Module 8

  1. State the class of each enzyme
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  2. Three enzymatic reaction are very exergonic, hence, thermodyncamically favorable reaction (Name them). These need to be REGULATED via _______ and ______ _______ mechanisms. Keq is GREATER than ____
    • Hexokinase (HXK), Phosphofructokinase 1 (PFK1) and Pyruvate kinase (PK) (all three are very exergonic)
    • allosteric and post-translational mechanisms
    • 1
  3. What about the REST of the non exergonic reactions? Additionally, what is the range of their ΔG, and what can be said of the rate of their forward reactions?
    • They are said to be at-equilibrium reactions
    • Single digit ΔG ranging from -6 to +6kJ/mol. for glycolysis
    • Rate of forward reaction = rate of reverse reaction
  4. How do you make the thermodynamically unfavorable reactions favorable? (Revisit pg 3 of the module)
    • a) Le Chatelier's Principle: On one side, you have incredibly high concentrations of molecule A, and on the other side you have an incredibly low concentration of molecule B. The system will always move away from A towards the direction of B, promoting the favor-ability of the reaction "all unfavorable are favorable"
    • b) "Because they are coupled": Take one unfavorable reaction and another that is VERY favorable, when combined we get a favorable reaction
    • **never used together always one or the other, waste of energy
  5. State the Keq and ΔG ranges and direction of reaction for the following reactions:
    • Exergonic reactions: Keq > 1 | ΔG ≤ -10 kJ/mol | to the right 
    • Equilibrium reactions: Keq ≥ 1 | ΔG -6 to +6 kJ/mol | in neither direction
    • Endergonic reactions: Keq < 1 | ΔG ≥ 10 kJ/mol | to the left
  6. Cells utilize diverse carbohydrate sources. However, ALL sugar molecules must convert to a singular sugar form, glucose. What is the ultimate fate of glucose?
    Glucose is ultimately oxidized to yield energy in a series of enzymatic steps
  7. Stage 1 of Glycolysis is the _______ step. What enzyme does the 1st reaction involve? What type of reaction is this?
    What is the story and formula?
    • Priming step
    • Hexokinase
    • This is a coupled reaction
    • Glycogen (broken down in the liver)
    • Glucose released into the bloodstream 
    • GLUT (Glucose Transporter) helps glucose enter cells
    • Hexokinase transfers phosphate groups, using ATP, unto glucose
    • Glucose + ATP → Glucose-6-phosphate + ADP +H+1 (equilibrium lies to the right)
  8. Define Kinase
    Kinase: an enzyme that catalyzes the transfer of a phosphate group from ATP to a specified molecule.
  9. Each phosphate group attached can be broken to release energy. Name the bond and rank each bond by energy released 
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    • Phosphodiester bond energy release
    • γ (30kJ) > β > α (10kJ)
  10. What is the purpose of the priming step? Why phosphorylate glucose? What is the advantage?
    • To phosphorylate glucose with hexokinase
    • Once phosphorylated, glucose is trapped inside the cell
  11. What are the activators and inhibitors of the 1st reaction
    • Acitivator: ADP, AMP
    • Inhibitor: Glucose-6-phosphate, ATP (competitive inhibitor)
  12. The six classes of enzymes
    • Hydrolase
    • Isomerase
    • Ligase
    • Lyase
    • Oxidoreductase
    • Transferase

    (Remember HILLOT)
  13. The hexokinase (transferase) reaction is a _______ REACTION. Hexokinase phosphorylates the ______ group on the 6th carbon of glucose. Draw Glucose getting phosphorylated
    • hydroxyl group
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  14. Explain Coupling strategy number 1 for glucose phosphorylation
    • Glucose + Pi ↔ glucose 6-phosphate + H2O ΔG = +13.8 kJ/mol
    • ATP + H2O ↔ ADP + Pi ΔG =-30.5 kJ/mol
    • Glucose + ATP ↔ glucose 6-phosphate + ADP ΔG = -16.7kJ/mol

    This is ATP hydrolysis with the gama phosphate
  15. The structure of (Acyl) CoA is DIFFERENT from ATP and has a tell tale reactive thiol group that will be pivotal in the biosynthesis of coenzyme A (CoA). In ATP hydrolysis we do _______ bond hydrolysis, however, in Acyl bond hydrolysis, we do ________ bond hydrolysis.
