Lectures 8and 9: Bioenergeticsandthe RegulationofGlycolysis Essential CellBiology FourthEdition

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Lectures 8 and 9: Bioenergeticsandthe Regulationof. Glycolysis Essential Cell. Biology Fourth. Edition Chapter 3Lectures 8 and 9: Bioenergeticsandthe Regulationof. Glycolysis Essential Cell. Biology Fourth. Edition Chapter 3 and

Energy Flow in Biota: Light energy from the sun is stored as Chemical energyEnergy Flow in Biota: Light energy from the sun is stored as Chemical energy in sugars and biomass By photosynthesis. In animals and other Organisms this chemical energy is released By respiration- a slow form of oxidation

This reaction yields energy due to breaking of chemical bonds.  The energy isThis reaction yields energy due to breaking of chemical bonds. The energy is expressed as a change in “Gibbs Free Energy: ENERGY OF REACTANTS – ENERGY OF PRODUCTS = CHANGE IN ENERGY = ΔΔ G G (“Gibbs Free Energy”)Glycolysis in all organisms breaks down sugars. How do chemical reactions allow the harvesting of energy stored in this chemical? The overall reaction in the breakdown of glucose is: CC 66 HH 1212 OO 66 + 6 O 22 6 CO 22 + 6 H 22 OO

Let’s look at the components of this change in free energy during a reactionLet’s look at the components of this change in free energy during a reaction by studying the following equation: ΔΔ G = ΔΔ H – T ΔΔ SS WHERE: FREE ENTHAPY ENTROPY ENERGY TEMPERATURE A CHANGE IN ENTHAPY IS REPRESENTED BY ΔΔ HH. . It represents a change in BOND ENERGY. . A CHANGE IN ENTROPY IS REPRESENTED BY ΔΔ SS. . It represents the degree of DISORDER. A greater number of of reaction products means greater disorder. The physical universe favors disorder in everything it does!

Now consider glycolysis! CC 66 HH 1212 OO 66  + 6 O 22Now consider glycolysis! CC 66 HH 1212 OO 66 + 6 O 22 6 CO 22 + 6 H 22 O O This reaction is highly favored because ΔG = — 686 Kilocal/Mole! (Note the minus sign!) First, ΔΔ H is negative ! WHY DO YOU THINK THIS IS? Second, ΔΔ S is positive ! WHY DO YOU THINK THIS IS? Highly favored reactions like this are called CATABOLIC. They release energy that can then be used for synthesis of new macromolecules by a cell.

Reactions in cells that make new macro- molecules (proteins, DNA, RNA, etc. ) areReactions in cells that make new macro- molecules (proteins, DNA, RNA, etc. ) are generally unfavorable These reactions are called ANABOLIC or SYNTHETIC. . They require an input of free energy to make them happen (go forward). Their ΔΔ G is positive! WHY DO YOU THINK THEIR ΔΔ G IS POSITIVE? Example of an anabolic reaction: 100 AMINO ACIDS PROTEIN + H 22 OO

How do we transfer the energy yield from highly favorable CATABOLIC reactions to unfavorableHow do we transfer the energy yield from highly favorable CATABOLIC reactions to unfavorable synthetic/Anabolic reactions that make macromolecules the cell needs?

ATP ADPΔG O = - 7 Kilocalories /mole. High Energy Compounds Transfer Energy ATP ADPΔG O = — 7 Kilocalories /mole. High Energy Compounds Transfer Energy

Polymerization of RNA requires a high energy intermediate formed by ATP hydrolysis Polymerization of RNA requires a high energy intermediate formed by ATP hydrolysis

A consequence of a chemical reaction being favorable is that it goes forward toA consequence of a chemical reaction being favorable is that it goes forward to form products. That is, the forward rate constant is greater than the reverse rate constant. CONSIDER THE GENERIC REVERSABLE REACTION: k (forward) A + B C + D k (reverse) Rate of forward reaction = k(f) [A] [B] Rate of reverse reaction = k(r) [C] [D] At equilibrium the reverse rate is equal to the forward rate; there is no further change.

The equilibrium state is described by the EQUILIBRIUM CONSTANT. At equilibrium,  k(f) [A]The equilibrium state is described by the EQUILIBRIUM CONSTANT. At equilibrium, k(f) [A] eqeq [B][B] eqeq = k(r) [C] eqeq [D][D] eqeq Thus, k(f)/k(r) = K eqeq (THE EQUILIBRIUM CONSTANT) Also, it is known that K( eqeq ) is related to ΔΔ GG OO by the following equation: ΔGΔG O O = — RT ln K eqeq = — RT ln [C] eq [D] eq [A] eq [B] eq

Thus, when a highly favored reaction goes to equilibrium: ΔGΔG OO   0,Thus, when a highly favored reaction goes to equilibrium: ΔGΔG OO <> 1 That is, lots of products have accumulated! Next, consider a metabolic pathway in which reactions are “coupled”, that is, the product of one is the reactant of the next. . [C] eq [D] eq [A] eq [B] eq

Or the simplest example: Assume that the second reaction is highly favored but theOr the simplest example: Assume that the second reaction is highly favored but the first reaction is not. Reaction 2 will go forward to produce lots of products! Product C will accumulate but reactant B will disappear. REACTION 1 WILL BE “PULLED” FORWARD! WHY? ?

