Mechanism of N2O reduction by the μ4-S tetranuclear Cuz cluster of nitrous oxide reductase

被引:308
作者
Gorelsky, SI [1 ]
Ghosh, S [1 ]
Solomon, EI [1 ]
机构
[1] Stanford Univ, Dept Chem, Stanford, CA 94305 USA
关键词
D O I
10.1021/ja055856o
中图分类号
O6 [化学];
学科分类号
0703 ;
摘要
Reaction thermodynamics and potential energy surfaces are calculated using density functional theory to investigate the mechanism of the reductive cleavage of the N-O bond by the mu(4)-sulfide-bridged tetranuclear Cu-Z site of nitrous oxide reductase. The Cu-Z cluster provides an exogenous ligand-binding site, and, in its fully reduced 4Cu(I) state, the cluster turns off binding of stronger donor ligands while enabling the formation of the Cu-Z-N2O complex through enhanced Cu-Z -> N2O back-donation. The two copper atoms (Cu-I and Cu-IV) at the ligand-binding site of the cluster play a crucial role in the enzymatic function, as these atoms are directly involved in bridged N2O binding, bending the ligand to a configuration that resembles the transition state (TS) and contributing the two electrons for N2O reduction. The other atoms of the Cu-Z cluster are required for extensive back-bonding with minimal a ligand-to-metal donation for the N2(O) activation. The low reaction barrier (18 kcal mol(-1\)) of the direct cleavage of the N-O bond in the Cu-Z-N2O complex is due to the stabilization of the TS by a strong Cu-IV(2+)-O- bond. Due to the charge transfer from the Cu-Z cluster to the N2O ligand, noncovalent interactions with the protein environment stabilize the polar TS and reduce the activation energy to an extent dependent on the strength of proton donor. After the N-O bond cleavage, the catalytic cycle consists of a sequence of alternating protonation/one-electron reduction steps which return the Cu-Z cluster to the fully reduced (4Cu(I)) state for future turnover.
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页码:278 / 290
页数:13
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