Photosystem II Flashcards
What is the overall equation for PSII’s catalytic activity?
4 Photons are needed to catalyse the following reaction:
2H2O + 2PQ + 4H+ (stroma)–>
O2 + 2PQH2 + 4H+ (lumen)
What is the active residue that allows PSII’s activity to occur?
A redox-active Tyrosine residue, a.k.a. Tyr161.
What are the major effects of PSII?
1) Transport of (four) protons against the concentration gradient (stroma to lumen), contributing to the PMF (∆p), which converts light energy to chemical energy.
2) Reduction of Plastoquinone (PQ) to Plastoquinol (PQH2). This happens twice per O2.
3) Output of an O2 molecule and 2 PQH2 molecules.
Where do electrons ultimately come from?
They ultimately come from water - with the manganese cluster ripping electrons off of water as it donates its own to P680.
The (water oxidation) S-state cycle
2H2O is overall converted to O2 (molecular oxygen) and 2H+. The Manganese cluster is reset in the process.
Equations to calculate efficiency
1) ∆G = N x h x c / wavelength (m)
Note - N = Avogadro’s constant, h = Planck’s constant, and c = the speed of light (constant)
2) ∆E = ∆E(acceptor) - ∆E(donor)
3) ∆G = -n x F x ∆E
Note - n = number of moles (of photons), F = Faraday’s constant, ∆E = result of equation 2
4) Harvested/Theoretical energies, or result of 3/(1xn)
Usually this leads to a result around 40% efficiency for photosynthesis.
Lowest redox potential in nature?
P700* at -1320mV!
What does PSII itself actually achieve?
Oxidation of H2O and Reduction of PQ.
What happens when P680 is excited?
The electron is immediately transported from P680* to accessory pigments - ChlA, then PheoA. From PheoA it goes to PQ-A, then PQ-B. This is called ‘charge separation’ and leaves an ‘electron hole’.
An electron is donated from the manganese cluster (thanks to Yz/Tyr161) to P680, leading to PQ-B gaining 2H+. (Since PQ is hydrophobic it can traverse the membrane, hence H+ transport).
Note - Pheo is chlorophyll but coordinates 2H+ not Mg2+. Hence brown (sensitive π - electron system).
Where are the electrons that get excited held?
In the π - electron system of the tetrapyrrole ring of chlorophyll, particularly those in the reaction centre of PSII.
Why doesn’t the electron hole get filled by the excited electron?
The charge separation occurs due to Marcus Theory’s postulation of inversion, where because the driving force back down to P680 is high, charge recombination through this route is slow.
Because the cofactor chain of ChlA/PheoA/PQa/PQb has much smaller driving force that matches the reorganisation energy much better, this occurs much faster and is favoured.
Note - energy lost in creating charge separation is worth it for the stability of the system (a.k.a. preventing charge recombination).
What does the PSII section of the redox Z-scheme look like? Where does the e- ultimately come from?
H2O –> P680/P680+ –> ChlA –> PheoA –> PQ-A –> PQ-B
In the simple Z-scheme model, PQ refers to PQ-B. ChlA and PheoA are omitted.
The e- comes from water ultimately, as it’s routed through the manganese cluster and further to P680 through Yz (Tyr161) catalysis
