lecture 13 .105 Glycolysis

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Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 1 Catabolism overview: Catabolism of glucose: C 6 H 2 O 6 + 6 O 2  6 CO 2 + 6 H 2 O ΔG°’ = -2,840 kJ/mol There is a lot of energy in glucose. If you just burned it, a lot of energy would be lost to the environment as heat. Metabolism is a slow controlled burn, so the energy liberated from glucose can be captured and put to use in the cell. GLYCOLYSIS is the first pathway wherein glucose energy is captured. Two phases: A priming or prepatory phase (Phase I) and a payoff phase (Phase II).

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 2 In the priming phase, energy is spent (2 ATP) converting glucose to a form that allows breakage of a C-C bond. In the payoff phase, make 4 ATP and 2 energy-rich pyruvate molecules. Also make 2 NADH + H + that can be used later if living in an O 2 containing environment. We will now go through each individual reaction. The molecular logic that applies here applies to everything in metabolism. PHASE I (5 reactions) Reaction 1. Glucose + ATP  G-6-P + ADP

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 3 ΔG°’ = -16.7 kJ/mol Glucose + Pi  G-6-P ΔG°’ = +13.8 kJ/mol ATP  ADP + Pi ΔG°’ = -30.5 kJ/mol Glucose + ATP  G-6-P + ADP ΔG°’ = -16.7 kJ/mol So equilibrium lies far to the right. Keq  700. Recall ΔG = ΔG°’ + RT ln([products] i /[reactants] i ) From Table 18.2, [glucose] = 5 mM [G-6-P] = 0.083 mM [ATP] = 1.85 mM [ADP] = 0.14 mM Put these in, we have ΔG = -33.9 kJ/mol So the reaction is even more favorable under cellular conditions than under standard state conditions. What does this step accomplish? That is, what is the cellular logic?

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Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 4 1. Glucose is carried into cells efficiently because it is uncharged. Phosphorylating it gives it a negative charge. Now it cannot cross the membrane as easily. 2. This is a site of regulation of glycolysis. The reaction is far from equilibrium. Cannot impose much regulation on reactions that operate close to equilibrium. A large ΔG means far from equilibrium, means regulation possible. Many (most?) reactions that we will talk about that are far from equilibrium are the steps that are highly regulated. Hexokinase: 1. A note about ATP. The real substrate for hexokinase, and for many ATP utilizing enzymes, is Mg-ATP. So without Mg 2+ the reaction does not go well. 2. Hexokinase is allosterically inhibited by G-6-P. What effect does this have on the reaction? As G-6-P rises, the activity of hexokinase goes down. This pulls [G-6-P] down, allowing hexokinase activity to rise again. 3. K M  0.1 mM. Normal blood levels of glucose are near 5 mM. So the enzyme operates efficiently. Glucokinase, and alternate enzyme for this reaction: 1. In liver 2. Much higher K M ,  10 mM KM. So this enzyme is only important at high [glucose].

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 5 Reaction 2. G-6-P  F-6-P This reaction is important for two reasons: 1. Converts C1 to an alcohol, which is required in the next step for phosphorylation. 2. Places a carbonyl at C2, which is required for cleavage at C3-C4 later on. Reaction operates near equilibrium, readily reversible. Reaction 3 – the second phosphorylation: F-6-P + ATP  FBP + ADP

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 6 Kinase is an enzyme that phosphorylates something. BIS means two phosphate groups attached to different places on the molecule. DI, TRI refers to phosphate groups strung together, as in ATP. F-6-P + Pi  FBP ΔG°’ = +16.3 kJ/mol When coupled to ATP hydrolysis, F-6-P + ATP  FBP + ADP ΔG°’ = -14.2 kJ/mol ΔG = -18.8 kJ/mol In erythrocytes, ΔG = -18.8 kJ/mol. So equilibrium is far to the right. This commits the cell to metabolizing glucose via glycolysis, so it is important in that regard. Also because the reaction is far from equilibrium, has the potential for regulation. In fact, it is highly regulated. 1. ATP is an allosteric inhibitor, so enzyme shuts off at high [ATP]. 2. AMP reverses inhibition by ATP. ADP + ADP  AMP + ATP Typically [ATP] = 10x [ADP] = 100x [AMP]. So a relatively small percentage change in [ATP] produces a large percentage change in [AMP]. This is felt by the enzyme. Reaction 4. FBP  DHAP + G-3-P ΔG°’ = 23.9 kJ/mol, hard to break the covalent bond. ΔG = -0.23 kJ/mol, essentially at equilibrium.

