Question

You collected 900 μL of 2 mg/mL BSA in water after a diafiltration step. Now you need to use this solution to make 200 μL of protein B at 1.25 mg/mL in 1X buffer. Using the 2 mg/mL BSA stock and the 10X buffer, make the desired solution.

Here is how far I've gotten:

C2= 1.25

V2= 200

C1= 2

V1= ?

V1=(V2)(C2)/(C1) = 125

PLEASE HELP. I see many of these on Chegg already but I had a difficult time following along!

Answer 1

Using C1V1=C2V2 you can calculate the volume of BSA protein to be added from the stock of 2mg/ml (you have 900ul of this- more than enough).

C1=2 mg/ml

C2=1.25 mg/ml

V1=?

V2=200 ul

V1 comes out to be 125 ul.

you have been provided with 10X buffer.

For 200 ul you will take 20 ul of 10X buffer to make it 1X. (200/10=20, simple math).

Now you have 125ul BSA protein +20ul of 10X buffer = 145 ul total volume. Make it up with water upto 200 ul. This is your desired solution.