What weights of Bovine Serum Albumin (molecular weight 66,000) are required to give solutions of 10ml and?

Question

What weights of Bovine Serum Albumin (molecular weight 66,000) are required to give solutions of 10ml and 100ml volumes containing 1μmole of BSA

Answer 1

1 mole corresponds to 66,000 g/liter

1 µmole corresponds to 66000*10^-6 = 0.066g/l = 66mg

100 ml = 0.1 l requires 66/10= 6.6 mg BSA

and 10 ml requires 0.066mg BSA


BSA = bovine serum albumin