Biochem Exam III

Question 1

What are the steps of the conversion of pyruvate to Acetyl-CoA?

Answer 1

1) Pyruvate OH- transporter transports pyruvate into mitochondrial matrix where pyruvate enters and OH- exits

2) Pyruvate is decarboxylated by the pyruvate dehydrogenase complex to form acetyl CoA. Pyruvate dehydrogenase complex is made up of pyruvate dehydrogenase (E1), Dihydroliopyl transacetylase (E2), and Dihydrolipoyl dehydrogenase (E3)

Question 2

Pyruvate dehydrogenase

Answer 2

E1 component of pyruvate dehydrogenase complex, catalyzes 2 reactions

1) pyruvate + TPP –> hydroxyethyl-TPP + CO2 (decarboxylation)

2) hydroxyethyl-TPP + lipoamide –> acetyllipoamide + TPP. hydroxyethyl-TPP is oxidized into TPP and the disulfide bridge of lipoamide (attached to E2) is reduced forming acetyllipoamide

Question 3

Dihydrolipoyl Transacetylase

Answer 3

E2 component of pyruvate dehydrogenase complex. Catalyzes the transfer of an acetyl group from acetyllipoamide to coenzyme A. Acetyl attaches to SH group in CoA molecule forming acetyl-CoA and acetyllipoamide is converted into dihydrolipoamide.

Question 4

Dihydrolipol dehydrogenase

Answer 4

E3 component of pyruvate dehydrogenase complex. Catalyzes the “reset” reaction where dihydrolipoamide is oxidized back into lipoamide after the disulfide bridge of lipoamide is reduced from the E1 oxidation reaction. FAD+ is reduced into FADH2 to reset this and then oxidized by NAD+ and this yields 1 NADH + H+ molecule.

Question 5

Pyruvate Dehydrogenase Complex Structure

Answer 5

Consists of a core of 8 E2 trimers. Each trimer has three alpha E2 subunits each (a3) so 24 E2 subunits total. Each trimer has three functional domains, a lipoamide binding domain, E3 interaction domain, and transacetylase domain.

Each E2 trimer is surrounded by three E1 subunits along the “corners” of the E2 core. E1 consists of 2 a-subunits and 2 B-subunits (a2B2), and a TPP prosthetic group so there are 96 subunits total.

The core of eight E2 trimers is also surrounded by 12 E3, 2 E3 on each “face” of the E2 core. E3 consists of an a-subunit and B-subunit (aB) and an FAD prosthetic group so there are 24 E3 subunits total. In total, there are 144 subunits.

Question 6

How are Beriberi and Arsenite poisoning linked with the pyruvate dehydrogenase structure?

Answer 6

Beriberi is a neurological condition that results when there is a deficiency of thiamine or Vitamin B1, which is a precursor to TPP. Since TPP serves as a cofactor for the pyruvate dehydrogenase complex, when there is a thymine deficiency, the pyruvate dehydrogenase complex cannot form acetyl-coA as readily.

Arsenite also inhibits the pyruvate dehydrogenase complex by inactivating the dihydrolipoamide component of E2 meaning the oxidation of hydroxyethyl-TPP to acetyllipoamide using a disulfide bridge cannot occur and acetyl-coA cannot be formed as readily.

Question 7

What are the stages of the citric acid cycle? What are the final products?

Answer 7

First the oxidation of two carbon atoms occurs as acetyl-CoA enters the citric acid cycle. Acetyl-CoA is initially coupled with oxaloacetate, a four carbon molecule, to produce a 6 carbon molecule citrate. Then citrate is oxidized and decarboxylated in a series of reactions to produce succinyl CoA (A 4 carbon molecule, 2 CO2, and 2 NADH).

Succinyl CoA then undergoes a series of reactions to regenerate oxaloacetate so that the cycle can continue. For each acetyl-CoA entering the cycle, 3 NADH + H+ is produced, 2 CO2 is produced, FADH2 is produced, and 1 GTP is produced

Question 8

What is the first step of the oxidation phase of the citric acid cycle? What enzymes and substrates are used, what intermediates are formed, and what products are formed?

