The Bohr effect - Human biology

Right, I require your services... if you'd be so kind to help

The Bohr effect is all to do with the ability of haemoglobin (Hb) to bind with oxygen, and the effect of O2 concentration and CO2 concentration on the affinity of Hb to associate or dissociate with O2. (correct me if I'm wrong )

The bit I'm stuck on is what prevents the O2 from binding to the Hb in acidic conditions - i.e. an environment where CO2 concentration is high, such as the muscles in respiration. I understand that as CO2 builds up in muscles, some of it diffuses into the blood and red blood cells (RBCs). RBCs have an enzyme called carbonic anhydrase which catalysis the reaction between CO2 and water to form carbonic acid. The carbonic acid the dissociates into H+ and hydrogencarbonate ions. Haemoglobinic acid then forms when hydrogen ions bind to Hb. It is then the hydrogen ions that prevent O2 from binding to the Hb - or so I think.

I've also been told that what prevents O2 from binding to Hb is the fact that a low pH (acidosis) breaks the ionic and disulphide bonds holding the tertiary and quaternary structure of the Hb together, so changing the shape of the Hb. It is that change in shape, which prevents the O2 from binding (as I’ve been told)

So which one prevents the binding of O2 to Hb; the presence of H ions on the Hb (and therefore increased blood acidity) or the change in structural shape of the Hb? I’m stuck!

I was hoping someone might know a little bit about this stuff? Oh the joys of human biology! By the way the exam board is OCR.

Reply 1

Revenged

15

H+ binding to oxygenated haemoglobin causes a change of shape of the protein (e.g. by breaking hydrogen bonds) so that oxygen no longer binds...

This is why the protein has less oxygen bound when there is high carbon dioxide concentration...

[I don't know if you do it at A-level but it is sort of similar to a 'non-competitive antagonism'... This is where binding of antagonist (not at active site) causes a reversible conformational change of the active site of a protein so no agonist can bind]

Reply 2

HenrySJA

OP

Revenged

H+ binding to oxygenated haemoglobin causes a change of shape of the protein (e.g. by breaking hydrogen bonds) so that oxygen no longer binds...

This is why the protein has less oxygen bound when there is high carbon dioxide concentration...

[I don't know if you do it at A-level but it is sort of similar to a 'non-competitive antagonism'... This is where binding of antagonist (not at active site) causes a reversible conformational change of the active site of a protein so no agonist can bind]

Thanks, that seems to all make sense. Our textbooks from OCR are, in places, badly written or are just plan wrong. For example the book says "The H readily combines with the Hb molecules inside RBCs (after dissociation from carbonic acid). This forms a compound called haemoglobinic acid. However, an Hb molecule that has done this cannot also bind with O2. The H ions make it drop its O2"

Is it just me, or does this sound badly written to anyone. There is absolutly no mention of H+ causing a change in shape of the Hb and also, why does the H+ just make the Hb drop its O2. It seems all too dumbed down to me!

Reply 3

bright star

1

if you do OCR bio, this is way more detail than you need. but i kno what you mean, it'd almost be easier if they stepped the level up a bit so that it at least makes logical sense. the "dropping O2" is to do with the structural change as a result of pH change. (do you understand that pH is just about concentration of H+ ions?) so the book does say the same thing, just ina stupid way!!

high CO2 levels cause the pH to fall (high conc of H+ ions). these bind to Hb, changing it's structure so it's no longer capable of binding to O2. the shape change (in simple terms) is due to the H+ binding in one place, twisting th molecule into a different shape so that the binding site for O2 is no longer specific enough. is this what you mean?

Reply 4

Wangers

16

Revenged

[I don't know if you do it at A-level but it is sort of similar to a 'non-competitive antagonism'... This is where binding of antagonist (not at active site) causes a reversible conformational change of the active site of a protein so no agonist can bind]

I haven't touched this for months, if memory serves at AS this is Non conpetitive inhibition at the allosteric site.

Reply 5

Wangers

16

HenrySJA

Thanks, that seems to all make sense. Our textbooks from OCR are, in places, badly written or are just plan wrong. For example the book says "The H readily combines with the Hb molecules inside RBCs (after dissociation from carbonic acid). This forms a compound called haemoglobinic acid. However, an Hb molecule that has done this cannot also bind with O2. The H ions make it drop its O2"

Is it just me, or does this sound badly written to anyone. There is absolutly no mention of H+ causing a change in shape of the Hb and also, why does the H+ just make the Hb drop its O2. It seems all too dumbed down to me!

