# User:Elizabeth Ghias/Notebook/Experimental Biological Chemistry/2012/02/07

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## Objective

To 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

1. The amount of Tris and the volume of HCl needed to make 250 mL of a 50mM tris buffer was calculated.
2. The amount of dye needed to react with 1 mL of BSA (one large scale reaction mixture) was calculated by determining the number of lysine residues in one molecule of BSA, the total number of lysine residues in 1 mL of BSA, and the number of dye molecules that are needed to react with the total number of lysine residues.

## Data

Tris 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

Mass of Dye for Reaction Calculations

• There are 60 lysine residues in one molecule of BSA
• MW of dye = 573.51 g/mol, MW BSA = 66463 Da
• We are going to add an excess of three dye molecules for every one lysine residue.

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