Answer 8

The formation of citrate first occurs, a 6 Carbon molecule.

Enzyme: Citrate synthase
Substrates: Oxaloacetate, Acetyl-CoA, H2O
Intermediate: Citryl-CoA
Products: Citrate, CoA

Question 9

What is the second step of the oxidation phase of the citric acid cycle? What enzymes and substrates are used, what intermediates are formed, and what product is formed?

Answer 9

The conversion of citrate to isocitrate occurs

Enzyme: Aconitase
Substrate: Citrate and H2O
Intermediate: cis-Aconitase
Product: Isocitrate

Citrate contains a prochiral carbon that doesn’t seem chiral initially but is chiral due to the spatial order of the groups. Aconitase dehydrates and rehydrates the pro-R-arm, that came from oxaloacetate, rather than the pro-S-arm that contains the acetyl group from acetyl-CoA.

Question 10

What is the third step of the oxidation phase of the citric acid cycle? What enzymes and substrates are used? What intermediates and products are formed?

Answer 10

The conversion of isocitrate (C6) to a-ketoglutarate (C5) occurs

Enzyme: Isocitrate dehydrogenase
Substrates: Isocitrate (C6) and NAD+
Intermediates: Oxalosuccinate (C6)
Products: a-ketoglutarate (C5), NADH, CO2

Question 11

What is the fourth step of the oxidation phase of the citric acid cycle? What enzymes and substrates are used? What intermediates and products are formed?

Answer 11

The conversion of a-ketoglutarate (C5) to Succinyl CoA (C4) occurs

Enzyme: a-ketoglutarate dehydrogenase complex
Substrate: a-ketoglutarate (C5), NAD+, CoA, TPP, Lipoamide
Product: Succinyl-CoA (C4), NADH, CO2

Question 12

What is the first step of the conversion of Succinyl-CoA to Oxaloacetate? What enzymes and substrates are used and what intermediates and/or products are formed?

Answer 12

Succinyl-CoA (C4) is converted to Succinate (C4)

Enzyme: Succinyl-CoA Synthetase
Substrates: Succinyl-CoA (C4), Pi, GDP
Intermediate: Succinyl Phosphate
Product: Succinate (C4), GTP, CoA

Question 13

What is the second step of the conversion of Succinyl-CoA to Oxaloacetate? What enzymes and substrates are used and what intermediates and/or products are formed?

Answer 13

Succinate (C4) is converted into Fumarate (C4)

Enzyme: Succinate Dehydrogenase
Substrates: Succinate, FAD
Product: Fumarate (C4), FADH2

Question 14

What is the third step of the conversion of Succinyl-CoA to Oxaloacetate? What enzymes and substrates are used and what intermediates and/or products are formed?

Answer 14

Fumarate (C4) is converted into Malate (C4)

Enzyme: Fumarase
Substrate: Fumarate (C4), H2O
Product: Malate (C4)

Question 15

What is the fourth and final step in the conversion of Succinyl-CoA to Oxaloacetate? What enzymes and substrates are used and what intermediates and/or products are formed?

Answer 15

Malate (C4) is converted into Oxaloacetate (C4)

Enzyme: Malate Dehydrogenase
Substrates: Malate (C4), NAD+
Product: Oxaloacetate (C4), NADH

Question 16

What is produced for one molecule of glucose via glycolysis, one molecule of pyruvate via the pyruvate dehydrogenase complex, and one molecule of acetyl-CoA via TCA? What is produced for 1 molecule of glucose overall?

Answer 16

For one molecule of glucose via glycolysis, 2 ATP, 2 NADH + H+, and 2 Pyruvate are produced

For one molecule of pyruvate via the pyruvate dehydrogenase complex, 1 NADH + H+ is produced and 1 Acetyl-CoA is produced

For one molecule of acetyl-CoA via TCA, 2 NADH + H+ is produced, 1 FADH2 is produced, and 1 GTP is produced

In total for 1 glucose molecule, 2 ATP, 2 GTP, 10 NADH + H+, and 2 FADH2 are produced. Glycolysis occurs in cytosol and pyruvate dehydrogenase complex and TCA acts in the mitochondria

Question 17

Where is the glycerol-3-phosphate shuttle prominent and how does it allow electrons from NADH + H+ produced in the cytosol to be used in the electron transport chain?

