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Does the size of a starter always dictate the cell count of yeast irregardless of how much yeast you started with (ignoring other factors such as aeration, wort density etc)?

For instance let's take two scenarios:

1) I have a 100 mL starter with the cell count of 2 billion. I created a new starter that is 600 mL. I now have a cell count of 8 billion.

2) 1) I have a 100 mL starter with the cell count of 1 billion. I created a new starter that is 600 mL. I now have a cell count of 8 billion.

The math isn't necessarily correct because I forget how it all works, but what I'm trying to illustrate is that no matter how much yeast you started with your always going to have a constant cell count based on the volume of wort. Is this true?

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No, it is not true. The "yield factor" of yeast is a function of both the starter volume and the inoculation rate.

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  • Cool link. By inoculation rate you mean the number of yeast cells present right? I.e in the article Billions/gram of extra.
    – fthinker
    Commented Nov 11, 2015 at 3:51
  • I really mean cells/volume, but Kai used cells/gram-extract for reasons he outlines in the article (and seems sound to me). For a given/fixed wort density (as you assume), they're going to be different values, obviously, but roughly comparable. For instance, most starter wort is somewhere around 1.035-1.042. Grams-extract/volume is going to vary a tiny amount in that range, so cells/mL vs. cells/gram-extract is going to vary by that same tiny amount. But, yes, cells/gram-extract is technically the more useful thing to track. At least, within some range of gravity.
    – jsled
    Commented Nov 11, 2015 at 18:54

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