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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

Find the concentration of [X] if the absorbance are 0.700 at 272 nm and

0.550 at 327 nm.

This problem can be worked with Equations 19-6 on a calculator or with the spreadsheet in Figure 19-4. Transferrin is the iron-transport protein found in blood. It has a molecular mass of 81 000 and carries twoFe3+ions. Desferrioxamine B is a chelator used to treat patients with iron overload (see the opening of Chapter 12). It has a molecular mass of about 650 and can bind oneFe3+Fe31. Desferrioxamine can take iron from many sites within the body and is excreted (with its iron) through the kidneys. Molar absorptivities of these compounds (saturated with iron) at two wavelengths are given in the table. Both compounds are colorless (no visible absorption) in the absence of iron.


(a) A solution of transferrin exhibits an absorbance of 0.463 at 470 nm in a 1.000-cm cell. Calculate the concentration of transferrin in milligrams per milliliter and the concentration of bound iron in micrograms per milliliter.

(b) After adding desferrioxamine (which dilutes the sample), the absorbance at 470 nm was 0.424, and the absorbance at 428 nm was 0.401. Calculate the fraction of iron in transferrin and the fraction in desferrioxamine. Remember that transferrin binds two iron atoms and desferrioxamine binds only one.

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 (x0)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∆Aat 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. The increase in absorbance when P is added is

The spreadsheet uses Solver to vary K and ∆Ein cells B10:B11 to minimize the sum of squares of differences between observed and calculatedin 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 ΔAcalc=ΔE[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. If half is reacted, then[X]=[PX]=2.85μMand[P]=P0−[PX]=40.4−2.85=37.55μMThe binding

constant is K=[PX]∧P][X])=[2.85μM]∧37.55μM][2.85μM])=2.7×104which we enter as our guess for K in cell B10. We estimatein cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, ΔA=Δε[PX].The measured

value ofin row 20 is 0.0291 and we just estimated that. Therefore, our guess for localid="1668328314124" Δεin cellB11isΔε=ΔA/[PX]=(0.0291)(2.85μM)=1.0×104in cell B11 is

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 andin cells B10:B11 to minimizeΣAoths−Acalc2in cell I21.

Fluorescence quenching in micelles. Consider an aqueous solution with a high concentration of micelles and relatively low concentrations of the fluorescent molecule pyrene and a quencher (cetylpyridinium chloride, designated Q), both of which dissolve in the micelles.


Quenching occurs if pyrene and Q are in the same micelle. Let the total concentration of quencher be [Q] and the concentration of micelles be [M]. The average number of quenchers per micelle isQ=[Q]/[M]. If Q is randomly distributed among the micelles, then the probability that a particular micelle has n molecules of Q is given by the Poisson distribution:

Probability of n molecules of Q in micelle =Pn=Qnn!e-Q

whereis n factorial(=n[n-1][n-2]....[1]). The probability that there are no molecules of Q in a micelle is

Probability ofmolecules of Q in micelle = Pn=Q00!e-Q=e-Q

because 0!=1

Let l0be the fluorescence intensity of pyrene in the absence of Q and let IQbe the intensity in the presence of Q (both measured at the same concentration of micelles). The quotient lQ/l0must be e-Qwhich is the probability that a micelle does not possess a quencher molecule. Substituting Q=[Q]/[M]gives

lQ/l0=e-Q=e-[Q]/[M]

Micelles are made of the surfactant molecule, sodium dodecyl sulfate. When surfactant is added to a solution, no micelles form until a minimum concentration called the critical micelle concentration (CMC) is attained. When the total concentration of surfactant, [S], exceeds the critical concentration, then the surfactant found in micelles is[S]-[CMC]. The molar concentration of micelles is

[M]=[S]-[CMS]Nav

where Nav is the average number of molecules of surfactant in each micelle.

Combining Equationsandgives an expression for fluorescence as a function of total quencher concentration, [Q]:

ln=l0lQ=[Q]Nav[S]-[CMS]

By measuring fluorescence intensity as a function of [Q] at fixed [S], we can find the average number of molecules of S per micelle if we know the critical micelle concentration (which is independently measured in solutions of S). The table gives data for 3.8μM

pyrene in a micellar solution with a total concentration of sodium dodecyl sulfate [S]=20.8mM

(a) If micelles were not present, quenching would be expected to follow the Stern-Volmer equation. Show that the graph of l0/lQversus [Q] is not linear.

(b) The critical micelle concentration is 8.1mM.Prepare a graph ofln(l0/lQ)versus [Q]. Use Equation 5 to find Nav, the average number of sodium dodecyl sulfate molecules per micelle.

(c) Find the concentration of micelles, [M], and the average number of molecules of Q per micelle,Q, when[Q]=0.200mM

(d) Compute the fractions of micelles containing,, andmolecules of Q when[Q]=0.200mM

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.
See all solutions

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