ARTICLE IN PRESS
Journal of Theoretical Biology 236 (2005) 397–421
www.elsevier.com/locate/yjtbi
Electric dipole theory and thermodynamics of actomyosin molecular
motor in muscle contraction
Markku J. Lampinen, Tuula Noponen
Laboratory of Applied Thermodynamics, Helsinki University of Technology, Sähkömiehentie 4 J, P.O. Box 4400 FIN-02015 HUT, Finland
Received 18 January 2005; received in revised form 14 March 2005; accepted 16 March 2005
Available online 24 May 2005
Abstract
Movements in muscles are generated by the myosins which interact with the actin filaments. In this paper we present an electric
theory to describe how the chemical energy is first stored in electrostatic form in the myosin system and how it is then released and
transformed into work. Due to the longitudinal polarized molecular structure with the negative phosphate group tail, the ATP
molecule possesses a large electric dipole moment (p0 ), which makes it an ideal energy source for the electric dipole motor of the
actomyosin system. The myosin head contains a large number of strongly restrained water molecules, which makes the ATP-driven
electric dipole motor possible. The strongly restrained water molecules can store the chemical energy released by ATP binding and
hydrolysis processes in the electric form due to their myosin structure fixed electric dipole moments (pi ). The decrease in the electric
energy is transformed into mechanical work by the rotational movement of the myosin head, which follows from the interaction of
the dipoles pi with the potential field V 0 of ATP and with the potential field C of the actin. The electrical meaning of the hydrolysis
reaction is to reduce the dipole moment p0 —the remaining dipole moment of the adenosine diphosphate (ADP) is appropriately
smaller to return the low negative value of the electric energy nearly back to its initial value, enabling the removal of ADP from the
myosin head so that the cycling process can be repeated. We derive for the electric energy of the myosin system a general equation,
which contains the potential field V 0 with the dipole moment p0 , the dipole moments pi and the potential field c. Using the
previously published experimental data for the electric dipole of ATP ðp0 ffi 230 debyeÞ and for the amount of strongly restrained
water molecules ðN ffi 720Þ in the myosin subfragment (S1), we show that the Gibbs free energy changes of the ATP binding and
hydrolysis reaction steps can be converted into the form of electric energy. The mechanical action between myosin and actin is
investigated by the principle of virtual work. An electric torque always appears, i.e. a moment of electric forces between dipoles p0
and pi ðjMjX16 pN nmÞ that causes the myosin head to function like a scissors-shaped electric dipole motor. The theory as a whole
is illustrated by several numerical examples and the results are compared with experimental results.
r 2005 Elsevier Ltd. All rights reserved.
Keywords: Electrical theory; Thermodynamics; Molecular motor; Muscle contraction
1. Introduction
Movements in muscles are generated by the myosins
that interact with the actin filaments. The energy
required for the mechanical work done by the myosin
head is taken from the hydrolysis reaction of adenosine
triphosphate molecules (ATP), which were first found in
Corresponding author. Tel.: +358 9 4513582; fax: +358 9 4553724.
E-mail addresses: markku.lampinen@hut.fi, [email protected].fi
(M.J. Lampinen).
0022-5193/$ - see front matter r 2005 Elsevier Ltd. All rights reserved.
doi:10.1016/j.jtbi.2005.03.020
muscle extracts by Lohmann (1929). During the muscle
contraction process, the ATP molecule is bound to the
myosin, hydrolysed and the chemical energy of ATP is
transformed into mechanical work and heat.
As the molecular motors act in the nanometer scale,
the Brownian motion introduces a stochastic nature into
the process as demonstrated by Feynman (1963) with
the thermal ratchet model. After that several advanced
models are presented (e.g. Astumian and Bier, 1996;
Mogilner et al., 1998; Shimokawa et al., 2003) to
describe the directional and stepping motion of a
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Brownian particle (the myosin head) in a periodic
potential. The role of force in the reaction kinetics of
ATP hydrolysis and its effect on the Michaelis reaction
rate model have recently been studied by, e.g. Lattanzi
and Maritan (2001), the purpose being to link chemical
kinetic parameters to mechanical parameters (force,
velocity).
The overall mean time for a chemical reaction
depends firstly on the time taken to bring the reactants
together (mass transfer), secondly on the time taken to
obtain proper mutual orientation of the reacting
molecules and to have sufficient energy to overcome
the activation energy barrier and finally on the reaction
time itself. As the chemical reaction time itself is actually
very short, of the magnitude of 1015 s as measured by
Zewail (1999), this means that in the myosin–actin
system there must be an instantaneous way of storing a
significant proportion of the chemical energy immediately upon its release. Otherwise the chemical energy
would simply be transformed into heat. The mechanical
deformation and displacement work takes time of the
magnitude of several milliseconds for one hydrolysis of
ATP as measured and demonstrated, e.g. by Nishiyama
et al. (2003), and Ishii and Yanagida (2002).
In this paper we present an electric theory to describe
how the chemical energy is stored in the myosin system
and how it is released and transformed into mechanical
work. The chemical energies released in different steps
of ATP binding and hydrolysis reactions are first stored
in electric fields, because the changes in the electric fields
take place at the speed of light in the medium. The
chemical energy stored in electrostatic form is released
and transformed into mechanical work which moves
actin filaments against the external loading force.
