Identify the type of glycosidic linkage present in the disaccharide below:O a (16)O a (1-4)08(1 - 6)03(1-4)НHOОНOHOHO. HННОНHOHНOHОН

Disaccharides are oligosaccharides composed of two monosaccharide units. The monosaccharide units in…

Answered: Identify the type of glycosidic linkage…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Copper ions present in Benedict’s reagent react with the free end of any reducing sugars, such as glucose, when heated. Originally blue in color, these copper ions are reduced by the sugar and produce an orange-red colored precipitate. Alternatively, iodine-potassium iodide (IKI) may also be used when working with starch. IKI contains special tri-iodine ions which interact with the coiled structure of a starch polymer. Prior to a reaction, the IKI displays a yellow-brown color; however, after reacting with starch, a dark purple or black color is presented. The molecule pictured below produced a blue color when tested with Benedict’s reagent, a yellow color when tested with IKI, and a violet color when tested with Biuret reagent. Based on the structure shown below and these chemical results, what kind of biomolecule is this?
Although the first two carbons of fructose and glucose are identical in structure to DHAP and GADP (from glycolysis), DHAP and GADP equilibriate on their in solution to favor the ketone over the aldehyde, while fructose and glucose do not. Why? a)The larger size of the molecule sterically hinders the isomerization b)The larger sugars have more OH groups which hydrogen bond and disrupt isomerization c)The larger sugars cyclize, and there is no carbonyl to isomerize in the cyclic form d)The larger sugars cyclize, and in the cyclic form the hydrogen bonding is very strong e)The larger sugars are less soluble in water than the smaller sugars
Which of the following statement regarding the ends of polysaccharides are true? for heads up 2 and 4 are not correct .. just pick only one answer 1) All polysaccharides have one, and only one, reducing end. 2) Some polysaccharides may have no reducing end. 3) Some polysaccharides may have no non-reducing ends. 4) All polysaccharides have a N-glycosidic bond at their reducing ends. 5) Some polysaccharides may have a functional group other than a carbonyl group at their reducing ends.
The sugar molecules shown below are a) Enantiomers b) Epimers c) Anomers d) Diastereomers
Classify the monosaccharides. H H-C- -OH H-C- -OH H- CH₂OH D-erythrose H- CH₂OH FO -OH H. CH₂OH D-erythrulose H- -OH -OH H-C OH CH₂OH H- D-ribose CH₂OH OH H-C OH CH₂OH D-ribulose H H-C- CH₂OH D-glyceraldehyde -OH HO- -C- H- CH₂OH FO -H -C- -OH H-C OH CH₂OH D-fructose CH₂OH C=O CH₂OH Dihydroxyacetone triose Answer Bank ketose hexose aldose tetrose pentose
Consider the trisaccharide below and answer the following questions: но. OH OH H. OH HO Но ОН ÓH H ÓH ОН C Name of monosaccharide A (Follow this format in typing your answer: ribose): Name the glycosidic linkage between monosaccharides A and B. (strictly follow this format, do not put spaces in between: beta-1,2): Is the trisaccharide a reducing sugar? Type Yes or No: I- I-
Name the types of glycosidic bonds found in this olygomer, from left to right (from 1 to 3) CH 20H CH 2OH CH 2OH но OH OH OH OH OH OH O alpha1-1, alpha1-4, n O alpha1-4, alpha1-4, n O alpha1-4, beta1-2 O alpha1-4, beta1-4 O beta1-4, alpha 1-4 « Previous Next Not saved Submit Quiz
For the following lipid, answer the questions listed below. 0- HC-o (CH)),CH=CH(CH,);CH; H,C-o-c-(CH2);CH=CHCH,CH=CHỊCH,),CH, Is this lipid considered a fat or oil? a. b. How many H2 molecules would be needed to go through a complete hydrogenation reaction? Would the hydrogenated product (after hydrogenation has occurred) have a higher or lower mp than the original starting material? С.
1- Please refer to the Fisher representation of the 5 monosaccharides below: ÇH2OH CHO CHO CHO CHO H- H- он H- OH OH HO HO но- H- OH HO Но- H- -OH H- H- -OH HO- OH HO H- -OH H- -OH HO HO H- OH ČH2OH ČH2CH ČH2CH ČH2CH ČH2CH D-Fructose D-Glucose L-Glucose L-Idose D-Galactose 1a- Which one(s) of the above molecules belong(s) to the subcategory of aldohexoses? and which one(s) of these molecules belong(s) to the subcategory of ketohexoses? Briefly explain your response. 1b- Which one of the above molecules is an enantiomer of D-Glucose? and which one is an epimer of L-Ildose? Briefly explain your response.
Consider the positively charged amino acid lysine Lys2+ 21 COOH I H&N-C-H I pH 14 12 10 8 6 4 2 0 CH₂ I CH₂ I CH₂ I CH₂ T NH₂+ 0 Nelson p85 2.18 = 2.18 PK₁ Lys+ COO™ I H₂N-C-H H₂N-C-H ī I -----) 8.95 Lysº 8.95 pK₂ pka carboxyl = 2.19 pkaamino = 9.67 pka sidechain = 4.25 COO™ I CH₂ I CH₂ I CH₂ I CH₂ I NH₂¹ 1.0 2.0 Equivalents of OH added- COO™ I H₂N-C-H I 10.79 1 10.79 pk Isoelectric point Lys CH₂2 I CH₂ I CH₂ I CH₂ T NH₂ 3.0 +H3N NH3+ T CH₂ T CH₂ CH₂ CH₂ -COO™ H Lysine (Lys, K) Physiological pH = 7.4 < pl → Amino acid is positively charged at physiological pH 1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution) 1) Draw all the forms of glutamate at various pH 2) Calculate the pl of this amino acid 3) Sketch a titration curve showing pH as a function of added [OH-] and locate the predominant forms of histidine in the curve STEPS: 1. Find the H atoms that can be removed on the molecule 2. Associated a pka value to each removable H. 3. Draw the Aa structure at:…
Classify the glycosidic bonds in each of the following disaccharides.
3) After watching the video convert the monosaccharides from a Fischer to a Haworth Projection. a) Convert D-mannose to its a-pyranose. CHO HOH HO H H OH -OH CH₂OH PH 3 D-mannose b) Convert L-galactose to its B-pyranose. H 1 CHO HOH OH H-OH HO-H 6CH₂OH L-galactose

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Identify the type of glycosidic linkage present in the disaccharide below: O a (1-6) O a (1-4) 03(1 - 6) 03(1-4) Н HO ОН OH ОН Н Н н н ОН ОН Н ОН Н ОН