User:Elizabeth Ghias/Notebook/Experimental Biological Chemistry/2012/02/07
Experimental Biological Chemistry | Main project page Previous entry Next entry |
ObjectiveTo determine the number of lysine residues in 1 mL of 17.7 and calculate the amount of Tris and HCl needed to create a Tris buffer with a pH of 8.5. Description
DataTris Buffer Calculations 8.5 = 8.06 + log[salt]/[acid] [salt]/[acid] = 2.754 (50-x)/x = 2.754 --> x = 13.318 mM = [HCl] (0.013318 M)(0.25 L) = (0.948 M)(x L) x = 3.512 mL HCl (0.050 M Tris)(0.25 L)(121.14 g/mol) = 1.514 g Tris
1.77 x 10-7 M BSA x 0.001 L = 1.77 x 10-8 mol BSA 1.77 x 10-8 mol BSA x 6.022 x 1023 molecules/mol = 1.066 x 1016 molecules BSA 1.066 x 1016 molecules BSA x 60 lysine residues = 6.393 x 1017 total lysine residues 6.393 x 1017 total lysine residues x 3 = 1.918 x 1018 dye molecules needed 1.918 x 1018 dye molecules x 1 mol/(6.022 x 1023 molecules) = 3.186 x 10-6 mol dye 3.186 x 10-6 mol dye x 573.51 g/mol = 1.83 mg dye
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