Lectures 8 and 9 Bioenergetics and the Regulation

Lectures 8 and 9: Bioenergetics and the Regulation of Glycolysis Essential Cell Biology Fourth Edition Chapter 3 and 13

Energy 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

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: C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O This 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 = ΔG (“Gibbs Free Energy”)

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

Now consider glycolysis! C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O This reaction is highly favored because ΔG = - 686 Kilocal/Mole! (Note the minus sign!) First, ΔH is negative! WHY DO YOU THINK THIS IS? Second, ΔS 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 macromolecules (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 ΔG is positive! WHY DO YOU THINK THEIR ΔG IS POSITIVE? Example of an anabolic reaction: 100 AMINO ACIDS PROTEIN + H 2 O

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

High Energy Compounds Transfer Energy ATP ΔGO = - 7 Kilocalories /mole ADP

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 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]eq[B]eq = k(r) [C]eq[D]eq Thus, [C]eq[D]eq k(f)/k(r) = [A] [B] = Keq (THE EQUILIBRIUM eq eq CONSTANT) Also, it is known that K(eq) is related to ΔGO by the following equation: ΔGO = - RT ln Keq [C]eq[D]eq = - RT ln [A]eq[B]eq

Thus, when a highly favored reaction goes to equilibrium: ΔGO << 0, That is, DELTA G IS NEGATIVE, and [C]eq[D]eq >> 1 [A]eq[B]eq 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. .

Or 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 AEROBIC GLYCOLYSIS in mitochondria (9 STEPS + Electron Transport) ANAEROBIC GLYCOLYSIS LOOKS LIKE THIS: THIS IS ANAEROBIC GLYCOLYSIS

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

The 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 ΔG = ~ - 8 kcal/mole STEP 3: (phosphofructokinase) FRUCTOSE-6 -PHOSPHATE + ATP FRUCTOSE 1, 6 – BISPHOSPHATE +ADP ΔG = ~ - 5 cal/mole STEP 10: (pyruvate kinase) PHOSPHOENOLPYRUVATE + ADP PYRUVATE + ATP ΔG = ~ - 4 kcal/mole

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

Enzymes go through the catalysis cycle shown thousands of times each second: ALLOSTERIC SITE ATP 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. ACTIVE SITE SUBSTRATE How can the speed of these enzymes be regulated by allosteric control?

Technique: 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-MENTEN EQUATION: Velocity of the Reaction Vmax 1/2 Vmax KM Substrate Concentration

The two parameters KM and VMAX are characteristic Of each enzyme-substrate combination. KM 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. VMAX 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 phosphofructokinase when it is feedback inhibited by ADP or feedback activated by AMP: Velocity of Phosphofructokinase When ATP is low Low ATP (0. 1 m. M) we need to make DECREASED more ATP and KM run glycolysis faster. Decreasing KM the KM for the INCREASED substrate at the High ATP (1 m. M) active site does this. Fructose-6 -Phosphate [m. M] DO YOU KNOW WHY? Suppose [fructose-6 -phosphate] = 1 m. M.

Anaerobic glycolysis summary: 1. It occurs in the cytoplasm in 10 steps 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. 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.