Question

Protein A stored at 0.8mg/mL

Protein B stored at 0.5mg/mL

Mixture is 1mL total volume, 0.5mg total protein, at a ratio of 1:2 (A:B)

How many uL of stock Protein A, stock Protein B, and buffer is needed to make this mixture?

Answer 1

total protein = 0.5 mg

ratio of amount of proteins is

A: B=1:2

so mass of protein A=total mass of protein required/3=0.5mg/3=0.167 mg

mass of protein B = total mass of protein- mass of protein A

= 0.5 - 0.167

= 0.333 mg

concentration of protein A is 0.8mg/mL

1 mL there is 0.8 mg of protein A,

1 mg is present in 1/0.8 mL

so 0.167 mg is present in (1/0.8)0.167=0.20875 mL of the solution.

volume of protein A required= 0.20875 mL

1 mL= 1000 uL

so volume of protein A required= 0.20875 1000=208.75 uL

concentration of protein B is  0.5mg/mL

1 mL there is 0.5 mg of protein B

1 mg is present in 1/0.5 mL

so 0.333 mg is present in (1/0.5)0.333=0.666 mL of the solution.

volume of protein A required= 0.666 mL

1 mL= 1000 uL

so volume of protein A required= 0.666 1000=666 uL

volume of buffer required= 1000-volume of protein A- volume of protein B

= 1000-280.75-666

= 53.25 uL