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Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 7 Reaction 5. DHAP  G-3-P ΔG°’ = +7.56 kJ/mol ΔG = +2.41 kJ/mol While this last step is endergonic, the entire reaction sequence 1 – 5 is highly exergonic, with ΔG = -53 kJ/mol, THIS IS THE END OF THE PREPARATORY PHASE Have now spent 2 ATP. 1 glucose + 2 ATP  2 G-3-P + 2 ADP PHASE II (Pay-off phase, 5 reactions.) Here we will realize a net gain of 2 ATP per each G-3-P

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 8 Reaction 6. G-3-P + NAD + + Pi  BPG + NADH + H + ΔG°’ = +6.3 kJ/mol; ΔG = -1.29 kJ/mol This is an oxidation of G-3-P to 1,3-BPG. Neither ATP nor ADP as the P i donor. Although the oxidation would be exergonic, formation of the high energy BPG and reduction of NAD+ make it endergonic. Close to equilibrium in cells. During oxidation of G-3-P, substrate is temporarily bound to a cysteine on the enzyme. Iodoacetic acid can bind to this cys and inactivate the enzyme. So, it’s inhibited by Iodoacetic Acid Reaction 7. BPG + ADP  3-PG + ATP ΔG°’ = -18.9 kJ/mol ΔG = +0.1 kJ/mol

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 9 Can think of this as energetically coupled to reaction 6. Then G-3-P + ADP + Pi + NAD+  3-PG + ATP + NADH+ + H + ΔG°’ = -12.6 kJ/mol This is an example of substrate level phosphorylation. Enzyme named for reverse reaction. Reaction 8. 3-PG  2-PG ΔG°’ = 4.4 kJ/mol ΔG = +0.83 kJ/mol Operates near equilibrium. A mutase is an enzyme that shifts a group from one position to another within a molecule. Reaction 9. 2-PG  PEP + H 2 O

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Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 10 ΔG°’ = +1.8 kJ/mol ΔG = +1.1 kJ/mol This is a dehydration. 2-PG has a relatively low free energy of hydrolysis, whereas PEP has a high free energy of hydrolysis. How to understand? Both have about the same potential metabolic energy with respect to complete decomposition to Pi, CO 2 and H 2 O. Enolase just rearranges 2-PG such that more of this energy is released during the hydrolysis (phosphate release) reaction. Reaction 10 PEP + ADP  pyruvate + ATP ΔG°’ = -31.4 kJ/mol ΔG = -23 kJ/mol Second substrate level phosphorylation. Here is the net gain for glycolysis. Site of regulation. Activated by AMP, F-1,6-BP; inhibited by ATP, acetyl CoA. This is the end of glycolysis. The total glycolysis reaction: Glucose + 2ADP + 2NAD +  2 pyruvate + 2ATP + 2NADH + 2H +

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 11 What are the fates of NADH and pyruvate made in glycolysis? NAD + is needed for oxidation of G-3-P, so it must be regenerated. 1. Under aerobic conditions, NAD + is regenerated during the operation of the electron transport chain in mitochondria. More on this later. 2. Under anaerobic conditions in muscle, pyruvate is reduced to lactate at the expense of NADH + H + , which regenerates NAD + . ΔG°’ = -25.2 kJ/mol ΔG = -14.8 kJ/mol Here, glucose  2 lactate ΔG°’ = -196 kJ/mol At the same time 2 ATPs are made, at 2x 30.5 kJ/mol, So energy captured / energy input = 61 / 196 = 0.31 or 31% efficiency 3. Under anaerobic conditions in yeast, pyruvate is reduced to EtOH at the expense of NADH ΔG°’ = -44.5* Beer and wine have EtOH and CO 2 made by yeast. CO 2 made by yeast causes bread to rise.

Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 12 Here, glucose  2EtOH + 2CO 2 ΔG°’ = -235 kJ/mol Efficiency of energy capture is (61 / 235) x 100 = 26% REGULATION Looking through your notes see that the steps with the large –ΔGs are #s 1 , 3, & 10 (and 11). All others are pretty close to equilibrium. ENERGETICS SUMMARY glucose + 6 O 2  6CO 2 + 6H 2 O ΔG°’ = -2840 kJ/mol glucose  2 pyruvate ΔG°’ = -146 kJ/mol (146 / 2840) x 100 = 5.2% of the energy available in glucose has been released in the formation of pyruvate. There is still a lot of energy available in pyruvate. So it makes sense that cells have evolved pathways to capture it. OTHER PATHWAYS INTO GLYCOLYSIS Just note that there are other ways to make some of the glycolytic intermediates, so you don’t have to start with glucose. In particular you can make G-6-P and F-6-P by other reactions. At the end of glycolysis we have made 2 molecules of pyruvate from glucose.

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Biological Sciences 105 Lecture 13, November 19, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 13 We had just finished looking at Glycolysis when we stopped last time ENERGETICS SUMMARY FOR GLYCOLYSIS glucose + 6 O 2  6CO 2 + 6H 2 O ΔG°’ = -2840 kJ/mol glucose  2 pyruvate ΔG°’ = -146 kJ/mol (146 / 2840) x 100 = 5.2% of the energy available in glucose has been released in the formation of pyruvate. There is still a lot of energy available in pyruvate. So it makes sense that cells have evolved pathways to capture it. OTHER PATHWAYS INTO GLYCOLYSIS Just note that there are other ways to make some of the glycolytic intermediates, so you don’t have to start with glucose. In particular you can make G-6-P and F-6-P by other reactions. At the end of glycolysis we have made 2 molecules of pyruvate from glucose.