Thermodynamics
Thermodynamics March-11-14 8:28 AM
Outline • Introduction • Free energy • Free energy and standard free energy • Energy coupling, coupled reactions • Thermodynamics of glycolysis
Introduction Hypothetical reaction: A ↔B Questions we often ask: • • •
•
Does the equilibrium favour the reactant or the product? How much energy was released? How much energy was conserved and in what form? o When pyruvate is converted to acetyl CoA energy is conserved through NADH and the thioesterbond How is the pathway reversed?
• •
How fast does the reaction go? How is a futile cycle avoided?
The first set of questions are thermodynamics and the second set of questions are kinetic questions Hypothetical reaction: A + B → C •
As A & B collide a transition molecule (AB) will be formed very briefly before C is produced •
AB is unstable and high energy, shown by an increase in energy level
•
The magnitude of the increase is known as the energy of activation (Ea) and acts as a barrier to prevent the production of C •
The higher Ea is the less likely it will be for A & B to form a transition state and successfully be converted to C •
Ea is the rate limiting step in the reaction, the larger its value the slower the reaction o Therefore plays a role in the rate of reaction
•
Enzymes work to reduce the activation energy of a reaction
•
ΔG determines the energy released in going from reactant to product •
Does value of ΔG depend on the pathway taken to get from A to B? o Nope
•
Does value of ΔG give us any indication of how fast a reaction will go? o Nope, that would be Ea • Activation energy barrier determines the rate of a reaction (how fast it will go, ie: kinetics), enzymes function to lower the activation energy barrier, speed up a reaction o
Do enzymes determine which direction a reaction will go in? No, enzymes only provide a place for the reaction; they can run in reverse because they grab onto the intermediates o
o o
What term determines the direction of a reaction? The value of ∆G
Laws of Thermodynamics First law: total energy of a system and its surroundings remains constant • •
energy is not created or destroyed Does that give us any information about the direction of a reaction?
No 2 Law: Spontaneous processes occur in directions which increase the overall disorder of the universe • Entropy increases (∆S) o
nd
•
Does indicate direction, but not easily measured.
•
Total energy change from a reaction (∆G) must equal the energy available to perform the reaction (ie energy for work; ∆H – enthalpy) and the energy that has gone to increasing disorder (∆S - entropy) o ∆G= ∆H-∆S •
Eg. Protein folding, can be a spontaneous event that would appear to be increasing order (ie decreasing ∆S) , not increasing disorder?
•
The criterion for spontaneity for biochemical processes which occur at constant press and temperature is a negative ΔG • Positive ΔG: not spontaneous, endergonic reaction (must be driven by input of energy to do work) • ΔG = 0, process is at equilibrium, forward and backward reactions are balanced • Change in free energy can be calculated from *Importantly, the temperature (ie movement of molecules in solution) must be accounted for. Therefore: ∆G= ∆H-T∆S •
This form of the equation for ΔG relates free energy change to heat (aka work) change (ΔH) and entropy change in a system exchanging energy with its surroundings at constant T and P • A negative ΔG may arise from? When ∆H > T∆S Example 1: oil in water, separates spontaneously but how if this this reduces disorder? • Negative ΔG due to an increase in ΔS • Why? Release of bound water from oil droplets as the oil comes together in one large drop • Large amounts of water surrounding the oil but more disorder when separated has less surface area Example 2: Protein folding is similar- keep in mind that in this case our assumption that ∆S is negative • hydrophobic effect • At low temperatures where T is small: o T* ∆S will be small o ∆H is more “likely” to be larger than T* ∆S o More likely that the reaction will proceed despite a perceived increase in order •
At higher temperatures, where T is large: T∆S will be large ∆H is more likely to be smaller than T∆S More likely that H will be overpowered by S, therefore reaction will NOT proceed o o o
Example 3: Glucose breakdown •
glucose + 6O2→ 6CO2 +6H2O
• • •
ΔH = -2880 kJ·mol-1 ΔS = 0.184 kJ·deg-1·mol-1 Temperature=25°C (298°K) Is this reaction spontaneous? = (-2888)-(298x0.184) = -2935 kJ/mol
•
• Does -ΔG tell us anything about the rate? No
•
• Does -ΔG tell us anything about the mechanism No
•
• Does -ΔG tell us anything about energy conservation No
Don't worry about deriving the equations just focus on using the equations given in class How do we relate free energy change to reactant/product concentrations?
