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# Introduction to Spectrophotometric Analysis - By using spectrophotometric analysis or spectrophotometry, one can determine the identity in terms of structure and species of a biomolecule as well as establish the concentration of a certain biomolecule

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Introduction

Name: Ng Yen Pheng Student ID: 22353046 Date: 6 March 2012 Demo: Title: Introduction to Spectrophotometric Analysis Introduction: Spectrophotometric Analysis which is one of the most common and valuable analysis techniques used to study the interaction of electromagnetic radiation with chemical compounds (Stewart & Ebel, 2000). By using spectrophotometric analysis or spectrophotometry, one can determine the identity in terms of structure and species of a biomolecule as well as establish the concentration of a certain biomolecule (Stewart & Ebel, 2000). Absorption spectrophotometry is the most common of spectrophotometry technique used by biochemist. The science behind this technique is when light passes through a solution, components or substances in the solution will absorb some of the light that passes through. This will result in the excitation of an electron to a higher energy level, and when the electron loses its energy and returns to its original energy state, the energy that is released is shown as emission of light in a different wavelength which results in a different colour, this process is known as fluorescence (Stewart & Ebel, 2000). Beer-Lambert`s Law describe the relationship between the substance concentration and the light absorption at a particular wavelength mathematically (Mikulecky, Gilman & Brutlag, 2009). According to this law, the light absorbed by the sample is directly proportional to the concentration of the sample. This is because when the concentration is higher, there are more particles present per unit of the sample to absorb the light emitted by spectrophotometer (Zijlstra, Buursma & Assendelft, 2000). ...read more.

Middle

A = 0.27mg/mL 0.27 = 0.0473x x = = 5.708 mg mL-1 To calculate the concentration of diluted sample A, using the equation above, Concentration of diluted protein A = 0.03mg/mL 0.03 = 0.0473x x = = 0.6342 mg mL-1 Since this sample has been diluted 10 times, the concentration of the undiluted sample using this value = 0.6342 mg mL-1 � 10 = 6.342 mg mL-1 To calculate the concentration of undiluted sample B, using the equation above, Concentration of undiluted protein B = 1.15mg/mL 1.15 = 0.0473x x = = 24.31 mg mL-1 To calculate the concentration of diluted sample B, using the equation above, Concentration of undiluted protein B = 0.17mg/mL 0.17 = 0.0473x x = = 3.594 mg mL-1 Since this sample has been diluted 10 times, the concentration of the undiluted sample using this value = 3.594 mg mL-1 � 10 = 35.94 mg mL-1 For each sample, two different dilutions are made, because according to the Beer-Lambert's Law, absorbance is directly proportional to concentration (Stewart & Ebel, 2000). Hence, in order to prove this law, the samples are made into two different concentrations through dilution to see its effects on the absorbance. Moreover, dilution was carry out For sample A, the undiluted sample would be considered to be more accurate. The discrepancy between the pairs of results can be seen in the difference of absorbance reading. This is because during the process of dilution, the diluted sample could be contaminated with other substances. ...read more.

Conclusion

As for the RNA, the components involve are the pyrimidine and purine bases (Boyer, 2009). (b) For RNA at 260nm, the absorbance is approximately 0.750 and the extinction coefficient is 25.0 cm-1(mg/mL)-1. Hence, using the Beer-Lambert Law, A = ?cl Where, A = Absorbance; ? = extinction coefficient c = Concentration; l = path length (1 cm) c = = = 0.03 mg/mL (c) Concentration = 1 mg/mL, Absorbance = 0.600. Hence, using the Beer-Lambert Law, A = ?cl Where, A = Absorbance; ? = extinction coefficient c = Concentration; l = path length(1 cm) ? = = = 0.600 cm-1(mg/mL)-1 (d) mg/mL is used as a concentration unit for protein and nucleic acid, because both molecules are very minute and can only be obtained in a small number. Hence, to use M would need the solution to be in a 1 liter volume which is too big of a ratio compared to mL. By using mg/mL, the molecules can be measured and calculated easily too as there will be less decimal place and zeros to consider. Conclusion: The absorbance spectra of haemoglobin, nucleic acids and protein have been obtained. Haemoglobin spectra will be shifted due to the reduction by sodium dithionite solution. Peaks in the spectra of proteins and nucleic acids are dependent on the nature of amino acid side chain and nitrogenous bases respectively. Concentration of inorganic phosphates from sample X and Y, too, have been obtained, which are 75�g mL-1 and 35�g mL-1 respectively.. ...read more.

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