BCH210 FINAL EXAM QUESTIONS
AND ANSWERS WITH CORRECT
SOLUTIONS 100% VERIFIED
Carbohydrate Cyclization - CORRECT ANSWES -- Monosaccharides can cyclize to
form rings. Due to ring strain, pyranose (6-C) and furanose (5-C)
result of hydroxyl group nucleophilic attack on aldehyde/ketone group within the same
sugar chain/group.
Glycosylation of membrane proteins - CORRECT ANSWES -- leads to increased
diversity and complexity
protein glycosylation - CORRECT ANSWES -- assists in folding and cellular recognition,
different sugar chains --> different interactions w antibodies allowing for cell-cell
recognition
Lectins are - CORRECT ANSWES -- proteins that bind to sugars
simplest monosaccharides - CORRECT ANSWES -- glyceraldehyde and
dihydroxyacetone
(C*H2O)n, has chiral central carbon
Aldoses - CORRECT ANSWES -- monosaccharides with an aldehyde group
Ketoses - CORRECT ANSWES -- monosaccharides with a ketone group
aldose numbering - CORRECT ANSWES -- relative to the aldehyde at C1
Ketose numbering - CORRECT ANSWES -- relative to the ketone at C2
L vs D - CORRECT ANSWES -- any protein enzyme that binds to a sugar looks for D
for binding capability, if L protein can't bind
l and d - CORRECT ANSWES -- capital for the last carbon (furthest from the carbonyl)
therefore capital D are the most important sugars
D vs L and placement of hydroxyls - CORRECT ANSWES -- D = right
L = left
,number of stereoisomers = 2^x
where x is the amount of chiral carbons
Carbohydrate Cyclization - CORRECT ANSWES -- Monosaccharides can cyclize to
form rings. Due to ring strain, pyranose (6-C) and furanose (5-C)
result of hydroxyl group nucleophilic attack on aldehyde/ketone group within the same
sugar chain/group.
hydroxyl = nucleophile, carbon = electrophile
hemiacetal - CORRECT ANSWES -- aldehyde derivative, only one alcohol addition
hemiketal - CORRECT ANSWES -- ketone derivative
OH positioning in cyclic saccharide - CORRECT ANSWES -- OH can point above or
below the ring
below = alpha --> more repulsions between OH groups
above = beta --> more stable
anomeric carbon - CORRECT ANSWES -- the new chiral center formed in ring closure;
it was the carbon containing the carbonyl in the straight-chain form
cyclic isomers - CORRECT ANSWES -- anomers that only differ in the position of the
new OH
(above or below the ring)
stereoisomers - CORRECT ANSWES -- same formula and order:
enantiomers = non-superimposable mirror images
diastereoisomers = not mirror images
Diastereoisomers - CORRECT ANSWES -- epimers - different at one chiral carbon
anomers - differ at the newly formed asymmetric C in the ring (OH positioning)
mutarotation - CORRECT ANSWES -- optical rotation of light resulting from a change at
the anomeric carbon (flipping from alpha to beta, etc.)
reducing sugars - CORRECT ANSWES -- have free aldehydes or ketones that react
with oxidizing agents
need to have the OH group next to the O in the ring
carbohydrate modification - CORRECT ANSWES -- phosphorylation, methylation, N-
containing functional group additions, hydroxyls/carbonyl can be removed, etc.
, For example: deoxy has no Oxygen on OH #2
glycosidic bonds - CORRECT ANSWES -- linkages found in polysaccharides
monosaccharides, disaccharides, oligosaccharides, polysaccharides - CORRECT
ANSWES -- 1, 2, 3-20, up to 1000s
mono are the basic energy units
Dissacharide Formation - CORRECT ANSWES -- newly anomeric carbon's OH
undergoes another attack from another OH on another monosaccharide, forming a
glycosidic linkage between the two sugars
could be any hydroxyl
carbon numbering in linkages - CORRECT ANSWES -- from the Oxygen in a clockwise
direction
condensation reaction in glycosidic linkages - CORRECT ANSWES -- H2O is lost (H
from C4 and OH from C1) the lone pair at 4 attacks the anomeric Carbon 1.
Both carbons are originally reducing sugars, but after the linkage, only the one on the
right is.
An alpha bond points - CORRECT ANSWES -- below the plain of the left sugar.
variations in glycosidic linkages - CORRECT ANSWES -- Methanol - O-glycosidic bond
Glycosylation - CORRECT ANSWES -- membrane proteins have sugar chains linked to
Asn (N-linked) or Ser/Thr (O-linked). Takes place in the ER and Golgi.
N-linked occurs - CORRECT ANSWES -- when nitrogen functional groups attack the
anomeric Carbon
amino acids link to sugars - CORRECT ANSWES -- through Asn and Ser/Thr residues
Naming Disaccharides - CORRECT ANSWES -- Put the non-reducing sugar on the left
- Identify the atom in the linkage (O or N)
- configuration of anomeric carbon (α/β)
- Name chiral form (D/L) and structure (pyranosyl/furanosyl)
- Location of glycosidic bond in brackets
- 2nd sugar anomer +name -ose or -oside
(sometimes drop the O and D because they are standard forms)
AND ANSWERS WITH CORRECT
SOLUTIONS 100% VERIFIED
Carbohydrate Cyclization - CORRECT ANSWES -- Monosaccharides can cyclize to
form rings. Due to ring strain, pyranose (6-C) and furanose (5-C)
result of hydroxyl group nucleophilic attack on aldehyde/ketone group within the same
sugar chain/group.
