Question

I am provided with the following stock solutions: 1M Tris, pH 8, 5 M NaCl, 1M MgCl₂, lysozyme, and DNase. Prepare a 25 mL solution of complete lysis buffer using the following recipe:

- 50 mM Tris (pH 8)

- 150 mM NaCl

- 2 mM MgCl₂

- 0.5 mg/mL lysozyme

- 0.04 uL/mL DNase

In order to perform ion-exchange chromatography, I also need to prepare elution buffers. Using the stock solutions listed above, I need to calculate how to make 10 mL of each of 50 mM Tris pH 8 with 0.1, 0.2, 0.3, and 0.4 M NaCl.

Thanks for the help, I appreciate it!

Answer 1

This is a dilution problem. We will use the dilution equation to find out the volumes of stock solutions required for preparing the lysis buffer. The dilution equation is

M₁*V₁ = M₂*V₂

where M₁ = concentration of stock solution

M₂ = concentration of the substance in the lysis buffer

V₁ = volume of stock solution required

V₂ = final volume of lysis buffer = 25.00 mL.

Substance	Concentration of stock solution provided	Concentration of substance in final lysis buffer	Volume of stock solution required
Tris, pH 8	1 M	50 mM = (50 mM)*(1 M/1000 mM) = 0.05 M	(1 M)*V1 = (0.05 M)*(25.00 mL) ===> V1 = 1.25 mL (ans)
NaCl	5 M	150 mM = (150 mM)*(1 M/1000 mM) = 0.15 M	(5 M)*V1 = (0.15 M)*(25.00 mL) ===> V1 = 0.75 mL (ans)
MgCl2	1 M	2 mM = (2 mM)*(1 M/1000 mM) = 0.002 M	(1 M)*V1 = (0.002 M)*(25.00 mL) ===> V1 = 0.05mL = (0.05 mL)*(1000 µL/1 mL) = 50 µL (ans)
Lysozyme	-	0.5 mg/mL	We have 25 mL buffer solution; hence, the mass of lysozyme required = (0.5 mg/mL)*(25.00 mL) = 12.5 mg; this can be weighed out on an analytical balance (ans).
DNAase	-	0.04 µL/mL	We have 25.00 mL buffer; hence the volume required is (0.04 µL/mL)*(25.00 mL) = 1.00 µL, which can be taken using as micropipette (ans).

Again use the dilution equation to find out the volume of Tris, pH 8 buffer required. The final volume is 10 mL now.

(1 M)*V₁ = (50 mM)*(10 mL)

===> V₁ = (50 mM)*(1 M/1000 mM)*(10 mL)/(1 M) = 0.5 mL (ans).

Next use the dilution equation to find out the volumes of NaCl required.

(5 M)*V₁ = (0.1 M)*(10 mL)

===> V₁ = (0.1 M)*(10 mL)/(5 M) = 0.2 mL (ans).

(5 M)*V₁ = (0.2 M)*(10 mL)

===> V₁ = (0.2 M)*(10 mL)/(5 M) = 0.4 mL (ans).

(5 M)*V₁ = (0.3 M)*(10 mL)

===> V₁ = (0.3 M)*(10 mL)/(5 M) = 0.6 mL (ans).

(5 M)*V₁ = (0.4 M)*(10 mL)

===> V₁ = (0.4 M)*(10 mL)/(5 M) = 0.8 mL (ans).