Chloroplasts have two photosystems (PSIl and PSI) that are connected by a series

Question

Chloroplasts have two photosystems (PSIl and PSI) that are connected by a series of electron carriers including pheophytin, plastoquinone, cytochrome f, and plastocyanin. Identify the effect of illuminating chloroplasts with 700 nm or 680 nm monochromatic light on the oxidation state of these electron carriers by moving the labels to the appropriate positions in the table. Oxidation State after 700 nm llumination Oxidation State after 680 nm Illumination Cytochrome Plastocyanin Plastoquinone Pheophytin oxidized reduced Choose the statement that best describes how illumination by monochromatic light affects the oxidation state of electron carriers. Scroll down. Illumination by 680 nm light activates electrons in PSll which are passed on to the electron carriers, reducing them. Illumination by 700 nm light activates electrons in PSlI which are passed on to the electron carriers, reducing them.

Explanation / Answer

Oxi. State after 700nm illumination

Oxi. State after 680nm illumination

Cytochrome F

Plastocynin

Plastoquinone

Pheophytin

So, according to the question we are assuming the instant at which the illumination is taking place. Therefore, accordingly we will assign the oxidation states to the following given molecules.

But before doing so we need to know the sequence in which these molecules are present for this please go through to Z- scheme. Their sequence is P680(PSII)-> Pheophytin->Plastoquionone->Cytochrome->Plastocynin ->P700(PSI)

Please note one important thing about the transfer of electrons. every transfer of electron from one enzyme to other enzyme is an oxidation/ reduction. The enzyme that is losing the electron is oxidised by the next enzyme (therefore the next enzyme is reduced).

Let’s assign oxidation state to the photosystem molecules in the table from the instant when the light is illuminated.

So first the P680 get reduced and excited when the light is illuminated, after that it will oxidised by Pheophytin (making it to be reduced) then it will oxidised by Plastoquinone (making it to be reduced) after which it will by oxidised Cytochrome f (making it to be reduced) which will again oxidised by Plastocynin (making it to be reduced). Plastocynin will be oxidised by P700(PSI).

Hope this will Help you solve the problem related to the table above.

In the process photosynthesis the electron moves from water through PSII to PSI and then to NADP+ ultimately therefore option 4 will be the answer

Oxi. State after 700nm illumination

Oxi. State after 680nm illumination

Cytochrome F

Plastocynin

Plastoquinone

Pheophytin

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