Myoglobin & Haemoglobin — Q-Bank
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Unit 4 Question Bank

Haem · cooperativity · T and R states · Bohr effect · BPG · haemoglobinopathies
25 MCQ · five options8 Definitions3 Written answersHarper's verified
Format note: the TMU Biochemistry paper gives five suggested answers (A–E), not four — these MCQs match that. Items tagged TMU 2019 or TMU 2020/21 come from the real papers. Answers are verified against Harper's Illustrated Biochemistry; the "marking schemes" in the source folder are other students' answer sheets, not official, so they are never used as the authority.
0 / 25 answered
1Which statement is CORRECT about myoglobin and haemoglobin?
A. The oxygen dissociation curve of myoglobin is sigmoidal whereas that of haemoglobin is hyperbolic
B. They both have quaternary structure
C. Haemoglobin has a higher affinity for oxygen than does myoglobin
D. They both take haem as a prosthetic group
E. Both exhibit the Bohr effect
Answer: D
Both contain haem as their prosthetic group. The others are all reversed or false: myoglobin's curve is hyperbolic and haemoglobin's sigmoid; only haemoglobin has quaternary structure; myoglobin has the higher affinity (which is exactly why it is a store and not a transporter); and myoglobin cannot show the Bohr effect because it is monomeric. This exact question appears on the TMU deck.TMU Lecture 4 Slide 24 · Harper's ch.6, pp.52–55
2Which statement about haem structure is NOT true?
A. The haem molecule is a cyclic tetrapyrrole consisting of four molecules of pyrrole
B. One atom of ferrous iron resides at the centre of the planar tetrapyrrole
C. Binding of the first O₂ to deoxyhaemoglobin shifts the haem iron towards the plane of the ring
D. The iron forms a total of six coordination bonds
E. The iron atom is coplanar with the tetrapyrrole ring in deoxymyoglobin
Answer: E
In deoxymyoglobin the iron lies about 0.03 nm OUTSIDE the plane, pulled towards His F8, so the ring puckers slightly. It moves into the plane only on oxygenation. Reversing the two states is the standard trap. This question closes the TMU deck.Harper's ch.6, pp.52–53 · TMU Lecture 4 Slide 25
3The fifth coordination position of the haem iron in myoglobin is occupied by ( ).
A. a nitrogen of the imidazole ring of histidine F8
B. a nitrogen of the imidazole ring of histidine E7
C. a molecule of molecular oxygen, bound end-on
D. a propionate side chain of the haem ring
E. a water molecule from the surrounding solvent
Answer: A
His F8, the proximal histidine, is bonded to the iron and anchors the haem to the protein. The sixth position is reserved for oxygen or left unoccupied. His E7, the distal histidine, lies on the opposite side of the ring and is NOT bonded to the iron — it guards the binding site.Harper's ch.6, p.52 · TMU Lecture 4 Slide 35
4Oxidation of the haem iron of haemoglobin from Fe²⁺ to Fe³⁺ ( ).
A. is essential for its function as an oxygen carrier
B. destroys its activity — it can no longer bind oxygen
C. converts haemoglobin into myoglobin entirely
D. increases its affinity for oxygen markedly
E. has no effect at all on oxygen binding
Answer: B
Methaemoglobin can neither bind nor transport O₂. Contrast with the cytochromes, where oxidation and reduction of the metal is essential to their function as electron carriers — same metal, opposite requirement. Methaemoglobin reductase normally reduces Fe³⁺ back to Fe²⁺.Harper's ch.6, pp.52, 57–58
5Myoglobin's polypeptide contains how many amino acid residues?
A. 141
B. 146
C. 153
D. 287
E. 574
Answer: C
153 residues, MW about 17 000, folded into eight right-handed α-helices named A–H, with roughly 75% of residues in helix. The molecule measures 4.5 × 3.5 × 2.5 nm.Harper's ch.6, p.52 · TMU Lecture 4 Slide 27
6With how many exceptions does the interior of myoglobin contain only non-polar residues?
A. None — the interior is entirely non-polar
B. Four — two histidines and two propionates
C. Eight — one per helix
D. Two — histidines E7 and F8
E. Twelve, all of them charged
Answer: D
His E7 and His F8 are the two polar residues buried in the otherwise non-polar interior, and they are buried precisely because they function in O₂ binding at the haem iron. When a protein breaks its own structural rule, that is where the function is.Harper's ch.6, p.52 · TMU Lecture 4 Slide 27
7Isolated haem binds carbon monoxide about 25 000 times more strongly than oxygen, yet CO does not displace O₂ in vivo because ( ).
A. the proximal histidine F8 occupies the sixth coordination position
B. carbon monoxide binds only to the ferric form of the iron
C. haemoglobin oxidises carbon monoxide to carbon dioxide
D. carbon monoxide cannot enter the erythrocyte at all
E. the distal histidine E7 blocks CO's preferred perpendicular angle
Answer: E
