lecture 24
DESCRIPTION
Lecture 24. Thermodynamics in Biology. A Simple Thought Experiment. Driving Forces for Natural Processes. Enthalpy Tendency toward lowest energy state Form stablest bonds Entropy Tendency to maximize randomness. Enthalpy and Bond Strength. Enthalpy = ∆H = heat change at constant pressure - PowerPoint PPT PresentationTRANSCRIPT
Lecture 24
Thermodynamics in Biology
A Simple Thought Experiment
1 E. coli cell (10-11 mL)
1 mL H2O
5 mg glucose
1 mg (NH4)SO4
Mg++, PO4=, Fe3+, etc...
48 hours
109 cells
1 mL H2O
0 mg glucose
<1 mg (NH4)SO4
CO2
Glucose + (NH4)SO4 Cells + CO2
Driving Forces for Natural Processes
• Enthalpy– Tendency toward lowest energy state
• Form stablest bonds
• Entropy– Tendency to maximize randomness
Enthalpy and Bond Strength• Enthalpy = ∆H = heat change at constant pressure
• Units– cal/mole or joule/mole
• 1 cal = 4.18 joule
• Sign– ∆H is negative for a reaction that liberates heat
Entropy and Randomness
153 freeamino acids
Decreasedrandomness
Myoglobin
Entropy and Randomness
• Entropy = S = measure of randomness– cal/deg·mole
• T∆S = change of randomness
• For increased randomness, sign is “+”
“System” Definition
System
Surroundings
Closed system:No exchange ofmass or energy
“System” Definition
Isolated system:Energy is exchanged
E
E
“System” Definition
E
EM
M
Open System:Mass and energyare exchanged
Cells and Organisms: Open Systems
• Material exchange with surroundings– Fuels and nutrients in (glucose)– By-products out (CO2)
• Energy exchange– Heat release (fermentation)– Light release (fireflies)– Light absorption (plants)
1st Law of Thermodynamics
• Energy is conserved, but transduction is allowed
• TransductionOne form
of EAnother form
of E
Light Plants Chemicalbonds
Mayer: 1842
2nd Law of Thermodynamics
• In all spontaneous processes, total entropy of the universe increases
2nd Law of Thermodynamics
• ∆Ssystem + ∆Ssurroundings = ∆Suniverse > 0
• A cell (system) can decrease in entropy only if a greater increase in entropy occurs in surroundings
• C6H12O6 + 6O2 6CO2 + 6H2O complex simple
Entropy: A More Rigorous Definition
• From statistical mechanics:– S = k lnW
• k = Boltzmann constant = 1.3810–23 J/K
• W = number of ways to arrange the system
• S = 0 at absolute zero (-273ºC)
Gibbs Free Energy• Unifies 1st and 2nd laws• ∆G
– Gibbs free energy– Useful work available in a process
• ∆G = ∆H – T∆S– ∆H from 1st law
• Kind and number of bonds
– T∆S from 2nd law• Order of the system
∆G• Driving force on a reaction• Work available distance from equilibrium• ∆G = ∆H – T∆S
– State functions• Particular reaction• T• P• Concentration (activity) of reactants and products
Equilibrium• ∆G = ∆H – T∆S = 0
• So ∆H = T∆S– ∆H is measurement of enthalpy– T∆S is measurement of entropy
• Enthalpy and entropy are exactly balanced at equilibrium
Effects of ∆H and ∆S on ∆G
Voet, Voet, and Pratt. Fundamentals of Biochemistry. 1999.
Standard State and ∆Gº
• Arbitrary definition, like sea level
• [Reactants] and [Products]– 1 M or 1 atmos (activity)
• T = 25ºC = 298K
• P = 1 atmosphere
• Standard free energy change = ∆Gº
Biochemical Conventions: ∆Gº
• Most reactions at pH 7 in H2O
• Simplify ∆Gº and Keq by defining [H+] = 10–7 M
• [H2O] = unity
• Biochemists use ∆Gº and Keq
Relationship of ∆G to ∆Gº
• ∆G is real and ∆Gº is standard• For A in solution
– GA = GA + RT ln[A]
• For reaction aA + bB cC + dD
– ∆G = ∆Gº + RT ln
– Constant Variable (from table)
º
[C]c [D]d
[A]a [B]b
}
Relationship Between ∆Gº and Keq
• ∆G = ∆Gº + RT ln
• At equilibrium, ∆G = 0, so
– ∆Gº = –RT ln
– ∆Gº = –RT ln Keq
[C]c [D]d
[A]a [B]b
[C]c [D]d
[A]a [B]b
Relationship Between Keq and ∆Gº
Keq ∆Gº (k /J mol)10-6 34.310-5 28.510-4 21.410-3 17.210-2 11.310-1 5.91 0.0101 -5.9102 -11.3103 -17.2
Will Reaction Occur Spontaneously?
• When:– ∆G is negative, forward reaction tends to occur– ∆G is positive, back reaction tends to occur– ∆G is zero, system is at equilibrium
∆G = ∆Gº + RT ln [C]c [D]d
[A]a [B]b
A + B C + D
A Caution About ∆Gº
• Even when a reaction has a large, negative ∆Gº, it may not occur at a measurable rate
• Thermodynamics– Where is the equilibrium point?
• Kinetics– How fast is equilibrium approached?
• Enzymes change rate of reactions, but do not change Keq
∆Gº is Additive (State Function)
Reaction
A B
B C
Sum: A C
Also: B A
Free energy change
∆G1º
∆G2º
∆G1º + ∆G2º
– ∆G1º
Coupling Reactions
Glucose + HPO42– Glucose-6-P
ATP ADP + HPO42–
ATP + Glucose ADP + Glucose-6-P
∆Gºkcal/mol kJ/mol +3.3
+13.8 –7.3 –
30.5 –4.0 –
16.7
Resonance Forms of Pi
–
–
–
–
P
O
OHO
O
P
O
OHO
O
P
O
OHO
O
P
O
OHO
O
So: resonance stabilization
etc...
P
O
OO
O
Phosphate Esters and Anhydrides
ROH + HO P
O
O
O
R O P
O
O
OH2O
H2OEsters:
R CO
OH
+ P
O
HO O
OH2O
H2O
R C
O
O P
O
O
O
Anhydrides:
= Hydrolysis
Hydrolysis of Glucose-6-Phosphate
∆Gº = –3.3 kcal/mol= –13.8 kJ/mol
O
CH2OP
O
O
HO
+ H2O O
CH2HO+ OHP
O
O
HOO
CH2OP
O
O
HO
+ H2O O
CH2HO+ OHP
O
O
HO
Ionization,resonance
Productstabilization