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Taylor Kuhlman
Quantitative Determinations of Proteins by Spectrophotometry
Introduction
If you have an unknown concentration and want to identify it you can measure how much the sample absorbs and then apply it to Beer’s Law1. With absorption spectroscopy, proteins that have high percent of residues will have a stronger absorption than one that has less. This can help us find its aromatic content2. With the Bradford’s Assay, it is used to measure the concentration of the total protein in a sample3. This works by the proteins binding to the dye in acidic conditions and results in a color change from brown to blue3. The purpose of this weeks lab was to use spectroscopy, as well as, colorimetic analysis to determine the proteins and nucleic acid and to find the concentration of them in a solution.

Methods
Results
Part A
1000 times dilution of Lz in 200 microliters
1/1000 * 200l = 0.2l of Lz
Graph 1: Absorbance vs. Wavelength of Lysozyme

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Graph 2: 280nm of lysozyme absorbance vs. concentration
Conc (g/ml) 100 50 20 10 4 1
Absorbance 0.152 0.108 0.091 0.79 0.82 0.079
Table 1: Values from Graph 2

Graph 3: Absorbance vs. wavelength for hemoglobin
Hemoglobin peaks around 400 nm which is above 280 nm. This could have been due to the heme group that can be absorbed in visible UV (>300 nm).

Graph 6: 280 nm for BSA absorbance vs conc.

Conc (g/ml) 100 50 20 10 4 1
Absorbance 0.081 0.08 0.069 0.062 0.07 0.063
Table 3: Values of graph 6

Graph 7: Aborbance vs wavelength DNA

Graph 8: Absorbance vs. conc DNA 260 nm
Conc (g/ml) 100 50 20 10 4 1
Absorbance 1.182 0.608 0.315 0.183 0.111 0.112
Table 4: Values of graph 8
Part A: Extinction Coefficient DNA calculation
A280 = E280 * C * L
1.182 ug/ml = E280 * (100 ug/ml)*(0.25cm/100ul)
E280 = (100ug/ml)(0.25cm/100ul)/1.182 nm
E280 = 0.212 cm/nm
Part B:

Graph 9: Absorbance vs Protein Standard Conc.
Microplate Well #
Row12345678
A0.3040.3090.3460.380.4170.4180.4760.476
B0.3010.3030.3530.4060.4260.4610.4960.497
C0.2930.2980.3450.3870.4230.450.4860.455
Average Absorbance0.2990.3030.3480.3910.4220.4430.4860.476
Table 5: Absorbance values
Microplate Well #
Row9101112
A0.3490.4150.3560.424
B0.360.4090.3530.413
C0.3430.4040.3550.413
Average Absorbance0.3506666670.4093333330.3546666670.416666667
Table 6: Average absorbance of unknown A and B
Discussion
Questions
The term “experimental error” refers to the natural variability inherent in carrying out any procedure, i.e. pipetting uncertainties, mixing errors, color development. This error relates to inherent assay reproducibility. Estimate the experimental error, as ± %, for the Bradford assay based upon the known concentrations of the analyzed proteins. Mention the basis of your estimate. A given protein may yield different apparent concentrations when analyzed by different assay procedures. These variations include experimental error plus other variations due to the nature of the unknown protein and to its behavior in the assay procedure. Based upon what you know about the Bradford assay, offer an explanation for any error in the determined concentrations.
BSA was used as a standard in the Bradford colorimetric assay. For comparison, use the extinction coefficient at 280 nm for BSA to calculate the concentration of hemoglobin and lysozyme from their A280 values. Compare these values to the actual concentrations of hemoglobin and lysozyme. What do you conclude from this comparison?
You have isolated the protein chymotrypsinogen from a cell lysate. Discuss the advantages and disadvantages of the following protein assays.
Absorbance at 215 nm-advantage is that the absorbance doesn’t change for the proteins with peptide bonds. Disadvantage is that buffering salts and solvents can also absorb at this wavelength.

Absorbance at 280 nm-At this wavelength a lot of buffering salts and solvents absorb at this wavelength which is a disadvantage. Advantage is amino groups that have aromatic side chains give a wavelength.
Bradford assay-Advantage is that the dye is measured at a light spectrum that is visible and binds to the proteins. Disadvantage is that is stains skin and glass.

Compare extinction coefficients for DNA (1 mg/ml, 260 nm) and the proteins 1 mg/ml, 280 nm). Explain any differences between the DNA and protein extinction coefficients in region of the UV spectrum.
-Mass attenuation coefficient of nucleic acids have larger mass, and have peaks at 260 nm and 280 nm when they are compared to protein. Proteins (even with high concentrations) display a small absorbance of 260 and 280 nm.
Suppose that you had a sample that contained equal concentrations of DNA and protein. Sketch the absorption spectrum you would expect to see in the 240-300 nm region.

BSA has the following amino acid profile compared with the average of known vertebrate proteins:
AA BSA Average Protein
Phe4.6% 4.0%
His 2.7% 2.9%
Lys 10.1% 7.2%
Arg3.9% 4.2%
Trp0.3% 1.3%
Tyr 3.4% 3.3%
Based upon this data, is BSA a good standard to use in the Bradford assay? Explain your answer.
-It is a good standard because BSA contains the amino acids phenylalanine, tryptophan, tyrosine, histidine, and lysine that have a minor reaction with the dye. Also BSA is easy to access.
References