At equilibrium, ΔG = 0 A↔ B
Forward reaction = K1 [A] Backward reaction = K2 [B] At equilibrium: K1 = K2 or K1/K2
Keq = [B] [A]
OR
Keq = [products] [reactants]
• The greater the displacement from Keq, the greater the free energy available →Any deviation from equilibrium starts a process that restores a balance to the system Delta G = 0 when? • At equilibrium Keq = [B]e/[A]e when delta G = 0 •
Intuitively: if [B] or [A] changes, resulting in changes to Keq, you can expect the delta G of the reaction to change as well. Example: add in more A to above equation A←--→B
Forward reaction is favoured (likely negative delta G) over reverse (likely positive delta G) Mass Action Ratio, Keq ratio and Free energy • Closed system, A↔ B Keq = [B]e/[A]e Balanced system, delta G is zero Neither reaction favoured • Mass Action Ratio = [B]o/[A]o (Γ) (aka gamma) Unbalanced system, delta G will have a value for forward/reverse reactions. One eaction will be favoured • Mass action ratio = Keq only at equilibrium • How does ΔG vary as mass action ratio is displaced from Keq?
The slope of the green curve will give us delta G
When you let in a ton of gas to the right then you will favour the movement left to right but not right to left When reaction is at equilibrium, ΔG = 0, slope of curve is zero, no change in free energy • If reaction is not functioning at equilibrium, (has not yet reached equilibrium) large [A], small mass action ratio, slope of line is steep and negative, giving -ΔG, reaction is spontaneous • For reaction going left to right, if reaction proceeds beyond Keq, slope is positive, ΔG is positive, reaction is not spontaneous
Summary: for reaction going left to right, the more the reaction is displaced to left from equilibrium, the more free energy will be available Example: ATP → ADP + Pi Keq = [ADP]e [P]e = [10-3][10-2] = 105 [ATP]e [10-10] Typical concentrations in mitochondria: 10-5x10-2/10-2 = 10-5 **in cell, reaction displaced left 10 orders of magnitude from equilibrium To the left or right? Left • this drastically increases delta G when making ATP, therefore drives rxn to the right Relating free energy change to Keq and mass action ratio ∆G= -2.303RT log[Keq/mass action ratio] R = gas constant = 0.0083 kJ·mol-1·K-1 T = absolute temperature °K (273 + C) 2.3 is conversion of natural logarithm (ln) to log Keq = equilibrium constant Γ = mass action ratio • • •
ΔG has a value that is a function of the displacement from equilibrium ∆G will me negative if marKeq
• ΔG is always calculated using actual in vivo concentrations and must always be negative for a real process –predicts direction • How does ΔG relate to ΔG° (standard free energy)? • ΔG° is standard free enry change for a reaction at standard conditions of 1 M concentrations of all reactants and products • May be postive or negative and cannot be used to predict reactions • It is related to the equilibrium constant as Relating it to delta G… =ΔG°+ 2.30RT log10 Keq Relating it to ΔG: ΔG = ΔG° + 2.303RTlogΓ At equilibrium, ΔG = ?
Γ=?
ΔG° = -2.303RTlogKeq • Standard free energy change at equilibrium • ΔG´ is calculated using actual in vivo concentrations and is always negative for a real spontaneous process (i.e. predicts direction.) • ΔG°´ is standard free energy change and may be positive or negative and can not predict direction Calculations ATP +H2O → ADP + Pi At pH = 7 , T = 25°C = 298°K , R=0.0083kJ·mol-1·K-1 Equilibrium concentrations [ATP]e=10-10M, [ADP]e=10-3M, [Pi]e=10-2M Observed concentrations in the mitochondria [ATP] o=10-2M, [ADP]o=10-5M, [Pi]o=10-2M Calculate : 1) ΔG°´ for hydrolysis of ATP? ΔG° = -2.303RTlogKeq = -5.7log (10-3x10-2)/10-10 = -5.7log(10…. Missed this •
ATP synthesis? (missed this)
1. ΔG´ for hydrolysis of ATP in the mitochondria ΔG´ = -2.303RTlog(Keq/Γ)
or
ΔG´ = ΔG°´ + 2.303RTlogΓ
Need to calculate mass action ratio first either way For ATP hydrolysis (Can't type this fast enough on my computer, need to ask Dr. Vesprini for the examples) 1. ΔG´ for ATP synthesis in the mitochondria • In previous slide mass action ratio was 10-5 Why is it 105 here? Hint: think about what mass action ratio is • •
ATP synthesis must be coupled to another reaction (or process) that releases large negative ΔG´ •
Do you know what that process is in mitochondria?
