Question

Question 20: B) & C)

(B) The purification protocol included application of 50mM Na,Na-buffer, pH 7.4 containing 10mM NaCl, 90mM KCL, 0.1% Tween-80, and 0.01% sodium azide. You have found following stock solutions in the lab previously prepared by a reliable lab member: a) 1M Na,Na,-phosphate buffer, pH 7.4 b) 2M NaCl c) 1M KCl d) 10% Tween 80 A small bottle of crystalline NaN3 was provided by your research supervisor. Calculate volumes (ml or µl) of each (a) through (d) stock solutions and the amount (mg or µg) of NaN3 need to be mixed together in order to prepare 250ml of required buffer.

(C) A purified protein is a metalloprotein which has a mononuclear zinc binding site. The purification procedure included several harsh conditions causing the protein to lose its metal, thus converting it into its apo-form. Calculate the amount of a 20mM ZnCl2 solution must be added to a 2ml aliquot of the protein in order to get a fully metallated form for further investigation.

Answer 1

(B) well here we have to be clear about the dilution formula.

Where:

Initial concentration or molarity.

Initial volume.

Final concentration or molarity.

Final volume

Then we apply this for every substance we are asked for to calculate.

Remember the transformation from mM to M is 10^-3

Our final volume will be 250 mL

a) From 1M Na,Na,-phosphate buffer, pH 7.4 we need 50mM Na,Na-buffer

VNa,Na-buffer. initial = 50x10^-3 M* 250 mL / 1M

VNa,Na-buffer. initial = 12.5 mL

This means we need 12.5 L of the 1M buffer to have 50mM in the final 250 mL solution.

b) From 2M NaCl prepare 250 mL of 10mM NaCl.

VNaCl. initial = 10x10^-3M * 250 mL / 2M

VNaCl. initial = 1.25 mL

c) From 1M KCl prepare 90mM KCl

VKCl initial = 90x10^-3M*250mL / 1M

VKCl initial = 22.5 mL

d) From 10% Tween 80 prepare 0.1% Tween-80

Vtween80.initial= 0.1% * 250mL / 10%

Vtween.initial = 0.25 mL

e) To prepare the 0.01 % of sodium azide in the 250 mL we need to calculate the molarity of this to be able to calculate the mass.

When we said there is a solution 0.01 % we mean that there are 0.01g of the solute for each 100mL of solution.

then to calculate the mas for 250 mL we do a simple calculation.

mNaN₃ to 250 mL= 0.01 g * 250 mL / 100 mL

m = 0.025 g = 25mg of NaN₃ to have the 0.01% concentration in the final 250 mL volume of solution.

C) To calculate this we need a concentration of the protein. I think there is that data missing because the complex formation depends on many variables, as the pH, therefore we could calculate the concentration of the protein..