Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For aspartic and glutamic acids, the isoelectric point occurs at a \(\mathrm{pH}\) where the net charge on the two carboxyl groups is \(-1\) and balances the charge of \(+1\) on the \(\alpha\)-amino group. Calculate pI for these amino acids.

Short Answer

Expert verified
Answer: The isoelectric point (pI) of aspartic acid is 2.77, and for glutamic acid, it is 3.22.

Step by step solution

01

Find the pK values for COOH, NH3+, and the side chain carboxyl group

Aspartic acid has the following pK values: pK1 (COOH) = 1.88, pK2 (NH3+) = 9.60, and pK3 (side chain carboxyl group) = 3.65. Glutamic acid has the following pK values: pK1 (COOH) = 2.19, pK2 (NH3+) = 9.67, and pK3 (side chain carboxyl group) = 4.25.
02

Identify the pH range between which the net charge of amino acid is zero

At the isoelectric point, we are looking for the pH when the COOH group has a -1 charge and the NH3+ group has a +1 charge. Since the side chain carboxyl group has a lower pK value than the NH3+ group, we will consider the pH range between the COOH group's pK value and the side chain carboxyl group's pK value as the range in which the net charge of the amino acid is zero.
03

Calculate the isoelectric point (pI) as the average of the two relevant pK values

For aspartic acid, the pI will be the average of pK1 (COOH) and pK3 (side chain carboxyl group). The isoelectric point can be calculated as: $$\mathrm{pI} = \frac{pK_1 + pK_3}{2}$$ $$\mathrm{pI} = \frac{1.88 + 3.65}{2}$$ $$\mathrm{pI} = \frac{5.53}{2}$$ $$\mathrm{pI} = 2.77$$ For glutamic acid, we will average pK1 (COOH) and pK3 (side chain carboxyl group) values. The isoelectric point can be calculated as:$$\mathrm{pI} = \frac{pK_1 + pK_3}{2}$$ $$\mathrm{pI} = \frac{2.19 + 4.25}{2}$$ $$\mathrm{pI} = \frac{6.44}{2}$$ $$\mathrm{pI} = 3.22$$
04

State the results

The isoelectric point (pI) for aspartic acid is 2.77, and for glutamic acid, it is 3.22.

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.

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

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amino Acids
Amino acids are fundamental units that build up proteins, which are crucial for numerous biological functions in living organisms. Each amino acid consists of a central carbon atom called the \( \alpha \-carbon \), attached to an amino group (\( -NH_3^+ \)), a carboxyl group (\( -COOH \)), a hydrogen atom, and a distinctive side chain, or R-group, that determines the properties and identity of the amino acid.

There are 20 standard amino acids, and among them, aspartic acid and glutamic acid are classified as acidic amino acids due to their additional carboxyl groups present in their side chains, which impart an acidic nature. These extra groups significantly contribute to the amino acid's behavior, especially when it comes to their isoelectric points—a concept intrinsic to understanding their biochemistry in aqueous solutions.
pK Values
The pK value is a critical concept in biochemistry that informs us about the acid dissociation constant of a group. It describes the pH at which a particular molecule or group, like an amino, carboxyl or side chain group, will exist half dissociated—meaning 50% will be in the protonated form and 50% in the deprotonated form.

For amino acids such as aspartic and glutamic acid, three important pK values are considered: the pK value for the \( \alpha \-COOH \) group (pK1), the pK value for the \( \alpha \-NH_3^+ \) group (pK2), and the pK value for the side chain carboxyl group (pK3). Knowing these values is essential for calculating the isoelectric point, which is the pH where the amino acid does not move in an electric field, commonly known as the point of zero net electric charge.
Biochemistry
Biochemistry is the study of chemical processes within and related to living organisms. It encompasses a vast array of chemical reactions, interactions, and signaling pathways that are foundational for life. A central topic of biochemistry is understanding how the structure of a molecule, such as an amino acid, relates to its function in the body.

Calculating the isoelectric point of amino acids is a biochemistry application that provides insight into how amino acids behave under varying pH conditions. This knowledge is crucial in areas such as protein purification, enzyme activity study, and understanding the physicochemical properties of proteins. It allows scientists and researchers to predict how amino acids and proteins will interact in different environments, affecting their biological roles and their interactions with other molecules.

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

Dinitrofluorobenzene, very often known as Sanger's reagent after the English chemist Frederick Sanger who popularized its use, reacts selectively with the \(N\)-terminal amino group of a polypeptide chain. Sanger was awarded the 1958 Nobel Prize in Chemistry for his work in determining the primary structure of bovine insulin. One of the few people to be awarded two Nobel Prizes, he also shared the 1980 award in chemistry with American chemists Paul Berg and Walter Gilbert for the development of chemical and biological analyses of DNA. Following reaction with 2,4 -dinitrofluorobenzene, all amide bonds of the polypeptide chain are hydrolyzed and the amino acid labeled with a 2,4-dinitrophenyl group is separated by either paper or column chromatography and identified. (a) Write a structural formula for the product formed by treatment of the \(N\)-terminal amino group with Sanger's reagent and propose a mechanism for its formation. (b) When bovine insulin is treated with Sanger's reagent followed by hydrolysis of all peptide bonds, two labeled amino acids are detected: glycine and phenylalanine. What conclusions can be drawn from this information about the primary structure of bovine insulin? (c) Compare and contrast the structural information that can be obtained from use of Sanger's reagent with that from use of the Edman degradation.

Although only L-amino acids occur in proteins, D-amino acids are often a part of the metabolism of lower organisms. The antibiotic actinomycin D, for example, contains a unit of D-valine, and the antibiotic bacitracin A contains units of D-asparagine and D-glutamic acid. Draw Fischer projections and three- dimensional representations for these three D-amino acids.

Account for the fact that the isoelectric point of glutamine (pI 5.65) is higher than the isoelectric point of glutamic acid (pI 3.08).

The BOC-protecting group may be added by treatment of an amino acid with di- tertbutyl dicarbonate as shown in the following reaction sequence. Propose a mechanism to account for formation of these products.

Several \(\beta\)-amino acids exist. There is a unit of \(\beta\)-alanine, for example, contained within the structure of coenzyme A (Problem 25.34). Write the structural formula of \(\beta\)-alanine.

See all solutions

Recommended explanations on Chemistry Textbooks

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