Cyclization of monosaccharides: A Aldopentoses can form both pyran and fu- ran rings. B] Pyran rings are more stable than furan rings and are thus more abundant; C Ketohexoses can only form furan rings;
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![Cyclization of monosaccharides:
A Aldopentoses can form both pyran and fu-
ran rings.
B] Pyran rings are more stable than furan
rings and are thus more abundant;
|C Ketohexoses can only form furan rings;
D Preferentially involves a primary alcohol
function, because it is more reactive than
a secondary alcohol function;
E Ketopentoses can only form pyran rings
and not furan rings;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13571319-5b25-421c-ab0c-069a433cb92a%2F80f76295-409a-4765-b40f-de93629288ef%2Flir9zt_processed.png&w=3840&q=75)
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- Assignment on KetohexosesConstruct the structure of the following enantiomers or mirror images:1. D and L Fructose2. D and L Allulose3. D and L Sorbose4. D and L TagatoseVitamin C is ascorbic acid. If you examine the structure of asorbic acid given in the background secton (equation 2) there is no -COOH group present. In the structure shown, the ascorbic acid is as a lactone or cyclic ester. Write the equation for the hydrolysis (cleaveg by addition of water) of the lactone to the carboxylic acid.Tridecane [CH3(CH2)11CH3] is a major component of the repellent that the stink bug Piezodonis guildinii releases from its scent glands when attacked. What fatty acid gives tridecane on decarboxylation?
- Thetealeavesthatcontainthecaffeinetobeisolatedarepredominantly composed of cellulose, a natural and extremely abundant polymer of glucose. Look up the chair structure of β-D-glucose (β-D-glucopyranose, the monomer from which the cellulose polymer is composed). Draw this structure, draw the flipped chair, and indicate which conformer you think is more stable, if either. BRIEFLY explain your choice5-phosphate) and an aldopentose (ribose-5-phosphate) to an aldotriose (glyceraldehyde-3-phosphate) and a ketoheptose (sedoheptulose-7-phosphate). Notice that the total number of carbons in the reactants andproducts is the same (5 + 5 = 3 + 7). Propose a mechanism for this reaction. xylulose-5-PCH2OH CH2OPO32−C OHOHHOH sedoheptulose-7-Pα -Amylose is an unbranched glucose polymer. Why would this polymer not be as effective a storage form of glucose as glycogen?
- In observing a Haworth or cyclohexane-chair representation of alpha-d-glucopyranose, the anomeric carbon can de best identified by: a) the carbon atom which is bonded to the most H atoms b) the carbon atom which is bonded to the least H atoms c) The carbon atom which has its OH group pointing down d) The carbon atom which is bonded to two oxygen atoms Cellulose differs from amylose in that: a) cellulose has 1-6 branches while amylose does not b) amylose has 1-6 branches while cellulose does not c) amylose has alpha glycosidic bonds while cellulose has beta glycosidic bonds d) cellulose contains sulfate while amylose does not. The glycosaminoglycan (mucopolysaccharide) which is not normally associated with joint tissue is: a) heparin b) keratan sulfate c) chondroitin sulfate d) hyaluronic acid In glycoproteins, which amino acid does not bond sugar molecules? a) Asn b) Ser c) Thr d) Gly Which feature do all lipid molecules share in common? a) fatty acid molecules…True or False: a. D-enantiomer is a mirror image of L-enantiomer b. In a D-carbohydrate, the hydroxyl group on the chiral carbon farther from the carbonyl group points to the right. c. L-carbohydrates are preferred by the humans when processing them after ingestion. d. Sugars that have the cyclic hemiacetal; groups on the anomeric carbon atom are reducing sugars e. In the beta-D –galactose the hydroxyl group on the anomeric carbon atom points up or above the plane of the paper. f. Maltose is a non-reducing sugar and does not have a hemi acetal group. g. Sucrose contain a hemiacetal group and hence a reducing sugar. h. Linen is prepared from cellulose. Separating Glycated Hb From Normal Hb (Integrates with Chapters 5and 6.) Human hemoglobin can react with sugars in the blood(usually glucose) to form covalent adducts. The a-amino groups ofN-terminal valine in the Hb b-subunits react with the C-1 (aldehyde)carbons of monosaccharides to form aldimine adducts, whichrearrange to form very stable ketoamine products. Quantitation ofthis “glycated hemoglobin” is important clinically, especially fordiabetic individuals. Suggest at least three methods by which glycatedHb (also referred to as HbA1c) could be separated from normal Hband quantitated
- Looking at the structure of alpha linolenic acid and knowing the electron carriers produced in each round of traditional beta oxidation for saturated fatty acid chains, how many rounds of beta oxidation will have diminished production of electron carriers due to the presence of double bonds?functional groups in palmitic acid, a saturated fatty acid and oleic acid, an unsaturated fatty acidGlycerophospholipids Phosphatidylethanolamine 3. In case the cell is in a state requiring large amount of ATP to support energy-requiring reactions/pathways, assuming that you have 1 mole of each of the said lipids are catabolized and complete oxidized, will the total net ATP yield from these two lipids be higher or lower than the sum of the net ATPs generated from each fatty acid components? Justify your answer in biochemical terms and using 5 sentences or less.