Chemistry Biochemistry and Bio-Medical Sciences
Aris Kaksis 2016.
Riga Stradin's University
The Oxidation Reduction of Biological Systems

BioThermodynamics.doc
OxRedBiologicalW.doc 
Biological Oxidation-Reduction Reactions
   The transfer of phosphoryl groups is a central feature of metabolism. Equally important is another kind of transfer,
electron transfer in oxidation-reduction(OxRed) reactions. These reactions involve the loss of electrons -e- by one
chemical species, which is thereby oxidized, and the gain of electrons +e - by another, which is reduced. The flow of
electrons => e- in oxidation-reduction (OxRed) reactions is responsible, directly or indirectly, for all work (W) done
by living organisms. In non-photo synthetic organisms, the sources of electrons e- are reduced compounds (foods);
in photo synthetic organisms, the initial electron -e-donor is a chemical species excited by the absorption A
of light ~hv = E energy. The path of electron - e-
flow =>  in metabolism is complex. Electrons e- move from => various metabolic intermediates to => specialized
electron e- carriers in enzyme - catalyzed reactions. The carriers e- in turn donate electrons e- to acceptors with higher
electron e- affinities, with the release of energy. Cells contain a variety of molecular energytransducers, which
convert the energy of electron e- flow into useful work: W = E•F•n ; where,
E potential between species (in volts V); F = 96485 C (coulomb) 1 mol of electrons e- electric charge in C;
n number of electrons e- involved in species OxRed reactions - between oxidised state (Ox) and
reduced state (Red) of compound. Established so called OxRed system with own RedOx potential E :

    Ox + n e- ArrowReverseReact Red ; E = +RTLn10_F•lgOx_Red.gif                               (1)

where - standard potential of given OxRed system measured at conditions when E = (as [Ox] = [Red]);
natural logarithm of number 10 - ln(10) = 2.302585093 ; universal gas constant - R = 8.3144 J/mol/K ;
absolute thermodynamics temperature T = 273.16° + 25° C = 298.16 K at standard temperature conditions measured:
as Kelvin scale value 273.16 K at zero 0°C point plus on Celsius scale measured 25°C but human body temperature
37°C that will be higher T = 273.16° + 37°(C) = 310.16 K non-standard conditions;
Faraday's constant - F = 96 485 C (coulomb) 1 mol of electrons e- electric charge in C;
number of electrons e- involved in OxRed system - n ; decimal logarithmic function - lg( ) of argument as ratio
([Ox]/[Red]) between oxidised form concentration - [Ox] as multiple over reduced form concentration [Red].
   We begin our discussion with a description of the general types of metabolic reactions in which electrons e- are
transferred. After considering the theoretical and experimental basis for measuring the energy changes DG in
oxidation reactions in terms of electromotive force (EMF), we discuss the relationship between this force,
expressed in volts V, and the free-energy change DG , expressed in joules J. We conclude by describing the
structures and oxidation-reduction chemistry of the most common of the specialized electron e- carriers =>,
which you will encounter repeatedly in later discussions.


    The Flow of Electrons can do Biological Work
         Every time we use a motor, an electric light or heater, or a spark to ignite gasoline in a car engine, we use
the flow => of electrons e- to accomplish work W=E•F•n, where n (in units mol) is the electron number
of moles moving from reduced form Red => to oxidised form Ox . In the circuit that powers a motor, the source
of electrons e- can be a battery containing two 2 chemical species that differ in affinity for electrons e- .
Electrical wires provide a pathway for electron e- flow => from the chemical species reduced form Red1
at one pole (-) of the battery, through the motor, to the chemical species oxidized form Ox2 at the other
second 2nd pole (+) of the battery.
(-) Red1 ArrowReverseReact Ox1 + n e- => Electrical wires number n electrons n e- - flow => Ox2 + n e- ArrowReverseReact Red2 (+)
    - E1     electron-carriers in biological systems n electrons n e- flow =>     E2
for closed circuit electric-motion force is calculated as sum  EMF = E1 - E2 in volts V .
Because the two 2 chemical species differ in their affinity for electrons e- , electrons e- flow => spontaneously
through the circuit, driven by a force proportional to the difference in electron e-affinity, the
electro-motive force (EMF ). The electromotive force (typically a few volts ±1÷3.5 V) can accomplish work
W=EMF•F•n if an appropriate energy transducer in this case a motor is placed in the circuit. The motor can be
coupled to a variety of mechanical devices to accomplish useful work W = EMF•F•n.
         Living cells have an analogous biological ''circuit,'' with a relatively reduced compound such as glucose as
the source of electrons e- . As glucose is enzymatic oxidized, the electrons e- released flow =>  spontaneously
through a series of electron-carrier intermediates to another chemical species, such as O2.
          This electron e- flow =>  is ex-ergonic because O2 has a higher affinity for electrons e- than do the
electron- -carrier intermediates. The resulting electro-motive force provides energy to a variety of molecular energy
transducers
(enzymes and other proteins) that do biological work W. In the mitochondrion, for example,
membrane-bound enzymes couple electron e- flow =>  to the production of a trans-membrane pH difference
DpH=8.36-5.00=3.36, accomplishing osmotic and electrical work. The proton H+ gradient thus formed has
potential energy, sometimes called the proton-motive force by analogy with electro-motive force. Another enzyme,
ATP synthase in the inner mitochondrial membrane, uses the proton-motive force Emembrane to do chemical work W:
synthesis of ATP from ADP and Pi as protons H+ flow => spontaneously across the membrane.
Similarly, membrane-localized enzymes in E. coli convert electro-motive force to proton-motive force, which is then
used to power flagella motion.
     The principles of electrochemistry that govern energy changes in the macroscopic circuit with a motor and battery
apply with equal validity to the molecular processes accompanying electron e- flow =>  in living cells. We turn now
to a discussion of those principles.

Oxidation-Reductions Can Be Described as Half-Reactions of two Ox ArrowReverseReact Red Systems
     Although oxidation and reduction must occur together, it is convenient when describing electron e- transfers =>
to consider the two 2 halves (OxRed systems) of an oxidation-reduction reaction separately. For example,
the oxidation of ferrous ion Fe2+ by cupric ion Cu2+,
Fe2+ + Cu2+ => Fe3+ + Cu+ can be described in terms of two 2 half-reactions (Ox ArrowReverseReact Red systems):
 (1) Fe2+        ArrowReverseReact Fe3+ + e- The electron-donating e- molecule in an oxidation-reduction reaction is called
(2) Cu2+ + e- ArrowReverseReact Cu+ the reducing agent or reductant; the electron-accepting + e- molecule is the oxidizing
agent or oxidant. A given agent, such as an iron cation existing in the ferrous (Fe2+) or ferric (Fe3+) state, functions
as a conjugate reductant-oxidant pair (RedOx pair), just as an acid and corresponding base function
as a conjugate acid-base pair. Recall from
Acid-Base Equilibrium that in acid-base reactions we can write a general equation:
   proton donor ArrowReverseReact H+ + proton acceptor.
In RedOx reactions we can write a similar general equation:
 electron donor ArrowReverseReact n e- + electron acceptor.
In the reversible half reaction (1) above, Fe2+ is the electron donor and (2) Fe3+ is the electron acceptor;
together, Fe2+ and Fe3+ constitute a conjugate RedOx pair.
     The electron e- transfers in the oxidation-reduction reactions of organic compounds are not fundamentally different
from those of inorganic species. In Reducing Sugars we considered the oxidation of a reducing sugar
(an free aldehyde or ketone) by cupric ion Cu2+(see reducing sugars):
  Aldehyde.gif + 4 OH- + 2 Cu2+ArrowReverseReact  CarbonicAcid.gif+ Cu2O + 2 H2O

This overall reaction can be expressed as two 2 half-reactions using OxRed systems:

(1) Aldehyde.gif+ 2 OH-ArrowReverseReactCarbonicAcid.gif+ 2 e- +  H2OBecause two 2 electrons 2 e- are removed from the

(2) 2 Cu2+      + 2 e- + 2 OH-ArrowReverseReact Cu2O + H2Oaldehyde carbon —(C=O)—H, the second 2nd half-reaction

(the one-electron reduction of cupric Cu2+ to cuprous ion Cu+) must be doubled 2 to balance the overall equation
two 2 electrons 2 e- are gained on two cupric Cu2+ cations converting to two cuprous ions Cu+.

Biological Oxidations Often Involve Dehydrogenation

The carbon in living cells exists in a range of oxidation states (Fig. 1). When a carbon atom shares an electron pair :
with another atom (typically H, C, S, N, or O), the sharing is unequal in favor of the more electronegative atom.
The order of increasing electronegativity: H < C < S < N < O respectively 2.2 < 2.55 < 2.58 < 3.04 < 3.44.
In over simplified but useful terms, the more electronegative atom ''owns'' the bonding electrons e- it shares with
another atom. For example, in methane (CH4) carbon C is more electronegative than the four 4 hydrogens H bonded
to it, and the C atom therefore ''owns'' all eight 8 bonding electrons 8 e- (Fig. 1). In ethane, the electrons e-
in the Tri.gifC-CTri.gif bond are shared equally, so each :::C:C::: atom owns only seven 7 of its eight 8 bonding electrons e-.
In ethanol, C-1 is less electronegative than the oxygen O to which it is bonded, and the O atom therefore ''owns''
both electrons 2 e- of the Tri.gifC-:O- bond, leaving ::.C-1 with only five 5 bonding electrons e-. With each formal loss
of electrons e-, the carbon C atom has undergone oxidation even when no oxygen O is involved, as in the conversion
of an alken (CH2=CH2) to an alkyne (CHTri.gifCH). In this case, oxidation (loss of electrons e-) is coincident with
the loss of hydrogen - H. In biological systems, oxidation is often synonymous, with dehydrogenation, and many
enzymes that catalyze oxidation reactions are dehydrogenases. Notice that the more reduced compounds in Figure 1
(top) are richer in hydrogen H than in oxygen O, whereas the more oxidized compounds (bottom) have more oxygen O
and less hydrogen H.

Not all biological oxidation-reduction reactions involve carbon C. For example, in the conversion of molecular
nitrogen N2 to ammonia NH3 : 6 H+ + 6 e- + °N2=> 2 (-3)NH3, the nitrogenN atoms are reduced.

Electrons e- are transferred => from one molecule (electron e-donor) to another (electron e-acceptor) in
one 1 of four 4 different ways:

1. Directly as electrons e-. For example, the Fe2+/Fe3+RedOx pair can transfer an electron e- to the
Cu+/Cu2+RedOx pair:             Fe2+ + Cu2+=>  Fe3+ + Cu+

2. As hydrogen H atoms. Recall that a hydrogen H atom consists of a proton H+ and a single electron e-.
In this case we can write the general equation:   AH2=> A + 2 e- + 2 H+

where AH2 is the hydrogen/electrondonor.
(Do not mistake the above reaction for an acid dissociation; the H+arises from the removal of a hydrogen atom
(H+ + e-).) AH2 and A together constitute a conjugate RedOx pair (A / AH2), which can reduce another compound
B (or RedOx pair, B / BH2) by transfer of hydrogen H atoms:                           AH2 + B => A + BH2

3. As a hydride ion (:H-), which has two 2 electrons e-.
                             This occurs m the case of NAD-linked dehydrogenases, described below.

