Photosystem I, Electron Flow Dynamics, and Photophosphorylation Stoichiometry

Photosystem I (PS I) and the P700 Reaction Center

• At Photosystem I (PS I), the reaction center utilized is designated as $P700$.

• The nomenclature "$P700$" indicates that the center absorbs light at a wavelength of $700\,nm$.

• Energy levels: Light at $700\,nm$ contains slightly less energy than the light utilized at Photosystem II (PS II), yet it provides "plenty of energy" to facilitate the necessary biochemical reactions.

• The primary process involves transferring an excited electron to a chlorophyll $a$ analog, designated as $A_0$.

• The oxidation-reduction process follows a sequence similar to Photosystem II:
      - Light excites the ground-state $P700$ to an excited state: $P700 \rightarrow P700^<em>$.     - The excited $P700^</em>$ and the chlorophyll analog ($A_0$) react such that $P700^*$ is oxidized to $P700^+$ and the chlorophyll analog is reduced (gaining an excess electron).

• Voltage and Reducing Power:
      - The change in voltage is significantly large, with a difference of $1.78\,V$ between $P700$ and the acceptor.
      - Consequently, $P700^*$ acts as an exceptionally effective reducing agent.

• Energetics:
      - The Gibbs free energy ($\Delta G$) for this excitation is approximately $171\,kJ/mol$.
      - This corresponds to the energy contained in light at $700\,nm$, which is approximately $171\,kJ/Einstein$.

Electron Flow and Carrier Proteins in Photosystem I

• The electron flow through Photosystem I shares similarities with Photosystem II but utilizes different specific carriers:
      - Antenna Molecules: Light excites an antenna molecule.
      - Exciton Transfer: Energy is transferred via exciton transfer to the "special pair" of chlorophyll molecules.
      - Oxidation: This energy transfer oxidizes the chlorophyll $a$.

• Carrier Sequence:
      - Phylloquinone: Electrons are passed to phylloquinone. Phylloquinone is characterized as a "two electron redox molecule."
      - Transition in Transfer Modes: The sequence transitions from single-electron transfers to a two-electron transfer at phylloquinone, and then reverts to single-electron transfers.
      - Iron-Sulfur Proteins: From phylloquinone, electrons move to iron-sulfur ($FeS$) proteins.
      - Ferredoxin: The ultimate receiver of these electrons in the PS I complex is ferredoxin. Ferredoxin is an iron-sulfur protein containing a $2Fe-2S$ center (designated as a $2FES$ center).

• Reduction of NADP+:
      - Ferredoxin transfers its electrons to $NADP^+$.
      - Stoichiometry of reduction: $NADP^+ + 2\text{ reduced ferredoxins} \rightarrow NADPH + 2\text{ oxidized ferredoxins}$.

• Electron Source:
      - The electrons required to reset the system are supplied by plastocyanin in its reduced state.
      - Overall reaction for the carriers: $PC_{red} + Fd_{ox} \rightarrow PC_{ox} + Fd_{red}$.
      - This specific step requires one photon because each of these carriers is a single-electron carrier.

Energetic Efficiency and Coupling with Cytochrome b6f

• Reduction Potentials: There is a substantial gap in reduction potential between plastocyanin and ferredoxin.

• Energy Capture:
      - The $\Delta G$ for the transfer of one electron from plastocyanin to ferredoxin is $74\,kJ$.
      - Given the photon energy is $171\,kJ/Einstein$: $\frac{74\,kJ}{171\,kJ} \approx 43\%$ energy capture efficiency.
      - This efficiency is slightly lower than that observed in Photosystem II.

• Coupling Mechanism:
      - Photosystem I and the cytochrome $b_6f$ complex are "relatively tightly coupled together."
      - The "extreme exergonicness" of the Photosystem I reaction helps drive the cytochrome $b_6f$ complex's reaction, which is described as being "somewhat inorganic" in nature.

Cyclic Electron Flow and Proton Gradient Management

• PS I has the capability to perform Cyclic Electron Flow.

• Mechanism: Instead of electrons flowing from ferredoxin to $NADP^+$ to create $NADPH$, electrons are transferred from ferredoxin back to plastoquinone ($PQ$).

• Flow Path: Ferredoxin $\rightarrow$ Plastoquinone $\rightarrow$ Cytochrome $b_6f$ complex $\rightarrow$ Plastocyanin $\rightarrow$ PS I.

• Utility: This mechanism allows the cell to pump protons into the lumen without producing "reduced electrons" (NADPH). This ensures the ratio of $ATP$ production is in line with $NADPH$ production, as the proton gradient drives $ATP$ synthesis.

• Cyclic Stoichiometry:
      - $2\,Fd_{red} + PQ \rightarrow 2\,Fd_{ox} + PQ_{red}$.
      - As the reduced plastoquinone ($PQ_{red}$) passes through the cytochrome $b_6f$ complex, four protons ($4\,H^+$) are moved from the stroma to the lumen.

