Glycolysis

Overview and regulation of pathway.

See chapter 15 of Fundamentals of Biochemisty: Life at the Molecular Level, 4th Ed by Voet, Voet, and Pratt.

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Overview

• Glycolysis: step by step• Stage I• Stage II

• Anaerobic Fates of Pyruvate• Homolactic fermentation • Alcoholic Fermentation

• Regulation of Glycolysis • T and R states of PFK • Substrate cycling

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Glycolysis Stage I

• Step 1: Phosphorylation of Glucose

• Step 2: Isomerization of G6P into F6P

• Step 3: Phosphorylation of F6P to FBP

• Step 4: Cleavage of FBP into GAP and DHAP

• Step 5: GAP and DHAP interconversion

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Step 1: Phosphorylation

• First step is the phosphorylation of glucose to produce glucose-6-phosphate (G6P).

• The reaction is catalyzed be hexokinase, a transferase, which uses ATP as the source of phosphate.

• In addition to G6P; ADP and H+ are also produced by the reaction.

• Magnesium is necessary to stabilize ATP, making the terminal phosphorous more accessible for nucleophilic attack.

• Binding of the sugar causes the enzyme to close, which excludes water from the active site.

• Hexokinase is a nonspecific enzyme that can phosphorylate a variety of hexoses.

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Q: Draw the structures of glucose and G6P.

Q: Write a mechanism for this phosphorylation. (Hint: it’s a concerted, one step mechanism; think SN2) So what is the purpose of the enzyme for such a simple reaction?

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Step 2: Isomerization

• G6P is converted to fructose-6-phosphate (F6P) by phosphoglucose isomerase (PGI).

• This reaction is the isomerization of an aldose to a hexose and the enzyme is an isomerase.

• The currently accepted mechanism is as follows:1. Substrate binds to active site.2. A proton donor catalyzes ring opening. 3. A base abstracts the C2 proton to form a cis enediolate

intermediate. 4. The captured proton is then placed onto C1 5. The ring then closes to form cyclic F6P

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Q: Write the enzymatic mechanism for the conversion of G6P to F6P via PGI.

Q: What are the likely residues involved in the mechanism?

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Step 3: Phosphorylation

• Phosphofructokinase (PFK) phosphorylates F6P to produce fructose-1,6-bisphosphate (FBP or F1,6P).

• This enzyme is categorized as a transferase and requires magnesium to properly function.

• This is a rate determining step and is plays a major role in control of the pathway. PFK is regulated by a number of compounds that will be covered later on.

• In addition to FBP, ADP and H+ are also produced by this reaction.

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Q: Draw the structure of F6P and FBP. Write a mechanism for their conversion (see hexokinase)

Q: What are some potential reasons that PFK is a control point of glycolysis? (Gotta think like a scientist here!)

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Step 4: Cleavage

• Aldolase catalyzes the cleavage of FBP into glyceraldehyde-3-phosphate (GAP) and dihydroxyacetone phosphate (DHAP).

• In plants and animals the mechanism of aldolase is as follows1. FBP binds to enzyme active site2. Carbonyl group reacts with amine of lysine to form a Schiff base.3. The C3-C4 bond is cleaved to form an enamine intermediate. GAP is

released 4. Protonation and tautomerization of enamine yields the iminium cation

from the Schiff base.5. The iminium cation is hydrolyzed to yield DHAP and free enzyme.

• Aldolase is a lyase (bond was cleaved and double bond formed was the carbonyl)

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Q: Draw the Fischer projections of FBP, GAP, and DHAP.

Q: Write the enzymatic mechanism of FBP cleavage via aldolase.

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Step 5: Interconversion of GAP and DHAP

• Triose phosphate isomerase (TIM) interconverts GAP and DHAP.

• TIM is a catalytically “perfect” enzyme. This means that the rate of the bimolecular reaction is diffusion controlled, so product formation occur as rapidly as enzyme and substrate can collide.

