Biology 482, fall 2017, Homework Problem Set 1 Name: 21. In the laboratory (Ener

Question

Biology 482, fall 2017, Homework Problem Set 1 Name: 21. In the laboratory (Energy Expenditure& Exercise), we measured heart rate, ventilation rate and volume, rate of oxygen consumption (VO2) reflects the difference in oxygen between inspired air and expired air, recall that for simplicity, V here refers to rate (and should have a dot over it) and not volume. The volume of oxygen consumed equals the volume of air times the percent of air comprised of oxygen. The rate of oxygen consumption takes time into account. oxygen consumption, and carbon dioxide production during exercise. In principle, the However, because t emperature, pressure, and water vapor affect air volumes, measured air are not all comparable to one another: to be comparable, they need to be corrected to a n set of conditions, such as STPD (standard temperature and pressure, dry). In simplified form, the equation for oxygen consumption is: where VE the rate of expired air (volume/time). oxygen, and FEO, fractional concentration of expired oxygen. FIO, is constant in outside afr (our lab software uses the value 0.2093. So: fractional concentration of inspired vò_VESTPD (0.2093-FE02). Outdoors, the fractional concentration of oxygen in the air is 0.2095. However, in a building in which people oxygen in the air are consuming oxygen, we will use 0.2093 as the fractional concentration of The formula for correcting VE to STPD is: --VESTPI)" (VEATPs / time) x ((P)" Pico) / 760)" (273 / (273 + Ta)) where VEATPS is the volume of expired air saturated and at ambient temperature and pressure, P is the barometric pressure, PhaD is the saturated water vapor pressure, 760 is the designated standard pressure (in mm Hg). 273 is degrees K, and T, is ambient temperature in degrees C. Units are not shown here for simplicity From these equations for VO, it Should be apparent that oxygen consumption is dependent Now use the equations above and the empirical values below (from our aerobic exercise lab VEATPs = 53.7 L on ventilation rate (VE) and oxygen extraction by the lungs (FIO -FEO2). the experiment) to ca calculations, including the units. Check that the units are right (it's easier than it might seem). tion). Be sure to show your Time interval -1 min P120 -19.8 mm Hg V5.mn . 3 2 L/mn (o.una)

Explanation / Answer

By putting all values in the formula given in question-

VESTPD= (53.7 ÷ 1) ×{(767-19.8)÷760}×{273 ÷ (273+22)}

VESTPD = 53.7× 0.98 × 0.93

VESTPD = 48.94 L min-1

Now VO2 = VESTPD (FIO2 – FEO2)

FIO2 = 0.2093

FEO2 = 0.163

By putting all values in the formula-

VO2 =48.94 (0.2093- 0.163)

VO2 =48.94 ×0.0463

VO2 =2.27 L min-1

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.