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  #1  
Old 06-22-2013
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Question How to solve this A-a gradient question?

Patient develops hypoxemia 35 minutes after ingesting a lethal dose of barbituates. What is the patients abg look like?

PO2; PCO2; A-a
40; 50; 35
40; 60; 40
50; 25; 10
50; 80; 10
60; 35; 25

I got the question wrong. Was hoping somebody could help me work this problem out.
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Old 06-22-2013
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Barbiturates depress respiratory center
Increase PaCO2 retention due to hypoventilation and decrease PaO2
A-a gradient remains normal cause theres no V/Q problem, no shunt and no diffusion problem

So my best guess would be 50-80-10
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Old 06-22-2013
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Quote:
Originally Posted by JJacobs152 View Post
Patient develops hypoxemia 35 minutes after ingesting a lethal dose of barbituates. What is the patients abg look like?

PO2; PCO2; A-a
40; 50; 35
40; 60; 40
50; 25; 10
50; 80; 10
60; 35; 25

I got the question wrong. Was hoping somebody could help me work this problem out.
Goljon says it the best.

in barbiturate OD u have depression of respiratory center so she wont be breathing as much so causing resp acidosis. so CO2 should b high and u can eliminate C and E. then in barbiturate OD also u have a normal A-a gradient, so you can eliminate A and B and left with D.
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Old 06-22-2013
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I also think its D.
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Old 06-22-2013
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yup, thanks for the responses fellas. The PO2 level on the far left, that's PaO2, correct?

Can't you solve for the PAO2, using 150 - PCO2/.8. In this case it came out to be 50.

My question is, how do I go about differentiating what causes a normal A-a gradient, an increase in A-a gradient, and everything else. I feel this was an easy concept, that I shouldn't have gotten wrong.

I got as far as realizing, barbs depress respiratory function, so C and E would be out, & from there I got stuck.
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Old 06-22-2013
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When the respiratory function is depressed, the O2 in the alveoli decrease. Similarly the O2 in the pulmonary capillaries also decrease. Since both O2 levels decrease in a similar and equal way, the gradient(difference) between the two does not change. Therefore the A-a gradient is normal.
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Old 06-23-2013
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Quote:
Originally Posted by shima View Post
Goljon says it the best.

in barbiturate OD u have depression of respiratory center so she wont be breathing as much so causing resp acidosis. so CO2 should b high and u can eliminate C and E. then in barbiturate OD also u have a normal A-a gradient, so you can eliminate A and B and left with D.
You did not answer his question.
He's asking why and how do we know that A-a gradient is normal in barbiturates overdose.
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Old 06-23-2013
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Quote:
Originally Posted by JJacobs152 View Post
Patient develops hypoxemia 35 minutes after ingesting a lethal dose of barbituates. What is the patients abg look like?

PO2; PCO2; A-a
40; 50; 35
40; 60; 40
50; 25; 10
50; 80; 10
60; 35; 25

I got the question wrong. Was hoping somebody could help me work this problem out.
50, 80, 10
Barbiturate ingestion causes hypoventilation, due to respiratory center depression. And this results in hypoxemia and hypercapnia. No breathing- no gas circulation between lungs and atmospheric air- CO2 accumulates in lungs and O2 decreases because perfusion isn't impaired. And this is also a reason why A-a remains normal.
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Old 06-24-2013
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Ok......let's concentrate on the question at first......here it says you have a pt. scenario who has got barbiturate poisoning that means pt. is under respiratory depression. As pt. is hypoventilating so it will invariably result in retention in Carbon di oxide i.e. Hypercapnia as well as decrease in Oxygen conc. i.e. Hypoxemia. But the key fact is that there will be no increase in normal Alveolar-arterial Oxygen gradient as respiratory center is itself depressed. Here respiratory membrane is intact & there is no shunt anomaly mentioned. So just look for the option which says normal A-a gradient..........now u have only two options C and D. Now concentrate which one shows both hypoxemia+hypercapnia>>>>> It's only option D)

So my answer is D).....Hope it helps.
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