• The [GAP] to [DHAP] ratio is 4.73x10-3 so at equilibrium there is significantly more DHAP, but as GAP is siphoned off for further breakdown, DHAP is converted into more GAP.

• TIM is an isomerase.

• This is the last step of Stage I of Glycolysis

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Q: Write the enzymatic mechanism of converting of GAP to DHAP.

Q: Write the enzymatic mechanism of converting of DHAP to GAP.

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Glycolysis Stage II

• Step 6: Phosphorylation of GAP

• Step 7: 1,3-BPG phosphorylates ADP to make first ATP

• Step 8: Isomerization of 3PG to 2PG

• Step 9: Enolase forms PEP

• Step 10: PEP phosphorylates ADP to make second ATP

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Step 6: 1st High Energy Intermediate

• Glyceraldehyde-3-phosphate dehydrogenase oxidizes and phosphorylates GAP using NAD+ and inorganic phosphate.

• The reaction produces 1,3-bisphosphoglycerate (1,3-BPG), NADH and hydronium.

• The currently accepted mechanism is as follows: 1. GAP binds to the enzyme2. Sulfur of cysteine acts as nucleophile to attack aldehyde carbonyl3. Hydride is transferred to NAD+ to form an acyl thioester. 4. Phosphate binds to enzyme-thioester-NADH complex. 5. Thioester intermediate is attacked by the phosphate. A proton

transfer to the sulfur results in the subsequent release of 1,3-BPG and NADH.

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Q: Write the enzymatic mechanism 1,3-BPG formation via GAPDH.

Q: Briefly explain the experiments that contributed to the elucidation of the GAPDH reaction mechanism.

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Step 7: First ATP

• Phosphoglycerate kinase (PGK), catalyzes the conversion of 1,3-BPG into 3-phosphoglycerate (3PG) with concurrent ATP synthesis from ADP.

• 3PG is a transferase that requires magnesium to properly function. The enzyme is called a kinase because the reverse reaction is a phosphoryl group transfer from ATP to 3PG.

• Upon binding, the domains swing close to allow a water free environment for the reaction.

• This reaction has a ΔG°’ value of -18.8 kJ/mol and is coupled to the previous reaction to help drive the reaction forward.

• The production of ATP in this reaction, does not involve oxygen and is an example of a substrate level phosphorylation.

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Q: Draw the structures of 1,3-BPG and 3PG. Indicate the high energy bonds that are broke and formed during steps 6 and 7.

Q: Write a balanced chemical equation for the reaction catalyzed by PGK.

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Step 8: Isomerization • Phosphoglycerate mutase (PGM) interconverts 3PG and 2-phosphoglycerate

(2PG). • This enzyme mutase (a type of isomerase) because it catalyzes the transfer

of a functional group from one position to another on a molecule. • The mechanism of PGM is as follows:

1. 3PG binds to the enzyme which has a phosphorylated histidine residue. 2. The histidine's phosphate group is transferred to the substrate resulting in a 2,3-

bisphosphoglycerate enzyme complex.3. The enzyme’s histidine residue is re-phosphorylated by the substrate’s 3-phosphate

group.4. The release of the product 2PG regenerates the active enzyme.

• Occasionally 2,3-bisphosphoglycerate is released from the active site to produce an inactive enzyme. So trace amounts of 2,3-BPG is necessary to regenerate the active site of the enzyme.

• Side note: 2,3-BPG specifically binds deoxyhemoglobin to decrease its oxygen affinity, so erythrocytes need substantially more 2,3-BPG than the amount needed for PGM.

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Q: Write the enzymatic mechanism for the conversion of 3PG into 2PG via PGM.

Q: Indicate where 2,3-BPG is in the mechanism above?

Q: Why might it be energetically favorable to use two phosphates during the mechanism instead of transferring just one around?

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Step 9: 2nd High Energy Intermediate

• 2PG is dehydrated to phosphoenolpyruvate (PEP) in a reaction catalyzed by enolase. Water is a byproduct of the reaction.

• The enzyme forms a complex with magnesium before the substrate binds.

