Monosaccharides - ANSWER-Soluble in water due to OH groups (always have either an aldehyde or
ketone)
Formula: (CH2O)n, where n is at least 3
Carbohydrate numbering - ANSWER-Always starts at the carbonyl; aldehyde is C1, ketone is C2
L vs. D isomers - ANSWER-Named based on left (l) vs. right (d...
BCH210 Runback Exam Questions|
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Monosaccharides - ANSWER-Soluble in water due to OH groups (always have either an aldehyde or
ketone)
Formula: (CH2O)n, where n is at least 3
Carbohydrate numbering - ANSWER-Always starts at the carbonyl; aldehyde is C1, ketone is C2
L vs. D isomers - ANSWER-Named based on left (l) vs. right (d) orientation of OH group
Capital lettering given to highest-number chiral carbon; D is physiologically relevant
Number of stereoisomers - ANSWER-2^x, where x is the number of chiral carbons
x = total number of carbons - 2 (aldehyde) or -3 (ketone)
Carbohydrate cyclization - ANSWER-Hydroxyl group initiates a nucleophilic attack on same-chain
carbonyl carbon, donating an electron pair
Must occur in solution as water donates/receives protons
Easily reversible
Furan - ANSWER-5-member ring (4C 1O)
Pyran - ANSWER-6-member ring (5C 1O); formed when attacking hydroxyl group is further down in the
sugar structure
Hemiacetal - ANSWER-Aldehyde derivative
Hemiketal - ANSWER-Ketone derivative
,Glucopyranose formation - ANSWER-Glucose's C5 hydroxyl attacks the aldehyde, forming a 6-member
ring with C1-C5 and the attacking hydroxyl's O; if a C2 ketone was attacked, C2-C5 would form a 5-
member ring
Alpha vs. beta glucopyranose - ANSWER-Alpha: OH points below ring
Beta: OH points above ring (more stable due to less repulsion between OH groups)
Refers to the OH (C1) formed when carbonyl O picks up an H+ from water, not the in-ring O from the
attacking hydroxyl
Honey - ANSWER-Fructose is the main sugar found in honey
Depending on which OH attacks the carbonyl, glucose and fructose can form furan (5) or pyran (6)
When honey is heated, reversible cyclization reactions can occur, so both of fructose's rings occur -->
affects sweetness
Stereoisomer - ANSWER-Same bonding patterns, but still not identical
a) Enantiomers: non-superimposable mirror images
b) Diastereomers: not mirror images
i) Epimer: differ at one chiral C (type of diastereomer)
ii) Anomer: differ at a newly chiral C in a ring (type of epimer)
, Reducing sugars - ANSWER-Sugars can be reducing agents if they have open aldehydes/ketones - cyclic
sugars need a free OH to convert back into a carbonyl to be reducing sugars
Look for free OH bound to a carbon next to O
Disaccharide formation - ANSWER-Any hydroxyl on sugar B attacks anomeric carbon of sugar A; water is
precipitated as glycosidic bond is formed
Only sugar B (attacking) remains a reducing sugar
N-glycosidic linkage - ANSWER-Nitrogen functional group attacks the anomeric carbon; N comes from
Asn side chain
O-glycosidic linkage - ANSWER-OH attacks anomeric carbon; O comes from Ser/Thr side chain
Disaccharide nomenclature - ANSWER-1. Atom in linkage (N or O)
2. Anomeric configuration
3. Sugar name + yl
4. Linked carbon numbers
5. Anomeric configuration
6. Sugar name + ose/oside
E.g. O-alpha-D-glycopyranosyl (1,4) alpha-D-glucopyranose
Glycosyl transferases - ANSWER-Transfer sugars onto acceptor molecules, with specificity for both
parties; sugars are first activated with UTP to form a high-energy UTP-sugar intermediate; acceptor
molecule uses OH or N to attack anomeric carbon on sugar, releasing UTP
Cleavage to monosaccharides - ANSWER-Cells can't use disaccharides like lactose/maltose/sucrose for
fuel or transport; enzymes in microvilli of the small intestine break down disaccharides into
monosaccharides
Enzyme specificity enabled by polar side chains in the active site, which form hydrogen bonds with
disaccharide substrates
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