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The 'Dark Phase' in photosynthesis as I know is not dependent on darkness, but is just 'Light Independent'. But the Warburg Flashing light experiment, makes me think again. Warburg obtained higher rates of photosynthesis in Chlorella, when he exposed it to rapid alternate periods of light and darkness, instead of illuminating it constantly. But if the so called dark phase is just 'Light Independent', then why did Warburg get higher photosynthesis rate in the setup? I think if continuous illumination was provided ATP and $\ce{NADPH}+ \ce{H+}$ would be present more, and thus should give higher rates of photosynthesis. Can anyone point where I'm mistaken?

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    $\begingroup$ Can you pls put a link to his paper? $\endgroup$ – Roni Saiba Oct 12 '17 at 2:32
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    $\begingroup$ I cannot comment so I write here. Besides details that I let to people with better knowledge than me, it seems you compare or "equate" amount of product and rate. That is not correct. Think of a sprinter (that actually stop, breath and eventually restart) and a marathon runner. Who is going faster and who is covering the longer distance? Indeed I suspect that a kind of fatigue is at play, like inhibition, saturation of some path. However this latter is just me speculating. I wanted to point out the difference above. It might help. $\endgroup$ – Alchimista Oct 12 '17 at 11:39
  • $\begingroup$ Your justification was indeed a help. I read this logic somewhere, but want to know precisely what the inhibitory factor is. Besides it may seem that I'm confusing rate and yield, due to my last line, but actually I'm talking about rate and not yield.. $\endgroup$ – jyoti proy Oct 12 '17 at 11:53
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according to my understanding, if we compare two situation 1. one hour continuous illumination and other 2. alternating period of light and dark (but the illumination time remain same). the rate of photosynthesis will be more in second case which indicate that even in dark, photosynthesis is taking place (ie. dark reaction)...just my thought...

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    $\begingroup$ Welcome to Biology SE! While this may answer the question it lacks support from references in the scientific literature. Un-referenced answers may be challenged or deleted. $\endgroup$ – vkehayas May 4 '18 at 10:41
  • $\begingroup$ A readily accessible paper explains, in detail, what @FunT is correctly stating. $\endgroup$ – Jim Young Jun 13 '18 at 17:09

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