This is not wrong per se however it is badly explained and personally without expanding on other factors they should have left this alone.

I've only read into this - so you might want to get someone else to check this, I'll try and take you through it step by step:

Before reading on make sure you are happy with the concept of Bohr shift and what happens with Fetal heamoglobin before you read on.

A working understanding of Le Chatelier's Principle is assumed from now on:

--------------------------------------------------------------------------

Consider a person in normal health undergoing strenuous excercise. Naturally the body compensates by elavating tidal flow and stroke volume through the sympathetic system. Bohr shift occurs as you have studied.

The important thing to note here is that the following equilibrium: (let this be EQn 1)

CO2 + H2O -> H2CO3 <-> H+ + HCO3-

occurs in plasma and RBS, as mediated by Carbonic Anhydrase, as you noted earlier. However the allosteric modification of the Hb molecule is somewhat less significant then you are led to believe at AS.

Before I go on some notation:

Hb - heamoglobin
Hb-02 - Oxyheamoglobin
Hb-(O2)n where n is 2-4
HbCo - carboxyheamoglobin

Note here that the hydrogen ions can bind to heamoglobin provided the heamoglobin isnt saturated- you may now be able to guess where this is headed...:

Now the Bohr shift can also be examined as a chemical equilibrium thus:

Eg - starting with 4 oxygens: Let this be EQn 2

Hb(O2)4 + H+ <---> Hb(O2)3 + O2

And so on...the last two oxygens usually stay on - because recall there are 2 alpha and 2 beta chains on a Hb molecule and hence they have different affinities for O2. The last 2 are sometimes known as a Oxygen reserve.

Now as you can see the high [H+] means the equilibrium position strongly lies to the right as the Hb(O2)3 binds with the H+ to form haemoglobinic acid [Hb(O3)3-H+] (incase you're wondering where the H+ "disappears off to", in the process liberating the oxygen molecule for use by needy cells.

Note that strictly the above notation is incorrect, however at A level im sure it will suffice.
-----------------------------------------------------------------------

Now although alot of the carbon dioxide exists as bicarbonate for homeostatic reasons, most of the remainder is bound to Hb - because the solubility of Co2 in blood is not high. Hence can you see how this allows transport of CO2 back into the lung enviroment?

And yes at the lungs a similar equilibrium exists where the CO2 is liberated and thus due to the low [H+] the heamoglobin is now free to associate with O2. thus establishing the ventilation cycle.

Do you see how living systems are stunningly elegant? If you want to read more the above process is called the Haldane Effect.

Enjoy biology, and try not to get too frustrated :smile:

Reply 6

HenrySJA

OP

bright star

if you do OCR bio, this is way more detail than you need. but i kno what you mean, it'd almost be easier if they stepped the level up a bit so that it at least makes logical sense. the "dropping O2" is to do with the structural change as a result of pH change. (do you understand that pH is just about concentration of H+ ions?) so the book does say the same thing, just ina stupid way!!

high CO2 levels cause the pH to fall (high conc of H+ ions). these bind to Hb, changing it's structure so it's no longer capable of binding to O2. the shape change (in simple terms) is due to the H+ binding in one place, twisting th molecule into a different shape so that the binding site for O2 is no longer specific enough. is this what you mean?

Yes, they could do with steping it up a bit. I see no point in the board just giving us stupid, oversimplified explanations for certain topics in human biology especially when you have something like the Bohr effect to try and get to grips with. Thanks for your explanation, it has made it a bit more clear - see, a little more detail can go a long way! :smile:

Reply 7

bright star

1

awww you're good!!! i couldn't be bothered..... just remember that there's no way you need ANYWHERE NEAR this kinda level for A level so if you don't understand, don't fret......

EDIT: i see that you do human bio, i jus did regular OCR bio, so i don't kno what specifically is on your sylabus actually.....

Reply 8

Wangers

16

You can now see why most books at AS gloss over this as conformation change (although it is true to an extent) The actual curve is affected by things like DPG,TEMP etc. These are superflous to need at A level....

