Jonathan B. Marder
MARDER at agri.huji.ac.il
Thu Jan 2 11:24:20 EST 1997
In article <email@example.com>,
> This is only a comment and a question, in addition to the very clear
>explanations of Dr. David Walker. I am not an expert in photosynthesis as a
>whole, but I wonder if this problem of chlorophyll fluorescence is not
>actually misleading. Most of the energy captured within chloroplasts by
>chlorophyll is channelled to an other chlorophyll molecule, or to a
>reaction center where it is converted to a low-potential redox potential,
>i.e. to a chemical potential. This in turn is used to create reductants
>that shall be used for the reductive assimilation of CO2 or other stuffs.
The problem of energy transfer is not much of a problem since the reaction
centres also contain chlorophyll and can also fluoresce. Once the reaction
centre chlorophyll becomes excited there is no way to tell if it was excited
by direct light absorption, or via the antenna.
>Some losses of energy during these transfers appear under the form of
>emitted light (fluorescence) or heat.
This needs to be clarified. Loss of the exciton can indeed occur via
fluorescence or heat. This reduces quantum yield of photosynthesis.
However, there is also loss of heat due to reduction of exciton energy e.g.
blue light photons have more energy than red light photons. When blue light is
absorbed by chlorophyll, the "extra" energy is immediately lost as heat so you
get the same excited state as if red light was absorbed. This type of heat
loss reduces *energetic* efficiency but *not* quantum yield. In the later
stages of photosynthesis, there are additional losses of heat which again
reduce energetic efficiency but have no effect on quantum yield.
>My question to the specialists is the following : what is the average
>proportion of energy re-emitted as fluorescence by chlorophyll in vivo in
>the course of photosynthesis ?
In healthy green leaves, just about all the fluorescence comes from
photosystem II. We normally consider that PSII has a quantum yield of about
85% i.e. that is the percentage of absorbed photons which drive the
photochemistry. Of the 15% quantum loss we would expect something like a third
(~5% of total) to occur via fluorescence. The "proportion of energy
re-emitted" (to quote your question) depends on the energy (wavelength) of the
incident photons. Thus, assuming 700 nm fluorescence, 5% fluorescence yield
would be equivalent to 4.64% energy loss via fluorescence for 650 nm (red)
excitation light, but only 3.2% energy loss via fluorescence for 450 nm (blue)
Jonathan B. Marder , Department of Agricultural Botany
E-mail: MARDER at agri.huji.ac.il | The Hebrew University of Jerusalem
Phone: (08 or +9728) 9481918 | /\/ Faculty of Agriculture
Fax: (08 or +9728) 9467763 |/ \ P.O.Box 12, Rehovot 76100, ISRAEL
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