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Please complete Table 2 and show the calculations for protein concentration and the concentration of the undiluted sample.  a). Using Excel software, enter data from Table 1 to plot a standard curve (corrected average absorbance on Y-axis and protein concentrations on X-axis). The standard curve must contain correct labels on both axes. b). From the standard curve, obtain a best fit line and equation. c). Use the best fit equation and data from Table 2 to find the concentrations of proteins in sample A2-D2 and E1.  [onto your plotted standard curve, indicate where A595 of A2-D2 and E1 are located]  d). Calculate the concentrations of proteins in undiluted samples A to E. Thank you very much.

Please complete Table 2 and show the calculations for protein concentration and the concentration of the undiluted sample.  a). Using Excel software, enter data from Table 1 to plot a standard curve (corrected average absorbance on Y-axis and protein concentrations on X-axis). The standard curve must contain correct labels on both axes. b). From the standard curve, obtain a best fit line and equation. c). Use the best fit equation and data from Table 2 to find the concentrations of proteins in sample A2-D2 and E1.  [onto your plotted standard curve, indicate where A595 of A2-D2 and E1 are located]  d). Calculate the concentrations of proteins in undiluted samples A to E. Thank you very much.

Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter9: Atomic Absorption And Atomic Fluorescence Spectrometry
Section: Chapter Questions
Problem 9.2QAP
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Please complete Table 2 and show the calculations for protein concentration and the concentration of the undiluted sample. 

a). Using Excel software, enter data from Table 1 to plot a standard curve (corrected average absorbance on Y-axis and protein concentrations on X-axis). The standard curve must contain correct labels on both axes.

b). From the standard curve, obtain a best fit line and equation.

c). Use the best fit equation and data from Table 2 to find the concentrations of proteins in sample A2-D2 and E1.  [onto your plotted standard curve, indicate where A595 of A2-D2 and E1 are located] 

d). Calculate the concentrations of proteins in undiluted samples A to E.

Thank you very much.

Table 1. Absorbances of protein standard solutions BSA Conc. (mg/mL) (From Part A) Sample A595 (Average from three wells) (0.232+0.248+0.218)/3= A595 (Corrected by blanks) ID Blank 0.233 Std. 1 0.003125 (0.227+0.224+0.218)/3= 0.223 -0.01 Std. 2 0.00625 (0.247+0.255+0.233)/3= 0.245 0.012 Std. 3 0.0125 (0.296+0.297+0.299)/3= 0.297 0.064 Std. 4 0.025 (0.429+0.418)/2 = %3D 0.424 0.191 Std. 5 0.05 (0.670+0.656+0.693)/3= 0.673 0.44 Std. 6 0.10 (1.096+1.073+1.124)/3= 1.097 0.864
Transcribed Image Text:Table 1. Absorbances of protein standard solutions BSA Conc. (mg/mL) (From Part A) Sample A595 (Average from three wells) (0.232+0.248+0.218)/3= A595 (Corrected by blanks) ID Blank 0.233 Std. 1 0.003125 (0.227+0.224+0.218)/3= 0.223 -0.01 Std. 2 0.00625 (0.247+0.255+0.233)/3= 0.245 0.012 Std. 3 0.0125 (0.296+0.297+0.299)/3= 0.297 0.064 Std. 4 0.025 (0.429+0.418)/2 = %3D 0.424 0.191 Std. 5 0.05 (0.670+0.656+0.693)/3= 0.673 0.44 Std. 6 0.10 (1.096+1.073+1.124)/3= 1.097 0.864
+ Table 2. Absorbances of samples A2 -D2 and E1 Final DF Concentration A595 A595 (Corrected by blanks) (From Part B) Sample Protein concentration of the undiluted (Average from three wells) ID (mg/mL) sample (mg/mL) Blank (0.232+0.248+0.218) /3= 0.233 (0.729+0.785+0.741) /3= 0.752 A2 1/500 A 0.519 1/500 (0.387+0.411+0.389) /3= 0.396 (0.255+0.269+0.269) /3= 0.264 (0.393+0.399+0.427) /3= 0.406 (0.287+0.292)/2= B2 В 0.163 C2 1/500 C 0.031 D2 1/500 D 0.173 E1 1/50 E 0.290 0.057
Transcribed Image Text:+ Table 2. Absorbances of samples A2 -D2 and E1 Final DF Concentration A595 A595 (Corrected by blanks) (From Part B) Sample Protein concentration of the undiluted (Average from three wells) ID (mg/mL) sample (mg/mL) Blank (0.232+0.248+0.218) /3= 0.233 (0.729+0.785+0.741) /3= 0.752 A2 1/500 A 0.519 1/500 (0.387+0.411+0.389) /3= 0.396 (0.255+0.269+0.269) /3= 0.264 (0.393+0.399+0.427) /3= 0.406 (0.287+0.292)/2= B2 В 0.163 C2 1/500 C 0.031 D2 1/500 D 0.173 E1 1/50 E 0.290 0.057
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