Answers A, C and E are true statements, whereas, B and D are false.
1. The coronary oxygen uptake must correspond to a total energy, which is 5-fold (100/20) larger than the energy necessary for the external work (16 J per s) of the heart.
The coronary oxygen uptake is: (16 * 5)/20 = 4 ml (STPD) of oxygen each second or 240 ml per min.
2. According to Ficks principle, the coronary bloodflow equals the coronary oxygen uptake divided by the arterio-venous oxygen content difference.
Coronary exercise-bloodflow: 240/(200 - 40) = 1.5 l per min. This bloodflow is 10-fold larger than the coronary bloodflow at rest.
3. The high lactic acid level in the blood reflects a considerable lactacide oxygen debth, which has to be paid back over the following hours. An oxygen debth of 10-15 l STPD is to be expected.
1.The oxygen debth is the extra volume of oxygen used in the 60 min post-exercise period. This is calculated as the difference between the total oxygen uptake and the uptake at rest (60 min * 0.3 l per min = 18 litres STPD of oxygen). The total oxygen uptake is (V°I * FIO2 ) minus (V°E *FEO2), or (V°E * (0.2093-0.1746)). This is (1090 * 0.0347) = 37.82. Subtraction of 18 litres provides us with a result of 19.82 litres STPD as oxygen debth.
2.The oxygen stores of the body is assumed to be exactly re-established after the 60 min period. The V°E of the athlete must be equal to the V°I in the 60 min period.
This is actually the case, since ( V°I * FIN2 ) is equal to (V°E *FEN2), and the two fractions are identical.
3.The total carbon dioxide output is ((1090*(0.035 - 0.0003)) or 37.82 litres, exactly identical with the total oxygen uptake in the post-exercise period. Accordingly, the ventilatory exchange quotient is maintained from rest at the value 1.
4.The lactacide oxygen debth following supramaximal work is used for oxidation of 75% of the lactate produced and for the conversion of the remaining lactate to glycogen in the liver. Restoration of phosphocreatine, ATP and muscle cells is part of the repayment of the oxygen debth.
1. The law of mass balance states the following: V°O2 = Q° × (CaO2 - Cv¯O2). The athlete can attain a maximal Q° of 40 l of blood min-1. Pulmonary bloodflow equals cardiac output, so to accommodate the increase during exercise, lung capillaries must open up. Thus unlike at rest, the apical parts of the lungs become well perfused, improving the (V°A / Q°) - ratio and increasing the area available for gas transport.
2. The fitness number is 6000 ml divided by the body weight of 74 kg, which means 81 ml STPD O2 kg-1 min-1 . This is an extremely high endurance capacity.
3. The mean oxygen tension gradient for pulmonary diffusion (DPO2 ) is equal to PAO2 minus the mean tension in the pulmonary capillaries. The relationship is as follows: V°O2 = (DL02 × DPO2). Accordingly, DPO2 is 6000/(9*60) = 11.1 kPa or 83 mmHg.
The substantial rise in DPO2 is caused by a substantial fall in the mean oxygen tension of the pulmonary capillaries due to more complete oxygen extraction in exercising muscles. A small increase in PAO2 also contribute, because V°A increases out of proportion to V°O2 above the AT.
1. Mass balance:
V°O2 = Cardiac output* (CaO2 - mixed venous CO2); V°O2 = 25 * 170 = 4250 ml STPD per min.
2. Assuming she is working at or above her V°O2 max we can calculate her fitness number to 4,250/62 = 69 ml O2 per kg and per min. This is an excellent result for a well-trained athlete.
1. V°O2 = (DLO2 × DPO2); V°O2 = (22 × 12) = 264 ml STPD each min.
The oxygen tension gradient can rise to a maximum value of 70 mmHg during exercise.
2. A rise in PAO2 and a fall in PvO2 during exercise cause the rise in the tension gradient from 12 to 70 mmHg. This rise alone is insufficient without a corresponding rise in the lung diffusion capacity. Thus the lung diffusion capacity must also increase to reach 4900. It must increase exactly to 70 ml STPD per min and per mmHg: (70 × 70 = 4900). The number of open capillaries increases, and they dilate. The cardiac output increases by a factor of 5-8. All these changes increase the total diffusion of oxygen, the diffusion barrier area and reduce the diffusion distance in the expanded lungs.
3. The renal bloodflow is 1200 ml per min and CaO2 is 200 ml per l. The renal oxygen consumption is 15 ml per min. The renal oxygen influx is (200* 1.2) = 240 ml per min. The renal oxygen outflux is (240 - 15) = 225 ml per min. The mixed venous CO2 is 225/1.2 = 187.5 ml per l. Thus, the renal oxygen content difference is (200 - 187.5) = 12.5 ml per l.
4. The small oxygen content difference obtained in question 3. is reasonable, because the high renal blood flow has a clearance purpose. The oxygen delivery to the kidney tissues is obviously redundant