Concept explainers
Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.
Determining the Branch Points and Reducing Ends of Amylopectin A 0.2-g sample of amylopectin was analyzed to determine the fraction of the total glucose residues, that are branch points in the structure. The sample was exhaustively methylated and then digested, yielding 50-μmol of 2,3-dimethylgluetose and 0.4 μmol of 1,2,3,6- letramethylglucose.
- What fraction of the total residues are branch points?
- I low many reducing ends does this sample of amylopectin have?
(a)
Interpretation:
The fraction of the total residues that are branch points in the structure of amylopectin is to be calculated.
Concept Introduction:
The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Polysaccharides are those types of sugars which contain more than ten units of monosaccharides.
Amylopectin is a polysaccharide and is one of the two forms of starch. It is composed of long chain of glucose attached by the
Answer to Problem 8P
The fraction of the total residues that are branch points in the structure of amylopectin is
Explanation of Solution
The given mass of amylopectin sample is
The given amount of
The conversion of
Thus, the moles of
The given amount of
The molecular weight of glucose is
The molecular weight of water is
If glucose unit losses a water molecule while forming a glycosidic linkage in amylopectin, then, the molecular weight of glucose in amylopectin is
The moles of a substance is calculated by the equation as,
Substitute the value of mass and molar mass of the sample in the above equation.
The moles of
Substitute the values of moles of
Thus, the fraction of the total residues that are branch points in the structure of amylopectin is
(b)
Interpretation:
The total number of reducing ends possessed by the sample of amylopectin is to be calculated.
Concept Introduction:
The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Polysaccharides are those types of sugars which contain more than ten units of monosaccharides.
Amylopectin is a polysaccharide and is one of the two forms of starch. It is composed of long chain of glucose attached by the
Answer to Problem 8P
The total number of reducing ends possessed by the sample of amylopectin is
Explanation of Solution
The given amount of
The moles of the reducing ends is calculated by the moles of
If one mole of compounds possesses
Thus, the total number of reducing ends possessed by the sample of amylopectin is
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Chapter 7 Solutions
Biochemistry
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- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Using Graphical Methods to Derive the Kinetic Constants for an Ordered, Single-Displacement Reaction The general rate equation for an ordered, single-displacement reaction where A is the leading substrate is v=Vmax[ A ][ B ](KsAKmB+KmA[ B ]+KmB[ A ]+[ A ][ B ])Write the Lineweaver-Burk (double-reciprocal) equivalent of this equation and from it calculate algebraic expressions for the following: a. The slope b. The y-intercepts c. The horizontal and vertical coordinates of the point of intersection when 1/v is plotted versus 1/[B] at various fixed concentrations of Aarrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Evaluation of -Helices in Proteins The hem agglutinin protein in influenza virus contains a remarkably long -helix, with 53 residues. How long is this -helix (in nm)? How many turns does this helix have? The typical residue in an -helix is involved in two H bonds. How many H bonds are present in this helix?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Solving the Sequence of an Oligopeptide From Sequence Analysis Data Amino acid analysis of a decapeptide revealed the presence of the following products: The following facts were observed: Neither car boxy peptidase A nor B treatment of the- decapeptide had any effect. Trypsin treatment yielded two tetrapcptides and free Lys. Clostripain treatment yielded a tetrapcptide and a hexapeptidc. Cyanogen bromide treatment yielded an octapeptide and a dipeptide of sequence NP (using the one-letter codes). Chymotrypsin treatment yielded two tripeptides and a telrapeptide. The N-terminal chymotryptic peptide had a net charge of — 1 at neutral pi I and a net charge of —3 al pH 12. One cycle of Ed man degradation gave the PTH derivative What is the ammo acid sequence of this decapeptide?arrow_forward
- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. The dissociation constant for a particular protein dimer is 1 micromolar. Calculate the free energy difference for the monomer-to-dimer transition.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Exploring the Dimensions of the α-Helix and Coiled Coils Imagine that the dimensions of the alpha helix were such that there were exactly 3.5 amino acids per turn instead of 3.6. What would be the consequences for coiled-coil structures?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Allosteric Regulation Versus Covalent Modification What are the relative advantages (and disadvantages) of allosteric regulation versus covalent modification?arrow_forward
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning