Bioinorganic Chemistry: Oxygen Carriers, Iron–Sulfur Proteins, Carbonic Anhydrase and Cu–Zn Superoxide Dismutase

This chapter answers Section XL0.4, “Bioinorganic Chemistry”, of XL0 Chemistry, the part of the Life Sciences (XL) paper that every candidate answers: “Oxygen transport and storage metalloproteins (hemoglobin, myoglobin, hemocyanin); Electron transport Fe-S metalloproteins; Carbonic anhydrase and Cu-Zn Superoxide dismutase.” It reads each metal site as coordination chemistry: which metal, which ligands, which oxidation and spin state, and why that combination does the job. It covers heme and O₂ binding in myoglobin and hemoglobin, with cooperativity and its effectors; the dicopper site of hemocyanin; rubredoxin, ferredoxins, the Rieske protein and HiPIP; zinc catalysis in carbonic anhydrase; and the copper cycle of superoxide dismutase. The numericals are oxygen-carrying capacity, saturation from the Hill equation, spin-only magnetic moments and the average oxidation state of iron in a cluster.

1. Metal sites in proteins: what to read off each one

A metalloprotein holds a metal ion with its own side chains, sometimes inside an organic cofactor such as a porphyrin. The protein supplies the ligands: the imidazole nitrogen of histidine, the thiolate sulfur of cysteine, carboxylate oxygens of aspartate and glutamate, and water or hydroxide at an open site. By choosing metal, ligands and geometry the protein decides what the site does. This heading has three jobs: reversible O₂ binding without net oxidation (hemoglobin, myoglobin, hemocyanin), one-electron transfer (Fe–S proteins) and catalysis, either as a Lewis acid (zinc in carbonic anhydrase) or by redox cycling (copper in superoxide dismutase).

For every site read off the oxidation state, the d-electron count and the spin state, then what they predict. Unpaired electrons make a site paramagnetic, and the spin-only moment is μ = √[n(n + 2)] BM for n unpaired electrons. A d¹⁰ ion (Zn²⁺, Cu⁺) has no d–d bands and no unpaired electrons, so it is colourless, diamagnetic and invisible to EPR; an odd-electron ion such as Cu²⁺ (d⁹) always has one unpaired electron and is EPR-active.

Spin-only moments of the ions met in this chapter
Ion and where it occursd electronsUnpaired electronsμ, spin-only (BM)
High-spin Fe(II): deoxy heme644.90
High-spin Fe(III): aquomet heme555.92
Low-spin Fe(III): cyanomet heme511.73
Cu(II): oxidised SOD, each Cu of oxy-hemocyanin before coupling911.73
Cu(I), Zn(II): deoxy-hemocyanin, reduced SOD, carbonic anhydrase1000

2. Heme in myoglobin and hemoglobin: binding O₂ without rusting

Heme is iron in protoporphyrin IX: four pyrrole nitrogens hold Fe in a square plane. In both proteins the fifth (axial) ligand is the proximal histidine (His F8); the sixth site, on the other face, is where O₂ binds, beside the distal histidine (His E7). Myoglobin (Mb) is a single chain with one heme that stores O₂ in muscle and speeds its diffusion to mitochondria. Hemoglobin (Hb) is an α₂β₂ tetramer with four hemes, packed in red cells, that carries O₂ from lungs to tissues.

In deoxy heme the iron is high-spin Fe(II) (d⁶, four unpaired electrons), five-coordinate and too large for the porphyrin hole, so it sits a few tenths of an ångström out of the plane towards the proximal His. When O₂ binds, the iron becomes six-coordinate and low-spin, shrinks and moves into the plane. Oxy-Hb and oxy-Mb bind O₂ end-on and bent (Fe–O–O near 120°) and are diamagnetic. Their O–O stretch, about 1105 cm⁻¹, is close to that of superoxide, so the bound unit is well described as Fe(III)–O₂⁻ with the two remaining spins paired by antiferromagnetic coupling; the older picture, low-spin Fe(II) with singlet O₂, also gives zero net spin. Either way the binding is reversible: O₂ leaves as O₂ and the iron returns to Fe(II).

🎯 Why the globin is needed
Heme alone in water does not carry O₂. A second Fe(II)-heme attacks the bound O₂, giving a peroxo-bridged Fe(III)–O–O–Fe(III) dimer that breaks down to a μ-oxo Fe(III)–O–Fe(III) dimer: the iron is oxidised irreversibly. The globin buries each heme in a hydrophobic pocket that no second heme can enter and little water reaches. A little Fe(III) still forms in red cells; this methemoglobin cannot bind O₂ (water occupies the sixth site) and is reduced back by methemoglobin reductase.

CO binds free heme tens of thousands of times more tightly than O₂, but in the proteins only a few hundred times more: the distal His hydrogen-bonds the bent Fe–O₂ unit but not CO, and the pocket disfavours the linear Fe–C–O that CO prefers. That margin still makes CO toxic.

3. Cooperativity, the Hill equation and oxygen-carrying capacity

Myoglobin has one site, so its fractional saturation θ follows a hyperbola, θ = pO₂/(P₅₀ + pO₂), with P₅₀ of only a few torr. Hemoglobin's curve is sigmoidal: binding one O₂ raises the affinity of the other hemes. The empirical Hill equation describes it: θ = pO₂ⁿ/(P₅₀ⁿ + pO₂ⁿ), or log[θ/(1 − θ)] = n log pO₂ − n log P₅₀. The slope n (Hill coefficient) is 1 for independent sites and about 2.8 to 3 for adult Hb, whose P₅₀ in blood is about 26 torr. n cannot exceed the number of interacting sites (four); it measures how strongly they interact, not how many there are.

Mechanism. Deoxy-Hb is in the low-affinity T (tense) state, held by salt bridges between subunits; oxy-Hb is in the high-affinity R (relaxed) state. When an iron moves into the porphyrin plane on binding O₂, it pulls the proximal His and its F helix, and the movement is relayed to the α₁β₂ interface. Each O₂ bound makes the switch to R easier, so later hemes bind more readily. Worked example: lungs at 100 torr, resting tissue at 40 torr. For Hb (P₅₀ = 26 torr, n = 2.8), θ falls from 0.978 to 0.770, so about 21% of sites unload. For Mb (P₅₀ = 2.8 torr, n = 1), θ falls from 0.973 to 0.935, only about 4%. The sigmoid places the steep part of Hb's curve in the tissue range, while Mb stays loaded until muscle pO₂ is very low, which suits a store.

  • Bohr effect: H⁺ and CO₂ lower O₂ affinity (right shift, P₅₀ rises): protonated His residues strengthen T-state salt bridges and CO₂ forms carbamates on N-terminal amino groups, so acidic working tissue receives more O₂. Heat also shifts the curve right.
  • 2,3-Bisphosphoglycerate (2,3-BPG): one molecule binds in the central cavity of deoxy-Hb, between the β chains, and stabilises T (right shift); its level rises at high altitude.
  • Fetal Hb (α₂γ₂): the γ chain has Ser where β has His143 in the BPG site, so it binds BPG weakly and has higher O₂ affinity (left shift), drawing O₂ from maternal blood.
  • CO: occupies some hemes and pushes the tetramer towards R (left shift), so the remaining O₂ is held more tightly and unloaded less, a double harm.

Oxygen-carrying capacity. A Hb tetramer (molar mass about 64,500 g mol⁻¹) carries four O₂, so one gram binds at most 4 × 22,400 mL/64,500 = 1.39 mL O₂ at STP (measured values are a little lower). Blood with 15 g Hb per 100 mL, fully saturated, carries about 20.8 mL O₂ per 100 mL, whereas O₂ dissolved at 100 torr is only about 0.3 mL per 100 mL: hemoglobin raises the capacity roughly seventy-fold.

⚠️ Reading shifts the right way
Right shift = higher P₅₀ = lower affinity = more unloading at a given pO₂. H⁺, CO₂, 2,3-BPG and heat shift right; fetal Hb, CO-bound Hb and myoglobin lie to the left. Affinity is read from P₅₀, not from the Hill slope.

4. Hemocyanin: two copper ions and a peroxide

Hemocyanin carries O₂ in many arthropods (crabs, lobsters, spiders) and molluscs (snails, octopus). It has no heme. Each O₂ site holds two copper ions, each bound by three histidines (a type 3 copper site), and the protein is dissolved in the haemolymph as large multi-subunit assemblies rather than packed in cells. Deoxy-hemocyanin is Cu(I)…Cu(I): d¹⁰, colourless and diamagnetic. Binding is a two-electron transfer, 2Cu(I) + O₂ → Cu(II)–O₂²⁻–Cu(II): O₂ becomes peroxide, bridging both coppers side-on (μ-η²:η²). Oxy-hemocyanin is intensely blue from peroxide-to-Cu(II) charge-transfer bands. Each Cu(II) is d⁹ with one unpaired electron, yet oxy-hemocyanin is diamagnetic and EPR-silent: the bridging peroxide couples the two spins strongly antiferromagnetically.

The O–O stretching frequency counts electrons. Electrons added to O₂ enter its π* orbitals and lower the bond order: O₂ (bond order 2, about 1555 cm⁻¹), superoxide O₂⁻ (1.5, about 1100 to 1150 cm⁻¹), peroxide O₂²⁻ (1, roughly 750 to 900 cm⁻¹). Oxy-Hb (about 1105 cm⁻¹) has therefore received about one electron, and oxy-hemocyanin (about 750 cm⁻¹) two.