Answer 17

It is prominent in skeletal muscle and there are three steps

1) NADH + H+ is oxidized into NAD+ while DHAP is reduced to Glycerol-3-phosphate by glycerol-3-phosphate dehydrogenase in the cytoplasm

2) Glycerol-3-phosphate is oxidized to DHAP while FAD is reduced to FADH2 by membrane bound Glycerol-3-phosphate dehydrogenase in the mitochondria

3) FADH2 transfers its electrons to coenzyme Q reducing it to QH2 which can be used in the electron transport chain

Question 18

Where is the Malate-Aspartate shuttle prominent and how does it allow electrons from NADH + H+ produced in the cytosol to be used in the electron transport chain?

Answer 18

It is prominent in the heart and liver tissue. This mechanism has 3 steps

1) Cytoplasmic malate dehydrogenase oxidizes NADH + H+ into NAD+ while oxaloacetate is reduced to malate

2) Malate-a-ketoglutarate antiporter allows malate to traverse the inner mitochondrial membrane where malate goes in and a-ketoglutarate goes out

3) Mitochondrial Malate Dehydrogenase oxidizes malate back into oxaloacetate as NAD+ is reduced to NADH + H+

Question 19

What is required for the Malate-Aspartate shuttle to continue working and how is this accomplished?

Answer 19

For the cycle to continue a-ketoglutarate must be present in the matrix and oxaloacetate must be present in the cytoplasm. An amine group is attached to alpha carbon.

1) Mitochondrial aspartate aminotransferase transfers an amine group and produces aspartate (C4) from oxaloacetate (C4) and a-Ketoglutarate (C5) from glutamate (C5)

2) Glutamate-Aspartate Antiporter shuttles aspartate into the cytoplasm where aspartate goes out and glutamate goes in

3) Cytoplasmic aspartate aminotransferase transfers an amine group and produces oxaloacetate (C4) from aspartate (C4) and glutamate (C5) from a-Ketoglutarate (C5)

Question 20

Complex I

Answer 20

NADH-Q Oxidoreductase. There are 4 steps to the mechanism

1) NADH reduces FMN to produce FMNH2
2) FMNH2 donates e- one at a time to iron-sulfur cluster proteins
3) Iron-sulfur proteins donate e- to Q one at a time
4) QH2 leaves the complex and travels to complex III
5) As electrons flow 4 protons are pumped

NADH + H+ + Q + 4 H+ –> NAD+ + QH2 + 4H+

Question 21

Complex II

Answer 21

Succinate-Q reductase. There are 4 steps to the mechanism

1) Succinate reduces FAD to produce FADH2
2) FADH2 donates e- one at a time to iron-sulfur cluster proteins
3) Iron-sulfur proteins donate e- one at a time to Q
4) QH2 leaves the complex and travels to Complex III

FADH2 + Q –> FAD + QH2

Question 22

Complex III

Answer 22

Q-Cytochrome c Oxidoreductase. There are 5 steps to this mechanism

1) QH2 donates 2 e- into the complex

2) One e- is sent to an iron-sulfur cluster which is then transported to Cyt c1 and then Cyt c which results in the transport of 2 protons out the mitochondria from QH2 to form Q which enters the Q pool

3) One e- is sent to Cyt bH and transported to Cyt bL then attaches back to a Q to generate a semi-quinone radical which is stabilized

4) Then a second QH2 enters the complex and is then oxidized delivering 2 e- One e- is sent to an iron-sulfur cluster which goes to Cyt c, one e- is sent to cyt bH, 2 H+ are pumped out, and the remaining Q becomes part of the Q pool again

5) But this time, the e- is sent to Cyt bH which is transported to the Q radical to form Q, and Q is reduced and regenerated to QH2 by accepting 2 protons from the matrix (not intermembrane space)

