Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
6th Edition
ISBN: 9798214167343
Author: GARRETT
Publisher: Cengage Learning US
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Chapter 21, Problem 1P

P700* Has the Most Negative Standard Reduction Potential Found in Nature In photosystem I. P700 in its ground state has an Δ ' = +04 V. Excitation of P700 by a photon of 700-nm light alters The Δ ' of P700* to -0.6 V. What is the efficiency of energy capture in this light reaction of P700?

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Interpretation Introduction

To propose:

In the P700 light reaction, energy efficiency should be determined.

Introduction:

The P700 is also known as primary donor photosystem I which is the chlorophyll that reacts at the center. This molecule associates with the photosystem I and has 700 nm the absorption spectrum.

Explanation of Solution

The equation of quantum light energy is given as,

  E=Nhcλ

We have,

  h=6.626×1034JsecN=6.02×1023c=3×108m/secλ=700nm=700×109 m

Substituting we have,

  E=(6.02×1023)(6.626×1034)(3×108)700×109

  =170950.8J=171 kJ

  ΔG0' Is calculated as,

  ΔG0'=nFΔE0'n=1F=96.485 kJ/Vmol

  ΔE0'=acceptorΔE0'donorΔE0'

  =0.6(0.4)=1.0V

So hence,

  ΔG0'=1(96.485)*1.0

  96.5 kJ/mol

P700 energy capture is calculated by the change of free energy of Gibb that is divided by one 700 nm photon for P700

  Efficiency=96.5171×100%

  =0.564=56.4%

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