Step I Aa gradient in V/Q defect

[jds112]

If there is airway obstruction, like a piece of food (as shown in the picture), why would the A-a gradient be increased? I understand what happens with a PE and why the listed values are occurring.

If there was airway obstruction, and the PAO2 is zero, how can there be an increased A-a gradient if PAO2 is zero to begin with?

 

[zhopv10]

A-a gradient is PA-Pa right? So PA=[fiO2*(barometric pressure 760 say-pressure h20 47)-PaCO2/.8] -PaO2. So the PA (alveolar) which is predicted is unchanged by the food (arguably maybe a little less co2 can diffuse and that would increase maybe? Meh not really as it says it would stay the same ). Now the Pa02 would be lower because there is hypoxemia so the PA would stay the same and then the measures Pa02 would be lower due to the decreased V/Q right. So then there is a bigger gradient. So I think the confusion is coming from thinking that PA is measured in the situation and thus would be zero but that is not the case. PA is calculated from what you see in the equation, the only actual physiological value taken is the pCO2. So I think of it as a gradient of what you would expect the paO2 should be given perfect transfer and what is happening. Some gradient is normal but too much shows a problem. So really the only things that change the PA part of the equation are barometric pressure, oxygen concentration of the air, and minute respirations (Co2 readily diffuses so it represents how hard you are breathing to achieve the PaO2 so a small number is increased resp effort to achieve the PaO2 and can increase the gradient. The other thing that affects the gradient is of course the Pao2 which so the other physiologically measured thing and that is what would go down here (and ostensibly the CO2 might be down and the PA increased too. This leads to the increased gradient seen. Hopefully that made some semblance of sense/is helpful. It's been a long day lol


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