For 1 NADH two net protons are pumped

2QH2 + Q + 2 Cytc (Fe+3) + 2 H+in —> 2Q + QH2 + 2 Cytc (Fe+2) + 4H+out

Question 23

Complex IV

Answer 23

Cytochrome c Oxidase, there are 7 steps to this mechanism

1) Two Cyt c donate an electron to 1 Cu-containing peptide (CuA) each. As 4 CytC become oxidized 4 protons are pumped out

2) The CuA reduces two Cyt a as 2 e- move through it

3) 2 Cyt a then reduces 2 Cyt a3. 1 Cyt a3 retains it’s e-, 1 Cyt a3 donates an e- to a CuB peptide, and this results in 1 reduced Cyt a3 and 1 reduced CuB peptide

4) When Cyt a3 and CuB are both reduced, they can now bind and reduce O2 forming a peroxide bridge. Two protons are then pumped out

5) Two more Cyt c donate one e- each to 1 CuA which then goes to the peroxide bridged Cyt a3-CuB pumping out 2 protons. The peroxide bridge is cleaved as 2 electrons are added and ion-oxygen adjuncts on Cyt a3(-OH) and CuB(-OH) are formed from the uptake a 2 protons from the matrix.

6) The uptake of 2 more protons from the matrix converts the ion-oxygen adjuncts to 2 H2O

7) This means for every 4 Cyt c or 2 NADH, 4 protons from the matrix produce 2 H2O and 4 protons are pumped out

For 1 NADH

2 Cytc (Fe+2) + 4H+(in) +(1/2)O2 –> 2 Cytc (Fe+3) + H2O + 2H+(out)

Question 24

Structure of ATP Synthase

Answer 24

F0 subunit is hydrophobic and found in the inner mitochondrial membrane.
- c-subunits make up the proton channel component, and this consists of two alpha-helices along with an aspartic acid residue.
- a subunit is located outside of the c-subunit ring and has two hydrophilic half-channels where one is open to the cytosol and the other is open to the matrix
- two b in a dimer peptides connect the F0 and F1 subunits and prevent the F1 subunit from rotating with the delta subunit of the F1

F1 subunit is hydrophilic in mitochondrial matrix
- made up of a a3B3 hexameric ring that consists of alternating a and B subunits and a y subunit core
- there is a central stalk with y and e subunits that connects the F1 subunit to the c subunit ring of the F0 subunit, this controls the spin of the c subunit ring
- each B subunit interacts with a different part of the y subunit core

Question 25

How is ATP produced from ATP synthase? What is the overall reaction and the steps?

Answer 25

Overall reaction: ADP(3-) + HPO4(2-) —> ATP(4-) + H2O

Steps

1) A single proton from the inner mitochondrial membrane space (cytosol) enters the ATP synthase complex via an F0 a-subunit, hydrophilic, cytoplasmic half-channel

2) The proton in the F0 a-subunit half channel protonates Asp 61 on one of the c-subunits on the c-subunit ring and neutralizes the negative charge

3) When the Asp 61 charge is neutralized, the uncharged Asp 61 can be exposed to a hydrophobic interior of the inner mitochondrial membrane meaning the c ring can rotate clockwise. This then exposes the next charged Asp 61 to a proton on the a-subunit from the cytosol

4) After all of the c subunits become protonated, when one c subunit rotates back to the F0 subunit, the Asp6 becomes deprotonated and a proton is released into the mitochondrial matrix via the other F0 a-subunit half channel. It is then protonated via the a-subunit channel containing a proton from the cytosol and the C ring can spin

5) The clockwise rotation of the c-subunit causes the y-subunit core to rotate within the a3B3 hexamer

6) As the y-subunit rotates the three B subunits can adopt one of three conformations.
- L conformation where the B-subunit can loosely bind ADP and Pi
- T-conformation where first the B-subunit can tightly bind ADP and Pi and then convert the bound ADP and Pi to ATP
- O conformation where the B-subunit can first release ATP and then allow ADP and Pi to enter the active site

Question 26

How can the amount of ATP produced per molecule of NADH/FADH2 be estimated?

Answer 26

For a ring with 10 subunits, 10 protons are required to complete a rotation and one complete rotation yields 3 ATP molecules

We can estimate 12 protons are required to complete a cycle so 4 protons are used to produce 1 ATP. NADH translocates 10 protons so it leads to the production of 2.5 ATP molecules. FADH2 translocates 6 protons so it leads to the production of 1.5 ATP molecules.