The apoprotein creates a hindered environment. CO prefers Fe, C and O all perpendicular to the haem plane; His E7 blocks that geometry while still allowing O₂ its favourable ~121° orientation. This reduces the haem–CO advantage from 25 000-fold to about 200-fold, and since O₂ is present in great excess it normally dominates.Harper's ch.6, p.53
8The oxygen dissociation curve of myoglobin is hyperbolic. The physiological consequence is that myoglobin ( ).
A. releases only a small fraction of its oxygen at tissue pO₂
B. fails to load oxygen fully in the lung at 100 mm Hg
C. releases essentially all of its oxygen at 40 mm Hg
D. cannot bind oxygen until pO₂ exceeds 100 mm Hg
E. delivers more oxygen per gram than haemoglobin does
Answer: A
Myoglobin loads readily at the 100 mm Hg of the lung, but at 40 mm Hg (venous) or 20 mm Hg (active muscle) it is still nearly saturated. Only when strenuous exercise drives muscle pO₂ to about 5 mm Hg does it release oxygen — precisely the behaviour required of an emergency reserve.Harper's ch.6, p.53
9P₅₀ is defined as ( ).
A. the percentage of haemoglobin saturated at arterial pO₂
B. the partial pressure of O₂ at which a given haemoglobin reaches half-saturation
C. the pressure at which haemoglobin releases half its carbon dioxide
D. the pH at which haemoglobin has no net charge
E. the partial pressure of O₂ in mixed venous blood
Answer: B
A measure of oxygen affinity. Crucially, a high P₅₀ means LOW affinity — more pressure is needed to half-fill the protein — and therefore better release at the tissues. P₅₀ always exceeds the pO₂ of the peripheral tissues.Harper's ch.6, p.55
10The P₅₀ values of HbA and HbF are respectively ( ).
A. 20 and 26 mm Hg
B. 100 and 40 mm Hg
C. 26 and 20 mm Hg
D. 40 and 20 mm Hg
E. 5 and 26 mm Hg
Answer: C
HbA 26, HbF 20 mm Hg. HbF's lower P₅₀ means higher affinity, which is what allows it to extract oxygen from the mother's HbA across the placenta. The same property is suboptimal after birth, because high affinity limits delivery to the tissues.Harper's ch.6, p.55
11Fetal haemoglobin has a higher oxygen affinity than HbA because ( ).
A. HbF contains four γ chains rather than two α and two γ
B. HbF carries a higher iron content per tetramer
C. HbF binds oxygen without cooperativity between chains
D. residue H21 of the γ chain is serine, so BPG binds weakly
E. the fetal haem group lacks propionate substituents
Answer: D
Serine cannot form a salt bridge, so BPG binds more weakly to HbF than to HbA. Less stabilisation of the low-affinity T state means higher oxygen affinity. This is the molecular explanation behind the P₅₀ difference and is worth an extra mark whenever HbF is discussed.Harper's ch.6, p.57
12The subunit composition of normal adult haemoglobin (HbA) and of fetal haemoglobin (HbF) are respectively ( ).
A. α₂γ₂ and α₂β₂
B. α₄ and β₄
C. α₂β₂ and ξ₂ε₂
D. α₂δ₂ and α₂β₂
E. α₂β₂ and α₂γ₂
Answer: E
HbA = α₂β₂; HbF = α₂γ₂. The developmental sequence runs ξ₂ε₂ in the early embryo → HbF (α₂γ₂) from the end of the first trimester → HbA (α₂β₂), whose completion is not reached until some weeks after birth.Harper's ch.6, p.55
13Cooperative binding of oxygen by haemoglobin means that ( ).
A. O₂ binds the tetramer more readily once other O₂ is bound
B. each subunit binds oxygen entirely independently of the others
C. oxygen and carbon dioxide compete for the same site
D. all four oxygen molecules bind simultaneously
E. oxygen binding requires the prior binding of BPG
Answer: A
This is Harper's exact wording. Cooperativity lets haemoglobin maximise both the quantity loaded at the pO₂ of the lungs and the quantity released at the pO₂ of the tissues. It is an exclusive property of multimeric proteins — which is why monomeric myoglobin cannot show it.Harper's ch.6, p.55
14During the transition from the T to the R state of haemoglobin, one pair of α/β subunits rotates through ( ) relative to the other pair.
A. 5°
B. 15°
C. 45°
D. 90°
E. 180°
Answer: B
15°, compacting the tetramer. The sequence is: O₂ binds → the haem iron moves into the plane → His F8 and its attached residues are pulled along → salt bridges between the C-terminal residues of all four subunits rupture → one α/β pair rotates 15°.Harper's ch.6, p.55 · TMU Lecture 4 Slide 10
15In haemoglobin, the T (taut) state is ( ).
A. the high-affinity, oxygenated form adopted in the lung
B. the conformation adopted only in the pulmonary capillary
C. the low-affinity, deoxygenated form stabilised by salt bridges
D. the form in which the haem iron has been oxidised to ferric
E. the monomeric form the protein adopts in solution
Answer: C
T = taut = low affinity = deoxy, held together by salt bridges between subunits. R = relaxed = high affinity = oxygenated. The same terms describe the low- and high-affinity conformations of allosteric enzymes — which is why this unit is essential preparation for Unit 7.Harper's ch.6, pp.55–56
16Oxygen affinity increases as successive oxygen molecules bind to haemoglobin because ( ).
A. BPG is progressively recruited into the central cavity
B. the pKa of histidine 146 progressively increases
C. the haem iron changes from ferrous to ferric
D. each later binding ruptures fewer salt bridges
E. the tetramer dissociates into free subunits
Answer: D
The first oxygen does the hard work of breaking salt bridges; each one after it has an easier job. Note also that the T→R transition does not occur after a fixed number of oxygens — it becomes progressively more probable with each, and unruptured bridges are progressively weakened.Harper's ch.6, p.56
17Which of the following does NOT stabilise the T state of haemoglobin?
A. Protons
B. Carbon dioxide
C. 2,3-bisphosphoglycerate
D. Chloride
E. Oxygen
Answer: E
Oxygen drives the T→R transition — it is the one thing in the list that stabilises R. Protons, CO₂, chloride and BPG all stabilise T, lowering affinity and enhancing delivery: the higher their concentration, the more oxygen must bind to trigger the transition. All four are signals that a tissue is working hard.Harper's ch.6, pp.56–57 · TMU Lecture 4 Slide 11
18Approximately what proportion of the CO₂ in venous blood is carried as haemoglobin carbamates?
A. 15%
B. 1%
C. 50%
D. 75%
E. 95%
Answer: A
About 15%, formed with the amino-terminal nitrogens of the polypeptide chains. Carbamate formation changes the charge on the amino terminals from positive to negative, which favours salt bridge formation between α and β chains — so it stabilises T. Most of the remaining CO₂ is carried as bicarbonate, formed via carbonic anhydrase.Harper's ch.6, p.56
19Deoxyhaemoglobin binds how many protons for every two molecules of oxygen released?
A. Two
B. One
C. Four
D. Eight
E. None
Answer: B
One proton per two O₂ released — a significant contribution to the buffering capacity of blood. The slightly lower pH of peripheral tissues, aided by carbamation, stabilises the T state and so enhances oxygen delivery. In the lungs the process runs in reverse.Harper's ch.6, p.56
20The protons responsible for the Bohr effect arise principally from ( ).
A. dissociation of the propionate groups of the haem
B. hydrolysis of 2,3-bisphosphoglycerate in the central cavity
C. rupture of salt bridges involving β-chain His146 when O₂ binds
D. oxidation of the haem iron from Fe²⁺ to Fe³⁺
E. the imidazole group of histidine F8 releasing H⁺
Answer: C
In the lungs, conversion to the R state breaks salt bridges involving β-chain His 146, and the protons dissociating from His 146 drive bicarbonate back to carbonic acid, which carbonic anhydrase dehydrates to exhaled CO₂. On releasing O₂ the T structure re-forms, raising the pKa of His 146 so it binds protons again.Harper's ch.6, p.56
21Myoglobin does NOT exhibit the Bohr effect because ( ).
A. it has no haem prosthetic group with which to bind protons
B. its haem iron is ferric rather than ferrous
C. it cannot bind carbon dioxide at physiological pH
D. it is monomeric, and the Bohr effect requires cooperativity
E. it contains no histidine residues in its sequence
Answer: D
The Bohr effect depends on cooperative interactions between the haems of the tetramer. Myoglobin's monomeric structure precludes it — the same structural reason it has a hyperbolic curve and no allosteric behaviour. One structural fact, three functional consequences.Harper's ch.6, p.56
222,3-bisphosphoglycerate binds to haemoglobin ( ).
A. as four molecules, one bound at each haem site
B. covalently to the amino terminal of each α chain
C. only when the protein is fully oxygenated in the lung
D. at the same site that binds carbon monoxide
E. as one molecule in the central cavity, only in the T state
Answer: E
One molecule, in the central cavity formed by the four subunits. The space between the H helices of the β chains is wide enough to accommodate BPG only in the T state. It forms salt bridges with three positive groups on each β chain — Val NA1, Lys EF6 and His H21 — which must be broken before R can form.Harper's ch.6, p.57 · TMU Lecture 4 Slide 18
23Elevated erythrocyte BPG, as occurs on prolonged exposure to high altitude, ( ).
A. lowers the affinity of HbA for oxygen and so enhances its release at the tissues
B. raises the affinity of HbA for oxygen and so improves loading in the lung
C. converts HbA into HbF
D. causes haemoglobin to bind carbon monoxide preferentially
E. has no effect on the oxygen dissociation curve
Answer: A
BPG stabilises the T state, so affinity falls and release at the tissues improves — the body trading loading efficiency for unloading efficiency, which is the right trade when ambient oxygen is scarce. Altitude also raises erythrocyte number and haemoglobin concentration.

