Fundamental Laboratory Approaches
for Biochemistry and Biotechnology

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by Alex J. Ninfa and David P. Ballou

Notes on Experiments Homework Answers Book Revisions Contact Us Authors Home Links		Individual Solutions for Chapter 2 Experiment 2-2, pp. 71-72 Problem1 Problem 2 Problem 3 Problem 4
		Problem 5 Problem 6 Problem 7 Experiment 2-3, p. 73 Problem 1 Problem 2 Chapter 2, Experiment 2-2, pp. 71-72  Problem1 1.0 mL of 0.1 mM PNP is used. (1x 10-3 L) (1x 10-4 M) = 1 x 10-7 moles Problem 2   Problem 3   [PNP] + [PNP-] = 0.1mM/4 Assume that the absorbance at 390 nm (max peak in Spec 20) is due to the PNP- , that the extinction coefficient for PNP- is 15 mM-1cm-1, and that the path length is 1 cm. Therefore A390 /e390 = [PNP-]. [PNP] = 0.1 mM/4 - [PNP-]  pH A410 [PNP-] {mM} [PNP-] {µM}  [PNP] {mM} [PNP] {µM} 5.00 0.008 5.33 e-4 0.053 0.0245 24.5 6.07 0.051 3.40 e-3 0.34 0.0216 21.6 6.6 0.137 0.009 9.0 0.016 16.0 7.4 0.180 0.012 12.0 0.013 13.0 8.0 0.215 0.014 14.0 0.011 11.0 8.5 0.270 0.018 18.0 0.007 7.0 9.0 0.323 0.022 22.0 0.003 3.0 10.0 0.338 0.023 23.0 0.002 2.0 Problem 4 At pH 8.5 we measured 0.270 absorbance. the total [PNP] = 0.025 mM. Therefore, 0.270/0.025 = 10.8. Note that this is the apparent extinction at pH 8.5, which is not the full extinction of the phenolate. The values you will use in Chapter 7 will also not be at full formation of the phenolate. Problem 5   A = (e410 )(conc)(Path); Path = 1 cm 0.282/(15,000 L/cm•mol)(1 cm) = 1.88 x 10-5 M = 1.88 X 10-2 mM = 18.8 µM Amount of PNP = 1.88 x 10-5 M)(7 x 10-3 L) = 1.32 x 10-7 moles Problem 6 2.8 x 10-3 g/140 g/mole = 2 x 10-5 moles (2 x 10-5 moles)/(2 x 10-2 L) = 1 e-3 M Dilute 0.1 mL to 5.0 mL, a 50-fold dilution. A = [(1 x 10-3 M)/ 50 ](15 x 103 L/cm•mol) = 0.3 Problem 7   Acidic form is shown below Experiment 2-3, p. 73 Problem 1   Refer to the spectrum you recorded in Experiment 2-1. Note that if you are at the wavelength of maximum absorbance, slight errors in selection of wavelength will have only small effects on the measured absorbance. In contrast, at other wavelengths (e.g., on the side of the peak), uncertainties in wavelength selection will cause large changes in absorbance. Moreover, the sensitivity is usually greatest at the wavelength of maximum absorbance. Problem 2 This question should be worded: Calculate the amount of PNP in nmol present in each of the above solutions in the 4 mL of the diluted PNP solutions. Example  For the 30 µM PNP solution (30 x 10-6 M)(4 x 10-3 L) = 1.2 x 10-7 moles = 120 nmoles Problem 3 Use the absorbance of the unknown and your standard curve to read the concentration in your unknown tube. Be sure to correct for any dilution to calculate the concentration in the original sample.