User:Elizabeth Ghias/Notebook/Experimental Biological Chemistry/2012/02/07: Difference between revisions

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==Data==
==Data==
<u>Tris Buffer Calculations</u>
<u>Tris Buffer Calculations</u>
8.5 = 8.06 + log[salt]/[acid]
8.5 = 8.06 + log[salt]/[acid]


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x = 3.512 mL HCl
x = 3.512 mL HCl


(0.050 M Tris)(0.25 L)(121.14 g/mol) = 1.514 g
(0.050 M Tris)(0.25 L)(121.14 g/mol) = 1.514 g Tris
 
 
<u>Mass of Dye for Reaction Calculations</u>
 
* 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.
 
 




==Notes==
==Notes==
* Three dye molecules were added for every one lysine residue.




Revision as of 10:47, 8 February 2012

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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.



Notes



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