M2: Pyruvate Dehydrogenase Complex L9

Question 1

What is the pyruvate dehydrogenase complex (PDC)?

Answer 1

Pyruvate dehydrogenase is the enzyme that catalyzes the reaction:
pyruvate + CoA + NAD+ –> acetyl-CoA + CO2 + NADH

Question 2

Where does glycolysis occur? PDC? CAC? Aerobic metabolism?

Answer 2

Glycolysis: cytoplasm

PDC, CAC & Aerobic metabolism: mitochondrion

Question 3

When does PDC act on pyruvate?

Answer 3

Once pyruvate is in the mitochondria.

Question 4

How does pyruvate get into the mitochondria?

Answer 4

Pyruvate gets into mitochondrion through the pyruvate translocase (H+ symport).

Question 5

What is the entrance point for carbons into the CAC?

Answer 5

Acetyl-CoA

Question 6

What is the goal of the CAC?

Answer 6

To generate reducing agents.

Question 7

What proteins make up the PDC?

Answer 7

(E1) Pyruvate dehydrogenase
(E2) Dihydrolipoyl transacetylase
(E3) Dihydrolipoyl dehydrogenase
E3-binding protein
Pyruvate dehydrogenase kinase
Pyruvate dehydrogenase phosphatase

Question 8

What 5 coenzymes/cofactors are required for the PDC? What proteins do they bind to?

Answer 8

1. Thiamine pyrophosphate (TPP): bound to E1
   Substrate=Pyruvatve
2. Lipoic acid: covalently linked to a Lys on E2 (lipoamide)
3. Coenzyme A (CoA): substrate for E2
4. Flavin adenine dinucleotide (FAD): bound to E3
5. Nicotinamide adenine dinucleotide (NAD+): substrate for E3.

Question 9

Why are the active sites of the 3 PDC enzymes in close proximity?

Answer 9

So that the product of E1 is passed on to the second, and so on. Substrate channeling.

Question 10

What is the role of PDC?

Answer 10

To generate Acetyl-CoA from pyruvate.

Question 11

Memorize Pyruvate dehydrogenase reaction.

Answer 11

L9 S6.

Question 12

What happens in step 1 of PDC? What enzyme is used? Is this step reversible or irreversible? Why?

Answer 12

E1 = pyruvate dehydrogenase

1. Nucleophilic attack by Thiamine Pyrophosphate (TPP) leading to a decarboxylation of pyruvate
2. C1 of pyruvate released as CO2
3. C2 and C3 attached to TPP as hydroxyethyl group = hydroxyethyl-TPP

Irreversible reaction because CO2 diffuses out of mitochondria.

Question 13

What is step 2 of the PDC? What enzyme is used? Is this step reversible/irreversible?

Answer 13

E2 = Dihydrolipoyl Transacetylase
Reversible

Step 2 regenerates activity of step 1:
The reactive center of lipoamide is a disulfide bond which can be readily and reversibly reduced to make acetyl-dihydrolipoamide. The 2 electrons removed in the oxidation reduce the S-S of the lipoyl group on E2 to SH. TPP is regenerated here!
Acetyl-dihydrolipoamide is a high energy intermediate bc it has a thio-ester bond.

Question 14

How does lipoamide affect substrate channeling?

Answer 14

Lipoamide is E2’s “swinging arm” to perform substrate channeling between E1 and E3

Question 15

What is step 3 of the PDC? What enzyme is used? Is this step reversible/irreversible?

Answer 15

E2 = Dihydrolipoyl Transacetylase
Reversible

Step 3: Trans-esterification to CoA =
Acetyl group from Acetyl-dihydrolipoamide gets put onto CoA to make acetyl-CoA. Dihydrolipoamide is produced (fully reduced lipoamide).

Question 16

What is the summary of the E2 products?

Answer 16

E2 generates acetyl-CoA and regenerates TPP for E1.

Question 17

Why is Acetyl-CoA a good entry point of the CAC?

Answer 17

Because it has an acetyl thio-ester bond which makes it a high energy compound (its hydrolysis = -31.5 kJ/mole).

Question 18

What is step 4 and 5 of the PDC? What enzyme is used? Is this step reversible/irreversible?

Answer 18

E3= Dihydrolipoyl Dehydrogenase
Reversible

Step 4: E3 re-oxidizes dihydrolipoamide to lipoamide (removes H’s and regenerates thio-ester bond). Since E3 contains a bound FAD prosthetic group, the oxidation of dihydrolipoamide is coupled to the reduction of FAD to FADH2.

