IChO 2013 · Theory problem 2

Break down photosynthesis – the Hill reaction

Download PDFIChO 2013

In the history of photosynthesis research, there were some breakthrough experiments which added much to our knowledge of this very complex process. One of such experiments was performed in 1930s by an English biochemist Robert Hill. In this problem, we consider some of his data together with the data of more recent experiments.

2.1 In plants, under illumination, carbon dioxide is reduced to carbohydrates (denote as \(\{\ce{CH2O}\}\)) and oxygen is produced. Write the overall equation of photosynthesis in plants.

Much of the photosynthesis takes place in chloroplasts – organelles found in plant cells and containing chlorophyll – the light-absorbing substance. Hill isolated chloroplasts from the cells by grinding the leaves in the sucrose solutions. The cell-free chloroplasts did not produce oxygen under illumination even in the presence of \(\ce{CO2}\). However, upon adding potassium ferrioxalate \(\ce{K3[Fe(C2O4)3]}\) (with the excess of potassium oxalate) to the chloroplast suspension Hill observed oxygen liberation under illumination even without \(\ce{CO2}\).

2.2 Hill’s experiment enabled to determine the source of oxygen during photosynthesis. Write the formulas of the oxidant and the reducing agent in the photosynthesis inside the plant cells and in the cell-free chloroplasts (the Hill reaction).

Hill measured the amount of evolved oxygen using muscle haemoglobin (Hill denoted it \(\ce{Hb}\)) which binds all molecular oxygen in a 1 : 1 ratio to form \(\ce{HbO2}\). The initial concentration of \(\ce{Hb}\) was \(0.6\cdot10^{-4}\ \mathrm{mol\,dm^{-3}}\). Kinetic curves corresponding to different ferrioxalate concentrations are shown in the figure (the upper curve corresponds to \(2.0\cdot10^{-4}\ \mathrm{mol\,dm^{-3}}\)).

Figure 1

The fraction of bound haemoglobin \(\ce{HbO2}\) (with respect to the initial amount of \(\ce{Hb}\)) as a function of time. Crosses denote the end of the reaction

(Figure 2a from the original Hill’s paper: R. Hill. Oxygen produced by isolated chloroplasts. – Proc. R. Soc. B, 1939, v. 127, pp. 192-210.)

2.3 a. From the figure, estimate the \(\ce{Fe}\) / \(\ce{O2}\) mole ratio at the end of the reaction. Do not take into account the iron from \(\ce{Hb}\).

b. Write the equation of Hill reaction assuming that it proceeds with a high yield.

c. Using the table of standard electrode potentials, determine the Gibbs energy of the Hill reaction at \(T=298\ \mathrm{K}\), oxygen pressure \(1\ \mathrm{mm\,Hg}\), \(\mathrm{pH}=8\) and standard concentrations of other species. Is this reaction spontaneous at such conditions?

Half-reaction \(E^\circ,\ \mathrm{V}\)
\(\ce{O2 + 4H+ + 4e -> 2H2O}\) \(+1.23\)
\(\ce{CO2 + 4H+ + 8e ->} \{\ce{CH2O}\} + \ce{H2O}\) \(-0.01\)
\(\ce{Fe^3+ + e -> Fe^2+}\) \(+0.77\)
\(\ce{Fe^3+ + 3e -> Fe^0}\) \(-0.04\)
\(\ce{[Fe(C2O4)3]^3- + e -> [Fe(C2O4)3]^4-}\) \(+0.05\)
\(\ce{[Fe(C2O4)3]^4- + 2e -> Fe + 3C2O4^2-}\) \(-0.59\)

Now, the name “Hill reaction” denotes photochemical oxidation of water by any oxidant other than carbon dioxide which is sensitized by plant cells or isolated chloroplasts.

