General, Organic, And Biological Chemistry, Hybrid (with Owlv2 Quick Prep For General Chemistry Printed Access Card)

7th Edition

ISBN: 9781305253070

Author: STOKER, H. Stephen

Publisher: Cengage Learning

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Neutral Amino Acids

Amino acids which do not have any charge on them are neutral amino acids.

Globular Protein

The globular proteins refer to the shape of protein specifically spherical in nature apart from spherical form fibrous, disordered and membrane-bound proteins exist. These globular proteins are miscible in water and form a colloidal solution rather than …

Dimer

Dimers are basic organic compounds, which are derivates of oligomers. It is formed by the combination of two monomers which could potentially be strong or weak and in most cases covalent or intermolecular in nature. Identical monomers are called homodime…

Dipeptide

A dipeptide is considered a mixture of two distinct amino acids. Since the amino acids are distinct, based on their composition, two dipeptide's isomers can be produced. Various dipeptides are biologically essential and are therefore crucial to industry.

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Textbook Question

Chapter 26.1, Problem 3QQ

The inactive form of pepsin is converted to its active form

a. in the small intestine
b. by hydrochloric acid
c. by gastrin
d. no correct response

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Chapter 26.1, Problem 2QQ

Chapter 26.1, Problem 4QQ

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Relative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/z

Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100

Part IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained. Draw the structures of these fragments.

Chapter 26 Solutions

General, Organic, And Biological Chemistry, Hybrid (with Owlv2 Quick Prep For General Chemistry Printed Access Card)

Ch. 26.1 - Which of the following statements about dietary...Ch. 26.1 - Dietary protein materials as they leave the...Ch. 26.1 - The inactive form of pepsin is converted to its...Ch. 26.1 - Which of the following is not a proteolytic...Ch. 26.2 - The dominant use for the amino acids of the amino...Ch. 26.2 - The most abundant amino acid in the amino acid...Ch. 26.2 - Prob. 3QQ Ch. 26.3 - Prob. 1QQ Ch. 26.3 - Prob. 2QQ Ch. 26.3 - The net effect of transamination is to collect the...

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The inactive form of pepsin is converted to its active form a. in the small intestine b. by hydrochloric acid c. by gastrin d. no correct response

Solution Summary: The author identifies the option by which the inactive form of pepsin is converted to its active form by hydrochloric acid.

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Chapter 26 Solutions