SPECTROSCOPY MINI RESULTS

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Georgia State University *

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3810

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Chemistry

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Feb 20, 2024

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SPECTROSCOPY MINI RESULTS Spectroscopy of the Known Sample A protein sample with a known concentration of 0.125 mM was stained with the optically active chromophore Fast Green FCF. The sample was diluted at 7 different concentrations using 0.1M Tris and the OD values of each sample was measured at 625nm. Each of the 7 diluted concentrations was placed into 7 different cuvettes, and the goal was to create a standard curve using the OD readings of these 7 different concentrations in order to observe the effect of increasing concentration on Optical Density (OD). It was found that the concentration of the protein sample was directly proportional to the OD. This result can be explained by the Lambert-Beer Law; OD = εcl, where y = OD and x = concentration (c) with a slope of 0.0361 which is equal to the extinction coefficient (ε) (Table 1, Figure 1). Table 1. Fast Green FCF Known Concentrations and its Optical Density The protein sample, which was stained with Fast Green FCF, was diluted into seven different concentrations and placed into seven separate cuvettes. The optical density (OD) of each sample was then measured and recorded in Table 1 using a spectrophotometer at a wavelength of 625nm. The results were used to plot a standard curve (Figure 1), which indicated that an increase in concentration led to an increase in the OD values. Cuvette # Concentration (µM) (Independent Variable) Optical Density at 625nm (Dependent Variable) 1 0.625 0.063 2 1.25 0.080 3 2.5 0.137 4 5 0.177 5 7.5 0.257 6 10 0.391 7 12.5 0.509 Figure 1. Fast Green FCF Standard Curve 0 2 4 6 8 10 12 14 0 0.1 0.2 0.3 0.4 0.5 0.6 f(x) = 0.04 x + 0.03 Fast Green FCP Standard Curve at 625nm Concentration (µM) OD at 625nm

A scatter plot was created to show the relationship between the OD values of the seven cuvettes from Table 1 and their corresponding wavelengths. The line of best fit was plotted as a dotted line with an equation, indicating a positive relationship between OD and concentration as per the Lambert-Beer Law. Spectroscopy of Unknown Samples The spectrophotometer was used to measure the OD values at wavelength 625nm of three stained samples: A, B, and C. Using the equation of the line generated from the standard curve (y= 0.0361x + 0.0273 in Figure 1), the concentrations of these three samples were calculated and recorded in Table 2. For samples B and C, a series of serial dilutions were performed because their respective OD values were above 1.0. The dilution factor was 1:2 for sample B and 1:10 for sample C (Table 2). Another concentration was calculated for the diluted samples of B and C and recorded in Table 2. Next, a manual spectrum analysis of Sample A was conducted to determine the peak wavelength at which Sample A absorbed the most visible light. The selected wavelength range was between 400nm to 700nm with an increment of 50nm (Figure 2). Two peaks were observed for sample A, the first one at 400nm with an OD value of 0.376 and the second one at 650nm with an OD value of 0.368 (Figure 2). Table 2. Unknown Samples A, B, and C The OD values of samples A, B, and C were obtained using the spectrophotometer. Sample B and C underwent a series of serial dilutions until their OD values were below 1.0. The diluted OD values were measured and recorded in Table 2 along with their dilution factors. Sample A did not undergo any dilution because its first OD reading was below 1.0. The obtained OD values for each sample were used to calculate their concentrations using the equation of the line found in Figure 1. Unknown Samples Undiluted OD at 625nm Dilution Ratio Diluted OD at 625 nm Undiluted Concentration (µM) Diluted Concentration (µM) A 0.836 None 0.836 22.40 22.40 B 1.391 1:2 0.689 37.78 18.33 C 4.157 1:10 0.328 114.40 8.33 Calculation for Concentration of Unknown Samples A, B, and C  Sample A  Using the equation, y = 0.0361x +0.0273, y= 0.836 and x is the concentration of A in µM  Hence, 0.836 = 0.0361(x) + 0.0273  x = ( 0.836 − 0.0273 ) 0.0361  x = 22.40 µM  Sample B o Undiluted:  Using the equation, y = 0.0361x +0.0273, for undiluted OD, y= 1.391 and x is the undiluted concentration of B in µM

 Hence, 1.391 = 0.0361(x) + 0.0273  x = ( 1.391 − 0.0273 ) 0.0361 z  x = 37.78 µM o Diluted:  Using the equation, y = 0.0361x +0.0273, for diluted OD, y= 0.689 and x is the undiluted concentration of B in µM  Hence, 0.689 = 0.0361(x) + 0.0273  x = ( 0.689 − 0.0273 ) 0.0361  x = 18.33 µM  Sample C o Undiluted:  Using the equation, y = 0.0361x +0.0273, for undiluted OD, y= 4.157 and x is the undiluted concentration of C in µM  Hence, 4.157 = 0.0361(x) + 0.0273  x = ( 4.157 − 0.0273 ) 0.0361  x = 114.40 µM o Diluted:  Using the equation, y = 0.0361x +0.0273, for diluted OD, y= 0.328 and x is the undiluted concentration of C in µM  Hence, 0.328 = 0.0361(x) + 0.0273  x = ( 0.328 − 0.0273 ) 0.0361  x = 8.33 µM

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