What does 50% Hb saturation mean?
Does it mean 50% Hb is fully saturated (4 oxygen molecules bound) and 50% Hb (no O2 bound)?
Does it mean 50% Hb is fully saturated (4 oxygen molecules bound) and 50% Hb (no O2 bound)?
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Each Hb has 4 heme groups and each heme group can combine with one O2 molecule.
Oxygen Saturation (SO2)
Four O2 molecules bound to a Hb molecule>>>>97% saturated (100 mmHg of O2)
Three O2 molecules bound to a Hb molecule>>>75% saturated (40 mmHg of O2)
Two O2 molecules bound to a Hb molecule >> 50% saturated (26 mmHg of O2)
One O2 molecule bound to a Hb molecule > 25% saturated (under physiological conditions, this oxygen remains attached to Hb)
Oxygen Saturation (SO2)
Four O2 molecules bound to a Hb molecule>>>>97% saturated (100 mmHg of O2)
Three O2 molecules bound to a Hb molecule>>>75% saturated (40 mmHg of O2)
Two O2 molecules bound to a Hb molecule >> 50% saturated (26 mmHg of O2)
One O2 molecule bound to a Hb molecule > 25% saturated (under physiological conditions, this oxygen remains attached to Hb)
If you're talking about a single Hb molecule, 50% sat is 2 oxygen molecules attached.
If you're talking about a persons blood as a whole, saying O2sat is 50% means that on average, each Hb molecule in the patients body has 2 molecules attached to it.
There won't be a situation where half of the patients Hb has zero oxygen molecules and the other half 4 to yield an average of 2.
If you're talking about a persons blood as a whole, saying O2sat is 50% means that on average, each Hb molecule in the patients body has 2 molecules attached to it.
There won't be a situation where half of the patients Hb has zero oxygen molecules and the other half 4 to yield an average of 2.
Sorry but I believe you are mistaken. I have just confirmed the answer with my lecturer: 50% sat Hb means that half the Hb has 4 O2 attached and half has none O2 attached due to cooperative binding. there is only two states Hb can be in deoxyHb and fully bound Hb 
Eh? If that's correct, I've managed to completely misunderstand the Hb-O2 dissociation curve.
The way I understand it, each Hb molecule can bind 0-4 O2 molecules. Hence the saturation of a single Hb molecule increases in 25% increments. Sigmoid kinetics means that the first O2 binding makes it easier for the second to bind, which makes it easier for the third to bind, and so on.
Furthermore, under normal circumstances, the first two O2 molecules never dissociate from Hb, so the baseline is 50% saturation. You'd expect arterial blood to be at 100% (a little less in reality), and for tissues to extract one O2 per Hb molecule, thus placing venous blood at roughly 75% saturation. I don't believe this means 75% of Hb molecules have 4 O2s and the other 25% have zero O2s attached to them to yield a net of 75% saturation. The partial pressure of oxygen in the blood is the force that determines how many Oxygen molecules remain attached to Hb, and this is a force that should apply equally to all the Hb molecules in the blood. I don't see how different Hb molecules can end up at 0% or 100% saturation while being exposed to the same PaO2.
Furthermore, I think you're implying sigmoid kinetics occurs between different Hb molecules, whereas my concept is that it occurs between different O2 binding sites on the same Hb molecule.
The way I understand it, each Hb molecule can bind 0-4 O2 molecules. Hence the saturation of a single Hb molecule increases in 25% increments. Sigmoid kinetics means that the first O2 binding makes it easier for the second to bind, which makes it easier for the third to bind, and so on.
Furthermore, under normal circumstances, the first two O2 molecules never dissociate from Hb, so the baseline is 50% saturation. You'd expect arterial blood to be at 100% (a little less in reality), and for tissues to extract one O2 per Hb molecule, thus placing venous blood at roughly 75% saturation. I don't believe this means 75% of Hb molecules have 4 O2s and the other 25% have zero O2s attached to them to yield a net of 75% saturation. The partial pressure of oxygen in the blood is the force that determines how many Oxygen molecules remain attached to Hb, and this is a force that should apply equally to all the Hb molecules in the blood. I don't see how different Hb molecules can end up at 0% or 100% saturation while being exposed to the same PaO2.
Furthermore, I think you're implying sigmoid kinetics occurs between different Hb molecules, whereas my concept is that it occurs between different O2 binding sites on the same Hb molecule.
Hey again,
I thought I understood Hb-O2 dissociation curve until I was told otherwise.
Cooperative binding: each oxygen increases the affinity for the next.
If you had two molecules of Hb, one empty (deoxyHb) and one with two oxygens already bound; if you add a new oxygen into this, to which Hb is it most likely to bind?
What then if you add another oxygen?
"normal circumstances, first two O2 moelcules never dissociate from Hb", this is what I thought initially.
(CRUCIAL POINT): Think of this example, if that were true, there would not be cyanosis!
