During light phase of photosynthesis, ____ is oxidized and ____ is reduced.

Question

Accepted Answer

During the light phase of photosynthesis, light energy is captured and converted into chemical energy in the form of ATP and NADPH. This process involves a series of redox reactions.

Let's analyze the options:

**Option 1: water, NADP**
*   **Water is oxidized:** In the light-dependent reactions, specifically at Photosystem II (PSII), water molecules undergo photolysis (splitting by light energy). The equation for water splitting is $2H_2O \rightarrow 4H^+ + 4e^- + O_2$. In this process, water loses electrons ($e^-$), which means it is oxidized. These electrons are then used to replace the electrons lost by PSII.
*   **NADP is reduced:** Nicotinamide Adenine Dinucleotide Phosphate (NADP$^+$) acts as the final electron acceptor in the non-cyclic electron transport chain. The electrons, which originated from water and passed through the electron transport chain, along with protons ($H^+$) from the stroma, are accepted by NADP$^+$ to form NADPH ($NADP^+ + 2e^- + H^+ \rightarrow NADPH$). The gain of electrons and protons by NADP$^+$ signifies its reduction.
*   Therefore, during the light phase, water is oxidized, and NADP$^+$ is reduced. This option correctly describes the primary redox events.

**Why other options are incorrect:**

**Option 2: $NADPH_{2}$, $CO_{2}$**
*   **$NADPH_{2}$ (NADPH):** NADPH is *formed* during the light phase as a product of reduction, not oxidized during the light phase. It is primarily used in the dark phase (Calvin cycle) as a reducing agent.
*   **$CO_{2}$:** Carbon dioxide ($CO_2$) is reduced during the *dark phase* (Calvin cycle) to form glucose, not during the light phase. It is the primary substrate for carbon fixation, not a reactant in the light-dependent redox reactions.

**Option 3: $CO_{2}$, water**
*   **$CO_{2}$:** As explained above, $CO_2$ is reduced in the dark phase, not oxidized in the light phase.
*   **Water:** Water is oxidized (loses electrons) in the light phase, not reduced (gains electrons).

**Option 4: $CO_{2}$, $NADPH_{2}$**
*   **$CO_{2}$:** As explained, $CO_2$ is reduced in the dark phase, not oxidized in the light phase.
*   **$NADPH_{2}$ (NADPH):** NADPH is formed by the reduction of NADP$^+$ in the light phase. It is not itself reduced further in the light phase; rather, it is a product that carries reducing power.

Based on the detailed explanation of the light-dependent reactions, Option 1 accurately identifies water as the molecule that is oxidized and NADP$^+$ as the molecule that is reduced.