Ebright R H, Wong J R
Proc Natl Acad Sci U S A. 1981 Jul;78(7):4011-5. doi: 10.1073/pnas.78.7.4011.
Binding analysis with purified bacterial receptor distinguishes two structural domains in cyclic AMP (cAMP). The first, the cyclic phosphate and furanose, constitutes a binding domain. This region is bound tightly to the receptor. The rest of cAMP is not bound; the adenine moiety of cAMP is exposed. Unlike binding, activity of cAMP requires the adenine moiety. To be active, cAMP must have in domain II the base adenine--specifically, its Watson--Crick atoms N-1 and N-6. Analysis of indoleacetic acid, a compound able to replace cAMP at the L-arabinose operon, indicates a similar distinction between binding and active domains. To be active, the indole must have substitution (carboxyl or amide) electronically comparable to the cAMP N-1 and N-6. On this basis, we propose a detailed mechanism for action of cAMP (or indoleacetic acid) in Escherichia coli. We propose that the exposed adenine of cAMP enters into the DNA. The adenine's N-1 and N-6 form hydrogen bonds to a thymine in DNA. This interaction destabilizes the DNA. It enhances transcription. Marked similarities indicate an identical mechanism for the steroid hormones in eukaryotes.
用纯化的细菌受体进行的结合分析区分了环磷酸腺苷(cAMP)中的两个结构域。第一个结构域是环磷酸酯和呋喃糖,构成一个结合域。该区域与受体紧密结合。cAMP的其余部分不结合;cAMP的腺嘌呤部分暴露在外。与结合不同,cAMP的活性需要腺嘌呤部分。要具有活性,cAMP在结构域II中必须含有碱基腺嘌呤——具体来说,是其沃森-克里克原子N-1和N-6。对吲哚乙酸(一种能够在L-阿拉伯糖操纵子处替代cAMP的化合物)的分析表明,结合域和活性域之间存在类似的区别。要具有活性,吲哚必须具有与cAMP N-1和N-6在电子性质上相当的取代基(羧基或酰胺)。在此基础上,我们提出了cAMP(或吲哚乙酸)在大肠杆菌中作用的详细机制。我们提出,cAMP暴露的腺嘌呤进入DNA。腺嘌呤的N-1和N-6与DNA中的胸腺嘧啶形成氢键。这种相互作用使DNA不稳定。它增强转录。显著的相似性表明真核生物中类固醇激素的作用机制相同。