Glycosylation of membrane proteins - CORRECT ANSWES -- leads to increased
diversity and complexity
protein glycosylation - CORRECT ANSWES -- assists in folding and cellular recognition,
different sugar chains --> different interactions w antibodies allowing for cell-cell
recognition
Lectins are - CORRECT ANSWES -- proteins that bind to sugars
simplest monosaccharides - CORRECT ANSWES -- glyceraldehyde and
dihydroxyacetone
(C*H2O)n, has chiral central carbon
Aldoses - CORRECT ANSWES -- monosaccharides with an aldehyde group
Ketoses - CORRECT ANSWES -- monosaccharides with a ketone group
aldose numbering - CORRECT ANSWES -- relative to the aldehyde at C1
Ketose numbering - CORRECT ANSWES -- relative to the ketone at C2
L vs D - CORRECT ANSWES -- any protein enzyme that binds to a sugar looks for D
for binding capability, if L protein can't bind
l and d - CORRECT ANSWES -- capital for the last carbon (furthest from the carbonyl)
therefore capital D are the most important sugars
D vs L and placement of hydroxyls - CORRECT ANSWES -- D = right
L = left
,number of stereoisomers = 2^x
where x is the amount of chiral carbons
Carbohydrate Cyclization - CORRECT ANSWES -- Monosaccharides can cyclize to
form rings. Due to ring strain, pyranose (6-C) and furanose (5-C)
result of hydroxyl group nucleophilic attack on aldehyde/ketone group within the same
sugar chain/group.
hydroxyl = nucleophile, carbon = electrophile
hemiacetal - CORRECT ANSWES -- aldehyde derivative, only one alcohol addition
hemiketal - CORRECT ANSWES -- ketone derivative
OH positioning in cyclic saccharide - CORRECT ANSWES -- OH can point above or
below the ring
below = alpha --> more repulsions between OH groups
above = beta --> more stable
anomeric carbon - CORRECT ANSWES -- the new chiral center formed in ring closure;
it was the carbon containing the carbonyl in the straight-chain form
cyclic isomers - CORRECT ANSWES -- anomers that only differ in the position of the
new OH
(above or below the ring)
stereoisomers - CORRECT ANSWES -- same formula and order:
enantiomers = non-superimposable mirror images
diastereoisomers = not mirror images
Diastereoisomers - CORRECT ANSWES -- epimers - different at one chiral carbon
anomers - differ at the newly formed asymmetric C in the ring (OH positioning)
mutarotation - CORRECT ANSWES -- optical rotation of light resulting from a change at
the anomeric carbon (flipping from alpha to beta, etc.)
reducing sugars - CORRECT ANSWES -- have free aldehydes or ketones that react
with oxidizing agents
need to have the OH group next to the O in the ring
carbohydrate modification - CORRECT ANSWES -- phosphorylation, methylation, N-
containing functional group additions, hydroxyls/carbonyl can be removed, etc.
, For example: deoxy has no Oxygen on OH #2
glycosidic bonds - CORRECT ANSWES -- linkages found in polysaccharides
monosaccharides, disaccharides, oligosaccharides, polysaccharides - CORRECT
ANSWES -- 1, 2, 3-20, up to 1000s
mono are the basic energy units
Dissacharide Formation - CORRECT ANSWES -- newly anomeric carbon's OH
undergoes another attack from another OH on another monosaccharide, forming a
glycosidic linkage between the two sugars
could be any hydroxyl
carbon numbering in linkages - CORRECT ANSWES -- from the Oxygen in a clockwise
direction
condensation reaction in glycosidic linkages - CORRECT ANSWES -- H2O is lost (H
from C4 and OH from C1) the lone pair at 4 attacks the anomeric Carbon 1.
Both carbons are originally reducing sugars, but after the linkage, only the one on the
right is.
An alpha bond points - CORRECT ANSWES -- below the plain of the left sugar.
variations in glycosidic linkages - CORRECT ANSWES -- Methanol - O-glycosidic bond
Glycosylation - CORRECT ANSWES -- membrane proteins have sugar chains linked to
Asn (N-linked) or Ser/Thr (O-linked). Takes place in the ER and Golgi.
N-linked occurs - CORRECT ANSWES -- when nitrogen functional groups attack the
anomeric Carbon
amino acids link to sugars - CORRECT ANSWES -- through Asn and Ser/Thr residues
Naming Disaccharides - CORRECT ANSWES -- Put the non-reducing sugar on the left
- Identify the atom in the linkage (O or N)
- configuration of anomeric carbon (α/β)
- Name chiral form (D/L) and structure (pyranosyl/furanosyl)
- Location of glycosidic bond in brackets
- 2nd sugar anomer +name -ose or -oside
(sometimes drop the O and D because they are standard forms)
