Calculating the Alveolar Oxygen & the A-a O2 gradient |

Created by Diane R. Karius, Ph.D. |

Calculating the Alveolar PO2

Earlier today, I gave you a very nice looking equation that was, sadly, an 'unsolvable' equation:

Alveolar O2 (PAO2) = Inspired Oxygen - Oxygen consumed

This equation makes sense - the amount of oxygen that is in the alveoli is determined by how much oxygen I take in minus the amount of oxygen consumed by my tissue. This equation, in fact, makes so much sense that clinically there are times you are going to need to solve the equation in order to figure out what is happening in the lungs. Afterall, x-rays and CAT scans tell me a lot about the structure of the lungs, but we can only infer what is happening functionally. Starting from this equation (and using readily available numbers relatively easy to get from even a very sick patient), you can begin to make some inferences about the pulmonary function. The following animation works through the same material that the text does.

First, I need to do something to make the equation solveable - after all, I did start the paragraph by telling you this was unsolveable!

Alveolar O

_{2}(PAO_{2}) = Inspired Oxygen - Oxygen consumedLet's start with the inspired oxygen levels - the first thing that I'll tell you is the 'proper' term for this variable is the "inspired partial pressure of oxygen' and we abbreviate that as P

_{i}O_{2}:P

_{A}O_{2 }= P_{I}O_{2}- Oxygen consumedNow - how do we figure out the P

_{i}O_{2}? Since we're getting the air from the atmostphere, we need to use the atmospheric number - so we start with the atmospheric pressure (which is 760 mm Hg at sea level). However, as we inhale the air, water is added to the airpriorto it reaching the alveoli - so we need to account for the water vapor that has been added by the airways. We account for this by subtracting the water vapor pressure (the partial pressure of the water that has been added) from the total atmospheric pressure. Luckily, you only need to know one number for this: at body temperature, the water vapor pressure is 47 mm Hg, so our calculation (which isn't complete yet) becomes:P(atm) - P

_{H20}(760 mm Hg - 47 mmHg)

713 mm Hg

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Now - oxygen does not make up all the air other than the water vapor pressure - it is only 21% of the air we breathe in, so we have more step to do at this point - we have to figure out how much oxygen is there:

P

_{i}O_{2}= (Patm - 47 mm Hg)x F_{i}O_{2}

(translation: FP_{i}O_{2}is the fraction of inspired oxygen (which is 21% or 0.21 if we are breathing room air)

_{i}O_{2}= (Patm - 47 mm Hg)0.21

= 713 x 0.21

= 149.7 mm HgSo we have identifed how much oxygen is being inspired - the second half of the equation asks us to identify how much oxygen is being taken away, but it does so in an indirect manner. In fact, it is so indirect that we are actually going to use the easily measured

CO(that is not a typo!!!!) to estimate the oxygen consumption. If you will recall from the gas exchange lecture, I told you that oxygen and carbon dioxide exchange were independent of one another in the lungs. Now, there is a place where they are tied together - and that's in the tissue where oxygen is being consumed and carbon dioxide being produced by metabolism. In fact, they are so tied together in the tissue that there is a very nice mathematical relationship between oxygen and carbon dioxide. This mathematical relationship is called_{2}the respiratory quotient (RQ)and is defined as the amount of CO2 produced divided by the oxygen consumption:

RQ = V(dot)CO2/V(dot)O2

The nice thing about this mathematical relationship is that you don't actually have to do any math to solve it (the best kind of math equation!). This is because, as a human, there are only three options for answers because there are only a couple of different fuel sources we are going to use (and therefore only three possible answers, which you can memorize easily). Our two options for fuel are glucose and fatty acids. If our cells are using exclusively glucose for energy production, 1 molecule of carbon dioxide is created for each oxygen burned, so the RQ is 1. We're actualy more efficient at using free fatty acids for energy production - for every 10 oxygens consumed, we only make 7 carbon dioxides, leading to an RQ of 0.7.

Now, the very quick among you are noticing that I've only given you two possible answers.. The third possible answer is due to the fact that, except for very specific circumstances (which I would have to tell you about), our cells are using a mix of gluose and free fatty acids at the time. When we do that, the RQ has been determined to be 0.8 (8 carbon dioxide produced for every oxygen consumed).