• Simplified Summary for Cyclic Flow:
      - $2\,H^+<em>{\text{stroma}} + 1\text{ photon (at PS I)} \rightarrow 2\,H^+</em>{\text{lumen}}$.
      - It is a highly effective method for converting light energy into a proton gradient without producing oxygen ($O_2$).

• Efficiency of Cyclic Flow:
      - Photon energy: $171\,kJ/Einstein$.
      - Energy for moving two protons: approximately $35\,kJ$.
      - While the efficiency is mathematically low, the speaker notes that sunlight is abundant beyond what any plant could possibly use.

Collective Summary of Linear Photosynthesis

• To produce one molecule of $NADPH$:
      - Photosystem II: Oxidation of water ($H_2O$) pumps two protons ($2\,H^+$).
      - Cytochrome b6f Complex: Transfer of electrons to plastoquinone results in four protons ($4\,H^+$) moving into the lumen.
          - Two protons are taken up by the initial reduction of plastoquinone A ($PQ_a$).
          - Two additional protons are taken up in the Q cycle.
      - Photon Cost: Requirements include two photons at PS II and two photons at PS I (Total: 4 photons).

• The Proton Pumping Count:
      - Stroma Removal: $2\,H^+$ (reduction of $PQ_a$) + $2\,H^+$ (Q cycle) + $1\,H^+$ (reduction of $NADP^+ \rightarrow NADPH$) = $5\,H^+$ total removed from stroma.
      - Lumen Accumulation: $2\,H^+$ (oxidation of water) + $4\,H^+$ (cytochrome $b_6f$) = $6\,H^+$ total produced in the lumen.

• Standard Stoichiometry for Calculations:
      - $6\,H^+/NADPH$
      - $12\,H^+/O_{2\text{ produced}}$

Photophosphorylation and ATP Synthase Structure

• The chloroplast $ATP$ synthase is very similar to the mitochondrial version, with the primary difference being the number of $c$ subunits.

• $c$ Subunits: Vary by species, typically in the range of $12$ to $15$ subunits.

• $ATP$ Production: Regardless of species, the synthase contains three $\alpha\beta$ subunits, meaning one full capital-gamma ($\gamma$) subunit rotation ($360^\circ$) always produces $3\,ATP$.

• Proton Cost per ATP:
      - If $15\,c$ subunits exist: $\frac{15\,H^+}{3\,ATP} = 5\,H^+/ATP$.
      - If $12\,c$ subunits exist: $\frac{12\,H^+}{3\,ATP} = 4\,H^+/ATP$.
      - The requirement is generally standardized between $4$ to $5\,H^+$ depending on the specific organism.

• Spatial Advantage: No transport of $ATP$ is required because it is produced directly in the stroma where carbon fixation (the dark reactions) occurs.

Integrated Calculation: Glucose Synthesis Example

• Scenario: Produce one molecule of glucose.

• Requirements for Glucose: $6\,CO_2$, $12\,NADPH$, and $18\,ATP$.

• System Variable: $13\,c$ subunits in the $ATP$ synthase.

• Step 1: Oxygen Production
      - Formula: $12\,NADPH \times \frac{1\,O_2}{2\,NADPH} = 6\,O_2$.
      - (Based on the fact that $2\,NADPH$ contain four electrons, matching the oxidation of $2\,H_2O \rightarrow 1\,O_2$).

• Step 2: Total Proton Requirement for ATP
      - With $13\,c$ subunits, $3\,ATP$ cost $13\,H^+$.
      - Calculation: $18\,ATP \times \frac{13\,H^+}{3\,ATP} = 78\,H^+$ required.

• Step 3: Protons Available from Linear Flow
      - Calculation: $12\,NADPH \times 6\,H^+/NADPH = 72\,H^+$.

• Step 4: Discrepancy and Cyclic Flow
      - Current deficit: $78\,H^+ - 72\,H^+ = 6\,H^+$.
      - Cyclic flow yields $2\,H^+$ per photon applied at PS I.
      - Required photons for deficit: $\frac{6\,H^+}{2\,H^+/\text{photon}} = 3\text{ photons}$.

• Step 5: Final Photon Tally
      - For Linear Flow: $12\,NADPH \times 4\,\text{photons/NADPH} = 48\text{ photons}$.
      - Distribution: $24\text{ photons}$ at PS II and $24\text{ photons}$ at PS I.
      - Adding Cyclic Requirement: Photosystem I requires the original $24\text{ linear photons} + 3\text{ cyclic photons} = 27\text{ photons}$.

• Final Answer Summary:
      - Oxygen Produced: $6\,O_2$
      - Photons at PS II: $24$
      - Photons at PS I: $27$
      - Total Water Molecules Used: $12\,H_2O$