In technical devices such as fuel cells or electric
accumulators, the energy from the electrochemical
reactions is stored by the charges in the conductive
electrodes of different potentials and then taken out as
electric current. In the myosin–actin system there are no
corresponding electrodes, but there are a large number
of suitable dipole molecules and polar groups which
make the storage of electrostatic energy possible. The
energy source and the most important dipole is the ATP
molecule, which has a large dipole moment ðp0 ffi
230 debye ðDÞÞ as measured experimentally by Mitsuo
(1956). When the ATP molecule binds with the myosin
and makes a complex with it, an electric potential field
(V 0 ) is generated around ATP, and when it is hydrolysed to adenosinediphosphate (ADP), the electric
dipole and its potential field are changed. Around the
ATP molecule there are polar groups of amino acids and
strongly restrained water molecules (with dipole moments pi ), which are in the electric field of ATP and can
store the electric energy. As these dipoles are fixed
within the structure of the myosin, their movements can
deform and rotate the myosin.
Suzuki et al. (1997) have studied by dielectric
spectroscopy with microwaves the hydration of the
myosin subfragment (S1). They found an interesting
result that there are a large number of strongly
restrained water molecules ðN ¼ 720Þ fixed in the
myosin head (S1), which do not follow the microwave
electric field as free water molecules. These strongly
restrained water molecules coupled with proteins are
able to store electric energy and also convert it into
mechanical work as their rotation is slow enough with
respect to the motion of the myosin head.
The actin filament is the other important electrical
system. With the molecular dynamics calculation, Bãnos
et al. (1996) have illustrated how the electrostatic
potential ðCÞ varies periodically along the actin filament.
The electric theory presented here is based on these
three elements: potential field V 0 with the dipole
moment p0 , the dipole moments pi and the potential
field C. The mechanical action between two electrical
systems, myosin and actin, is investigated by the
principle of virtual work with the aid of the electrostatic
energy function. We will derive mathematical formulae
for the forces and torques that move and deform the
myosin head. We calculate the numerical values for
them and for the electric energies to see how the theory
can explain the experimental results reported in the
literature, which are collected and briefly presented in
Section 2.
2. Description of the muscle system
2.1. Force of myosin– actin interaction
The muscle consists of parallel bundles of muscle
fibers, each of which is a single muscle cell. When the
muscle contracts, the length of the muscle cell decreases
and its diameter increases while the volume of the cell
remains more or less the same. The shortening of the
muscle is understood well by the internal structure of the
sarcomeres, in which the myosin and actin protein
filaments slide past one another (first presented by
Huxley and Hanson, 1954; Huxley and Niedergerke,
1954). The system of a hexagonal sarcomere subunit
together with its basic dimensions and geometry is
shown in Fig. 1(a).
The force for shortening the sarcomere is developed
by the interaction of the myosin heads with the actin
filaments. There are about 300 myosin heads per half
myosin filament in one geometrical sarcomere unit
(Squire, 1981), which, being in contact with the actin
units, finally generate the contraction force of the muscle
cell. The principle of the molecular motor process of the
myosin head with the actin is shown in Fig. 1(b). The
fraction of the time that each myosin head spends in its
attached phase is called a duty ratio, which depends on
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399
Incoming
action potential
Myosin
head
Longitudinal
tubules
Ca2+
ATP
Tropomyosin
Tropomyosin
Actin
Troponin
Actin
Troponin
Actin filament
(Ca2+)ICS
=1 mol/lµ
mol/l
Myosin heads (300)
Pi
ATP
base
Resting position
ADP
(Ca2+)ICS
Ca 2+
=1 - 10µmol/l
10 mol/l
Formation of
actin-myosin complex
Relaxing effect
of ATP
ATP
a= 5
.5n m
Myosin heads (300)
Myosin filament
Major
movement of A
T
2-
3µ
ted
trac
con
m(
ed)
Myosin heads resume
original position
lax
- re
Pi
A
~1.2µm
Sarcomere
Loss of Pi
Actin
filament
ATP
Z-plate
Minor
movement of A
Myosin
filament
Z-plate
Binding of ATP
2
A=1620 nm
2
Vmax = 1620 nm
•
3 µm = 4.9•10 m
-21
3
With
ATP
AD
P
Without
ATP
25 nm
(a)
Stable “rigor complex”
cannot be broken:
rigor mortis
End position of the head
Loss of ADP
(b)
Fig. 1. (a) Structure of the sarcomere; (b) principles of molecular mechanisms involved in the sliding of actin and myosin filaments (slightly modified
from Despopoulos and Silbernagl (1991) with permission).
the velocity of the muscle, the greatest being at zero
velocity and smallest at maximum velocity (Howard,
2001). Theoretically, the maximum force F max is
generated if all available actin–myosin crossbridges are
in use simultaneously and when the velocity is zero, i.e.
when we have an isometric tension. If the maximum
tension of the skeletal myofibrils is smax ¼ 2:0 105 N=m2 (Bagshaw, 1994) and if we assume that the
duty ratio is 0.14, i.e only 14% of the myosin heads
(300) are simultaneously generating force at the
maximum load with zero velocity (Howard, 2001), we
obtain the following estimation (see Fig. 1) for the
maximum force per myosin head: F max ¼ smax A=
ð0:14 300Þ ¼ 7:7 pN. The values for F max differ due
to the different tension data for smax and due to the
different estimations for the proportion of attached
actin–myosin crossbridges. Morel (1991) has presented
an estimation of 8 1 pN for the maximum force.