I thought I understood Hb-O2 dissociation curve until I was told otherwise.
Cooperative binding: each oxygen increases the affinity for the next.
If you had two molecules of Hb, one empty (deoxyHb) and one with two oxygens already bound; if you add a new oxygen into this, to which Hb is it most likely to bind?
What then if you add another oxygen?
"normal circumstances, first two O2 moelcules never dissociate from Hb", this is what I thought initially.
(CRUCIAL POINT): Think of this example, if that were true, there would not be cyanosis!
Well cyanosis isn't a normal circumstance.
I'm not getting your explanation. Sure, the Hb molecule with two O2s already bound to it will have greater affinity to pick up another two molecules, but whats stopping the Hb molecule with zero O2s from picking up 4? The PaO2 is clearly enough to push Hb to 100% sat in this case.
I'm not getting your explanation. Sure, the Hb molecule with two O2s already bound to it will have greater affinity to pick up another two molecules, but whats stopping the Hb molecule with zero O2s from picking up 4? The PaO2 is clearly enough to push Hb to 100% sat in this case.
Hey again,
i think the pao2 is irrelevant in this case.
What I believe is that there are only two states of Hb fully saturated and non-saturated (i.e. deoxyHb).
As to "whats stopping the Hb molecule with zero O2s from picking up 4?", it is energetically more favourable for a Hb molecule with some oxygen bound already to bind more so that it becomes fully saturated than a deoxy Hb binding an molecule of oxygen.
50% Hb sat of course is not possible in real life as patient would be dead, lets try a more realistic example-90% Hb sat:
90% Hb molecules have 4o2 bound and 10% have no oxygen bound:-why does 10% have no o2 bound?-this doesnt depend on pao2, more likely a V/Q mismatch
i think the pao2 is irrelevant in this case.
What I believe is that there are only two states of Hb fully saturated and non-saturated (i.e. deoxyHb).
As to "whats stopping the Hb molecule with zero O2s from picking up 4?", it is energetically more favourable for a Hb molecule with some oxygen bound already to bind more so that it becomes fully saturated than a deoxy Hb binding an molecule of oxygen.
50% Hb sat of course is not possible in real life as patient would be dead, lets try a more realistic example-90% Hb sat:
90% Hb molecules have 4o2 bound and 10% have no oxygen bound:-why does 10% have no o2 bound?-this doesnt depend on pao2, more likely a V/Q mismatch
I agree with Dr.Picard. 50% Hb saturation means that at a time 2 O2 molecules are bound to a molecule of Hb. If you were to randomly pick any Hb molecule from arterial blood at 50% saturation, you'd find 2 molecules of O2 attached to it. It doesn't mean that half of Hb molecules are carrying 4 O2 and the other half are without any O2.
This picture may help you to understand:
This picture may help you to understand:
Attachments
Hey canary,
"If you were to randomly pick any Hb molecule from arterial blood at 50% saturation, you'd find 2 molecules of o2 attached"
-the key to this is cooperative binding: you wouldnt find all Hb molecules with 2 molecules oxygen attached as it is energetically less favourable than Hb with 4o2 attached and Hb with no oxygen attached.
"If you were to randomly pick any Hb molecule from arterial blood at 50% saturation, you'd find 2 molecules of o2 attached"
-the key to this is cooperative binding: you wouldnt find all Hb molecules with 2 molecules oxygen attached as it is energetically less favourable than Hb with 4o2 attached and Hb with no oxygen attached.
It does affect all Hb molecules equally, how much it affects Hb molecules depends on the severity of V/Q mismatch.
I think we are deviating too much from the original question i made; regarding the real meaning of what saturation is referring to.
I think we are deviating too much from the original question i made; regarding the real meaning of what saturation is referring to.
a v/q mismatch doesnt mean there is a mismatch in the whole lung, only a certain (probably small part of the lung)-the thing about fully saturated hb and deoxyhb is due to cooperative binding- that is why there are only two states Hb can be in. So at a certain point in the lung if there is less ventilation, so less o2 to ensure all Hb molecules are fully saturated
A V/Q mismatch can produce Hb with less than the normal amount of O2 molecules attached, but this has nothing to do with cooperative kinetics. I think you're mixing up multiple concepts. Shunted blood, eg, won't reach 100% sat because its not exposed to high O2 tension; not because of cooperative kinetics.
Hb has four possible states, since it can bind 4 O2 molecules. Hb passing through the lung will hit 100% sat, but a small amount remains at venous sat (75%), because theres a small shunt across the lungs even under normal circumstances. This Hb remains at lower sat because it isn't exposed to the gas exchanging areas of the lung. It has nothing to do with cooperative kinetics, which is the source of your confusion, and mine, since I assumed it was central to this discussion. This small shunt hitting the arterial circulation will then decrease the net saturation of the blood to be a bit under 100%.