Question 27

What are the components of the B-adrenergic receptor pathway and what is the mechanism?

Answer 27

Components: 7TM Receptor, G-protein complex, Adenylate Cyclase, cAMP, Protein Kinase A. Found in skeletal muscle cells

1) The 7TM receptor binds to a G-protein complex that is “off” when epinephrine or another ligand is not bound

2) When epinephrine binds, the alpha subunit of the G protein releases GDP and GTP binds to it, meaning the GTP-bound a-subunit is on. It then dissociates from the 7TM receptor and the B/y subunits of the G protein also dissociate.

3) When the GTP-bound a-subunit is activated, it binds to adenylate cyclase causes adenylate cyclase to be activated. Adenyl-cyclase can then convert ATP to cyclic AMP or cAMP

4) Elevated levels of cAMP activate Protein Kinase A (PKA) which consist of 4 subunits, 2 regulatory subunits and 2 catalytic subunits

5) When PKA is bound to cAMP and activated, the R-subunits release the C-subunits and the C subunits phosphorylate their target proteins on Ser and Thr

6) Phosphodiesterase then hydrolyzes cAMP to AMP which then allows the R and C subunits to reassociate inactivating PKA

7) The a-subunit of the G protein then hydrolyzes GTP to produce GDP and Pi and the GDP-bound a-subunit can then reassociate with the B and y subunits which then reassociates with the B-adrenergic receptor

Epinephrine binds in the skeletal muscle

Question 28

What are the components of the a-adrenergic receptor pathway and what is the mechanism?

Answer 28

Components: 7TM Receptor, G-protein complex, Phosphilipase C, diacylglycerol (DAG), inositol-1,4,5-triphosphate (IP3), Protein Kinase C (PKC). Found in liver or smooth muscle cells

1) When epinephrine binds to the 7TM receptor, the alpha subunit of the G protein releases GDP and binds to GTP. The GTP-bound a-subunit then dissociates from the receptor and activates phospholipase C on the membrane

2) Phospholipase C then hydrolyzes phosphotidylinositol-4,5-bisphosphate (PIP2) to diacylglycerol (DAG) and inositol-1,4,5-triphosphate (IP3)

3) IP3 causes a rapid release of CA(+2) by opening IP3-regulated calcium ion channels in the endoplasmic reticulum

4) DAG and IP3 then activate PKC which phosphorylates the target proteins that are on Ser and Thr. Calcium is also a partial activator of PKC

Epinephrine binds in the liver or smooth muscle

Question 29

What are the components of the glucagon receptor pathway and what is the mechanism?

Answer 29

It is secreted by a-cells in the pancreas, binds to the glucagon receptor, and acts via cAMP/PKA which is similar to the B-adrenergic receptor pathway

Question 30

What are the components of the insulin receptor pathway and what is the mechanism?

Answer 30

1) Insulin receptor is phosphorylated upon binding of insulin

2) This phosphorylates the IRS-1 protein

3) This recruits phosphoinositide-3-kinase that adds a phosphate group to the 3 carbon of Phosphatidylinositol 4,5-bisphosphate (PIP2) to form PIP3

4) PIP3 dependent protein kinase (PDK1) then becomes activated and phosphorylates Protein Kinase B (Akt) which phosphorylates the target proteins

Question 31

What are 4 types of glucose transporters and when are they activated?

Answer 31

1) GLUT1 & GLUT3 are found in almost all mammalian tissue and used for glucose uptake at a constant rate. Active when concentration is above KM, and normal serum glucose is 4-8 mM (KM = 1 mM)

2) GLUT2 is found in liver and pancreatic B cells and found when KM = 15-20 mM

3) GLUT4 is found in muscle and fat cells and used when KM = 5 mM. Insulin stimulates the release of vesicles containing pre-made GLUT4, and endurance training increases the amount of GLUT4 present

4) GLUT5 is found in the small intestine and serves as a fructose transporter

Question 32

How does the regulation of glycolysis in the skeletal muscle occur?

