In bohr effect if H binds to Hb-forms T state,then how affinities is d

In bohr effect if H binds to Hb-forms T state,then how affinities is decided with O2 and offloading at tissue,can’t understand.

Reading File

Finding Sources

Reading File

Loading Image

I have everything needed for a complete, well-illustrated answer. Here it is:

The Bohr Effect: How H⁺ Drives O₂ Offloading at Tissues

Your confusion is totally understandable - this concept involves a circular-seeming logic. Let me break it down step by step.

Step 1: What happens at the tissues (where O₂ needs to be delivered)

Metabolically active tissues produce CO₂ as a waste product. When this CO₂ enters the RBC:

CO₂ + H₂O → H₂CO₃ → HCO₃⁻ + H⁺

(catalyzed by carbonic anhydrase inside RBCs)

This generates free H⁺, lowering pH inside the RBC. This is the key trigger.

CO₂ entering RBC generates H⁺, which displaces O₂ from HbO₂

Step 2: Why does H⁺ cause O₂ to be released? (The molecular mechanism)

The key insight from Lippincott's Biochemistry:

Deoxyhemoglobin has a GREATER affinity for H⁺ than does oxyhemoglobin.

Why? Because specific histidine residues (particularly His 146 on the beta-chain) have a higher pKa in deoxy-Hb (T state) than in oxy-Hb (R state). When H⁺ concentration rises at the tissues:

H⁺ binds to these histidine residues
This protonation creates ionic bonds / salt bridges between subunits
Those salt bridges lock Hb into the T state (taut, low-affinity conformation)
In the T state, Hb has LOW affinity for O₂
So O₂ is released to the tissues

The reaction is written as:

HbO₂ + H⁺ ⇌ Hb-H⁺ + O₂

Shift to the right happens at tissues (high H⁺, low pO₂).

Step 3: The "chicken and egg" logic - resolved

Your real question seems to be: "Does H⁺ binding cause the T state, or does T state just happen to bind H⁺?"

Answer: It's a two-way allosteric relationship.

Condition	What shifts	Result
High H⁺ (tissues, low pH)	Favors T state	Low O₂ affinity → O₂ released
High pO₂ (lungs)	Favors R state	Low H⁺ affinity → H⁺ released
Low pO₂ (tissues)	Favors T state	Low O₂ affinity → O₂ released

At tissues: high CO₂ → high H⁺ → H⁺ binds Hb → T state stabilized → O₂ affinity drops → O₂ is offloaded to tissues
At lungs: high pO₂ → O₂ binds Hb → R state stabilized → H⁺ affinity drops → H⁺ released → pH rises

So H⁺ binding is not a consequence of being in the T state; it is an allosteric effector that actively drives the R→T transition, reducing O₂ affinity.

Step 4: The dissociation curve shifts RIGHT

Right shift of O₂-Hb dissociation curve at lower pH - Lippincott Biochemistry

At pH 7.2 (tissues), the curve shifts right. This means:

At the same pO₂ of ~40 mmHg found in tissues, Hb is less saturated with O₂ than it would be at pH 7.4
More O₂ is unloaded to the tissues - exactly what active metabolism needs

As [Medical Physiology] quantifies it: the binding of approximately 0.7 moles of H⁺ causes the release of 1 mole of O₂ under physiological conditions.

Summary in one sentence

H⁺ (produced by tissue CO₂ metabolism) binds to histidine residues on Hb, stabilizes the T state via salt bridges, which has inherently low O₂ affinity, so O₂ is released to where it is needed most - the metabolically active tissues.

The T state does not cause O₂ offloading by coincidence - H⁺ actively shifts Hb from R (high affinity) to T (low affinity), making the whole system self-regulating: the harder a tissue works, the more CO₂/H⁺ it makes, the more O₂ it gets.

Sources: Lippincott Illustrated Reviews: Biochemistry 8th ed, pp. 104-106; Basic Medical Biochemistry 6th ed, pp. 201-203; Medical Physiology (Boron & Boulpaep), p. 956

Tell me if i am right-COHb-more affinity-means-left shift -more R state?

Finding Sources

Reading File

Loading Image

You are mostly right - but the "more R state" part needs a careful correction. Here's the full picture:

Your logic: COHb → more affinity → left shift → more R state

Your step	Correct?	Verdict
COHb has more affinity for O₂	✅ Yes	CO binds Hb ~250x more tightly than O₂
This causes a LEFT shift	✅ Yes	P50 decreases (O₂ held more tightly)
The remaining heme sites are in "more R state"	✅ Partially correct - with a nuance	See below

Where you are correct

CO binds one or more heme sites on Hb. When it does, it locks the whole Hb tetramer into a conformation resembling the R state (high-affinity). This allosterically makes the remaining free heme sites bind O₂ with higher affinity too - hence the left shift of the curve.

The textbook says it plainly ([Costanzo Physiology, 7th ed]):

"Those heme groups not bound to CO have an increased affinity for O₂. Thus P50 is decreased, making it more difficult for O₂ to be unloaded in the tissues."

So yes - the free heme sites are functionally in an R-state-like high-affinity configuration.

The important nuance you must add

CO poisoning is doubly dangerous because of TWO simultaneous problems:

CO poisoning curve - saturation capped and left-shifted - Costanzo Physiology

CO occupies heme sites → fewer sites available for O₂ → O₂ carrying capacity reduced (the curve's ceiling is lower, e.g. 50% max in the graph above)
CO locks remaining sites in R-state-like high affinity → left shift → the O₂ that IS carried refuses to be released at tissues

So your chain is right, but it's not "more R state" in the literal sense of "all 4 subunits snap into R state." It is more accurate to say: CO allosterically increases O₂ affinity of the remaining heme groups, mimicking/promoting the R-state conformation. The Bohr effect (T-state promotion by H⁺) is also blunted, making tissue delivery even worse.

Summary

COHb → high affinity → left shift → remaining hemes behave like R state → O₂ not released at tissues

✅ Your reasoning is correct. Just add: this is on top of CO physically blocking heme sites, making the net effect catastrophic - less O₂ loaded AND less O₂ offloaded.

Source: Costanzo Physiology 7th ed, p. 228; Harrison's Principles of Internal Medicine 22E

This is a shared conversation. Sign in to Orris to start your own chat.