4. Through direct combination with oxygen O2. In this case, oxygen O2 combines with an organicreductant and is
covalently incorporated in the product, as in the oxidation of a hydrocarbon to an alcohol by
1/2 O2 presented as O:                                                        R-CH2-H + .:O:. => R-CH2-::O::-H

The hydrocarbon is the electron e-donor and the oxygen O atom is the electron e-acceptor.
All four 4 types of electron e- transfer occur in cells. The neutral term reducing equivalent is commonly used
to designate a single 1 electron e- equivalent participating in an oxidation-reduction reaction, no matter whether
this equivalent is an electron e- per se, a hydrogen H (H+ + e-) atom, or a hydride ion :H-, or whether the electron e-
transfer takes place in a reaction with oxygen O to yield an oxygenated product. Because biological fuel molecules are
usually enzymatic dehydrogenated to lose two 2reducing equivalents at a time, and because each oxygen O
atom can accept two 2reducing equivalents, biochemists by convention regard the unit of biological oxidations as
two 2 reducing equivalents passing from substrate => to oxygen O.

Methane

Methane.gif


8


Figure 1. Oxidation
states of carbon C occurring in the biosphere. The oxidation states are illustrated with some representative compounds.
   Focus on the black carbon C atom and its bonding electrons e-. When this carbon C is bonded to the less electronegativeH atom, both bonding electrons
(blue - : •) are assigned to the carbon C.
   When carbon C is bonded to another carbon C, bonding electrons e- are shared equally, so one blue of the two 2 electrons e-e- is assigned to the black carbon C. When the black carbon C of our interest is bonded to the more electronegativeO atom, the bonding electrons e- are assigned to the oxygen O:.
   The number to the right of each compound is the number n of electrons e- ''owned'' by the black carbon C of our interest, a rough expression of the oxidation state of that carbon C.
   When the black carbon C undergoes oxidation (loses electrons e-), the number n |<= gets smaller.
   Thus the order of increasing ­ oxidation state is alkane <

methane <

          < alkane <

           < alkene <

           < alcohol <

           < alkyne <

           < aldehyde <

           < ketone <

                        < formic acid <

                          < carbon mondioxide.

                        < carboxylic acid <

                          < carbon dioxide.

Ethane (alkane)

Ethane.gif


7

Ethene (alkene)

Ethene.gif


6


Ethanol 7
(alcohol)

Ethanol.gif


5

Acetylene (alkYne)


Ethyne.gif


5


Formaldehyde

Formaldehyde.gif


4

 
Acetaldehyde
 
(aldehyde) 7

AcetAldehyde.gif



3


Acetone (ketone) 7 2

Acetone.gif



2


Formic acid (carboxylic acid)

FormicAcid.gif



2


Carbon monoxide

CarbonMonOxide.gif


2


Acetic acid (carboxylic   
    7 1
     
acid)

AceticAcid.gif



1


Carbon dioxide

CarbonDiOxide.gif


0

Reduction Potentials Measure Affinity for Electrons

When two 2 conjugate RedOx pairs are
together in solution, electron e- transfer from the electron e-donor of one RedOx pair to the electron e-acceptor of
the other may occur spontaneously. The tendency for such a reaction depends on the relative affinity of the electron
e- acceptor of each RedOx pair for electrons e-.

The standard reduction potential, E° a measure (in volts V) of this affinity, can be determined in an experiment such
as that described in Figure 2. Electrochemists have chosen as a standard of reference the half-reaction

H+ + e-ArrowReverseReact  H |¬(Pt) , (1/2 H2) Hydrogen saturated platinum (Pt) H2is a solid metal - first I type electrode n=1 .

E=E°+RTLn10_F.giflg([H3O+]); because of Brønsted water protonation H+ + H2O=> H3O+       (2)

E = -0.05916•pH ; because -lg([H3O+]) = pH , and RTLn10_F.gif = 0.05916 V at standard conditions (3)

   The electrode at which this half-reaction occurs (called a half-cell) is arbitrarily assigned a standard reduction potential
of 0.00 V. When this hydrogen electrode is connected through an external circuit to another half-cell in which an
oxidized species and its corresponding reduced species are present as pure H2 compound at standard concentrations C
(each solute at 1 M, each gas at 101.3 kPa or 1 atm), electrons e- tend to flow through the external circuit from
the half-cell of lower standard reduction potential E°1 to the half-cell of higher ­ standard reduction potential E°2.
   By convention, the half-cell with the stronger ­ tendency to acquire electrons e- is assigned a positive value of 2 > 0.00 V.

EMF.gif
hydrogen electrode          Test cell containing
pure H2 gas at standard      pH=7 concentration
pressure saturates              of the oxidized [H+]
platinum Pt metal with      and reduced H(Pt)
H hydrogen forming          (like solide as left)
reference cell with              species of the RedOx
known E = 0.00V               pair to be examined

   Voltmeter the device for measuring emf conected to refrence snap and test snap Figure 2. Measurement of the standard reduction      potential (Eo) of a RedOx pair at pH=7 condition.       Electrons e- flow => from the test electrode to the reference electrode if standard reduction potential is negative Eo < 0.00 V or vice versa if positive Eo >  0.00 V. The ultimate reference half-cell is the hydrogen electrode, as shown here. Electro-chemists have chosen as a standard of reference the half-reaction which this half-reaction occurs is arbitrarily assigned a standard reduction potential E=E° of 0.00 V pH=0. The electro-motive force (EMF=Eo-E) at pH=7, E of or the hydrogen electrode RedOx system H++ e-ArrowReverseReact H |¬(Pt) is -0.41412V=Eo=-0.05916•pH and EMF=-0.41412-0.0V .

   The direction of electron e- flow => depends on the relative electron ''pressure'' or potential of the two 2 cells. A salt bridge containing a saturated KCI solution provides a path for counter-ion movement between the test cell and the reference cell. From the observed emf and the known E = 0.00 V for reference cell, the Eoof the test cell containing the RedOx pair is obtained as calculation EMF + E = Eo = -0.41412 V. The cell that gains electrons has, by convention, the more positive ­ E > Eoreduction potential and so as well as E = 0.00 V >  -0.41412 V =  Eo.

   The reduction potential E of a half-cell depends not only on the chemical species present but also on their activities,
approximated by their concentrations C. About a century ago, Walther Nernst derived an equation that relates standard
reduction potential
(E') to the reduction potential (E) at any concentration of oxidized [Ox] and reduced [Red] species
in the cell:
Table 1. Standard Reduction Potentials Eo and EM of Some Biologically Important Half-Reactions, at 37 °C for
pH=7.36 and 8.37 (in mitochondria), at standard conditions 298.16 K, pH=0
for H+/ H reference electrode E°=0.00 V, E°H2O corrected to water concentration [H2O] = 997.07/18.0153 = 55.3457 M
from equations where involved, and E°37. at body temperature conditions 310.16 °K (37 °C) calculated from E°H2O

Half-reaction - OxRed systems      Data source

 Eo(V)

 EM(V)

 E°(V)

 E°H2O(V)

 E°37(V)