The two O₂ carriers side by side
FeatureHemoglobin and myoglobinHemocyanin
Metal siteOne Fe in a heme, axial proximal HisTwo Cu, each with three His; no porphyrin
Deoxy formHigh-spin Fe(II), paramagnetic, dark redCu(I)…Cu(I), d¹⁰, diamagnetic, colourless
Oxy formFe(III)–superoxide-like, diamagnetic, bright redCu(II)–peroxide–Cu(II), coupled, diamagnetic, blue
O₂ per metal; binding mode1 O₂ per Fe; end-on, bent, ν(O–O) ≈ 1105 cm⁻¹1 O₂ per 2 Cu; side-on bridge, ν(O–O) ≈ 750 cm⁻¹
Where it worksRed cells (Hb), muscle (Mb)Dissolved in haemolymph

5. Iron–sulfur proteins: one-electron carriers

In Fe–S proteins high-spin iron sits in a roughly tetrahedral set of sulfur donors: cysteine thiolates, usually with bridging inorganic sulfide (S²⁻). Each cluster accepts or donates one electron at a time, however many iron atoms it has. Little changes in geometry between the two oxidation states and the added electron is spread over the cluster, so electron transfer is fast. The protein surroundings (hydrogen bonds to the sulfurs, solvent exposure, charged residues) tune the reduction potential from strongly negative to positive.

The Fe–S electron carriers (potentials are typical values, not fixed constants)
ProteinSite and ligandsCouple usedTypical E°′ and role
RubredoxinOne Fe, four Cys; no inorganic sulfideFe(III)/Fe(II)Near 0 V; bacteria
Plant-type ferredoxin[2Fe–2S], two S²⁻, four Cys[2Fe–2S]²⁺/⁺About −0.4 V; carries electrons from photosystem I towards NADP⁺
Rieske protein[2Fe–2S]; one Fe has two Cys, the other two His[2Fe–2S]²⁺/⁺About +0.3 V; complex III (bc₁) and cytochrome b₆f
Bacterial ferredoxin[4Fe–4S] cube, four S²⁻, four Cys[4Fe–4S]²⁺/⁺About −0.4 V; also in complexes I and II
HiPIP (high-potential iron–sulfur protein)The same [4Fe–4S] cube[4Fe–4S]³⁺/²⁺About +0.35 V; photosynthetic bacteria

Oxidation-state bookkeeping. The core charge counts only Fe and S²⁻: total Fe charge = core charge + 2 × (number of S²⁻). [2Fe–2S]²⁺ is two Fe(III); [2Fe–2S]⁺ is Fe(III) + Fe(II). For the cube, [4Fe–4S]³⁺ has total Fe charge 11 (average 2.75: three Fe(III), one Fe(II)), the resting 2+ state averages 2.50 and the 1+ state 2.25. The electrons are delocalised, so these are formal counts. This is the three-state picture: all [4Fe–4S] proteins share the 2+ state, but HiPIP buries its cube in a hydrophobic pocket with few hydrogen bonds to the sulfurs, which favours the 3+/2+ couple, while ferredoxins use 2+/1+.

ℹ️ Telling clusters apart in the laboratory
Acid-labile sulfide: acidifying the protein releases H₂S from bridging S²⁻; rubredoxin and heme proteins give none. EPR: the high-spin irons couple antiferromagnetically, so [2Fe–2S]²⁺ and [4Fe–4S]²⁺ are S = 0 and silent, while [2Fe–2S]⁺, [4Fe–4S]⁺ (reduced ferredoxins) and [4Fe–4S]³⁺ (oxidised HiPIP) are S = ½ and EPR-active. Not every cluster carries electrons: in aconitase one iron of a [4Fe–4S] cube lacks a Cys ligand and binds citrate as a Lewis acid.

6. Carbonic anhydrase: zinc as a Lewis acid

Carbonic anhydrase catalyses CO₂ + H₂O ⇌ HCO₃⁻ + H⁺, which is slow uncatalysed. The fast human isozyme CA II turns over about 10⁶ CO₂ per second, with kcat/KM near the diffusion limit. In red cells it turns tissue CO₂ into bicarbonate for transport and reverses the reaction in the lungs. The active site is Zn(II) bound tetrahedrally by three histidine imidazoles and one water.

  • Activation: Zn(II) polarises the bound water and lowers its pKa by several units, to about 7, so at physiological pH much of the site is Zn–OH⁻, a strong nucleophile.
  • Attack: CO₂ sits in an adjacent hydrophobic pocket; Zn–OH⁻ attacks its carbon to give zinc-bound HCO₃⁻.
  • Release: water displaces bicarbonate, giving Zn–OH₂ again.
  • Regeneration (rate-limiting): Zn–OH₂ loses a proton to re-form Zn–OH⁻; His64, at the mouth of the cavity and not a zinc ligand, shuttles the proton to buffer in the solution.
🎯 Why zinc?
Zn(II) is d¹⁰: it is redox-inert, so it cannot reduce O₂ or be oxidised in the cell; it has no ligand-field stabilisation, so it moves easily between four- and five-coordination during the cycle; and it is a strong Lewis acid that binds histidine nitrogen firmly. Being d¹⁰ it is spectroscopically silent, so chemists substitute Co(II), which gives an active, coloured enzyme whose spectrum reports on the site. Sulfonamides such as acetazolamide inhibit by binding zinc in place of hydroxide; the drug lowers aqueous humour formation in glaucoma.