The dependence of the force F on the speed v can be
roughly approximated by a parabola (see e.g. Howard,
2001, Fig. 16.5—force velocity curve for a myosin
crossbridge in skeletal muscle; or Despopoulos and
Silbernagl, 1991), which gives the following working
power P for a myosin head:
v 2 v
P ¼ Fv ¼ F max vmax 1 .
vmax vmax
(1)
The minimum power P ¼ 0 is obtained by v ¼ 0 and by
v ¼ vmax , and the maximum power by v=vmax ¼ 1=3
ðdP=dz ¼ 0; z ¼ v=vmax Þ, which means v ¼ 2:0 mm=s for
the fast skeletal muscle with vmax ¼ 6 mm=s (Howard,
2001). The corresponding mean force giving the maximum power is thus F ¼ F max ð1 1=3Þ2 ¼ 4=9F max ¼
4=9 7:7 pN ¼ 3:4 pN, which is in agreement with the
studies of Finer et al. (1994). The maximum value for P
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defined by Eq. (1) is thus Pmax ¼ 3:4 pN 2:0 mm=s ¼
6:8 1018 W.
2.2. Kinematics of the myosin motion and moments of the
power stroke force
The myosin molecule can be divided into three parts:
head, neck and tail. The head and neck together are
called a subfragment 1 (S1). The myosin head has an
actin-binding site and a catalytic site which receives ATP
(Rayment et al., 1993). These two sites are located on
opposite sides of the head. The portion of the head
separated by these two sites can rotate around a narrow
junction O (Masuda, 2003). The myosin subfragment
(S1) is connected to the neck via a flexible joint N and
the neck is joined to the tail by a joint T. The tail is
connected to the myosin filament as shown in Fig. 1.
With these notations we can describe the kinematics of
the myosin motion according to Fig. 2.
Let us suppose that the myosin head is in contact with
the actin at point A. The joint O of the rotatable portion
is at a distance l OA from A, the joint N of the neck is at a
Fig. 2. Kinematics of scissors-shaped motion of the myosin head.
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distance l NO from O and the joint T of the tail is at a
distance l TN from N. According to the pictures modeled
from crystallographic data (Geeves and Holmes, 1999),
the distances are approximately l OA ¼ l NO ¼ 4 nm and
l TN ¼ 6:5 nm. (It is interesting to note that here, too, in
nanoscale biological structures, the ratio of the golden
section
can be found in the lever arms ð6:5: 4 ¼ 1:625 ffi
pffiffiffi
ð 5 þ 1Þ=2Þ that rectify the motion.)
As the tail is fixed with the myosin filament, which is
not moving (because of the symmetry, see Fig. 1), it
means that the motion of joint T is approximately only
in the vertical direction of the actin. To simplify the
mathematics, we assume here that the joint T is fixed.
Hence, if the x-coordinate is parallel with the actin and
its positive direction is the same as the power stroke
motion, the motion of point A, then we may write the
kinematics of the motion as
xT ¼ const:;
xN xT ¼ l TN sin jTN ,
xO xN ¼ l NO sin jNO ;
xA xO ¼ l OA sin jOA ,
where the angles jTN , jNO and jOA are the angles that
the lines of the bodies TN, NO and OA make with the
perpendicular direction of x-axis. The clockwise direction is taken as the positive angle, as shown in Fig. 2.
Hence,
xA xT ¼ l TN sin jTN þ l NO sin jNO þ l OA sin jOA
and the displacement of A in the differential form is
dxA ¼ l TN cos jTN djTN þ l NO cos jNO djNO
þ l OA cos jOA djOA .
ð2Þ
Let us suppose that at some instant t during the power
stroke the myosin head draws the actin by a force F. As
the mass of the myosin head (S1) is only 120 000 Da, i.e.
m ¼ 1:990 1022 kg (and the mass of strongly
restrained water molecules fixed in it 2:15 1023 kg),
the inertia forces are very small compared with the
acting mean force and they can be neglected. The
moments of the force F at the joints N and O, and at
the contact point A, can then be calculated from the
static equilibrium conditions:
M N ¼ Fl TN cos jTN ;
M O ¼ M N þ Fl NO cos jNO ,
M A ¼ M O þ Fl OA cos jOA .
ð3Þ
The angles jTN , jNO and jOA change during the power
stroke process. Assuming that when the power stroke
starts the angles are jOA ¼ 30 , jNO ¼ 40 and
jTN ¼ 50 , the working distance of point A will be
5.5 nm (see Fig. 2). Substituting those angles into Eq. (3)
and assuming that the force at that moment is
F ¼ 3:4 pN, as discussed in Section 2.1, we get M N ¼
14 pN nm, M O ¼ 25 pN nm and M A ¼ 36 pN nm.
These torques are necessary to explain the power
stroke process in terms of mechanics. With the
electric theory we will study how these torques arise
401
from the electric dipoles of the acto-myosin system and,
particularly, what is the torque M that produces the
scissors-shaped motion of the myosin head shown in
Fig. 2.