In another experiment (1952), quinone in an acid solution was used as an oxidant in the Hill reaction initiated by light flashes in the Chlorella algae. Experimental data are shown in the figure. The volume of oxygen (in \(\mathrm{mm^3}\), at temperature \(10\,^\circ\mathrm{C}\) and pressure \(740\ \mathrm{mm\,Hg}\)) per one gram of chlorophyll per one flash was determined as a function of light intensity for natural photosynthesis and for isolated chloroplasts. It was found that the maximum yield of oxygen is the same for natural photosynthesis and the Hill reaction.

Figure 2

(Figure 1 from: H. Ehrmantraut, E. Rabinovitch. Kinetics of Hill reaction. – Archives of Biochemistry and Biophysics, 1952, v. 38, pp. 67-84)

2.4 a. Determine the reaction order of a photochemical Hill reaction with respect to light intensity at low and high intensity. For each case choose one of the following three values: 0, 1 or 2.

b. How many chlorophyll molecules participate in the formation of one oxygen molecule in the saturation limit of the Hill reaction? (The molecular mass of chlorophyll is about 900 Da).

The quantum requirement of the light redox reactions is defined as the average number of light photons (not necessarily integer) needed for the transfer of one electron from a reducing agent to an oxidant. The isolated chloroplasts were irradiated during 2 hours by a monochromatic light (wavelength \(672\ \mathrm{nm}\)) with the energy input \(0.503\ \mathrm{mJ\,s^{-1}}\), and the total volume of oxygen formed was \(47.6\ \mathrm{mm^3}\) (under the same conditions as in question 4).

2.5 Calculate the quantum requirement for the Hill reaction.

2.6 Try to make conclusions from the above experiments (questions 2.2 – 2.5). Tick the relevant boxes.

Yes No
In natural photosynthesis, water oxidation and \(\ce{CO2}\) reduction are separated in space.
In chloroplasts, \(\ce{O2}\) is produced from \(\ce{CO2}\).
Oxidation of water in chloroplasts requires light illumination.
Most of chlorophylls in chloroplasts participate directly in the photochemical \(\ce{O2}\) production.
In isolated chloroplasts, every absorbed photon causes transfer of one electron.

Official English text of the problem.

Answers

  • 2.1 \[ \boxed{\ce{CO2 + H2O ->[light]} \{\ce{CH2O}\}+\ce{O2}} \]

  • 2.2 In plant cells: oxidant \(\ce{CO2}\), reducing agent \(\ce{H2O}\). In Hill’s experiment: oxidant \(\ce{[Fe(C2O4)3]^3-}\), reducing agent \(\ce{H2O}\).

  • 2.3(a) The graph gives approximately \(\boxed{n(\ce{Fe})/n(\ce{O2})=4.4}\). The ideal stoichiometric ratio is \(4:1\).

  • 2.3(b) \[ \boxed{\ce{4[Fe(C2O4)3]^3- + 2H2O ->[light] 4[Fe(C2O4)3]^4- + O2 + 4H+}} \]

  • 2.3(c) \(\boxed{\Delta_{\mathrm r}G\approx+256\ \mathrm{kJ\,mol^{-1}}}\), per mole of reaction as written, producing one mole of \(\ce{O2}\). The reaction is not thermodynamically spontaneous under the stated conditions.

  • 2.4(a) Low light intensity: \(\boxed{1}\). High light intensity: \(\boxed{0}\).

  • 2.4(b) Approximately \(\boxed{2.2\times10^3}\) chlorophyll molecules per oxygen molecule produced per saturating flash.

  • 2.5 \(\boxed{2.55\ \text{photons per transferred electron}}\), interpreting the stated energy input as absorbed light energy.

  • 2.6 In the order printed: \(\boxed{\text{Yes; No; Yes; No; No}}\), with the interpretive qualifications stated below.

Names and SMILES for the molecular species in these answers are provided at the start of the solution.