•
At equilibrium, what is ΔG´ for ATP hydrolysis? 0
How far is Glycolysis from equilibrium? Table 17-1 in text compares ΔG´ and ΔG°´ for glycolytic reactions in heart muscle Which 3 reactions occur far from equilibrium? 1,3,10 How would you determine how far these reactions were from equilibrium? Compare Keq to mar Calculate how far these reactions are from equilibrium at 25C
Calculate how far these reactions are from equilibrium at 25C HK: ΔG = -2.303RTlog (Keq/Γ)
-27.2=-27.205.7(log Keq/mar) 4.8=4.8 log (keq/mar) 104.8=Keq/mar 6.3x104 = Keq/r PFK: ΔG = -2.303RTlogKeq/Γ -25.9 = 4.5=logKeq/Γ … Keq/mar= 3.1x104
PK: ΔG = -2.303RTlogKeq/Γ -13.9 = -5.7(logKeq/Γ) 2.4 = logKeq/Γ Keq/mar= 251 To get a small mass action ratio you need lots of reactants and small amount of products
Overall what does this tell us? 3 out of 10 rxns in glycolysis are exergonic and are NOT at equilibrium. These are NOT reversible despite the (↔indication) and are in fact under kinetic control 7 out of 10 rxns in glycolysis are close to equilibrium. These are INDEED reversible and are under mass action control. These rxns are reversible under the right conditions How far is glycolysis overall from equilibrium? Do we have the overall ΔG´ for glycolysis? Glucose +2NAD+ + 2ADP+2Pi→ 2pyruvate +2NADH +2ATP +2H2O +4H+ We can add up all the ΔG´s from Table 17-1 ΔG´=(-27.2)+(-1.4)+(-25.9)+(-5.9)+0+ 2(-1.1) + 2(-0.6) +2(-2.4) +2(-13.9) =96.4 ΔG´ = -2.303RTlog (Keq/Γ) … Keq/r= 7.9x1016 16 orders of magnitude! Energy Coupling and Coupling reactions Thermodynamically unfavourable reactions (ΔG´ positive) can be driven forward by coupling them to a favourable reaction i)A + B ↔C + D ii)D + E ↔F+G iii)A + B + E ↔C + F + G
ΔG´ = ΔG´ = ΔG´ =
+10 kJ/mol -20 kJ/mol -10 kJ/mol
Overall, reaction (iii)proceeds L to R with an overall favourable ΔG´ Reaction (ii), by removing a small amount of D that was formed in reaction (i) pulls reaction (i) L to R Energy coupling requires a common intermediate between the 2 processes, in this case “D” was the common intermediate Eg glycolysis vs gluconeogenesis ΔG´ for glycolysis is –96.4 kJ/mol
To reverse glycolysis, ΔG´ = +96.4kJ/mol Gluconeogenesis requires input of energy from hydrolysis of 4 ATP equivalents In vivo hydrolysis of ATP releases 57kJ/mol (ΔG´ = -57kJ/mol) 4 X 57kJ/mol = -228 kJ/mol Reversal of glycolysis = +96.4 kJ/mol ΔG´ for gluconeogenesis –131.6 kJ/mol
Kinetic Control Review of differences between ΔG and kinetics Flux through a pathway • Reciprocal Kinetic control Glycolysis vs gluconeogenesis Hormonal control of glycogen metabolism Review: • How many reactions in glycolysis 10 • How many reactions function close to equilibrium (ΔG´ ≈ 0) 7 • These reactions are called equilibrium reactions – They are reversible reactions – Enzymes function at maximal turnover rate, bring the reaction to equilibrium – Reaction directions are determined by concentrations of reactants or products ΔG´=ΔG°´ + 2.303 RTlog10Γ The above are all mass action ratio They are not the rate limiting steps of a pathway, both their forward and reverse reactions function faster than the rate limiting step of the pathway Have no impact on the flux (rate of flow) of intermediates through metabolic pathways because there are slower steps in the pathway 3 of reactions in glycolysis are nonequilibrium reactions • • • •
Large negative ΔG´ values, Not reversible Under kinetic control
Typically found at beginning and end of pathway By regulating the enzymes of these reactions, the rate of flow through a pathway is contolled
•
The flux (rate of flow) through a pathway will vary depending on the physiological
need •
Define flux through a pathway as J- Vf-Vr
•
At equilibrium, rate of forward reaction = rate of reverse reaction, J = ? 0
•
At equilibrium, there is no flux through a pathway even through vf and vr may be
large •
When a reaction is spontaneous and far from equilibrium, vf >>vr and J ≈ vf
