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

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==Objective==
==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==
==Description==
# The amount of Tris and the volume of HCl needed to make 250 mL of a 50mM tris buffer was calculated.
# 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==
==Data==
* Add data and results here...
<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 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.
 
1.77 x 10<sup>-7</sup> M BSA x 0.001 L = 1.77 x 10<sup>-8</sup> mol BSA
 
1.77 x 10<sup>-8</sup> mol BSA x 6.022 x 10<sup>23</sup> molecules/mol = 1.066 x 10<sup>16</sup> molecules BSA
 
1.066 x 10<sup>16</sup> molecules BSA x 60 lysine residues = 6.393 x 10<sup>17</sup> total lysine residues
 
6.393 x 10<sup>17</sup> total lysine residues x 3 = 1.918 x 10<sup>18</sup> dye molecules needed
 
1.918 x 10<sup>18</sup> dye molecules x 1 mol/(6.022 x 10<sup>23</sup> molecules) = 3.186 x 10<sup>-6</sup> mol dye
 
3.186 x 10<sup>-6</sup> mol dye x 573.51 g/mol = 1.83 mg dye
 
 
 
 


==Notes==
==Notes==
This area is for any observations or conclusions that you would like to note.


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 10:55, 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.

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



Notes



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