H2O2+2 H3O++ 2 e-   =  4 H2O                Suchotina

1.7356

1.6734

1.776

1.9821

1.9742

O2-+2 H3O++ e-   =    H2O2+ 2 H2O       David Harris

0.305

0.1806

1.2764

1.48246

1.4251

O2g+4 H3O+++ 4 e-= 6 H2O                     Suchotina

0.813

0.751

1.2288

1.38334

1.3732

NO3-+3H3O++2e-=HNO2+4H2O    University Alberta

0.2889

0.1957

0.9275

1.13355

1.1291

NO3-+ 2 H3O++2e-= NO2-+ 3 H2O          David Harris

0.3913

0.3291

0.8351

0.98967

0.95138

p-quinone+2H3O+ +2e-=Hydroquinone+2H2O

0.2336

0.1714

0.6994

0.80243

0.79365

O2aq+2H3O++2e-=H2O2aqua+ 2H2O University Alberta

0.2336

0.1715

0.6945

0.7975

0.7937

Fe3+ + e- = Fe2+      University Alberta

0.783

0.783

0.7690

0.7690

0.7830

Ubiquinone+2H3O+ +2e-=Ubiquinol+2H2O

0.0197

0.0819

0.4591

0.56215

0.5404

Fumarate2-+2H3O+ +2e-=Succinate2-+2H2O

0.0332

0.0953

0.4451

0.54815

0.52695

CrotonylCoA+2H3O++2e-=ButyrylCoA+2H2O

-0.0774

-0.1395

0.3991

0.50215

0.48273

C6H6O6+2H3O+ +2e-=AscorbicAcid+2H2ODC.Harris

-.0862

-.1483

0.3900

0.4930

0.47395

Glyoxylate+2H3O++2e-=glycolate+2H2OD.C.Harris 25°C

-0.111

-0.171

0.324

0.42715

0.42715

Cytochrome F Fe3+ + e-= Fe2+      David Harris

0.3509

0.3509

0.3650

0.3650

0.3509

[FeIII(CN)6]3-+ e-= [FeII(CN)6]4-    University Alberta

0.3258

0.3258

0.3557

0.3557

0.3258

Oxalo-acetate2-+2H3O+ +2e-=Malate2-+2H2O

-0.2225

-0.2847

0.2481

0.35115

0.33757

Cytochrome a3 Fe3++ e-= Fe2+

0.3365

0.3365

0.3500

0.3500

0.3365

Pyruvate-+2H3O+ +2e-=lactate-+2H2O   

-0.2408

-0.3030

0.2291

0.33215

0.3193

FADfree+2H3O+ +2e-=FADH2 +2H2O                   *

-0.2735

-0.3356

0.1951

0.29815

0.28662

CH3CHO+2H3O++2e-=CH3CH2OH+2H2OKortly Sh

-0.2784

-0.3406

0.1900

0.2930

0.28169

Cytochrome a Fe3+ +e-= Fe2+

0.2788

0.2788

0.290

0.290

0.2788

GlutaS-Sthione+2H3O+ +2e-=2GlutathSH+2H2O

-0.2841

-0.3462

0.1841

0.28715

0.27604

Srhb+2H3O+ +2e-=HSH+2H2O       University Alberta

-0.2859

-0.3480

0.1739

0.27693

0.27424

Cytochrome c Fe3+ + e- =  Fe2+

0.2442

0.2442

0.254

0.254

0.2442

LipoicAcidS-S+2H3O++2e-=LipSHSH+2H2O

-0.3417

-0.4039

0.1241

0.22715

0.21837

Cytochrome c1 Fe3+ + e- = Fe2+

0.2115

0..2115

0.220

0.220

0.2115

AcetoAcetate-+2H3O+ +2e-=-OHButyrate-+2H2O

-0.3956

-0.4577

0.0681

0.17115

0.16453

-Ketoglutarate2-+CO2+2H3O++2e-=isocitrate2-+2H2O

-0.4283

-0.4904

0.0341

0.13715

0.13185

  H3O++ e-=H(Pt) + H2O   

-0.4611

-0.5232

0.000

0.10303

0.09904

Cytochrome b Fe3+ + e- = Fe2+

0.074

0.074

0.077

0.077

0.074

CH3COOH+2H3O++2e-=CH3CHO+3H2OSuchotina

-0.5784

-0.6407

-0.118

0.03654

0.03513

13PGlycerate4-+2H3O++2e-=Glycaldeh3-P2-+2H2O+Pi2-

-0.5873

-0.6496

-0.1314

-0.0284

-0.0273

NADP++H3O+ +2e-=NADPH+ H2O   

-0.3429

-0.3740

-0.117

-0.0654

-0.0629

NAD+ +H3O+ +2e-=NADH + H2O     David Harris

-0.3391

-0.3702

-0.113

-0.0614

-0.0590

O2g+ e-  =  O-2aq                  Suchotina

-0.2355

-0.2355

-0.245

-0.245

-0.2355

Ferredoxin Fe3+ + e- = ferredoxin Fe2+

-0.415

-0.415

-0.432

-0.432

-0.415

2C3H4O3 + 4 H3O+ + 4e- = C6H12O6 + 4 H2O Stryer

-.9975

-1.060

-0.5427

-0.4397

-0.4373

  H2O + e- = H(Pt) + OH-     Suchotina

-0.5938

-0.6559

-0.828

-0.9311

-0.8951

Data mostly from: 1. Loach, P.A. (I 976) In Handbook of Biochemistry and Molecular Biology,
2. 3rd edn (Fasman, G.D. ed.), Physical and Chemical Data, Vol. 1, pp. 122-130 e, CRC Press,
3. A.M.Suchotina, Handbook of Electro-Chemistry, Petersburg ,1981."Chimia"© 
4. S.Kortly and L.Shucha. Handbook of chemical equilibria in analytical chemistry. 1985.Ellis Horwood Ltd.©
5. University Alberta Data Tables Molar Thermodynamic Properties of Pure Substances,
 http://www.vhem.ualberta.ca/courses/plambeck/p101/p00403.htm
6. Boca Raton, FL.''This is the value for free FAD;
 FAD bound to a specific flavo-protein (for example succinate dehydrogenase) has a different
7. David A. Harris, "Bio-energetic at a Glance". b Blackwell Science Ltd ©, 1995, p.116.
8. Daniel C.Harris, "Quantitative chemical analysis". W.H.Freeman and Company ©, 5th ed.1999, p545

     E = E°+ RTLn10_F.gif•lgOx_Red.gif 

where - standard potential of given OxRed system measured at conditions when E=E° (as [Ox]=[Red]);
natural logarithm of number 10 - ln(10)=2.302585093 ; universal gas constant - R=8.3144 J/mol/K ;
absolute thermodynamics temperature T=273.16°+ 25°(C)=298.16 K at standard temperature conditions measured:
as Kelvin scale value 273.16 K at zero 0°C point plus on Celsius scale measured 25°C but human body temperature
37°C that will be higher T = 273.16° + 37°(C) = 310.16 K non-standard conditions;
Faraday's constant - F = 96 485 C (coulomb) 1 mol of electrons e- electric charge in C;
At 298 K (25°C) and at 310.16 K (37°C), this expression reduces to respectively following expressions:

E = E°+ 0_0591.gif•lgOx_Red.gif ; E = E°+ 0_6154.gif•lgOx_Red.gif

Many half-reactions of interest to biochemists involve protons H+or thermodynamically corrected reality
hydronium ion H3O+. As in the definition of D. biochemists define the standard state for oxidation-reduction
reactions as pH 7.36 and express the standard reduction potential as Eo, the standard reduction potential at pH 7.36.
The standard reduction potentials given in Table 1 and used throughout this book are values for Eo and are therefore
only valid for systems at neutral pH. Each value represents the potential difference when the conjugate RedOx pair,
at equal concentrations [Ox]=[Red] and pH=7.36, is connected with the standard (pH=0) hydrogen electrode.
Notice in Table 1 that when the conjugate pair H+/H at pH 7 is connected with the standard hydrogen electrode
(pH 0), electrons e- tend to flow from the pH 7 cell => to => the standard (pH 0) cell; the measured Eo for the
H+/ H pair according (3) is -0.05916*7=- 0.41412 V

Standard Reduction Potentials Can Be Used to Calculate the Free-Energy Change

The usefulness of reduction potentials stems from the fact that when reduction potentials E1 and E2 values have
been determined for any two 2 half-cells, relative to the standard hydrogen electrode potential E°=0.00 V as
reference, their EMF=EOx2-ERed1 values  relative to each other are also known. 'We can then predict the direction
Red1=> Ox2 in which electrons e- will tend => to flow when the two 2 half-cells are connected through an
external circuit or when components Red1=>  Ox2 of both half-cells are present in the same solution. Electrons e-
tend =>  to flow to the half-cell with the more positive and the strength of that tendency is proportional to the
difference in reduction potentials, D.

The energy DG made available by this spontaneous electron e- flow => by work action DG = W
(the free-energy change DG for the oxidation-reduction reaction) is proportional ~ to DE. According RedOx system
positive charged oxidized form Oxn+ - formed with lost electrons ne- - flows =>  from metal site (mostly used
platinum Pt and bearing free electron ne- gas) to solution. In this movement process RedOx system are
accomplished the chemical work W = E•F•n by spending given RedOx system free energy
-DG = -(GOx - GRed) in conversion of reduced form Red to oxidized form Oxn+:

                 Red ArrowReverseReact Oxn+ + ne-;  W = E•F•n = DG   (4)

Here n represents the number of electrons ne- transferred in the reaction.

Chemical Potential of Species µ

Professor Ilya Prigogine and co has shown that chemical potentialµ of compound A show, how much change
of free energy DGA brings into system of our interest when adding of 1 mol amount of compound A .

In a fact: how great amount of free energy belongs to one 1 mol of compound.

It means how much free energy DGA has itself per 1 mol compound A ,
chemical potential µ of compound A if amount of compound in molar numbers is DnA = 1 mol
    µA = DGA_DnA.gif ; µA = DG° + R•T•ln(XA) ,                                                             (5)
where XA is concentration of A unitles mol fraction XA = dnA_Dntotal.gif 

For pure compound A when nA = ntotal mol fraction is XA = 1 so ln(1) = 0 and µ = D that present
standard free energy of formation the 1 mol compound A from elements.

Reaction proceeds completely until end only when products of reaction have hardly little disposition to reverse change
back.
In other words these products of reaction have trifling or zero value of chemical potential.

Data from: I. Prigogine, R. Defey. "Chemical Thermodynamics". 1954, Longmans Green and co ©.

Conditions of chemical equilibrium.

Provided chemical potential of reaction products is taking into consideration (it has anything remarkable level
of value), then reaction proceeds not completely until end, go not on but one can observe the setting in equilibria.
In state of equilibria sum of chemical potentials for initial compounds is equal to sum of chemical potentials
for products – according chemical equation

            Red DirectReverseR.gif  Oxn+ + ne- ;  W = E•F•n = DG

initial compounds       products          work accomplished by movement of positively (n+) charged Oxn+
form from metal surface, comprising electron ne- gas , and touching Red form, to solution of RedOx system.

For RedOx system due to electric work of charged Oxn+ movement between metal and RedOx system solution
sides are not equal µRedµOxn+ + n µe-  but coefficient ne- means e-+e-+e-…n times electron e- takes a part
in reaction as is seen in expression of equilibrium (4). Free energy change DG for any chemical reaction is
to calculate as chemical potential sum subtraction: the productSum.gifµproduct minus initialSum.gifµinitial :

DG = (Sum.gifµOxn+ + n µe- ) - Sum.gifµRed=E•F•n , and equilibrium establishes when electric work is compensated
by free energy change DG=W=E•F•n and formation on metal (Pt) electrode reduction potential E .


                                    
Sum.gifµRed+ E•F•n= Sum.gifµOxn+ + e-                                                        (6)

At equilibrium the chemical potential sum of initial compounds and products are equal and reduced form includes
the compensating free energy change DG=W=E•F•n forming on metal (Pt) electrode reduction potential E .
Becomes obvious that chemical potential sum of oxidised form has the number n additional chemical potential
of free electrons e- those values for all known RedOx systems are different and mostly laying in side interval
between -90 ÷ +90 kJ/mol. Electrons ne- are occupied metal (Pt) free electron gas solid phase and as pure solid
compound has mol fraction concentration Xe- = 1. Expressing above mentioned meaning of chemical potentials (5)
we can calculate the free energy change DG in given RedOx system for Red form:

DRed + R•T•ln(XRed) + E•F•n= DOxn+ + R•T•ln(XOxn+) + n De- + n•R•T•ln(Xe-)
DRed + R•T•ln(XRed) + E•F•n= DOxn+ + R•T•ln(XOxn+) + n De-                                             (7)
DG = E•F•n = DOxn+ +n De-  - DRed + R•T•lnXOx_XRed.gif

DG° = E°•F•n = DOxn+ +n De-  - DRed                                                                                 (8)

Standard free energy change RedOx system for reduced form Red is expressed as DRed=E°•F•n and
for oxidized form Oxn+ as                                        DOxn+ = - E°•F•n

DGRed = E•F•n = E°•F•n + R•T•lnXOx_XRed.gif ; DGOxn+ = -E•F•n = -E°•F•n - R•T•lnXOx_XRed.gif   (9)

With this equation we can calculate the standardfree-energy change D for any RedOx system from the values
of in a table of reduction potentials (Table 1) and the free-energy change DG according (9) at known
concentrations XRedand XOxn+ of the each species form (DGRed and DGOxn+) participating in the RedOx system.

Considerable oxidation-reduction reaction is composed from two 2RedOx systems (half-reactions) using
compounds reaction equivalence law |+m'•ne-| = |-n'•me-| we have balanced oxidation-redaction reaction and
can get the summary reaction of both half-reactions :

(-)  Red1  ArrowReverseReact  Ox1n++ ne-   |•m'

(+) Ox2m+ + me-ArrowReverseReact Red2   |•n'

m'•Red1 + n'•Ox2m+=>m'•Ox1n++ n'•Red2 ; initial compounds => forming => products direction of reaction.

With this equation we can calculate the standardfree-energy change D for equi-molar amount of any
oxidation-reduction reaction from the values of in a table of reduction potentials (Table 1)

DG° = m'D1Red - n'D2Oxn+ = m'•E°1Red•F•n - n'•E°2Oxn+•F•m = (1Red -E°2Oxn+)•F•(m'n=n'm), where nm
is equivalent - common number of electrons e- involved in RedOx reaction can be less n'm' ≤ n•m .

and the free-energy change DG according (9) at known concentrations XRedand XOxn+ of the each species
(DGRed and DGOxn+) participating in the reaction.

DG=m'DG1Red-n'DG2Oxn+=m'•E1Red•F•n -n'•E2Oxn+•F•m =(E1Red-E2Oxn+)•F•(m'n=n'm)=

= (1Red-E°2Oxn+)F•(m'n=n'm)+R•T•lnlnK-X_X.gif, where equilibrium        (10)

constant Keq = K-X_X.gif is the ratio as a multiple products over initial compounds concentrations.

