Answer:
1st: Light excites an electrom from photosystem II
2nd: Light excites an electrom from photosystem I
3rd: Electrons pass through an electron chain, which generates a H+ gradient used to make ATP
4th: Electron reduce NADP+ to NADPH
Explanation:
While the light simultaneously excites both photosystems, it must first occur in photosystem II and then be able to transfer the e-energized to photosystem I
1st Step:
Within the photosystems we find different photosynthetic pigments, that is, capable of absorbing light. These pigments are classified according to the maximum absorption wavelengths.
When the light hits the photosystems they absorb it and the delocalized -e (electrons) are energized or "excited".
Then these energized ones are transferred to molecules within the membrane that houses the pigments.
The e- that takes the photosystem I are provided by the photosystem II
2nd and 3rd Step:
The -e energized from photosystem II are transferred to a transport chain of -e within the membrane containing the pigments. As these -e circulate, they lose energy that is used to translocate H + (protons).
The accumulation of H + within the membrane generates an electrochemical gradient.
H + return to the stroma through the enzyme ATP synthase. This operation is called chemosmosis.
This enzyme uses H + to catalyze the synthesis of ATP (ADP + Pi), a process called phosphorylation.
4th Step:
The e-energized of photosystem I are used to reduce NADP + and generate NADPH that are used in conjunction with ATP to generate "light independent reactions"
Those lost from photosystem I are replaced by e-de-energized from photosystem II, while those lost from photosystem II are replaced by e-released from water by photolysis.
Water is divided by the energy of light into H + (used in chemosmosis) and oxygen (released as a byproduct)