2.3. Free energy changes of the ATP hydrolysis reaction
steps
Each reaction step in the myosin system can be
described mathematically as ðn1 ; . . . ; nm Þ ! ðn01 ; . . . ; n0m Þ,
where nk is the amount of species (k) before the reaction
and n0k its amount after the reaction. When the reaction
takes place at constant temperature T and at constant
pressure p, a small change in the Gibbs free energy of the
system can be written with the derivatives as
DG ¼ GðT; p; n01 ; . . . ; n0m Þ GðT; p; n1 ; . . . ; nm Þ
m
X
qG 0
ðnk nk Þ
¼
qn
k
k¼1
¼
m
X
k¼1
mk ðn0k nk Þ ¼
m
X
½m0k ðTÞ þ RT ln ak Dnk ,
ð4Þ
k¼1
where mk ¼ qG=qnk and Dnk ¼ n0k nk . The chemical
potential of species (k) at the reference state is
m0k ðTÞ ¼ mk ðT; p0 ; m0k ; c0 ; pH0 ; I 0 Þ, where the reference
state is defined as p0 ¼ 1 bar, m0k ¼ 1 mol=kg,
c0 ½Mg2þ ðaqÞ ¼ 103 mol=l, pH0 ¼ 7 and I 0 ¼ 250 mM
(the ionic strength of the solution). The universal gas
constant is R ¼ 8:314 J=molK. The activity ak for the
dissolved species in the cell solution can be estimated by
the ideal solution model as ak ¼ mk =m0k , where mk is the
molality (mol/kg) of species (k) in the cell solution.
If the system is in the thermodynamic equilibrium
state ðn1 ; . . . ; nm Þ, then the Gibbs energy is at the
minimum and hence its change is zero:
DG ¼ GðT; p; n01 ; . . . ; n0m Þ GðT; p; n1 ; . . . ; nm Þ
m
X
½m0k ðTÞ þ RT ln ak Dnk ¼ 0,
¼
ð5Þ
k¼1
where Dnk ¼ n0k nk , and, according to the ideal
solution model, ak ¼ mk =m0k .
The overall hydrolysis reaction is ATPðaqÞ !
ADPðaqÞ þ Pi ðaqÞ, where ATP is decomposed into
adenosine diphosphate (ADP) and inorganic phosphate
(Pi ).
If the amount of the hydrolysis reactions is Dn
(moles), then DnATP ¼ Dn; DnADP ¼ Dn and DnP ¼
Dn. Hence, we get from Eq. (4)
DG=Dn ¼ ½ðm0ADP ðTÞ þ RT ln aADP Þ
þ ðm0P ðTÞ þ RT ln aP Þ
½ðm0ATP ðTÞ þ RT ln aATP Þ
aADP aP
,
¼ DG 0 ðTÞ þ RT ln
aATP
ð6Þ
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where DG 0 ðTÞ ¼ ½m0ADP ðTÞ þ m0P ðTÞ m0ATP ðTÞ. From
Eqs. (5) and (6) we find that in the equilibrium state
DG 0 ðTÞ ¼ RT ln K, from which we can calculate
DG 0 ðTÞ when the equilibrium constant K ¼ aADP
aP =aATP is experimentally determined. Kodama (1985)
gives the value K ¼ 5 105 (at T ¼ 298:15 K), which
implies DG 0 ðTÞ ¼ RT ln K ¼ 32:5 103 J=mol at
T ¼ 298:15 K.
The molality mATP (mole/kg water) refers to the sum
of all the species in the muscle cell solution containing
ATP (MgATP2 , MgATP , MgATP, ATP4 , ATP3 ,
etc.), and similarly mADP and mP refer to the sum of all
existing species of ADP and Pi in the solution of the
muscle cell. The resting molality of ATP is
mATP ¼ 224 mmol=kg, but as soon as the muscle
contraction process starts, the molality mATP decreases
and mADP and also mP increase so that the values of
mATP , mADP and mP actually change with time. Without
reproduction of ATP in the mitochondrium inside the
muscle cell, the term ðDGÞ would soon be zero, meaning
that no more work could be extracted from the system,
which corresponds to the equilibrium state of Eq. (5).
However, because of the reproduction of ATP, there will
be a balance of reproduction and consumption, and
hence the molalities mATP , mADP and mP will depend on
the consumption rate. In red blood cells, typical steadystate values, which are almost independent of the stress
and therefore perhaps good estimations, are mATP ¼
1:4 103 mol=kg, mADP ¼ 0:2 103 mol=kg and
mP ¼ 103 mol=kg (Heinrich and Schuster, 1996). In
skeletal muscle cells of the rabbit, the molalities found
(likely at rest) are mATP ¼ 4 103 mol=kg, mADP ¼
20 106 mol=kg and mP ¼ 2 103 mol=kg (Howard,
2001). Substituting both these approximations into
Eq. (6) we obtain estimates
DG=Dn ¼ 54:4 103 J=mol
ðaATP ¼ 1:4 103 ; aADP ¼ 0:2 103 ; aP ¼ 103 Þ,
ð7Þ
DG=Dn ¼ 61:0 103 J=mol
ðaATP ¼ 4 103 ; aADP ¼ 20 106 ; aP ¼ 2 103 Þ
for the Gibbs free energy change of the ATP hydrolysis
reaction ATPðaqÞ ! ADPðaqÞ þ Pi ðaqÞ at T ¼ 298:15 K.
During the hydrolysis reaction of ATP (T), the
myosin (M) can be in four different chemical states: in
the free state (M), in the ATP-bound state ðM TÞ, in the
hydrolysis products-bound state ðM D PÞ and in the
ADP-bound state ðM DÞ. After the state ðM DÞ the
adenosine diphosphate (D) is removed from the myosin
and the next hydrolysis cycle can start in the myosin.
