# How to make µg/ml concentrations of proteinase-K?

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.

• 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. – March Ho Mar 24 '15 at 18:45
• @MarchHo Yes it is a buffer used to break down the flies protiens via the protienase. Since I will be doing PCR on flies. – Ro Siv Mar 24 '15 at 18:56
• @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. – Ro Siv Mar 24 '15 at 18:59
• Using your agar example, your agarose is 10g/L = 10mg/mL. – March Ho Mar 24 '15 at 19:13
• What is NaCl₂ ?? – WYSIWYG Mar 25 '15 at 7:10

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:

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