During photosynthesis, a process that occurs in plants and in cyanobacteria and also in purple photosynthetic bacteria, light energy is converted into chemical energy. The overall process is described by the chemical equation:
CO2 + H2O → (CH2O) +O2 (requires photons, hv) (1)
Notice that the oxidation state of the carbon atom has changed from +4 to 0, and so an overall reduction reaction has occurred. Here, "(CH2O)" represents a carbohydrate. You'll also notice that there's been a change in the oxidation state of oxygen, since its oxidation state is almost always "-2" when combined with other atoms, but it is "0" in its natural state of O2. So, oxygen (water) is oxidized in the overall process.
It turns out that the overall reaction of photosynthesis takes place in two steps. In the first step, the oxygen in water is oxidized by the light energy:
2H2O → O2 + 4 [H.] (requires photon energy) (2)
Here, [H.] represents a reducing agent. In the second step, the [H.] reduces the carbon in CO2:
4 [H.] + CO2 → (CH2O) + H2O (3)
When one adds these two reactions together, the overall reaction (1) results. [H.], the reducing agent, is an "intermediate" in the overall reaction. It looks like two H2O molecules are needed in reaction (2) but then you get one of them back in reaction (3).
Reaction (2) is known as the "light reaction" and reaction (3) is known as the "dark reaction" but both can take place in the light. However, (2) is light-dependent, whereas (3) is not.
You have already studied the "dark reaction" and I will refer you to Dr. Diwan's notes on the subject. As the overall process of photosynthesis involves a series of electron transfer reactions, we are in the realm of oxidation-reduction chemistry, and it would help to review the basics of these processes because we will be going into this topic in greater depth. There is a direct analogy to electron transfer in the mitochondrion, in which clumps of energy are transferred from one electron carrier to another along a "chain" and H+ ions are translocated out, across the mitochondrial membrane, thus generating an electrochemical gradient. The energy inherent in this gradient is used to synthesize ATP in the process of "oxidative phosphorylation." The same processes occur in photosynthesis and the chloroplast, the site of photosynthesis in plants and blue-green algae (but not in photosynthetic bacteria), is the analog of the mitochondrion in eukaryotes.
Chapter 22 ("Electron Transport and Oxidative Phosphorylation") in Voet & Voet (3rd Edition) is one of the most important chapters in the entire text (at least in my opinion) and it would help to reread it as you look at the light reaction of photosynthesis in more detail over the next two lectures.
A Brief Review of the Interaction Between Light (Photons) and Matter
The first step in photosynthesis is the interaction of light with chlorophyll molecules. The chemical structures of the various chlorophyll molecules are based upon the cyclic tetrapyrrole that is also seen in the heme group of globins and cytochromes. Various modifications of this group, namely ring saturation characteristics and substitutions on the rings provide a series of pigment molecules that, as a group, absorb effectively over the wavelength range of 400 nm - 700 nm, the spectrum of visible light. It is the high degree of conjugation of these molecules that makes them so efficient as absorbers of visible light.
The first point, then, is that photons are absorbed by molecules when their electronic spectra have strong absorption bands around the frequency of the incoming photon, as the chlorophylls do. In the process of "absorption" of a photon, a molecule is "excited" to a state of higher energy because an electron has moved from a ground state molecular orbital energy level to a higher energy level. Only certain transitions can occur, however, because the frequency corresponding to the energy difference must match the frequency of the photon. This is a statement of the energy conservation law and can be written as DE=hv, and the peaks that you see in the absorption spectrum of a molecule correspond to the possible energy level differences that allow absorption of photons of corresponding frequency.
A molecule in an excited electronic state must dissipate the extra energy in some way, and we can talk about different modes of "decay" of that energy to a lower electronic state (which does not necessarily have to be the ground state). There are three types of dissipation that can occur: "non-radiative," "radiative," and chemical reaction.
Non-radiative Dissipative Processes: See Energy Level Diagram
(1) Vibrational Cascades: If the excited state is one in which the vibrational energy level is also excited, then this vibrational energy can be transferred to other molecules via collisions, and the energy appears as heat as the molecule relaxes to its ground vibrational state in the excited electronic state.
(2) Internal Conversion: If two molecular potential energy curves intersect, then an isoenergetic transition can occur from a higher electronic energy state to an upper vibrational energy level of a lower electronic state. This can then be followed by a vibrational cascade, the energy again appearing as heat. This is what we will see in the case of chlorophyll, but, in this instance, the relaxation is to the lowest excited state.
