Presentations text content in Thermodynamics of Biological Systems
Thermodynamics of Biological Systems
Champion Deivanayagam
Center for Biophysical Sciences and Engineering
University of Alabama at Birmingham.
Slide2Outline:
Laws of thermodynamics
Enthalpy
Entropy
Gibbs free energy
some examples
Slide3What you need to know for your exam:
Definition of a thermodynamic system
Three laws of Thermodynamics
Definitions of Enthalpy, Entropy
Gibbs Free energy
ATP’s ionization states and its potential
Slide4Energy: 1. Kinetic Energy
2. Potential Energy
Energy is the capacity to do work
Kinetic energy is the form of energy expended by objects in motion
A resting object still possess energy in the form of potential energy
Energy can be converted from one form into another
Slide5Thermodynamics:
A study of energy changes in systems:
System:
1. Isolated
2.
Closed
3. Open
Slide6First law of thermodynamics:
Energy is neither created nor destroyed; the energy of the universe is a constant
The total internal energy of an isolated system in conserved.
E = E
2
– E
1
= q + w
q – heat absorbed by the system from surroundings
w – work done on the system by the surroundings
Mechanical work is defined as movement through some distance caused by the application of force
Internal energy is independent of path
and represents the present state of the
system and is referred to as a State function
Slide7Mechanical Work:
w
= P
V where V = V
2
– V
1
Work may occur in multiple forms:
1. Mechanical
2 . Electrical
3. magnetic 4. Chemical
The calorie (cal, kcal) are traditional units
Joule’s is the recommended SI unit.
Table of important thermodynamic units and constants
At constant pressure
work can be defined as
Slide8Enthalpy:
H = E + PV
In a
constant pressure system
(as in most biological systems)
one can then define it as
H = E + PV
When you expand on this equation:
H = q (simply put the heat energy of the system)
Enthalpy changes can be measured using a calorimeter:
For a system at equilibrium for any process where A
B
the standard enthalpy can be determined from the temperature dependence
using:
R is the gas constant
R= 8.314 J/mol . K
Notice the ° sign: These are used to denote standard state:
For solutes in a solution, the standard state is normally unit activity (simplified to 1M concentration)
Slide9Protein
denaturation
:
Slide10Study of temperature induced reversible denaturation
of
chymotrypsinogen
At pH 3.0
T(K) 324.4 326.6 327.5 329.0 330.7 332.0 333.8
K
eq
0.041 0.12 0.27 0.68 1.9 5.0 21.0
Native state (N)
Denatured state (D)
K
eq = [D] / [N]H° at any given temperature is the negative
of the slope of the plot:
H° = [14.42]/[0.027] x 103 = +533 kJ/mol
van’t
Hoff Plot
Positive values for
H° would be expected to break bonds and expose hydrophobic groups
During the unfolding process and raise the energy of the protein in solution.
Slide11Second law of thermodynamics:Every energy transfer increases the entropy (disorder) of the universe
System tends to proceed from ordered states to disordered states
Some definitions:
Reversible: a process where transfer of energy happens in both directions
Irreversible: a process where transfer of energy flows in one direction
Equilibrium: A
B
(all naturally occurring process tend to equilibrium)
Slide12Entropy:
Entropy changes measure the dispersal of energy in a process.
Relationship between entropy and temperature
S = k
ln
W
fina
l
– k
ln
W
initial
k is the Boltzmann’s constant
W – number of microstates
dS
reversible
=
dq
/T
S = k
ln
W
Slide13
Third law of thermodynamics:
Entropy of any crystalline substance must approach zero as temperature approaches 0° K
The absolute entropy can be calculated from this equation:
Cp is the heat capacity, defined as the amount of heat 1 mole of it can store as the
temperature of that substance is raised by 1 degree.
For biological systems entropy changes are more useful than absolute entropies
Slide14Gibb’s free energy ‘G’
Determines the direction of any reaction from the equation:
G = H – TS
For a
constant pressure and temperature
system (as most biological systems) then the
Equation becomes easier to handle
G = H  TS
The enthalpy and entropy are now defined in one equation.