If you were to pull a Hb molecule at random from arterial blood that has yet to supply tissue, the overwhelming odds are that it will have 4 O2 molecules attached. Passing through tissues, Hb molecules with 4 O2 attached will give up one O2 and hit 75%. The small fraction that was shunted across the lungs and was already at 75% before reaching tissues, won't give up any O2, since tissue tension isn't low enough for a second O2 to be lost. And if it were, all Hb molecules would be equally prone to hitting 50%.
The blood entering the lungs will be at 75%, and the cycle repeats.
Hb has four possible states, since it can bind 4 O2 molecules. Hb passing through the lung will hit 100% sat, but a small amount remains at venous sat (75%), because theres a small shunt across the lungs even under normal circumstances. This Hb remains at lower sat because it isn't exposed to the gas exchanging areas of the lung. It has nothing to do with cooperative kinetics, which is the source of your confusion, and mine, since I assumed it was central to this discussion. This small shunt hitting the arterial circulation will then decrease the net saturation of the blood to be a bit under 100%.
If you were to pull a Hb molecule at random from arterial blood that has yet to supply tissue, the overwhelming odds are that it will have 4 O2 molecules attached. Passing through tissues, Hb molecules with 4 O2 attached will give up one O2 and hit 75%. The small fraction that was shunted across the lungs and was already at 75% before reaching tissues, won't give up any O2, since tissue tension isn't low enough for a second O2 to be lost. And if it were, all Hb molecules would be equally prone to hitting 50%.
The blood entering the lungs will be at 75%, and the cycle repeats.
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It would be exceedingly rare for zero O2 to be sitting on any given Hb molecule. 50% saturation refers to the fraction of O2 binding sites occupied overall. And the curve will show what pO2 is needed to achieve 50% of site saturation.
Pulse oximetry can tell the degree to which Hb molecules are oxygen bound by using different wavelengths of light. If oximetry said 50% saturation, for instance, then it means, on average, 50% of sites are unoccupied. That isn't synonymous with 50% of Hb molecules have zero bound sites. When four O2 are bound, the Hb is in the R state, which allows for easier unloading; after it unloads, it transitions to the T form, which makes it harder to unload. Other circulating Hb in the R state have a greater proclivity to unload vs those circulating nearby now in the T state.
From UpToDate:
"The heme-heme interaction, a function of the tetrameric nature of hemoglobin, implies that the four heme groups do not undergo simultaneous oxygenation or deoxygenation; instead, the state of each individual heme with regard to associated oxygen influences the binding of oxygen to the other heme groups."
Pulse oximetry can tell the degree to which Hb molecules are oxygen bound by using different wavelengths of light. If oximetry said 50% saturation, for instance, then it means, on average, 50% of sites are unoccupied. That isn't synonymous with 50% of Hb molecules have zero bound sites. When four O2 are bound, the Hb is in the R state, which allows for easier unloading; after it unloads, it transitions to the T form, which makes it harder to unload. Other circulating Hb in the R state have a greater proclivity to unload vs those circulating nearby now in the T state.
From UpToDate:
"The heme-heme interaction, a function of the tetrameric nature of hemoglobin, implies that the four heme groups do not undergo simultaneous oxygenation or deoxygenation; instead, the state of each individual heme with regard to associated oxygen influences the binding of oxygen to the other heme groups."
Hey i also have read about this further and spoken to more people about it.
e.g. venous blood Hb is saturated at 75% and an article states it to be each Hb carry 3O2.
Most articles and books oversimplify this; this ignores the concept of cooperative binding and you couldnt explain for example for any saturation that is not 0, 25, 75, 100.
thus, 50% saturation refers to 50% 4O2 bound and 50% no oxygen bound to Hb
(ps pO2 is irrelevant here, we are just talking about the meaning of a % saturation)
e.g. venous blood Hb is saturated at 75% and an article states it to be each Hb carry 3O2.
Most articles and books oversimplify this; this ignores the concept of cooperative binding and you couldnt explain for example for any saturation that is not 0, 25, 75, 100.
thus, 50% saturation refers to 50% 4O2 bound and 50% no oxygen bound to Hb
(ps pO2 is irrelevant here, we are just talking about the meaning of a % saturation)
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On average, yes. I don't think anyone is in disagreement with that.
I disagree with this because according to that definition cyanosis cannot occur even at 25% Hb saturation as each Hb would have 1 O2 molecule-of course this is not true and the definition of cyanosis is inadequate oxygenation of blood, specifically referring to a high conc. of deoxyHb in capillaries (according to this costanzo definition-there is no deoxyHb hence no cyanosis which is false)
I think many physiology textbooks and articles oversimplify this concept
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You do realize that this is Step 1 forum AND saying Costanzo is engaging in falsification?
Please read what i said again, I never said Costanzo is engaging in falsification, i am saying cyanosis is referring to conc. of deoxyHb and that saying if 50% sat meant 2O2 on average per Hb, this would not make sense.
The thoughts about are my own thoughts about the meaning and this is a discussion for students, please do not take anything personally
