Glycolysis

Question 1

Name the steps in the “preparatory”/energy investment phase

Answer 1

1) Phosophorylation of glucose to produce glucose-6-phosphate – 1st ATP INVESTED
2) Isomerization reaction to produce fructose-6-phosphate
3) Phosphorylation to produce Fructose 1,6 bisphosphate – 2nd ATP INVESTED
4) Retro-aldol reaction to produce glyceraldehyde 3-phosphate and dihydroxyacetone phosphate. DHAP is isomerized to G3P, yielding 2 molecules of G3P

Question 2

Name the steps in the “payoff” phase of glycolysis

Answer 2

1) Phosphorylation of G3P, coupled to production of NADH, to produce 1,3-bisphosphoglycerate
2) Substrate-level phosphorylation of ADP to form ATP and 3-phosphoglycerate
3) Isomerization to produce 2-phosphoglycerate
4) Elimination of water to produce phosphoenolpyruvate
5) Substrate-level phosphorylation of ADP to produce ATP and pyruvate

Question 3

What are the three main fates of pyruvate?

Answer 3

1) Oxidation, through loss of CO₂, to acetyl CoA
2) Lactic acid fermentation – in low-oxygen conditions, NADH builds up, and without NAD+, glycolysis cannot proceed, so NAD+ must be regenerated through transfer of a hydride to pyruvate, generating lactic acid
3) Decarboxylation and reduction to ethanol

Question 4

What is the overall equation for glycolysis?

Answer 4

There is a net production of 2 ATP, 2 NADH

Glucose + 2 NAD+ + 2 ADP + 2Pi ⇔ 2 pyruvate + 2 NADH + 2 ATP + 2H+ + 2 H₂O

ΔG^°= -85 kJ/mol; largely irreversible, very negative

Question 5

Step 1 of glycolysis: hexokinase reaction

Answer 5

Coverts glucose into glucose 6-phosphate

• Prevents glucose from being moved out of cell by glucose transporter
• ATP is used to phosphorylate glucose; negative free energy of ATP hydrolysis is greater than energy needed to phosphorylate glucose from P_i, so reaction proceeds spontaneously in the forward direction; irreversible

This means that, to do the reverse reaction, need to find an alternate reaction pathway, because the G6P is lower-energy than ATP, and hydrolysis alone cannot supply the necessary energy for the formation of ATP

Question 6

Step 2 of glycolysis: phosphohexose isomerase

Answer 6

Converts G6P to F6P

• This is a reversible reaction; is not coupled to a highly exothermic reaction like the hexokinase reaction is
• Mechanism: keto-enol tautomerization, with an enediol intermediate

Question 7

Step 3 of glycolysis: Phosphofructokinase-1

Answer 7

Catalyzes phosphorylation of F6P to fructose 1,6 bisphosphate

• Requires input of ATP; negative ΔG makes it essentially irreversible in cellular conditions
• First committed step of glycolysis; G6P and F6P have other possible fates, F1,6BP does not
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Question 8

Step 4 of glycolysis: fructose 1,6 bisphosphate aldolase

Answer 8

Catalyzes cleavage of F 1,6 BP into G3P and DHAP

• Reversible retro-aldol reaction

Question 9

Name the steps of the retro-aldol reaction for cleaving F 1,6 BP

Answer 9

1) . Nucleophilic attack of lysine residue on C2 carbonyl of fructose, forming an imine and kicking off water, covalently linking molecule to the enzyme
2) Retro-aldo reaction: Base deprotonates C4 OH, electrons drop onto C4, forming carbonyl; electrons from C4-C3 move onto C3-C2, first molecule of G3P released
3) C3 grabs a proton
4) Schiff base is hydrolyzed, generating DHAP

Question 10

Step 5 of glycolysis: triose phosphate isomerase

Answer 10

• DHAP is converted into G3P
• Mechanism: keto-enol tautomerization reaction
• C3 becomes the carbonyl carbon

Question 11

After glucose has been split into two 3-carbon molecules, which atoms from the original glucose are equivalent?