G is negative for
exergonic
reactions (release energy in the form of work) is positive for endergonic
reactions (absorbing energy in the form of work)
Consider a reaction: A + B
C + D
At Equilibrium:
G° = RT
ln
K
eq
and
K
eq
= 10

G°/2.3RT
Slide15Example of chymotrypsinogen
denaturation
From the
van’t
Hoff plot we calculated
H° = +533 kJ/mol
At pH 3.0
T(K) 324.4 326.6 327.5 329.0 330.7 332.0 333.8
K
eq
0.041 0.13 0.27 0.68 1.9 5.0 21.0
The equilibrium constant at 54.5 °C (327.5K) is 0.27
Then
G
°
= (8.314 J/
mol
·
K
) (327.5K)
ln
(0.27) =  3.56 kJ/mol
Similarly calculating
S
° =  (
G  H°) / T = 1620 J/
mol
·K
Slide16For a process to occur spontaneously the system must either give up energy (decrease H) or give up order (increase in S)
or both
In general for the process to be spontaneous:
G must be negative
The more negative the G value, the greater the amount of work the process can perform
Exergonic
reactions:
G is negative and the reaction is spontaneous
Example: Cellular respiration
C
6
H12O6 + 6O2 6C02 + 6 H2
0
G = 686 kcal/mol (2870 kJ/mol)
(For each molecule of glucose broken 686 kcal energy is made available for work)
Endergonic
reactions:
G is positive and requires large input of energy:
Example: Photosynthesis where the energy is derived from the sun.
Slide17Table: Variation of Reaction Spontaneity (Sign of
D
G
) with the signs of
D
H
and
DS
.
Slide18ATP: Adenosine triphosphate
A cell does three kinds of work:
1. Mechanical work: beating of cilia, muscle contraction etc.
2. Transport work: Moving substances across membranes
3. Chemical work: Enabling nonspontaneous reactions to occur spontaneously
e.g.
Protein synthesis.
The molecule that powers most kinds of work in the cell is ATP
Slide19Energy is released when one or more phosphate
groups are hydrolyzed
ATP + H
2
O
→ ADP + Pi (
G
°
= 35.7 kJ/mol)
G° = RT
ln
Keq
Slide20The activation energies for phosphoryl
grouptransfer
reactions
(200 to 400 kJ/mol) are substantially larger than the free energy of hydrolysis of ATP (30.5 kJ/mol).
ΔG
o
` = 30.5 kJ/mole
= 7.3 kcal/moleΔG` = 52 kJ/mole = 12.4 kcal/mole
Cellular conditions
:
[ADP][P
i] / [ATP] = 1/850
Slide21Ionization States of ATP
ATP has five dissociable protons
pK
a
values range from 01 to 6.95
Free energy of hydrolysis of ATP is relatively constant from pH 1 to 6, but rises steeply at high pH
Since most biological reactions occur near pH 7, this variation is usually of little consequence
Slide22The pH dependence of the free energy of hydrolysis of ATP. Because pH varies only slightly in biological environments, the effect on G is usually small.
Slide23
The free energy of hydrolysis of ATP as a function of total Mg2+ ion concentration at 38°C and pH 7.0
.
(
Adapted from Gwynn
, R. W.,
and
Veech
, R. L., 1973. The equilibrium constants of the adenosine triphosphate hydrolysis and the adenosine
triphosphate
citrate
lyase
reactions. Journal of Biological Chemistry 248:6966–6972.)
Slide24The free energy of hydrolysis of ATP as a function of concentration at 38°C, pH 7.0. The plot follows the relationship described in Equation (3.36), with the concentrations [C] of ATP, ADP, and Pi assumed to be equal.
Slide25
Slide26
What is the Daily Human Requirement for ATP?
The average adult human consumes approximately 11,700 kJ of food energy per day
Assuming thermodynamic efficiency of 50%, about 5860 kJ of this energy ends up in form of ATP
Assuming 50 kJ of energy required to synthesize one mole of ATP, the body must cycle through 5860/50 or 117 moles of ATP per day
This is equivalent to 65 kg of ATP per day
The typical adult human body contains 50 g of ATP/ADP
Thus each ATP molecule must be recycled nearly 1300 times per day
Slide27Isothermal titration calorimetry
Reference and experimental cell
Heat energy required to maintain both of them
At the same level is measured and integrated.
This allows for the measurement of
Kd
,
G and S
Slide28A1 A2 A3
P1 P2 P3
S
Vregion
W M C
1
39
201
448
578
828
840
960
1486
1561
A1 A2 A3
P1 P2 P3
Vregion
ITC studies on
A123 +
VP3 regions
1:1
Stoichiometric
ratio
(
N=0.946
±
0.006) and
K
d
~ 40 nm
Large
release of Heat energy indicates
Ordering of structures
Hydrogen bonding
The
K
d
and the energy released indicate that the interaction between these two regions is strong.
Slide29What happens in a cell if G = 0 ?
Thermodynamics of Biological Systems
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