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Rutherford > Photosystem II > Flashcards

Photosystem II Flashcards

1
Q

What is the enzyme name for PSII

A

Water/ plastaquinone photo oxidoreductase

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2
Q

Overall reaction for PSII

A

2H2O + 2PQ (plastaquinone) —> 2PQH2 (plastaquinol) + O2

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3
Q

Water oxidation half reaction

A

2H2O —> 4H+ + 4e- + O2 (Em +820mV)

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4
Q

Quinone reduction half reaction

A

PQ + 2H+ + 2e- —> PQH2 (Em +100 mV)

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5
Q

Delta E for PSII

A

-720 meV (delta G = 72kJ/mol) UNFAVOURABLE - energy input = light

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6
Q

How many volts does a 680nm photon correspond to

A

1.82 eV

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7
Q

Components of PSII

A
  1. D1 and D2 2. CP43 and CP43 3. Cytb559 4. Extrinsic polypeptides 5. Dozens of other small subunits
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8
Q

What are the two central proteins that are next to each other

A

D1 and D2

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9
Q

D1 and D2

A

form pseudo homodimer each of 5 TM helices and symmetrical - location of all redox cofactors needed for enzyme function

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10
Q

CP43 and CP43

A

peripheral antenna (light collection) - both 6 Tm helices and symmetrical - contain chlorophyll and carotenoid (also provide AAs to active site)

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11
Q

Cytb559

A

2 TM helices with Hemeb

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12
Q

Extrinsic polypeptides

A

insulate active site

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13
Q

How many chlorophylls does PSII contain

A

6

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14
Q

What absorbance for PSII

A

680 nm (nm at which it loses its colour - bleach)

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15
Q

Main cofactors

A
  1. Chlorophyll (4) 2. Pheophytin (2) 3. Quinone (2) 4. Manganese ions (4) 5. Calcium ion (1)
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16
Q

What’s the redox active amino acid in PSII

A

Tyrosine

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17
Q

Structure of pheophytin

A

Free base form of chlorophyll i.e. no metal - 2 H+ instead (easier to reduce)

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18
Q

Chl/Chl(minus radical)

A

Em = -1V (very reducing)

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19
Q

Chl/Chl (plus radical)

A

Em = +0.85V (very oxidising)

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20
Q

PD1+/ PD1

A

Em = 1.25V

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21
Q

Ph/Ph-

A

Em = -500mV

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22
Q

Q/QH2

A

Em = 100mV

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23
Q

Is semiquinone a strong oxidant or reductant

24
Q

Qa

A

1 e- chemistry at fairly low potential (no H+)

25
Q

Qb

A

2 e- chemistry - stabilised by protonation of nearby amino acid and its electrostatic environment

26
Q

Tyr(radical)/Tyr

A

Em = 1.2V (very oxidising)

27
Q

How is Tyr radical stabilised

A

Histidine H-bonding partner and by H-bonds to waters

28
Q

Manganese cluster

A

Mn4CaO5 Em = 1V

29
Q

What are the ligands for the manganese cluser

A

Carboxylic acids and histidine

30
Q

So how does PSII work

A

Excitation by light (ground state to excited)

31
Q

What happens to the electron when there’s excitation by blue light

A

Electron enters 2nd unoccupied molecular orbital (2nd excited singlet state)

32
Q

What happens when there’s excitation by light in light collecting proteins

A

Energy transfer (because the pigments are close together)

33
Q

What happens in reaction centres when there’s excitation by light

A

Charge separation (electron transfer)

34
Q

Energy conversion

A

Wavelength (nm) = Plancks constant x speed of light / electron volts (eV)

35
Q

What is hc

A

1240 nm/eV

36
Q

Wavelength of blue light

37
Q

Kinetic control

A

To avoid going back and recombination

38
Q

What controls electron transfer rate

A
  1. Distance (r) 2. Driving force (delta G or E) 3. Reorganisation energy (lambda) 4. Intervening medium (beta)
39
Q

How much is lost due to back reactions

40
Q

In what step is there a big loss in energy

A

Electron transfer from Ph anion to quinone

41
Q

What is the role of tyrosine

A

Electron carrier between chlorophyll (PD1) and the Mn4Ca active site

42
Q

What is the role of His

A

To accept proton when tyrosyl radical formed

43
Q

The S state cycle

A

Flash dependence of water splitting. Every fourth flash, electrons ripped out and O2 comes out. Move up oxidation each flash

44
Q

What does the S state cycle show from is trend

A

Dampening effect (10% loss)

45
Q

What is the most reduced S state

46
Q

What is the confusion in the redox cycle for water oxidation

A

Whether Mn5+ is made or Tyr is oxidised AND where does water come in

47
Q

Main features of redox cycle

A
  1. In most oxidising state (S4) - 2 molecules of H2O are oxidised in a 4 electron process with no intermediates
  2. Other steps in the cycle (S0-S1-S2-S3-S4) involve oxidation of active site components (Mn, Tyr)
  3. Charge accumulation steps are mainly compensated by H+ release to the lumen - these are ‘chemical protons’ that came from the water that is oxidised
48
Q

Why is water oxidised in one step

A

To prevent formation of reactive oxygen species

49
Q

How many volts produced by PSII per photochemical turnover

50
Q

What is special about manganese cluster

A

Variation in position of oxygen 5 (2 states - Mn3+ or Mn4+) - acid base chemistry or radical chemistry

51
Q

Overall how much is lost due to back reactions in PSII

A

600meV (but probably lose another 150-200 due to e- from 4 states and QH2)

52
Q

What are the real life conditions

A

pH in lumen = 6. pH in stroma = 7.8

53
Q

Real Em values

A

H20 half RX Em= 879mV not 820mV. PQ half RX Em=60mV not 100mV (total = -819meV)

54
Q

Total energy used

A

600meV + 150/200meV + 820 meV = 1.62eV (1.82 in from red photon)

55
Q

What could be used for the extra energy

A

Oxygen release and extra driving force

Rutherford (4 decks)

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