Week 8 Johnson Lecture 4

Question 1

What does PSII do?

Answer 1

A biological enzyme that is capable of oxidising H2O to O2 and It reduces PQ to PQH
the special pair chlorophylls P680 are the primary electron donor

Question 2

What is the redox scheme for PSII?

Answer 2

To achieve the splitting of water a redox potential (the difference) of greater than +820 mV is required. The P680+/P680 redox couple has a redox potential of +1200 mV while Yz+/ Yz has a redox potential of +950 mv, thus they can act as oxidants H2O/O2 couple.The P680*/P680+ redox couple has a redox potential of -630 mV

Question 3

How much energy does light contain?

Answer 3

Calculation;
ΔG=(Nhc)/λ
(gives us the energy of one mole of photons)
N = Avogadro’s number = 6.02 × 1023
h = Planck’s constant = 6.63 × 10-34 j × s
c = speed of light in a vacuum = 3 ×108 m s-1 λ = wavelength of the light
e.g. 1 mole of 680 nm (680 × 10-9 m) photons
E=((6.02 × 1023) × (6.63 × 10-34) × (3 × 108))/(680 × 10-9) ΔG = 176085 J mol-1
ΔG = 176 kJ mol-1

Question 4

What are the energetics of the PSII reaction?

Answer 4

Overall reaction:
2H2O + 2PQ + 4H+stroma —–> O2 + 2PQH2 + 4H+lumen
oxidises H2O to O2 (E0’ =+820 mV) it reduces PQ to PQH2 (E0’ =+80 mV)
ΔE0’ = E0’(acceptor)- E0’(donor) ΔE0’ = +80 - (+820)
ΔE0’ = -740 mV
ΔG = -nFΔE0’ (n = moles of electrons transferred, F= Faraday constant) ΔG = -4 × 96.5 × 10-3 ×-740
ΔG = +285 kJ mol-1
This free energy input is provided by 4 x 680 nm photons
(176 kJ mol-1 each = 704 kJ mol-1) 285/704 = 40% efficiency

Question 5

What do the cofactors that take part in the electron transfer reactions in PSII do? and what are they?

Answer 5

At the heart of the PSII reaction centre are several key co- factors which take part in electron transfer reactions from water to plastoquinone.
they are ;
-Fe2+
-Plastoquinone (Qa)
-Plastoquinone (Qb)
-Pheophytin (PheoA)
-Pheophytin (Pheob)
-Chlorophyll a a 
-Chlorophyll a b
-Chlorophyll a P680A/P680B
-Tyrosine161 (Yz)
-Manganese cluster (Mn4CaO5)

Question 6

How does PSII work?

Answer 6

Basically

• Chlorophyll a P680A/P680B is excited, charges separate to positive and negative
• The negative electron is donated from the special pair to accessory pigments; Chlorophyll a a to Pheophytin A to Plastoquinone A to Plastoquinone B
• Positive charge is still in special pair
• Manganese cluster attracts water molecule and an electron from water is taken and transferred to special pair, transfer occurs via tyrosine161
• P680 is now regenerated
• Proton from manganese cluster leaves stroma
• -It takes 2 turnovers (2 photons) to make 1 PGH2
• -it takes 4 turnovers (4 photons) to male 1 O2

Question 7

What is plastoquinone?

Answer 7

Plastoquinone, a lipid soluble electron carrier shares a very similar structure to ubiquinone, which is present in mitochondria and bacteria. The 2 e- are used to reduce the two C=O carbon atoms from +2 to +1 oxidation state, forming two -OH groups.

Question 8

What ensures that the charges are kept separate?

Answer 8

To ensure this, a chain of electron acceptors which are energetically ‘downhill’ from P680* ensures e- and hole (+) are rapidly separated spatially (electron transfer efficiency decays exponentially with distance), slowing down the reverse reactions and stabilising charge separation. This minimises energy loss as heat via recombination of ChlA- and P680+