⚠ Note: lower affinity means P₅₀ increases (Harper's p.57). TMU Slide 20 says “decreases P₅₀”, which is an error — the two halves of that sentence contradict each other. Answer in terms of affinity and you are safe either way.Harper's ch.6, p.57 — corrects TMU Lecture 4 Slide 20
24In haemoglobin S, the mutation is ( ).
A. glutamate replaces valine at position 6 of the β chain
B. valine replaces glutamate at position 6 of the β chain
C. tyrosine replaces histidine F8
D. serine replaces histidine at position H21 of the γ chain
E. lysine replaces glutamate at position 6 of the α chain
Answer: B
The non-polar valine replaces the polar surface residue Glu6 of the β subunit, generating a hydrophobic “sticky patch” present in both oxyHbS and deoxyHbS. Option C is haemoglobin M and option D is the γ-chain feature that explains HbF's high affinity — both are real, but different, facts.Harper's ch.6, p.58 · TMU Lecture 4 Slide 30
25Deoxyhaemoglobin A terminates the polymerisation of HbS fibres because ( ).
A. HbA binds BPG more tightly and so remains in solution
B. HbA is present at a much lower concentration
C. HbA lacks the second sticky patch needed to bind on
D. HbA has a higher oxygen affinity than HbS
E. HbA cannot adopt the T state at all
Answer: C
Both HbA and HbS carry the complementary patch exposed in the T state, so deoxyHbA can join a growing fibre — but it lacks the Val6 sticky patch, so nothing can attach beyond it and polymerisation stops. This is why heterozygotes are largely protected, and why inducing HbF expression is an emerging therapy.Harper's ch.6, p.58
1 Prosthetic group — 2′+
A small non-protein molecule that forms a permanent part of a protein and is necessary for its function.

