question from USMLEasy.com site --- someone pleas -

[ArchivalUser]

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

[ArchivalUser]

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.

[ArchivalUser]

Aussie0157

Speciic mention is made that pH is normal, this implies that in THIS patient PaCO2 is probably also normal

[ArchivalUser]

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.

[ArchivalUser]

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