Cytochromes in Electron-Transfer Chains
Heme redox cycling and stepwise biological electron transport
Lesson 3798 of 4,500 · Bioinorganic Chemistry
Learning objectives
- Distinguish cytochromes a, b and c by haem structure, attachment and axial ligation
- Relate reduction potentials to the direction and free-energy change of electron flow
- Explain why low-spin six-coordinate haem suits fast one-electron transfer
Introduction
Respiration and photosynthesis release energy in small steps rather than one explosive reaction. Electrons pass along chains of redox centres, each slightly more oxidising than the last, and the energy released is captured as a proton gradient. Cytochromes are key links in these chains. Unlike haemoglobin, which keeps its iron as Fe(II), a cytochrome shuttles its iron between Fe(III) and Fe(II), accepting one electron and passing it on.
Core explanation
Types of cytochrome. Cytochromes are classified by their haem:
- b-type contain protohaem (the same haem as in haemoglobin), held by non-covalent interactions and axial ligands. - c-type have haem covalently attached by two thioether bonds to cysteines in a Cys–X–X–Cys–His (CXXCH) sequence; the His becomes an axial ligand. - a-type carry heme a, which has a formyl group and a long hydrophobic farnesyl-derived tail; they are found in cytochrome c oxidase.
Each type absorbs light at slightly different wavelengths, which is how they were first distinguished in cell extracts.
Axial ligation and spin state. Electron-transfer cytochromes are usually six-coordinate and low-spin in both oxidation states. Mitochondrial cytochrome c uses histidine and methionine; cytochrome b₅ uses two histidines. With no open coordination site, the iron cannot bind O₂ or other small molecules, which prevents unwanted side reactions. Low-spin Fe(III) (d⁵) and Fe(II) (d⁶) have similar Fe–N bond lengths, because the added electron enters a non-bonding t₂g-type orbital. The reorganisation energy is therefore small, and electron transfer is fast.
Tuning the potential. The same Fe(III)/Fe(II) couple spans a wide range in different cytochromes, from roughly −0.1 V to above +0.3 V. Factors include:
- Axial ligands: a soft methionine thioether stabilises Fe(II) and raises E°′ relative to a bis-histidine site. - Haem exposure: burying the haem in a hydrophobic protein disfavours the extra positive charge of Fe(III), raising E°′. - Nearby charges and hydrogen bonds, and haem substituents.
Mitochondrial cytochrome c has E°′ ≈ +0.25 V, positioned between complex III and complex IV.
The respiratory chain. Electrons from NADH (E°′ ≈ −0.32 V) flow through complex I, ubiquinone and complex III (containing cytochromes b and c₁ and a Rieske iron–sulfur centre) to cytochrome c. This small, water-soluble protein moves in the intermembrane space and delivers one electron at a time to complex IV, where CuA, heme a and the heme a₃–CuB site reduce O₂ (E°′ ≈ +0.82 V for O₂/H₂O). The total span of about 1.1 V is divided into steps, each coupled to proton pumping.
Electron tunnelling. Electrons tunnel through protein between centres that are often 10–15 Å apart. The rate falls roughly tenfold for every additional 1.7 Å of separation, so redox centres in a chain are spaced close enough for fast transfer but kept apart to prevent short circuits. Marcus theory links the rate to the driving force (−ΔG°) and the reorganisation energy.
Formulae
ΔG°′ = −nFΔE°′, where ΔE°′ = E°′(acceptor) − E°′(donor), n = electrons transferred, F = 96 485 C mol⁻¹.
Step-by-step reasoning
1. List the redox centres and their E°′ values. 2. Electrons flow spontaneously from lower (more negative) to higher (more positive) E°′. 3. Calculate ΔE°′ = E°′(acceptor) − E°′(donor) for each step. 4. Convert to free energy with ΔG°′ = −nFΔE°′; negative values indicate favourable steps. 5. Check distances and reorganisation energy to judge whether the step can be fast.
