Reactions of the Oxygen Cycle

Question 1

What does this diagram show and what can we deduct from it?

Answer 1

• nEθ = measures energy and hence shows the energy of the different oxygen intermediates
• The normal metabolism of dioxygen releases the greatest amount of free energy
• This metabolic pathway avouds the release of O₂⁻ and H₂O₂
• (only really note the last reaction)

Question 2

Cytochrome c deliver in electrons from the electron transport chain into a protein caalled cytochrome c oxidase
what does cyctochrome c do?

Answer 2

transport electrons which results in the eventual reduction of oxygen

Question 3

What characteristics of the Cyt a₃ - CuB make it good at transporting electrons?

Answer 3

It has an average oxidation state
Cu₂(II) + e- → Cu₂(1.5)
meaning rapid electron transfer

Question 4

What characteristics of Cuₐ make it a good transport protein?

Answer 4

• Mixture of hard and soft ligands
• Compromise between Cu(I) and Cu(II) states
• Again, allowing for rapid electron transfer

Question 5

What techniques would we use to find out the electronic structure of the following centre

Answer 5

• EPR - As Cu(II) is paramagnetic and Cu(I) is diamagnetic
• XAS - to look at oxidation state and metal-ligand covalency

Question 6

Describe the properties of the following Cyt a redox shuttle

Answer 6

• Cyt a is a low spin Iron²⁺⁄ ³⁺ haem centre
• It is a regular cytochrome
• It is a 6 coordinate haem with a low spin
• Allows rapid electron transfer: Fe(II) + e⁻ → Fe(II)

Question 7

Describe the properties of the following Cyt a₃ - CuB

Answer 7

• The top centre This centre is similar to hemocyanin complex (phenol side-chain called tyrosine as part of a post-translational modificationto the histinine)
• The bottom centre is reminiscent of the iron centre in deoxy haemoglobin
• The oxygen goes between the Iron and Copper centres

Question 8

O₂ + 4H⁺ + 4e ⇌ 2H₂O
is the reaction which occurs at the active site
Why must 4e- be added in one go?

Answer 8

• If we want to break the oxygen-oxygen bond and get a bond order of zero
• We have to put in 4 electrons all in one go
• This avoids the formation of a reactive O₂⁻ and O₂²⁻ species

Question 9

How is 4e- added in a row to oxygen at

Answer 9

• 1 electron from: Cu(I) → Cu(II) +e⁻
• 2 electrons from: Fe(II) → Fe(IV) + 2e⁻
• Tyrosine to tyrosol radical = 1 electron
• = 4 electrons total given to the O₂ unit in one step and breaks the O₂ bond

Question 10

How is the resting state of the enzyme recovered?

Answer 10

• Needs to gain 4e- and 4H+

Question 11

The first proton is gained by the copper centre
Forming a

Answer 11

Forming a copper hydroxide unit
Cu(II) – O⁻ + H⁺ → Cu(II) – OH

Question 12

The second electron is gained by tyrosol to form tyrosine again
The next proton and electron are gained by Iron(II) to form

Answer 12

Form Iron(III) hydroxide
Fe(IV)=O + e⁻ + H⁺ → Fe(III)-OH

Question 13

What happens once Iron(III) hydroxide and Copper(II) hydroxide are formed

Answer 13

They react together to form Fe(II), Cu(I) and 2x water
Fe(IV)-OH + Cu(II)-OH + 2e⁻ + 2H⁺ → Fe(II) + Cu(I) + 2H₂O
Original centres for the next dioxygen to react with

Question 14

What is Cu,Zn-Superoxide Dismutase do?

Answer 14

• This enzyme protects against the formation of superoxide (O₂*⁻)
• It takes two superoxides with two protons and generates O₂ and hydrogen peroxide
• 2H⁺ + 2 O₂*⁻ → O₂ + H₂O₂
• (Disporportion reaction: oxygen is oxidised and hydrogen peroxide is reduced)
• Reaction is catalysd by copper-zinc centre (zinc has no role in oxidation however)

Question 15

What is the role of the zinc in Cu,Zn-Superoxide Dismutase?

Answer 15

• Holds the imidazole unit in place close to the Cu(II)
• The Cu(II) is bound to a imidazolate (deprotonated)
• A H⁺ binds to imidazolate to from imidazol and then Cu(II) takes an e⁻ from superoxide to make oxygen
• Cu(II)-imidazolate⁻ + H⁺ + O₂*⁻ → Cu(I) + H⁺-imidazolate⁻

Question 16

How does the Cu,Zn Superoxide dismutase reform after dioxygen is formed from superoxide

Answer 16

• Cu(I) + H⁺-imidazolate⁻ + H⁺ + O₂*⁻ → H₂O₂ + Cu(II)-imidazolate⁻
• Reforms the Cu(II)-imidazolate⁻ from Cu(I)
• Called a Cu(II)↔ Cu(I) “ping-pong” mechanism

Question 17

The Cu,Zn Superoxide dismutase forms a side product of hydrogen peroxide when removing superoxide
How does your body combat this?

Answer 17

• Catalase protects against hydrogen peroxide
• 2H₂O₂ ⇌ O₂ + 2H₂O

Question 18

• The disproportionation of H₂O₂ occurs in two steps:
• Fe(III) +H₂O₂ ⇌ Compound I + H₂O
• Compound I + H₂O₂ ⇌ Fe(III) + H₂O + O₂
• What is compound 1?

Answer 18

• Compound I can be considered a haem species with a Fe(V)=O centre
• (charge is donated from the porphyrin ring to stabilise this high oxidation state of Fe(V))

Question 19

What techniques could we use to probe the compound I centre

Answer 19

• EPR - using dₙ electron account and high/low spin
• XAS