(3) Intersystem Crossing: An isoenergetic transition between states of different spin angular momentum, followed by a vibrational relaxation cascade.
(1) Fluorescence: Light (photons) emitted when an electronically excited molecule decays to a lower state of the same multiplicity. About 3-6% of the light absorbed by plants is dissipated in this way.
(2) Phosphorescence: Same idea as fluorescence, but the transition is between states of different multiplicity, and the "lifetime" is longer in phosphorescence.
(3) Exciton Transfer (Resonance Energy Transfer): Transfer of energy to a nearby unexcited molecule with similar electronic properties. This can happen because the molecular orbital energy levels of the molecules overlap. This mechanism will play an important role in photosynthesis.
(3) Reaction with another molecule, including transfer of its energy to another molecule, which can then react
(4) Photooxidation: Transfer of an electron to an acceptor molecule. In photosynthesis, an excited chlorophyll molecule, Chl*, donates an electron, thereby becoming oxidized to the cationic free radical, Chl+.
Photosynthesis in Purple Photosynthetic Bacteria
We'll look at a simpler example of photosynthesis first, and use it as an introduction to photosynthesis in plants and cyanobacteria (blue-green algae). Although the primary reactions of photosynthesis take place at "photosynthetic reaction centers," the first level of interaction of light with an organism that carries out photosynthesis is at an assembly of chlorophyll molecules that "harvest" light (the "light-harvesting complex"). Such an assembly results in a greater chance that photons will be captured and, because of the strategic arrangement of the individual chlorophyll and other accessory light-absorbing molecules, the transfer of energy to the photosynthetic reaction center is very fast (< 10-10 s) and very efficient (>90%).
NOTE TO SPRING 2007 STUDENTS: INSTEAD OF 1LGH, VIEW THE STRUCTURE 1KZU, THE INTEGRAL MEMBRANE LHC FOR Rh. acidophilus, WHICH IS SIMILAR IN STRUCTURE TO 1LGH. www.RCSB.org
The ring of helices
The light-absorbing molecules (e.g., chlorophyll)
The Mg(2+) ions
Additional ligands of the Mg(2+) ions (make some educated guesses as to what side chains might provide electron pair donors)
Both the LHC and the reaction centers are membrane bound structures but there are no chloroplasts in the purple photosynthetic bacteria. The electron transfer processes occur within the cell membrane and the overall process is a cyclic one (i.e., there is no net oxidation-reduction). Protons are transferred across the membrane, from the cytoplasmic side to the outside, establishing a proton gradient whose dissipation drives ATP synthesis. A similar situation holds for the cyanobacteria and plants, but in these organisms, the process occurs in chloroplasts and the overall reaction is not a cyclic one.
Describe the geometric relationship between the two BChl molecules of 2RCR at the 12:00 position. These two BChl molecules constitute the "special pair" and light of wavelength 870 nm is preferentially absorbed here; hence the alternate designation as "P870."
The two symmetrically-related sets of molecules do not function in the same way, however. The flow of electrons is from P870 to the right-sided set, which is within the L subunit, and then over to the left-sided set, which is within the M subunit. The intervening Fe(II) atom does not directly participate in the redox reactions but it is part of the overall "circuit" that eventually ends where it started, at the special pair.
Specifically, the reactions through the L-subunit are:
(1) Excitation of an electron at the special pair as a result of absorption of a photon channeled to it by the LHC.
(2) Delocalization of the excited electron over both BChl a molecules of the special pair.
(3) Transfer of the electron in the excited state of P870 (P870*) to BPheo a, through an intervening accessory BChl a molecule, resulting in P870 +BPheo a-. The net result here is that the BChl a of the special pair has been photo-oxidized while the BPheo a molecule is reduced. This occurs on a timescale of 10-12 s, thus preventing return of the electron back to the special pair. If the latter had occurred, the excitation energy would be dissipated by the release of heat due to the process of internal conversion.
(4) Transfer of the electron to QA (or to a second QB in some cases) to form QA- (an anionic semiquinone), which is located in a hydrophobic pocket.
The electron flow through the M-subunit is as follows:
(1) QB, which is more solvent exposed than QA, accepts the excited electron to form QB-. QA is now back in its oxidized state. Fully reduced QA can hold two electrons, which QA- is the "semiquinone" form. Even though two electrons will pass from the special pair through the entire cycle for one complete photosynthetic cycle, the QA never is in its fully reduced form.
(2) QB- does accept a second electron before the next step, though, to become QB-2, which then accepts two protons from the cytoplasmic side of the membrane to form QBH2. The net effect, then, is that two electrical excitation events have been transduced to a two-electron chemical reduction event.
(3) Both electrons are carried back to P870+ through a membrane-bound electron transport chain that includes:
1. A membrane-bound pool of ubiquinone molecules,
2. A cytochrome bc1 complex, and
3. Cytochrome c2.
During this process, QH2 passes its two protons back across the plasma membrane into the periplasm when its two electrons are passed on to cytochrome bc1. However, cytochrome c2 is only a one-electron acceptor, and the transfer from QH2 occurs in a two-stage "Q-cycle" with the result that, for every two electrons that reach cytochrome c2, 4 H+ are translocated into the periplasm from the cytoplasm, thereby generating an H+ gradient. Dissipation of this chemical gradient powers ATP synthesis.
For a review of the "Q-cycle," including an animation, click here:
(4) When an electron reduces P870+, it can now absorb another photon.
What causes the overall process to be unidirectional? To understand this, we will have to say a bit more about redox reactions and irreversibility.
A Brief Review of the Thermodynamics of Redox Reactions:
Oxidation-reduction reactions ("redox reactions") involve electron transfers between reactants and, in the process, the oxidation states of atoms in those reactants change. One atom might lose one or more electrons and become oxidized, which the other will gain one or more electrons and become reduced. A balanced equation of the overall process assures that an equal number of electrons are lost and gained.
It is customary to break an overall redox reaction into two "half-reactions," one being the reduction half-reaction and the other being the oxidation half-reaction. To do this, you must be able to recognize the atoms whose oxidation states are changing. If you do not remember how to assign and/or determine oxidation states, this would be a good topic to review now in any introductory chemistry text.
Let's use the "light reaction" as an example of the thermodynamic calculations. The overall reaction is:
2 NADP+ + 2H2O → 2 NADPH + O2 + 2H+ (4)
There are two half reactions here, one involving H2O and O2 (oxidation, as written), while the other involves NADP+ and NADPH ( reduction, as written). How does one determine whether this reaction is spontaneous in the direction written? To do this, each half reaction is written as a reduction reaction and the biochemical standard (i.e., pH= 7.35-7.45) reduction potentials (E°') of each are compared. These standard reduction potentials are tabulated in most biochemistry texts, and their units are "Volts." In this example the half reactions and their corresponding standard reduction potentials are:
O2 + 4 e- + 4H+ → 2H2O E°' = .815 V (5)
NADP+ + H+ + 2 e- → NADPH E°' = -.324 V (6)
Notice that there are 4 electrons transferred in the first half-reaction, but only 2 electrons transferred in the second. In order to balance charge we had to multiply the second half-reaction by "2" in order to get (4). To determine the DE°' for reaction (4), we change the sign on E°' for reaction (5) and add it to that for reaction (6).
ΔE°' = -.324 V + (-.815 V) = -1.139 V
We did not multiply the E°' for reaction (6) by "4" because E°'s are intensive properties and they do not, therefore, depend upon amounts of material.
A (physiologic) process is spontaneous under standard conditions in the direction as written if the (physiologic) standard Gibbs Free Energy change (DG°') for the process is less than zero. Since DG°'= -nFDE°', the Gibbs Free Energy change here is :
DG°'= -(4)(96,485 J V-1 mol-1)(-1.135 V) = +439.6 kJ mol-1
Here, "n" is the number of electrons transferred and "F" is the Faraday Constant
Then reaction (4) as written is not spontaneous under standard conditions! Instead, it is driven by the energy input of photons.
Look closely at the "Table of Standard Reduction Potentials of Some Biochemically Important Half-Reactions" in your text. You will notice that, for a given half-reaction, a half reaction above it in the table has an E°' value more positive than a half reaction below it. This means that any half reaction above it will oxidize (take electrons away from it) and it, in turn, will oxidize any half reaction below it. When we follow the trail of electrons in the two schemes of the light reactions that we are about to study, you will notice that electrons will spontaneously flow from algebraically lower to algebraically higher E°' values. For each step, then, the DE°' values will be >0 and the DG°' values will be <0 (each step is spontaneous under standard conditions). If you understand this, then it is easy to understand the spontaneity (irreversibility) of each of the steps.
Link here for Power Point presentation: PHOTOSYNTHESIS.ppt
Link here for Photosynthesis Study Questions: STUDY QUESTIONS FOR PHOTOSYNTHESIS LECTURES