Some proteins consist of polypeptide alone; others, such as myoglobin and haemoglobin, require a prosthetic group — in their case haem, held in a hydrophobic pocket by non-covalent bonding.TMU Lecture 4 Slide 34
2 Haem — 3′+
A cyclic tetrapyrrole consisting of four molecules of pyrrole linked by methyne bridges, with one atom of ferrous iron (Fe²⁺) at the centre of the planar ring, bonded to all four pyrrole nitrogens.

The β-position substituents are methyl, vinyl and propionate. The iron forms six coordination bonds: four to the pyrrole nitrogens, a fifth to the proximal histidine F8, and a sixth reserved for oxygen or left unoccupied. Its network of conjugated double bonds absorbs visible light and colours haem deep red.Harper's ch.6, p.52 · TMU Lecture 4 Slides 4, 35
3 Cooperative binding — 3′ · classic Section I term+
The phenomenon whereby a molecule of O₂ binds to a haemoglobin tetramer more readily if other O₂ molecules are already bound.

It permits haemoglobin to maximise both the quantity of O₂ loaded at the pO₂ of the lungs and the quantity released at the pO₂ of peripheral tissues, and is the reason its dissociation curve is sigmoid rather than hyperbolic. Cooperative interactions are an exclusive property of multimeric proteins, and are critically important to aerobic life.Harper's ch.6, p.55
4 P₅₀ — 3′+
The partial pressure of oxygen at which a given haemoglobin reaches half-saturation — a measure of its oxygen affinity.

A high P₅₀ indicates LOW affinity, and therefore better release of oxygen at the tissues. In all instances P₅₀ exceeds the pO₂ of the peripheral tissues. HbA: 26 mm Hg. HbF: 20 mm Hg — the lower fetal value allowing HbF to extract oxygen from maternal HbA across the placenta.Harper's ch.6, p.55
5 T and R states — 3′+
T (taut) is the low-affinity, deoxygenated conformation of haemoglobin, stabilised by salt bridges between the carboxyl-terminal residues of the four subunits. R (relaxed) is the high-affinity, oxygenated conformation.

The transition is triggered when O₂ binding pulls the haem iron into the plane of the ring, moving His F8, rupturing the salt bridges and rotating one α/β pair through 15°. The same terms describe the low- and high-affinity conformations of allosteric enzymes.Harper's ch.6, pp.55–56
6 The Bohr effect — 3′ · very likely Section I term+
The reciprocal coupling of proton and oxygen binding by haemoglobin.

In peripheral tissues: CO₂ combines with water to form carbonic acid, which dissociates into protons and bicarbonate. Deoxyhaemoglobin binds one proton for every two O₂ released, acting as a buffer; the resulting lower pH stabilises the T state and enhances O₂ delivery.

In the lungs: uptake of oxygen releases protons, which combine with bicarbonate to form carbonic acid; carbonic anhydrase dehydrates this to CO₂, which is exhaled. Oxygen binding therefore drives CO₂ exhalation.