• Fluoride ions inhibit glycolysis by blocking enolase activity. Fluoride blocks substrate binding to enolase by forming a complex with magnesium at the active site.

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Q: Consider the following portion of glycolysis:

GAP 1,3-BPG 2,3-BPG 3PG 2PG

We know that 2,3-BPG affects the oxygen affinity of hemoglobin. What effect would each of the following have on its affinity? Show this with a graph.

A. Hexokinase deficiency in erythrocytes

B. Pyruvate kinase (last step of glycolysis) deficiency in erythrocytes

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Step 10: 2nd ATP formation

• The phosphate from PEP is transferred to ADP to form ATP. This reaction is catalyzed by pyruvate kinase (PK).

• The starting materials are PEP, ADP and H+ and the products are pyruvate and ATP.

• The PK reaction requires both potassium and magnesium and its mechanism is as follows:

1. The beta phosphoryl oxygen of ADP nucleophilically attacks the PEP phosphorous atom, displacing enolpyruvate and forming ATP.

2. Enolpyruvate tautomerizes to pyruvate.

• This reaction is another example of substrate level phosphorylation.

• The driving force of this reaction is the last step of the mechanism .

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Q: Write the enzymatic mechanism of pyruvate kinase, make sure to show the function of both potassium and magnesium.

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Q: Write a balanced equation for each step of glycolysis.

Write the net balanced chemical equation for glycolysis.

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Fates of Pyruvate

• Pyruvate has essentially three possible fates, it can either be passed to the citric acid cycle or it can undergo fermentation to either ethanol or lactate. The pathway that’s followed is based on the energy needs of the cell at the time as well as the type pf cell.

• When oxygen is present (aerobic conditions) pyruvate will be oxidized via the citric acid cycle.

• When little to no oxygen is present (anaerobic conditions), pyruvate will be converted to a reduced product to re-oxidize the NADH produced by glycolysis.

• In muscle cells pyruvate is reduced to lactate in homolactic fermentation and in yeast cells pyruvate is reduced to ethanol via a process known as alcoholic fermentation.

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Homolactic Fermentation

• Lactate dehydrogenase (LDH) converts pyruvate, NADH, and H+

into L-Lactate and NAD+.• This reaction is freely reversible and so pyruvate and lactate can

be equilibrated based on cellular needs. • The proposed mechanism is as follows: a hydride ion is

stereospecifically transferred from C4 of NADH to C2 of pyruvate with concomitant transfer of a proton from histidine.

• The arginine residues and the histidine residue help to properly orient the substrate to ensure a stereospecific process.

• The lactate produced in muscle cells is mostly taken to the liver to synthesize glucose.

• Contrary to popular belief its not the lactate that causes muscle fatigue but rather the excess acid from glycolysis. Muscles won’t fatigue if lactate increases but pH is held constant.

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Q: Draw the structure of the active site that allows lactate dehydrogenase to be stereospecific in its reaction.

Q: Write the net balanced chemical equation for the overall process of anaerobic glycolysis.

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Alcoholic Fermentation

• This pathway is irreversible and occurs in two steps. The first step is decarboxylation followed by reduction.

• The first reaction is catalyzed by pyruvate decarboxylase (this enzyme is not found in animals) and produces carbon dioxide and acetaldehyde. It requires thiamine pyrophosphate (TPP) as a cofactor to help stabilize the anion intermediate.

• The second reaction is catalyzed by alcohol dehydrogenase and uses NADH oxidation to turn acetaldehyde into ethanol. Yeast alcohol dehydrogenase is a tetramer that binds zinc. The zinc ions polarizes the carbonyl group of acetaldehyde to help facilitate the hydride transfer.

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Decarboxylation

• Pyruvate decarboxylase removes a molecule of carbon dioxide from pyruvate using the enzyme pyruvate decarboxylase.

• The coenzyme thiamine pyrophosphate (TPP) is synthesized from vitamin B1 and binds tightly but non covalently. Its catalytically active functional group is the thiazolium ring.

