Grabolle Markus, Dau Holger
Freie Universität Berlin, FB Physik Arnimallee 14, D-14195 Berlin, Germany.
Biochim Biophys Acta. 2005 Jun 30;1708(2):209-18. doi: 10.1016/j.bbabio.2005.03.007. Epub 2005 Apr 2.
Photon absorption by one of the roughly 200 chlorophylls of the plant Photosystem II (PSII) results in formation of an equilibrated excited state (Chl200*) and is followed by chlorophyll oxidation (formation of P680+) coupled to reduction of a specific pheophytin (Phe), then electron transfer from Phe- to a firmly bound quinone (QA), and subsequently reduction of P680+ by a redox-active tyrosine residue denoted as Z. The involved free-energy differences (DeltaG) and redox potentials are of prime interest. Oxygen-evolving PSII membrane particles of spinach were studied at 5 degrees C. By analyzing the delayed and prompt Chl fluorescence, we determined the equilibrium constant and thus free-energy difference between Chl200* and the [Z+,QA-] radical pair to be -0.43+/-0.025 eV, at 10 mus after the photon absorption event for PSII in its S(3)-state. On basis of this value and previously published results, the free-energy difference between P680* and [P680+,QA-] is calculated to be -0.50+/-0.04 eV; the free-energy loss associated with electron transfer from Phe to QA is found to be 0.34+/-0.04 eV. The given uncertainty ranges do not represent a standard deviation or likely error, but an estimate of the maximal error. Assuming a QA-/QA redox potential of -0.08 V, the following redox-potential estimates are obtained: +1.25 V for P680/P680+; +1.21 V for Z/Z+ (at 10 mus); -0.42 V for Phe-/Phe; -0.58 V for P680*/P680+.
植物光系统II(PSII)中约200个叶绿素中的一个吸收光子,导致形成平衡激发态(Chl200*),随后叶绿素氧化(形成P680+),同时特定的去镁叶绿素(Phe)被还原,然后电子从Phe-转移到紧密结合的醌(QA),随后由一个称为Z的氧化还原活性酪氨酸残基将P680+还原。所涉及的自由能差(ΔG)和氧化还原电位是主要关注点。在5摄氏度下研究了菠菜的放氧PSII膜颗粒。通过分析延迟和即时叶绿素荧光,我们确定了平衡常数,从而确定了Chl200与[Z+,QA-]自由基对之间的自由能差在PSII处于S(3)状态、光子吸收事件后10微秒时为-0.43±0.025电子伏特。基于该值和先前发表的结果,计算出P680与[P680+,QA-]之间的自由能差为-0.50±0.04电子伏特;发现与电子从Phe转移到QA相关的自由能损失为0.34±0.04电子伏特。给定的不确定范围不代表标准偏差或可能误差,而是最大误差的估计值。假设QA-/QA氧化还原电位为-0.08伏,可得到以下氧化还原电位估计值:P680/P680+为+1.25伏;Z/Z+(在10微秒时)为+1.21伏;Phe-/Phe为-0.42伏;P680*/P680+为-0.58伏。