The protons arise from rupture of salt bridges involving β-chain His 146. The effect depends on cooperativity and is therefore absent in myoglobin.Harper's ch.6, p.56
7 Haemoglobinopathy — 2′+
A condition in which a mutation in the gene encoding an α or β subunit of haemoglobin compromises its biological function.

Of the over 1100 known mutations affecting human haemoglobins, almost all are extremely rare and benign and present no clinical abnormality; the term is reserved for those that impair function. More than 7% of the world's population carry a haemoglobin disorder.Harper's ch.6, p.57
8 Methaemoglobin — 3′+
Haemoglobin in which the haem iron is ferric (Fe³⁺) rather than ferrous, and which can therefore neither bind nor transport oxygen.

Normally the enzyme methaemoglobin reductase reduces Fe³⁺ back to Fe²⁺. Methaemoglobinaemia arises from oxidation by agents such as sulfonamides, from hereditary haemoglobin M (in which His F8 is replaced by tyrosine, whose phenolate anion stabilises Fe³⁺), or from reduced methaemoglobin reductase activity.Harper's ch.6, pp.57–58
1 Describe the structure of haemoglobin and myoglobin and explain the linkage between protein structure and biological function. 8′ — TMU study question, past paper

The central statement

Myoglobin and haemoglobin have different primary structures but nearly identical secondary and tertiary structures. They differ fundamentally in quaternary structure — and every difference in their biological function follows from that one structural difference.

Myoglobin — structure

  • Primary: a single polypeptide of 153 aminoacyl residues, MW about 17 000.
  • Secondary: about 75% of residues lie in eight right-handed α-helices of 7–20 residues, named A–H from the amino terminal.
  • Tertiary: a compact, roughly spherical molecule of 4.5 × 3.5 × 2.5 nm, with a polar surface and an interior of non-polar residues — with two exceptions, His E7 and His F8, which lie close to the haem iron and function in oxygen binding.
  • Quaternary: none — myoglobin is monomeric.

Haemoglobin — structure

  • A tetramer of two subunit types: α₂β₂ in HbA. Subunits are held together by non-covalent forces, importantly salt bridges.
  • Each subunit closely resembles myoglobin: rich in α-helix, highly compact, hydrophilic residues outside and hydrophobic inside, the hydrophobic interior forming a pocket that binds one haem by non-covalent bonding. The α subunit has seven rather than eight helical regions.
  • Four haem groups, four oxygen molecules — one per subunit.

The prosthetic group

Haem is a cyclic tetrapyrrole of four pyrroles linked by methyne bridges, with central Fe²⁺. The iron forms six coordination bonds: four to the pyrrole nitrogens, a fifth to the proximal histidine F8, and a sixth reserved for oxygen. The distal histidine E7 lies on the opposite face and hinders the binding of carbon monoxide.

Structure determines function

MyoglobinHaemoglobin
LocationRed skeletal muscleErythrocytes
FunctionOxygen storageOxygen transport, plus CO₂ and protons
CooperativityNone — monomericPresent — a multimeric property
O₂ curveHyperbolicSigmoid
Bohr effectAbsentPresent

The mechanism linking the two. In deoxymyoglobin the iron lies about 0.03 nm outside the plane of the haem. On oxygenation it moves into the plane, and because it is bonded to His F8 the histidine and its attached residues are pulled with it. In monomeric myoglobin this motion has no further consequence. In the haemoglobin tetramer it ruptures salt bridges between the carboxyl-terminal residues of all four subunits, so that one α/β pair rotates 15° relative to the other, converting the molecule from the low-affinity T (taut) state to the high-affinity R (relaxed) state. Because subsequent binding events require the rupture of fewer salt bridges, the affinity of the remaining haems rises — cooperativity, and hence the sigmoid curve.