Visual explanation
Draw a vertical potential axis from −0.4 V at the top to +0.9 V at the bottom. Place NADH near the top, then ubiquinone, cytochrome b, cytochrome c₁, cytochrome c, cytochrome a and a₃, and O₂/H₂O at the bottom. Arrows running downhill show the electron path; beside each large drop, draw protons crossing a membrane.
Real-world analogy
A bucket brigade passes water hand to hand down a hillside. Each person stands slightly lower than the previous one, so every pass is easy, and no single person has to carry the bucket the whole way. Cytochromes each hold one electron briefly, and the downhill arrangement of potentials keeps the flow moving in one direction.
Real-world example
Cyanide and carbon monoxide inhibit cytochrome c oxidase by binding the heme a₃ site, blocking the final step of the chain. Electrons then back up through all earlier cytochromes, and cells cannot use oxygen even when plenty is available. Separately, release of cytochrome c from mitochondria into the cytoplasm is a signal that triggers programmed cell death.
Why?
Why are electron-transfer haems six-coordinate while oxygen-binding haems are five-coordinate? A cytochrome's job is to move electrons without reacting with anything else. Filling the sixth position blocks O₂ binding and locks in a low-spin state with minimal structural change on reduction. An oxygen carrier or oxygenase needs that open site.
Common misconception
"Electrons travel along a continuous metal wire through the protein." There is no conduction band in a protein. Electrons tunnel through bonds and space between discrete centres, and the rate depends sharply on distance, driving force and reorganisation energy.
Worked example
Question: Calculate ΔG°′ for transfer of one electron from cytochrome c (E°′ = +0.25 V) to the O₂/H₂O couple (E°′ = +0.82 V).
Reasoning: ΔE°′ = 0.82 − 0.25 = 0.57 V. ΔG°′ = −1 × 96 485 × 0.57 = −55 000 J mol⁻¹.
Answer: About −55 kJ per mole of electrons; reducing one O₂ requires four such electrons, releasing about −220 kJ mol⁻¹ under standard biochemical conditions.
Quick check
1. Why does cytochrome c have a small reorganisation energy for its Fe(III)/Fe(II) change? Answer: Both states are low-spin six-coordinate, and the added electron occupies a non-bonding orbital, so bond lengths barely change.
Exam focus
Classify cytochromes a, b and c by haem type and attachment. Use E°′ values to predict electron flow and calculate ΔG°′ = −nFΔE°′, keeping signs consistent. Explain how axial ligands and haem burial tune potential, and why six-coordinate low-spin haem suits electron transfer.
Advanced insight
Complex III uses the Q cycle, in which a two-electron quinol donates one electron to the high-potential Rieske–cytochrome c₁ branch and the other to the low-potential b hemes. This bifurcation doubles the protons translocated per electron pair delivered to cytochrome c. The Rieske protein domain physically moves between sites during turnover, showing that conformational change can gate electron transfer as well as distance.
Summary
Cytochromes are haem proteins that transfer single electrons by Fe(III)/Fe(II) cycling. They are classified as a, b or c by haem structure and attachment. Six-coordinate, low-spin sites give fast transfer with small reorganisation energy and no side reactions. The protein tunes E°′ through axial ligands, burial and local charge, arranging centres so electrons flow downhill from NADH to O₂ with energy captured in steps.
Practice questions
1. How is haem attached in c-type cytochromes? Answer: Covalently, through two thioether bonds to cysteines in a CXXCH sequence, with the histidine acting as an axial ligand. 2. Predict the effect on E°′ of replacing the methionine ligand of cytochrome c with histidine. Answer: E°′ decreases, because histidine stabilises Fe(III) more than the soft thioether does. 3. Calculate ΔG°′ for one electron moving from a centre at −0.10 V to one at +0.25 V. Answer: ΔE°′ = 0.35 V; ΔG°′ = −96 485 × 0.35 ≈ −34 kJ mol⁻¹. 4. Why are redox centres in a chain usually within about 15 Å of each other? Answer: Tunnelling rates fall steeply with distance, so closer spacing is needed for electron transfer fast enough to support respiration.