Solution

The relevant species are:

Formula Name SMILES
\(\ce{CO2}\) Carbon dioxide O=C=O
\(\ce{H2O}\) Water O
\(\ce{O2}\) Dioxygen O=O
\(\ce{H+}\) Hydrogen ion, or proton [H+]
\(\ce{C2O4^2-}\) Oxalate ion [O-]C(=O)C(=O)[O-]
\(\ce{[Fe(C2O4)3]^3-}\) Tris(oxalato)ferrate(III), or ferrioxalate [Fe+3]123(<-[O-]C(=O)C(=O)[O-]->1)(<-[O-]C(=O)C(=O)[O-]->2)<-[O-]C(=O)C(=O)[O-]->3
\(\ce{[Fe(C2O4)3]^4-}\) Tris(oxalato)ferrate(II) [Fe+2]123(<-[O-]C(=O)C(=O)[O-]->1)(<-[O-]C(=O)C(=O)[O-]->2)<-[O-]C(=O)C(=O)[O-]->3

The extended SMILES for the iron complexes use arrows for oxygen-to-iron coordinate bonds. Each oxalate ligand is bidentate, giving six oxygen donors in an approximately octahedral arrangement. Neither a particular \(\Delta\) nor \(\Lambda\) configuration is specified by the problem or encoded here. These bond arrows do not represent the electron flow of the redox reaction.

The notation \(\{\ce{CH2O}\}\) is a carbohydrate composition unit, not a uniquely specified molecule, so it has no unique SMILES. Likewise, the unspecified chlorophyll of approximate molecular mass \(900\ \mathrm{Da}\) does not identify a unique molecular structure.

2.1 — Overall photosynthesis equation

Carbon dioxide is reduced to carbohydrate, while water is oxidized to oxygen. The balanced half-reactions are

\[ \ce{CO2 + 4H+ + 4e^- ->} \{\ce{CH2O}\}+\ce{H2O} \]

and

\[ \ce{2H2O -> O2 + 4H+ + 4e^-}. \]

Adding them and cancelling the electrons, protons and one water molecule gives

\[ \boxed{\ce{CO2 + H2O ->[light]} \{\ce{CH2O}\}+\ce{O2}}. \]

This equation is written per carbon atom incorporated into carbohydrate.

2.2 — Oxidants and reducing agents

An oxidant accepts electrons and is reduced; a reducing agent supplies electrons and is oxidized.

System Oxidant Reducing agent
Overall photosynthesis in plant cells \(\ce{CO2}\) \(\ce{H2O}\)
Hill reaction with ferrioxalate \(\ce{[Fe(C2O4)3]^3-}\) \(\ce{H2O}\)

In photosynthesis, carbon’s average oxidation state decreases from \(+4\) in carbon dioxide to \(0\) in the carbohydrate composition unit. Oxygen in water increases in oxidation state from \(-2\) to \(0\).

In Hill’s experiment, the iron centre in ferrioxalate accepts an electron, changing from oxidation state \(+3\) to \(+2\). Water still supplies the electrons and the evolved oxygen. Oxygen evolution without carbon dioxide shows that carbon dioxide is not required as the oxygen source.

2.3(a) — Experimental iron-to-oxygen ratio

At the endpoint, the upper curve shows approximately \(75\%\) of the initial haemoglobin bound to oxygen. Because binding is \(1:1\),

\[ c(\ce{O2}) \approx0.75\,c_0(\ce{Hb}) =0.75(0.6\times10^{-4}) =4.5\times10^{-5}\ \mathrm{mol\,dm^{-3}}. \]

Each ferrioxalate ion contains one iron atom, so the initial concentration of iron supplied by ferrioxalate is

\[ c(\ce{Fe})=2.0\times10^{-4}\ \mathrm{mol\,dm^{-3}}. \]

Thus

\[ \boxed{ \frac{n(\ce{Fe})}{n(\ce{O2})} \approx \frac{2.0\times10^{-4}}{4.5\times10^{-5}} =4.4 }. \]

The graph therefore supports an ideal ratio of approximately four iron centres per oxygen molecule. For a \(4:1\) stoichiometry, the theoretical oxygen concentration is \(5.0\times10^{-5}\ \mathrm{mol\,dm^{-3}}\); the graph corresponds to about \(90\%\) of that amount.