•
Flux through a pathway is set by nonequilibrium reaction (rate determining step)
•
Rate determining step • • •
Flux through A to B = vf – vr = J It is the same as the flux through the rate determining step If the rate through S to A increases by ΔJ, then [A] increases, which increases the rate of the forward reaction • Flux through a pathway can vary by 100 fold or more • The change in flux through the rate determining step is due to a change in enzyme activity not by the change in [S] •
Regulation of glycolysis and gluconeogenesis is very interesting in that – High flux through each of these pathways – Direction of flux is often reversed, but only occurs in one direction at one time
• How is this accomplished? – Reciprocal kinetic control of opposing pathways at rate limiting steps • What are the mechanisms of control of these rate limiting steps? Allosteric control of enzymes Covalent modification Substrate cycles- If Vf and Vr are rates from 2 different enzymes then they can be independently varied Genetic control
• What are the three enzymes that catalyze non-equilibrium reactions in glycolysis ? HK, PFK, PK • We can define these three steps as 3 small pseudocycles • All three pseudocycles have: – Large -ΔG (catalyze non-equilibrium reactions) – Opposing reactions are catalyzed by different enzymes – Opposing reactions offer independent control of rate of flux through glycolysis or gluconeogenesis by controlling activity of distinct enzyme The first step of glycolysis that commits the carbon in glucose to further catabolism by glycolysis is what step? Glu→G6P What enzyme catalyzes the first committed step of glycolysis and is defined as the rate determining step of glycolysis?
Allosteric Enzymes • Enzymes that are activated or inhibited by substrates, products or coenzymes in the pathway • • •
Allosteric enzymes are multimeric (n subunits) Contain n binding sites for substates Contain n allosteric binding sites for activator or inhibitor
•
Binding at the allosteric site produces a conformational change in the enzyme that changes o Km for a substrate which is the [S] giving 1/2Vmax (changes the affinity of the enzyme for its substrate) o Vmax (maximal velocity) •
Let’s look at V vs [S] graph where the affinity for substrate is modulated
Phosphofructokinase F-6-P +ATP → Fructose 1,6 bisphosphate + ADP • Enzyme is under allosteric control • Means that activity of enzyme is increased or inhibited by effectors that are substrates, products or co-enzymes of the pathway but not necessarily of the enzyme itself
PFK is a tetramer with 2 conformations, R and T, in equilibrium
R (active) • • • • •
F6P + ATP
ADP + ATP
T (inactive)
ATP is both a substrate and an allosteric inhibitor of PFK Each subunit has 2 binding sites for ATP Substrate site binds ATP equally well in the R or T conformation Allosteric site binds ATP almost exclusively in the T state, stabilizes that T state F-6-P (the other substrate) binds preferentially to the R state (has higher affinity for R state than T state) • Question: What is the effect of high [ATP] ? • [ATP]hi inhibits PFK by binding to the T state, shifting the equilibrium to favour the T state, the state that F-6-P has a lower affinity for • ATP is high as a result of a low metabolic demand, then PFK is inhibited and flux through the pathway is low
•
•
So, when [ATP]low, PFK active, flux through pathway is increased to replenish ATP pool
Outline • Introduction • Free energy • Free energy and standard free energy • Energy coupling, coupled reactions • Thermodynamics of glycolysis
Introduction Hypothetical reaction: A ↔B Questions we often ask: • • •
•
Does the equilibrium favour the reactant or the product? How much energy was released? How much energy was conserved and in what form? o When pyruvate is converted to acetyl CoA energy is conserved through NADH and the thioesterbond How is the pathway reversed?
