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首页> 外文期刊>Applied and Environmental Microbiology >Role of D-ribose as a cometabolite in D-xylose metabolism by Saccharomyces cerevisiae.
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Role of D-ribose as a cometabolite in D-xylose metabolism by Saccharomyces cerevisiae.

机译:D-核糖作为代谢物在酿酒酵母D-木糖代谢中的作用。

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The influence of D-ribose as a cosubstrate on the uptake and metabolism of the non-growth substrate D-xylose by Saccharomyces cerevisiae ATCC 26602 was investigated. Xylose was taken up by means of low- and high-affinity glucose transport systems. In cells exposed for 2 days to a mixture of xylose and ribose, only the high-affinity system could be detected. Glucose strongly inhibited the transport of xylose by both systems. Starvation or exposure to either xylose or ribose resulted in inactivation of xylose transport, which did not occur in the presence of a mixture of ribose and xylose. A constitutive non-glucose-repressible NADPH2-dependent xylose reductase with a specific activity of ca. 5 mU/mg of protein that converted xylose to xylitol was present in a glucose-grown culture. No activity converting xylitol to xylulose or vice versa was found in crude extracts. Both xylose and ribose were converted to their corresponding polyols, xylitol and ribitol, as indicated by 13C nuclear magnetic resonance spectroscopy. Furthermore, ethanol was detected, and this implied that pathways for the complete catabolism of xylose and ribose exist. However, the NADPH2 required for the conversion of xylose to xylitol is apparently not supplied by the pentose phosphate pathway since the ethanol produced from D-[1-13C]xylose was labelled only in the C-2 position. Acetic acid was produced from ribose and may assist in the conversion of xylose to xylitol by cycling NADPH2.
机译:研究了D-核糖作为共底物对酿酒酵母ATCC 26602对非生长底物D-木糖的摄取和代谢的影响。木糖通过低亲和力和高亲和力的葡萄糖转运系统吸收。在暴露于木糖和核糖混合物2天的细胞中,只能检测到高亲和力系统。葡萄糖强烈地抑制了两个系统对木糖的运输。饥饿或暴露于木糖或核糖会导致木糖运输失活,这在存在核糖和木糖的混合物时不会发生。一种非葡萄糖可抑制的组成型NADPH2依赖性木糖还原酶,具有约1的比活性。在葡萄糖生长的培养物中存在5 mU / mg的将木糖转化为木糖醇的蛋白质。在粗提取物中未发现将木糖醇转化为木酮糖或反之亦然的活性。如13C核磁共振波谱所表明的,木糖和核糖都转化为它们相应的多元醇,木糖醇和核糖醇。此外,检测到乙醇,这意味着存在木糖和核糖完全分解代谢的途径。但是,由于从D- [1-13C]木糖产生的乙醇仅被标记在C-2位置,因此木糖转化为木糖醇所需的NADPH2显然不是由戊糖磷酸途径提供的。乙酸是由核糖产生的,可通过循环NADPH2协助木糖转化为木糖醇。

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