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首页> 外文期刊>Genes and Development: a Journal Devoted to the Molecular Analysis of Gene Expression in Eukaryotes, Prokaryotes, and Viruses >Specific functional interactions of nucleotides at key -3 and +4 positions flanking the initiation codon with components of the mammalian 48S translation initiation complex.
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Specific functional interactions of nucleotides at key -3 and +4 positions flanking the initiation codon with components of the mammalian 48S translation initiation complex.

机译:起始密码子两侧的关键-3和+4位置上的核苷酸的特定功能相互作用与哺乳动物48S翻译起始复合物的成分有关。

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摘要

Eukaryotic initiation factor (eIF) 1 maintains the fidelity of initiation codon selection and enables mammalian 43S preinitiation complexes to discriminate against AUG codons with a context that deviates from the optimum sequence GCC(A/G)CCAUGG, in which the purines at (-)3 and (+)4 positions are most important. We hypothesize that eIF1 acts by antagonizing conformational changes that occur in ribosomal complexes upon codon-anticodon base-pairing during 48S initiation complex formation, and that the role of (-)3 and (+)4 context nucleotides is to stabilize these changes by interacting with components of this complex. Here we report that U and G at (+)4 both UV-cross-linked to ribosomal protein (rp) S15 in 48S complexes. However, whereas U cross-linked strongly to C(1696) and less well to AA(1818-1819) in helix 44 of 18S rRNA, G cross-linked exclusively to AA(1818-1819). U at (-)3 cross-linked to rpS5 and eIF2alpha, whereas G cross-linked only to eIF2alpha. Results of UV cross-linking experiments and ofassays of 48S complex formation done using alpha-subunit-deficient eIF2 indicate that eIF2alpha's interaction with the (-)3 purine is responsible for recognition of the (-)3 context position by 43S complexes and suggest that the (+)4 purine/AA(1818-1819) interaction might be responsible for recognizing the (+)4 position.
机译:真核起始因子(eIF)1保持了起始密码子选择的保真度,并使哺乳动物43S预起始复合物能够区分AUG密码子,且其背景不同于最佳序列GCC(A / G)CCAUGG,其中嘌呤在(-) 3和(+)4位置最为重要。我们假设eIF1通过拮抗48S起始复合物形成过程中核糖体复合物在密码子-反碱基碱基配对时发生的构象变化而起作用,并且(-)3和(+)4上下文核苷酸的作用是通过相互作用来稳定这些变化与这个复杂的组件。在这里,我们报告在(+)4处的U和G都通过紫外线与48S复合物中的核糖体蛋白(rp)S15交联。但是,尽管U在18S rRNA的螺旋线44中与C(1696)牢固地交联,而与AA(1818-1819)的交联却不太好,而G仅与AA(1818-1819)交联。在(-)3处的U交联到rpS5和eIF2alpha,而G仅交联到eIF2alpha。紫外线交联实验和使用α-亚基缺陷的eIF2完成的48S复合物形成的测定结果表明,eIF2alpha与(-)3嘌呤的相互作用是由43S复合物识别(-)3上下文位置的原因,并表明(+)4嘌呤/ AA(1818-1819)相互作用可能是识别(+)4位置的原因。

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