Simultaneously with the chemical reaction cycle the
myosin makes conformational changes and it has two
main mechanical states. The myosin can be attached to
the actin (A) filament ðA MÞ or it can be detached
ðA þ MÞ. Obviously there are some couplings of
chemical changes to the conformational changes during
the hydrolysis cycle. All the Gibbs free energy changes
for the different reaction steps shown in Fig. 3 are
calculated as in Eq. (6). The equilibrium constants are
taken from the work of Kodama (1985).
2.4. Efficiency and displacement distances
In the mitochondrium inside the muscle cell ATP is
reproduced by recombining ADP and Pi back together
with the aid of metabolic energy. In this circulation
process, which was first realized by Lipmann (1941),
altogether Dn ¼ 30232 mol ATP can be regenerated by
one mole of glucose (Nelson and Cox, 2003) if the
energy is produced aerobically. As the oxidation
reaction enthalpy for glucose (a-glucose) is DH G ¼
2803 kJ=mol at 25 C (Goldberg and Tewari, 1988),
the minimum amount of metabolic energy required to
reproduce ATP aerobically from ADP and Pi is thus
DH G =Dn ¼ 2803=32 ¼ 88 kJ=mol of ATP.
A part of the chemical energy released in the
hydrolysis reaction ATP ! ADP þ Pi can be converted
into mechanical work W. If the amount of hydrolysis
reactions completed is Dn, the maximum amount of
work can be estimated by the change of the Gibbs free
energy of the system as
aADP ap
0
W p DG ¼ DG ðTÞ þ RT ln
Dn,
(8)
aATP
where we have used Eq. (6) for DG. Hence, we get an
estimation for the maximum theoretical efficiency of the
muscle contraction process:
work produced by hydrolysis of ATP
metabolic energy for regeneration of ATP
DG=Dn
¼
DH G =Dn
54:4 kJ=mol
¼ 0:62 ¼ 62%,
¼
88 kJ=mol
Z¼
ð9Þ
where we have used the lower value of Eq. (7). With the
greater value of Eq. (7) the corresponding thermodynamic efficiency would be 69%.
The real mechanical total efficiency for a man walking
uphill at a gradient of about 20% (or more) has been
measured to be 25% (Margaria, 1976). The efficiency of
the working myosin head must, of course, be higher
than the total efficiency, which means that the work
done by the myosin head per one ATP hydrolysis
reaction in these experiments is at least W min X0:25
88 103 =ð6:023 1023 Þ ¼ 3:6 1020 J. Hence, the
working distance of the myosin head at the force F ¼
3:4 pN is 3:6 1020 J=3:4 1012 N ffi 11 nm, which
means two steps of 5.5 nm in length. This is in agreement
with the studies of Finer et al. (1994), who measured
stepwise movements averaging 11 nm ð2 5:5 nmÞ under
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403
G
A•M+ATP(aq)
-33.4 kJ
mol
M•T
kJ
5.7mol
A•M•T
-5.7 kJ
mol
M•D•P
kJ
-5.7
mol
kJ
∆G = -54.4 mol
A•M•D•P
kJ
-22.8 mol
A•M+ADP(aq) +Pi(aq)
7.4 kJ
mol
A•M•D
Time
Fig. 3. Free-energy diagram for acto-myosin ATP hydrolysis with the dissociating/reassociating pathway presented by Kodama (1985). The
molalities are assumed as mATP ¼ 1:4 mmol=kg; mADP ¼ 0:2 mmol=kg and mP ¼ 1 mmol=kg and the Gibbs free energy changes are calculated
accordingly: ðT ¼ ATP; D ¼ ADP; P ¼ Pi ; M ¼ myosin; A ¼ actinÞ.
conditions of low load (single force transients averaging
3–4 pN were measured under isometric conditions).
Under small loading forces, Ishii and Yanagida (2002)
have observed that the myosin moves the actin forward
several steps of 5.5 nm in length. (The total attachment
time t per myosin head with the actin for one completed
ATP hydrolysis reaction is at least tXW min =Pmax ¼
3:6 1020 J=6:8 1018 W ¼ 5:3 ms.)
3. Electric theory
3.1. Description of the electric thermodynamic system
The thermodynamic system is composed of the
myosin head and neck with the aqueous cell solution
in it, and of the ATP molecule bound to the myosin. The
important part of the thermodynamic surroundings of
the system is the actin.
The myosin system contains the dipole moment p0 of
ATP at the ATP hydrolysis point and around it there are
polar groups of amino acids and strongly restrained
water molecules, whose dipole moments are marked as
pi . The myosin is in contact with or at least very close to
the actin system, which has an electrostatic potential
field C. The thermodynamic system with the notations is
shown schematically in Fig. 4.