Consider the reaction in which acet-aldehyde is reduced by the biological electron carrier NADH:

H3C-CH=O + NADH + H3O+ArrowReverseReact H3C-CH2-OH + NAD+ + H2O                                        (11)
acet-aldehyde                                     ethanol

The relevant half-reactions and their E37 values at pH = 7.36 , (pH-log(1/aHOH)) are:

(1) NADH      + H2O             ArrowReverseReact  NAD+ + H3O+ + 2e-       E137 = -0.0590 V   (David Harris)

(2) CH3CHO+2H3O+ + 2e- ArrowReverseReact CH3CH2OH+ 2 H2O        E237 =  0.28169 V  (Suchotina)

By convention (10), n=2=m number of electrons 2e- DE37 is expressed as E137 of the electron donor minus E237
of the electron acceptor. Because acet-aldehyde is accepting electrons from NADH in our example
DE37 = E137-E237 = -0.0590 V- 0.28169 V = -0.34069 V, and n is 2. Therefore -0.0590- 0.28625     -0.34525*2*96485 = -66622.89250/1000 = -66.62289 kJ/mol

DG° =  DE37•F•n = -0.34069 V • 2 mol • 96485 C/mol = -65.7429495kJ/mol

This is the free-energy change for the oxidation-reduction reaction at equilibrium state when DG = 0 and so :

DG°= -R•T•ln(Keq);NADacetAcetat.gif=Keq=eDG_RT.gif== 1.18065•1011
equilibrium is shifted far to products as shows Equilibrium constant Keq=1.181•1011. Anaerobic fermentation conditions NAD+ concentration exceeds NADH ratio 70 times at pH = 7.36. At presence of air oxygen O2 ratio [NAD+]/[NADH] is ten times higher 700 times over concentration NADH, what cause reaction condition to oxidize ethanol and acet-aldehyde as well known aerobic fermentation forms acetic acid. If ethanol ratio of concentrations is one as produced ethanolequal the same acet-aldehyde amount in aerobic fermentation: than calculated free energy change is positive
DG=5195 J/mol=(-65.7429495+70.938)27.51•8.3144•310.15• ln(8.85352•1011•1);
DG= -742 J/mol=(-65.7429495+65.00071)25.21•8.3144•310.15• ln(8.85352•1011•0.1)
produces 10% ethanol over acetaldehyde and reaction shifted toward acetic acid:
Anaerobic DG shifted to ethanol negative:

DG = (E1Red -E2Oxn+)F•(m'n=n'm) + R•T•lnlnK-X_X.gif 

DG = (NADH - E°Acetaldehyde)F•2 + R•T•ln lnNADacetAldehyde.gif 

DG= -65.7429495 +8.3144•310.15•ln (8.85352•1010•0.1)= -6680 J/mol

= -66.62289kJ/mol + 8.3144•310.16•ln(1.31368) =  -66.62289*1000+703.574102 = -66622.89000+703.574102 =-65919.315898/= -65.7429495kJ/mol+ 8.3144•310.16•ln(8.85352•1010•0.1) = -65.7429495+59.063= -6.680
Calculation the free-energy changes DG show possible modulation biological RedOx reaction at any
concentrations X for the reaction driving forces and as regulation direction for favorable products formation.
BioThermodynamics.doc page 13 erythrocyte -53.47kJ/mol T=310,16 K .

Cellular Oxidation of Glucose to Carbon Dioxide Requires Specialized Electron Carriers

The principles of oxidation-reductionenergetic described above apply to the many metabolic reactions that
involve electron e- transfers. For example, in many organisms, the oxidation of glucose supplies energy for the
production of ATP. The complete oxidation of glucose:       C6H12O6 + 6 O2==> 6 CO2 + 6 H2O

has a D of -3049,55 kJ/mol. This is a much larger release of free energy than is required for ATP synthesis
(erythrocyte pH = 7.36 use -53.47kJ/mol 55,4% of 100% 120 kJ/mol. Cells do not convert glucose
to CO2aqua in a single, high­-energy-releasing reaction, but rather in a series of controlled reactions, some of which are
oxidations. The free energy released in these oxidation steps is of the same order of magnitude as that required for
ATP synthesis from ADP, with some energy to spare. Electrons e- removed in these oxidation steps are transferred
to coenzymes specialized for carrying electrons e-, such as NADH and FADH2 (described below).

A Few Types of Coenzymes and Proteins Serve as Universal Electron Carriers

The multitude of enzymes that catalyze cellular oxidationschannel electrons e- from their hundreds 100 of different
substrates into just a few types of universal electron carriers. The reduction of these carriers in catabolic processes
results in the conservation of free energy released by substrateoxidation. NAD+, NADP+, FMN, and FAD are
water-soluble coenzymes that undergo reversible ArrowReverseReact oxidation and reduction in many of the electron-transfer e-
reactions of metabolism. The nucleotides NAD+ and NADP+ move readily from one enzyme to => another; the flavin
nucleotides FMN
and FAD are usually very tightly bound to the enzymes, called flavo-proteins, for which they
serve as prosthetic groups. Lipid-soluble quinones such as ubiquinone and plasto-quinone act as electron carriers
and protondonors in the non-aqueous environment of membranes. Iron-sulfur proteins and cytochromes, which
have tightly bound prosthetic groups that undergo reversible ArrowReverseReact oxidation and reduction, also serve as electron e-
carriers in many oxidation-reduction reactions. Some of these proteins are water-soluble, but others are peripheral
or integral membrane proteins.

We conclude this chapter by describing some chemical features of nucleotide coenzymes and some of the enzymes
(dehydrogenases and flavo-proteins) that use them. The oxidation- reduction chemistry of quinones, iron-sulfur
proteins, and cytochromes is discussed in Oxidative Phosphorylation and Photo-Phosphorylation.

NADH and NADPH Act with Dehydrogenases as Soluble Electron Carriers

Nicotin-amide adenine dinucleotide (NAD+ in its oxidized form) and its close analog nicotin-amide adenine
dinucleotidephosphate
(NADP+) are composed of two 2nucleotides joined through their phosphate groups by
a phospho-anhydride bond (Fig. 3). Because the nicotinamide ring resembles pyridine, these compounds are
sometimes called pyridinenucleotides. The vitamin niacin is the source of the nicotin-amide moiety in
nicotin-amide nucleotides.

Both coenzymes undergo reversible ArrowReverseReact reduction of the nicotinamide ring (Fig. 3). As a substrate molecule
undergoes oxidation (dehydrogenation), giving up two 2 hydrogen H atoms, the oxidized form of the nucleotide
(NAD+ or NADP+) accepts a hydride ion (:H- the equivalent of a proton H+ and two 2 electrons e-) and is
transformed into the reduced form (NADH or NADPH). The second 2nd proton H+ removed from the substrate is
released to the aqueous solvent H2O. The half-reaction for each type of nucleotide is therefore

(1) NADH + H2OArrowReverseReactNAD+ + H3O+ + 2e-  E137 = -0.0590 V (David Harris)

(2) NADPH + H2OArrowReverseReactNADP+ + H3O+ + 2e-  E237 = -0.0629 V   (CRC)

Reduction of NAD+ or NADP+ converts the benzenoidring of the nicotin-amide moiety (with a fixed positive (+)
charge on the ring nitrogen N) to the quinonoid form (with no charge ° on the nitrogen °N). Note that the reduced
nucleotides absorb light at 340 nm: the oxidized forms do not (Fig. 13). The plus sign in the abbreviations NAD+ and
NADP+ does not indicate the no charge on these molecules (they are both negatively (-) charged), but rather that
the nicotin-amide ring is in its oxidized form, with a positive (+) charge on the nitrogen N+ atom. In the abbreviations
NADH and NADPH, the "H" denotes the added hydride ion. To refer to these nucleotides without specifying their
oxidation state, we use NAD and NADP.

The total concentration of NAD++ NADH in most tissues is about 10-5M; that of NADP+ + NADPH is about
10 times lower . In many cells and tissues, the ratio of NAD+ (oxidized) to NADH (reduced) is high, favoring
hydride H- transfer from a substrate to NAD+ to form NADH. By contrast, NADPH (reduced) is generally present
in greater ­ amounts than its oxidized form, NADP+, favoring hydride H- transfer from NADPH to a substrate.
This reflects the specialized metabolic roles of the two 2coenzymes: NAD+ generally functions in oxidations - usually
as part of a catabolic reaction; and NADPH is the usual coenzyme in reductions nearly always as part of anabolic
reaction. A few enzymes can use either coenzyme. but most show a strong preference for one over the other.
This functional specialization allows a cell to maintain two 2 distinct pools of electron carrier, with two 2 distinct
functions, in the same cellular compartment.

     More than 200 enzymes are known to catalyze reactions in which NAD+ (or NADP+) accepts a hydride :H- ion
from a reduced substrate AH2, or NADPH (or NADH) donates a hydride :H- ion to an oxidized substrate A.
The general reactions are 

Red substrate  AH2 + NAD+ + H2OArrowReverseReactA + NADH + H3O+                                             (11)

Ox substrate  A + NADPH + H3O+ArrowReverseReactAH2 + NADP+ + H2O                                          (11)

where AH2is the reduced substrate and A the oxidized substrate. The general name for an enzyme of this type is
oxidoreductase; they are also commonly called dehydrogenases. For example, alcohol dehydrogenase catalyzes
the first 1st step in the catabolism of ethanol, in which ethanol is oxidized to acet-aldehyde:
H3C-CH2-OH + NAD+ + H2OArrowReverseReactH3C-CH=O + NADH + H3O+                                           (11)

        Ethanol                                  Acet-aldehyde

Notice that one of the carbon atoms C in ethanol has lost a hydrogen H; the compound has been oxidized from
 an alcohol to an aldehyde (Fig. 1).

When NAD+ or NADP+is reduced. the hydride :H- ion could in principle the transferred to either side of
the nicotin-amide ring: the front (A side) or the back (B side) as represented in Figure 3. Studies with isotopically
labeled * substrates have shown that a given enzyme catalyzes either an A-type or B-type transfer, but not both.
For example, yeast alcohol dehydrogenase and lactate dehydrogenase of vertebrate heart transfer a hydride
:H- ion to (or remove a hydride :H- ion from) the A side of the nicotin-amide ring: they are classed as type A
dehydrogenases
to distinguish them from another group of enzymes that transfer a hydride :H- ion to
(or remove a hydride : H- ion from) the B side of the nicotin-amide ring (Table 2).

The association between a dehydrogenase and NAD or NADP is relatively loose; the coenzyme readily diffuses
from one enzyme to another, acting as a water-soluble carrier of electrons e- from one 1 metabolite to another.
For example, in the production of alcohol during fermentation of glucose by, yeast cells, a hydride :H- ion is removed
from glycer-aldehyde 3-phosphate by, one 1 enzyme (glycer-aldehyde 3-phosphate dehydrogenase, a type B enzyme)
and transferred to NAD+. The NADH produced then leaves the enzyme surface and diffuses to another enzyme
(alcohol dehydrogenase, a type A enzyme), which transfers a hydride :H- ion to acet-aldehyde, producing ethanol. Reduced (calculated) At T=310 K (37° C) human glyceraldehyde3phosphate:

Reduced      (calculated)
OHC-CHOH-CH2OPO3H-+2H2O+HPO42-ArrowReverseReact-2O3POOC-CHOHCH2OPO3H-+2e-+2H3O+E1=-0.0273V
(Ox) NAD+ + H3O+ + 2e-  ArrowReverseReactNADH + H2O  E2H2O= -0.0614 V (David Harris)

E137 = 0.0317 V + E2o = 0.0317 -0.059= -0.0273
glyceraldehyde3phosphate <=>1,3-PhosphoGlycerate
?G° = 6.117 kJ/mol = F•2•(E1o - E2o) ; (E1o - E2o) = ?G° /2/F = 6.117*1000/2/96485 = 0.0317 V
(1) OHC-CHOH-CH2OPO3H- + NAD+ + H2O + HPO42-=>

  Glyceraldehyde 3-phosphate =>-2O3POOC-CHOHCH2OPO3H-+NADH+H3O+DG°=6.276 kJ/mol      acetaldehyde                                 3-phospho-glycerate                                (Carnegie Mellon Univ.)

