Question

Which of the following is the correct pathway of the electron during light reaction?

• A. Fd→ PC→ PQ→ Fe-S
• B. PC→ PQ→ Fd→ Fe-S
• C. PQ→ PC→ Fe-S→ Fd
• D. Fd→ Fe-S→ PC→ PQ

Answer 1

The correct option is C PQ→ PC→ Fe-S→ Fd

When PS II receives a light of wavelength 680nm, the electron is extruded. It passes sequentially through a series of electron carriers namely phaeophytin, plastoquinone, cytochrome $b_{6}f$ complex and plastocyanin. The electron loses an adequate amount of energy while passing over the cytochrome complex. This energy is used in the synthesis of ATP. The electron is then passed onto PS I. When PS I absorbs light of wavelength 700nm, it expels the electron which is received by a modified chlorophyll molecule $A_0$, $A_1$, Fe-S centre, ferredoxin (Fd) and finally to ferredoxin ${\mathrm{NADP}}^{+}$ reductase (FNR) which passes on the electron to ${\mathrm{NADP}}^{+}$ which gets converted to NADPH. So the pathway of electrons is P680→ pheophytin→ PQ→ cyt $b_{6}f$→ PC→ P700→ $A_0$→ $A_1$→ Fe-S→ Fd→ FNR→ ${\mathrm{NADP}}^{+}$