BCH210 EXAM| 341 QUESTIONS| WITH COMPLETE SOLUTIONS
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BCH210
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University Of Toronto (U Of T
)
agarose and agar correct answer: polysaccharides used to run DNA gels and for microbial plates
bacterial cell walls correct answer: contain polysaccharides in their peptidoglycan layer
lectins correct answer: are proteins that bind to sugars
protein glycolsylation correct answer: assis...
bch210 exam| 341 questions| with complete solution
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University of Toronto (U of T
)
BCH210
BCH210
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BCH210 EXAM| 341 QUESTIONS| WITH
COMPLETE SOLUTIONS
agarose and agar correct answer: polysaccharides used to run
DNA gels and for microbial plates
bacterial cell walls correct answer: contain polysaccharides in
their peptidoglycan layer
lectins correct answer: are proteins that bind to sugars
protein glycolsylation correct answer: assists in folding and
cellular recognition, e.g. ABO blood groups
-ose correct answer: indicates a molecule is some kind of carb
blood group antigens correct answer: glycoproteins containing
different oligosaccharides
glucose and fructose correct answer: C6H12O6, they are
isomers, functional groups organized differently, both
monosaccharides
sucrose correct answer: a disaccharide composed of glucose and
fructose linked by an O-glycosidic bond
simplest monosaccharide formula correct answer: (C • H2O)n
n is any number >= 3
,ketose vs aldose correct answer: ketose sugars have a ketone
group (C=O on Carbon 2)
aldose sugars have an aldehyde group ( C=O on Carbon 1)
D or L correct answer: assigned to the asymmetric chiral carbon
furthest away to the carbonyl (last hydroxyl group on the sugar
that is on a chiral carbon determines the L or D isomeric form)
-D OH on right
-L OH on left
formula for different linear hexoses correct answer: 2^n
n is # of chiral carbons
-a ketoheptose has 7 carbons, 4 of which are chiral (ignore last
Carbon and first 2 carbons) so 2^4=16, 16 linear stereoisomers
D sugars correct answer: biologically important
carbohydrate cyclization correct answer: aldehyde/ketone
carbonyl undergo nucleophilic attack by hydroxyl groups
-forms a hemiacetal bond (aldehyde derivative)
-or a hemiketal bond (ketone deriative)
hemiacetal bond correct answer: aldehyde derivative
hemiketal bond correct answer: ketone derivatice
hemiacetal formation correct answer: 1. lone pair of electrons
on an alcohol's oxygen attacks the aldehyde group
2. aldehyde grabs a proton from water
3. original alcohol that attacked donates a proton back to water
,4. new alcohol group is formed & alcohol's side chain + O is
attached
Glucose Cyclization correct answer: C5 OH reacts with C1
aldehyde group to form pyranose ring: Glucopyranose
can be:
1. α -D-glucopyranose (hydroxyl on C5 attacks from top and
new hydroxyl is below the plane of the ring)
2. β-D-glucopyranose (hydroxyl on C5 attacks from below and
new hydroxyl is above the plane of the ring)
α -D-glucopyranose correct answer: glucose cyclization,
hydroxyl on C5 attacks from top and new hydroxyl is below the
plane of the ring
β-D-glucopyranose correct answer: glucose cyclization,
hydroxyl on C5 attacks from below and new hydroxyl is above
the plane of the ring
MORE STABLE FORM as not directly beside other OH
isomer correct answer: same molecular formula, diff structure
constitutional isomer correct answer: diff order of functional
group bonding
stereoisomers correct answer: same formula and order but
-enantiomer
-diastereoisomer
enantiomer correct answer: Non-superimposable mirror image
-invert all chiral centres
, diastereoisomers correct answer: isomers that are not mirror
images but have same formula and order
-epimer
-anomer
diastereoisomer epimer correct answer: differ at one
asymmetric carbon (D vs. L)
-invert some chiral centres but not all
diastereoisomers anomer correct answer: differ at newly formed
asymmetric C in the ring structure
chair conformation: new OH axial correct answer: α form
chair conformation: new OH equatorial correct answer: β form
mutarotation correct answer: the optical rotatio of light resulting
from a change at the anomeric carbon
reducing sugars correct answer: open chain aldoses if they can
react w/ oxidizing agents (have a free aldehyde)
oxidizing agent = copper, aldehyde is converted to a sugar acid
identifying reducing sugars in cyclic sugars correct answer:
look for a free hydroxyl by the anomeric carbon
carbohydrate modification correct answer: sugars can be
phosphorylated, methylated or N-containing functional groups
maybe be added
hydroxyls may be removed
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