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Outside of 'purposefully' removing mitochondria to diffuse cytochrome C's apoptotic threat, I would assume that lung carcinomas, and their proximity to high oxygen levels, would have the lowest rate of cells entering the Warburg effect. Would this be a correct assumption?

Thanks

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As far as I can tell, there are no data to support your hypothesis. Someone correct me if I'm wrong.

There's actually some data to support that aerobic glycolysis occurs in lung carcinomas despite their proximity to oxygen (ref). There's even an article on the prolifertative pathway behind lung microvascular endothelial cells (ref), alluding to this requirement for anaerobic glycolysis in rapid proliferation.

T cells exhibit a similar pathway, switching to aerobic glycolysis in the effector phase where T cells are rapidly expanding (see clonal expansion). The general trend is that the Warburg effect lends itself to the rapid accumulation of biomass, supporting rapid cell division, instead of energy production to support cell survival (ref). You could rationalize it like cancer is already a failure in cell death, and so the cancer cells need to worry more about carbon compounds that support rapid division. There was also a paper by Cai et al. in 2014 that described their hypothesis behind aerobic glycolysis in cancer, whereby the Warburg effect helps tumors resist anoikis and thereby promote metastasis (ref).

It may be that the Warburg effect is a means to a required end for cancer cells, but as far as I know there's is no correlation between proximity to oxygen and incidence of Warburg metabolism. There are papers out there that continue to ask: is the Warburg effect a cause or consequence of cancer (ref? My thought is that cancer cells will undergo the Warburg effect at a given stage regardless of histology or behavior because it's beneficial.

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