Question

1. (4pts) The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl.

The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW = 58.44 g/mole) and 350ml of 1M imidazole (68.08 g/mole). Show your calculations for credit.

Answer 1

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula . It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration             Volume to be             Final Concentration

                                               added            

1 M Tris                                     2.5 mL                         10 mM

5 M NaCl                                  7.5 mL                        150 mM

1 M Imidazole                           75 mL                         300 mM    

. is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

. $1*V_1= 0.01 M *250mL\\V_1 = 2.5mL$

The stock concentration of NaCl (5 M ) is as follows:

. $1*V_1= 0.15 M *250mL\\V_1 = 7.5mL$

The stock concentration of Imidazole (1 M ) is as follows:

. $1*V_1= 0.03 M *250mL\\V_1 = 75mL$

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.