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Chapter 19: Q1TY (page 464)
Find the concentration of [X] if the absorbance are 0.700 at 272 nm and
0.550 at 327 nm.
The concentration of [X] if the absorbance are 0.700 at 272 nm and
0.550 at 327 nm is .
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Figure 19-6 is a Scatchard plot for the addition of 0-20nM antigen X to a fixed concentration of antibodyPrepare a Scatchard plot from the data in the table and find K for the reaction. The table gives measured concentrations of unbound X and the complex PX. It is recommended that the fraction of saturation should span the range ,-0.2-0.8. What is the range of the fraction of saturation for the data?
19-C. The protein bovine serum albumin can bind several molecules of the dye methyl orange. To measure the binding constant K for one dye molecule, solutions were prepared with a fixed concentration \(\left( {{X_0}} \right)\) of dye and a larger, variable concentration of protein (P). The equilibrium is Reaction 19-18, with X 5 methyl
orange.
Experimental data are shown in cells A16-D20 in the spreadsheet on the next page. The authors report the increase in absorbance \((\Delta A)\) at 490 nm as P is added to X. X and PX absorb visible light, but P does not. Equilibrium expression 19-20 applies and (PX) is given by Equation 19-21. Before P is added, the absorbance is \({\varepsilon _X}{X_{0 - }}\). The increase in absorbance when P is added is
The spreadsheet uses Solver to vary K and \(\Delta E\) in cells B10:B11 to minimize the sum of squares of differences between observed and calculated \(\Delta A\) in solutions with different amounts of P. Cell E16 computes (PX) from Equation 19-21, which is Equation A on line 6 of the spreadsheet. Cells F16 and G16 find (X) and (P) from mass balances. Cell H16 computes \(\Delta {A_{calc}} = \Delta E(PX)\)which is Equation B on line 7.
To estimate a value of K in cell B10, suppose that 50% of X has reacted in row 20 of the spreadsheet. The total concentration of X is\({X_0} = 5.7\mu M\). If half is reacted, then \((X) = (PX) = 2.85\mu M\) and\((P) = {P_0} - (PX) = 40.4 - 2.85 = 37.55\mu M\). The binding
constant is \(K = (PX)/(P)(X)) = (2.85\mu M)/(37.55\mu M)(2.85\mu M)) = \)\(2.7 \times 1{0^4}\) which we enter as our guess for K in cell B10. We estimate \(\Delta \varepsilon \)in cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, \(\Delta A = \Delta \varepsilon (PX)\).The measured
value of \((\Delta A)\)in row 20 is 0.0291 and we just estimated that \((PX) = \)\(2.85\mu M\). Therefore, our guess for \(\Delta \varepsilon \)in cell B11 is \(\Delta \varepsilon = \Delta A/\)\((PX) = (0.0291)/(2.85\mu M) = 1.0 \times 1{0^4}\)
Your assignment is to write formulas in columns E through J of the spreadsheet to reproduce what is shown and to find values in cells E17:J20. Then use Solver to find K and \(\Delta \varepsilon \)in cells B10:B11 to minimize \(\Sigma {\left( {{A_{oths\;}} - {A_{calc\;}}} \right)^2}\)in cell I21.
The spreadsheet gives the productfor four pure compounds and a mixture at infrared wavelengths. Modify Figure 19-4 to solve four equations and find the concentration of each compound. You can treat the coefficient matrix as if it were molar absorptivity because the path length was constant (but unknown) for all measurements.
A study was conducted with derivatives of the DNA nucleotide bases adenine and thymine bound inside micelles () in aqueous solution.
Sodium dodecyl sulfate forms micelles with the hydrocarbon tails pointed inward and ionic headgroups exposed to water. It was hypothesized that the bases would form ahydrogen-bonded complex inside the micelle as they do in DNA:
To test the hypothesis, aliquots of 5.0 mMadenine derivative were mixed with aliquots of 5.0 mMthymine derivative in proportions shown in the table. Each solution also contained 20mMsodium dodecyl sulfate. The concentration of product measured by nuclear magnetic resonance also is shown in the table. Are the results consistent with formation of a 1:1complex? Explain your answer.
Two ways to analyze a mixture. Figure 19-5 shows the spectrum of the indicator bromothymol blue adjusted to several pH values. The spectrum at pHis that of the pure blue form and the spectrum at pH 1.8is that of the pure yellow form. At other pHvalues, there is a mixture of the two forms. The total concentration isand the path length isin all spectra. For the purpose of calculation, assume that there are more than two significant digits in concentration and path length. Absorbance at the dots on three of the curves in Figure 19-5 is given in the table.
(a) Prepare a spreadsheet like Figure 19-3 to use absorption at all six wavelengths to findin the mixture. Comment on the sum.
(b) Fromin the mixture, and fromfor HIn,calculate theof the mixture. (This calculation is the source of pH labels in the figure.)
(c) Use Equations 19-6 at the peak wavelengths ofto findin the mixture. Compare your answers to those in (a). Which answers, (a) or (c), are probably more accurate? Why?
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