To summarize the last two paragraphs:

If we were burning primarily glucose, the RQ would be 1.0

while if we were burning fats, the RQ would be 0.7

Since we usually burn a mix of fuels, the measured RQ is usually near 0.8For those of you who are wondering how I would indicate to you what RQ to use, I can summarize it easily:

If I say nothing, assume that the RQ is 0.8(the patient is using a mix of fuels)

If I want you to use anRQ of 1.0,I would specify that the patient is receiving anIV glucose solution

If I want you to use anRQ of 0.7, the patient ishypoglycemic or a diabetic and relying on fatty acid metabolism.

Back to our equation:We can in fact measure O2 consumption (the V(dot)O2), but in a critically ill patient, it is not desireable to do that. Instead, we are going to use the respiratory quotient, some simple math, and one major detail about carbon dioxide to help us figure this out. The major detail about carbon dioxide that is important for us right here is the fact that the all of the CO2 dissolved in our blood is the result of cellular metabolism - we don't breathe any in (or enough for us to worry about in this setting). Therefore, we can measure the PaCO2 (the arterial concentration of carbon dioxide) and plug it into the equation for RQ as an estimate of carbon dioxide production. We then rearrange the equation to find out what our oxygen consumption needs to be to produce that much carbon dioxide:RQ = [V(dot)CO2]/[V(dot)O2]

Step 1: We are going to substitute the PaCO2 in for V(dot)CO2:

RQ = PaCO2/[V(dot)O2]

Step 2: Rearrange the equation to get V(dot)O2 where we want it:

V(dot)O2 (oxygen consumed) = PaCO2/RQStep 3: Substitute this into the Alveolar gas equation:

P_{A}O_{2 }= P_{I}O_{2}- Oxygen consumed

P_{A}O_{2 }= P_{I}O_{2}- [PaCO_{2}/RQ]Using the following values:

PaCO_{2}= 40 mm Hg (the normal value)

RQ = 0.8 (based on the assumption that we are using a mix of metabolic fuels)This becomes:

P

_{A}O_{2 }= P_{I}O_{2}- [40 mm Hg /0.8]= P

_{I}O_{2}- [50]We'd previously discovered that the P

_{I}O_{2}was 149.7 mm Hg, so the equation now becomes:P

_{A}O_{2 }= 149.7- [50]P

_{A}O_{2}= 99.7 mm Hg

The Alveolar-arterial O_{2}gradient (A-a O_{2}gradient)Now, if you're sensible, you are wondering why I dragged you through that big long calculation... after all, although it is nice to know what the alveolar partial pressure of oxygen is, it probably doesn't seem very useful right now. That's because we have one more (thankfully simple) step to go through.

As we talked about in the gas exchange lecture, the whole point of the alveolus is to bring the blood and the air together in such a way that

the alveolar oxygen and the arterial oxygen come to equilibrium with each other.In other words, in an ideal world, the PAO_{2}should be the same as the PaO_{2}. If there is a large difference between the PAO_{2}and the PaO_{2}, there is a problem with gas exchange.

So now for some simple math:

A-a O

_{2}gradient = PAO_{2}- PaO_{2}Step 1: Get the PAO

_{2}from the alveolar gas equation:

A-a O_{2}gradient = 99.7 mm Hg - PaO_{2}Step 2: Get the PaO

_{2}from the arterial blood gases:

A-a O_{2}gradient = 99.7 mm Hg- 96 mmHg

= 3.7 mm HgThere are a number of ways of arriving at a normal value for the A-a O

_{2}gradient. For a physiologist, we like a normal A-a value of 0 mm Hg, but we have to do things that are REALLY invasive to get that value, so clinically that is NOT the normal value. One common normal range is 8 + (20% of the patient's age). Other clinicians use consider <12 mm Hg normal. In any case, the number we've just calculated indicate that gas exchange is occuring normally.If the number had been high, it would indicate that the process of gas exchange has been impaired by some disease process. This set of calculations is very useful for distinguishing between hypoxia due to an alveolar process and the hypoxia produced by hypoventilation (inadequate respiratory efforts) - in the latter case, the gradient is within the normal range. Since the normal physiological value is 0, there is no clinical entity associated with a LOW A-a O

_{2}value.

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