Why this matters physiologically. A sigmoid curve is steep across the range between lung (pO₂ 100 mm Hg) and tissue (40, or 20 mm Hg in active muscle), so haemoglobin loads almost fully in the lung and unloads a large fraction in the tissues. Myoglobin's hyperbolic curve leaves it nearly saturated at tissue pO₂, releasing oxygen only when muscle pO₂ falls to about 5 mm Hg — exactly the behaviour required of a reserve store.

Marking guide: myoglobin structure across four orders 1.5 · haemoglobin as α₂β₂ tetramer with four haems 1 · statement that primary structures differ but secondary/tertiary are near-identical 1 · haem and the six coordination bonds with His F8 and E7 1.5 · mechanism of the T→R transition 1.5 · cooperativity linked to the sigmoid curve 1 · physiological interpretation of the two curves 0.5.
2 Describe the linkage between primary structure and molecular disease, based on sickle cell anaemia. 8′ — TMU study question, past paper

The principle

Sickle cell anaemia is the classic molecular disease: a change in a single amino acid of the primary structure produces, through a chain of structural consequences, a clinical illness. The gene, the protein, the cell and the patient can be connected in one continuous argument.

1 · The mutation

In haemoglobin S, the non-polar valine replaces the polar surface residue glutamate at position 6 of the β subunit:

HbA  Val-His-Leu-Thr-Pro-Glu-Glu-Lys
HbS  Val-His-Leu-Thr-Pro-Val-Glu-Lys

A single base change in the β-globin gene; a single residue changed in a chain of 146.

2 · The consequence for the protein surface

Substituting a hydrophobic residue for a charged one at a surface position generates a hydrophobic “sticky patch” on the surface of the β subunit. This patch is present in both oxyHbS and deoxyHbS — it is a permanent feature of the mutant protein.

3 · The complementary patch

Both HbA and HbS possess a complementary sticky patch elsewhere on their surface. Critically, this complementary patch is exposed only in the deoxygenated T state.

4 · Polymerisation

Consequently, at low pO₂, the sticky patch of one deoxyHbS molecule binds the complementary patch of the next, and deoxyHbS polymerises into long, insoluble, twisted helical fibres. In the oxygenated R state the complementary patch is hidden and no polymer forms — which is why the disease is episodic and precipitated by hypoxic stress such as high altitude, infection or exertion.

5 · The cell and the patient

The fibres distort the erythrocyte into the characteristic sickle shape, converting a normally flexible cell into a stiff one. Such cells are vulnerable to lysis in the interstices of the splenic sinusoids, producing the haemolytic anaemia, and cause multiple secondary clinical effects through vascular occlusion.

Why heterozygotes are protected

Binding of deoxyHbA terminates fibre polymerisation, because HbA possesses only the complementary patch and lacks the second sticky patch needed to bind a further haemoglobin molecule. HbA therefore acts as a chain terminator. The same reasoning underlies an emerging treatment — inducing HbF expression to inhibit HbS polymerisation — alongside stem cell transplantation and, in future, gene therapy.

The wider lesson

Compare HbS with haemoglobin M, in which His F8 is replaced by tyrosine. There the substituted residue is bonded to the haem iron itself: the phenolate anion stabilises Fe³⁺ and oxygen binding is abolished. Two single-residue substitutions, two entirely different mechanisms of disease — because what matters is not only which amino acid changes but where it sits in the folded structure.

Marking guide: the mutation stated precisely as Val for Glu6 of the β chain 1.5 · sticky patch generated on the β subunit surface 1.5 · complementary patch exposed only in the deoxy T state 1.5 · polymerisation into insoluble fibres at low pO₂ 1.5 · sickling and lysis in the splenic sinusoids 1 · HbA terminating polymerisation, or a second example such as HbM 1.
3 Elucidate the roles of the Bohr effect and 2,3-bisphosphoglycerate in oxygen delivery. 8′ — 'elucidate'

The problem both solve

A transport protein must do two contradictory things: bind oxygen tightly in the lung and release it readily in the tissues. Cooperativity provides part of the answer. The rest comes from ligands that shift the T⇌R equilibrium according to local conditions — and the general rule is that anything stabilising the T state lowers affinity and improves delivery.

