Question concerning PO2 calculation??

[yu-gi-oh]

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100ml normal blood with PO2 of 2mmHg and another 100ml of normal blood with PO2 of 98mmHg. Both blood is mixed anaerobically to equilibrium. What is the PO2 of resultant mixture?

Answer: 27mmHg

Anyone get this calculation????

 

[principe]

I don't think you need a formula for this one. Just look at the oxyhemoglobin dissociation curve. Mixing the two vials would give you an SaO2 of 50%, which corresponds exactly to 27mmHg (P50).

 

[yu-gi-oh]

I also was thinking of this after looking at the answer..

I guess this must be the way to get this question..

 

[Planktonmd]

I think you are confusing PO2 and Hb saturation.
How is mixing a vial with PO2 of 98 mmhg and a vial with PO2 of 2 mmhg going to give you a PO2 of 27%?
Maybe the OP wanted to say that the 2 vilas had a saturation of 98% and 2 % not PO2??

I don't think you need a formula for this one. Just look at the oxyhemoglobin dissociation curve. Mixing the two vials would give you an SaO2 of 50%, which corresponds exactly to 27mmHg (P50).

 

[principe]

Just look at the graph, dude. At both ends of the curve PaO2 is pretty much equal to SaO2. This doesn't apply to the middle part of the curve, which I think was the reason the question used 98mmHg and 2mmHg values (at the extreme ends of the curve).

 

[2win]

Just look at the graph, dude. At both ends of the curve PaO2 is pretty much equal to SaO2. This doesn't apply to the middle part of the curve, which I think was the reason the question used 98mmHg and 2mmHg values (at the extreme ends of the curve).

You are right:
Vial 1 : CaO2= HB(15)X1.34XSaO2+0.003XPao2(2)=almost 0 because of the Pa02=2 (see HgB dissociation curve)

Vial 2 CaO2=HB (15)X1.34XSa02+0.003xPaO2(98) = 100.2

Mixing them in anaerobic conditions PO2=0

CaO2=HgB(15)X1.34XSao2 +0.003xPa02(0)=20.1XSa02
It was mentioned that the blood is "normal" - HgB=15
Vial1Ca02+Vial2Ca02=0+100.2=100.2 (total content of O2)
100.2=20.1xSa02
Sa02=50% Therefore from the HgB dissociation curve PaO2=27mmHG

 

[Planktonmd]

You are right! 👍

Just look at the graph, dude. At both ends of the curve PaO2 is pretty much equal to SaO2. This doesn't apply to the middle part of the curve, which I think was the reason the question used 98mmHg and 2mmHg values (at the extreme ends of the curve).

 

[pgg]

You are right:
Vial 1 : CaO2= HB(15)X1.34XSaO2+0.003XPao2(2)=almost 0 because of the Pa02=2 (see HgB dissociation curve)

Vial 2 CaO2=HB (15)X1.34XSa02+0.003xPaO2(98) = 100.2

Mixing them in anaerobic conditions PO2=0

CaO2=HgB(15)X1.34XSao2 +0.003xPa02(0)=20.1XSa02
It was mentioned that the blood is "normal" - HgB=15
Vial1Ca02+Vial2Ca02=0+100.2=100.2 (total content of O2)
100.2=20.1xSa02
Sa02=50% Therefore from the HgB dissociation curve PaO2=27mmHG

100.2 / 20.1 = 5 not 50, and in these equations SpO2 of 50% gets entered as 0.5 not 50 anyway.

But this is a red herring; even if perfect, the math is unnecessary and just makes the problem harder (if not impossible) to solve since ultimately you have to eyeball the end point somewhere in the steep part of the dissociation curve. What I mean is that a general solution for varying Hb would be

CaO2 vial 1 = 1.34 * 1 * Hb + .003 * 98 (in 1 mL)
CaO2 vial 2 = 1.34 * 0 * Hb + .003 * 2 (in 1 mL)

CaO2 (vial 1+2) = 1.34 * Hb + .3 (in 2 mL)

CaO2 (1 mL of the 1+2 mix) = .67 * Hb + .15

1.34 * SpO2 * Hb + .003 * PO2 = .67 * Hb + .15

PO2 = (.67 * Hb + .15 - 1.34 * SpO2 * Hb) / .003

Assuming Hb is 15 (normal blood)

PO2 = 3400 - 6700 * SpO2

And from here you need to look at the dissocation curve to find a PO2 and SpO2 that lie on it and satisfy this equation. (We can't exactly solve it, because there's no formula for the dissociation curve ... so we're stuck eyeballing it.) As it turns out this equation produces reasonable solutions with an SpO2 around 50% ... but very small changes yield very different PO2s (eg, valid pairs are SpO2 .5 & PaO2 of 50, and SpO2 of .5034 & PaO2 of 27) ... but no human can look at a graph and give that precise an answer.


But again this math is unnecessary; if you mix equal volumes of 100% saturated and 0% saturated blood, regardless of Hb and regardless of the shape/position of the dissociation curve, you're going to get a 50% saturated mix. From there it's easy to take the appropriate curve and plot a rough PO2.

If we were starting with vials that had unequal Hb concentrations, different volumes, or PO2 values NOT at the edges of the dissociation curve, you're stuck doing the math ... but you likely won't end up with as precise an answer as 27 (unless the endpoint is itself at the edge of the curve).

There, I think I've pedanted that to death. 🙂

 

[fakin' the funk]

But again this math is unnecessary; if you mix equal volumes of 100% saturated and 0% saturated blood, regardless of Hb and regardless of the shape/position of the dissociation curve, you're going to get a 50% saturated mix. From there it's easy to take the appropriate curve and plot a rough PO2.

I think what you're really saying is, if you take mix equal volumes of blood where one contains 20 ml O2/dl and the other contains 0 ml O2/dl, then you get 10 ml O2/dl and the resultant SaO2 and pO2.

 

[pgg]

I think what you're really saying is, if you take mix equal volumes of blood where one contains 20 ml O2/dl and the other contains 0 ml O2/dl, then you get 10 ml O2/dl and the resultant SaO2 and pO2.

Yes.