Expt3 Bradford Assay- Sample Journal Article Report F09

# 0004385x 00125865 328 00004385x 00125865 x 7193

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the original concentration was: (708 μg/mL)(10) = 7078 μg/mL B1: Y = 0.0004385x + 0.0125865 .328 = 0.0004385x + 0.0125865 x = 719.3 μg/mL Since it is a 10x dilution, the original concentration was: (719.3 μg/mL)(10) = 7193 μg/mL A2: Y = 0.0004385x + 0.0125865 .094 = 0.0004385x + 0.0125865 x = 185.7 μg/mL Since it is a 50x dilution, the original concentration was: (185.7 μg/mL)(50) = 9283 μg/mL B2: Y = 0.0004385x + 0.0125865 .088 = 0.0004385x + 0.0125865 x = 172.0 μg/mL Since it is a 50x dilution, the original concentration was: (172.0 μg/mL)(50) = 8600 μg/mL Mean: (7078 μg/mL + 7193 μg/mL + 9283 μg/mL + 8600 μg/mL)/4 = 8038 μg Standard Deviation: ∑ (x n ∑ (x n −x ) 2 (n − 1) 4 1 − 8038) (n − 1) 2 , using the four above values = 1080 μg/mL. So the deviation is 8038 μg/mL ± 1080 μg/mL. Discussion: The overall objectives of this experiment were met. From the experiment, it was possible to observe how the Bradford method could be used to both draw the standard curve for standard solutions of bovine serum albumin, and to determine the concentration of a solution of bovine serum albumin of unknown concentration. Using the standard curve that was drawn using the data found using the Bradford method, the concentrations of certain dilutions of the unknown, and hence the concentration of the unknown, could be found. Normally, as the concentration of a protein increases, the absorbance also increases. This occurs because...
View Full Document

## This note was uploaded on 03/23/2014 for the course CHEMISTRY 4141 taught by Professor Jonklaas during the Spring '13 term at Baylor.

Ask a homework question - tutors are online