    • phosphodiester bond hydrolysis
    • THIOESTER bond hydrolysis
  16. Explain coupling strategy #2 Acyl-Coenzyme A (CoA) bond hydrolysis
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    • 1) As the Coenzyme A begins to accept an alkyl/acyl group, an acetyl group is formed with a HIGH ENERGY THIOESTER bond 
    • 2) The idea is to couple unfavorable reactions with the breaking of about 1 mol of the THIOESTER bonds (Acyl CoA hydrolysis) which releases a lot of energy, making the reaction favorable

    **Hydrolysis meaning we have a chemical breakdown of a molecule due to a reaction with water
  17. State an example of the Acyl-coenzyme A bond hydrolysis
    • The addition of myristoyl or palmitoyl group on to proteins is actually unfavorable
    • However, upon the hydrolysis of a myristoyl-CoA or palmitoyl-CoA, a lot of energy is released and coupled to protein addition reaction, making it favorable
  18. Not all hexokinase isoforms have the same level of enzymatic behavior. Hexokinase I has a _____ Km value for glucose and is active when there are _____ levels of glucose. Hexokinase IV has a _____ Km value for glucose. It is present in ______ tissue and active ONLY when there are _____ levels of glucose. What is the difference in affinity?
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    • low 
    • LOW
    • high 
    • hepatic (liver)
    • HIGH
    • Hexokinase IV has 100 times lower affinity for glucose
  19. Most kinases use ATP as a ______ _______ _____
    **Revisit slide 7
    phosphate donor cofactor
  20. What is happening here?
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    • Reaction number 2 in Glycolysis
    • Isomerization of Glucose-6-phosphate into Fructose-6-phosphate 
    • Later cleavage generates two 3-carbon molecules
    • Glucose-6-phosphate (cyclical) → opens to non-cyclical form → isomerize →cyclical fructose
  21. Reaction 3 involves Phosphofructokinase phosphorylating ________. This is a very _______ reaction. What is the purpose of dual phosphorylation (2-story)? State the activators and inhbitors
    NT = Not going to be tested
    • fructose
    • favorable
    • (1) To generate Fructose-1-6-bisphosphate which can be isomerized into Fructose-2,6-bisphosphate (F-2,6-BP) by PFK-2. 
    • (2) F-2,6-BP increases glycolytic activity by activating PFK-1. 
    • Activators: AMP, ADP and Fructose-2-6-bisphosphate (NT: also Insulin)
    • Inhibitor: ATP (NT: also High citrate concentration)
  22. Which molecule contains more energy:
  23. As it pertains to the phosphorylation of Fructose-6-phosphate, PFK1 activity is _____ with low intracellular ATP levels, and _____ with high intracelluar ATP levels.
    • higher
    • lower
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    • ***The sigmoidal curve should have you thinking higher Km and multimeric, while the hyperbolic curve should have you thinking lower Km and monomeric scavenger
  24. _______ function of two mammalian hormones controls glycolytic flux through PFK-2. What is the general function of each domain and which portion is the PFK-2 and which portion is the FBPase-2 
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    • Antagonistic
    • The kinase portion adds a phosphate group to a molecule, while a phosphotase removes a phosphate group from a molecule
    • The 32-250 domain is the PFK-2 portion and the 250-470 domain is the FBPase-2 portion