7. Cu–Zn superoxide dismutase: copper does the redox, zinc holds the frame

Superoxide, O₂⁻, forms when single electrons leak to O₂, for example from the respiratory chain. Superoxide dismutase (SOD) removes it by dismutation (disproportionation): 2O₂⁻ + 2H⁺ → O₂ + H₂O₂. Oxygen at −½ in superoxide goes partly to 0 (O₂) and partly to −1 (H₂O₂); catalase and peroxidases then remove the H₂O₂. The reaction runs by a ping-pong cycle at copper:

  • Cu(II)–SOD + O₂⁻ → Cu(I)–SOD + O₂
  • Cu(I)–SOD + O₂⁻ + 2H⁺ → Cu(II)–SOD + H₂O₂

Both steps are downhill because the Cu(II)/Cu(I) potential of the site lies between that for reducing O₂ to superoxide (negative) and that for reducing superoxide to H₂O₂ (strongly positive). Cytosolic Cu,Zn-SOD (SOD1) is a dimer; each subunit holds one Cu and one Zn. Cu(II) is bound by four histidines plus water; Zn(II) by three histidines and an aspartate. One histidine, deprotonated to imidazolate, bridges Cu and Zn. When Cu(II) is reduced, the Cu–N bond to this bridge breaks and the histidine takes up a proton, which helps form H₂O₂. Zinc never changes oxidation state: it holds the bridge and the active-site loops in place. Positively charged residues around the channel steer the anionic substrate to copper, so the rate constant, about 2 × 10⁹ M⁻¹ s⁻¹, is close to diffusion control.

⚠️ Zinc in two different jobs
In carbonic anhydrase zinc is the catalyst (a Lewis acid); in Cu,Zn-SOD it is structural and copper is the catalyst. Neither zinc changes oxidation state. Other SODs use Mn (mitochondrial matrix) or Fe (many bacteria, chloroplasts). The Cu,Zn enzyme is inhibited by cyanide, which binds copper, a test that separates it from the Mn and Fe enzymes.

Key takeaways

  • Deoxy heme is high-spin Fe(II), five-coordinate and out of plane; oxy heme is low-spin, in-plane and diamagnetic, best described as Fe(III)–superoxide. The globin blocks μ-oxo dimer formation; met-Hb, Fe(III), cannot bind O₂.
  • Hb binds cooperatively (T → R, Hill n ≈ 2.8 to 3 and never above 4, P₅₀ ≈ 26 torr). H⁺, CO₂, 2,3-BPG and heat shift the curve right; fetal Hb, CO and myoglobin lie left. Capacity: at most 1.39 mL O₂ per gram of Hb.
  • Hemocyanin: 2Cu(I) + O₂ gives a side-on Cu(II)–peroxide–Cu(II) bridge, blue yet diamagnetic through antiferromagnetic coupling; one O₂ per two Cu. The O–O stretch counts the electrons O₂ has received.
  • Fe–S proteins carry one electron per cluster: rubredoxin (no sulfide), [2Fe–2S] ferredoxins, Rieske (two His), [4Fe–4S] ferredoxins (2+/1+, about −0.4 V) and HiPIP (3+/2+, about +0.35 V). Average Fe oxidation state = (core charge + 2 × number of S²⁻)/number of Fe.
  • Carbonic anhydrase: Zn(II)(His)₃–OH₂ with pKa ≈ 7 gives nucleophilic Zn–OH⁻; proton transfer through His64 limits the rate. Zinc is a redox-inert, flexible Lewis acid, and sulfonamides inhibit by binding it.
  • Cu,Zn-SOD: 2O₂⁻ + 2H⁺ → O₂ + H₂O₂ by a Cu(II)/Cu(I) ping-pong cycle; zinc is structural, an imidazolate bridges the metals, and electrostatic guidance gives a near-diffusion-limited rate.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In deoxyhemoglobin, the iron of each heme is best described as:

    1. Low-spin Fe(II), six-coordinate, lying in the porphyrin plane with water as its sixth ligand
    2. High-spin Fe(III), six-coordinate, with a water molecule bound trans to the proximal histidine
    3. High-spin Fe(II), five-coordinate, displaced out of the porphyrin plane towards the proximal His
    4. Low-spin Fe(III), five-coordinate, sitting in the porphyrin plane with an empty sixth site for O₂
    Show answer

    Answer: C — High-spin Fe(II), five-coordinate, displaced out of the porphyrin plane towards the proximal His

    Deoxy heme iron is high-spin Fe(II) (d⁶, four unpaired electrons) bound only to the four porphyrin nitrogens and the proximal His; the large high-spin ion sits out of the plane. High-spin Fe(III) with bound water is aquomethemoglobin, which cannot bind O₂, so that tempting choice describes the oxidised, non-functional form. Low-spin, in-plane iron is what O₂ binding produces.