2.3(b) — Balanced Hill reaction

Formation of one oxygen molecule from water releases four electrons:

\[ \ce{2H2O -> O2 + 4H+ + 4e^-}. \]

The inferred four iron centres per oxygen molecule therefore each accept one electron:

\[ \ce{[Fe(C2O4)3]^3- + e^- -> [Fe(C2O4)3]^4-}. \]

Multiplying this reduction half-reaction by four and adding gives

\[ \boxed{ \ce{4[Fe(C2O4)3]^3- + 2H2O ->[light] 4[Fe(C2O4)3]^4- + O2 + 4H+} }. \]

Both sides have total charge \(-12\), and all atoms are balanced.

2.3(c) — Gibbs energy under the specified conditions

The relevant standard reduction potentials are

\[ E^\circ_{\text{ferrioxalate}}=+0.05\ \mathrm{V}, \qquad E^\circ_{\ce{O2}/\ce{H2O}}=+1.23\ \mathrm{V}. \]

For ferrioxalate reduction coupled to water oxidation,

\[ E^\circ_{\mathrm{cell}}=0.05-1.23=-1.18\ \mathrm{V}. \]

Four electrons are transferred per reaction as written. Using \(F=96485\ \mathrm{C\,mol^{-1}}\),

\[ \Delta_{\mathrm r}G^\circ =-4FE^\circ_{\mathrm{cell}} =+455.4\ \mathrm{kJ\,mol^{-1}}. \]

The reaction quotient is

\[ Q= \frac{ a(\ce{[Fe(C2O4)3]^4-})^4 a(\ce{H+})^4 (p_{\ce{O2}}/p^\circ) }{ a(\ce{[Fe(C2O4)3]^3-})^4 a(\ce{H2O})^2 }. \]

Under the stated conditions, the complex activities and water activity are unity, while

\[ a(\ce{H+})=10^{-8}, \qquad \frac{p_{\ce{O2}}}{p^\circ}=\frac{1}{760} \]

using \(p^\circ=1\ \mathrm{atm}\). Therefore,

\[ Q=\frac{10^{-32}}{760}. \]

With \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\),

\[ \begin{aligned} \Delta_{\mathrm r}G &=\Delta_{\mathrm r}G^\circ+RT\ln Q\\ &=455.4+ \frac{8.314(298)}{1000} \ln\left(\frac{10^{-32}}{760}\right)\\ &=\boxed{+256.4\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

The positive Gibbs energy means that the chemical reaction is not spontaneous under these conditions. Light supplies energy to drive it.

2.4(a) — Dependence on light intensity

For flashes of fixed duration, the oxygen yield indicates how the photochemical reaction responds to intensity \(I\).

At low intensity, the rising portion of the Hill-reaction curve is approximately linear:

\[ Y_{\ce{O2}}\propto I, \]

so the apparent order is 1.

At high intensity, the curve approaches a plateau. Increasing intensity produces essentially no further increase in oxygen yield:

\[ Y_{\ce{O2}}\propto I^0, \]

so the apparent order is 0.

2.4(b) — Chlorophyll molecules per oxygen molecule

The saturation yield is approximately

\[ V_{\ce{O2}}=12\ \mathrm{mm^3} \]

per gram of chlorophyll per flash.