  25. What are the 2 pancreatic endocrine signaling cells of interest and at which glucose levels do they intervene
    • α cell aka glucagon: intervenes when blood sugar/glucose levels are too low
    • β cell aka insulin: intervenes when blood sugar/glucose levels are too high
  26. Glycolysis is ______ active with Fructose-6-phosphate and ______ active with Fructose-2,6-bisphosphate (Why?)
    • less
    • more
    • PFK-1 is activated
  27. How do we transition from low blood sugar to higher blood sugar? (5-story)
    • Pancreatic endocrine signaling directs increase in glucagon concentrations 
    • Glucagon (agonist and peptide hormone) binds to a glucagon receptor, activating Protein Kinase A (PKA)
    • Protein kinase A consumes an ATP resulting in a phosphorylated (PFK-2)-FBPase-2 complex at its Protein kinase A recognition motif (RRXS/T) (**A kinase-kinase reaction)
    • This phosphorylation activates the FBPase-2 portion of the complex 
    • FBPase-2 then converts Fructose-2,6-bisphosphate into Fructose-6-phosphate, increasing the activity of gluconeogenesis (making of glucose) and the overall glucose content of the blood
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  28. Where is the Protein kinase A recognition motif (state the motif sequence) and which residues are targeted
    • In the PFK-2 portion of the PFK-2-FBPase-2 complex, there is usually an RRXS/T motif
    • Most of the time Serine is targeted, however Threonine can be targeted as well
  29. How do we transition from high blood sugar to low blood sugar? (5-story)
    • Pancreatic endocrine signaling directs increase in insulin concentrations
    • Insulin (antagoinst) binds to an insulin receptor activating protein phosphotase-1
    • Protein phosphotase-1 dephosphorylates the PFK-2 on the (PFK-2)-FBPase-2 complex resulting in an active PFK-2
    • PFK-2 proceeds to phosphorylate Fructose-6-phosphate converting it into Fructose-2,6-bisphosphate
    • Fructose-2,6-bisphosphate is an allosteric activator of PFK-1 which signals an increase in glycolysis activity and a decrease glucose content in the blood
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  30. What exactly does Fructose-2,6-bisphosphate do to the enzymatic behavior of PFK1? How?
    • Fructose-2,6-bisphosphate activates PFK1
    • It is a potent positive allosteric effector/regulator of PFK1 activity
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  31. F-2,6-BP is an ISOMER of _______, which is a product of _____. Draw Fructose 2,6-bisphosphate
    • F-1,6-BP 
    • PFK1
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  32. What happens to the product of PFK1 after allosteric activation by F-2,6-BP? Specify how this affects glycolysis
    • The highly charged F-1,6-BP molecule is opened and cleaved by aldolase and the Glyceraldehyde-3-phosphate (GAP or G3P) can go through glycolysis
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  33. This is an overview of the mechanism aldolase uses to cleave F-1,6-BP. What types of catalytic mechanisms are being utilized by aldolase? 
    Account for: Lys-229 and Asp-33 in both Schiff base formation and hydrolysis
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    • Schiff base formation: Lys-229 is covalent catalysis and Asp-33 is using general base
    • Schiff hydrolysis: Lys-229 is general base (it picked up a proton) and Asp-33 is general acid
  34. Schiff bases are also called ______
    • imines 
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  35. What happens after Aldolase generates GAP and DHAP? Which direction does equilibrium lie in? Why is that direction beneficial? What is the enzyme responsible for making that benefit possible
    • Isomerization of DHAP to GAP
    • Equilibrium lies to the left: ~96% DHAP and ~4% GAP
    • This direction is beneficial because it invokes Le Chatelier's principle, the tendency for larger concentrations of one isomer moving in the direction of the isomer in a smaller concentration
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  36. After triosephosphate isomerase moves the equilibrium converts DHAP, how many GAP molecules are present?
  37. Phosphoglucoisomerase converts _______ to _______ while Triosephosphate isomerase converts _______ to _______
    • aldose to ketose
    • ketose to aldose
  38. Draw the mechanism:
    DHAP to GAP
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  39. What happens after we get both GAP molecules? Name the enzyme involved
    • 2 GAP molecules get oxidized by glyceraldehyde-3-dehydrogenase with the help of an inorganic phosphate group and 2 NAD+
    • The result is 2 molecules of 1,3-Bisphophoglycerate
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  40. What does NAD stand for and what is its function? State which of the following is the reduced NADH and which is the oxidized NAD+
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    • NAD: Nicotinamide Adenine Dinucleotide, a cofactor in reduction-oxidation (redox) reactions, in this case GAPDH (DeHydrogenase)
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  41. Draw the 5 stage mechanism:
    GAPDH (dehydrogenase) takes GAP to 1,3 BPG
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  42. What is the role of NAD in GAPDH?