The Bohr effect

Definition: the reciprocal coupling of proton and oxygen binding by haemoglobin.

In the peripheral tissues. CO₂ generated by respiration combines with water to form carbonic acid — a reaction catalysed in the erythrocyte by carbonic anhydrase — which dissociates into bicarbonate and protons. Deoxyhaemoglobin binds one proton for every two O₂ molecules released, contributing significantly to the buffering capacity of blood. The resulting lower pH, aided by carbamate formation, stabilises the T state and thereby enhances oxygen delivery.

In the lungs. The process reverses. As O₂ binds to deoxyhaemoglobin, protons are released; they combine with bicarbonate to form carbonic acid, and dehydration of H₂CO₃ by carbonic anhydrase yields CO₂, which is exhaled. The binding of oxygen thus drives the exhalation of CO₂.

The molecular origin. The protons arise from the rupture of salt bridges when O₂ binds to T-state haemoglobin — specifically those involving β-chain His 146. On release of oxygen the T structure and its salt bridges re-form, which increases the pKa of His 146 so that it binds protons once more.

Note: the Bohr effect depends on cooperative interactions between the haems of the tetramer, so the monomeric structure of myoglobin precludes it.

Carbon dioxide transport

Haemoglobin also carries CO₂ directly, as carbamates formed with the amino-terminal nitrogens of the chains, accounting for about 15% of the CO₂ in venous blood. Carbamate formation changes the charge on the amino terminals from positive to negative, favouring salt bridge formation between α and β chains — so it, too, stabilises T. Most of the remaining CO₂ travels as bicarbonate.

2,3-bisphosphoglycerate

Synthesis. A low pO₂ in peripheral tissues promotes the synthesis of BPG in erythrocytes from the glycolytic intermediate 1,3-bisphosphoglycerate.

Binding. The tetramer binds one molecule of BPG in the central cavity formed by its four subunits. The space between the H helices of the β chains lining that cavity is wide enough to accommodate BPG only when haemoglobin is in the T state.

Mechanism. BPG forms salt bridges with three positively charged groups on each β chain — the terminal amino group via Val NA1, and Lys EF6 and His H21. These are additional salt bridges that must be broken before conversion to the R state. BPG therefore stabilises deoxygenated T-state haemoglobin and lowers oxygen affinity.

Two consequences worth stating.

  • Fetal haemoglobin. Residue H21 of the γ subunit is serine rather than histidine. Serine cannot form a salt bridge, so BPG binds more weakly to HbF; the T state is less stabilised, and HbF therefore has a higher oxygen affinity (P₅₀ 20 versus 26 mm Hg) — enabling it to extract oxygen from maternal HbA across the placenta.
  • Adaptation to altitude. Prolonged exposure raises erythrocyte number, haemoglobin concentration and BPG synthesis. Elevated BPG lowers the affinity of HbA for oxygen, enhancing release at the peripheral tissues — a trade of loading efficiency for unloading efficiency, which is the correct trade when ambient oxygen is scarce.

The unifying statement

Protons, carbon dioxide, chloride and BPG all stabilise the T state, and the higher their concentration the more oxygen must bind to trigger the transition to R. All four are signals that a tissue is metabolically active — and all four act by the same means, preserving or adding the salt bridges that hold the taut state together.

Marking guide: Bohr effect defined as reciprocal proton/O₂ coupling 1 · tissue limb with carbonic anhydrase and 1 H⁺ per 2 O₂ 1.5 · lung limb, oxygen binding driving CO₂ exhalation 1 · His 146 and the pKa change 1 · BPG synthesis from 1,3-BPG under low pO₂ 0.5 · BPG binds one molecule in the central cavity, T state only 1.5 · mechanism via extra salt bridges lowering affinity 1 · one application — HbF or altitude 0.5.