• •
How fast does the reaction go? How is a futile cycle avoided?
The first set of questions are thermodynamics and the second set of questions are kinetic questions Hypothetical reaction: A + B → C •
As A & B collide a transition molecule (AB) will be formed very briefly before C is produced •
AB is unstable and high energy, shown by an increase in energy level
•
The magnitude of the increase is known as the energy of activation (Ea) and acts as a barrier to prevent the production of C •
The higher Ea is the less likely it will be for A & B to form a transition state and successfully be converted to C •
Ea is the rate limiting step in the reaction, the larger its value the slower the reaction o Therefore plays a role in the rate of reaction
•
Enzymes work to reduce the activation energy of a reaction
•
ΔG determines the energy released in going from reactant to product •
Does value of ΔG depend on the pathway taken to get from A to B? o Nope
•
Does value of ΔG give us any indication of how fast a reaction will go? o Nope, that would be Ea • Activation energy barrier determines the rate of a reaction (how fast it will go, ie: kinetics), enzymes function to lower the activation energy barrier, speed up a reaction o
Do enzymes determine which direction a reaction will go in? No, enzymes only provide a place for the reaction; they can run in reverse because they grab onto the intermediates o
o o
What term determines the direction of a reaction? The value of ∆G
Laws of Thermodynamics First law: total energy of a system and its surroundings remains constant • •
energy is not created or destroyed Does that give us any information about the direction of a reaction?
No 2 Law: Spontaneous processes occur in directions which increase the overall disorder of the universe • Entropy increases (∆S) o
nd
•
Does indicate direction, but not easily measured.
•
Total energy change from a reaction (∆G) must equal the energy available to perform the reaction (ie energy for work; ∆H – enthalpy) and the energy that has gone to increasing disorder (∆S - entropy) o ∆G= ∆H-∆S •
Eg. Protein folding, can be a spontaneous event that would appear to be increasing order (ie decreasing ∆S) , not increasing disorder?
•
The criterion for spontaneity for biochemical processes which occur at constant press and temperature is a negative ΔG • Positive ΔG: not spontaneous, endergonic reaction (must be driven by input of energy to do work) • ΔG = 0, process is at equilibrium, forward and backward reactions are balanced • Change in free energy can be calculated from *Importantly, the temperature (ie movement of molecules in solution) must be accounted for. Therefore: ∆G= ∆H-T∆S •
This form of the equation for ΔG relates free energy change to heat (aka work) change (ΔH) and entropy change in a system exchanging energy with its surroundings at constant T and P • A negative ΔG may arise from? When ∆H > T∆S Example 1: oil in water, separates spontaneously but how if this this reduces disorder? • Negative ΔG due to an increase in ΔS • Why? Release of bound water from oil droplets as the oil comes together in one large drop • Large amounts of water surrounding the oil but more disorder when separated has less surface area Example 2: Protein folding is similar- keep in mind that in this case our assumption that ∆S is negative • hydrophobic effect • At low temperatures where T is small: o T* ∆S will be small o ∆H is more “likely” to be larger than T* ∆S o More likely that the reaction will proceed despite a perceived increase in order •
At higher temperatures, where T is large: T∆S will be large ∆H is more likely to be smaller than T∆S More likely that H will be overpowered by S, therefore reaction will NOT proceed o o o
Example 3: Glucose breakdown •
glucose + 6O2→ 6CO2 +6H2O
• • •
ΔH = -2880 kJ·mol-1 ΔS = 0.184 kJ·deg-1·mol-1 Temperature=25°C (298°K) Is this reaction spontaneous? = (-2888)-(298x0.184) = -2935 kJ/mol
•
• Does -ΔG tell us anything about the rate? No
•
• Does -ΔG tell us anything about the mechanism No
•
• Does -ΔG tell us anything about energy conservation No
Don't worry about deriving the equations just focus on using the equations given in class How do we relate free energy change to reactant/product concentrations?
