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Chapter 19: Q22P (page 488)

The end of Box 19-2 states that “the advantage of up conversion for biomedical probes is that low energy near-infrared (800 to 1000 nm) incident radiation stimulates little background emission from the complex biological matrix that can be highly fluorescent under visible radiation.” Suggest why near-infrared radiation stimulates less emission than visible radiation and why this behavior is useful.

Short Answer

Expert verified

The reason for the promotion of infra-red radiation shows less emissions than visible radiation and the significance of these behaviors was explained.

Step by step solution

01

Reason for less stimulation:

The reason for the stimulation of infra-red radiation shows less emissions than visible radiation because fewer biological molecules draw near infrared radiation at a distance of 800-1000nm..

02

STEP-2: Reason for its useful behavior:

The importance of this behavior is due to the fact that a small matrix fluorescence is promoted

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Most popular questions from this chapter

Now we use Solver to find Kfor the previous problem. The only absorbing species at 332 nmis the complex, so, from Beer’s law $[complex] = A / ε (becausepathlength = 1.000 cm) . I_{2}$ is either free or bound in the complex,so $[I_{2}] = [I_{2}]_{tot} - [complex] .$ There is a huge excess of mesitylene, so $[mesitylene]$ $\approx [mesitylene]_{tot}$

$K = \frac{[c o m p l e x]}{[l_{2}] [m e s i t y l e n e]} = \frac{A / ε}{({[l_{2}]}_{t o t} - A / ε) {[m e s t i t y l e n e]}_{t o t}}$

The spreadsheet shows some of the data. You will need to use all the data. Column A contains [mesitylene] and column B contains ${[l_{2}]}_{tot}$ _. Column C lists the measured absorbance. Guessa value of the molar absorptivity of the complex, $ε, in cell A 7 .$ Then compute the concentration of the complex $(= A / ε)$ in column D. The equilibrium constant in column $E is given by E 2 =$ $[complex] / ([I_{2}] [mesitylene]) = (D 2) / ((B 2 - D 2) * A 2) .$

should we minimize with Solver? We want to varyεin cell A7 until the values of Kin column E are as constant as possible. We would like to minimize a function like $\sum (K_{i} - K_{average})^{2}$ , where K_iis the value in each line of the table and Kaverage is the average of all computed values. The problem with $\sum (K_{i} - K_{average})^{2}$ is that we can minimize this function simply by making $K_{i}$ very small, but not necessarily constant. What we really want is for all the $K_{i}$ to be clustered around the mean value. A good way to do this is to minimize the relative standard deviationof the K, which is (standard deviation)/average. In cell E5we compute the average value of Kand in cell E6the standard deviation. Cell E7contains the relative standard deviation. Use Solver to minimize cell E7by varying cell A7. Compare your answer with that of Problem 19-13.

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.

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 find $[{In}^{-}] and [HIn]$ in the mixture. Comment on the sum $[{In}^{-}] + [HIn]$ .

(b) From $[{In}^{-}]$ in the mixture, and from $p K_{a} = 7.10$ for 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?

Spectroscopic data for the indicators thymol blue (TB),semithymol blue (STB), and methylthymol blue (MTB) are shown in the table. A solution ofTB,STB,MTB in a1.000-cm cuvet had absorbances of 0.412at455nm,0.350 at485nm, and 0.632 at545nm. Modify the spreadsheet in Figure 19-4 to handle three simultaneous equations and find $[TB], [STB]$ and $[MTB]$ in the mixture.

Scatchard plot for binding of estradiol to albumin. Data in the table come from a student experiment to measure the binding constant of the radioactively labeled hormone estradiol (X)to the protein, bovine serum albumin $(P) . Estradiol (7.5 nM)$ was equilibrated with various concentrations of albumin for $30 \min at 37 ° C .$ A small fraction of unbound estradiol was removed by solid phase microextraction ( $Section 24 - 4$ ) and measured by liquid scintillation counting. Albumin is present in large excess, so its concentration in any given solution is essentially equal to its initial concentration in that solution. Call the initial concentration of estradiol ${[X]}_{0}$ and the final concentration of unbound estradiol [X]. Then bound estradiol is $[X]_{0} ― [X]$ and the equilibrium constant is

$X + P ⇌ PX K = \frac{[PX]}{[X] [P]} = \frac{[X]_{0} - [X]}{[X] [P]}$

which you can rearrange to
localid="1663648487221" $\frac{[X]_{0}}{[X]} = K [P] + 1$

A graph of ${[X]}_{0} / [X]$ versus [P]should be a straight line with a slope of K.The quotient $[X]_{0} / [X]$ is equal to the counts of radioactive estradiol extracted from a solution without albumin divided by the counts of estradiol extracted from a solution with estradiol. (b) What fraction of estradiol is bound to albumin at the first and last points?

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