Oxygen Dissociation Curve - AS
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Oxygen Dissociation Curve - ASHi,
Im a little bit confuzzled about the oxygen dissociation curve and why it is sigmoid.So I realise WHY it is physically and all that malarky; because the first oxygen molecule is hard to attach, 2nd and 3rd easy etc, but how is this an advantage in oxygen transport, anyone know?
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If you decrease the partial pressure of O2 by a small amount (just above when it is steep), then you will notice there is a large drop in the saturation of Hb with O2. So a small decrease in pO2 leads to a lot of O2 becoming dissociated with Hb and available for respiration.
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Basically, the oxygen dissociation curve means that where the oxygen is in a very high concentration, for example in the lungs, haemoglobin has a very high affinity for oxygen and so it has a very high saturation of oxygen. This allows as much oxygen as possible to be taken up into the haemoglobin in the red blood cells from the alveoli in the lungs.
Where the oxygen concentration is low, for example in the muscles, haemoglobin has a low affinity of oxygen. This allows oxygen to be relased at the muscles where it is needed for aerobic respiration.
Therefore the shape of the oxygen dissocation curve allows efficient transportation of oxygen.
Hope that explains things!
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That's a much better explanation, nice one(Original post by Revenged)
Basically, the oxygen dissociation curve means that where the oxygen is in a very high concentration, for example in the lungs, haemoglobin has a very high affinity for oxygen and so it has a very high saturation of oxygen. This allows as much oxygen as possible to be taken up into the haemoglobin in the red blood cells from the alveoli in the lungs.
Where the oxygen concentration is low, for example in the muscles, haemoglobin has a low affinity of oxygen. This allows oxygen to be relased at the muscles where it is needed for aerobic respiration.
Therefore the shape of the oxygen dissocation curve allows efficient transportation of oxygen.
Hope that explains things!
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(Original post by endeavour)
This part of biology becomes awesome after you do buffers and pH in chemistry
MMM...if you say so...much prefer health and disease myself...much easier...
Anyway thanks for that you are all stars of the highest variety.
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So...let me clarify.. the advantage of the sigmoid curve is that its easy to get oxygen when the partial pressure is high (lungs) and easy to lose when the partial pressure is low (Respiring tissues) BUT inbetween its just normal? -
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Yeah,(Original post by onlylittleme)
So...let me clarify.. the advantage of the sigmoid curve is that its easy to get oxygen when the partial pressure is high (lungs) and easy to lose when the partial pressure is low (Respiring tissues) BUT inbetween its just normal?
inbetween the curve is steep... therefore a small increase in the amount of oxygen around would greatly increase haemoglobin affinity of oxygen -
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Is it to do with the equilibria:
H-Hb + O2(g) <--> HbO2- + H3O+(aq)
and
CO2(g) + 2H2O(l) <--> HCO3-(aq) + H3O+(aq)
...?
So protonated Hb is present only when there is a high pp of CO2 and a lower p of O2, due to the forward reaction in that second equation. High [CO2] pushes Eqb of eqn 2 to the right, so higher [H3O+] is apparant. This then pushes the Eqb of the 1st eqn to the left, so the protonated Hb is more dissociated???
I'd be chuffed if I'm anywhere near being right - this isnt on my syllabus! -
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No - the first eqation is actually wrong as this is not how O2 binds. In fact it just acts as a ligand on the Fe2+ ion found in the Haem group, so no H+ is lost when O2 binds.(Original post by corkskrew)
Is it to do with the equilibria:
H-Hb + O2(g) <--> HbO2- + H3O+(aq)
and
CO2(g) + 2H2O(l) <--> HCO3-(aq) + H3O+(aq)
...?
So protonated Hb is present only when there is a high pp of CO2 and a lower p of O2, due to the forward reaction in that second equation. High [CO2] pushes Eqb of eqn 2 to the right, so higher [H3O+] is apparant. This then pushes the Eqb of the 1st eqn to the left, so the protonated Hb is more dissociated???
I'd be chuffed if I'm anywhere near being right - this isnt on my syllabus!
What happens is that H+ can protonate a histidine amino acid in part of the protein. This gives it a postive charge and enables it to Hydrogen bond with an aspartate amino acid in one of the other subunits. This favours a form of the protein in which the O2 binding site is blocked (T Form) so lowers the affinity for Oxygen. -
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I'm sure your right but you really are complicating things.(Original post by oxymoron)
No - the first eqation is actually wrong as this is not how O2 binds. In fact it just acts as a ligand on the Fe2+ ion found in the Haem group, so no H+ is lost when O2 binds.
What happens is that H+ can protonate a histidine amino acid in part of the protein. This gives it a postive charge and enables it to Hydrogen bond with an aspartate amino acid in one of the other subunits. This favours a form of the protein in which the O2 binding site is blocked (T Form) so lowers the affinity for Oxygen.
All you need to know for edexcel bio is that in a high concentration of CO2 causes the Bohr effect.
This confused me at AS because I wasn't really wasn't told the reason behind it - all I was told was that it causes the oxygen dissociation graph to shift to the left.
This is caused by CO2 dissolving in water in the blood plasma, forming carbonic acid (H2CO3-). This carbonic acid then dissociates into hydrocatbonate ions (HCO3-) and H+.
CO2(g) + 2H2O(l) <--> H2C03- <--> HCO3-(aq) + H3O+(aq)
That's what the reaction above shows. Don't worry about it being a revisible reaction you only need to consiser it going forwards. H30+ is just H+.
This links in nicely to what you learn since the HCO3- ions are the main form that CO2 is transported in the body. Also, the H+ ions are taken up by the haemoglobin, causing haemoglobin to realease oxygen. This is the Bohr shift.
To help clarify a few things. A buffer is something that resists changes in pH (concentration of H+ ions). Haemoglobin acts as an acidic buffer since it resists increases in the H+ concentration of the blood by taking up H+ ions.
The importance of the Bohr shift is that in areas of high carbon dioxide concentraion, for example in respiring muscles, it causes more oxygen to be released by the method above. -
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CO2(g) + 2H2O(l) <--> H2C03- <--> HCO3-(aq) + H3O+(aq)(Original post by corkskrew)
....Le Chatelier's Principle......???????........
Yeah, i never thought of that myself.
If the Haemoglobin takes up the H+ (H30+) ions, the reaction will try to oppose the increasing in the forward reaction by causing equilibrium to shift to the right. Therefore causing more H+ and HC03- to be in solution.
Therfore, more CO2 to be to be will be carried in the blood plasma in the form of HC03-.
So by haemoglobin acting as a buffer, it not only causes more oxygen to be released at areas of high areas of high CO2 concentration it also causes more CO2 to be carried in the blood as well. -
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(Original post by oxymoron)
No - the first eqation is actually wrong as this is not how O2 binds. In fact it just acts as a ligand on the Fe2+ ion found in the Haem group, so no H+ is lost when O2 binds.
What happens is that H+ can protonate a histidine amino acid in part of the protein. This gives it a postive charge and enables it to Hydrogen bond with an aspartate amino acid in one of the other subunits. This favours a form of the protein in which the O2 binding site is blocked (T Form) so lowers the affinity for Oxygen.
Don't give me chance to answer then 
And Revenged, yes this is more than A level, degree level biochemistry and medicine to be precise. Interesting nonetheless
I hate oxygen. I'm going to have to understand sometime this week..
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