Selected Solutions to End of Chapter 19 (Part 3) Problems
Class 3
27. Data: Photosynthesis data 1804: total weight of oxygen and dry organic matter produced by
plants is more than the weight of CO2 consumed. Where did the extra weight come from? A basic
chemical question.
Water. Water releases protons and electrons that get incorporated into plant biomass.
28. There are several types of photosynthetic bacteria. One of the classics are the Purple Sulfur
Bacteria. Instead of green they are purple: a color that comes from two pigments. One type of
bacterial chlorophyll is blue and that is mixed in these cells with red carotenoid pigments to produce a
beautiful purple color.
These bacteria use H2S rather than H2O as the electron donor for non-cyclic
photophosphorylation. These bacteria are anaerobes and only do photosynthesis
anaerobically. Otherwise they do photosynthesis electron transport producing ATP and
NADPH for CO2 fixation. If they used H2O as an electron donor, they would kill themselves off
because for many anaerobes O2 is toxic.
30. Phaeophytin level inhibition of photosynthetic electron flow. Pheophytin is simply a regular
chlorophyll that lacks Mg++ in the tetrapyrole ring. Such an inhibitor would block non-cyclic
photophosphorylation (photosystem-2) but not photosystem-1 which is the cyclic
photophosphorylation. The effect of this is that the cells would be able to produce ATP from the
proton motive force of photosystem-1, but could not produce NADPH.
31. Spinach chloroplasts illuminated in a buffer without ADP and P i , then immediately put in the dark
will start making ATP when ADP and Pi are added. Why?
Light generated the proton motive force which could not be used to make ATP because in the
light they were artificially starved of ADP and Pi. Then in the dark, when ADP and Pi are
added, the proton motive force was used to make ATP until the proton motive force was
consumed.
35. Light energy for non-cyclic photophosphorylation: generation of NADH from H2S in purple sulfur
bacteria. These are all based on molar amounts. Data: Eo’ for H2S is -0.243 v and Eo’ for NADH is
-0.320 v. So all we need to do is get the ΔGo’ for electron transfer from H2S to NAD+ and then match
that to the energy of different colors of light:
So lets get the ΔGo’ for electron transfer to NAD+ from H2S.
Remember: ΔEo’ = Eo’ electron acceptor – Eo’ electron donor
ΔEo’ = -0.320 v – (- 0.243 v) = - 0.077 v
ΔGo’ = - n F ΔEo’ = - (2)(96.5 kJ/ v mole)(-0.077 v) = + 14.9 kJ/mole
The Figure above shows that all visible light, even the less energetic red to far red has enough
energy to make this endothermic electron transport work.
37. Energetics of photoreduction of NADP+ in chlorophyll-a type photosynthesis:
2 H2O + 2 NADP+ + 8 photons  2 NADPH + O2
Does 700 nm light have enough energy for this? From the figure above you can see that 700
nm light has 170 kJ/Einstein (we calculated 168 in the lecture using the Planck-Einstein
equation). And, an einstein is a “mole” of light ( = 6.02 x 1023 photons).
First lets calculate the chemical energetics beginning with Eo’ and get the ΔEo’.
ΔEo’ = Eo’ electron acceptor – Eo’ electron donor = -.324 v – (0.816 v) = - 1.14 v
ΔGo’ = - (4)(96.5 kJ/ v mole)(-1.14 v) = 440 kJ/mole
On a mole per mole basis, this took 8 moles or 8 einsteins of light.
8 einsteins of light at 700 nm = 8 einsteins x 170 kJ/enistein = 1360 kJ
Subtract the 440 kJ required for the chemistry, and you have 920 kJ left over…that light is
plenty enough energetic to reduce NADP+ using protons from water.
39. How often does a spinach chlorophyll react (absorb a photon). Lots of photosynthesis research
uses spinach, which is not only a wonderful vegetable to eat, but the very first photochemistry of
plants were done with spinach (just about the easiest leaves to get to the biochemistry – easily
crushed to isolate clean preparations of chloroplasts, mitochondria, etc.) Data: MW of chlorophyll-a is
892 daltons. There is 20 μg chlorophyll-a / cm2 of leaf surface. Noonday sun reaching the plant has
energy of 5.4 J/cm2 .min at the speed of light. The leaf absorbs about 50% on the incident light. The
average life time of a chlorophyll excited state is 1 ns.
This question is incomplete, it lacks an essential data component: the wavelength(s) of light. So, one
way to do this is take a “sort of” average wavelength (which of course means energy content) in the
visible range: from the figure above, take violet (400 nm) to red (700 nm) the average is 550 nm.
So using the Planck-Einstein equation E = hv to calculate the energy of a photon will help. But first we
need to calculate “v” :
v is the vibrational frequency = c/λ = (3 x 108 m/s) / (550 x 10-9 m) = 5.45 x 1014 s-1
Now E = hv = (6.6 x 10-34 J . s) x 5.45 x 1014 s-1 = 3.6 x 10-19 J per photon.
Now lets get to how many chlorophyll molecules are near the surface of the leaf, part 1 how
many moles/cm2 :
20 μg/cm2
892 μg/μmole
= 0.022 μmole / cm2 ….. now get this to molecules / cm2.
(0.022 x 10-6 moles / cm2 ) (6.02 x 1023 molecules / mole) = 1.32 x 1016 molecules/cm2
Now lets get how many photons are arriving at the leaf surface:
And take into account 50% photons getting to chlorophyll (5.4 J/cm2.min) x 0.4 = 2.7 J/cm2.min
(2.7 J/cm2 . min) / (3.6 x 10-19 J / photon) = 7.5 x 1018 photons / cm2 . min
= 1.25 x 1017 photons / cm2 . sec
So: 1.25 x 1017 photons are being absorbed per second by 1.32 x 1016 molecules of
chlorophyll
(1.32 x 1016 molecules) / (1.25 x 1017 photons/sec) = 0.1 sec for each photon to strike a
chlorophyll molecule.
Being the reaction time for the excited state is 1 ns, there is plenty of time for the reaction to go
and chlorophyll to “rest” until the next photon strikes it. Which means that only one in 108
molecules of chlorophyll react per 0.1 sec.
41. Function of photosystem-1 (cyclic photophosphorylation). During times when [NADPH] is high,
the non-cyclic photosystem-2 is off, but photosystem-1 is running, what is it’s function? During this
time O2 is not evolved, PS-2 is off and there is no need to make NADPH. PS-1 is still running to only
supply ATP by generating the proton motive force.