Talk:M9 salts - Revision history

Damian Ekiert: /* Miller Vs. Sambrook */

2007-06-10T02:39:39Z

Miller Vs. Sambrook

←Older revision		Revision as of 02:39, 10 June 2007
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	In the end, the two recipes are only slightly different when you take the difference in hydration into account. When you make 1x medium from 10x Miller M9, you end up with ~42mM Na<sub>2</sub>HPO<sub>4</sub>, while 1x medium made from 5x Sambrook M9 leaves you with ~48mM Na<sub>2</sub>HPO<sub>4</sub>. Presumably, this is close enough.		In the end, the two recipes are only slightly different when you take the difference in hydration into account. When you make 1x medium from 10x Miller M9, you end up with ~42mM Na<sub>2</sub>HPO<sub>4</sub>, while 1x medium made from 5x Sambrook M9 leaves you with ~48mM Na<sub>2</sub>HPO<sub>4</sub>. Presumably, this is close enough.
		+	(dcekiert)

	==References==		==References==

Damian Ekiert: /* Miller Vs. Sambrook */

2007-06-09T18:50:55Z

Miller Vs. Sambrook

←Older revision		Revision as of 18:50, 9 June 2007
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	== Miller Vs. Sambrook ==		== Miller Vs. Sambrook ==

-	At first I thought that this discrepancy may have arisen due to a miscalculation somewhere down the line brought about by the difference in molecular mass between Na<sub>2</sub>HPO<sub>4</sub><math>\cdot</math>7H<sub>2</sub>O and Na<sub>2</sub>HPO<sub>4</sub>, but I did a little digging and it seems that the real difference is who you trust more: Sambrook or Miller. The [http://openwetware.org/wiki/M9_salts[Sambrook protocol]] for 5x M9 salts uses 64g Na<sub>2</sub>HPO<sub>4</sub><math>\cdot</math>7H<sub>2</sub>O while the [Miller protocol for 5x calls for 30g Na<sub>2</sub>HPO<sub>4</sub> (hence 60g Na<sub>2</sub>HPO<sub>4</sub> in most 10x recipes).	+	At first I thought that this discrepancy may have arisen due to a miscalculation somewhere down the line brought about by the difference in molecular mass between Na<sub>2</sub>HPO<sub>4</sub><math>\cdot</math>7H<sub>2</sub>O and Na<sub>2</sub>HPO<sub>4</sub>, but I did a little digging and it seems that the real difference is who you trust more: Sambrook or Miller. The [http://openwetware.org/wiki/M9_salts[Sambrook protocol]] for 5x M9 salts uses 64g Na<sub>2</sub>HPO<sub>4</sub><math>\cdot</math>7H<sub>2</sub>O while the [http://www.changbioscience.com/protocols/recipe/M9salts10X.htm[Miller protocol]] for 5x calls for 30g Na<sub>2</sub>HPO<sub>4</sub> (hence 60g Na<sub>2</sub>HPO<sub>4</sub> in most 10x recipes).

	In the end, the two recipes are only slightly different when you take the difference in hydration into account. When you make 1x medium from 10x Miller M9, you end up with ~42mM Na<sub>2</sub>HPO<sub>4</sub>, while 1x medium made from 5x Sambrook M9 leaves you with ~48mM Na<sub>2</sub>HPO<sub>4</sub>. Presumably, this is close enough.		In the end, the two recipes are only slightly different when you take the difference in hydration into account. When you make 1x medium from 10x Miller M9, you end up with ~42mM Na<sub>2</sub>HPO<sub>4</sub>, while 1x medium made from 5x Sambrook M9 leaves you with ~48mM Na<sub>2</sub>HPO<sub>4</sub>. Presumably, this is close enough.