At equilibrium, ΔG = 0 A↔ B
Forward reaction = K1 [A] Backward reaction = K2 [B] At equilibrium: K1 = K2 or K1/K2
Keq = [B] [A]
OR
Keq = [products] [reactants]
• The greater the displacement from Keq, the greater the free energy available →Any deviation from equilibrium starts a process that restores a balance to the system Delta G = 0 when? • At equilibrium Keq = [B]e/[A]e when delta G = 0 •
Intuitively: if [B] or [A] changes, resulting in changes to Keq, you can expect the delta G of the reaction to change as well. Example: add in more A to above equation A←--→B
Forward reaction is favoured (likely negative delta G) over reverse (likely positive delta G) Mass Action Ratio, Keq ratio and Free energy • Closed system, A↔ B Keq = [B]e/[A]e Balanced system, delta G is zero Neither reaction favoured • Mass Action Ratio = [B]o/[A]o (Γ) (aka gamma) Unbalanced system, delta G will have a value for forward/reverse reactions. One eaction will be favoured • Mass action ratio = Keq only at equilibrium • How does ΔG vary as mass action ratio is displaced from Keq?
The slope of the green curve will give us delta G
When you let in a ton of gas to the right then you will favour the movement left to right but not right to left When reaction is at equilibrium, ΔG = 0, slope of curve is zero, no change in free energy • If reaction is not functioning at equilibrium, (has not yet reached equilibrium) large [A], small mass action ratio, slope of line is steep and negative, giving -ΔG, reaction is spontaneous • For reaction going left to right, if reaction proceeds beyond Keq, slope is positive, ΔG is positive, reaction is not spontaneous
Summary: for reaction going left to right, the more the reaction is displaced to left from equilibrium, the more free energy will be available Example: ATP → ADP + Pi Keq = [ADP]e [P]e = [10-3][10-2] = 105 [ATP]e [10-10] Typical concentrations in mitochondria: 10-5x10-2/10-2 = 10-5 **in cell, reaction displaced left 10 orders of magnitude from equilibrium To the left or right? Left • this drastically increases delta G when making ATP, therefore drives rxn to the right Relating free energy change to Keq and mass action ratio ∆G= -2.303RT log[Keq/mass action ratio] R = gas constant = 0.0083 kJ·mol-1·K-1 T = absolute temperature °K (273 + C) 2.3 is conversion of natural logarithm (ln) to log Keq = equilibrium constant Γ = mass action ratio • • •
ΔG has a value that is a function of the displacement from equilibrium ∆G will me negative if marKeq
• ΔG is always calculated using actual in vivo concentrations and must always be negative for a real process –predicts direction • How does ΔG relate to ΔG° (standard free energy)? • ΔG° is standard free enry change for a reaction at standard conditions of 1 M concentrations of all reactants and products • May be postive or negative and cannot be used to predict reactions • It is related to the equilibrium constant as Relating it to delta G… =ΔG°+ 2.30RT log10 Keq Relating it to ΔG: ΔG = ΔG° + 2.303RTlogΓ At equilibrium, ΔG = ?
Γ=?
ΔG° = -2.303RTlogKeq • Standard free energy change at equilibrium • ΔG´ is calculated using actual in vivo concentrations and is always negative for a real spontaneous process (i.e. predicts direction.) • ΔG°´ is standard free energy change and may be positive or negative and can not predict direction Calculations ATP +H2O → ADP + Pi At pH = 7 , T = 25°C = 298°K , R=0.0083kJ·mol-1·K-1 Equilibrium concentrations [ATP]e=10-10M, [ADP]e=10-3M, [Pi]e=10-2M Observed concentrations in the mitochondria [ATP] o=10-2M, [ADP]o=10-5M, [Pi]o=10-2M Calculate : 1) ΔG°´ for hydrolysis of ATP? ΔG° = -2.303RTlogKeq = -5.7log (10-3x10-2)/10-10 = -5.7log(10…. Missed this •
ATP synthesis? (missed this)
1. ΔG´ for hydrolysis of ATP in the mitochondria ΔG´ = -2.303RTlog(Keq/Γ)
or
ΔG´ = ΔG°´ + 2.303RTlogΓ
Need to calculate mass action ratio first either way For ATP hydrolysis (Can't type this fast enough on my computer, need to ask Dr. Vesprini for the examples) 1. ΔG´ for ATP synthesis in the mitochondria • In previous slide mass action ratio was 10-5 Why is it 105 here? Hint: think about what mass action ratio is • •
ATP synthesis must be coupled to another reaction (or process) that releases large negative ΔG´ •
Do you know what that process is in mitochondria?