Step 5: enzyme-bound FADH2 reoxidized to FAD by NAD+, thus generating NADH.

Question 19

What is the aim of E3 (steps 4 and 5)?

Answer 19

E3 restores E2 and its own FAD for another round of generation of acetyl-CoA (for CAC).
E3 generates NADH (for Ox-phos).

Question 20

Why is the PDC reaction irreversible?

Answer 20

Because the first step is irreversible.

Question 21

What are the mechanistic advantages of multienzyme complexes?

Answer 21

1. Minimized distances for substrates in between active sites: increased reaction rate without having to maintain large pools of intermediates
2. Metabolic intermediates are channeled between successive enzyme sites: 2 important effects:
   - side reactions are minimized
   - protection for chemically labile intermediates
3. Coordinated control of reactions: shutting off one enzyme effectively shuts the system down.

Question 22

Why is the regulation of PDC so important?

Answer 22

• No other pathway in mammals for synthesis of acetyl-CoA from pyruvate
• Animals cannot synthesize glucose from acetyl CoA. Thus, conversion of pyruvate to acetyl CoA commits glucose carbons to oxidation in CAC or fatty acids synthesis.

Question 23

What are the two levels of control of the PDC?

Answer 23

• Product inhibition by acetyl-CoA and NADH

- Covalent modification (phophorylation of E1 Pyruvate dehydrogenase).

Question 24

How is the PDC regulated by product inhibition?

Answer 24

1. At E3, NADH competes for the binding of NAD and at E2, Acetyl-CoA competes for the binding of CoA. Since step 3 and 5 are reversible and there’s more NADH or Acetyl-CoA the reaction goes backwards.
2. Since the reaction is going backwards, the substrate for E2 is Acetyl-CoA (instead of CoA), it can no longer accept the Hydroxyethyl-TPP intermediate from E1 to regenerate TPP. This causes a buildup of the Hydroxyethyl-TPP intermediate which causes less TPP.
3. Since TPP is required for pyruvate to be decarboxylated (E1). This backs up the whole system and you end up with an accumulation of pyruvate.

Question 25

What happens to pyruvate when the PDC is inhibited due to product inhibition?

Answer 25

In these conditions, the pyruvate would undergo gluconeogenesis in liver cells. In muscle cells, they would be converted into lactate and sent back to liver through blood (cori cycle). Prevents the useless consumption of pyruvate.

Question 26

What 2 enzymes regulate PDC by phosphorylation?

Answer 26

1. Pyruvate dehydrogenase kinase (PDK) | 2. Pyruvate dehydrogenase phosphatase (PDP)

Question 27

How is the PDC de-activated by phosphorylation? How is PDK inhibited?

Answer 27

Inactivation of PDC via PDK (inhibitor): 1) PDK is allosterically activated by acetyl CoA and NADH, 2) PDK phosphorylates E1 3) E1 becomes inactive => Shuts down the entire PDC PDK is allosterically inhibited by pyruvate, ADP, Ca2+ => reverses inhibition

Question 28

How is the PDC activated by phosphorylation?

Answer 28

Reactivation of PDC via PDP: 1) PDP is allosterically activated by Ca2+ (perfect when you exercise!) 2) PDP dephosphorylates E1 3) E1 becomes active

Question 29

What 2 enzymes control the utilization of pyruvate?

Answer 29

``` Pyruvate Dehydrogenase Complex (PDC): aerobic glycolysis Lactate dehydrogenase (LDH): anaerobic glycolysis (generate lactate). ```

Question 30

What happens to PDC in cancer cells?

Answer 30

In cancer cells PDC activity is suppressed by phosphorylation through PDK. Therefore, Pyruvate is converted to lactate (Warburg effect). Warburg: Even in the presence of normal amounts of oxygen, cancer cells will prefer to use aerobic glycolysis instead of oxidative phosphorylation.

Question 31

What happens to PDC regulation when you stop running?

Answer 31

When you stop running: (muscles go back rest) - No need to produce excess ATP (CAC not needed) - Acetyl-CoA and NADH accumulate - E3 and E2 are OFF through product inhibition - E1 is OFF through phosphorylation

Question 32

What happens to PDC regulation when you start running?

Answer 32

``` When you start running: Need to produce ATP Activated glycolysis produce pyruvate: - Blocks PDK; activates E1 Ca2+ increases for muscle contraction: - Activates PDP and inhibits PDK - E1 = ON because dephosphorylation of E1 is favoured. ```