Question

For lab, I need to prepare 25 mL of complete lysis buffer. I am provided with 1 M Tris, 5 M NaCl, 1 M MgCl_2, lysozyme, and DNase.

Our complete lysis buffer contains all of the following components: 50 mM Tris, 150 mM NaCl, 2 mM MgCl₂, 0.5 mg/mL lysozyme, and 0.04 uL/mL DNase.

Plan and calculate exactly how to make the complete lysis buffer.

*My instructor recommended getting the amount of lysozyme needed into a solution to make it easier to combine with the other components.

Answer 1

Ans. Use the following equation-               C1V1 = C2V2            - equation 1

            Where, C1 = concentration; V1 = Volume           - stock solution

                        C2 = Concertation;   V2 = Volume           - final/diluted solution

V2 in all cases is to be made upto 25.0 mL.

#1. Required amount of Tris buffer

Given,

            C1 = 1.0 M    

            C2 = 50 mM = 0.050 mM                           ; [1 mM = 0.001 M]

Putting the values in equation 1 -

1. M x V1 = 0.050 M x 25.0 mL

Or, V1 = (0.050 M x 25.0 mL) / 1.0 M = 1.25 mL

So, required volume of 1.0 M tris buffer stock = 1.25 mL

#2. Required amount of NaCl

Given,

            C1 = 5.0 M

C2 = 150 mM = 0.150 mM                        

Putting the values in equation 1 -

V1 = (0.150 M x 25.0 mL) / 5.0 M = 0.75 mL

#3. Required amount of MgCl2

Given,

            C1 = 1.0 M

C2 = 2 mM = 0.002 mM                             

Putting the values in equation 1 -

V1 = (0.002 M x 25.0 mL) / 1.0 M = 0.05 mL

#4. Required amount of lysozyme

Note: [Lysosome] in its stock is NOT mentioned. Let it be X mg/mL

Given,

            C1 = X mg/ mL

C2 = 0.5 mg/ mL                 

Putting the values in equation 1-

V1 = (0.5 mg mL^-1 x 25.0 mL) / X mg mL^-1 = (12.5 / X) mL

## If lysosome is provided in form of lyophilized powder:

Mass of lysozyme required = desired [lysozyme] x Vol. of lysis buffer

                                                = (0.5 mg/ mL) X 25 mL

                                                = 12.5 mg

So, add 12.5 mL of lyophilized lysozyme to the lytic buffer.

Then don’t calculate “V1” or required volume of lysozyme solution in V_total (see below).

#5. Required amount of DNAse

Note: [DNAse] in its stock is NOT mentioned. Let it be Y uL/mL

Given,

            C1 = Y uL/ mL

C2 = 0.04 uL/ mL                 

Putting the values in equation 1 -

V1 = (0.04 uL mL^-1 x 25.0 mL) / Y uL mL^-1 = (1 / Y) mL

# Total (summed) volume of all reagents, V_total

V_total= 1.25 mL (Tris) + 0.75 mL (NaCl) + 0.05 mL (MgCl₂) + (12.5 / X) mL of lysosome + (1/ Y) mL of DNAse

If lysozyme is provided in solid form, then-

Vfinal = 1.25 mL (Tris) + 0.75 mL (NaCl) + 0.05 mL (MgCl₂) + (1/ Y) mL of DNAse

Since, X and Y are unknown, we proceed with V_total.

# Amount of water required:

We proceed assuming solid lysozyme is provided.

Amount of water required = Final volume of buffer – V_final

                                                = 25.0 mL – V_final

# Preparation of 25 mL lysis buffer:

Transfer the calculated amount of all the reagents to a 50.0 mL sterile tube. Add the required amount of water to make the final volume to 25.0 mL. It is the desired lysis buffer.