  2. Free Fe(II)-heme dissolved in water binds O₂ only briefly and is soon converted into an Fe(III) species that no longer binds O₂. What does the globin chain of myoglobin do that prevents this?

    1. It buries each heme in a hydrophobic pocket, so a second heme cannot reach the bound O₂ to form a bridged Fe(III)–O–O–Fe(III) dimer
    2. It donates two electrons from the proximal histidine to each bound O₂, so that the oxygen is released as peroxide rather than as O₂
    3. It holds the iron permanently in the Fe(III) state, which binds O₂ far more reversibly than Fe(II) does in free solution
    4. It converts the high-spin iron into low-spin iron before O₂ arrives, so that O₂ can never accept an electron from the metal
    Show answer

    Answer: A — It buries each heme in a hydrophobic pocket, so a second heme cannot reach the bound O₂ to form a bridged Fe(III)–O–O–Fe(III) dimer

    In water a second Fe(II)-heme attacks the Fe–O₂ unit, giving a peroxo-bridged dimer that breaks down to the μ-oxo Fe(III)–O–Fe(III) dimer: irreversible oxidation. The globin isolates each heme in a hydrophobic pocket, blocking this bimolecular route. Fe(III) heme (the met form) does not bind O₂ at all, deoxy iron is high-spin until O₂ binds, and myoglobin releases O₂ as O₂, not as peroxide.

  3. Which of the following changes shift the O₂-binding curve of adult human hemoglobin in red cells to the right (raise P₅₀)?

    1. A fall in blood pH from 7.4 to 7.2 at constant pCO₂
    2. A rise in red-cell 2,3-bisphosphoglycerate after some days at high altitude
    3. Occupation of one of the four hemes of each tetramer by carbon monoxide
    4. Replacement of the β chains by the γ chains found in fetal hemoglobin
    Show answer

    Answer: A — A fall in blood pH from 7.4 to 7.2 at constant pCO₂; B — A rise in red-cell 2,3-bisphosphoglycerate after some days at high altitude

    H⁺ (Bohr effect) and 2,3-BPG both stabilise the T state and lower O₂ affinity, raising P₅₀. CO bound to one heme pushes the tetramer towards R and raises the affinity of the remaining hemes, a left shift. Fetal γ chains bind 2,3-BPG weakly (Ser in place of His143), so fetal Hb has higher affinity, also a left shift.

  4. O₂ binding by hemoglobin in blood follows the Hill equation θ = pO₂ⁿ/(P₅₀ⁿ + pO₂ⁿ) with P₅₀ = 26 torr and n = 3.0. Blood leaves the lungs at pO₂ = 100 torr and passes through exercising muscle where pO₂ = 20 torr. The percentage of hemoglobin's O₂-binding sites that unload O₂ in the muscle (lung saturation minus muscle saturation, × 100), to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 67

    θ(100) = 100³/(26³ + 100³) = 1,000,000/1,017,576 = 0.9827. θ(20) = 20³/(26³ + 20³) = 8000/25,576 = 0.3128. Unloaded fraction = 0.9827 − 0.3128 = 0.6699, i.e. 67%. A non-cooperative curve (n = 1) with the same P₅₀ would give only 0.794 − 0.435 = 0.36, so cooperativity nearly doubles the delivery.

  5. Blood contains 15.0 g of hemoglobin per 100 mL. Take the molar mass of hemoglobin as 64,500 g mol⁻¹, four O₂ bound per tetramer at full saturation, and 22.4 L mol⁻¹ as the molar volume of O₂ at STP. If arterial hemoglobin is 98% saturated, the volume of O₂ (mL at STP) bound to hemoglobin in 100 mL of arterial blood, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 20.4

    Moles of Hb = 15.0/64,500 = 2.326 × 10⁻⁴ mol. Moles of O₂ at full saturation = 4 × 2.326 × 10⁻⁴ = 9.302 × 10⁻⁴ mol, which is 9.302 × 10⁻⁴ × 22,400 = 20.84 mL. At 98% saturation: 0.98 × 20.84 = 20.42, i.e. 20.4 mL. Forgetting the factor of four gives 5.1 mL.

