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Bioenergetics and Metabolism: Energy Input Restrains Entropy Metabolic Flux is Highly-Regulated . Bioc 460 Spring 2008 - Lecture 24 (Miesfeld). Life on earth requires energy from the sun. The flux of people through the DC metro is similar to the flux of metabolites in cells.
Bioenergetics and Metabolism:Energy Input Restrains Entropy Metabolic Flux is Highly-Regulated
Bioc 460 Spring 2008 - Lecture 24 (Miesfeld)
Life on earth requires
energy from the sun
The flux of people through the DC metro is similar to the flux of metabolites in cells
Entropy wins every time
Energy is required to keep the level of glucose higher inside the saguaro than outside in the desert. Similarly, energy is required to keep the level of NaCl lower inside the whale than in the sea water.
What is the source of energy for the saguaro, what about the whale?
Photosynthetic autotrophs use solar energy to oxidize water and generate chemical energy that us used to convert CO2 into carbohydrate (C6H12O2).
Heterotrophs are dependent on photosynthetic autotrophs, or other heterotrophs, to provide the chemical energy needed for life.
The byproducts of aerobic respiration are CO2 and water.
• Oxidation Is Loss of electrons and
Reduction Is Gain of electrons (OIL RIG).
Nicotinamide adenine dinucleotide (NAD), flavin adenine dinucleotide (FAD), and quinone (Q) are electron carriers in numerous biochemical reactions.
• First Law of Thermodynamics:
The total amount of energy in the universe does not change; energy can neither be created or destroyed. Energy conversion in biological systems is never 100%.
• Second Law of Thermodynamics:
In the absence of energy input, all natural processes in the universe tend toward disorder, the measure of this disorder, or randomness, is called entropy (S).
• Gibbs Free Energy (G) and Equilibrium Constant (Keq):
In the reaction A-->B, ΔG = ΔGº’+ RT • ln [B]equilibrium [A]equilibrium
Energy cannot be created or destroyed – it can only be converted between forms.
Antoine Laurent Lavoisier and the guinea pig experiment, 1783. The amount of water collected from melting ice was a measure of metabolic heat produced by the guinea pig.
The energy available for oxidation of a reduced compound in a biological system can be determined precisely using a device called a bomb calorimeter. A bomb calorimeter measures the change in temperature that occurs when a compound is completely oxidized. For example, the amount of stored energy in glucose.
C6H12O6 + 6 O2→ 6 CO2 + 6 H2O + heat
Total energy potential of 1 gram of glucose is the same regardless of the metabolic path taken, i.e., 15.7 kJ of energy. We can use this experimental value to calculate metabolic energy available from glucose oxidation. Note that the total amount of energy available does not tell you if glucose oxidation is favorable or not, for that, you need to know the change in Gibbs Free Energy (G).
All natural processes in the Universe tend towards disorder, the measure of which is entropy (S). When S>0, then randomness is increasing.
Energy is required to restrain the natural tendency toward disorder, i.e., to keep S from increasing in a system, energy must be provided.
An increase in entropy happens when a cell dies and its contents equilibrate with the environment. To stay alive, one must maintain order and prevent the system (cell) from reaching equilibrium with the environment.
The H2O molecules become more and more disordered as heat is added to the system which increases the entropy.
Ice melting at room temperature is spontaneous however, this process can be restrained by keeping the ice in a freezer that is kept below 0º C using the energy of electricity.
Since the absolute values for enthalpy (H) and entropy (S) cannot be easily determined, the Gibbs Free Energy term (G) is defined as the change between two states:
G = H - TS
At equilibrium, under constant temperature and pressure, the Keq is also a measure of reaction directionality since a Keq > 0 means product formation is favored, whereas, a Keq < 0 means that reactant formation is favored.
A + B C + D Keq = [C] [D]
Gibbs defined the relationship between G and Keq using a term called the standard free energy change, Gº (kJ/mol), in terms of the gas constant (R) and absolute temperature (T):
G = Gº + RT ln Keq
The Gº’ value (used by biochemists) is experimentally determined by setting up the reaction under standard physiological conditions (pH 7, 55.5 M H2O, 298ºK, 1 atmosphere pressure, and 1 M of each component), in the presence of enzyme, and letting the reaction go to equilibrium. Once the reaction has reached equilibrium, G is 0, and Gº’ can be calculated by measuring [reactant] and [product]:
G = Gº’ + RT ln Keq = 0
Gº’ = -RT ln Keq
If the Gº’ < 0, the reaction is favorable as written, if Gº’ > 0, the reaction is unfavorable in the direction written.
The G of a reaction in a cell is calculated using the Gibbs Free Energy equation and the mass action ratio which is a measure of the actual concentration of reactants and products under physiological conditions. Remember that energy available for work comes from the fact that most chemical reactions inside cells are not at equilibrium (equilibrium is death).
G = Gº’ + RT ln [products]actual
How can the G for a reaction in a cell be favorable even though the Gº’ is unfavorable?
