Biochemistry: Concepts and Connections (2nd Edition)
2nd Edition
ISBN: 9780134641621
Author: Dean R. Appling, Spencer J. Anthony-Cahill, Christopher K. Mathews
Publisher: PEARSON
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Chapter 9, Problem 2P
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α-D-Galactopyranose rotates the plane of polarized light, but the productof its reduction with sodium borohydride (galactitol) does not. Explain thedifference.
The visible spectrum of B-carotene (C40oHs, MW 536.89, the orange pigment in
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Calculate the concentration in milligrams per milliliter of B-carotene that gives
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Chapter 9 Solutions
Biochemistry: Concepts and Connections (2nd Edition)
Ch. 9 - Prob. 1PCh. 9 - -D-Galactopyranose rotates the plane of polarized...Ch. 9 - Provide an explanation for the fact that a-...Ch. 9 - Why is a type O individual considered a universal...Ch. 9 - The disaccharide , -trehalose differs from the ,...Ch. 9 - A reducing sugar will undergo the Fehling...Ch. 9 - Prob. 7PCh. 9 - Prob. 8PCh. 9 - Indicate whether the structures shown are R and S...Ch. 9 - Prob. 10P
Ch. 9 - Draw (using Haworth projections) the fragments of...Ch. 9 - One or more of the compounds shown below will...Ch. 9 - Why do you suppose that the influenza virus...Ch. 9 - Prob. 14PCh. 9 - Are mannose and galactose epimers? Allose and...Ch. 9 - Prob. 16PCh. 9 - Explain in about one sentence why it is important...Ch. 9 - Prob. 18PCh. 9 - Prob. 19PCh. 9 - Prob. 20PCh. 9 - Prob. 21PCh. 9 - Prob. 22PCh. 9 - Prob. 23P
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- Are fatty acid propyl esters (biodiesel) formed in the acid-catalysed esterification/transesterification reaction on waste oil more or less polar than glycerol? Briefly explain.arrow_forwardLithium ion inhibits the synthesis of inositol trisphosphate by inhibiting a reaction in the breakdown of inositol trisphosphate.Explain this apparent paradox.arrow_forwardLithium ion inhibits the synthesis of inositol trisphosphate by inhibiting a reaction in the breakdown of inositol trisphosphate. Explain this apparent paradox.arrow_forward
- Define about X-gal (technically 5-bromo-4-chloro-3-indolyl-b-D- galactopyranoside) ? Explain importance of this ?arrow_forwardWill the synthesis of 9,10-dihydroanthracene-9,10-α,β-succinic anhydride via Diels-Alder reaction occur faster under photochemical conditions? Explain.arrow_forwardAt equilibrium, D-galactose exists almost exclusively in its α and β pyranose forms. Aqueous solutions are freshly prepared for the α and β forms, both at the same concentration and temperature. The solution of the α form rotatesplane-polarized light +150.7°, whereas the solution of the β form rotates the light +52.8°. Over time, both solutions have the same measured rotation of +80.2°. How much of the equilibrated solution does each form account for?arrow_forward
- Sulfur compounds give onions their unique flavor and properties. Compound 1 is the starting material for the majority of these flavor components and is converted to Compound 2 under the action of the enzyme alliinase (Reaction 1, unbalanced). +NH3 Compound 1. Reaction 1 Alliinase is a pyridoxal phosphate (PLP, vitamin B6) dependent enzyme. PLP is a planar molecule with a 6-membered ring, a hydroxyl group, an aldehyde group, and a phosphate. PLP is covalently attached to alliinase by forming an aldimine with the side chain of the amino acid at position 251. During catalysis the aldimine is replaced by one formed with the substrate, which facilitates Reaction 1. OH Compound 2 Compound 2 is a very reactive substance known as a sulfenic acid, with an estimated pK, of 12.5. Compound 2 spontaneously and rapidly condenses to form Compound 3, which is one of the main flavor components of onions. 'S OH Compound 2 alliinase Reaction 2 When an onion is cut, the cell components are allowed to mix…arrow_forwardThe molecular structure of vitamin A is conjugated with N = 10. Calculatethe wavelength of light in the first transition of vitamin A.arrow_forwardAn inhibitor that specifically labels chymotrypsin at histidine 57 is N-tosylamido-l-phenylethyl chloromethylketone. How would you modify the structure of this inhibitor tolabel the active site of trypsin?arrow_forward
- The amount of branching (number of (α1→6) glycosidic bonds) in amylopectin can be determined by the following procedure. A sample of amylopectin is exhaustively methylated—treated with a methylating agent (methyl iodide) that replaces the hydrogen of every sugar hydroxyl witha methyl group, converting —OH to —OCH3 . All the glycosidic bonds in the treated sample are then hydrolyzed in aqueous acid, and the amount of 2,3-di-O-methylglucose so formed is determined.(a) Explain the basis of this procedure for determining the number of (α1→6) branch points in amylopectin. What happens to the unbranched glucose residues in amylopectin during the methylation and hydrolysis procedure?(b) A 258 mg sample of amylopectin treated as described above yielded 12.4 mg of 2,3-di-O-methylglucose. Determine what percentage of the glucose residues in the amylopectin contained an (α1→6) branch. (Assume that the average molecular weight of a glucose residue in amylopectin is 162 g/mol.)arrow_forwardAn inhibitor that specifically labels chymotrypsin at histidine 57 is Ntosylamido-l-phenylethyl chloromethyl ketone. How would you modify the structure of this inhibitor to label the active site of trypsin?arrow_forwardDraw the sturucture Turanose: -D-glucopyranosyl-(13)--D-fructofuranosearrow_forward
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