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Cooperativity of haemoglobin and oxygen dissociation curve of haemoglobin
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Noor Muhammad Khan
Cooperativity of haemoglobin and oxygen dissociation curve of haemoglobin
I agree with everything in the answer from Bryan Krause but, in case it helps, here is the explanation that I came up with.
One way to think about this is in terms of the familiar MWC two-state model, which is essentially what you have used to frame your question.
If the concentration of oxygen is high enough for 50% saturation then, to a first approximation, most of the haemoglobin molecules have bound some oxygen and in terms of the whole population of molecules they are mostly in the relaxed, high affinity state in which any unoccupied binding sites will have the dissociation constant that is characteristic of that state. So all you are seeing in the right half of the curve is the usual hyperbolic binding curve for the relaxed form. For any saturation curve this is the behaviour that falls out of the mathematical description of the binding.
In other words, in terms of your word 'easy' it is always less easy to bind the next increment of ligand as you approach saturation because the concentration of occupied binding sites has increased so the rate of dissociation has also increased.
I agree with everything in the answer from Bryan Krause but, in case it helps, here is the explanation that I came up with.
One way to think about this is in terms of the familiar MWC two-state model, which is essentially what you have used to frame your question.
If the concentration of oxygen is high enough for 50% saturation then, to a first approximation, most of the haemoglobin molecules have bound some oxygen and in terms of the whole population of molecules they are mostly in the relaxed, high affinity state in which any unoccupied binding sites will have the dissociation constant that is characteristic of that state. So all you are seeing in the right half of the curve is the usual hyperbolic binding curve for the relaxed form. For any saturation curve this is the behaviour that falls out of the mathematical description of the binding.
In other words, in terms of your word 'easy' it is always less easy to bind the next increment of ligand as you approach saturation because the concentration of occupied binding sites has increased so the rate of dissociation has also increased.
As you may notice from the graph you provided, there actually is no such thing as true 100% saturation: there is always an asymptotic approach to 100%, regardless of cooperativity.
Let's think, though, of what is actually plotted in that graph: $p_{O_2}$ effectively oxygen concentration, versus number of total percentage of binding sites that have an $O_2$ bound.
However, this also means that as you go up on the y-axis, the number of sites available to bind also goes down. So even if those sites have a high affinity for oxygen, there aren't as many of them open!
That's one big problem with the type of plot you show: it doesn't highlight the cooperativity very well. Therefore, people sometimes plot this information in a "Hill plot", with an algebraic shuffling and log scaling like this:
...where $\theta$ represents the fraction of receptors bound to ligand, and $L$ is the (unbound) ligand concentration. $n$ is the Hill Coefficient, i.e., the cooperativity.
When plotted this way, you can easily see the effect of the cooperativity of hemoglobin versus myoglobin (which has cooperativity of 1, because it has a single binding site):
As you may notice from the graph you provided, there actually is no such thing as true 100% saturation: there is always an asymptotic approach to 100%, regardless of cooperativity.
Let's think, though, of what is actually plotted in that graph: $p_{O_2}$ effectively oxygen concentration, versus number of total percentage of binding sites that have an $O_2$ bound.
However, this also means that as you go up on the y-axis, the number of sites available to bind also goes down. So even if those sites have a high affinity for oxygen, there aren't as many of them open!
That's one big problem with the type of plot you show: it doesn't highlight the cooperativity very well. Therefore, people sometimes plot this information in a "Hill plot", with an algebraic shuffling and log scaling like this:
...where $\theta$ represents the fraction of receptors bound to ligand, and $L$ is the (unbound) ligand concentration. $n$ is the Hill Coefficient, i.e., the cooperativity.
When plotted this way, you can easily see the effect of the cooperativity of hemoglobin versus myoglobin (which has cooperativity of 1, because it has a single binding site):
I agree with everything in the answer from Bryan Krause but, in case it helps, here is the explanation that I came up with.
One way to think about this is in terms of the familiar MWC two-state model, which is essentially what you have used to frame your question.
If the concentration of oxygen is high enough for 50% saturation then, to a first approximation, most of the haemoglobin molecules have bound some oxygen and in terms of the whole population of molecules they are mostly in the relaxed, high affinity state in which any unoccupied binding sites will have the dissociation constant that is characteristic of that state. So all you are seeing in the right half of the curve is the usual hyperbolic binding curve for the relaxed form. For any saturation curve this is the behaviour that falls out of the mathematical description of the binding.
In other words, in terms of your word 'easy' it is always less easy to bind the next increment of ligand as you approach saturation because the concentration of occupied binding sites has increased so the rate of dissociation has also increased.
I agree with everything in the answer from Bryan Krause but, in case it helps, here is the explanation that I came up with.
One way to think about this is in terms of the familiar MWC two-state model, which is essentially what you have used to frame your question.
If the concentration of oxygen is high enough for 50% saturation then, to a first approximation, most of the haemoglobin molecules have bound some oxygen and in terms of the whole population of molecules they are mostly in the relaxed, high affinity state in which any unoccupied binding sites will have the dissociation constant that is characteristic of that state. So all you are seeing in the right half of the curve is the usual hyperbolic binding curve for the relaxed form. For any saturation curve this is the behaviour that falls out of the mathematical description of the binding.
In other words, in terms of your word 'easy' it is always less easy to bind the next increment of ligand as you approach saturation because the concentration of occupied binding sites has increased so the rate of dissociation has also increased.
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As you may notice from the graph you provided, there actually is no such thing as true 100% saturation: there is always an asymptotic approach to 100%, regardless of cooperativity.
Let's think, though, of what is actually plotted in that graph: $p_{O_2}$ effectively oxygen concentration, versus number of total percentage of binding sites that have an $O_2$ bound.
However, this also means that as you go up on the y-axis, the number of sites available to bind also goes down. So even if those sites have a high affinity for oxygen, there aren't as many of them open!
That's one big problem with the type of plot you show: it doesn't highlight the cooperativity very well. Therefore, people sometimes plot this information in a "Hill plot", with an algebraic shuffling and log scaling like this:
See Wikipedia
...where $\theta$ represents the fraction of receptors bound to ligand, and $L$ is the (unbound) ligand concentration. $n$ is the Hill Coefficient, i.e., the cooperativity.
When plotted this way, you can easily see the effect of the cooperativity of hemoglobin versus myoglobin (which has cooperativity of 1, because it has a single binding site):
http://cbc.arizona.edu/classes/bioc462/462a/NOTES/hemoglobin/hemoglobin_function.htm via Fig. 7-13, Nelson & Cox Principles of Biochemistry, 3rd ed., 2000
As you may notice from the graph you provided, there actually is no such thing as true 100% saturation: there is always an asymptotic approach to 100%, regardless of cooperativity.
Let's think, though, of what is actually plotted in that graph: $p_{O_2}$ effectively oxygen concentration, versus number of total percentage of binding sites that have an $O_2$ bound.
However, this also means that as you go up on the y-axis, the number of sites available to bind also goes down. So even if those sites have a high affinity for oxygen, there aren't as many of them open!
That's one big problem with the type of plot you show: it doesn't highlight the cooperativity very well. Therefore, people sometimes plot this information in a "Hill plot", with an algebraic shuffling and log scaling like this:
See Wikipedia
...where $\theta$ represents the fraction of receptors bound to ligand, and $L$ is the (unbound) ligand concentration. $n$ is the Hill Coefficient, i.e., the cooperativity.
When plotted this way, you can easily see the effect of the cooperativity of hemoglobin versus myoglobin (which has cooperativity of 1, because it has a single binding site):
http://cbc.arizona.edu/classes/bioc462/462a/NOTES/hemoglobin/hemoglobin_function.htm via Fig. 7-13, Nelson & Cox Principles of Biochemistry, 3rd ed., 2000
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