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Unformatted text preview: 1 PS#7 20.110/2.772 Fall 2008 2.772/20.110 Problem Set #7 Due Wednesday November 5 in 16-429 at 3pm NO LATE PSETS WILL BE ACCEPTED Please submit the problems as: 1,2,3 4,5,6 7,8, 9, 10 1.) EDR 19.8 An important biological application of absorption spectroscopy is the determination of the concentrations of solutions of nucleic acids. The π → π * electronic transitions within the purine and pyrimidine bases of nucleic acids have absorption maxima near a wavelength of 260 nm. Assume the extinction coefficient of a nucleic acid is 1.00 × 10 4 M –1 cm –1 at 260 nm. If the concentration of a nucleic acid solution is 5.00 × 10 –4 M, calculate the absorbance of this solution in a 1.00 cm cell at 260 nm. 2.) EDR 19.9 The absorbance of a nucleic acid solution at 260 nm is called A 260 . The OD (i.e., optical density) is the amount of nucleic acid in a volume of 1.00 mL in a 1.00-cm path length cell for which A 260 =1.00. How many moles of nucleotide are contained in a 1.00 mL solution of a double-stranded nucleotide for which A 260 = 2.50, assuming the extinction coefficient per nucleotide is 7.000 × 10 3 M –1 cm –1 . Express this quantity in OD’s. Assume a 1.00 cm path length. 3.) EDR 19.10 Because of interactions between transition dipoles of the constituent nucleotides, the extinction coefficient for a single strand polynucleotide is not simply the sum of the extinction coefficients for the individual nucleotides. These dipole-dipole interactions depend on 1/ r 3 where r is the distance between bases, so for the purpose of calculating the extinction coefficient for a single-stranded polynucleotide, only nearest neighbor interactions need be considered. For a hypothetical polynucleotide strand GpCpUp ... ApG the extinction coefficient is ε ( GpCpUp ... ApG ) = 2 [ ε ( GpC ) + ε ( CpU ) + ε ( ApG )] – [ ε ( Cp ) + ε ( Up ) + ......
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This note was uploaded on 11/12/2008 for the course 20 20.110 taught by Professor Griffith during the Fall '08 term at MIT.
- Fall '08