BIBC102.SU05

Lecture 10

August 22, 2005

PHOTOSYNTHESIS

BIBC 102 SU05.2

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1

2

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6

7

8

9

Normalized Score

No

. of

Stu

den

ts

50 60 70 80 90 100 % 60 72 86 96 108 120 points

F (<60 pts) D (>60 pts) C (>72 pts) B (>86 pts) A(>102 pts)

MIDTERM

PHOTOSYNTHESIS

LNC Chapter 19 Section 19.6

LNC Fig. 19-36

LNC Fig. 19-37

LNC Fig. 19-38

The Light Reactions

Photophosphorylation

LNC Fig. 19-39

The Physics of Light Absorption:

E = h = hc/

h = Planck’s constant (6.626 x 10-34 J.sec)

c = speed of light (2.998 x 108 m/sec in vacuum)

= wavelength (380 - 700 nm for visible region)

1 mole of photons (1 Einstein) with = 700 nm has the energy of

171 kJ

CHLOROPHYLL

LNC 19-40a

PHYCOERYTHROBILIN

LNC 19-40b

-CAROTENE

LNC 19-40c

Nature 416, 807 - 808 (2002)

Coprophagy: An unusual source of essential carotenoids

J. J. NEGRO*, J. M. GRANDE*, J. L. TELLA*, J. GARRIDO†, D. HORNERO†, J. A. DONÁZAR*, J. A. SANCHEZ-ZAPATA‡, J. R. BENÍTEZ§ & M. BARCELL§

* Department of Applied Biology, Estación Biológica de Doñana, CSIC, Pabellón del Perú, Avda María Luisa, s/n 41013 Seville, Spain† Food Biotechnology Department, Instituto de la Grasa, CSIC, Avda Padre García Tejero 4, 41012 Seville, Spain‡ Area de Ecología, Departamento de Biología Aplicada, Universidad Miguel Hernández, 03312 Alicante, Spain§ Zoo de Jerez, Taxdirt s/n 11404 Jerez de la Frontera, Spain

e-mail: negro@ebd.csic.es

The rare Egyptian vulture (Neophron percnopterus) stands out among the Old World vultures (Family Accipitridae ) because of its brightly ornamented head, which is coloured yellow by carotenoid pigments, and its practice of feeding on faeces. Here we show that Egyptian vultures obtain these pigments from the excrement of ungulates . To our knowledge,

LUTEIN (XANTHOPHYLL)

LNC 19-40d

LNC 19-41

LNC 19-44

Fate of a photon absorbed by a molecule:

1. Internal conversion: kinetic energy; rotations and vibrations

2. Fluorescence: a photon is re-emitted, but at longer

3. Exciton transfer:

resonance energy transfer from one molecule to another nearby molecule

4. Photooxidation:

the excited molecule actually loses an electron to form acationic free radicalthe free radical becomes a strong oxidizing agent

LNC Fig. 19-42

A light-harvesting complex with 7 chlorophyll a, 5 chlorophyll band 2 lutein molecules (a monomer is shown)the functional unit of a LHC is a trimer

LNC 19-45

Exciton transfer

Photo-oxidation

LNC 19-51b

A trimeric photosystem I viewed from the thylakoid lumen perpendicular to the membrane

LNC 19-51c

Photosystem I (monomer) with antenna chlorophylls and carotenoids

Reaction center

LNC 19-49

pheophytin

plastoquinone

mobilequinone

Cyt b6fcomplex

Plastocyanin(Cu)

special acceptorchlorophyll

phylloquinone

iron-sulfur center

Ferredoxin ([Fe-S]

flavoprotein-ferredoxinNADP+oxidoreductase

PHYLLOQUINONE = A1 (in previous Figure)

redox center

hydrophobic membrane anchor

LNC 19-51a

Photosystem I

FX , FA , FB and ferredoxinare

[Fe-S] proteins

From PHOTOSYSTEM II

NADP+

From:

A. Zouni et al. Nature 409, 739-743 (2001)

Structure of Photosystem II of Synechococcus elongatus

LNC 19-56

LNC 19-49

LNC 19-54b

The cytochrome b6f complex

LNC 19-54c

The cytochrome b6f complex

To reaction center of PS I

LNC 19-57

LNC 19-53

LNC 19-52

cyclic photophosphorylation

No oxygen producedNo NADP+ reduced

ATP is synthesized

bacteriorhodopsin

LNC Fig. 19-59

LNC Fig. 19-59b

2 H2O + 2 NADP+ O2 + 2 NADPH + 2H+

½ O2 + 2 H+ + 2e- H2O E’o = + 0.816 volts

NADP+ + H+ + 2e- NADPH E’o = - 0.324 volts

G’o = - 2 x 96.5 kJ/Vmol x (-1.240volts) = + 239 kJ/mol i.e. we need 2 x 239 kJ to make 2 moles of NADPH

One mole of photons has ~180 kJ/EinsteinFrom 8 moles of photons we get ~ 1440 kJ

We also pump protons to set up a protonmotive force that can be used to make ATP

From measurements: we make ~ 3 moles ATP per mole of O2 produced

End of lecture 10Aug. 22, 2005