-1
$\begingroup$

How does one prepare concentrations in the mass/volume (weight/volume) form, for substances like nucleic acids or in this case, proteinase? A detailed example would be helpful.

I need to prepare squishing buffer for a DNA extraction /PCR exercise to help me learn the materials better for a in class "internship".

The squishing buffer recipe I found from a laboratory class manual describes it as: 10 mM of Tris-Cl, pH 8.2, 1 mM of EDTA, 25 mM of NaCl , and 200 µg/ml of proteinse K (freshly diulted).

I understand molarity concentrations well enough. One only has to determine the total molar mass of a substance, in this case NaCl, and with that molar mass (g/mol) then use the given desired molarity (in this case 25 mM of NaCl, which equals 0.025 mol /one liters) to get the number of grams of the substance needed. In this case, since we need 0.025 moles / 1 Liter, we would need to multiply that by the molar mass to get: (g/mol) X (mol/l) = g/l of NaCl.

I understand percent solutions (to a degree) as well. If I needed 1% agrose gel, I would take 1 gram of agarose solid powder and bring that to 100 ml total to obtain a 1%, either using water or buffer?

I am having trouble understanding the concepts of concentration, in reference to mg/ml.

From what I understand, µg/ml is a weight (mass)/volume type of concentration. Therefore, should it be the case that if I wanted to prepare 200 µg/ml of proteinase (freshly diluted), how would I go about it? Would I add 5 ml of deionized water to get a working volume of 1000 µg of proteinase per 5 ml of water, which would be equivalent to 1 mg of proteinase per 5 ml of water. From that 1mg of proteinase to 5ml of water, should I then draw 0.2 µl (or 200 ml) of solution to then have the 200 µg/ml? Or am I totally wrong? I tried googling for more information, but I became more and more confused.

$\endgroup$
  • $\begingroup$ What is a "squishing buffer"? The composition seems to indicate some kind of protease lysis buffer, but I have never heard of anything by that name. $\endgroup$ – March Ho Mar 24 '15 at 18:45
  • $\begingroup$ @MarchHo Yes it is a buffer used to break down the flies protiens via the protienase. Since I will be doing PCR on flies. $\endgroup$ – Ro Siv Mar 24 '15 at 18:56
  • $\begingroup$ @canadianer So lets say I have some of that lyophilized powder. Could you give an example of doing so? I just cant seem to picture doing the process for some reason. $\endgroup$ – Ro Siv Mar 24 '15 at 18:59
  • $\begingroup$ Using your agar example, your agarose is 10g/L = 10mg/mL. $\endgroup$ – March Ho Mar 24 '15 at 19:13
  • 2
    $\begingroup$ What is NaCl₂ ?? $\endgroup$ – WYSIWYG Mar 25 '15 at 7:10
2
$\begingroup$

It depends on what form your proteinase K is in.


You may have a stock solution of some concentration, in which case you just add a specific volume according to $C_1V_1=C_2V_2$.

Example: Let's say you want to prepare 5 mL ($V_2$) of 200 µg/mL ($C_2$) proteinase K from a stock of 1000 µg/mL ($C_1$). You're looking for the volume of stock solution to add ($V_1$).

$$V_1=\frac{5\ mL\ \cdot \ 200\ \mu g/mL}{1000\ \mu g/mL}=1\ mL$$

So you take 1 mL of the stock and bring it up to 5 mL.


Or you may have a lyophilized powder. In this case, you have two options:

1) You can make a stock solution from the powder and, again, use $C_1V_1=C_2V_2$. This has the advantage of not having to weigh the powder every time you need to use it. This may not be a viable option, however, as I'm not sure of the stability of proteinase K in solution.

2) You can just weigh out the amount that will give you the desired concentration (which depends on the volume of buffer you're making).

Example: Again, you want to prepare 5 mL of 200 µg/mL proteinase K. The total mass you need is:

$$200\ \mu g / mL\ \cdot \ 5\ mL=1000\ \mu g$$

So weigh out a milligram and bring it up to 5 mL.


There's nothing special about mass concentration; you can convert it to molarity given the molecular weight of proteinase K. According to Wikipedia, proteinase K is 28.9 kDa or 28900 g/mol. If you are making this conversion, however, make sure you find the actual molecular weight of the protein you're using. 200 µg/mL is equal to:

$$\frac{0.2\ g/L}{28900\ g/mol}=6.92\ \mu M$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.