(2) H3C-CH=O+NADH+H3O+ArrowReverseReact H3C-CH2-OH + NAD+ + H2O DG°= -58.856 kJ/mol (calculated)

DG° = 6.276 kJ/mol = F•2(E1o - E2o) ; (-0.113-0.192)•96485•2 = -0.305*2*96485/1000 = -58.856+6.276

Sum H3CCHO+OHCCHOHCH2OPO3H-+H2O+HPO42-=>H3CCH2OH+-2O3POOCCHOHCH2OPO3H-
Sum DG° = 6.276 kJ/mol +  -58.856 kJ/mol =  -52.580 kJ/mol

Notice that in the overall reaction there is no net production or consumption of NAD+ or NADH; the coenzymes
function catalytically and are recycled repeatedly without a net change in the concentration C of
[NAD+]+[NADH].
NAD++ 2e- + H+=> NADHright.gifor=> NADHleft.gif
 (a) NAD+ oxidized                                                 A side   NADH reduced    B side
       In NADP+  is esterified this hydroxyl with phosphate                 
  Absorbance
Absorbance.gif
                 220            240              260              280               300                320               340                 360                     380 => nm

(b)                          Wavelength (nm)        =>
Figure 3.NAD and NADP (a) Nicotin-amide adenine di-nucleotide (NAD+) and its phosphorylated analog
NADP+ undergoes reduction to NADH and NADPH, accepting a hydride :H- ion (two 2 electrons e- and
one 1 proton H+) from an oxidizablesubstrate. The hydride :H- ion is added to either the front (the A side) or
the back (the B side) of the planar nicotin-amide ring (see Table 2). (b) The UV absorption spectra of NAD+ and
NADH. Reduction of the nicotin-amide ring produces a new, broad absorption band with a maximum at 340 nm.
The production of NADH during an enzyme-catalyzed reaction can be conveniently followed by observing
the appearance of the absorbance at 340 nm; a=6200M-1cm-1
Table 2. Stereo specificity of Dehydrogenases That Employ NAD+ or NADP+ as Coenzymes

Enzyme

Coenzyme

Stereo chemical specificity for nicotin-amide ring  (A or B)

Iso-citrate dehydrogenase

NAD+

A

a-Keto-glutarate dehydrogenase

NAD+

                 B

Glucose 6-phosphate dehydrogenase

NADP+

                 B

Malate dehydrogenase

NAD+

A

Glutamate dehydrogenase

NAD+

or NADP+                B

Glyceraldehyde 3-phosphate dehydrogenase

NAD+

                   B

Lactate dehydrogenase

NAD+

A

Alcohol dehydrogenase

NAD+

A

Table 3. Some Enzymes (Flavo-proteins) That Employ Flavin Nucleotide Coenzymes

Enzyme

Flavin nucleotide

Fatty acyl-CoA dehydrogenase

FAD

Di-hydro-lipoyl dehydrogenase

FAD

Succinate dehydrogenase

FAD

Glycerol 3-phosphate dehydrogenase

FAD

Thio-redoxin reductase

FAD

NADH dehydrogenase (Complex 1)

FMN

Glycolate dehydrogenase

FMN

     flavin mono-nucleotide (FMN)                                                   isoalloxazine ring

FAD.gif+e-+H+ FMNH.gif+e-+H+ FMNHH.gif

 ­Flavin adenine dinucleotide (FAD)             =>    FADH* (FMNH*)      =>   FADH2 (FMNH2)

                                                                       =>    (FMN) (semi-quinone)=> (fully reduced)

Figure 4. Structures of oxidized and reduced FAD and FMN. FMN consists of the structure above the dashed line
shown on the  oxidized (FAD) structure. The flavin nucleotides accept two 2 hydrogen H atoms (two 2 electrons e-
and two 2 protons H+), both of which appear in the flavin ring system. When FAD or FMN accepts only
one 1 hydrogen H atom, the semi-quinone, a stable free radical, forms.

Flavin Nucleotides Are Tightly Bound in Flavo-proteins

Flavo-proteins (Table 3) are enzymes that catalyze oxidation-reduction reactions using either flavin
mono-nucleotide
(FMN) or flavin adenine dinucleotide (FAD) as coenzyme (Fig. 4). These coenzymes are
derived from the vitamin riboflavin. The fused ring structure of flavin nucleotides (the isoalloxazine ring)
undergoes reversible reduction, accepting either one 1 or two 2 electrons e- in the form of one 1 or two 2 hydrogen H
atoms (each atom an electron e- plus a proton H+) from a reduced substrate. The fully reduced forms are abbreviated
FADH2 and FMNH2. When a fully oxidizedflavin nucleotide accepts only one 1 electron e- (one hydrogen H atom),
the semi-quinone form of the isoalloxazine ring is produced, abbreviated FADH* and FMNH*. Because
flavo-proteins can participate in either one-1 or two-2 electron e- transfers, this class of proteins is involved in
a greater diversity of reactions than the pyridine nucleotide-linked dehydrogenases.

Like the nicotin-amide coenzymes (Fig. 14-15), the flavin nucleotides undergo a shift in a major absorption band
on reduction. Oxidized flavo-proteins generally have an absorption maximum near 570 nm; when reduced,
the absorption maximum shifts => to about 450 nm. This change can be used to assay reactions involving
a flavo-protein.

The flavin nucleotide in most flavo-proteins is bound rather tightly to the protein, and in some enzymes, such
as succinate dehydrogenase, it is bound covalently. Such tightly bound coenzymes are properly called prosthetic
groups. They do not transfer electrons e- by diffusing from one 1enzyme to another second 2nd; rather, they provide
a means by which the flavo-protein can temporarily hold electrons e- while it catalyzes electron e- transfer from
a reducedsubstrate to an electron e- acceptor. One important feature of the flavo-proteins is the variability in
the standard reduction potential () of the bound flavin nucleotide. Tight association between the enzyme and
prosthetic group confers on the flavin ring a reduction potential E typical of that particular flavo-protein, sometimes
quite different from that of the free flavin nucleotide. FAD bound to succinate dehydrogenase, for example, has an
EM
= 0.0953 V compared with -0.3356 V for free FAD. Flavo-proteins are often very complex: some have, in addition
to a flavin nucleotide. tightly bound inorganic ions (iron Fen+ or molybdenum Mon+, for example) capable
of participating in electron e- transfer.

Summary

Living cells constantly perform work W and thus require energy E for the maintenance of highly organized.
structures, for the synthesis or cellular components, for movement, for the generation of electric currents,
for the production of light, and for many other processes. Bio-energetic is the quantitative study of energy E
relationships and energy E conversions in biological systems. Biological energy E transformations obey the laws
of thermodynamics. All chemical reactions are influenced by two 2 forces: the tendency to achieve the most
stable bonding state (for which enthalpy, H, is a useful expression to show minimum reach at the most stable
bonding state ) and the tendency to achieve the highest^degree of dispersed energy, T•S (called bound energy),
which measure is the entropy, S. The production of entropy in a reaction as positive difference DS = Sproducts - Sreactants> 0
is entropy increase^from Sreactants to Sproducts due to dispersion of energy among the members of reaction. The net
driving force in a reaction is the free-energy G decrease => from Greactants to=> Gproducts and negative difference
DG = Gproducts - Greactants< 0, which represents the net effect of these
two 2 factors: T•DS+DG = DH . Cells require sources of free energy G to perform work W.

The standard transformed free-energy change, D, is a physical constant characteristic for a given reaction and
can be calculated from the equilibrium constant Keq for the reaction: DG°= -R•T•In(Keq). The actual free-energy
change, DG, is a variable, which depends on, D and on the concentrations C of reactants and products:
DG= DG°+ R•T•ln([products]/[reactants]). When DG is large and negative DG < 0, the reaction tends to go in
the forward => direction; when it is large and positive DG > 0, the reaction tends to go in the reverse <= direction;
and when DG = 0, the system is at equilibrium. The free-energy change DG for a reaction is independent of
the pathway by which the reaction occurs only on initial(Greactants) and final (Gproducts) states. Free-energy changes DG
are additive; the net chemical reaction that results from the successive occurrence of reactions sharing a common
intermediate has an overall free-energy change DGreaction that is the sum of the DGreaction = DGreaction1 + DGreaction2 values
for the individual reactions reaction1 and reaction2.

ATP is the chemical link between catabolism and anabolism. It constitutes the energy currency of the living cell.
Its ex-ergonic conversion to ADP and Pi, or to AMP and PPi, is coupled to a large number of endergonic reactions
and processes. In general, it is not ATPhydrolysis, but the transfer of a phosphoryl, pyro-phosphoryl, or adenylyl
group from ATP to a substrate or enzyme molecule that couples the energy of ATP breakdown to endergonic
transformations of substrates. By these group transfer reactions, ATP provides the energy for anabolic reactions,
including the synthesis of informational molecules, and for the transport of molecules and ions across membranes
against <=> concentration C gradients and electrical potential E gradients. Muscle contraction is one of several
exceptions to this generalization; the conformational changes that produce muscle contraction are driven by ATP
hydrolysis directly.

Cells contain other metabolites with large, negative DG < 0, free energies of hydrolysis, including
phospho-enol-pyruvate, 1,3-bis-phospho-glycerate, and phospho-creatine. These high-energy compounds,
like ATP, have a high phosphoryl group transfer potential; they are good donors of the phosphoryl group.
Thio-esters also have high ­ free energies G of hydrolysis.

Biological oxidation-reduction reactions can be described in terms of two 2 half-reactions (called RedOx systems),
each with a characteristic standard reduction potential, E°. When two 2 electro-chemical half-cells, each containing
the components (oxidized and reduced  forms) of a half-reaction, are connected, electrons e- tend to flow => to
the half-cell with the higher^reduction potential E. The strength of this tendency is proportional to the difference
between the two 2reduction potentials (DE) and is a function of the concentrations C of oxidized [Ox] and
reduced [Red] species. The standard free-energy change D for an oxidation-reduction reaction is directly
proportional to the difference in standard reduction potentials D of the two 2 half-cells: DG° = F•n•D.

Many biological oxidation reactions are dehydrogenation in which one 1 or two 2 hydrogen H atoms (electron e-
and proton H+) are transferred from a substrate to a hydrogen Hacceptor. Oxidation-reduction reactions in cells
involve specialized electron e-carriers. NAD and NADP are the freely diffusible coenzymes of many
dehydrogenases. Both NAD+ and NADP+accept two 2 electrons e- and one 1 proton H+. FAD and FMN, the flavin
nucleotides
, serve as tightly bound prosthetic groups of flavo-proteins. They can accept either one 1 or
two 2 electrons e-. In many organisms, a central energy-conserving process is the stepwiseoxidation of glucose
to CO2, in which some of the energy of oxidation is conserved in ATP as electrons e- are passed to O2.

