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The question is in the title, but I'll explain why the question arose. I'm curious about the rates that various cells in the body divide, and have found various information relating to this, but nowhere can I find how often the cells divide in hair follicles (only that hair grows about 1 cm per 28 days, and that hair matrix epithelial cells divide very rapidly, which results in chemo patients' hair falling out). I figure that I can calculate a back-of-the envelope answer if I know the average size of hair cells. For example if they are 10 $\mu$m, and hair grows 1 cm per 28 days, then I would estimate that the relevant cell turnover time in the follicle is on the order of 1.5 per hour, assuming that the cellular division leading to the outward growth of the hair proceeds in a serial fashion. This sounds like a roughly plausible estimate, since it is significantly faster than the turnover of other cells in the human body with a high turnover (like stomach or blood neutrophils) which have a turnover of a few days.

In any case I would appreciate any insight, whether the size of hair follicle matrix epithelial cells or the cellular turnover rate in the follicle during the anagen phase.

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  • $\begingroup$ 10uM sounds reasonable, thouhg perhaps a bit low for a keratinocyte when "flattened out". But 1.5 cell divisions / hour sounds way too high. The fastest dividing human cells I'm aware of are activated lymphocytes, which can reach about 10h doubling time. The fastest cancer cell lines are around 15h. Probably there is a larger dividing mass in the follice which then "thins out" into the hair, so that comparing hair growth rate to cell diameter underestimates the doubling time. $\endgroup$ – Roland Sep 27 '15 at 9:24

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