Mitsuo (1956) measured by the impedance bridge
method (in water and in solvent dioxane) for the electric
dipole moment of ATP at room temperature the value
(with the accuracy of 20 D) p0 ¼ 230 D ¼ 7:68
1028 Asm, where 1 D ¼ 1018 esu cm ¼ 3:341030 Asm.
For adenosine (A) Mitsuo (1956) measured the value
p0 ðAÞ ¼ 42 D. Using the same vectorial model for the
dipole moments as Mitsuo (1956), we have estimated the
dipole moment of ADP as p0 ¼ 90 D and the direction
angles as shown in Fig. 5. (The charge of the phosphate
tail is decreased from ð2Þ to ð1Þ and the length of
( which changes
the molecule is shortened by 3:5 A,
the longitudinal component of the dipole moment
from 194 D to ½ð14:4 þ 19:0Þ=2=½ð17:9 þ 22:5Þ=2 ð1Þ=ð2Þ 194 D ¼ 80 D) (Fig. 5).
The water molecule has an oxygen atom with two
( ð0:0955 nmÞ and
hydrogen atoms, at a distance 0:955 A
with an angle 104:5 , which are covalently bonded to the
oxygen atom. The permanent dipole moment of an
isolated water molecule in the gas phase has been
measured experimentally to be pðH2 OðgÞÞ ¼ 1:855 D
(Lovas, 1978). In the condensed phase the water
molecules interact with one another and the dipole
moment is greater than the value of the isolated water
molecule. Applying theoretical perturbation theory, Tu
and Laaksonen (2000) studied the dipole moment of
water in liquid phase with 256 water molecules at
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404
Fig. 4. Electric thermodynamic system with notations.
ATP 230 ± 20D
H
C N
N
C N+H2
C C
+
C
C
N
CH
O (HCOH)2
CH
CH
P O
O
O P
O
O
(b)
P
O
N-
CH
O
42D
CH2
63 °
O
O P O 80D
90D
O
O P O
230D
O
O
+
C N+H2
C
CH H
199D
O
O P
C
N
Mg
Mg
O
HO
C
O (HCOH)2
CH2 194D
O
O P O
22.5Å
N
42D
78 °
C
H
CH
O (HCOH)2
CH2
17.9 Å
O
HO P O
C N+H2
N-
+
C
H
(a)
N
ADP
H
C N
N
N-
N
H
C
O
O
(c)
Fig. 5. (a) Dipole moment of ATP, (b) its vectorial form, measured and presented by Mitsuo (1956) and (c) estimation of the dipole moment of ADP.
temperature T ¼ 298:15 K and for the average dipole
moment they obtained pi ¼ 2:65 D ¼ 8:85 1030 Asm.
Suzuki et al. (1997) found in microwave experiments
that there are a large number of strongly restrained
water molecules N ¼ 720 ð50Þ fixed in the myosin head
(S1) that do not follow the microwave electric field as
free water molecules. (If the water molecules are stuck in
the proteins of the helix-type periodical structure at the
same distances d as the ascending of a-helices, then d ffi
0:57 nm (Geeves and Holmes, 1999)—a distance that is
about six times the size of the water molecule.) They also
found that, in addition to the strongly restrained water
molecules N ¼ 720 ð50Þ, the myosin head (S1) also
contains N f ¼ 1410 ð80Þ weakly restrained water
molecules. The weight ratio of total restrained water
to S1 protein is thus ½ð720 þ 1410Þ 0:018=6:023
1023 =1:99 1022 ¼ 0:32, which is in the same range
as the values (0.3–0.4) found for other proteins.
To analyse the electric energy between dipoles p0 and
pi ði ¼ 1; 2; . . . ; NÞ we need to define their mutual
position and the directional axes h0 and hi of the dipoles
p0 and pi , respectively. The dipole p0 makes an angle
a ¼ ðh0 rÞ with the position vector r of the dipole pi , and
the dipole pi makes an angle b ¼ ðrhi Þ with the position
ARTICLE IN PRESS
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vector, and the angle between axes h0 and hi is ðh0 hi Þ.
The cosines of the angles are marked as m0 ¼ cosðh0 rÞ,
mi ¼ cosðrhi Þ and m0i ¼ cosðh0 hi Þ. When the dipole
moments p0 and pi are in the same plane, then
ðh0 hi Þ ¼ ðh0 rÞ þ ðrhi Þ ¼ a þ b, i.e. m0i ¼ cosða þ bÞ (see
Fig. 4). When we study the interaction of dipoles pi and
pj we define the cosines of the angles correspondingly as
mi ¼ cosðrhi Þ, mj ¼ cosðrhj Þ and mij ¼ cosðhi hj Þ.
As our thermodynamic system is a small nano-scale
system, the results have to be understood either as mean
values of many similar myosin systems or as timeaveraged mean values for cycles of one myosin system
repeated several times.
3.2. Electrostatic potential of the actin– troponin/
tropomyosin system without and with Ca2þ -binding
The two actin filaments with the two tropomyosins
coiling around it (see Fig. 1) make the system electrically
almost cylindrically symmetrical. The electric potential
C arising from the charges distributed along the
actin–troponin/tropomyosin system satisfies Laplace’s
equation outside the system (where there are no charges
of the actin–tropomyosin system). In the cylindrical
symmetric form, this equation is
q2 c 1 qc q2 c
þ
þ
¼ 0,
(10)
qr2 r qr qx2
where x is the coordinate along the actin filament and r
is the radial coordinate perpendicular to the x-axis. If
the effective radius of the actin–tropomyosin is R0 , then
Eq. (10) is valid when rXR0 .
The solution to Eq. (10) can be found in the following
form:
cðr; xÞ ¼ f ðrÞgðxÞ þ hðrÞ
(11)
with the boundary condition qC=qr ! 0 when r ! 1.
