BCH210 FINAL QUESTIONS WITH COMPLETE SOLUTIONS
UDP glucose pyrophosphorylase - Answer--transfers a UMP onto G1P on the non-reducing ends to form UDP-glucose
-requires a UTP
-2 phosphates is the leaving group
phosphoglycomutase - Answer--isomerase
glucose 1,6 biphosphate converted into G1P w...
BCH210 FINAL QUESTIONS WITH
COMPLETE SOLUTIONS
UDP glucose pyrophosphorylase - Answer--transfers a UMP onto G1P on the non-
reducing ends to form UDP-glucose
-requires a UTP
-2 phosphates is the leaving group
phosphoglycomutase - Answer--isomerase
glucose 1,6 biphosphate converted into G1P which could then be added onto the
glycogen during glycogen synthase.
-has a phosphate in its active site, and donates the phosphate to C1 and accepts the
phosphate on C6
mitochondria structure - Answer-outer membrane- has porins that are made up of beta
barrel proteins that allow small molecules to pass through
inner membrane-regulates what gets inside matrix- import into the matrix gets regulated
so stuff is only broken down when you need energy
-has cristae (folds) which increase the surface area of the inner membrane.
-has membrane proteins- selectively permeable
-where the ETC is
Matrix-pyruvate gets imported there
pyruvate dehydrogenase complex - Answer-converts pyruvate to acetyl-CoA
-links the production of pyruvate in the cytoplasm and the CAC in the matrix (soluble
enzyme)
-oxoreductase- catalyzes an oxidative decarboxylation reaction producing CO2 and
NADH
-complex with 3 different subunits-there are multiple copies of each subunit and each
subunit associates non covalently
-requires 5 cofactors for function- 2 are cosubstrates (Coenzyme A and NAD) and TPP,
lipoamide and FAD are prosthetic groups (intermediate roles in the reaction)
-not a metabolon
-in mitochondrial matrix (soluble)
PDC reaction and 3 subunits - Answer--negative delta G because of the
decarboyxlation reaction that takes place in the first subunit. these are highly favourable
reactions because CO2 is unstable and wants to come off pyruvate. Release of CO2 is
driving the reaction in the fwd direction (increase in the entropy of the system)
-irreversible reaction. because of this, PDC must be regulated
-catalyzes an oxidative decarboxylation reaction
,-2 pyruvates gets oxidized to form 2 acetyl CoA
-NAD gets reduced to NADH
E1- pyruvate dehydrogenase
1. decarboxylation reaction- removed carbon dioxide from pyruvate. This remains the
acetyl group which TPP accepts. Without TPP, this reaction wouldn't happen because
there wouldn't be anything to accept the acetyl group
E2-dihydrolipoyl transacetylase
-TTP transfers the acetyl group on CoA. The lippoamide accepts the acetyl group from
TPP forming acetyl lipoamide then puts the acetyl on CoA forming acetyl CoA
E3-dihydrolipoamide dehydrogenase
-resets the enzyme into its starting state through redox reactions (so the thiol groups on
E2 could go back to the oxidized form)
-FAD is being reduced into FADH2 and then oxidized again by using NAD which gets
reduced to NADH
regulation of PDC - Answer-1. covalent modification by PD kinase and PD phosphatase
-PD kinase adds a phosphate onto E1 of PDC to inactivate enzyme.
-PD kinase gets activated by higher concentrations of acetyl CoA, NADH and
inactivated by higher concentration of pyruvate and ADP
-PD phosphatase removed the phosphate from E1 of PDC to activate the enzyme. it
gets activated by calcium
2. allosteric control- buildup of reaction products
-feedback inhibition by increase in acetyl CoA on E2 and NADH on E3
-acetyl CoA could come from the breakdown of fatty acid chains (getting energy from
another form or lots of glucose)
-increase in ATP inhibits PDC
-when theres lots of acetyl coA that means theres increase in pyruvate= increase in
alanine, could inhibit glycolysis- glucose will be stored at glycogen
What happens when the muscles are at rest (PDC) - Answer--there is lots of ATP
present but not being use
-that means lots of NADH, Acetyl CoA is present. This activates the PD kinase which
phosphorylates the PDC to turn off pathway so no more ATP could be made
what happens when to the muscles while running - Answer--using up ATP so ADP is
being made
-high pyruvate because the ATP is being used up
-PD kinase is inhibited so it won't phosphorylate PDC
sucrose - Answer-glucose + fructose with a glycosidic bond
simple monosaccharide - 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 - 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 - 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 - Answer-can form a furan ring and pyran ring
-main sugar found in honey
carbs modification - 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 - Answer-same molecular formula, different structure (ie fructose and glucose)
constitutional isomers - Answer-different order of functional group binding. same
molecular formula (ie fructose and glucose- different location of carbonyl group)
stereoisomers and types - Answer-same formula and order
they could be enatiomers or diastereoisomers
Enatiomers- non superimposable mirror images
Diastereoisomers- not mirror images
types of diasterosiomers - 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 - 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
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