Thermodynamics of Biological Systems

Champion DeivanayagamCenter for Biophysical Sciences and EngineeringUniversity of Alabama at Birmingham.

Outline:

Laws of thermodynamics

Enthalpy Entropy Gibbs free energy

some examples

What you need to know for your exam:

1. Definition of a thermodynamic system2. Three laws of Thermodynamics3. Definitions of Enthalpy, Entropy4. Gibbs Free energy5. ATP’s ionization states and its potential

Energy:

1. Kinetic Energy2. 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

Thermodynamics:

A study of energy changes in systems:

System:1. Isolated2. Closed3. Open

First law of thermodynamics:

Energy is neither created nor destroyed; the energy of the universe is a constantThe total internal energy of an isolated system in conserved.

E = E2 – E1 = q + w

q – heat absorbed by the system from surroundingsw – 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 pathand represents the present state of the system and is referred to as a State function

Mechanical Work:

w = -PV where V = V2 – V1

Work may occur in multiple forms:1. Mechanical2 . Electrical3. magnetic4. 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

Enthalpy: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 AB the standard enthalpy can be determined from the temperature dependenceusing:

R- is the gas constantR= 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)

Protein denaturation:

Study of temperature induced reversible denaturation of chymotrypsinogen

At pH 3.0

T(K) 324.4 326.6 327.5 329.0 330.7 332.0 333.8Keq 0.041 0.12 0.27 0.68 1.9 5.0 21.0

Native state (N) Denatured state (D)

Keq = [D] / [N]

H° at any given temperature is the negativeof the slope of the plot:

H° = -[14.42]/[-0.027] x 10-3 = +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.

Second 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)

Entropy:

Entropy changes measure the dispersal of energy in a process.

Relationship between entropy and temperature

S = k ln Wfinal – k ln Winitial

k- is the Boltzmann’s constantW – number of microstates

dSreversible = dq/T

S = k ln W

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

Gibb’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 theEquation 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[ ][ ]

ln[ ][ ]

C DG G RT

A B

At Equilibrium: G° = RT ln Keq

and Keq = 10 - G°/2.3RT

Example of chymotrypsinogen denaturation

From the van’t Hoff plot we calculated H° = +533 kJ/mol

At pH 3.0

T(K) 324.4 326.6 327.5 329.0 330.7 332.0 333.8Keq 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

For a process to occur spontaneouslythe 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

C6H12O6 + 6O2 6C02 + 6 H20 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.

Table: Variation of Reaction Spontaneity (Sign of DG) with the signs of DH and DS.

ATP: Adenosine triphosphate

A cell does three kinds of work:

1. Mechanical work: beating of cilia, muscle contraction etc.2. Transport work: Moving substances across membranes3. Chemical work: Enabling non-spontaneous reactions to occur spontaneously

e.g. Protein synthesis.

The molecule that powers most kinds of work in the cell is ATP

Energy is released when one or more phosphategroups are hydrolyzed

ATP + H2O → ADP + Pi (G° = -35.7 kJ/mol)

G° = RT ln Keq

The activation energies for phosphoryl group-transfer reactions (200 to 400 kJ/mol) are substantially larger than the free energy of hydrolysis of ATP (-30.5 kJ/mol).

ΔGo` = -30.5 kJ/mole

= -7.3 kcal/mole

ΔG` = -52 kJ/mole

= -12.4 kcal/mole

Cellular conditions:

[ADP][Pi] / [ATP] = 1/850

Ionization States of ATP

• ATP has five dissociable protons• pKa values range from 0-1 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

The 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.

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.)

The 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.

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

Isothermal titration calorimetry

Reference and experimental cell

Heat energy required to maintain both of themAt the same level is measured and integrated.

This allows for the measurement of Kd, G and S

0.0 0.5 1.0 1.5

-200

-150

-100

-50

0-14

-12

-10

-8

0 10 20 30 40 50 60 70

Titration of 12 M VP3by 10 L x 25 injections of 96 M A

1-3

at 25oC

Time (min)

µcal/s

ec

N 0.946 + 0.006

Ka

(2.6 + 0.4) x 107 M-1

G -10.1 kcal/molH (-212 + 2) kcal/molTS -202 kcal/mol

Molar Ratio

kcal/m

ole

of in

jecta

nt

A1 A2 A3 P1 P2 P3S V-region W M C

1 39 201 448 578 828 840 960 1486 1561

A1 A2 A3

P1 P2 P3V-region

ITC studies on A123 + VP3 regions

1:1 Stoichiometric ratio (N=0.946 ± 0.006) and Kd ~ 40 nm

Large release of Heat energy indicatesOrdering of structuresHydrogen bonding

The Kd and the energy released indicate that the interaction between these two regions is strong.

What happens in a cell if G = 0 ?