    • 2 NAD molecules, our GAPDH cofactor, carry electrons in form of their hydrogens (1 hydrogen = 1 electron)
    • When in the mitochondria, the NADHs (mitochondrial) activate complex 1
    • Complex 1, a mitochondrial protein complex, pumps protons from the mitochondrial matrix to the intermembrane space
    • The protons are used to generate ATP at a consistent rate, so as a result, each REDUCED NADH is the equivalent of 2.5-3 ATPs
    • We have 2 NADHs so that means 5-6 ATPs
  43. What are the positive and negative effectors of GAPDH
    • NADH and ATP reduce the affinity for the cosubstrate
    • AMP and ADP increase affinity
  44. How are NADH equivalents native to the cytosol, converted into mitochondrial NADH equivalents? (6-story) What is the strategy called?
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    • The Malate-Aspartate Shuttle strategy
    • NADH can't go across the membrane so Malate dehydrogenase converts Oxaloacetate (reduced) into Malate and the reaction is coupled with NADH getting oxidized into cytosolic NAD+
    • Malate can go into the matrix but cannot exit, so it activates Oxaloacetate by reducing NAD+ of the matrix to NADH of the matrix
    • Oxaloacetate is then converted into Aspartate by a transaminase (matrix)
    • Aspartate is capable of leaving the matrix and enters cytosol
    • Once in the matrix, aspartate can undergo a reverse reaction with transaminase to generate a cytosolic oxaloacetate which can be converted into cytosolic Malate again by Malate dehydrogenase restarting the process
  45. What happens after our two 1,3 BPG molecules are generated? What is the importance of this reaction?
    • Both 1,3-BPG with ADP and H+ are converted into two 3-phosphoglycerate molecules and two ATP molecules
    • The importance is that ATP is finally generated (substrate level phosphorylation)
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  46. What are the last 3 reaction after both of our 3-PGs are formed? What is the Net ATP gained?
    • 1) They are converted into 2-PG by phosphoglycerate mutase
    • 2) The 2-PGs are converted into phosphoenolpyruvate by enolase
    • 3) Both phosphoenolpyruvates are converted first into pyruvate (enol form) by pyruvate kinase (cpld with ATP formation) and eventually pyruvate be resonance
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    4 ATP made - 2 ATP used = 2 ATP
  47. What are the steric consequences of going from 3-PG to 2-PG?
    • Less steric repulsion to more steric repulsion
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  48. What happens to pyruvate after it is generated as the last metabolic intermediate in glycolysis? (4-story)
    • In Eukaryotes, pyruvate is translocated into the mitochondria using a translocase.
    • Pyruvate is decarboxylated (removal of CO2) using a multienzyme complex called pyruvate dehydrogenase
    • After decarboxylation, pyruvate forms acetyl
    • Acetyl is combined with CoA to form acetyl-CoA
  49. How does regulation of pyruvate kinase get us to a LOW blood glucose level? How can it take us to HIGH blood glucose levels? What are the targets, the transition and results?
    • Pyruvate kinase is phosphorylated by pyruvate kinase kinase and therefore less active allowing for a low blood glucose levels
    • Pyruvate kinase is dephosphorylated by Pyruvate kinase phosphotase which makes more active allowing for high blood glucose levels
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    • The targets are serine, threonine and tyrosine and they transition from:
    • (ROH) → (RO-) → (ROPO32-)
    • Which results in:
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  50. In the pyruvate dehydrogenase complex and its regulation cycle, what are E1, E2, & E3 attached to?
    • E1 to TPP (thiamine pyrophosphate
    • E2 to Lipoamide
    • E3 FAD (oxidized) flavin adenine dinucleotide FADH2 when reduced
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  51. In one round of successful PDH complex reaction cycle we generate _____ _______ ______ equivalents
    two NADH reducing equivalents
  52. T or F PDH does not go through phosphorylation and dephosphorylation
    False PDH goes through both
  53. PDH is _______ when active and ______ when inactive
    • dephosphorylated
    • phosphorylated
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  54. Name and label the following structures
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    (cut of single bond goes to H and double bond to O)

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    • E1PDH
  55. Lipoamide is a ________ linked cofactor to a ______ side chain of ____ of PDH
    • covalently
    • lysine
    • E2
Card Set
Module 8
Module 8