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 ΔG°' for hydrolysis of ATP under conditions existing in the chloroplast is about -50 kJ/mol, calculate the corresponding percent efficiency of cyclic photophosphorylation using light of 700nm.

Possibly Helpful Information:

E=h(c/λ)

h=Planck's Constant=6.626176 x 10^-34 J·s

C=Speed of Light=3x10⁸ m/sec

Answer 1

First, we need to calculate the total energy, E, produced by 700 nm light as given below:

We know,

E = h c/λ

Where

h = Planck’s constant = 6.626176 x 10^-34 J·s

c = Speed of Light = 3x10⁸ m/sec

λ = wavelength of light used, given in the problem, i.e. 700 nm = 7.0 x 10^-7 m

Substituting the above values into the equation, we will get

E = (6.626176 x 10^-34) x (3x10⁸) / 7.0 x 10^-7

= 19.878528 x 10^-26 / 7.0 x 10^-7

= 19.878528 x 10^-19 / 7.0

E = 2.8398 x 10^-19 J/photon

We need to calculate the energy produced by 1 mole of photons = 6.022×10²³ photons as given below:

E = 2.8398 x 10^-19 x 6.022×10²³ J/mole of photons

E = 17.1012756 x 10⁴ J/ mole of photons

E = 171.013 kJ/ mole of photons

Since two electrons must be passed to generate 1 ATP, we need to calculate energy produced by two moles of photons (1 per electron),

Therefore the total light energy produced by 2 mole of photons = 171.013 x 2 = 342.026 kJ

The free energy, G^o, produced during hydrolysis of 1 mole of ATP = -50 kJ

We can calculate percent efficiency of cyclic photophosphorylation by dividing the above value (-50 kJ) with energy produced by 2 mole of photons (342.026 kJ) as shown below:

The percent efficiency: 50/342.026 = 0.14618 x100 = 14.618% = 14.6%

So percent efficiency of cyclic photophosphorylation using light of 700nm in the chloroplast = 14.6%.