lecture 13

.pdf

School

University of California, Davis *

*We aren’t endorsed by this school

Course

BIS105

Subject

Biology

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by CommodoreLightningBarracuda37

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?

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

lecture 13

Browse Popular Homework Q&A