  6. Two purified O₂-binding proteins are studied by plotting log[θ/(1 − θ)] against log pO₂. Protein X gives a straight line of slope 1.0 over the whole range. Protein Y gives a slope close to 1 at very low and very high pO₂ and a maximum slope of 2.8 near half-saturation. Which conclusion do these data alone justify?

    1. Protein Y carries 2.8 O₂-binding sites per molecule on average, so it must be a mixture of dimers and tetramers
    2. Protein Y has at least three interacting O₂ sites per molecule, and binding at some of them raises the affinity of the rest
    3. Protein X must be a monomer with a single heme, because any oligomer would show a Hill slope above 1
    4. Protein Y binds O₂ more tightly than protein X at every pO₂, because its Hill plot is the steeper of the two
    Show answer

    Answer: B — Protein Y has at least three interacting O₂ sites per molecule, and binding at some of them raises the affinity of the rest

    The Hill coefficient is a lower bound on the number of interacting sites: n = 2.8 needs at least three sites with positive cooperativity (adult Hb has four). The slope is not a site count, so reading it as an average of 2.8 sites is wrong. An oligomer whose sites are independent also gives slope 1, so X need not be a monomer. Affinity is read from P₅₀ (where the line crosses log[θ/(1 − θ)] = 0), not from the slope.

  7. Using the spin-only formula μ = √[n(n + 2)] BM, where n is the number of unpaired electrons, the magnetic moment of the high-spin Fe(III) heme in aquomethemoglobin minus that of the high-spin Fe(II) heme in deoxyhemoglobin, in BM to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.02

    High-spin Fe(III) is d⁵ with five unpaired electrons: μ = √35 = 5.916 BM. High-spin Fe(II) is d⁶ with four unpaired electrons: μ = √24 = 4.899 BM. Difference = 5.916 − 4.899 = 1.017, i.e. 1.02 BM. Oxyhemoglobin, by contrast, is diamagnetic.

  8. In oxyhemocyanin, the bound dioxygen is present as:

    1. Superoxide bound end-on to a single Cu(II) ion, while the second copper ion remains Cu(I)
    2. Hydroperoxide bound end-on to one iron of a non-heme diiron centre bridged by an oxo group
    3. Neutral O₂ bound end-on and bent to an iron(II) held in a porphyrin by a proximal histidine
    4. Peroxide bridging two Cu(II) ions side-on, each copper also being bound by three histidines
    Show answer

    Answer: D — Peroxide bridging two Cu(II) ions side-on, each copper also being bound by three histidines

    Each of the two Cu(I) ions gives one electron, so O₂ becomes peroxide bridging both Cu(II) ions side-on (μ-η²:η²), with ν(O–O) near 750 cm⁻¹. The hydroperoxide on a diiron site describes hemerythrin, a different invertebrate carrier, and the porphyrin description fits hemoglobin. A superoxide on one copper would leave unpaired spin, but oxyhemocyanin is diamagnetic.

  9. Oxyhemocyanin contains two Cu(II) ions, each d⁹, yet it shows no EPR signal and is diamagnetic at room temperature. The best explanation is:

    1. Each Cu(II) adopts a low-spin configuration in its square-planar field, in which a d⁹ ion has all of its electrons paired
    2. The copper is really Cu(I) in the oxy form, and the bound dioxygen stays a neutral molecule whose two spins cancel
    3. The bridging peroxide couples the single unpaired electron on each Cu(II) antiferromagnetically, giving zero net spin
    4. The unpaired electrons move onto the six histidine imidazole rings, where they pair with one another
    Show answer

    Answer: C — The bridging peroxide couples the single unpaired electron on each Cu(II) antiferromagnetically, giving zero net spin

    A d⁹ ion always keeps one unpaired electron, whatever its ligand field, so low-spin pairing is impossible. The two Cu(II) spins interact through the orbitals of the bridging peroxide and pair up (antiferromagnetic coupling), giving S = 0. A Cu(I) with neutral O₂ description contradicts the Cu(II) state given in the stem and the peroxide-like O–O stretch near 750 cm⁻¹. Histidine rings do not take up the copper electrons.

  10. Which of the following statements comparing hemocyanin with hemoglobin are correct?

    1. Hemocyanin binds one O₂ per two copper ions, whereas hemoglobin binds one O₂ per iron
    2. The O–O stretching frequency is lower in oxyhemocyanin than in oxyhemoglobin, because more electron density has been transferred to the bound O₂
    3. Like hemoglobin, hemocyanin holds its metal inside a porphyrin ring, but with copper in place of iron
    4. Deoxyhemocyanin is blue because of d–d transitions of its two Cu(I) ions
    Show answer

    Answer: A — Hemocyanin binds one O₂ per two copper ions, whereas hemoglobin binds one O₂ per iron; B — The O–O stretching frequency is lower in oxyhemocyanin than in oxyhemoglobin, because more electron density has been transferred to the bound O₂

    Two Cu(I) together supply the two electrons that turn one O₂ into peroxide, so the stoichiometry is one O₂ per two Cu; heme iron binds one O₂ each. Peroxide (bond order 1, about 750 cm⁻¹) has received more electrons than the superoxide-like O₂ of oxy-Hb (about 1105 cm⁻¹). Hemocyanin has no porphyrin; its coppers are held by histidines. Cu(I) is d¹⁰, has no d–d transitions, and deoxyhemocyanin is colourless; the blue colour belongs to the oxy form.

