# Talk:M9 salts

(Difference between revisions)
 Revision as of 12:48, 9 June 2007 (view source) (→$Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$)← Previous diff Revision as of 14:07, 9 June 2007 (view source) (→$Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$)Next diff → Line 1: Line 1: The 5x M9 salts recipe here seems to be almost identical to the 10x M9 salts protocol we use in our lab and which I also found elsewhere (see e.g. http://www.changbioscience.com/protocols/recipe/M9salts10X.htm and http://www.atcc.org/mediapdfs/2057.pdf). It could be that bacteria don't mind if they get the double of salts, but it might nevertheless be interesting to know what conc. of salts would be best to use... The 5x M9 salts recipe here seems to be almost identical to the 10x M9 salts protocol we use in our lab and which I also found elsewhere (see e.g. http://www.changbioscience.com/protocols/recipe/M9salts10X.htm and http://www.atcc.org/mediapdfs/2057.pdf). It could be that bacteria don't mind if they get the double of salts, but it might nevertheless be interesting to know what conc. of salts would be best to use... (MDolinar) (MDolinar) - - == $Na_2HPO_4 \cdot 7H_2O$  Versus  $Na_2HPO_4$ == - - I think that this may be explained by the additional mass of 7 waters associated with each $Na_2HPO_4$.  The molecular weight of $Na_2HPO_4$ is about 142g/mol, while that of $Na_2HPO_4 \cdot 7H_2O$ is about 268g/mol.  So 5x M9 salts could be made with either 64g/L $Na_2HPO_4 \cdot 7H_2O$ (according to Sambrook) or 33.9g/L $Na_2HPO_4$--the concentration will be the same.  Then if you double this for 10x salts, you would end up needing 67.8g/L, which is close to the 60g/L given in the standard 10x recipes.  I can only wave my hands at the 7.8g discrepancy and say that someone goofed or altered the recipe slightly. (dcekiert)