Sugars cyclic molecule formation

The sugar specificity of RNase Tx appears to require a 2 -hydroxyl group for the substrate because DNA is not attacked by RNase Tx. This is consistent with the intermediary formation of 2, 3 -cyclic phosphate and also with the finding that 2 -0-methylated guanylyl bonds in tRNA is resistant to the enzyme (48)- Holy and Sorm (49) found that RNase Tx did not attack L-guanosine 2, 3 -cyclic phosphate and L-inosine 2, 3 -cyclic phosphate. They found further that RNase Tx split 9-(a-L-lyxo-furanosyl)-hypoxanthine 2, 3 -cyclic phosphate but not the D-lyxofura-nose derivative, and they concluded that the substrate molecule was fixed at least to three regions of RNase Tx (50). 

The cyclodextrins (CD s) have a proven capability for the formation of complexes with many hydrophobic structures 114.15.16). The a, P, and y cyclodextrins are cyclic oligomers of 6, 7, and 8 glucose (anylose) rings, respectively. The center cavity of the P form can easily include a large portion of a steriod molecule. The a cyclodextrin cavity can capture at best a small portion. Each cyclodextrin sugar unit has three hydroxyl groups, and no other substituents. These materials are not known, by themselves, to have any particular biological functions. 

The simultaneous presence of isothiocyanale and free hydroxyl groups in a sugar molecule may lead to formation of intramolecular cyclic thiocarba-... 

Two hydroxyls of a carbohydrate molecule may react with an aldehyde with the formation of a cyclic acetal. These products are known as 0-aryli-dene or 0-alkylidene derivatives of sugars. Acetone condenses similarly to give 0-isopropylidene (acetone) derivatives 2a). 

In an aqueous solution, the open-chain form of o-glucose is in equilibrium with the two cyclic hemiacetals. Because formation of the cyclic hemiacetals proceeds nearly to completion (unlike formation of acyclic hemiacetals), very little glucose is in the open-chain form (about 0.02%). Even so, the sugar still undergoes the reactions discussed in previous sections (oxidation, reduction, imine formation, etc.) because the reagents react with the small amount of open-chain aldehyde that is present. As the open-chain compound reacts, the equilibrium shifts to produce more open-chain aldehyde, which can then undergo reaction. Eventually, all the glucose molecules react by way of the open-chain form. 

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