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The following data was obtained from a patient complaining of shortness of breath.
HR = 120 breaths/min
Blood pressure = 145/95
PaO2 = 60 mmHg
H+ concentration = 25 nM
Respiratory rales
A tentative diagnosis of left heart failure is made. Which one of the following values will be below normal?
A.
Left atrial pressure
B.
Pulmonary artery resistance
C.
Pulmonary artery pressure
D.
A-a gradient
E.
PaCo2
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The answer is: E
The tentative diagnosis of left heart failure is based on the shortness of breath and the accompanying respiratory rales. The low PaO2 most likely results from the pulmonary edema accompanying the left heart failure, which interferes with gas exchange. The low PaO2 stimulates the chemoreceptors, which, in turn, stimulate ventilation. The increased ventilation reduces PaCo2 to below normal. The presence of hyperventilation is confirmed by the low H+ ion concentration. The heart failure is responsible for the increased left atrial pressure, and the low PO2 is responsible for the increased pulmonary artery resistance and pressure. Providing oxygen will overcome the diffusional barrier and increase arterial PO2. The high PO2 will reduce pulmonary artery resistance. In order to restore PaO2 toward normal, the alveolar PO2 must be raised more than the arterial PO2 will increase, and, therefore, the A-a gradient will be increased. Restoring the arterial PO2 toward normal will decrease the chemoreceptor respiratory drive, and, therefore, both the PCo2 and the H+ ion concentration will increase toward normal.
