This preview shows page 1. Sign up to view the full content.
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
 Spring '13
 Jonklaas
 Chemistry

Click to edit the document details