                                             Further Reading

                                         Bio-energetic and Thermodynamics

1.  Prigogine, R. Defey. "Chemical Thermodynamics". 1954, Longmans Green and co ©.
Correct basic concepts for Biochemical Thermodynamics.

2.  S.Kortly and L.Shucha. Handbook of chemical equilibria in analytical chemistry.
                                                                                            1985.Ellis Horwood Ltd.©

3.  3rd edn (Fasman, G.D. ed.), Physical and Chemical Data, Vol. 1, pp. 130 e,
                                                       The Chemical Rubber Publishing Co. CRC Press ©

4.  Atkins, P.W. (1984) The Second Law, Scientific American Books, Inc., New York.
A well-illustrated and elementary discussion of the second law and its implications.

5.  Becker, W.M. (1977) Energy and the Living Cell: An Introduction to Bio-energetics,
                                                                  J.B. Lippincott Company, Philadelphia.
A, clear introductory account of cellular metabolism, in terms of energetics.

6.  Bergethon, P.R. (1998) The Physical Basis of Biochemistry, Springer Verlag, New York.
The excellent general references for physical biochemistry, with good discussions of the
application of thermodynamics to biochemistry.

7.  Edsall, J.T. & Gutfreund, H. (198.'3) Bio-thermodynamics: The Study of Biochemical
Processes at Equilibrium, John Wiley & Sons, Inc., New York.

8.  Harold, F.M. (1986) The Vital Force: A Study of Bio-energetics,
                                                                           W.H. Freeman and Company, New York.
A beautifully clear discussion of thermodynamics in biological processes.

9.  Harris, D.A. (1995) Bio-energetics at a Glance, Blackwell Science, Oxford.
A short, clearly written account of cellular energetics, including basic concepts on thermodynamics.

10.       Morowitz, H.J. (1978) Foundations of Bio-energetics, Academic Press, Inc., New York.
Clear, rigorous description of thermodynamics biology. Out of print.

11.       Tinoco, L, Jr., Saner, K., & Wang, J.C. (1996)
Physical Chemistry: Principles and Applications in Biological Sciences,
3rd edn, Prentice-Ha1l, Inc., Upper Saddle River, NJ.
Thermodynamics.

12.  van Holde, K.E., Johnson, W.C., & Ho, P.S. (1998) Principles of Physical Biochemistry,
                                                                               Prentice-Hall, Inc., Upper Saddle River N.J.

Phosphoryl Group Transfers and ATP

1.  Alberty, R.A. (1994) Biochemical thermodynamics. Biochim. Biophys. Acta 1207, 1-11.
Explains the distinction between biochemical and chemical equations, and the calculation and
meaning of transformed thermodynamic properties for ATP and other phosphorylated compounds.

2.  Bridger, W.A. & Henderson, J.F. (1983) Co/I ATP, .John Wiley & Sons, Inc., New York.
The chemistry of ATP, its role in metabolic regulation, and its catabolic and anabolic roles.

3.  Frey, P.A. & Arabshahi, A. (1995) Standard free-energy change for the hydrolysis of
the a-b-phospho-anhydride bridge in ATP. Biochemistry 34, 1 1,307-11,310.

4.  Hanson, R.W. (1989) The role of ATP in metabolism. Biochem. Educ. 17, 86-92,
Excellent summary of the chemistry and biology of ATP

5.  Lipmann, F. (1941) Metabolic generation and utilization of phosphate bond energy.
Adv. enzymol. 11,96-162.
The classic description of the role of high-energy phosphate compounds in biology.

6.  Pullman, B. & Pullman, A. (1960) Electronic structure of energy-rich phosphates.
Radiat. Res. Suppl. 2, pp. 160-181.
An advanced discussion of the chemistry of' ATP and other ''energy-rich'' compounds.

7.  Veech, R.L., Lawson, J.W.R., Cornell, N.W., & Krebs, H.A. (1979)
Cytosolic phosphorylation potential. J Biol. Chem. 254, 6538-6547.
Experimental determination of ATP, ADP and Pi concentrations in brain, muscle, and liver, and
a discussion of the problems in determining the real free-energy change for ATP synthesis in cells.

8.  Westheimer, F.H. (1987) Why nature chose phosphates. Science 235, 1173-1178.
A chemist's description of the unique suitability of phosphate esters and anhydrides
for metabolic transformations.

                                                                             Biological Oxidation- Reduction Reactions

1.  Dolphin, D., Avramovic, O., & Poulson, R. (eds) (1987) Pyridine Nucleotide Coenzymes:
Chemical, Biochemical, and Medical Aspects, John Wiley & Sons, Inc., New York.
An excellent two-volume collection of authoritative reviews. Among the most useful are
the problems by Kaplan, Westheimer, Veech, and Ohno and Ushio.

                                                                                   Problems

1. Entropy Changes during Egg Development

Consider a system consisting of an egg in an incubator. The white and yolk of the egg contain proteins,
carbohydrates, and lipids. If fertilized, the egg is transformed from a single meiotic cell to a complex mitotic cells
in organism. Discuss this irreversible => process in terms of the entropy changes DS in the system, surroundings,
and universe. Be sure that you first clearly define the system and surroundings-environment.

2. Calculation of DG° from Equilibrium Constants

Calculate the standard free-energy changes Dthe following metabolically important enzyme-catalyzed reactions
at 25°C and pH 7.0 from the equilibrium constants Keq given.

(a) DG° = - R•T•ln(Keq) = -8.3144*298.16•ln(6.8) = -2479.0215*1.916923= -4752.093331  = -4.752 kJ/mol

      Glutamate+oxalo-acetate<=> aspartate amino-transferase <=> aspartate + -keto-glutarate Keq=6.8

(b) DG° = - R•T•ln(Keq) = -8.3144*298.16•ln(0.0475) = -2479.0215*-3.04703= 7553.65288 =  7.553 kJ/mol

      Di-hydroxy-acetone phosphate <=>triose phosphate isomerase <=> glyceraldehyde 3-phosphate Keq= 0.0475

(c) DG° = - R•T•ln(Keq) = -8.3144*298.16•ln(254) = -2479.0215*5.537334= -13727.170039  = -13.727 kJ/mol

      Fructose 6-phosphate- +ATP4-<=>phospho-fructo-kinase<=>fructose 1,6-bisphosphate2- + ADP3- Keq=254

DG° = - R•T•ln(Keq) ; DG = DG° + R•T•ln(Keq)

3. Calculation of Equilibrium Constants Keq from D

Calculate the equilibrium constants Keq for each of the following reactions at pH 7.0 and 25°C, using
the D values of Table 1-1 Keq = EXPDG_RT.gif= -13.8*1000/8.3144/298.16 =  EXP(--5.5667) = 261.573 corrected by D = -R•T•ln(eq) = -8.3144*298.16•ln(299.9981•10-7
) = -2479.0215*-15.0195 = 37233.6634
DG°=Do+G°HPO4+G°H3O+-G°H2PO4-G°H2O=
= -13.8 +
(-1282) + (-284.7) - (-1323) - (-306.7) = 49.306 kJ/mol

D = -R•T•ln(299.9981•10-9)=37.2336kJ/mol ; eq= 1.1469*261.573•10-9 = 299.9981•10-7eq=3;GluGlu6P.gif=3•10-7=K°H2PO4Keq=1.15•10-9261.573= eq;Deq= 37.2 kJ/mol

   H2PO4- +  H2O<=>HPO42- + H3O+ ; H2PO4 = 1.1469•10-9 (Kortly Shucha)

(a) Glucose 6-phosphate-+H2O<=>glucose6-phosphatase<=>glucose+H2PO4-;Keq=261.573;Do=-13.8 kJ/mol

Glucose 6-phosphate-+2H2O<=>glucose6-phosphatase<=>glucose+HPO42-+H3O+;K°eq=3•10-7;DG°= 49.3 kJ/mol

Equilibrium is shifted toward formation of Glucose 6-phosphate- <= and is affected by H3O+ concentration.

(b) eq =LactoseGluGal.gif =  610.221 ; DG° = -15.9 kJ/mol

(b) Lactose + H2O <=>-galactosidase<=>glucose + galactose     KeqH2O = 33773

(c) Malate<=>fumarase<=>fumarate + H2O              KeqH2O = 0.004615

(c) eq =FumaratMalate.gif =  0.28636  ; DG° =  3.1 kJ/mol

4. Experimental Determination of K°eq and D

If a 0.1 M solution of glucose 1-phosphate is incubated with a catalytic amount of phospho-gluco-mutase,
the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of
the reaction components are:

Glucose 1-phosphate-<=>phospho-gluco-mutase<=>glucose 6-phosphate-

[Glu6P-] = 4.5•10-3 M                                           9.6•10-2 M = [Glu6P-]

Calculate eq=[Glu6P-]/[Glu1P-]=21.3 and D=-R•T•ln(21.33)=-7.586 kJ/mol for this reaction at 25°C.

5. Experimental Determination of DG° for ATP Hydrolysis

A direct measurement of the standard free-energy change D associated with the hydrolysis of ATP is
technically demanding because the minute amount of ATP remaining at equilibrium is difficult to measure
accurately. The value of D can be calculated indirectly, however, from the equilibrium constants of two 2
other enzymatic reactions having less favorable equilibrium constants:
D1=Do+G°HPO4+G°H3O+-G°H2PO4-G°H2O=49.306 kJ/mol

   H2PO4- +  H2O<=>HPO42- + H3O+  ; H2PO4 = 1.1469•10-9 (Kortly Shucha)

Glucose-6-phosphate- + H2O => glucose + H2PO4- ; Keq= 270 ;        Deq= -13.879 kJ/mol

Glucose 6-phosphate- + H2O Arrow206.gif glucose + H2PO4- ; o = 261.573 ;    Do= -13.8 kJ/mol

eq•H2PO4=K°eq1GluGlu6P.gif=3.1•10-7; = 4.7262•10-7 = 1.1469*261.573•10-9 =Deq1=37.16 kJ/mol

(1) Glucose-6-phosphate- + 2 H2O => glucose + HPO42-  + H3O+ ; eq1= 3.097•10-7;D1= 49.3 kJ/mol

(2) ATP4- + glucose => ADP3- + glucose 6-phosphate- ; Keq2= 890

eq2 =ADPglu6PATP.jpg = 890 ; Deq2= -16.836 kJ/mol

Using this information, calculate the standard free energy D of hydrolysis of ATP at 25°C.

eq3= =eq1eq2 =GluGlu6P.gifADPglu6PATP.gif

D3=Do+G°HPO4+G°H3O+-G°H2PO4-G°H2O= -30.5 + (-1282) + (-284.7) - (-1323) - (-306.7) = 32.606 kJ/mol

(3)ATP4-+2H2O=>ADP3-+HPO42-+H3O+; eq3=0.0002756;Deq3=20.32 kJ/mol;D123=32.464 kJ/mol

Keq°o==eq3/H2PO4=  240300 ; DGeq°o = -30.714 kJ/mol

(3) ATP4- + H2O => ADP3- + H2PO4- ; o = 220409 ;Do = -30.500 kJ/mol

6. Difference between DG° and DG Consider the following inter conversion, which occurs in glycolysis :

Fructose 6-phosphate-<=>glucose 6-phosphate- ; eq= 1.97

eq= Glu_Fruc.gif= 1.97  = 531 331  ; DG°= -R•T•ln(eq) = -1.5399 kJ/mol

(a) What is D for the reaction (assuming that the temperature is 25°C)?