Answer 32

Hexokinase - regulated by negative feedback via Glucose-6-phosphate

PFK - regulated by lowering PFK’s affinity for fructose-6-phosphate. AMP competes with ATP for allosteric binding so PFK activity is increased when a low ATP/AMP ratio is present, and PFK activity is decreased when a high ATP/AMP ratio is present. Decrease in pH also inhibits PFK

Pyruvate kinase - high ATP allosterically inhibits pyruvate kinase and fructose-1,6-bisphosphate activates pyruvate kinase. Unphosphorylated pyruvate kinase also has increased activity, so glucagon also decreases pyruvate kinase activity by phosphorylating pyruvate kinase via the PKA/cAMP pathway. This is because glucagon signals low blood sugar and inhibits glucose metabolism

Question 33

How does the regulation of glycolysis in the liver occur?

Answer 33

Glucokinase - primary enzyme that phosphorylates glucose in the liver. Fructose-6-phosphate inhibits glucokinase by lowering the enzyme’s affinity for glucose

Pyruvate kinase - regulated allosterically by high ATP (which inhibits it) and fructose-1,6-bisphosphate (which activates it)

Question 34

How does PFK regulation occur in the liver?

Answer 34

PFK-1 - regulated allosterically

citrate: inhibitor, indicates citric acid cycle is active and glycolysis can slow

low pH: inhibitor, acidic metabolites produced by citric acid cycle are present indicating glycolysis can slow down

high ATP: inhibits it by lowering the affinity of PFK to fructose-6-phosphate, high ATP indicates glycolysis can slow

fructose-2,6-bisphosphate: synthesized from fructose-6-phosphate and activates PFK-1 by increasing PFK’s affinity for fructose-6-phosphate and also diminishes the inhibiting effects of ATP

PFK-2 - regulated allosterically and through phosphorylation

kinase activity: stimulated when PFK-2 is unphosphorylated (inhibited when phosphorylated). When kinase activity is activated fructose-6-phosphate is phosphorylated into fructose-2,6-bisphosphate which activates PFK-1 and stimulates glycolysis

activated by inorganic phosphate and inhibited by citrate allosterically. high fructose-6-phosphate levels also stimulate phosphoprotein phosphatase which dephosphorylates PFK-2 and stimulates kinase activity/glycolysis.

phosphatase activity: stimulated when PFK-2 is phosphorylated (inhibited when unphosphorylated). When phosphatase activity is activated fructose-2,6-bisphosphate is dephosphorylated into fructose-6-phosphate which inhibits glycolysis

fructose-6-phosphate inhibits PFK-2 phosphatase activity allosterically. Also activated by glucagon since PFK-2 is phosphorylated by PKA/cAMP pathway stimulating phosphatase activity/gluconeogenesis.

Question 35

What are three ways gluconeogenesis is regulated?

Answer 35

All of these three enzymes are activated when gluconeogenesis is activated

Pyruvate carboxylase: activated by acetyl-CoA and inhibited by ADP

Phosphoenol pyruvate carboxykinase: inhibited by ADP

Fructose-1,6-bisphosphatase: inhibited by fructose-2,6-bisphosphate and AMP, activated by citrate

Question 36

How is the pyruvate dehydrogenase complex regulated?

Answer 36

E2: high acetyl-CoA concentrations inhibit it and high HS-CoA concentrations activate it (substrate stimulation product inhibition)

E3: High NADH/NAD+ ratio inhibits, low NADH/NAD+ ratio activates

E1: regulation through phosphorylation occurs where phosphorylated pyruvate dehydrogenase is inactive and unphosphorylated pyruvate dehydrogenase is active

pyruvate dehydrogenase kinase: phosphorylates pyruvate dehydrogenase (E1) and renders it inactive, regulated allosterically. Activated by high acetyl-CoA, high NADH/NAD+ ratio, high ATP and inhibited by high ADP and high pyruvate

pyruvate dehydrogenase phosphatase: dephosphorylates pyruvate dehydrogenase and renders it active. Activated by high Ca(+2) as a byproduct of the a-adrenergic signal transduction pathway

Question 37

How is the citric acid cycle regulated? How is this different between mammalian and bacterial cells?

