Question

Extrusion of electron from reaction centre of PS II leaves a hole which is filled by electron released from

• A. $H_{2}O$
• B. $C{O}_2$
• C. Chlorophyll
• D. Light

Answer 1

The correct option is A $H_{2}O$

Non-cyclic phosphorylation involves both Photosystem I and Photosystem 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 synthesized 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 the formation of $NADPH$. Extrusion of the electron from reaction centre of PS II leaves a hole which is filled by electron released from $H_{2}O$. This is achieved by electrons available due to the 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 $H^{+}$, $O$ 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.

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