HB saturation increases at any PO2 in the presence of CO due to positive co-oper

[peroxidase]

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so what is this trying to say and this graph trying to show at the bottom of page 563 in the bible 2012?

it says hb saturation increases at any given PO2 in the presence of CO due to positive cooperativity.

So CO exerts a positive cooperativity effect allowing O2 to bind more readily? I dont really see this being shown by the Blue line in the graph though.

Any ideas? Thankyou

 

[Phloston]

so what is this trying to say and this graph trying to show at the bottom of page 563 in the bible 2012?

it says hb saturation increases at any given PO2 in the presence of CO due to positive cooperativity.

So CO exerts a positive cooperativity effect allowing O2 to bind more readily? I dont really see this being shown by the Blue line in the graph though.

Any ideas? Thankyou

Cooperativity and propensity for O2 to bind both increase as you shift along the curve up and to the right because of increased relaxed-, vs taut-, configuration. This is not the same as shifting the curve to the right, where the taut-form would become predominate.

CO induces an Hb-transition from the taut- to relaxed-configuration more strongly than O2 because of greater p4-electron overlap with iron's LUMO based on the second oxygen atom not being present to more readily draw e-'s away (thereby raising the HOMO of the available oxygen atom). This means a lesser pO2 is needed for subsequent Fe2+-O2-binding since the relaxed-form predominates with concurrent CO-binding.

 

[usmlesuccess]

phloston, I am sure that whatever you wrote is right, but there is a muuuuuuuuuuch simpler way of explaining it I think, lol 🙂 Since CO has a much higher affinity it functions to take away the binding sites that were originally going to be occupied by oxygen. Thus "Vmax", in a sense has decreased,even though you can argue that 100% oxygen will help resolve that.


--->Since the "Vmax" has decreased the saturation at any given o2 has increased!!!!

 

[Phloston]

phloston, I am sure that whatever you wrote is right, but there is a muuuuuuuuuuch simpler way of explaining it I think, lol 🙂 Since CO has a much higher affinity it functions to take away the binding sites that were originally going to be occupied by oxygen. Thus "Vmax", in a sense has decreased,even though you can argue that 100% oxygen will help resolve that.


--->Since the "Vmax" has decreased the saturation at any given o2 has increased!!!!

Careful. CO is considered a competitive inhibitor of haemoglobin. Therefore it increases Km and leaves Vmax unchanged.

The only real exception here is that while most competitive inhibitors are reversible (and CO technically is, which is why you give 100% oxygen to "outcompete" it), I've seen a practice question where it was considered a "near-irreversible" competitive inhibitor. Non-competitive inhibitors (e.g. phenoxybenzamine) are purely irreversible and decrease Vmax.

 

[usmlesuccess]

Right phloston, which is why I put Vmax in "quotation marks". If you look at the graph that the poster was referring to in FA, you will see that they showed vmax decreased.

 

[Phloston]

Right phloston, which is why I put Vmax in "quotation marks". If you look at the graph that the poster was referring to in FA, you will see that they showed vmax decreased.

The graph doesn't depict decreased Vmax. The y-axis isn't fully representative of the reaction rate. It's merely just showing that at atmospheric O2, CO isn't overcome readily. I see what you're saying though as far as how the graph appears in FA. Just be aware that this isn't a strict rxn rate vs [substrate] curve because Vmax is achieved as [substrate] increases further along the x-axis (not depicted).

 

[peroxidase]

thank you both very much!

 

[docord]

Deleted in 2013.