bio98a_l13

advertisement
100%
90%
80%
70%
http://www.youtube.com/watch?v=EF_xdvn52As
Bio 98 - Lecture 13
Energetics
Thermodynamics
The 2nd Law
Unsustainable
“All natural and technological processes proceed in such a way that the availability of the
remaining energy decreases. In all energy exchanges, if no energy enters or leaves an isolated
system, the entropy of that system increases. Energy continuously flows from being
concentrated, to becoming dispersed, spread out, wasted, and useless. New energy cannot be
created and high-grade energy is being destroyed. An economy based on endless growth is
unsustainable.”
Overview
1. What is the relationship between the free energy
change (G) of a reaction and its equilibrium
constant?
2. How can compounds like ATP be used to drive
metabolic reactions that are unfavorable?
3. What is it about the structure of ATP and related
compounds that make them so “full of energy”?
http://dwb4.unl.edu/Chem/CHEM869P/CHEM869PLinks/www.fordham.edu/Biochem_3521/lect15/glycolysis.html
Simple Thermodynamics
[products]
Keq =
[reactants]
aA + bB
Keq
cC + dD
c
d
a
b
[C]
[D]
=
[A] [B]
If a process is spontaneous then the reaction
moving to the right (products) is favored and
Keq > 1 or Greactants > Gproducts
Example: an important reaction in biology is the
inter-conversion of various sugars. The
equilibrium shown below is the inter-conversion
of glucose-6-phosphate (G6P) and fructose-6phosphate (F6P):
[F6P]
Keq =
= 0.504
[G6P]
Glucose-6-phosphate
G6P
Fructose-6-phosphate
F6P
[F6P]
Keq =
= 0.504
[G6P]
Suppose you dissolve 1 mole of G6P and 0.3 moles of F6P into a liter of
water. What is the concentration of each at equilibrium?
The sum is 1.3 moles in a liter or 1.3 M, so
[F6P]+[G6P] = 1.3 M, thus [G6P] = 1.3 M – [F6P]
[F6P]
Keq =
= 0.504
1.3M- [F6P]
[F6P] = 0.504(1.3M -[F6P])
Solve for [F6P] = 0.44 M.
And the concentration of G6P = [G6P] = 1.3 - 0.44 M = 0.86 M.
Free Energy
A measure of the spontaneity of a reaction is called the
Gibbs free energy (G).
The reactants and products have a free energy. We can only
observe the change () in free energy as we go from
reactants to products:
Gºreaction = Gproducts - Greactants
For a spontaneous process Greactants > Gproducts, so
Gºreaction < 0
Enzyme-catalyzed reaction:
S
E
P
G (free energy)
S‡
ES‡
G‡
E+S
E+P
G
Reaction coordinate
1. Enzymes do not alter the equilibrium or G.
2. They accelerate reaction rates by decreasing G‡.
3. They accomplish this by stabilizing the transition state(s).
Free Energy
G° is called the standard Gibbs free energy change for
a given reaction under standard conditions of pressure
(1 atm), temperature (25 °C or 298 K) and [H+] (pH 7.0
or 0.1 mM). Under standard conditions the concentration
of every reactant and product is 1 M (and water is 55 M!).
It basically tells us whether, at equilibrium, a reaction
lies to the right or left. However, inside living systems
reactions are seldom at equilibrium, so a more useful
quantity is G under non-equilibrium conditions.
1. Gibbs free energy (G) - free energy needed to
convert reactants to products under a defined set of nonequilibrium conditions.
aA + bB
cC + dD
Which direction is favored?
[C]c[D]d
G = ∆G° + RT ln –––––––– = ∆G° + RT ln Q
[A]a[B]b
ln(x) is the logarithm
base e not base 10!
∆G°: Standard free energy of a reaction. Depends on free
energies inherent in the structures of the chemicals.
RT ln Q: accounts for differences in the concentrations of
reactants and products (law of mass action); R is the gas
constant (8.31 J K-1 mol-1), T is the temperature in Kelvin.
aA + bB
cC + dD
Which direction is favored?
[C]c[D]d
∆G = ∆G° + RT ln –––––––– = ∆G° + RT ln Q
[A]a[B]b
Consider special cases to “test” this equation
(1) [A], [B], [C], [D] = 1 M
Then ∆G = ∆G° (by definition!)
(2) Q is a very small number (eg. 1x10-6): [reactants] >> [products]
Will make ∆G more negative/favorable with respect to ∆G°.
Follows law of mass action, ie. low [products] will pull balance to right.
(3) Q is a very large number (eg. 1x106): [products] >> [reactants]
Will make ∆G more positive/unfavorable with respect to ∆G°.
Follows law of mass action, ie. high [products] will push balance to left.
The Equilibrium State
∆G = ∆G° + RT ln Keq
At equilibrium, neither direction of the reaction is favored,
therefore ∆G = 0 (by definition).
With ∆G° + RT ln Keq = 0
ln Keq = -∆G°/RT
Keq = e-∆G°/RT
Example
Glucose + HPO4-(Glu)
(Pi)
Glu-6-phosphate
(G6P)
∆G° =
+14 kJ/mol
Keq = e[-∆G°/RT] = e(-14,000 J/mol)/(8.314 J/K/mol x 298 K)
=
e-5.65
= 3.5 x
10 -3
[G6P]
= ––––––––––
[Glu][HPO4--]
The equilibrium lies far to the left; thus the reaction is not