• Pyruvate decarboxylase has the following mechanism:1. The ylid form of TPP, a nucleophile, attacks the carbonyl carbon of

pyruvate. 2. Carbon dioxide leaves, creating a resonance stabilized carbanion adduct

in which the thiazolium ring of the coenzyme acts as an electron sink.3. Carbanion is protonated 4. The TPP ylid is eliminated to form acetaldehyde and regeneration of the

active enzyme.

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Q: Draw the structure of TPP and identify all the relevant functional groups.

Q: Write the enzymatic mechanism of pyruvate decarboxylation.

Q: Use unambiguous structural formulas to show the stereospecificity of alcohol dehydrogenase.

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Regulation

• How scientists determine flux control points:• Measure in vivo ΔG for each reaction, enzymes with large ΔG values are

often possible control points• In vitro identification of allosteric modifiers of the enzymes catalyzing

the rate determining steps. • Measurement of the in vivo levels of potential regulators to determine if

the concentration changes are consistent with proposed control mechanism.

• Each tissue regulates glycolysis differently, but in heart muscle the three major control points are the enzymes: hexokinase, phosphofructokinase and pyruvate kinase.

• The other steps in the pathway operate at near equilibrium and so these steps are sensitive to changes in concentration of intermediates, but aren’t major control points.

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Regulation

• In muscles the major source of G6P is from glycogen and not glucose and so hexokinase is not a major player in the control of glycolysis in muscle cells.

• Pyruvate kinase controls the last step of the pathway and is unlikely to control the entire pathway.

• In muscle cells PFK is the likely control point for the pathway. • PFK is a tetrameric enzyme with two conformational states (T and

R). ATP is both a substrate and allosteric inhibitor of PFK. PFK has two sites for ATP, a substrate site and an inhibitor site.

• AMP, ADP, and FBP reverse the effects of ATP and are activators of PFK.

• ATP binds the substrate site equally well in both sites but binds the inhibitor site almost exclusively in the T state.

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T and R states of PFK

• PRK is a tetrameric protein that has two ATP binding sites, one substrate and one inhibitor site.

• ATP can always bind the substrate site but will only bind the inhibitor state when in the T state.

• F6P preferentially binds PFK when its in the R state. • When there are large amount of ATP present in the cell, ATP binds

to the T state of the PFK and shifts the T/R equilibrium towards the T state and this causes a decrease affinity for F6P (similar to how 2,3-BPG lowers deoxyhemoglobins oxygen affinity.)

• An activator such as ADP or AMP counters the effects of ATP by binding to the R state which shifts the equilibrium to favor the R state.

• The enzyme is active in the R state and inactive in the T state.

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T and R states of PFK

• ATP may initially seem to be the major control factor for PFK, but this is not true. The metabolic flux through glycolysis may be 100 fold or more yet the intracellular concentration of ATP only varies by about 10% between rest and vigorous exertion.

• So what actually controls PFK?

• The inhibition of PFK by ATP is reversed by AMP and ADP.

• ATP concentration is buffered by creatine kinase and adenylate cyclase.

• In muscle, [ATP] is about 50 times greater than [AMP] and about 10 times greater than [ADP]. So minor changes in [ATP] have large effects on [AMP] and [ADP] due to adenylate cyclase action, which in turn will have large effects on the flux of glycolysis.

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Q: Draw a hyperbolic and sigmoidal curve.

Q: Show graphically what happens to PFK activity as ATP concentration varies. Show as F6P varies.

Q: Briefly explain the control of PFK in terms of its R and T states and indicate what compounds perform each function.

Q: Why is PFK a good control point for glycolysis?

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Substrate Cycling • Only near equilibrium reactions can undergo large changes in flux.

For reactions with a large ΔG that is not the case, since the rate of the reverse reaction is negligible.

• However, near equilibrium like conditions may be imposed on a reaction by having a forward reaction enzyme and a reverse reaction enzyme.

• The reverse reaction of PFK (hydrolysis of FBP to F6P) is catalyzed by Fructose-1,6-bisphosphatase (FBPase) a completely different enzyme.