Substituting Eq. (11) into Eq. (10) gives
1 00
1
g00 ðxÞ
f ðrÞ þ f 0 ðrÞ ¼ k2 ¼ const.
(12)
f
r
gðxÞ
and
1
h00 ðrÞ þ h0 ðrÞ ¼ 0.
r
Solving the unknown functions f, g and h from
(12)–(13) with the boundary condition and
substituting them into Eq. (11), we obtain
K 0 ðkrÞ
r
cos kx þ c1 1 ln
cðr; xÞ ¼ c0
,
K 0 ðkR0 Þ
R0
(13)
Eqs.
then
(14)
where K 0 is the modified Bessel function of order zero,
and c0 and c1 are the integration constants. The actin
monomers are spaced periodically at distance a ¼
5:5 nm along the F-actin (Geeves and Holmes, 1999).
Hence, if we set k ¼ 2p=a in Eq. (12), the potential c
405
also varies periodically according to the actin’s structure. According to the studies of Bãnos et al. (1996), c0
can be approximated as c0 ¼ 12:5 mV corresponding to
0:5 kB T at T ¼ 298:15 K for a unit charge. The radius
R0 is quite close to the diameter of the actin monomer
and on the basis of the pictures of Narita et al. (2001),
we take R0 ¼ a ¼ 5:5 nm.
The radial component of the electric field is
E r ¼ qc=qr. By integrating it along actins over N
periodical monomers the periodical term ðcos kxÞ
disappears and from Eq. (14) we obtain for the
integrated mean value of radial component E r ¼ c1 =r.
Applying Gauss’s theorem (Maxwell’s first equation in
integrated form) for the actin–troponin/tropomyosin
system we get
Z
Sqi
¼
E n dA
0 r
Z
ZA
qc
dA
¼ along
around symmetric
qr r¼R0
N actins N actintrop system
½0; Na
½at distance R0 c
¼ 1 2pR0 Na
R0
ð15Þ
from which we get
c1 ¼
Sqi =ð0 r Þ
,
2pNa
(16)
where 0 ¼ 8:854 1012 As=ðVmÞ and r ¼ 78:5.
In the activated state Ca2þ is bound to troponin. Each
tropomyosin is associated with seven actin subunits
(Pollard and Earnshaw, 2002), i.e. 1 tropomyosin
controls 7 actin monomers. So if we take the system as
seven actin monomers long ðN ¼ 7Þ, but having two
parallel coiled actin monomers (see Fig. 1), it contains
two Ca2þ -ions, meaning that the net charge in the
system is Sqi ¼ 2 2 1:6 1019 As. Substituting all
the numerical values above into Eq. (16), we get c1 ¼
0:004 V.
As the numerical values of kr are large
ðkrXkR
0 ffi¼ 2pÞ, we can use the approximation K 0 ðzÞ ffi
pffiffiffiffiffiffiffi
ez = 2pz and rewrite Eq. (14) as
pffiffiffiffiffiffiffiffiffiffi
cðr; xÞ ¼ c0 ½ekðrR0 Þ = r=R0 r
cos kx þ c1 1 ln
,
ð17Þ
R0
where the parameters are c0 ¼ 12:5 mV, c1 ¼ 4 mV,
R0 ¼ a ¼ 5:5 nm and k ¼ 2p=a ¼ 1:142 nm1 . The equipotential surfaces with the function Cðr; xÞ given by Eq.
(17) are shown in Fig. 6.
When we study the electric field very close to the
actin–tropomyosin system, we can use the distance s r R0 as a variable. When s is very small
compared
pffiffiffiffiffiffiffiffiffiffi
with R0 we can use the approximations r=R0 ffi 1;
ln r=R0 ¼ lnð1 þ s=R0 Þ ffi s=R0 , and Eq. (17) can be
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406
nm
5.0
4.5
4.0
3.5
3.0
2.5
-1 mV
1 mV
-1 mV
-2 mV
2 mV
3 mV
4 mV
5 mV
-2 mV
2.0
1.5
-3 mV
-4 mV
-5 mV
1.0
r-Ro
-3 mV
-4 mV
-5 mV
0.5
0
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5 10.0 10.5 11.0 nm
9.0
9.5 10.0 10.5 11.0 nm
Relaxed state (a)
nm
5.0
4.5
4.0
2 mV
3.5
3.0
3 mV
2.5
ψ = 4 mV
2.0
1 mV
1.5
1.0
r-Ro
5 mV
6 mV
7 mV
8 mV
Distance of ATP
0 mV
-1 mV
-2 mV
-3 mV
0.5
1 mV
0 mV
-1 mV
-2 mV
-3 mV
0
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
Contraction state (b)
Fig. 6. Equipotential surfaces cðr; xÞ ¼ const. outside the actin–tropomyosin system based on Eq. (17). The curves (a) are without calcium ions
ðc1 ¼ 0Þ and (b) with calcium ions in troponin ðc1 ¼ 4 mVÞ.
approximated as
s
ks
cðs; xÞ ¼ c0 e cos kx þ c1 1 .
R0
(18)
In the x– s coordinates Laplace’s equation is
q2 c=qx2 þ q2 c=qs2 ¼ 0, which the function cðs; xÞ given
by Eq. (18) satisfies.
