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]
[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




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


Use categories like tags. Change the "Course" category to the one corresponding to your course. The "Miscellaneous" tag can be used for particular experiments, as instructed by your professor. Please be sure to change or delete this tag as required so that the categories remain well organized.


[[Category:Course]]
[[Category:Course]]

Revision as of 10:44, 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


Notes

  • Three dye molecules were added for every one lysine residue.



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