Name: Instructor & Section: Protein Analysis Lab
Worksheet
x=0.1781
6.
Adjust the final concentration of the unknown samples determined in Step 4 by multiplying the concentration by the dilution factor used. For example, sample diluted 1:50 gives a reading of 0.224 absorbance units, which gives a concentration of M mg/mL (this is an example, not the actual values you are calculating). (2 points)
The final concentration of unknown sample is M x 50 = mg/mL .
-0.1894(50)=-9.4728
-0.1782(50)=-8.9089
7.
Determine the final concentration of the unknown samples and record in Table 3. (4 points)
Table 3
Final Concentration of Unknown Samples
Sampl
e
A595
Concentration
Read from Standard
Curve (mg/mL)
Dilution Factor
Final Concentrati
on (mg/mL)
White Meat Sample (Sample A)
0.559
-0.1894
50
-9.4728
Dark Meat Sample (Sample B)
0.563
-0.1782
50
-8.9089
8.
Analysis:
Discuss this experiment. Was the hypothesis supported? Discuss each step – cell fractionation, spectrophotometry, standard curve, calculation of protein. What were the potential sources of error? Treat this like a lab report analysis.
No personal pronouns. (10 points)
The hypothesis was possibly supported because there was less mitochondrial protein in the white meat than the dark meat. There was less concentration from the white meat for the standard curve.
Might have lost a protein due to a possible contamination in the test which could have affected the results. Or there might not have been enough chicken homogenized in the test.
S22 4