favored. G6P has a higher inherent free energy than do
glucose and Pi together.
Energy Coupling - an important role for enzymes
and for high-energy compounds like ATP.
(1) A
B
∆G° = +14 kJ/mol (unfavorable)
(2) C
D
∆G° = -30 kJ/mol (favorable)
---------------------------------------------------------------------------A+C
Enz
B+D
∆G° = -16 kJ/mol
(favorable)
By coupling (1) and and (2), an enzyme can “tap” the
energy of a highly favorable reaction to drive an otherwise
unfavorable reaction!
ADP vs. ATP
ADP, net charge: -3
ATP, net charge: -4
Mitochondria and ATP
Movie: Powering the Cell
http://multimedia.mcb.harvard.edu/
Example of energy coupling: hexokinase reaction
∆G° (kJ/mol)
Reaction
Glu + Pi
ATP
G6P
ADP + Pi
+14 (unfavored)
-31 (favored)
------------------------------------------------------------hexokinase
Glu + ATP
G6P + ADP
-17 (favored)
Using Keq = e[-∆G°/RT] with R = 8.314 J/K/mol and T = 298 K
we obtain Keq = 955
2. ATP hydrolysis is highly exergonic
O O O
Ad-Rib-O-P-O-P-O-P-O + H2O
O O O
ATP
O O
O
Ad-Rib-O-P-O-P-O + O-P-O + H+
O O
OH
ADP
(Pi)
∆G° = -31 kJ/mol
Why is ATP hydrolysis so highly exergonic?
1. Electrostatic repulsion - less with products
2. Resonance forms - more with products
3. Proton release
• [H+] is kept very low in the cell (~10-7 M or 0.1 mM)
• This favors reactions that release protons
(mass action effect)
4. Hydration stabilization - more with products
Similar considerations explain why other phosphate
ester compounds also store energy; for example
phosphoenolpyruvate, phosphocreatine, etc.
Why is ATP hydrolysis so highly exergonic?
1. Electrostatic repulsion - less with products
-
2. Resonance forms - more with products
ATP turnover
Each of the approximately one hundred trillion human cells contains
about one billion ATP molecules, or about 1023 ATP molecules in the
body. For each of these ATP “the terminal phosphate is added and
removed 3 times each minute” (Kornberg, 1989). This amount is
sufficient for only a few minutes and must be rapidly recycled.
The total human body content of ATP is only about 50 grams. The
ultimate source of energy for generating ATP is food; ATP is simply the
carrier or storage unit of energy. The average daily intake of 2,500
food calories translates into a turnover of a whopping 180 kg (400 lbs)
of ATP (Kornberg, 1989).
Enthalpy and Entropy
G = H - TS
H or enthalpy is the heat given off in a reaction.
H is defined as Hproducts – Hreactants. When heat is given
off then H < 0. This means a given reaction is
enthalpically favorable.
Suppose we have a reaction going on in a beaker. If the
beaker heats up then the reaction is giving off heat so
H < 0.
Enthalpy and Entropy
G = H - TS
S or entropy is a measure of disorder in a system.
Similar to H, S is defined as Sproducts - Sreactants.
A spontaneous or favorable process tends toward higher
disorder so S > 0 is favorable.
Enthalpy and Entropy
G = H - TS
In this container the gas
molecules are confined to a
small space. The order in the
system is relatively high, so
the entropy, S, is small.
In this container the same
number of gas molecules have
more space. Each molecule has
more “freedom” and the
system is more disordered.
Here the entropy, S, is larger.
Enthalpy and Entropy
G = H - TS
Consider what happens to the thermodynamic quantities in a reaction
between 2 molecules to form 1 molecule:
A+B
AB
The formation of a chemical bond is usually a heat-generating
process so H < 0. However, in going from 2 molecules to 1
molecule, the system will become more ordered. Therefore, S < 0
(decrease in entropy). Whether or not a reaction is favorable
depends on the balance between enthalpy and entropy. It is typical
for enthalpy and entropy to balance or compensate for one
another.
A classic example is the reaction between oxygen and hydrogen
to form water
O2 + 2 H2
2 H 2O
Take a balloon filled with hydrogen gas. Touch a lit match to
the balloon. What happens? A big boom and lots of heat
given off. Is H favorable or unfavorable??
Add solid urea to a beaker of water. The beaker gets very
cold. Is H favorable or unfavorable???
BIOCHEMISTRY
Bio97
Bio98
Bio99
THE END
Good luck with the midterm!
My lab is looking for Bio199 students
(GPA of 3.5 or better, A- or better in 97, 90% or better in 98 midterm, 12-16 hours
per week / 3-4 units per quarter, 2 year commitment and participation in Excellence
in Research.
Please email after the midterm with “199” in the subject line if you’re interested.)
Download
Related flashcards

Physical chemistry

44 cards

Speed skating venues

16 cards

Equilibrium chemistry

14 cards

Catalysis

15 cards

Create Flashcards