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Figure 19-6 is a Scatchard plot for the addition of 0-20nM antigen X to a fixed concentration of antibodyP=(Po=10nM)Prepare a Scatchard plot from the data in the table and find K for the reactionP+XPX. 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?

Short Answer

Expert verified

The range of the fraction of saturation for the data is from 0.29 to 0.85 and the value of K for the reactionP+XPXisK=4.109M-1.

Step by step solution

01

Find the range of fraction of saturation and the value of for the given reaction:

Hence, the spreadsheet is,

Thus, the column D is the range of the fraction of saturation (from0.29 to 0.85).

Plot the graph as,

Since, the slope is K , the value ofK=4.109M-1Therefore, the value of K for the reaction isK=4.109M-1 .

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Most popular questions from this chapter

Here is an immunoassay to measure explosives such as trinitrotoluene (TNT) in organic solvent extracts of soil. The assay employs a flow cytometer, which counts small particles (such as living cells) flowing through a narrow tube past a detector. The cytometer in this experiment irradiates the particles with a green

laser and measures fluorescence from each particle as it flows past the detector.

1. Antibodies that bind TNT are chemically attached to 5mmdiameter latex beads.

2. The beads are incubated with a fluorescent derivative of TNT to saturate the antibodies, and excess TNT derivative is removed. The beads are resuspended in aqueous detergent.

3. 5mlof the suspension are added to 100mlof sample or standard. TNT in the sample or standard displaces some derivatized TNT from bound antibodies. The higher the concentration of TNT, the more derivatized TNT is displaced.

4. An aliquot is injected into the flow cytometer, which measures fluorescence of individual beads as they pass the detector. The figure shows median fluorescence intensity 6 standard deviation. TNT can be quantified in the ppb to ppm range.

Draw pictures showing the state of the beads in steps 1, 2, and 3and explain how this method works.

Method of continuous variation. Make a graph of absorbance versus mole fraction of thiocyanate from the data in the table.

(c)Why does one solution containacid and the otheracid?

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.

Iodine reacts with mesitylene to form a complex with an absorption maximum at 332 nm in CCl4 solution:

(b)Spectrophotometric data for this reaction are shown in the table.Because[mesitylene]tot>>[I2],we can say that [mesitylene][mesitylene]tot. Prepare a graph ofA/([mesitylene][I2]tot)versusA/[I2]totand find the equilibrium constant and molar absorptivity of the complex.

The figure shows spectra of1.00×10-4MMnO4-,1.00×10-4and an unknown mixture of both, all in1.000cmpath length cells. Absorbances are given in the table. Use the least squares procedure in Figure 19-3 to find the concentration of each species in the mixture.

Visible spectrum ofMnO4-,Cr2O72-, and an unknown mixture containing both ions.

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