Answer 11

C1 = C6; both are phosphorylated

C2 = C5; both are bound to a hydroxyl

C3 = C4; both are carbonyls

Question 12

Step 6 of glycolysis: glyceraldehyde 3-phosphate dehydrogenase

Answer 12

Oxidizes G3P to 1,3 BPG, accompanied by formation of NADH

• Aldehyde group of G3P is oxidized to an acyl phosphate
• Energy of oxidation is used to drive the formation of relatively high-energy acyl phosphate, which is then used to phosphorylate ADP – another example of energetic coupling

Question 13

Mechanism of glyceraldehyde 3-phosphate dehydrogenase: steps

Answer 13

1. Cysteine does a nucleophilic attack on the aldehyde, acyl substitution reaction kicks off hydride, which is accepted by NAD+
2. A molecule of Pi does a phosphorolysis reaction, releasing 1,3 BPG

Question 14

Step 7 of glycolysis: phosphoglycerate kinase

Answer 14

Transfers phosphate from 1,3 BPG to an ADP

• Reversible reaction; made possible by relatively high energy of acyl phosphate group, which is similar in energy to ATP bond energy

Question 15

What is the difference between ΔG° and ΔG, and how is it relevant to glycolysis?

Answer 15

• ΔG° refers to the free energy change in going from reactants to products when everything is starting at a concentration of 1 M
• ΔG reflects the position of the equilibrium/tendency to drive forward under ACTUAL cellular conditions
  
  • For example: ΔG° might be negative if there are slightly more products than reactants at equilibrium, but if there is a relativel build-up of products in the cell, ΔG of the reaction could be positive, because it wants to drive backwards

Question 16

Step 8 of glycolysis: phosphoglycerate mutase

Answer 16

Catalyzes reversible transfer of phosphoryl group from C3 to C2, producing 2-phosphoglycerate

Question 17

Step 9 of glycolysis: enolase

Answer 17

Dehydrate 2-phosphoglycerate to produce phosphoenolpyruvate

• Reversible reaction
• Coverts compound into one with a high phosphoryl group transfer potential

Question 18

Step 10 of glycolysis: pyruvate kinase

Answer 18

Transfers phorporyl group from phosphoenolpyruvate to ADP, producing ATP and pyruvate

• Pyruvate first appers in the enol form, then tautomerizes to the keto form
• Has a large, very negative ΔG – one of the three steps in glycolysis that is highly regulated

Question 19

What is the relationship between Q and K?

Answer 19

• Think of the equilibrium concentrations (where Q = K) as being an energy minimum; at this point, ΔG of the reaction will be 0, because there will be no driving force pushing the reaction forwards or backwards
• Whenever Q is anything other than K (lowest point in the picture) it will be driven towards whatever direction is downhill; ΔG at pink point is positive, negative at blue. Adding reactants will always push Q to the left, and will therefore make ΔG more negative
• For the glycolysis reactions in which ΔG is near 0, Q is near K, so forward and reverse reactions will be happening at a similar rate, making them reversible
• For the glycolysis reactions in which ΔG is very negative, the driving force favors the forward reaction, so the reverse reaction happens very slowly

Question 20

What does it mean to say that organisms are in a dynamic steady state?

Answer 20

• Refers to the fact that the concentrations of metabolites in the cell do not change appreciably over time
• FLUX (rate of metabolite flow) through the pathways may change, but all steps will adapt along with it so as not to cause a build-up or shortage of products at any one step

Question 21

What is flux and what are some general strategies for regulating it?

Answer 21

• Flux is the rate of metabolite flow through a pathway
• Flux may change, but if it changes for one step, it must change for all of them to avoid a build-up of anything along the way
• May modulate flux by changing the amount of enzyme or catalytic activity (allosteric regulation, covalent modification), or by sequestering the enzyme in one part of the cell (i.e. glucose transporter sequestered until insulin is available)

Question 22

Glycolytic reactions that are near equilibrium in vivo

Answer 22

The closer the reactions are to equilibrium, the weaker the driving force on them, and the closer that the forward and reaction rates will be to one another