•
At equilibrium, what is ΔG´ for ATP hydrolysis? 0
How far is Glycolysis from equilibrium? Table 17-1 in text compares ΔG´ and ΔG°´ for glycolytic reactions in heart muscle Which 3 reactions occur far from equilibrium? 1,3,10 How would you determine how far these reactions were from equilibrium? Compare Keq to mar Calculate how far these reactions are from equilibrium at 25C
Calculate how far these reactions are from equilibrium at 25C HK: ΔG = -2.303RTlog (Keq/Γ)
-27.2=-27.205.7(log Keq/mar) 4.8=4.8 log (keq/mar) 104.8=Keq/mar 6.3x104 = Keq/r PFK: ΔG = -2.303RTlogKeq/Γ -25.9 = 4.5=logKeq/Γ … Keq/mar= 3.1x104
PK: ΔG = -2.303RTlogKeq/Γ -13.9 = -5.7(logKeq/Γ) 2.4 = logKeq/Γ Keq/mar= 251 To get a small mass action ratio you need lots of reactants and small amount of products
Overall what does this tell us? 3 out of 10 rxns in glycolysis are exergonic and are NOT at equilibrium. These are NOT reversible despite the (↔indication) and are in fact under kinetic control 7 out of 10 rxns in glycolysis are close to equilibrium. These are INDEED reversible and are under mass action control. These rxns are reversible under the right conditions How far is glycolysis overall from equilibrium? Do we have the overall ΔG´ for glycolysis? Glucose +2NAD+ + 2ADP+2Pi→ 2pyruvate +2NADH +2ATP +2H2O +4H+ We can add up all the ΔG´s from Table 17-1 ΔG´=(-27.2)+(-1.4)+(-25.9)+(-5.9)+0+ 2(-1.1) + 2(-0.6) +2(-2.4) +2(-13.9) =96.4 ΔG´ = -2.303RTlog (Keq/Γ) … Keq/r= 7.9x1016 16 orders of magnitude! Energy Coupling and Coupling reactions Thermodynamically unfavourable reactions (ΔG´ positive) can be driven forward by coupling them to a favourable reaction i)A + B ↔C + D ii)D + E ↔F+G iii)A + B + E ↔C + F + G
ΔG´ = ΔG´ = ΔG´ =
+10 kJ/mol -20 kJ/mol -10 kJ/mol
Overall, reaction (iii)proceeds L to R with an overall favourable ΔG´ Reaction (ii), by removing a small amount of D that was formed in reaction (i) pulls reaction (i) L to R Energy coupling requires a common intermediate between the 2 processes, in this case “D” was the common intermediate Eg glycolysis vs gluconeogenesis ΔG´ for glycolysis is –96.4 kJ/mol
To reverse glycolysis, ΔG´ = +96.4kJ/mol Gluconeogenesis requires input of energy from hydrolysis of 4 ATP equivalents In vivo hydrolysis of ATP releases 57kJ/mol (ΔG´ = -57kJ/mol) 4 X 57kJ/mol = -228 kJ/mol Reversal of glycolysis = +96.4 kJ/mol ΔG´ for gluconeogenesis –131.6 kJ/mol
Kinetic Control Review of differences between ΔG and kinetics Flux through a pathway • Reciprocal Kinetic control Glycolysis vs gluconeogenesis Hormonal control of glycogen metabolism Review: • How many reactions in glycolysis 10 • How many reactions function close to equilibrium (ΔG´ ≈ 0) 7 • These reactions are called equilibrium reactions – They are reversible reactions – Enzymes function at maximal turnover rate, bring the reaction to equilibrium – Reaction directions are determined by concentrations of reactants or products ΔG´=ΔG°´ + 2.303 RTlog10Γ The above are all mass action ratio They are not the rate limiting steps of a pathway, both their forward and reverse reactions function faster than the rate limiting step of the pathway Have no impact on the flux (rate of flow) of intermediates through metabolic pathways because there are slower steps in the pathway 3 of reactions in glycolysis are nonequilibrium reactions • • • •
Large negative ΔG´ values, Not reversible Under kinetic control
Typically found at beginning and end of pathway By regulating the enzymes of these reactions, the rate of flow through a pathway is contolled
•
The flux (rate of flow) through a pathway will vary depending on the physiological
need •
Define flux through a pathway as J- Vf-Vr
•
At equilibrium, rate of forward reaction = rate of reverse reaction, J = ? 0