Convert the experimental conditions:

\[ T=283.15\ \mathrm{K}, \qquad p=\frac{740}{760}(101325) =9.866\times10^4\ \mathrm{Pa}, \]

\[ V_{\ce{O2}}=12\times10^{-9}\ \mathrm{m^3}. \]

The ideal-gas equation gives

\[ n(\ce{O2}) =\frac{pV}{RT} = \frac{(9.866\times10^4)(12\times10^{-9})} {8.314(283.15)} =5.03\times10^{-7}\ \mathrm{mol}. \]

One gram of chlorophyll contains approximately

\[ n(\text{chlorophyll}) =\frac{1\ \mathrm{g}}{900\ \mathrm{g\,mol^{-1}}} =1.11\times10^{-3}\ \mathrm{mol}. \]

The molecular ratio equals the molar ratio:

\[ \frac{N(\text{chlorophyll})}{N(\ce{O2})} = \frac{1.11\times10^{-3}}{5.03\times10^{-7}} \approx \boxed{2.2\times10^3}. \]

Thus about \(2200\) chlorophyll molecules are associated with the production of one oxygen molecule per saturating flash. This is an effective pigment-to-output ratio; it does not mean that all these pigments directly perform the oxygen-forming redox chemistry.

2.5 — Quantum requirement

The total light energy supplied is

\[ E_{\mathrm{light}} =(0.503\times10^{-3}\ \mathrm{J\,s^{-1}}) (2\times3600\ \mathrm{s}) =3.6216\ \mathrm{J}. \]

Using

\[ h=6.626\times10^{-34}\ \mathrm{J\,s}, \quad c=2.998\times10^8\ \mathrm{m\,s^{-1}}, \quad N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}, \]

the energy of one \(672\ \mathrm{nm}\) photon is

\[ \varepsilon_\gamma =\frac{hc}{\lambda} = \frac{(6.626\times10^{-34})(2.998\times10^8)} {672\times10^{-9}} =2.956\times10^{-19}\ \mathrm{J}. \]

The amount of photons is consequently

\[ n_\gamma =\frac{E_{\mathrm{light}}}{N_{\mathrm A}\varepsilon_\gamma} =2.034\times10^{-5}\ \mathrm{mol}. \]

At the same pressure and temperature as in 2.4,

\[ n(\ce{O2}) = \frac{(9.866\times10^4)(47.6\times10^{-9})} {8.314(283.15)} =1.995\times10^{-6}\ \mathrm{mol}. \]

Each oxygen molecule corresponds to four electrons transferred from water:

\[ n_e=4n(\ce{O2}) =7.979\times10^{-6}\ \mathrm{mol}. \]

Hence the quantum requirement is

\[ q=\frac{n_\gamma}{n_e} = \frac{2.034\times10^{-5}}{7.979\times10^{-6}} = \boxed{2.55\ \text{photons per electron}}. \]

Equivalently, approximately \(10.2\) photons are required per oxygen molecule formed. The reciprocal electron yield is about \(0.392\) net electrons transferred per photon.

2.6 — Conclusions

Statement Box to tick Reason
In natural photosynthesis, water oxidation and \(\ce{CO2}\) reduction are separated in space. Yes This is the intended biological interpretation: water oxidation occurs at the thylakoid membrane, while carbon fixation occurs in the stroma. Hill’s experiment demonstrates that oxygen evolution can operate with an alternative electron acceptor.
In chloroplasts, \(\ce{O2}\) is produced from \(\ce{CO2}\). No Oxygen evolution occurs without carbon dioxide. Water is the oxygen source and electron donor.
Oxidation of water in chloroplasts requires light illumination. Yes The observed process is photochemical; the positive Gibbs energy calculated in 2.3(c) is consistent with its need for an energy input.
Most of chlorophylls in chloroplasts participate directly in the photochemical \(\ce{O2}\) production. No The large pigment-to-oxygen ratio is consistent with many chlorophyll molecules collecting and transferring excitation energy to a much smaller number of reaction centres.
In isolated chloroplasts, every absorbed photon causes transfer of one electron. No On the absorbed-energy interpretation of 2.5, approximately \(2.55\) photons are required per net electron transferred, giving only about \(0.392\) electron per photon.

The solution was checked against the official solution.