simrin
A patient with reduced VC, FRC, and RV is found to have a normal pH. A tentative diagnosis of diffuse interstitial fibrosis is made. Which of the following characteristics are consistent with this disease?
A. An increase in lung compliance
B. A decrease in respiratory rate
C. An increase in the V/Q ratio
D. A decrease in PaCo2
E. An increase in the FEV1/FVC ratio
simrin
The answer to the above question according to the question bank is E - which I completely understand. But my question is, shouldnt choice D also be a correct answer?????????? Thats where I get lost. Choice D says decreased CO2 -- which is true in all restrictive pattern diseases.
For all others - the explination given for the correct answer was:
Interstitial lung disease is a restrictive-type lung disease in which all the lung volumes, including VC, FRC, and RV, are reduced. The decrease in lung volumes is caused by a decrease in lung compliance. The low compliance produces a high recoil force so that a higher than normal fraction of the forced vital capacity (FEV1/FVC) can be exhaled during expiration. The normal pH is consistent with a normal alveolar ventilation. Because lung compliance is reduced, the normal alveolar ventilation is maintained with a breathing pattern consisting of a high rate and a low volume.
Aussie0157
Speciic mention is made that pH is normal, this implies that in THIS patient PaCO2 is probably also normal
Mom
The only reason I can see why D would not be correct is a compensatory process. As Co2 is blown off, resp alkalosis occurs, the kidneys compensate w/ metabolic acidosis by reabsorping H+, therby bring pH comes back toward norml in the process.
amb
the concept most often tested with restrictive lung diseases in the usmle...is that there is increased recoil of the lungs...and so the fev1/ fvc increase