•
At equilibrium, there is no flux through a pathway even through vf and vr may be
large •
When a reaction is spontaneous and far from equilibrium, vf >>vr and J ≈ vf
•
Flux through a pathway is set by nonequilibrium reaction (rate determining step)
•
Rate determining step • • •
Flux through A to B = vf – vr = J It is the same as the flux through the rate determining step If the rate through S to A increases by ΔJ, then [A] increases, which increases the rate of the forward reaction • Flux through a pathway can vary by 100 fold or more • The change in flux through the rate determining step is due to a change in enzyme activity not by the change in [S] •
Regulation of glycolysis and gluconeogenesis is very interesting in that – High flux through each of these pathways – Direction of flux is often reversed, but only occurs in one direction at one time
• How is this accomplished? – Reciprocal kinetic control of opposing pathways at rate limiting steps • What are the mechanisms of control of these rate limiting steps? Allosteric control of enzymes Covalent modification Substrate cycles- If Vf and Vr are rates from 2 different enzymes then they can be independently varied Genetic control
• What are the three enzymes that catalyze non-equilibrium reactions in glycolysis ? HK, PFK, PK • We can define these three steps as 3 small pseudocycles • All three pseudocycles have: – Large -ΔG (catalyze non-equilibrium reactions) – Opposing reactions are catalyzed by different enzymes – Opposing reactions offer independent control of rate of flux through glycolysis or gluconeogenesis by controlling activity of distinct enzyme The first step of glycolysis that commits the carbon in glucose to further catabolism by glycolysis is what step? Glu→G6P What enzyme catalyzes the first committed step of glycolysis and is defined as the rate determining step of glycolysis?
Allosteric Enzymes • Enzymes that are activated or inhibited by substrates, products or coenzymes in the pathway • • •
Allosteric enzymes are multimeric (n subunits) Contain n binding sites for substates Contain n allosteric binding sites for activator or inhibitor
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Binding at the allosteric site produces a conformational change in the enzyme that changes o Km for a substrate which is the [S] giving 1/2Vmax (changes the affinity of the enzyme for its substrate) o Vmax (maximal velocity) •
Let’s look at V vs [S] graph where the affinity for substrate is modulated
Phosphofructokinase F-6-P +ATP → Fructose 1,6 bisphosphate + ADP • Enzyme is under allosteric control • Means that activity of enzyme is increased or inhibited by effectors that are substrates, products or co-enzymes of the pathway but not necessarily of the enzyme itself
PFK is a tetramer with 2 conformations, R and T, in equilibrium
R (active) • • • • •
F6P + ATP
ADP + ATP
T (inactive)
ATP is both a substrate and an allosteric inhibitor of PFK Each subunit has 2 binding sites for ATP Substrate site binds ATP equally well in the R or T conformation Allosteric site binds ATP almost exclusively in the T state, stabilizes that T state F-6-P (the other substrate) binds preferentially to the R state (has higher affinity for R state than T state) • Question: What is the effect of high [ATP] ? • [ATP]hi inhibits PFK by binding to the T state, shifting the equilibrium to favour the T state, the state that F-6-P has a lower affinity for • ATP is high as a result of a low metabolic demand, then PFK is inhibited and flux through the pathway is low
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So, when [ATP]low, PFK active, flux through pathway is increased to replenish ATP pool
