In a vastly simplified view of human metabolism, we can assume that The body is

Question

In a vastly simplified view of human metabolism, we can assume that The body is deriving energy primarily from The following two reactions that are happening at The same time: C15H31COOH + 23 O2 rightarrow 16 CO2 + 16 H2O C6H12O6 + 6 O2 rightarrow 6 CO2 + 6 H2O The first reaction more or less corresponds to "burning fat" while The second one corresponds to "burning sugar." In a metabolic experiment, an individual running on a treadmill is fitted with a mask delivering air and analyzing exhaled gas. Fresh air is delivered to The individual at a rate of 10 L/min, and is assumed to contain 77.4% N2, 20.6% O2, and 2.0% H2O only. The exhaled gas steady-state composition is measured to be 15.1% O2, 3.7% CO2, 6.2% H2O, and 75.0% N2. Calculate The rate of production of CO2 and The rate of consumption of O2 in L/min. Using The ideal gas law, calculate The corresponding rates in moles/min assuming that The temperature of all gases is 25 Degree C and The total pressure is 1 atm at time of measurement. What is The fractional conversion of oxygen due palmitic acid oxidation and due to glucose oxidation, respectively?

Explanation / Answer

a)
In 1 min,
O2 enetering= 20.6% of 10 L   = 2.06 L
CO2 entering = 0 L

O2 coming out= 15.1% of 10 L = 1.51 L
CO2 coming out = 3.7% of 10 L = 0.37 L

rate of production of CO2 is 0.37 L/min

rate of consumption of O2= (2.06-1.51) L/min   =0.55 L/min

use:
P*V=n*R*T
1*0.37 = n* 0.08205*298
n= 0.015 mol
rate of production of CO2 is 0.015 mol/min

use:
P*V=n*R*T
1*0.55 = n* 0.08205*298
n= 0.022 mol
rate of consumption of O2=0.022 mol/min

b)
O2 is used up at the rate of 0.022 mol/min
in 1min.
total oxygen used = 0.022 mol

oxygen used in 1st reaction = 23*0.022/(23+6) =0.017mol
rate of palmitic acid oxidation = 0.017/23 mol/min = 7.6*10^-4 mol/min

oxygen used in 2nd reaction = 6*0.022/(23+6) =4.55*10^-3 mol
rate of glucose oxidation = 4.55*10^-3 /6 mol/min = 7.6*10^-4 mol/min

c)
oxygen used in 1st reaction = 0.017mol
oxygen used in 2nd reaction = 4.55*10^-3 mol

fraction due to palmitic acid = 0.017 / (0.017+4.55*10^-3) = 0.79
fraction due to glucose = 1-0.79= 0.21

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