Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Q20P (page 488)

Explain how signal amplification is achieved in enzyme linked immunosorbent assays.

Short Answer

Expert verified

In enzyme- linked immunosorbent assay (ELISA), enzyme catalyzes the reaction in which colored or fluorescent product is formed. So, many molecules of colored product are created for each analyte molecule.

Step by step solution

01

ELISA.

It stands for enzyme-linked immunoassay. It is a common method used for the detection of antibodies from the blood.

02

Explanation.

In enzyme- linked immunosorbent assay (ELISA) the antibody 1 is bound to a polymeric support. Then, analyte is incubated with the polymer-bound antibody to form a complex. The surface is then washed to remove unbound substances. Then, the antibody-antigen complex is treated with antibody 2. Antibody 2 was prepared by covalent attachment of an enzyme. Each enzyme catalyzes the reaction in which colored or fluorescent product is formed. So, many molecules of colored product are created for each analyte molecule.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When are isosbestic points observed and why?

The graph shows the effect of pH on quenching of luminescence of tris(2,2'-bipyridine) Ru(II) by 2,6-dimethylphenol. The ordinate, KSV, is the collection of constants, kq /(ke + kd), in the Stern-Volmer equation. The greater KSV, the greater the quenching. Suggest an explanation for the shape of the graph and

estimate pKa for 2,6-dimethylphenol.

What is the advantage of a time-resolved emission measurement with Eu3+versus measurement of fluorescence from organic chromophores?

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 frompKa=7.10for 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?

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

0.550 at 327 nm.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free