What can be said about rate of a reaction based on the value of Gº’?
For example, the product of endergonic reaction (A ↔ B) is the substrate for an exergonic reaction (B ↔ C) to yield an overall favorable reaction that converts reactant A to product C.
The ΔG of this coupled reaction is the sum of the ΔG values for each individual reaction and is favorable, thereby, driving the reaction A ↔ C. The same is true if we were using Gº’ values for these reactions.
A ↔ B ΔG = +4 kJ/molB ↔ C ΔG = -10 kJ/mol
A ↔ C ΔG = -6 kJ/mol
One type of shared reaction is one in which ATP hydrolysis (phosphoryl transfer) is used as a source of energy to drive unfavorable reactions
ATP hydrolysis: ΔGº’ = - 30.5 kJ/mol
The first step in glycolysis is catalyzed by the enzyme hexokinaseand utilizes phosphoryl transfer from ATP to drive the unfavorable reaction of glucose phosphorylation in a coupled reaction
Glucose + Pi ↔ glucose 6-phosphate + H2O ΔGº’ = +13.8 kJ/molATP + H2O ↔ ADP + Pi ΔGº’ = -30.5 kJ/molGlucose + ATP ↔ glucose 6-phosphate + ADP ΔGº’ = -16.7 kJ/mol
By combining endergonic reactions with exergonic reactions, some of which can be regulated by controlling enzyme activities, and by altering the concentrations of reactants and products as a result of dietary intake, the flux of metabolites through linked reactions can vary. The two most important considerations for metabolic flux are:
1) the concentration of reactants and products in the cell
2) enzyme activity levels.
A ↔ B (G < 0) the rate of B is favored when [A] is high
B ↔ C (G = 0) the reaction is sensitive to [B] and [C]
C ↔ D (G > 0) the enzyme X is highly regulated
A --> D under conditions where [A] is high and enzymes X is active
A combination of two reactions in the glycolytic pathway show how conversion of glucose-6-phosphate to fructose-1,6-bisphosphate (BP) occurs in cells even though the Gº’ for each uncoupled reaction is unfavorable (not counting the Gº’ for ATP hydrolysis).
Glucose-6-P ↔ Fructose-6-P + ATP ↔ Fructose-1,6-BP + ADP
Glucose-6-P ↔ Fructose-6-P Gº’ = +1.7 kJ/mol
Fructose -6-P + Pi ↔ Fructose-1,6-BP Gº’ = +16.3 kJ/mol
ATP ↔ ADP + Pi Gº’ = -30.5 kJ/mol
Glucose-6-P + ATP ↔ Fructose-1,6-BP + ADP Gº’ = -12.5 kJ/mol
The actual concentration of glucose-6-P in the cell under homeostatic conditions is 8.3 x 10-5M, and that of fructose-6-P is 1.4 x 10-5M.
Using Gibbs Free Energy equation, and the mass action ratio, the G for the conversion of glucose-6-P to fructose-6-P by the enzyme phosphoglucose isomerase can be calculated.
Glucose-6-P ↔ Fructose-6-P
G = Gº’ + RT ln [Fructose-6-P]actual
G = +1.7 kJ/mol + RT ln 1.4 x 10-5M
8.3 x 10-5M
G = -2.9 kJ/mol
Since the phosphofructokinase 1 reaction uses phosphoryl transfer energy available from ATP, it is a highly favorable reaction that uses up fructose-6-P and “pulls” the phosphoglucose isomerase reaction to the right.
How would metabolic flux be affected if more glucose-6-P were added?
The adenylate system describes the relationship between ATP, ADP, and AMP and can be described as the “energy charge” in the cell. The energy in the two ATP phosphoanhydride bonds ( and ) can be used not only to stimulate catalysis and metabolic flux, but also to induce protein conformational changes and to alter protein-protein interactions sites.
Cleavage of the phosphoanhydride bonds in ATP is favored for several reasons:
• unstable due to PO4-
• entropy increases
• increased solvation
• release of H+
ATP is continually being hydrolyzed and resynthesized from ADP + Pi.
The average person cycles through 50 kg of ATP every day!
Converting AMP to ADP to provide substrate for phosphorylation reactions is the job of adenylate kinase. The primary phosphorylating systems in nature are oxidative phosphorylation (aerobic respiration) and photophosphorylation (photosynthesis). Both of these processes require energy input to drive the ATP synthase reaction.
AMP + ATP ↔ 2 ADP (Adenylate kinase; G = ~0 kJ/mol)2 ADP + 2 Pi ↔ ATP + ATP (ATP synthase; G = +61 kJ/mol) AMP + 2 Pi ↔ ATP
Where does the energy come from in aerobic respiration and photosynthesis to drive the ATP synthase reaction forward?
Relative metabolic flux through catabolic and anabolic pathways responds to changes in the Energy Charge in the cell to maintain homeostasis
What do you think the primary mechanism is that regulates metabolic flux through catabolic and anabolic pathways in response to EC?