Lidocaine: an inhibitor in the free-radical-induced hemolysis of erythrocytes

J Biochem Mol Toxicol. Mar-Apr 2009;23(2):81-6. doi: 10.1002/jbt.20267.

Abstract

Lidocaine was reported to protect erythrocytes from hemolysis induced by 2,2'-azobis(2-amidinopropane) dihydrochloride (AAPH). Since AAPH-induced hemolysis was a convenient in vitro experimental system to mimic erythrocytes undergoing peroxyl radicals attack, the aim of this work was to investigate the antioxidant effect of lidocaine on AAPH-induced hemolysis by chemical kinetics. As a result, one molecule of lidocaine can only trap 0.37 radical, much lower than melatonin. Meanwhile, lidocaine cannot protect erythrocytes from hemolysis induced by hemin, which the mechanism of hemolysis was due to the erythrocyte membrane destroyed by hemin. Accordingly, lidocaine protected erythrocytes by scavenging radicals preferentially rather than by stabilizing membrane. Moreover, the interactions of lidocaine with two radical species, including 2,2'-azinobis(3-ethylbenzothiazoline-6-sulfonate) radical cation (ABTS(+*)) and 2,2'-diphenyl-1-picrylhydrazyl (DPPH), indicated that lidocaine can reduce ABTS(+*) with 260 microM as the 50% inhibition concentration (IC(50)) and cannot react with DPPH. Thus, lidocaine served as a reductant rather than a hydrogen donor to interact with radicals. Finally, the quantum calculation proved that, compared with the melatonin radical, the stabilization of N-centered radical of lidocaine was higher than the amide-type N-centered radical but lower than the indole-type N-centered radical in melatonin. These results provided basic information for lidocaine to be an antiradical drug.

Publication types

  • Research Support, Non-U.S. Gov't

MeSH terms

  • Amidines / toxicity
  • Erythrocytes / drug effects*
  • Free Radicals*
  • Hemin / toxicity
  • Hemolysis / drug effects*
  • Humans
  • Lidocaine / pharmacology*

Substances

  • Amidines
  • Free Radicals
  • 2,2'-azobis(2-amidinopropane)
  • Hemin
  • Lidocaine