in class the clearance of inulin is equal to Glomerular Filtration Rate (GFR). Per unit body weight, shrews (the smallest mammals) have the highest metabolic rates and the highest GFR per unit body weight. Calculate the GFR of a 4 gram shrew if: Plasma conc. 2 mg/l urine conc 80 mg/l urine flow rate is 3.5 µl /min. If total blood volume is 7% of body mass (as it is in nearly all mammals) how long does it take a shrew to filter 100% of its blood volume?
GFR = Clearance of Inulin = urine flow rate (vol/time) * Urine concentration/plasma concentration