• FBPase and PFK are controlled by factors that represent the current needs of the cell. And while there is an energetic cost to having these two enzymes working, the benefit is that the cell can rapidly change between glucose breakdown and glucose synthesis.

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Q: Use a picture to explain substrate cycling and explain why its useful to the cell?

Q: What are the costs to the cell to use a substrate cycle? (Hint: why was it once called a futile cycle?)

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Q: Define each of the following acronyms and draw the structure of everything that isn’t an enzyme.

1. PFK2. G6P3. F6P4. PGI5. FBP6. GAP7. DHAP8. TIM9. GAPDH10. 1,3-BPG11. NAD+/NADH

12. 2,3-BPG13. PGM14. 3PG15. 2PG16. PEP17. Pyruvate18. LDH19. Lactate20. TPP21. YADH22. LADH

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Q: Write the reaction catalyzed by each enzyme and classify each enzyme.

1. Aldolase

2. Enolase

3. Glyceraldehyde-3-phosphate dehydrogenase

4. Hexokinase

5. Phosphofructokinase

6. Phosphoglucose isomerase

7. Phosphoglycerate kinase

8. Phosphoglycerate mutase

9. Pyruvate kinase

10.Triose phosphate isomerase

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Q: If glucose was isotopically labeled at C1, where would the label be in pyruvate?

Q: If glucose was isotopically labeled at C5, where would the label be in PEP?

Q: If glucose was isotopically labeled at C3, where would the label be in pyruvate?

Q: If glucose was isotopically labeled at C6, where would the label be in PEP?

Q: If glucose was isotopically labeled at C4, where would the label be in pyruvate?

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Q: If the concentration of ATP increases, what effect will this have on the flux through glycolysis? Explain how this works.

Q: What would be the yield of ATP if G6P was the starting material for glycolysis?

Q: Write the net balanced chemical equation for glycolysis.

Q: In hypothetical situation, imagine that glycolysis was allowed to continue indefinitely with no other cellular processes occurring. What effect would this have on the pH?

Q: What is the biological function of fermentation? Reed

Q: Fill in the blanks: words only, no acronyms.

The first reaction of glycolysis is the transformation of _________ into _________, a reaction catalyzed by _________. G6P is then isomerized into _________, which is catalyzed by _________. F6P is then cleaved into _________ and _________by the enzyme _________. GAP and DHAP can then be interconverted by the enzyme _________. From here, GAP undergoes substrate level phosphorylation with concurrent _________ reduction that results in _________, this reaction is catalyzed by _________. _________ can then be used to synthesized ATP from ADP in a reaction catalyzed by _________. The resulting _________ is then isomerized into _________ by the enzyme _________. The resulting 2PG is then dehydrated into _________ in a reaction catalyzed by _________. PEP then donates its phosphate group to ADP to make ATP in a reaction that produces _________. This reaction is catalyzed by _________.

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Glycolysis summary

The first reaction of glycolysis is the transformation of glucose into glucose-6-phosphate, a reaction catalyzed by hexokinase. G6P is then isomerized into fructose-6-phosphate, which is catalyzed by phosphoglucose isomerase. F6P is then cleaved into glyceraldehyde-3-phosphate and dihydroxyacetone phosphate by the enzyme aldolase. GAP and DHAP can then be interconverted by the enzyme triose phosphate isomerase. From here, GAP undergoes substrate level phosphorylation with concurrent NAD+ reduction that results in 1,3-bisphosphoglycerate, this reaction is catalyzed by GAP dehydrogenase. 1,3-bisphospoglycerate can then be used to synthesized ATP from ADP in a reaction catalyzed by phosphoglycerate kinase. The resulting 3-phosphoglycerate is then isomerized into 2-phosphoglycerate by the enzyme phosphoglycerate mutase. The resulting 2PG is then dehydrated into phosphoenolpyruvate in a reaction catalyzed by enolase. PEP then donates its phosphate group to ADP to make ATP in a reaction that produces pyruvate. This reaction is catalyzed by pyruvate kinase.

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