Hirao Hajime, Kumar Devesh, Shaik Sason
Department of Chemistry and the Lise Meitner-Minerva Center for Computational, Quantum Chemistry, The Hebrew University of Jerusalem, 91904 Jerusalem, Israel.
J Inorg Biochem. 2006 Dec;100(12):2054-68. doi: 10.1016/j.jinorgbio.2006.09.001. Epub 2006 Sep 19.
Density functional calculations show that in the absence of Compound I, the primary oxidant species of P450, the precursor species, Compound 0 (FeOOH), can effect double bond activation of 5-methylenylcamphor (1). The mechanism is initiated by homolytic cleavage of the O-O bond and formation of an OH radical bound to the Compound II species by hydrogen bonding interactions. Subsequently, the so-formed OH radical can either activate the double bond of 1 or attack the meso position of the heme en route to heme degradation. The calculations show that double bond activation is preferred over attack on the heme. Past the double bond activation, the intermediate can either lead to epoxidation or to a glycol formation. The glycol formation is predicted to be preferred, although in the P450(cam) pocket the competition may be closer. Therefore, in the absence of Compound I, Compound 0 will be capable of epoxidizing double bonds. Previous studies [E. Derat, D. Kumar, H. Hirao, S. Shaik, J. Am. Chem. Soc. 128 (2006) 473-484] showed that in the case of a substrate that can undergo only C-H activation, the bound OH prefers heme degradation over hydrogen abstraction. Since the epoxidation barrier for Compound I is much smaller than that of Compound 0 (12.8 vs. 18.9kcal/mol), when Compound I is present in the cycle, Compound 0 will be silent. As such, our mechanism explains lucidly why T252A P450(cam) can epoxidize olefins like 5-methylenylcamphor but is ineffective in camphor hydroxylation [S. Jin, T.M. Makris, T. A. Bryson, S.G. Sligar, J.H. Dawson, J. Am. Chem. Soc. 125 (2003) 3406-3407]. Our calculations show that the glycol formation is a marker reaction of Compound 0 with 5-methylenylcamphor. If this product can be found in T252A P450(cam) or in similar mutants of other P450 isozymes, this will constitute a more definitive proof for the action of Cpd 0 in P450 enzymes.
密度泛函计算表明，在不存在细胞色素P450的主要氧化剂物种化合物I的情况下，前体物种化合物0（FeOOH）能够实现5-亚甲基樟脑（1）的双键活化。该机制由O-O键的均裂引发，并通过氢键相互作用形成与化合物II物种结合的OH自由基。随后，如此形成的OH自由基既可以活化1的双键，也可以在血红素降解的过程中攻击血红素的中位。计算表明，双键活化比攻击血红素更占优势。在双键活化之后，中间体既可以导致环氧化，也可以导致二醇的形成。预计二醇的形成更占优势，尽管在细胞色素P450（樟脑）口袋中竞争可能更为接近。因此，在不存在化合物I的情况下，化合物0将能够使双键环氧化。先前的研究[E. Derat，D. Kumar，H. Hirao，S. Shaik，J. Am. Chem. Soc. 128 (2006) 473 - 484]表明，对于仅能进行C-H活化的底物，结合的OH优先选择血红素降解而非氢提取。由于化合物I的环氧化势垒比化合物0的小得多（12.8对18.9千卡/摩尔），当循环中存在化合物I时，化合物0将不起作用。因此，我们的机制清楚地解释了为什么T252A细胞色素P450（樟脑）能够使5-亚甲基樟脑等烯烃环氧化，但在樟脑羟基化方面无效[S. Jin，T.M. Makris，T. A. Bryson，S.G. Sligar，J.H. Dawson，J. Am. Chem. Soc. 125 (2003) 3406 - 3407]。我们的计算表明，二醇的形成是化合物0与5-亚甲基樟脑的标记反应。如果在T252A细胞色素P450（樟脑）或其他细胞色素P450同工酶的类似突变体中能发现该产物，这将为化合物0在细胞色素P450酶中的作用提供更确凿的证据。