(b) If the concentration of [Fruc6P-] is adjusted to 1.5 M and that of
                                          [Glu6P-] is adjusted to 0.5 M, what is DG? -1539.9+R•T•ln(0.3) =

(c) Why are D and DG different? DG= DG° + R•T•ln(Glu_Fruc.gif) = -1539.9+-2723.54 =-4.263.4 kJ/mol

7. Dependence of DG on pH. The free energy DG released by the hydrolysis of ATP under standard conditions
at pH=7 is Do=-30.5 kJ/mol. If ATP is hydrolyzed under standard conditions but at pH=5.0, is more or less
free energy released? Why?

eq3==1.94•10-6;K°eq3/H2PO4=K°o==220409

D3=Do+G°HPO4+G°H3O+-G°H2PO4-G°H2O= -30.5 + (-1282) + (-284.7) - (-1323) - (-306.7) = 32.606 kJ/mol

DG=DG°+R•T•ln=32606+R•T•ln(0.0001/0.01/aoHOH/aoHOH*0.01•10-pH)== 19397.7 -7981.3= -10.11 kJ/mol (pH=0); at T=25°C

-38.65 kJ/mol (pH=5); -50.06 kJ/mol (pH=7); -52.12 kJ/mol (pH=7.36); -57.88 kJ/mol (pH=8.37)

-11.83 kJ/mol (pH=0); at T = 37°C in mitochondria

-41.52 kJ/mol (pH=5); -53.39 kJ/mol (pH=7); -55.53 kJ/mol (pH=7.36); -61.53 kJ/mol (pH=8.37)

8. The DG° for Coupled Reactions
Glucose 1-phosphate- is converted into fructose 6-phosphate- in two 2 successive reactions:

Glucose 1-phosphate- => glucose 6-phosphate- ; D1 = -7.3 kJ/mol

Glucose 6-phosphate- =>. fructose 6-phosphate- D2 = +1.7 kJ/mol

Using the D values in Table 1.1, calculate the equilibrium constant,
DG°= D1+D2 = -7.3+1.7 = -5.6 kJ/mol for the sum of the two 2 reactions at 25°C:

Glucose 1-phosphate- => fructose 6-phosphate- ; eq= Keq1•Keq2 = EXP(5600/8.314400/298.16) =
EXP(2.258956) =
9.57309

9. Strategy for Overcoming an Unfavorable Reaction: ATP-Dependent Chemical CouplingThe
phosphorylation of glucose to glucose 6-phosphate- is the initial step in the catabolism of glucose. The
direct phosphorylation of glucose by H2PO4- and HPO42- is described by the equation at T = 310.16 K:

(a) Glucose + H2PO4-=>. glucose 6-phosphate- + H2O, DGo° =13.8 kJ/mol

K°a=GluGlu-6P.gif; a2H2PO4 =GluGlu-6PH.gifH2PO4_HPO4.gif

(a2) Glucose + HPO42- + H3O+ =>glucose 6-phosphate- + 2H2O, DG°=-49.306 kJ/mol

K°a2=GluGlu-6PH.gif=Kglu_Glu6PH.gif=2.0119•10+8

H2PO4=H2PO4_HPO4.gif; Pi=4.8 mM=[H2PO4-]+[HPO42-];[HPO42-]=4.8 - KH2PO4.gif=
=[HPO42-]+[HPO42-]KH2PO4_HPO4.gif=4.8=[HPO42-]•(1+KH2PO4_HPO4.gif);

[HPO42-]=HPO4_48.gif ; [Glu6P-] =Glu6P_48.gif= 1.0574•10-5M

(a) Calculate the equilibrium constant a for the above reaction. In the rat hepatocytepH=7.36 and at pH=7
the physiological concentrations of glucose and [H2PO4-]+[HPO42-] are maintained at approximately 4.8 mM. What is
the equilibrium concentration of glucose 6-phosphate- obtained by the direct phosphorylation of glucose by
H2PO4-+HPO42-? Respectively [Glu6P-] = 8.5•10-8M and 1.275•10-7M (pH 7.36 and 7)
Does this reaction represent a reasonable metabolic step for the catabolism of glucose? Explain.

(b) In principle, at least, one way to increase the concentration of glucose 6-phosphate- is to drive the equilibrium
reaction to the right by increasing the intracellular concentrations of glucose and

H2PO4- + HPO42-. Assuming a fixed concentration of H2PO4- + H2PO42- at 4.8 mM, how high would the
intracellular concentration of glucose have to be to give an equilibrium concentration of glucose 6-phosphate-
of [Glu6P-]=250 µM (normal physiological concentration)'? Would this route be physiologically reasonable, given
that the maximum solubility of glucose is less than 1 M?

[Glu]=Glu_48.gif= 23.64M at pH=7.36 23.64259868

(c) The phosphorylation of glucose in the cell is coupled to the hydrolysis of ATP; that is, part of the free energy
of ATP hydrolysis is utilized to effect the endergonic phosphorylation of glucose at T = 310.16 K:

(1) Glucose + H2PO4-<=>glucose6-phosphate- + H2O      ; Do1 =   13.8     kJ/mol

(2) ATP4- + H2O<=>ADP3- + H2PO4- ; o = 220409       ; Do2 =  -30.500 kJ/mol

Sum: ATP4- + glucose<=>ADP3- + glucose 6-phosphate-    ; Do   =   -16.7    kJ/mol at To = 298.16 K

o= 842.63 <= EXP(-DG°/R/T) = 649.3 = K°; K°eqo= ADPglu6PATP.gif =890 ; Deqo= -16.836 kJ/mol

(1) glucose + HPO42-  + H3O+<=>Glucose-6-phosphate- + 2H2O     ;D1  = -49.3 kJ/mol

(2) ATP4-+2H2O<=>ADP3-+HPO42-+H3O+                                         ; D2=  32.606 kJ/mol

Sum: ATP4- + glucose<=>ADP3- + glucose 6-phosphate-                     ;D  =  -16.694 kJ/mol

D2=Do2+G°HPO4+G°H3O+-G°H2PO4-G°H2O= -30.5 + (-1282) + (-284.7) - (-1323) - (-306.7) = 32.606 kJ/mol

[Glu]=Glu10-7notH.gif =Glu10-7KnotH.gif = 1.504•10-7M not depend on concentration [H3O+]

Calculate K for the overall reaction. For the ATP-dependent phosphorylation of glucose, what concentration of
glucose is needed to achieve a 250 µM intracellular concentration of glucose 6-phosphate when the concentrations
of ATP and ADP are 3.38 and 1.32 mM, respectively? Does this coupling process provide a feasible route, at least
in principle, for the phosphorylation of glucose in the cell? Explain.

(d) Although coupling ATP hydrolysis to glucosephosphorylation makes thermodynamic sense, how this coupling
is to take place has not been specified. Given that coupling requires a common intermediate, one conceivable route is
to use ATP hydrolysis to raise the intracellular concentration of H2PO4- + HPO42- and thus drive the unfavorable
phosphorylation of glucose by H2PO4- + HPO42-. Is this ~i reasonable route? (Think about the solubility products
of metabolic intermediates.)

(e) The ATP-coupled phosphorylation of glucose is catalyzed in hepatocytes by the enzyme glucokinase. This
enzyme binds ATP and glucose to form a glucose-ATP-enzyme complex, and the phosphoryl group is
transferred directly from ATP to glucose. Explain the advantages of this route.

10. Calculations of DG° for ATP-Coupled Reactions From data in Table 1-2 calculate the D value for
the reactions:

(1o) PhosphCreatine- + H2O<=>Creatine + H2PO4-                            ;D1o = -43.095 kJ/mol

D1`=D1o+G°HPO4+G°H3O+-G°H2PO4-G°H2O=  43.095 + (-1282) + (-284.7) - (-1323) - (-306.7) = 20.011 kJ/mol

(1) PhosphCreatine- + 2 H2O<=>Creatine + HPO42-  + H3O+           ;D1   =   20.011 kJ/mol

(2) ADP3-+HPO42-+H3O+<=>ATP4-+2H2O                                    ; D2  =   -32.606 kJ/mol

(a) Phosphocreatine- + ADP3-<=>creatine + ATP4-                           ;D=  -12.595 kJ/mol

Do =-79.005+-32.606= -111.611 ; 3.8*4.184 = 15.899; 

(1bo) fructose + H2PO4-<=>fructose 6-phosphate- + H2O               ;D1bo = -15.899 kJ/mol

D1b`=D1bo+G°HPO4+G°H3O+-G°H2PO4-G°H2O=  -15.899 + (-1282) + (-284.7) - (-1323) - (-306.7) = -79.005 kJ/mol

(1) fructose 6-phosphate- + 2 H2O<=>fructose + HPO42-  + H3O+   ;D1b   =     -79.005 kJ/mol

(2) ADP3-+HPO42-+H3O+<=>ATP4-+2H2O                                  ; D2b  =     -32.606 kJ/mol

(b) ATP4-+ fructose<=>ADP3- + fructose 6-phosphate-                   ;Db    =   -111.6 kJ/mol

11. Coupling ATP Cleavage to an Unfavorable Reaction.

To explore the consequences of coupling ATP hydrolysis under physiological conditions to a thermodynamically
unfavorable biochemical reaction, consider the hypothetical transformation X => Y, for which DG° 20 kJ/mol.

(a) What is the ratio [Y]/[XI at equilibrium? o =  3.135•10-4

(b) Suppose X and Y participate in a sequence of reactions during which ATP4- is hydrolyzed to ADP3- and H2PO4-,
The overall reaction is :

(1) X<=>Y                                                        ; Do1 =   20        kJ/mol

(2) ATP4- + H2O<=>ADP3- + H2PO4- ; o = 220409      ; Do2 =  -30.500 kJ/mol

X + ATP4- + H2O<=>Y + ADP3- + H2PO4- ; Do = Do1+G°o2 = Do = 20+(-30.5) = -10.500 kJ/mol

eqo=ADP_ATP-Y_X.gif=69.1 ;eqo•[H2O] =Y_X.gif = 3810

Y_X.gif=ATP_ADP-X_Y20.gif=1.994•106;[H2PO4-]=

Pi=8.05 mM=[H2PO4-]+[HPO42-] ; [H2PO4-]=8.05/1000-H2PO4_HPO4-805.gif ;

Calculate [Y]/[X] for this reaction at equilibrium. Assume that the concentrations of [ATP4-], [ADP3-], and
([H2PO4-]+ [HPO42-])are all 1 M when the reaction is at equilibrium T = 310.16 °K.

(c) We know that [ATP4-], [ADP3-], and [H2PO4-l are not 1 M under physiological conditions. Calculate
[Y]/[X] for the ATP-coupled reaction when the values of [ATP4-], [ADP3-], and [H2PO4-l are those found
in rat myocytes (Table 1-3).

12. Calculations of DG at Physiological Concentrations.

Phosphocreatine- + ADP3-<=>creatine + ATP4-        ;D = -12.5 kJ/mol  -43+30.5

DG=DG°+R•T•lnADP-PCr_Cr-ATP.gif= -12500+RF*T*ln(2.6*1/0.73/4.7)= 13215.2 -13.215 kJ/mol at T = 310.16 K

at 37°C as it occurs in the cytosol of neurons, in which phosphocreatine- is present at [PCr-]=4.7mM,
creatine at [Cr]=1.0 mM, ADP3- at 0.73 mM, and ATP4- at 2.6 mM.