  11. Which electron-transfer protein contains a [2Fe–2S] cluster in which one iron is bound by two cysteines and the other iron by two histidines?

    1. The Rieske protein of the cytochrome bc₁ complex (complex III) of mitochondria
    2. Rubredoxin of anaerobic bacteria, which holds a single iron with four cysteines
    3. Plant-type ferredoxin, which passes electrons from photosystem I towards NADP⁺
    4. High-potential iron–sulfur protein (HiPIP) of photosynthetic purple bacteria
    Show answer

    Answer: A — The Rieske protein of the cytochrome bc₁ complex (complex III) of mitochondria

    The Rieske cluster has two His ligands on one iron, which raises its reduction potential to about +0.3 V. Plant ferredoxin has the same Fe₂S₂ core but four Cys ligands and a potential near −0.4 V, so it is the closest wrong choice. Rubredoxin has one iron and no sulfide, and HiPIP contains a [4Fe–4S] cube.

  12. In the oxidised form of a high-potential iron–sulfur protein (HiPIP) the cluster core is [Fe₄S₄]³⁺, the charge counting only the four Fe and four inorganic S ions and excluding the cysteine ligands. Treating each inorganic sulfur as S²⁻, the average oxidation state of iron, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2.75

    Four S²⁻ contribute −8. Total iron charge = +3 − (−8) = +11, shared by four Fe: 11/4 = 2.75, formally three Fe(III) and one Fe(II). The resting [Fe₄S₄]²⁺ state averages 2.50 and the reduced ferredoxin state [Fe₄S₄]⁺ averages 2.25.

  13. A newly purified bacterial protein contains four iron atoms per molecule and releases H₂S when acidified. As isolated it gives no EPR signal; after a one-electron oxidation it shows an S = ½ signal. Its reduction potential is about +0.35 V. The protein is most likely:

    1. A bacterial ferredoxin whose single [4Fe–4S] cluster cycles between the 2+ and 1+ states
    2. A high-potential iron–sulfur protein whose [4Fe–4S] cluster cycles between the 3+ and 2+ states
    3. A protein with four rubredoxin-type centres, each a single iron held by four cysteines with no sulfide
    4. A Rieske protein whose [2Fe–2S] cluster cycles between its oxidised 2+ and reduced 1+ states
    Show answer

    Answer: B — A high-potential iron–sulfur protein whose [4Fe–4S] cluster cycles between the 3+ and 2+ states

    Acid-labile sulfide rules out rubredoxin-type sites. A Rieske [2Fe–2S] cluster that is EPR-silent as isolated is already in its oxidised 2+ state and cannot be oxidised further, and it has two irons, not four. Ferredoxin and HiPIP share a [4Fe–4S] cube that is EPR-silent in the 2+ state, but the ferredoxin becomes EPR-active on reduction (2+ → 1+) near −0.4 V, whereas HiPIP becomes EPR-active on oxidation (2+ → 3+) at a positive potential. Oxidation to S = ½ at +0.35 V identifies HiPIP.

  14. Which of the following statements about iron–sulfur proteins are correct?

    1. The [4Fe–4S] cluster of a ferredoxin transfers one electron at a time, although it contains four iron atoms
    2. In rubredoxin and in ferredoxins, each iron is roughly tetrahedrally surrounded by four sulfur donors
    3. Rubredoxin releases H₂S on acidification because its iron is bridged by inorganic sulfide
    4. The [4Fe–4S] cluster of aconitase carries electrons from citrate to NAD⁺ in the citric acid cycle
    Show answer

    Answer: A — The [4Fe–4S] cluster of a ferredoxin transfers one electron at a time, although it contains four iron atoms; B — In rubredoxin and in ferredoxins, each iron is roughly tetrahedrally surrounded by four sulfur donors

    Fe–S clusters are one-electron carriers whatever their nuclearity, and in rubredoxin (four Cys sulfurs) and ferredoxins (Cys sulfurs plus bridging S²⁻) iron is tetrahedral with sulfur donors. Rubredoxin has no inorganic sulfide, so acid releases no H₂S from it. Aconitase's cluster acts as a Lewis acid that binds citrate for its isomerisation to isocitrate; it passes no electrons to NAD⁺.

