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BCH210 Final| 170 QUESTIONS| WITH COMPLETE SOLUTIONS

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sucrose correct answer: glucose + fructose with a glycosidic bond simple monosaccharide correct answer: formula: (CH2O)n -simplest carbohydrates -have an aldehyde or ketone group sugars with a aldehyde have a prefix aldo- sugars with a ketone group have a prefix keto- -if a ketose, the ketone group is always at carbon 2- this decreases the amount of isomers that could form -made up of chiral carbons- allowing the sugars to form different isomers, leading to structural diversity (aldose could form more because they have more chiral carbons) -aldehyde and ketone groups are reactive which allow them to cyclize (formation of 2 additional structures- beta or alpha). this also means that they are reducing sugars when they react with oxidizing agents carbohydrate numbering and configuration correct answer: a sugar with an aldehyde- start counting the carbons at the aldehyde group (ie aldehyde is c1) a sugar with a ketone- ketone is always at C2 count based on that -each chiral carbon could form 2 possible isomers, d or l -aldehydes and ketones are either D or L isomers based on the orientation of the OH on the chiral carbon furthest from the aldehyde or ketone group -D isomer= OH on the right -L isomer=OH is on the left -D sugars are biologically important- L isomers are found in bacteria

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BCH210 Final| 170 QUESTIONS| WITH
COMPLETE SOLUTIONS
sucrose correct answer: glucose + fructose with a glycosidic
bond

simple monosaccharide correct answer: formula: (CH2O)n
-simplest carbohydrates
-have an aldehyde or ketone group
sugars with a aldehyde have a prefix aldo-
sugars with a ketone group have a prefix keto-
-if a ketose, the ketone group is always at carbon 2- this
decreases the amount of isomers that could form
-made up of chiral carbons- allowing the sugars to form different
isomers, leading to structural diversity (aldose could form more
because they have more chiral carbons)
-aldehyde and ketone groups are reactive which allow them to
cyclize (formation of 2 additional structures- beta or alpha). this
also means that they are reducing sugars when they react with
oxidizing agents

carbohydrate numbering and configuration correct answer: a
sugar with an aldehyde- start counting the carbons at the
aldehyde group (ie aldehyde is c1)
a sugar with a ketone- ketone is always at C2 count based on
that

-each chiral carbon could form 2 possible isomers, d or l
-aldehydes and ketones are either D or L isomers

,based on the orientation of the OH on the chiral carbon furthest
from the aldehyde or ketone group
-D isomer= OH on the right
-L isomer=OH is on the left

-D sugars are biologically important- L isomers are found in
bacteria
-D or L isomers could be alpha or beta when cyclized
to calculate the number of possible linear steroisomers, 2^x
where x= number of chiral carbons that could isomers

carbohydrate cyclization correct answer: -intramolecular
reaction that takes place within the sugar molecule that coverts
the sugar from linear to cyclic form in solution (in solution this
reaction is reversible)
-possible because the functional group of the sugar (OH,
aldehydes and ketone) is very reactive
-the aldehyde or ketone carbonyl undergoes a nucleophilic
attack by the farthest hydroxyl group from it to form a cyclic
structure
-reactions for a member furan ring or 6 member pyran ring
-the Oxygen in the cyclized form comes from the hydroxyl that
attacked the ketone or aldehyde
-the chirality of cyclic structures can be detected by
mutarotation, a change in the optical rotation of polarized light
following the interconversion between the alpha and beta
anomer. this is how we could measure how much beta and alpha
rings there is in solution based on how the polarized light is
being rotated

The new bond may be:

,-hemiacetal: aldehyde derivative (OH attacking an aldehyde)
-hemiketal: ketone derivative (OH attacking a ketone)

-final product is an alpha when the hydroxyl is below the
Oxygen or a beta when the hydroxyl is above
-the beta structure is more stable because the OH are opposite
from each other (less repulsion). Because of this, more would be
found in solution.

fructose correct answer: can form a furan ring and pyran ring
-main sugar found in honey

carbs modification correct answer: sugars can be
phosphorylated, methylated or N-containing functional groups
may be added
-Hydroxyls (or even carbonyls) may be removed
-this increases the complexity of carbohydrate structure
-this is not a PTM because only proteins get translated. But this
changes the property of the sugar (ie will no longer obey the
(CH2O)n formula- will now be recognized by different proteins

isomers correct answer: same molecular formula, different
structure (ie fructose and glucose)

constitutional isomers correct answer: different order of
functional group binding. same molecular formula (ie fructose
and glucose- different location of carbonyl group)

stereoisomers and types correct answer: same formula and order
they could be enatiomers or diastereoisomers

, Enatiomers- non superimposable mirror images
Diastereoisomers- not mirror images

types of diasterosiomers correct answer: Epimers: different at
one chiral carbon

Anomers: differ at a newly formed, chiral carbon in the ring
structure (ie beta-D-glucose and alpha-D-glucose)

how are reducing sugars formed correct answer: -in solution
-a sugar is a reducing sugar if a hydroxyl group is free and
beside the oxygen in a ring to form back into the linear structure
(yield a positive reducing test)
-the sugar must be in linear form so the aldehyde is free
-the aldehyde group becomes oxidized to reduce copper (the
hydrogen on the aldehyde gains an oxygen)
-the copper 2+ gets reduced to copper + to form Cu2O- this will
be detected in solution which will allow us to determine if a
sugar is a reducing sugar
-aldehyde becomes a carboxylic acid, the free H picked up an
oxygen

-aldehyde= reducing agent, gets oxidized (from a carbonyl
group to carboxylic acid group)
-copper= oxidizing agent, gets reduced

-these sugars could react with anything in the body so it is not
good for too many to be found in the blood

alpha 1,4 glycosidic bond correct answer: ie between 2 alpha
glucose molecules

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