Oxygen → Pyridoxal
Reaktionsgleichung
O=O.[CH3:3][C:4]1[C:9]([OH:10])=[C:8]([CH:11]=[O:12])[C:7]([CH2:13][O:14]P(O)(O)=O)=[CH:6][N:5]=1>P([O-])([O-])([O-])=O>[CH3:3][C:4]1[N:5]=[CH:6][C:7]([CH2:13][OH:14])=[C:8]([CH:11]=[O:12])[C:9]=1[OH:10]Komponenten
4
Edukt 2
Produkt 1
Reagenzien, Katalysatoren, Lösungsmittel 1
Versuchsvorschrift
A 3.87 g (0.06 mmol) portion of the thus obtained hemoglobin was dissolved in 0.1 M phosphate buffer (pH 8.0), and the solution was adjusted to a total volume of 25 ml. After de-oxidation of the solution by vigorous blowing of argon until the partial pressure of oxygen reached 2 mmHg or less, addition of pyridoxal 5′-phosphate was effected in accordance with a known method (R. Benesch et al., J. Biol. Chem., 257 (3), 1320-1324, 1982) to obtain pyridoxal-modified hemoglobin. The thus obtained pyridoxal-modified hemoglobin solution was adjusted to a total volume of one liter by adding 0.1 M borate buffer (pH 8.2).