  15. Why can the zinc-bound water of carbonic anhydrase act as a nucleophile towards CO₂ at physiological pH?

    1. Zn(II) cycles to Zn(I) and back, passing an electron to CO₂ that makes it reactive towards free water
    2. Zn(II) has a large ligand-field stabilisation energy that locks the bound water into a reactive geometry
    3. Zn(II) binds CO₂ through its carbon atom and turns it into a carbonyl group that free water then attacks
    4. Zn(II), a strong Lewis acid, lowers the pKa of its bound water to about 7, so much of it is present as Zn–OH⁻
    Show answer

    Answer: D — Zn(II), a strong Lewis acid, lowers the pKa of its bound water to about 7, so much of it is present as Zn–OH⁻

    Coordination to Lewis-acidic Zn(II) polarises the bound water and lowers its pKa by several units to about 7, so at pH 7.4 much of the site is Zn–OH⁻, a strong nucleophile next to the CO₂ pocket. Zn(II) is d¹⁰: it has no Zn(I) state accessible in biology and no ligand-field stabilisation, so the redox and LFSE choices fail; CO₂ is attacked at carbon by Zn–OH⁻, not bound to zinc through carbon.

  16. In human carbonic anhydrase II, replacing histidine-64 (a residue at the mouth of the active-site cavity, not one of the three zinc ligands) by alanine lowers kcat for CO₂ hydration many-fold. Activity is largely restored when imidazole or 4-methylimidazole is added to the buffer. These results best support the conclusion that His64:

    1. is a ligand of the zinc ion, and the added imidazole takes its place in the coordination sphere of the metal
    2. transfers a proton from the zinc-bound water to the buffer, regenerating Zn–OH⁻ for the next turnover
    3. binds the CO₂ molecule in the hydrophobic pocket and positions it next to the zinc-bound hydroxide
    4. attacks CO₂ directly as a nucleophile, and the added imidazole performs the same attack in its place
    Show answer

    Answer: B — transfers a proton from the zinc-bound water to the buffer, regenerating Zn–OH⁻ for the next turnover

    After bicarbonate leaves, Zn–OH₂ must lose a proton to regenerate Zn–OH⁻; this step limits the rate, and His64 relays the proton to buffer. A small external base such as imidazole can take over the relay, which is why it rescues the mutant. The stem says His64 is not a zinc ligand, so the ligand-replacement idea fails; CO₂ sits in a hydrophobic pocket and is attacked by Zn–OH⁻, not by His64.

  17. During catalysis by Cu,Zn-superoxide dismutase, which metal ion changes its oxidation state, and between which states?

    1. Zinc, cycling between Zn(II) and Zn(I), while copper holds the active-site histidines in their places
    2. Copper, cycling between Cu(II) and Cu(III), with zinc supplying the second electron in each turnover
    3. Copper, cycling between Cu(II) and Cu(I), while zinc stays Zn(II) throughout and plays a structural role
    4. Both metals, copper between Cu(II) and Cu(I) and zinc between Zn(II) and Zn(I), in alternating steps
    Show answer

    Answer: C — Copper, cycling between Cu(II) and Cu(I), while zinc stays Zn(II) throughout and plays a structural role

    Cu(II) oxidises one O₂⁻ to O₂ and becomes Cu(I); Cu(I) reduces the next O₂⁻, with two protons, to H₂O₂ and returns to Cu(II). Zn(II), d¹⁰, has no oxidation state accessible in biology; it anchors the imidazolate bridge and the active-site loops. Cu(III) plays no part in the cycle.

  18. Which of the following statements about carbonic anhydrase and Cu,Zn-superoxide dismutase are correct?

    1. In carbonic anhydrase zinc is the catalytic Lewis acid, whereas in Cu,Zn-superoxide dismutase zinc is structural and copper is catalytic
    2. Superoxide dismutase converts two superoxide ions, together with two protons, into one O₂ and one H₂O₂
    3. Sulfonamide drugs such as acetazolamide inhibit carbonic anhydrase by binding to its zinc ion
    4. The zinc sites of native zinc enzymes are routinely studied through the d–d absorption bands of Zn(II)
    Show answer

    Answer: A — In carbonic anhydrase zinc is the catalytic Lewis acid, whereas in Cu,Zn-superoxide dismutase zinc is structural and copper is catalytic; B — Superoxide dismutase converts two superoxide ions, together with two protons, into one O₂ and one H₂O₂; C — Sulfonamide drugs such as acetazolamide inhibit carbonic anhydrase by binding to its zinc ion

    Zinc is the catalyst in carbonic anhydrase and a structural ion in Cu,Zn-SOD, where copper cycles. Dismutation is 2O₂⁻ + 2H⁺ → O₂ + H₂O₂. Sulfonamides bind the active-site zinc in place of hydroxide. Zn(II) is d¹⁰ and has no d–d bands, which is exactly why chemists replace it with Co(II) to obtain a spectroscopic probe.