a) [21]+[19] = [40] = [4] {since 40 when divided ny 12 leaves a remainder of 4 } b) [37][27] = [ 999] = [9] { Since now the remiander will be 9 when divided by 45} c) [33]([83] - [67]) = [33] [16] = [528] = [28] d) [18] + [8] = [26] = [0]. Hence additive inverse of [18] is [8]. Solution a) [21]+[19] = [40] = [4] {since 40 when divided ny 12 leaves a remainder of 4 } b) [37][27] = [ 999] = [9] { Since now the remiander will be 9 when divided by 45} c) [33]([83] - [67]) = [33] [16] = [528] = [28] d) [18] + [8] = [26] = [0]. Hence additive inverse of [18] is [8]..

1. Answer:
The net equation of the reaction catalyzed by the protein complex of photosystem I for plants as
mentioned below-
6H2O + 6CO2 ----------> C6H12O6+ 6O2
Formation of NADPH-
In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons
from water. The electrons travel through the chloroplast electron transport chain to photosystem I
(PSI), which reduces NADP+ to NADPH. The electron transport chain moves protons across the
thylakoid membrane into the lumen.
The enzyme ferredoxing-NADP+-reductase, a flavoprotein containing protein, reduces NADP+
to NADPH. Photosystem I can run in two modes. First, it promotes the reduction of NADP+ to
provide reductive power for biosynthetic purposes. Second, photosystem I can switch to a cyclic
operation where cytochrom bf complex serves as the electron acceptor instead of ferredoxin-
NADP-reductase.
Role of FAD (flavin adenine dinucleotide):
FAD is a redox cofactor, more specifically a prosthetic group, involved in several important
reactions in metabolism. It can exist in three different redox states- quinone, semiquinone, and
hydroquinone. FAD is converted between these states by accepting or donating electrons.
FAD, in its fully oxidized form, or quinone form, accepts two electrons and two protons to
become FADH2 (hydroquinone form). The semiquinone (FADH·) can be formed by either
reduction of FAD or oxidation of FADH2 by accepting or donating one electron and one proton
respectively.
Solution
Answer:
The net equation of the reaction catalyzed by the protein complex of photosystem I for plants as
mentioned below-
6H2O + 6CO2 ----------> C6H12O6+ 6O2
Formation of NADPH-
In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons
from water. The electrons travel through the chloroplast electron transport chain to photosystem I
(PSI), which reduces NADP+ to NADPH. The electron transport chain moves protons across the
thylakoid membrane into the lumen.
The enzyme ferredoxing-NADP+-reductase, a flavoprotein containing protein, reduces NADP+

2. to NADPH. Photosystem I can run in two modes. First, it promotes the reduction of NADP+ to
provide reductive power for biosynthetic purposes. Second, photosystem I can switch to a cyclic
operation where cytochrom bf complex serves as the electron acceptor instead of ferredoxin-
NADP-reductase.
Role of FAD (flavin adenine dinucleotide):
FAD is a redox cofactor, more specifically a prosthetic group, involved in several important
reactions in metabolism. It can exist in three different redox states- quinone, semiquinone, and
hydroquinone. FAD is converted between these states by accepting or donating electrons.
FAD, in its fully oxidized form, or quinone form, accepts two electrons and two protons to
become FADH2 (hydroquinone form). The semiquinone (FADH·) can be formed by either
reduction of FAD or oxidation of FADH2 by accepting or donating one electron and one proton
respectively.