13. Free Energy Required for ATP Synthesis under Physiological Conditions.

In the cytosol of rat hepatocytes, the mass-actionratio is :

o=ADP-H2PO4-HPO4_ATP.gif = 5.33•10-2 M-1                    ; at 37°C T = 310.16 K

Pi = [H2PO4-]+[HPO42-] ; [H2PO4-] =ADP-R_ATP.gif -H2PO4_HPO4-805.gif ;

[H2PO4-]=H2PO4_R.gif;DG=DG°+R•T•lnDGratATP.gif=36.99 kJ/mol

ATP4- + H2O<=>ADP3- + H2PO4- ; K°= 220409 ;DG° = -30.500 kJ/mol

 +30500+RF*T*ln(0.0533*aHOH*(1+2.543489E-9*aHOH/10^(-7.36)))= 36988.76 

Calculate the free energy DG required to synthesize ATP4- in a rat hepatocyte.

14. Daily ATP Utilization by Human Adults.

(a) A total of 30.5 kJ/mol of free energy DG is needed to synthesize ATP4- from ADP3- and H2PO4- when the
reactants and products are at 1 M concentration (standard state). Because the actual physiological concentrations
of ATP4-, ADP3-, and H2PO4-are not 1 M, the free energy DG required to synthesize ATP4- under physiological
conditions is different from D. Calculate the free energy DG required to synthesize ATP4- m the human
hepatocyte when the physiological concentrations of
ATP4-, ADP3-,  (H2PO4-+ HPO42-) are 3.5,1.50, 5.0 mM and pH=7.36, respectively, at 37°C.

H2PO4=H2PO4_HPO4.gif;Pi = 5 mM=[H2PO4-]+[HPO42-];[H2PO4-]=5 -H2PO4_HPO4-805.gif

H2PO425= 1.1469•10-9; DG° = -R•T•ln(H2PO4) = 51.034 kJ/mol at T = 298.16 K

H2PO437 = EXP(-51033.6/T/RF) = 2.543489•10-9 at T = 310.16 K

[H2PO4-]=H2PO4hepatoCyte.gif; DG°= -R•T•ln at T = 310.16 K

DG=DG°+R•T•lnDG-ATPhepatoCyte.gif=60.3976 kJ/mol  atT = 310.16 K

ATP4- + H2O<=>ADP3- + H2PO4- ; K°= 220409 ; DG°= -30.500 kJ/mol

(b) A 68 kg (150 lb) adult requires a caloric intake of 2 000 kcal (8 360 kJ) of food per day (24 h). The food
is metabolized and the free energy DG is used to synthesize ATP4-, which then provides energy DG for the body's
daily chemical and mechanical work W=-DG. Assuming that the efficiency of converting food energy E into ATP4-
is 50%, calculate the weight mATP of ATP4- used by a human adult in 24 h. What percentage of the body weight
does this represent?

nATP = 8360/60.397598/2 = 69.208 mol ; mATP = nATP•MATP = 69.208050*506.91 = 35082 g

(c) Although adults synthesize large amounts of ATP4- daily, their body weight, structure, and composition do not
change significantly during this period. Explain this apparent contradiction.

15. Rates of Turnover of and Phosphates of ATP4- A-O-OPO--O-OPO--O-OPO--O- (A----O-).

If a small amount of ATP4- labeled with radioactive phosphorus in the terminal position, [-32P] ATP4-, is added to
a yeast extract, about half 1/2 of the 32P activity is found in H2PO4- within a few minutes, but the concentration
of [ATP4-]= const remains unchanged. Explain. If the same experiment is carried out using ATP4- labeled with 32P
in the central position, [-32P] ATP4-, the 32P does not appear in H2PO4- within such a short time. Why?

16. Cleavage of ATP to AMP and PPi during Metabolism

The synthesis of the activated form of acetate (acetyl-CoA) is carried out in an ATP-dependent process:

Acetate + CoA4-<=>acetyl-CoA4- + H2O                           ; D1 =  31.4 kJ/mol

ATP4- +   H2O<=>AMP2-          + PPi2-                          ; D2 = -45.6 kJ/mol

Acetate + CoA + ATP4-<=>acetyl-CoA + AMP2- + PPi2- ; DG° = -45.6+31.4 = -14.2= -14.2 kJ/mol

(a) The D for the hydrolysis of acetyl-CoA to acetate and CoA is -31.4 kJ/mol and that for hydrolysis
of ATP4- to AMP2- and PPi2- is -45.6 kJ/mol. Calculate D for the ATP-dependent synthesis of acetyl-CoA.

(b) Almost all cells contain the enzyme inorganic pyro-phosphates, which catalyzes the hydrolysis of PPi2- to H2PO4-.
What effect does the presence of this enzyme have on the synthesis of acetyl-CoA? Explain!

H2PO3-O-O3PH22- + H2O<=>2 H2PO4-; DPP = -19 kJ/mol ;Dtotal = DG° + DPP = -45.6+31.4 = -14.2-19= -33.2 kJ/mol

H2PO3-O-O3PH22- + 3 H2O<=>2 HPO42- + 2 H3O+ ; DPPH =  107.21 kJ/mol

17. Energy for H3O+ Pumping The parietal cells of the stomach lining contain membrane ''pumps'' that transport
hydrogen ions H3O+ from the cytosol of these cells (pHplasma 7.36) into the stomach, contributing to the acidity
of gastric juice (pHstomach 1.2). Calculate the free energy required to transport 1 mol of hydrogen H3O+ ions through
these pumps. (Hint: See Oxidative Phosphorylation.)
Assume a temperature of 37°C or T = 310.16 °K. 1445440

H3O+plasma<=>H3O+stomach ; Keq = [H3O+stomach]/[H3O+plasma]=10-pHplasma/10-pHstomach=1.445•106

          DG° =  -R•T•ln(Keq) =-36577 = -36.577 kJ/mol

18. Standard Reduction Potentials The standard reduction potential, , of any RedOx pair is defined for the
half-cell reaction in equilibrium of each RedOx system:

Oxidizing agentn+ + n electrons-<=>reducing agent

The values for the NAD+/NADH and pyruvate/lactate conjugate RedOx pairs are -0.113 and 0.2291 V, respectively
but at 37° C degree (310.15 K) 37: -0.059 and 0.3193 V.

(a) Which conjugate pair has the greater tendency to lose electrons? Explain.

(b) Which is the stronger oxidizing agent? Explain.

(c) Beginning with 1 M concentrations of each reactant and product at pH 7.36, in which direction will the
following reaction proceed?

Pyruvate-+H3O+ +NADH=>lactate-+NAD++ H2O; (NADH-E°pyruvate)•F•2=DG°= -73.000 kJ/mol

DG= DG° + R•T•lnpyruvateLactate.gif= -18.957 kJ/mol favorable direction of reaction

(d) What is the standard free-energy change (D) at 37 °C for the conversion of pyruvate to lactate

(e) What is the equilibrium constant (Keq) for this reaction?

Keq=pyruvateLactate.gif=6.149•1012 ; pH•Keq=KHpyruvateLactate.gif=268417

19. Energy Span of the Respiratory Chain Electron e- transfer in the mitochondrial respiratory chain
may be represented by the net reaction equation

2 NADH + 2 H3O+ + O2=> 2 NAD+ + 4 H2O (NADH- E°O2)F•4 = DG° = -552.74 kJ/mol

(a) Calculate the value of D for the net reaction of mitochondrial electron e- transfer at 37°C.

(b) Calculate D for this reaction.          DE° = E°NADH- E°O2 = -0.059-1.3732 = -1.4322 V

(c) How many nATP molecules can theoretically be generated by this reaction if the free energy of ATP synthesis
under cellular conditions is 52 kJ/mol? n = 10.63

20. Dependence of Electromotive Force on Concentrations

Calculate the electromotive force EMF (in volts V) registered by an electrode immersed in a solution containing
the following mixtures of NAD+and NADH at pH 7.36 and 37 °C, with reference to a half-cell of E° 0.00 V.

NADH      + H2O<=>NAD+   + H3O+ + 2e-             E37 = -0.0590 V   (David Harris)

EMF = E = E37 + R•T/F/2•lnNADH_NAD.gif= -0.36983 V

(a) 1.00 mM NAD+ and 10.0 mM NADH; E = -0.36983 V

(b) 1.00 mM NAD+ and 1.00 mM NADH ; E = -0.33906 V

(c) 10.0 mM NAD+ and 1.00 mM NADH ; E = -0.30829 V

21. Electron Affinity of Compounds List the following substances in order of increasing ­ tendency to accept
electrons e- at pH=7.36 by RedOx potential Eo values:

O2g+4 H3O++ 4 e-= 6 H2O Suchotina;                                Eo =   0.8130 V ; E37 =   1.3732 V

O2aq+2H3O++2e-=H2O2aqua+ 2H2O University Alberta; Eo =   0.2336 V ; E37 =   0.7937 V

oxalo-acetate2-+2H3O+ +2e-= Malate2-+2H2O CRC ;     Eo = -0.2225 V ; E37 =   0.3376 V

O2g+ e-  =  O-2aq Suchotina;                                                Eo = -0.2355 V ; E37 = -0.2355 V

NADP++H3O+ +2e-= NADPH + H2O CRC ;                     Eo = -0.3429 V ; E37 = -0.0629 V

a-Ketoglutarate+CO2+2H3O++2e-= isocitrate+2H2O ; Eo = -0.4283 V ; E37 =   0.13185 V

(a) a-keto-glutarate + CO2 (yielding iso-citrate); (b) oxalo-acetate; (c) O2 ; (d) NADP+.

22. Direction of Oxidation-Reduction Reactions

Which of the following reactions would you expect to proceed in the direction shown under standard conditions
pH=7.36
and 37°C, assuming that the appropriate enzymes are present to catalyze them?

(Red-E°Ox)•F•n = DG° kJ/mol ;

DG= DG°+ R•T•lnNADoxaloAcetatMalat.gif= 10.053 kJ/mol unfavorable direction => for (a)

(a) Malate2-+ NAD++ H2O => oxalo-acetate2-+NADH+H3O+;(malate-E°NAD+)•F•2 = DG°= 53.756 kJ/mol

DG= DG° - R•T•ln(10-7.36) = 0.568 kJ/mol <= direction
favorable to left for (b)

(b) aceto-acetate- + NADH + H3O+ => b-hydroxy-butyrate- + NAD+ + H2O ; DG°= -43.135 kJ/mol

DG= DG° - R•T•ln(10-7.36) = -29.298 kJ/mol => direction
favorable to right

(c) Pyruvate- + NADH + H3O+ => lactate- + NAD+ + H2O         ; DG°= -73.001 kJ/mol => direction

(d) Pyruvate- + b-hydroxy-butyrate-  => lactate- + aceto-acetate- ; DG°= -29.866 kJ/mol => direction

(e) Malate- + pyruvate- => oxalo-acetate- + lactate-                       ; DG°= 3.526 kJ/mol <= direction

(f ) Acetaldehyde + succinate2-=> ethanol + fumarate2-                ; DG°= 46.447 kJ/mol <= direction