• Does NOT mean reactant and product concentrations are equal – recall U-shaped energy graph
• Because they are close to equilibrium, small changes in concentration can produce large changes in net rate – they are near the bottom of the curve, where it doesn’t take much to change direction
• If Q and K are within 1-2 orders of magnitude of one another, then the reaction is near equilibrium

6/10 glycolysis reactions are maintained near equilibrium: 1) Aldolase, 2) Triose phosphate isomerase, 3) G3P dehydrogenase and phosphoglycerate kinase, 4) Phosphoglycerate mutase, 5) enolase, 6) Phosphoglucose isomerase

Question 23

Non-equilibrium reactions in glycolysis

Answer 23

Reactions: hexokinase, PFK-1, pyruvate kinase

• These reactions are maintained far from equilibrium by keeping the cellular concentrations of the metaoblites at appropriate levels
• The forward reaction occurs at a much faster rate than the reverse reaction, because the driving force pushes it forward
• Non-equilibrium reactions limit flux through a pathway – when you add more hexokinase or PFK, flux increase through pathway, whereas addition of one of the equilibrium enzymes does not (think pipes)

Question 24

Why can’t the cell allow highly exergonic reactions to go to equilibrium?

Answer 24

• Example: PFK
  
  • If allowed to go to equilibrium, concentration of F 16 BP would wreak osmotic havoc on the cell
  • If the concentration of ATP was allowed to drop to equilibrium concentrations, it would no longer be an exergonic reaction and would be useless for driving cellular processes – must maintain relatively high concentration of ATP to keep Q < K
• Therefore, when there are metabolic changes, the entire downstream pathway must adjust to ensure that the reaction stays far from equilibrium

Question 25

Why does the cell need to maintain a constant concentration of ATP?

Answer 25

1. Kinetic effect: The enzymes that depend on ATP are often saturated, so a decrease in the concetration of ATP would cause a reduction in rate, which could be catastrophic
2. Thermodynamic effect: the breakdown of ATP to ADP and P_iis only energetically favorable at certain concentrations; if the concentration of the products were to get too high,and Q > K, then ΔG would be negative for the production of ATP and ATP could no longer be used to drive endergonic reactions

When there are sudden changes in ATP, it’s actually AMP that experiences a relatively greater change in concentration, so many regulatory mechanisms are sensitive to AMP

Question 26

What is the flux control coefficient?

Answer 26

• Indicates relative contribution of a particular enzyme to control of flux
• For glycolysis: 0.79 hexokinase, 0.21 for PFK-1

Question 28

Regulation of hexokinase – difference between liver and muscle isozymes

Answer 28

• Humans have 4 hexokinase isozymes
• Liver isozyme, hexokinase 4, has a half saturation much higher than muscle isozymes, allowing it to be directly regulated by the level of blood glucose; when blood glucose concentration low, it’s not captured by phosphorylation, and instead just leaves the cell
• Hexokinase 1-3 are allosterically inhibited by product, G6P, but H4 is not

Question 29

Mechanisms of regulating PFK-1

Answer 29

1. High cellular ATP inhibits by binding to allosteric site and lowering affinity for substrate
2. Citrate inhibits it when other sources of energy are meeting the cell’s energetic needs
3. High ADP and AMP concentrations activate it, by relieving inhibition by ATP

The curve shifts to the left when there is low ATP, shifts to the right when there is high ATP

Question 30

What is the function of fructose 2,6 bisphosphate?

Answer 30

• Regulates PFK-1 and F16 bisphosphatase – activates PFK-1 (shifts curve to the left, higher activity at lower concentrations) and inhibits F 1,6 bisphosphatase (shifts curve to right, lower activity at same concentration)

Question 31

How is fructose 2,6 bisphosphate regulated?

Answer 31

• Relative concentrations of glucagon and insulin regulate it, so that when glucose is low, it is distributed into the bloodstream rather than going through glycolysis
• A single bifunctional enzyme catalyzes its formation and degradation from F6P; when glucagon high, enhances FBPase activity, lowering concentration of F26BP and inhibiting PFK1
• Insulin activates PFK-2 activity, leading to prodution of F26BP, stimulating PFK-1

Question 32

How is pyruvate kinase regulated?

Answer 32

• Allosterically inhibited by acetyl CoA and ATP
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