Problem 2 - Biochemistry: Concepts and Connections

Question

In cyclic photophosphorylation, it is estimated that two electrons must be passed through the cycle to pump enough protons to generate one ATP. Assuming that the \(\Delta G\) for hydrolysis of ATP under conditions existing in the chloroplast is about \(-50 \mathrm{~kJ} / \mathrm{mol}\), what is the corresponding percent efficiency of cyclic photophosphorylation, using light of \(700 \mathrm{~nm}\) ?

Step-by-Step Solution

Answer

The percent efficiency of cyclic photophosphorylation is approximately 14.62%.

Step 1: Calculate Energy of a Photon

To find the energy of a photon of light with a given wavelength, use the equation \[ E = \frac{hc}{\lambda} \]where \( E \) is the energy of a photon in joules, \( h \) is Planck’s constant \((6.626 \times 10^{-34} \text{ J s})\), \( c \) is the speed of light \((3.00 \times 10^8 \text{ m/s})\), and \( \lambda \) is the wavelength of light in meters. We need to convert 700 nm to meters: \[ 700 \text{ nm} = 700 \times 10^{-9} \text{ m} \]Plug in these values to find the energy:\[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{700 \times 10^{-9}} \approx 2.84 \times 10^{-19} \text{ J} \]

Step 2: Calculate Energy Per Mole of Photons

Use Avogadro's number \( (6.022 \times 10^{23} \text{ mol}^{-1}) \) to calculate the energy per mole of photons:\[ E_{\text{per mole}} = 2.84 \times 10^{-19} \times 6.022 \times 10^{23} \approx 171 \text{ kJ/mol} \]

Step 3: Calculate Energy Required to Form ATP

Given that two electrons are required to produce one ATP, we calculate the energy input for two moles of photons:\[ E_{\text{input}} = 2 \times 171 = 342 \text{ kJ/mol} \]One ATP requires \(-50 \text{ kJ/mol}\) for hydrolysis.

Step 4: Calculate Efficiency of Photophosphorylation

Efficiency \( \eta \) is calculated using the formula:\[ \eta = \left(\frac{E_{\text{useful}}}{E_{\text{input}}}\right) \times 100\%= \left(\frac{50}{342}\right) \times 100\% \approx 14.62\% \]

Step 5: Conclusion on Efficiency

Thus, the corresponding percent efficiency of cyclic photophosphorylation using light of \(700 \text{ nm}\) is approximately **14.62\%**.

Key Concepts

Efficiency of Photophosphorylation

Photophosphorylation is a process where light energy is converted into chemical energy, particularly in the form of ATP. In cyclic photophosphorylation, electrons follow a cyclic pathway that helps generate ATP without producing NADPH, crucial for processes in chloroplasts. The efficiency of this process is a measure of how effectively the light energy absorbed is used to produce ATP.

In the exercise provided, approximately 14.62% efficiency is calculated for cyclic photophosphorylation when using light with a wavelength of 700 nm. Efficiency (eta) is determined by the ratio of the useful energy (needed to synthesize ATP) to the total energy input (supplied by light). Here, two moles of photons are required, calculated to input around 342 kJ/mol, while the ATP hydrolysis energy is -50 kJ/mol.

This efficiency indicates that not all absorbed light energy gets converted into a usable form for biological processes, leading to energy losses often dissipated as heat. However, this efficiency is quite significant given the complexity and demands of cellular work.

ATP Synthesis

ATP, or adenosine triphosphate, is a vital energy carrier in cells, providing the necessary power for various biological functions. During cyclic photophosphorylation, light-driven electron transfers within chloroplasts create a proton gradient across the thylakoid membrane, which in turn drives ATP synthesis.

In this process, electrons pass through photosystem complexes and return to the starting point, pumping protons into the thylakoid lumen. This proton gradient, a form of potential energy, powers ATP synthase, an enzyme facilitating the conversion of ADP and inorganic phosphate into ATP.

For every two electrons cycling through the system, sufficient protons are translocated to synthesize one ATP molecule. This systematic cycle is efficient for ATP production without needing additional substrates like water or NADP+, unlike non-cyclic photophosphorylation. Cyclic photophosphorylation is particularly vital when the cell's demand for ATP outpaces its need for reducing power.

Photon Energy Calculation

The energy contained in a single photon of light is crucial for driving biological processes like photophosphorylation. To calculate this energy, the key formula is \( E = \frac{hc}{\lambda} \), where:

• \( E \) is the energy in joules,
• \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s} \)),
• \( c \) is the speed of light (\(3.00 \times 10^8 \, \text{m/s} \)), and
• \( \lambda \) is the wavelength in meters.

In this instance, light with a 700 nm wavelength is used, equivalent to \(700 \times 10^{-9} \) meters. By substituting these into the equation, the calculated energy of a photon is approximately \(2.84 \times 10^{-19}\) joules.

Further, using Avogadro's number, you can expand this calculation to determine the energy per mole of photons, which translates to 171 kJ/mol. This high precision in energy calculation helps elucidate the photophosphorylation efficiency and subsequentially the ability of cells to convert light into stored chemical energy. The better we understand photon energetics, the better we can comprehend the intricacies of cellular energy transformations.