Abstract

Hyperglycemia is a medical condition when the blood glucose levels increase exceeds the reasonable limit. Hyperglycemia is a typical sign of diabetes mellitus (DM). Indonesia is the sixth-ranked country in the world after China, India, United States, Brazil, and Mexico. Treatment with herbs is currently being developed. Pare (Momordica charantia) and Belimbing wuluh (Averrhoa bilimbi) are plants that found around us. Some studies state that each of these plants can be anti-diabetic. The hyperglycemia can cause an immune system disorder characterized by pancreatic β cell death involving IL-1β, kappa B (NF)-κB nuclear factor, and Fas. The ability of NF-kB activation will affect the number of cytokines expressed by T cells, namely TNF-α, and IFN-γ. The purpose of this study is to determine the effect of NF-kB activation on blood glucose levels in hyperglycemia mice. The results showed that the positive control treatment showed an increase in the number of NF-kB activations in CD4 and CD8 cells. EPBW (combination of Averrhoa bilimbi extract and Momordica charantia) administration results at a dose of 10 mg.kg-1 BW showed a reduction in the amount of activated NF-kB as a substitute for the reduction. In addition, that dose can reduce blood sugar levels in mice hyperglycemia model.

Highlights

  • RESULTThe positive control group is a hyperglycemia (K+) mouse model of the relative number of CD4+ cells that express the highest nuclear factor kappa B (NF-kB) compared to the healthy control group (K-), which is 7.45% (p

  • INTRODUCTION* Hyperglycemia is a typical sign of diabetes mellitus (DM)

  • Diabetogenic effects on streptozotocin administration will be initiated by reactive oxygen species (ROS) through direct toxic effects on GLUT 2, the cytokine action of TNF-α, and INF-γ due to stimulation of dependent T cells

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Summary

RESULT

The positive control group is a hyperglycemia (K+) mouse model of the relative number of CD4+ cells that express the highest NF-kB compared to the healthy control group (K-), which is 7.45% (p

Discussion
CONCLUSION
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