Answer 37

Allosteric regulation of three enzymes in mammalian cells

citrate synthase: inhibited by ATP

isocitrate dehydrogenase: activated by high [Ca(2+)] (under stress) and ADP and inhibited by ATP and NADH

a-ketoglutarate dehydrogenase: inhibited by ATP, Succinyl-CoA, and NADH

In E. Coli isocitrate dehydrogenase is active when unphosphorylated and inactive when phosphorylated and a kinase and phosphatase phosphorylates (deactivates)/dephosphorylates (activates) respectively.

isocitrate dehydrogenase kinase is inhibited and phosphatase is activated by the same 5 molecules via allosteric regulation activating TCA: isocitrate, oxaloacetate, pyruvate, 3-phosphoglycerate, phosphoenolpyruvate

Question 38

How are the “building blocks” of glycogen formed in glycogen synthesis?

Answer 38

UDP-glucose pyrophosphorylase synthesizes UDP-glucose from UTP and Glucose-1-phosphate via the reaction

Glucose-1-phosphate + UTP –> UDP-glucose + PPi

This reaction is irreversible since PPi spontaneously breaks down into 2Pi immediately in the presence of H2O even though it is thermodynamically reversible.

Uridine triphosphate (UTP) is made up of uracil, ribose, and pyrophosphate (3 phosphate groups) and UDP glucose can be formed when terminal phosphates of UTP get removed and initial phosphate can be added to phosphate at 1 C of glucose

Question 39

How are glycogen chains formed?

Answer 39

Glycogen synthase catalyzes the overall reaction where a glycosyl molecule from UDP-glucose is transferred to the hydroxyl group at C4 of the terminal glucose in a polymer forming a a-1,4-glycosidic bond. It requires a glycogen molecule with a minimum of 4 residues

UDP-glucose + Glycogen (n residues) –> UDP + Glycogen (n+1 residues)

Glycogenin is an enzyme with it’s own substrate (autocatalysis) that is made up of two subunits, and each subunit catalyzes the addition of glycosyl subunit from UDP-glucose to Tyr-194 of another UDP-glucose subunit forming an a-1,4 glycosidic bond. It acts as a primer by polyermizing ~8 glucose molecules and after the primer is formed, glycogen synthase can then further polymerize the glucose molecules into glycogen

Question 40

What is the purpose of branching and how does it occur in glycogen synthesis?

Answer 40

When a “string” of at least 11 glycosyl residues are formed, a branching enzyme breaks the chain of glycogen and transfers a string of residues (typically 7) from the non-reducing end to the presumptive branch point and a linkage is formed through an a-1,6-glycosidic bond. A branch is typically seen every 10 residues.

Branching allows the number of terminal glucose residues to increase, and since these terminal residues are sites of action for glycogen phosphorylase and synthase, branching increases the rate of glycogen synthesis and degradation.

Question 41

How does glycogen degradation occur?

Answer 41

Glycogen phosphorylase phosphorylates glucose-1-phosphate on the terminal glucose residue (non-reducing end) and protonates the O at the 4 carbon. It uses a phosphate group to break the a-1,4 linkage

Glycogen (n residues) + Pi –> glucose-1-phosphate + glycogen (n-1)

In order to debranch, glucose residues are removed to the limit dextran by phosphorylase (4th glucose residue from branch glucose) and three enzymes are used

Transferase: transfers the 3 terminal glucose residues from outer branch to the center (main and linear branch)

a-1,6-glucosidase hydrolyzes the a-1,6-glycosidic bond to release a free glucose molecule

Phosphorylase can then continue to remove glucose residues from the non-reducing end producing Glucose-1-phosphate molecules

Phosphoglucomutase converts glucose-1-phosphate to glucose-6-phosphate for further metabolism and for export out of the liver to other tissues making glucose-1,6-bisphosphate as an intermediate

Question 42

What are the 4 states of glycogen phosphorylase?

