Question: In photosynthesis, photolysis of water is used in A. Reduction of \[NAD{P^ + }\] B. Oxidation of...

Question

In photosynthesis, photolysis of water is used in
A. Reduction of $NAD{P^ + }$
B. Oxidation of $NADP$
C. Oxidation of $FAD$
D. None of the above

Answer 1

Solution

Photolysis means the splitting of the water molecule into hydrogen and oxygen in the presence of light. During the initial stage of photosynthesis, the photosystem $2$ absorbs the light energy and the chlorophyll electrons get excited. When the electron falls to the ground state, due to the presence of photons the water molecule breaks. Photolysis has its role in non-cyclic photophosphorylation.

Complete explanation:
Option A reduction of NADP+: During non-cyclic photophosphorylation the electrons produced during photolysis of water cycles between $PS - 1$ and $PS - 2$ . After one cycle the electrons given out by $P700$ of PS-I are taken up by the primary pigment molecule and passed to $NADP$ through Ferredoxin. The electrons along with ions reduce $NADP$ to $NADPH$ .
Option A is correct.
Option B Oxidation of $NADP$ : $NADP$ is an abbreviation for Nicotinamide adenine dinucleotide phosphate. Oxidation of $NADP$ reaction occurs in mitochondria during oxidative phosphorylation. Hence this option is not correct.
Option B is incorrect.
Option C Oxidation of $FAD$ : $FAD$ is a redox active coenzyme. $FAD$ is an abbreviation for flavin adenine dinucleotide. It is involved in several enzymatic reactions in metabolism. It does not have any relation with photolysis of water.
Option C is incorrect.
Option D None of the above: Since Photolysis of water helps in reduction of $NAD{P^ + }$ , this option cannot be correct.
Option D is incorrect.

Hence, Option A. Reduction of $NAD{P^ + }$ is the correct answer.

Note:
During non-cyclic photophosphorylation the electrons released from $P680$ of PS-II are accepted by the primary electron acceptor. The electrons pass through a series of electron acceptors like PQ- cytochrome complex, PC and finally to $P700$ of PSI. The electrons given out by $P700$ are taken up by primary pigment molecules and are ultimately passed to $NADP$ through Ferredoxin. The electrons along with ions reduce $NADP$ to $NADPH$ .