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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.

Short Answer

Expert verified

When bead passes in front of the detector, it excites, and the fluorescence is measured. According to the graph, the more TNT in the standards, the less fluorescence related to the beads.

Step by step solution

01

Binding of TNT antibodies to the latex beads


Antibodies that bind TNT are attached to the latex beads.

02

Saturation of antibody with fluorescent labeled TNT

The antibody is saturated with fluorescent labelled TNT, the excess TNT is removed.

03

Displacement of fluorescence derivatized TNT from bound antibodies

TNT in the standard or sample displaces some fluorescence derivatized TNT from bound antibodies

04

Injection of aliquot

Then an aliquot is injected into the flow cytometer. When bead passes in front of the detector, it excites, and the fluorescence is measured. According to the graph, the more TNT in the standards, the less fluorescence related to the beads.

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The spreadsheet gives the productฮตbfor 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.

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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.

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