Answer 42

In muscle, unphosphorylated glycogen phosphorylase (b form) is inactive and phosphorylated glycogen phosphorylase (a form) is active. Each form has two states where T state is less active (less likely to be phosphorylated) and R state is more active (more likely to be phosphorylated). a-form favors the R-state and b-form favors the T-state

Question 43

What are the 4 states of glycogen phosphorylase and what is the level of activity between them?

Answer 43

The lowest level of activity is the b form of the T state (1), the second lowest level of activity is the b form of the R state (2), the second highest level of activity is the a form of the T state (3), and the highest level of activity is the a form of the R state (4)

Question 44

How is glycogen phosphorylase allosterically regulated in the muscle?

Answer 44

Allosteric regulation only occurs on the b form

When ATP is low, AMP levels are high and AMP binds to an allosteric state to stabilize the R-state meaning glycogen phosphorylase activity is elevated

When ATP is high, ATP stabilizes the T state making glycogen phosphorylase less active

Glucose-6-phosphate stabilizes the T state (less active)

Question 45

How is glycogen phosphorylase allosterically regulated in the liver?

Answer 45

Allosteric regulation only occurs on the a form

Glucose mediates the transition of the a-form from the R to the T state reducing glycogen phosphorylase activity

Question 46

What is the function of phosphorylase kinase and how is it regulated?

Answer 46

It phosphorylates glycogen phosphorylase at Ser-14 activating it and allowing glycogen to be broken down

It contains a y subunit that serves as the catalytic subunit and aBd subunits that serve as regulatory units. The d subunit binds to a Ca(+2) protein called calmodulin, a regulatory protein that activates the enzyme

It becomes fully activated in the presence of PKA/cAMP via the B-adregenic receptor (epinephrine) in skeletal muscle, PKA/cAMP via glucagon pathway in the liver, and Ca(+2) via the a-adregenic receptor (epinephrine) in the liver or smooth muscle

Question 47

What is the function of glycogen synthase and how is it regulated?

Answer 47

Glycogen synthase synthesizes glycogen

Unphosphorylated glycogen synthase (a form) is active and phosphorylated glycogen synthase (b form) is inactive

PKA phosphorylates glycogen synthase inhibiting it in the muscle and liver, PKB causes it to be activated

Question 48

How is glycogen synthase regulated by insulin in the liver?

Answer 48

Insulin increases the presence of protein kinase B. This phosphorylates glycogen synthase kinase (generally only present in the liver) rendering it inactive. When glycogen synthase kinase is inactive, it cannot phosphorylate glycogen synthase, and since unphosphorylated glycogen synthase (the a form) is active, this means that insulin stimulates glycogen synthesis, and this makes sense since insulin is secreted when blood glucose levels are high

Question 49

How do protein phosphatases regulate glycogen synthesis and degradation in the muscle?

Answer 49

In muscle, protein phosphatatase 1 (PP1) removes the phosphate group of an enzyme regulated by phosphorylation.

The GM protein allows PP1 to associate with its target when it is not phosphorylated, and the phosphorylation of the GM protein slows the binding of PP1 with its target. Protein Kinase A phosphorylates the GM protein preventing the binding of an enzyme with PP1 and the removal of a phosphate

The phosphorylation of an inhibitor protein by Protein Kinase A also binds to the active site of PP1 and stops it from removing phosphate groups of enzymes

PP1 functions to “turn off” both glycogen synthesis and degradation, and epinephrine and glucagon inhibits this process via PKA

Question 50

How do protein phosphatases regulate glycogen synthesis and degradation in the liver?

Answer 50

The presence of phosphoryated glycogen phosphorylase (a-form) in the R state (low glucose) inhibits PP1 activity and allows glycogen degradation to occur

PP1 can dephosphorylate glycogen phosphorylase when in the T state, but in the R state, phosphate groups are hidden after they form a dimer/inhibitory complex

When high levels of glucose are present, the T state is stabilized since glucose induces the transition of phosphorylated glycogen phosphorylase (a-form) from R to T state allowing PP1 to become active.

When PP1 is active glycogen phosphorylase a can be dephosphorylated (converted to b form) reducing glycogen degradation and glycogen synthase can be dephosphorylated (converted to the active a-form) and stimulating glycogen synthesis