Colorimetric reagents. Proteins will intensely absorb 280 nm light even in the absence of a colorimetric reagent (See figure). Why is it advantageous to use the Bradford reagent to r protein concentration rather than just measuring protein absorbance (280 nm) directly?
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- Why do not all proteins have the same absorption coefficient (ɛ) at 280 nm? Pick one: a.lt is mainly due to the presence of Asp, Glu, Asn, Gln, Arg and His in the protein b.lt is mainly due to the presence of phenylalanine in the protein c.It is mainly due to the presence of tryptophans and tyrosines in the protein d.The statement is incorrect; all proteins have the same (E) at 280 nmA.) Which parameter between "solvent polarity function" and dielectric constant show better correlation with the peak positions? B.) When a chromophore binds to proteins, the binding site is generally more hydrophobic than the solution. Then, what kind of shift do you expect to observe in the absorption peak position compared to the peak observed when the chromophore is free in solution?Draw the most likely ion fragment for the signals at m/z 111, 139, and 156.
- Answer Q26, 27, 28 showing detailly all explanationsConsider the following pairs of mitcules. Start either weshared spectroscopy & GC MS 6 destyl which molecule which is MAA Join or 3Using the mass spectrum below select the most likely fragment that produc peak at 57: 100 Relative Intensity 80- 40 20 0- 10 15 20 25 30 35 OC₂H4O+ C3HO+ C₂HO₂+ CH₂O3+ 40 45 50 m/z 55 60 65 70 75
- When we perform Bradford Assay to measure protein concentration, the spectrophotometer is set 595 nm because: a. the excitation peak of Coomassie Blue is best observed at 595 nm b. when Coomassie Blue binds to proteins, its maximum absorbance shifts to 595 nm c. the reducing effects of Coomassie Blue shifts the protein's maximum absorbance to 595 nm d. myoglobin is best observed at 595 nmAll ethyl esters of long-chain aliphatic acids (for example, ethyl tetradecanoate, C13H27COOCH2CH3) show significant fragment ions at m/z 88, 73, and 45. Draw the structure of the ion that gives the m/z 73 peak. • In cases where there is more than one answer, just draw one. • You do not have to include lone pairs in your answer. • Include all valence radical electrons in your answer. • Do not use the square brackets tool in your answer. ChemDoodleApplying the best-fit method you produce the following plot. Please select the correct statement. Group of answer choices: 1. The Bradford assay gives a hyperbolic plot for absorbance versus protein concentration. 2. The Bradford assay is measured at at absorbance of 600 nm 3. Within a range of relatively low protein concentrations, the hyperbolic curve can be approximated reasonably well by a straight line.
- 1. The EI-MS of methylcyclohexane is shown below. Locate the following information in it: CH3 (a) The molecular ion. (b) The base peak. (c) The Mt - C3H7 fragment ion. 83 100 55 80 60 98 40 27 69 20 60 70 80 90 100 20 30 40 501. The two mass spectra below correspond to two isomers of C5H10O2: methyl butanoate and valeric acid. Match the spectrum with the appropriate compound. Place the m/z ratio and the structures (show all atoms) for the labeled fragments in the table below. 100 - Compound MS-NU-1708 Methyl butanoate 80 - 1 60 - 40 - CH3 2 H3C 20 - 10 20 30 40 50 60 70 80 90 100 m/z 100 - Compound MS-NU-3125 3 80 - Valeric acid 60 - 4 40 - H3C 20 - 10 20 30 40 50 60 70 80 90 100 m/z Fragment 1 Fragment 2 Fragment 3 Fragment 4 m/z 74 87 Fragment Relative IntensityWhich method, AES or XPS, is better for resolving chemical shifts and why?