Abstract
In a recent paper, we showed the difference between the first stage of the one-substrate and the two-substrate transketolase reactions – the possibility of transfer of glycolaldehyde formed as a result of cleavage of the donor substrate from the thiazole ring of thiamine diphosphate to its aminopyrimidine ring through the tricycle formation stage, which is necessary for binding and splitting the second molecule of donor substrate [O.N. Solovjeva et al., The mechanism of a one-substrate transketolase reaction, Biosci. Rep. 40 (8) (2020) BSR20180246]. Here we show that under the action of the reducing agent a tricycle accumulates in a significant amount. Therefore, a significant decrease in the reaction rate of the one-substrate transketolase reaction compared to the two-substrate reaction is due to the stage of transferring the first glycolaldehyde molecule from the thiazole ring to the aminopyrimidine ring of thiamine diphosphate. Fragmentation of the four-carbon thiamine diphosphate derivatives showed that two glycolaldehyde molecules are bound to both coenzyme rings and the erythrulose molecule is bound to a thiazole ring. It was concluded that in the one-substrate reaction erythrulose is formed on the thiazole ring of thiamine diphosphate from two glycol aldehyde molecules linked to both thiamine diphosphate rings. The kinetic characteristics were determined for the two substrates, fructose 6-phosphate and glycolaldehyde.
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