Glycolysis is broken up into ANAEROBIC GLYCOLYSIS in the cytoplasm (10 STEPS) and AEROBICGlycolysis is broken up into ANAEROBIC GLYCOLYSIS in the cytoplasm (10 STEPS) and AEROBIC GLYCOLYSIS in mitochondria (9 STEPS + Electron Transport) ANAEROBIC GLYCOLYSIS LOOKS LIKE THIS: THIS IS ANAEROBIC GLYCOLYSIS

The highly favored reactions are steps 1, 3 and 10. These reactions, by havingThe highly favored reactions are steps 1, 3 and 10. These reactions, by having a large change in free energy, “pull” all the other steps forward. Thus, it is these steps that determine the speed of glycolysis. It makes sense to regulate these steps. How is that done? THE THREE BIG CHANGES IN FREE ENERGY DURING ANAEROBIC GLYCOLYSIS

The highly energetically favorable steps in anaerobic glycolysis are shown below.  They areThe highly energetically favorable steps in anaerobic glycolysis are shown below. They are regulated by allosteric binding of metabolites STEP 1 : : (hexokinase) GLUCOSE + ATP GLUCOSE 6 – PHOSPHATE + ADP ΔΔ G = ~ — 8 kcal/mole STEP 3: (phosphofructokinase) FRUCTOSE-6 -PHOSPHATE + ATP FRUCTOSE 1, 6 – BISPHOSPHATE +ADP ΔΔ G = ~ — 5 cal/mole STEP 10: (pyruvate kinase) PHOSPHOENOLPYRUVATE + ADP PYRUVATE + ATP ΔΔ G = ~ — 4 kcal/mole

ATP and ADP are important allosteric regulators!Speed of anaerobic glycolysis is regulated by feedbackATP and ADP are important allosteric regulators!Speed of anaerobic glycolysis is regulated by feedback (red) and feed forward (green) allosteric control. * * ** = steps 1, 3, and 10 which are highly regulated

Enzymes go through the catalysis cycle shown thousands of times each second:  TheEnzymes go through the catalysis cycle shown thousands of times each second: The speed at which an enzyme can do this is dependent on the concentration of substrate, how tightly it binds the substrate, and how fast the enzyme works. ALLOSTERIC SITE ACTIVE SITE ATP How can the speed of these enzymes be regulated by allosteric control? SUBSTRAT

Technique: Measuring Enzyme Reaction Velocity:  One can do an experiment to measure theTechnique: Measuring Enzyme Reaction Velocity: One can do an experiment to measure the speed (velocity) of an enzyme reaction and find how it depends on substrate concentration: The experiment requires a series of incubations each with the same amount of enzyme but different substrate concentrations in each one. Then the rate at which product is formed in each Incubation is measured.

The velocity data from this experiment conform to a parabola described by the MICHAELIS-MENTENThe velocity data from this experiment conform to a parabola described by the MICHAELIS-MENTEN EQUATION : Substrate Concentration. V e lo c ity o f th e R e a c tio n. Vmax 1/2 Vmax K M

The two parameters KK MM and VV MAXMAX are characteristic Of each enzyme-substrate combination.The two parameters KK MM and VV MAXMAX are characteristic Of each enzyme-substrate combination. KK MM is the substrate concentration required to obtain a half maximal velocity and is related to how tightly the enzyme binds the substrate at its active site. VV MAXMAX is the maximal rate that the enzyme can work at and is related to how many enzyme proteins are present and how fast each one works. Allosteric regulators can change these parameters!

Here is what happens to the enzyme phospho- fructokinase when it is feedback inhibitedHere is what happens to the enzyme phospho- fructokinase when it is feedback inhibited by ADP or feedback activated by AMP: When ATP is low we need to make more ATP and run glycolysis faster. Decreasing the K M for the substrate at the active site does this. DO YOU KNOW WHY? Suppose [fructose-6 -phosphate] = 1 m. M. High ATP (1 m. M) Low ATP (0. 1 m. M) Fructose-6 -Phosphate [m. M]V e lo city o f P h o sp h o fru cto kin ase K M INCREASEDDECREASED K M

Anaerobic glycolysis summary: 1. 1. It occurs in the cytoplasm in 10 steps 2.Anaerobic glycolysis summary: 1. 1. It occurs in the cytoplasm in 10 steps 2. 2. Steps 1, 3 and 10 are most favorable and pull the pathway forward. They are the steps that are most important to regulate. 3. Regulation is by ALLOSTERIC CONTROL. 4. 4. Glycolysis requires the input of ATP in early steps in order to yield more ATP at later steps. 5. Overall, the yield of ATP is very modest. Only 4 ATP molecules per molecule of glucose processed.

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