Damian Ekiert: Miller Vs. Sambrook

2007-06-09T18:50:06Z

Miller Vs. Sambrook

←Older revision		Revision as of 18:50, 9 June 2007
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)
		+
		+	== Miller Vs. Sambrook ==
		+
		+	At first I thought that this discrepancy may have arisen due to a miscalculation somewhere down the line brought about by the difference in molecular mass between Na<sub>2</sub>HPO<sub>4</sub><math>\cdot</math>7H<sub>2</sub>O and Na<sub>2</sub>HPO<sub>4</sub>, but I did a little digging and it seems that the real difference is who you trust more: Sambrook or Miller. The [http://openwetware.org/wiki/M9_salts[Sambrook protocol]] for 5x M9 salts uses 64g Na<sub>2</sub>HPO<sub>4</sub><math>\cdot</math>7H<sub>2</sub>O while the [Miller protocol for 5x calls for 30g Na<sub>2</sub>HPO<sub>4</sub> (hence 60g Na<sub>2</sub>HPO<sub>4</sub> in most 10x recipes).
		+
		+	In the end, the two recipes are only slightly different when you take the difference in hydration into account. When you make 1x medium from 10x Miller M9, you end up with ~42mM Na<sub>2</sub>HPO<sub>4</sub>, while 1x medium made from 5x Sambrook M9 leaves you with ~48mM Na<sub>2</sub>HPO<sub>4</sub>. Presumably, this is close enough.
		+
		+	==References==
		+
		+	J. Sambrook, D.W. Russell, ''Molecular Cloning: A Laboratory Manual'' (Cold Spring Harbor Laboratory Press, New York, ed. 3, 2001) pg. A2.2
		+	<br>J. Miller, ''Experiments in Molecular Genetics'' (Cold Spring Harbor Laboratory Press, New York, 1974) pg. 431

Damian Ekiert: /* $Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$ */

2007-06-09T18:07:07Z

<math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math>

←Older revision		Revision as of 18:07, 9 June 2007
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	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)
-
-	~~== <math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math> ==~~
-
-	I think that this may be explained by the additional mass of 7 waters associated with each <math>Na_2HPO_4</math>. The molecular weight of <math>Na_2HPO_4</math> is about 142g/mol, while that of <math>Na_2HPO_4 \cdot 7H_2O</math> is about 268g/mol. So 5x M9 salts could be made with either 64g/L <math>Na_2HPO_4 \cdot 7H_2O</math> (according to Sambrook) or 33.9g/L <math>Na_2HPO_4</math>--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)

Damian Ekiert: /* $Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$ */

2007-06-09T16:48:07Z

<math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math>

←Older revision		Revision as of 16:48, 9 June 2007
Line 4:		Line 4:
	== <math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math> ==		== <math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math> ==

-	I think that this may be explained by the additional mass of 7 waters associated with each <math>Na_2HPO_4</math>. The molecular weight of <math>Na_2HPO_4</math> is about 142g/mol, while that of <math>Na_2HPO_4 \cdot 7H_2O</math> is about 268g/mol. So 5x M9 salts could be made with either 64g/L <math>Na_2HPO_4 \cdot 7H_2O</math> (according to Sambrook) or 33.9g/L <math>Na_2HPO_4</math>--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.	+	I think that this may be explained by the additional mass of 7 waters associated with each <math>Na_2HPO_4</math>. The molecular weight of <math>Na_2HPO_4</math> is about 142g/mol, while that of <math>Na_2HPO_4 \cdot 7H_2O</math> is about 268g/mol. So 5x M9 salts could be made with either 64g/L <math>Na_2HPO_4 \cdot 7H_2O</math> (according to Sambrook) or 33.9g/L <math>Na_2HPO_4</math>--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)

Damian Ekiert: $Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$

2007-06-09T16:45:45Z

<math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math>

←Older revision		Revision as of 16:45, 9 June 2007
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)
		+
		+	== <math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math> ==
		+
		+	I think that this may be explained by the additional mass of 7 waters associated with each <math>Na_2HPO_4</math>. The molecular weight of <math>Na_2HPO_4</math> is about 142g/mol, while that of <math>Na_2HPO_4 \cdot 7H_2O</math> is about 268g/mol. So 5x M9 salts could be made with either 64g/L <math>Na_2HPO_4 \cdot 7H_2O</math> (according to Sambrook) or 33.9g/L <math>Na_2HPO_4</math>--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.

MDolinar at 09:53, 30 March 2006

2006-03-30T09:53:22Z

New page

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)

Talk:M9 salts - Revision history

Damian Ekiert: /* Miller Vs. Sambrook */

Damian Ekiert: /* Miller Vs. Sambrook */

Damian Ekiert: Miller Vs. Sambrook

Damian Ekiert: /* Na_2HPO_4 \cdot 7H_2O Versus Na_2HPO_4 */

Damian Ekiert: /* Na_2HPO_4 \cdot 7H_2O Versus Na_2HPO_4 */

Damian Ekiert: Na_2HPO_4 \cdot 7H_2O Versus Na_2HPO_4

MDolinar at 09:53, 30 March 2006

Damian Ekiert: /* $Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$ */

Damian Ekiert: /* $Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$ */

Damian Ekiert: $Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$