Field of Science

Two mistakes discovered

My GFAJ-1 cells grow on the designated medium (AML60) if it is solidified by agar, but not if it is solidified with agarose.  In response to a tweet I sent out about possible nutrient differences between agar and agarose, Mark Martin suggested potassium.  This led me to the discovery of two errors, one by me and one by Wolfe-Simon et al.

My error was that my stock phosphate solution was sodium phosphate, not potassium phosphate as specified by Wolfe-Simon et al.  Because the specified no-phosphate AML60 base recipe does not include a source of potassium, my liquid medium had no potassium.  Agar does contain potassium as a contaminant; the analysis of Bacto agar lists 0.121% potassium.  My back-of-the-envelope (literally) calculation converts this to about 0.2 mM potassium in medium solidified with 1.5% agar.  That's not a lot. but probably enough, and tenfold more than is present in agarose.

Wolfe-Simon et al's error makes their growth results even harder to interpret.  The originally published recipe for AML60 medium ('artificial Mono Lake') includes 1.5 mM potassium phosphate, which provides both potassium and phosphate.  (The lake water is about 24 mM potassium.)  This medium was modified by Wolfe-Simon et al, who initially removed the phosphate and replaced it with arsenate.  Although they used potassium phosphate for their +P version, they used sodium arsenate, not potassium arsenate, for their +As versions. Thus their arsenate-grown cells were starved for both arsenate and potassium.

Anyway, I've now added 2.5 mM potassium chloride to my AML60 medium.


  1. Cool. The real tests should move pretty quickly now.

  2. Nice! Could the growth in +As seen in the paper thus be caused by traces of potassium in the sodium arsenate? Presumably it wouldn't take a lot of potassium, and they added a lot of Na-arsenate. This would also easily explain the poor growth in -As/-P.

    Justin Cayce


Markup Key:
- <b>bold</b> = bold
- <i>italic</i> = italic
- <a href="">FoS</a> = FoS