Question

Being provided are:

• 50mM Tris Buffer

• 50mg/ml Lysozyme

• 20mg/ml DNase

• 0.5M NaCl

How will you prepare a 0.25L of 10mM Tris buffer pH 7.2 containing?

 200ug/ml Lysozyme

 0.1mg/ml DNase

 100mM NaCl

Answer 1

Step-1: Find the moles of Tris present in the required tris buffer.

V = 0.25 L

M = 10 mM * (1M / 1000 mM) = 1.0*10^-2 M

Moles of Tris present in 0.25 L of 10 mM tris buffer = M*V = 1.0*10^-2 mol/L * 0.25 L = 2.5*10^-3 mol

Step-2: Find the volume of 50 mM tris buffer equivalent to 2.5*10^-3 mol.

M = 50 mM * (1M / 1000 mM) = 5.0*10^-2 M

V = ?

2.5*10^-3 mol = M*V = 5.0*10^-2 mol/L * V

=> V = (2.5*10^-3 mol) / (5.0*10^-2 mol/L) = 0.05 L = 50 mL 50 mM tris buffer

Lysozyme:

mg of lysozyme present in the desired solution = (200 ug / mL) * 250 mL*(1 g / 10⁶ ug)*(10³ mg / 1g)

= 50 mg

Since the concentration of given lysozyme solution is 50 mg/mL, we have to add 1 mL of 50 mg/mL

lysozyme

DNase:

mg of DNase present in the desired solution = (0.1mg/mL)* 250 mL = 25 mg

Since the concentration of given DNase solution is 20 mg/mL we have to add = 25 mg * (1 mL/20 mg)

= 1.25 mL DNase

NaCl:

M1 = 100 mM = 0.1 M V1 = 0.250 L

M2 = 0.5 M, V2 = ?

Applying law of dilution:

M1*V1 = M2*V2

=> 0.1 M * 0.250 L = 0.50 M * V2

=> V2 = 0.05 L = 50 mL of 0.5 M NaCl

So we have to add 50 mL of 0.5 M NaCl

Take

50 mL 50 mM tris buffer, 1 mL of 50 mg/mL lysozyme, 1.25 mL of 20 mg/mL DNase and 50 mL of 0.5 M NaCl. Then mix 1M HCl till the pH becomes 7.4. Then dilute the solution with deionized water till the volume reaches 0.25 L (or 250 mL)