Write a short summary with conclusions about the trypan blue and SDH results. Do the class results agree with the expected effects of the various treatments?
Trypan blue exclusion is a method in which the blue dye is a crucial tool used to stain the nucleus of a cell that is non-viable or damaged. This shows early stages of toxicity whereas, the SDH leakage indicates a later onset of toxicity. The SDH leakage method requires the cell to have larger holes in the membrane to allow the sorbitol dehydrogenase enzyme molecules to leak out. This is because the enzyme molecules are much larger in comparison to metal ions.
Menadione (MD) readily undergoes a redox cycle in hepatocytes because they have high metabolic activity. When MD was present …show more content…
This active process would only occur in a living organism since requires energy, in the form of ATP. This process requires reducing agents, NADPH in the cytosol and NADH from the nucleus, to reduce into red coloured products and cause alterations in metabolic activity. Iodonitrotetrazolium reduction would most likely be observed in the later stages of toxicity since it requires energy.
What effects would you expect to see from menadione exposure if the hepatocytes were preincubated with: ascorbic acid (Vitamin C), buthionine sulfoximine (BSO, an inhibitor of glutathione synthesis), trolox (tocopherol analogue)
Exposure to menadione (Vitamin K3) will lead to the generation of reactive oxygen species and free radicals produced from redox cycling. High concentrations of reactive oxygen species can result in damage to tissues, mitochondrial DNA damage as well as cell death by necrosis or apoptosis (Loor et al., 2010). Glutathione (GSH) is essential in combating oxidative stress, due to the sulfhydryl group present on the cysteine residue. However, the levels of GSH are depleted in the presence of the reactive oxygen species produced by
Trypan blue exclusion is a method in which the blue dye is a crucial tool used to stain the nucleus of a cell that is non-viable or damaged. This shows early stages of toxicity whereas, the SDH leakage indicates a later onset of toxicity. The SDH leakage method requires the cell to have larger holes in the membrane to allow the sorbitol dehydrogenase enzyme molecules to leak out. This is because the enzyme molecules are much larger in comparison to metal ions.
Menadione (MD) readily undergoes a redox cycle in hepatocytes because they have high metabolic activity. When MD was present …show more content…
This active process would only occur in a living organism since requires energy, in the form of ATP. This process requires reducing agents, NADPH in the cytosol and NADH from the nucleus, to reduce into red coloured products and cause alterations in metabolic activity. Iodonitrotetrazolium reduction would most likely be observed in the later stages of toxicity since it requires energy.
What effects would you expect to see from menadione exposure if the hepatocytes were preincubated with: ascorbic acid (Vitamin C), buthionine sulfoximine (BSO, an inhibitor of glutathione synthesis), trolox (tocopherol analogue)
Exposure to menadione (Vitamin K3) will lead to the generation of reactive oxygen species and free radicals produced from redox cycling. High concentrations of reactive oxygen species can result in damage to tissues, mitochondrial DNA damage as well as cell death by necrosis or apoptosis (Loor et al., 2010). Glutathione (GSH) is essential in combating oxidative stress, due to the sulfhydryl group present on the cysteine residue. However, the levels of GSH are depleted in the presence of the reactive oxygen species produced by