Question

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

Answer 1

Step-by-step explanation:

From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer

0.25 L elution buffer = 250 mL elution butter

The breaking buffer that we use this week contains

10mM Tris = 0.01 M

150mM NaCl = 0.15 M

300mM imidazole. = 0.3 M

The stock concentration of Tris in 1M

Therefore ; by using the formula:
`M_1V_1 = M_2 V_2`; we can determine the volume in the preparation; so;

`1*V_1 = 0.0 1 \ M * 250 \ mL`

`V_1 = (0.0 1 \ M * 250 \ mL)/(1 )`

`V_1 = 2.5 \ mL`

In NaCl, The amount of stock concentration is 5 M

so; using the same formula; we have:

`5*V_1 = 0.15 \ M * 250 \ mL`

`V_1 = (0. 15 \ M * 250 \ mL)/(5 )`

`V_1 = 7.5 \ mL`

From Imidazole ; the amount of stock concentration is

`1*V_1 = 0.3 \ M * 250 \ mL`

`V_1 = (0. 3 \ M * 250 \ mL)/(1 )`

`V_1 = 75 \ mL`

Thus; we can have a table as shown as :

Stock concentration volume to be added Final concentration

1 M of Tris 2.5 mL 10 mM

5 M of NaCl 7.5 mL 150 mM

1 M of Imidazole 75 mL 300 mM

In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.