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Hint: It is generated in non-cyclic phosphorylation by the reduction of NADP+. This type of photophosphorylation is absent in bacterias due to the absence of a certain photosystem.
Complete answer:
$NADP{ H }_{ 2 }$ is formed in the z-scheme or non-cyclic photophosphorylation where both photosystem-I and photosystem-II are present, by reduction of $NADP^{ + }$. The actual reduction of NADP+ to $NADP{ H }_{ 2 }$ takes place in photosystem-I
The steps involved in non-cyclic photophosphorylation are as follows:
-Absorption of light energy of specific wavelength by the chlorophyll and accessory pigments. These pigments absorb energy and transfer it to the reaction center of PS-II P680. This results in photoexcitation of the P680 reaction center and it discharges one electron which is passed to a pigment named pheophytin.
-This loss of electron is compensated by absorbing electrons released during the photolysis of water.
-After passing through a series of carriers plastoquinone, cytochromes b6-f complex, and plastocyanin the electron is passed to photo center P700 of PS-1 by plastocyanin.
-P700 releases this electron on absorbing light energy of a suitable wavelength.
-This electron is transferred through several quinones, FeS complexes, Ferredoxin, and NADP reductase.
-This NADP reductase reacts with NADP and reduces it to $NADPH+{ H }^{ + }$. This H+ is obtained from the stroma.
So, the correct answer is, “ $NADP{ H }_{ 2 }$ is generated through Photosystem-I.”
Note: It should be noted that the PS-I and PS-II photosystems are named just in the order in which they were discovered. In cyclic photophosphorylation of PS-I is involved but both photosystems are involved in the non-cyclic type of photophosphorylation. In the non-cyclic type of photophosphorylation both ATP and NADPH are formed, which are considered as assimilatory power, but in the cyclic type of photophosphorylation only ATP formation takes place.
Complete answer:
$NADP{ H }_{ 2 }$ is formed in the z-scheme or non-cyclic photophosphorylation where both photosystem-I and photosystem-II are present, by reduction of $NADP^{ + }$. The actual reduction of NADP+ to $NADP{ H }_{ 2 }$ takes place in photosystem-I
The steps involved in non-cyclic photophosphorylation are as follows:
-Absorption of light energy of specific wavelength by the chlorophyll and accessory pigments. These pigments absorb energy and transfer it to the reaction center of PS-II P680. This results in photoexcitation of the P680 reaction center and it discharges one electron which is passed to a pigment named pheophytin.
-This loss of electron is compensated by absorbing electrons released during the photolysis of water.
-After passing through a series of carriers plastoquinone, cytochromes b6-f complex, and plastocyanin the electron is passed to photo center P700 of PS-1 by plastocyanin.
-P700 releases this electron on absorbing light energy of a suitable wavelength.
-This electron is transferred through several quinones, FeS complexes, Ferredoxin, and NADP reductase.
-This NADP reductase reacts with NADP and reduces it to $NADPH+{ H }^{ + }$. This H+ is obtained from the stroma.
So, the correct answer is, “ $NADP{ H }_{ 2 }$ is generated through Photosystem-I.”
Note: It should be noted that the PS-I and PS-II photosystems are named just in the order in which they were discovered. In cyclic photophosphorylation of PS-I is involved but both photosystems are involved in the non-cyclic type of photophosphorylation. In the non-cyclic type of photophosphorylation both ATP and NADPH are formed, which are considered as assimilatory power, but in the cyclic type of photophosphorylation only ATP formation takes place.
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