Spoiler

Reply 9

Wangers

16

HenrySJA

Yes, they could do with steping it up a bit. I see no point in the board just giving us stupid, oversimplified explanations for certain topics in human biology especially when you have something like the Bohr effect to try and get to grips with. Thanks for your explanation, it has made it a bit more clear - see, a little more detail can go a long way! :smile:

Remember the A level has to be doable across the whole country and spectrum of interest/ability. to be confortable with most of the deeper explainations demands subject specific graduates, this is not always possible.

Reply 10

HenrySJA

OP

Wangers, bright star and everyone else; many thanks for your input on this. I supppose I can understand why they can't go into a load of detail otherwise they couldn't fit the A2 content into 1 year. But as I've said before, a little more detail about why such events occur in the body can actually help to get across the basics and I suppopse in a (twisted) way, ensuring the book is a bit more "basic" encourages self-directed learning in students... well it has for me anyway!

Again, thanks for the help! :smile:

Reply 11

bright star

1

you mean it's encouraged you to ask us :wink:

to be fair, all i did was rewrite what you were told in an actual sentence.....

Reply 12

HenrySJA

OP

bright star

you mean it's encouraged you to ask us :wink:

to be fair, all i did was rewrite what you were told in an actual sentence.....

As opposed to me not understanding what I've been told / read in the textbook, and then being fobbed off by my lecturer when I ask them to explain what I've read / been told... we are therefore not encouraged to find out things for ourselves and delve into a fraction more detail to gain an understanding of the topic!

Reply 13

Toiletpaper8

13

I believe that it's the H+ ions binding to the delta- within the protein itself and repels the delta+ causing the shape of the protein complex to alter.

Not sure though :p:

Reply 14

Wangers

16

bright star

you mean it's encouraged you to ask us :wink:

to be fair, all i did was rewrite what you were told in an actual sentence.....

I must admit that was also revision - I need to keep the wheels turning....besides the carnage a question that actually requires understanding would bring...remember you can make it - learn,markscheme,regutgitate. You understand as much or as little as you like.

Reply 15

Wangers

16

Toiletpaper8

I believe that it's the H+ ions binding to the delta- within the protein itself and repels the delta+ causing the shape of the protein complex to alter.

Not sure though :p:

(pedant mode)

What Delta -? When referring to a complex olecule please be more specific, do you mean to Alpha chain, beta chain or central heam?

I'm not trying to have a go, but it makes things easier to understand at higher levels.

Reply 16

GregoryJL

Forgive me, Wangers, but whilst I'm sure that there are differing affinities for the two chains, I was under the impression there was very strong cooperativity between the subunits, and so individual haemoglobin proteins would tend to be entirely saturated and unsaturated - and not half saturated with the two chains with the higher affinity keeping their oxygen. Could I trouble you to point me where you found that out? Wiki doesn't mention it, and I recall a number of reasonable textbooks which didn't mention anything about this oxygen reserve, and indeed pretty explicitly noted that the 'middling' steps between complete saturation and desaturation were short.

But as I've been a medic for a grand total of a week, you're probably right. :wink:

Reply 17

Wangers

16

Unfortuneatly the last time I looked at this was over a year ago in my L6 after a friend was asked about it at interview. But:

From what I remember the middleing steps are not that discreate, I wrote that equilibrium position without considering the effects of DPG,TEMP et al, these wouldof course affect the equilibrium position. Of course since the equilibriums between saturation and unpaired heamoflobin are dependant on so many factors and dynamic I dont think molecules specificlly stay in various states, more likely that it more unlikely to find Hb completly unpaired. (if that makes sense). I think in this case groups are considered rather then individual molecules - I'm sorry if my explaination implied that degrees of affinity would definitively isolate individual molecules to certain saturation states.

I do admit I am rather lacking in proof - my the scientific method this should be burned! I think what I remember as a "reserve" scientificlly strictly does not exist - mearly in most enviroments undergoing normal respiration the usual unloading is only about 50%

I refer you to this PDF, which is more knowledgeable then myself.

http://www.heckgrammar.kirklees.sch.uk/index.php?p=10312

Its the module 3 - page 15.

I do stress that this itself is not scientificlly rigourous (strictly speaking). I apologise if my speil has garbled empirical truths through my own misunderstandings, and do concede your point entirely :smile:

The Bohr effect - Human biology

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