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Evolution of enzyme catalytic power. Characteristics of optimal catalysis evaluated for the simplest plausible kinetic model

机译:酶催化能力的演变。最简单合理的动力学模型评估的最佳催化特性

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摘要

1. Evolutionary changes in the structure of an enzyme that provide an increase in its Km value are considered. Provided that Km increases as a result of increases in the forward rate constants of the catalysis relative to the reverse rate constants, the enzyme catalyses the conversion of a fixed concentration of its substrate more rapidly when its structure provides that Km>[S] than when Km<[S]. 2. Catalytic efficiency of enzymes is discussed in terms of the simplest plausible model, the Haldane [(1930) Enzymes, Longmans, London] reversible three-step model: [Formula: see text] The rate equation for the forward reaction of this model (formation of P) may be written in the simple form: [Formula: see text] Keq. is the equilibrium constant (=[P]eq./[S]eq.), and kcat.=V/[E]T, where [E]T is the total enzyme concentration. 3. To assess the effectiveness of an enzyme, it is necessary only to determine the extent to which the constraints of a particular kinetic mechanism permit v2 (v when Km»[S]) to approach vd (the diffusion-limited rate). 4. The value of the optimal rate of catalysis (vopt., the maximal value of v2) is dictated by the equilibrium constant for the reaction, Keq.; v2=vd/a, where [Formula: see text] when k+1 is assumed equal to k−3, and vopt.=vd/amin.. When Keq.≥1, it is necessary that k+2»k−1 for a to take its minimum value, amin.; when Keq.«1, it is necessary only that k+2»Keq.·k−1, i.e. a can equal amin. even if k+2<k−1. When Keq.»1, vopt.=vd; when Keq.=1, vopt.=vd/2, and when Keq.«1, vopt.=Keq.·vd. 5. The analysis, together with predicted effects of evolutionary pressure, suggests that in practice the rates of the fastest enzyme-catalysed freely reversible reactions might be expected to be lower than the value of k+1[E]T[S] by about an order of magnitude, particularly if Keq.<1. 6. The existing literature suggests that, in general, appropriate values of Km have evolved for the provision of high rates of catalysis but that many values of kcat. are not large enough to provide optimal rates of catalysis unless the value of k+1 in vivo is lower than its value in free solution.
机译:1.考虑了一种酶结构的进化变化,该变化会增加其Km值。假设Km是由于催化的正向速率常数相对于反向速率常数增加而增加,则当其结构提供Km> [S]时,该酶可以更快地催化固定浓度底物的转化。 Km <[S]。 2.根据最简单的合理模型Haldane [(1930)酶,Longmans,伦敦)可逆的三步模型讨论了酶的催化效率:[公式:请参见文本]该模型正向反应的速率方程(P的形式)可以用以下简单形式编写:[公式:参见文字] Keq。是平衡常数(= [P] eq ./ [S] eq。),kcat。= V / [E] T,其中[E] T是总酶浓度。 3.为了评估酶的有效性,仅需要确定特定动力学机制的约束条件允许v2(当Km»[S]时为v)接近vd(扩散限制率)的程度。 4.最佳催化速率的值(vopt。,v2的最大值)由反应的平衡常数Keq。决定; v 2 = v d / a ,其中[公式:参见文字]当 k +1 等于 k -3 v opt。 = v d / a min。。当 K eq。≥1时,必须 k +2 » k −1 表示 a 的最小值 a min。;当 K eq。«1时,只需要 k +2 » K eq。· k −1 ,即 a 可以等于 a < sub> min。,即使 k +2 em> k -1 。当 K eq。»1时, v opt。 = v d ;当 K eq。 = 1时, v opt。 = v d / 2,并且当 K eq。«1时, v opt。 = K eq。· v d 。 5.分析以及对进化压力的预测影响表明,在实践中,最快的酶催化的可逆反应的发生率可能低于 k +的值1 [E] T [S]大约一个数量级,特别是如果 K eq。 <1。 6.现有文献表明,总的来说,为了提供高催化速率,已经发展出适当的 K m 值,但是 k < / em> cat。的大小不足以提供最佳的催化速率,除非体内 k +1 的值低于免费解决方案中的值。

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