Question

In non-cyclic photophosphorylation the electron released by $P_{680}$ is replaced by

• A. Plastoquinone
• B. Water
• C. $C{O}_2$
• D. $NAD{P}^{+}$.

Answer 1

Answer: (B)

Water

Non-cyclic phosphorylation involves both Photosystem I and Photsystem II. These two photosystems work in series, first PS II and then PS I. The two photosystems are connected through an electron transport chain. $NADPH$ is synthesised by this kind of electron flow. Electrons are transferred from the PS II and then passed on to the pigments of PS I. Then, electrons get excited and transferred to another acceptor. These electrons are then moved downhill again to a molecule of $NAD{P}^{+}$. The addition of these electrons results in formation of $NADPH$. The electron released from PS II by $P_{680}$ must be replaced. This is achieved by electrons available due to splitting of water. The water splitting complex is associated with PS II, which itself is physically located on the inner side of the membrane of the thylakoid. Water is split into $H^{+}$, $\left[O\right]$ and electrons. The protons and oxygen formed by splitting of water is released within the lumen of the thylakoids. The oxygen produced is released as one of the net products of photosynthesis.

$2H_{2}O$ $⟶$ $4H^{+}$ + $O_2$ + $4e^{-}$