3.3. Electrostatic energy of the myosin system
The electric potential at the point ðx; y; zÞ due to a
point charge ðq0 Þ placed at the origin ð0; 0; 0Þ is
V ðx; y; zÞ ¼
q0 1
¼ q0 f ðx; y; zÞ,
4p0 r r
(19)
where r ¼ ½x2 þ y2 þ z2 1=2 and f ðx; y; zÞ ¼ ð1=4p0 r Þ
ðx2 þ y2 þ z2 Þ1=2 . If we place at a distance h0 from the
origin another point charge ðþq0 Þ, then the potential due
to the pair of charges (þq0 at ð‘0 h0 ; m0 h0 ; n0 h0 Þ and q0
at ð0; 0; 0Þ) will be
V 0 ðx; y; zÞ
¼ q0 f ðx ‘0 h0 ; y m0 h0 ; z n0 h0 Þ q0 f ðx; y; zÞ
qf
qf
qf
ð‘0 h0 Þ þ ðm0 h0 Þ þ ðn0 h0 Þ
¼ q0
qx
qy
qz
þ q0 Oðh20 Þ
d
¼ q0 h0
f ðx; y; zÞ þ q0 Oðh20 Þ,
dh0
ð20Þ
where ‘0 , m0 and n0 are the direction cosines of the axis
h0 with the coordinate axis x; y and z, respectively. The
differentiation of f into the direction of the axis h0 can
be written as df =dh0 ¼ ‘0 qf =qx þ m0 qf =qy þ n0 qf =qz.
ARTICLE IN PRESS
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407
The notation Oðh20 Þ in Eq. (20) means that it contains h20
or higher terms of h0 in Taylor’s expansion.
We can now diminish h0 and increase q0 in such a
manner that their product remains the same,
p0 ¼ q0 h0 ¼ constant, which is the dipole moment of
the pair of charges. Mathematically, when the two point
charges coincide in this manner, we get an ideal electric
dipole (Maxwell, 1873). As q0 Oðh20 Þ ! 0, we see from
Eq. (20) that the potential at the point ðx; y; zÞ will be
d
f ðx; y; zÞ
dh0
p0 d 1
pm 1
¼ ¼ 0 0 2,
4p0 r dh0 r
4p0 r r
V 0 ðx; y; zÞ ¼ p0
ð21Þ
where m0 ¼ cosðh0 rÞ is the cosine of the angle between
the axis h0 (the direction axis of the dipole moment p0 )
and the position vector r. The projection of the change
dh0 in the direction of the radius vector r is dr ¼ m0 dh0 ,
and hence dð1=rÞ=dh0 ¼ ð1=r2 Þ dr=dh0 ¼ m0 =r2 , which
is used in Eq. (21).
In the two-dimensional case, and if h0 is parallel to the
x-axis, we have m0 ¼ x=r and Eq. (21) takes the form
V 0 ðx; yÞ ¼
p0 x
,
4p0 r r3
(22)
2 1=2
where r ¼ ½x2 þ y . We have located the origin of the
potential V 0 ðx; yÞ of Eq. (22) in the ATP hydrolysis
reaction point presented by Suzuki et al. (1997) and
calculated some equipotential curves V 0 ðx; yÞ ¼ const.
The results are drawn on the same scale as the myosin
head and they are shown in Fig. 7.
Let us then assume that there is a dipole molecule
with a dipole moment pi in the electric potential field of
V 0 and that it is located so that its positive charge ðqi Þ is
at ðx; y; zÞ and its negative charge at ðx ‘i hi ; y mi hi ;
z ni hi Þ, where ‘i ; mi and ni are the direction cosines of
the axis hi with the coordinate axes x; y and z,
respectively. The electrostatic energy of the dipole
molecule is
ui0 ¼ qi V 0 ðx; y; zÞ qi V 0 ðx ‘i hi ; y mi hi ; z ni hi Þ
d
V 0 ðx; y; zÞ þ qi Oðh2i Þ
¼ qi hi
dhi
and, in a way similar to Eq. (20), when hi becomes very
small so that pi ¼ qi hi ¼ constant, we get
dV 0
pi p0
d2
1
ui0 ¼ pi
¼
dhi
4p0 r dh0 dhi r
pi p0 d m0 ¼
4p0 r dhi r2
pp 1
¼ i 0 3 ½m0i 3m0 mi ,
ð23Þ
4p0 r r
where m0i ¼ cosðh0 hi Þ is the cosine of the angle between
the axes h0 (direction of p0 ) and hi (direction of pi ) and
Fig. 7. Illustration of the equipotential curves of the dipole moment of
ATP based on Eq. (22). The curves are calculated with the dipole
moment p0 ¼ 230 D ¼ 7:68 1028 Asm and with r ¼ 78:5.
mi ¼ cosðrhi Þ is the cosine of the angle between the
position vector r and the axis hi .1
The electrostatic energy of the dipole pi with respect
to the dipole pj can be written analogously with Eq. (23)
in the form
pi pj
dV j
d2
1
uij ¼ pi
¼
dhi
4p0 r dhi dhj r
pi pj 1
¼
½m 3mi mj ,
ð24Þ
4p0 r r3 ij
where mi is the cosine of the angle between the position
vector r and the direction of pi , mj is the cosine of the
angle between the direction of r and pj , and mij is the
cosine of the angle between the direction axes of pi
and pj .
In the myosin system the dipole moments p0 and pi
ði ¼ 1; 2; . . . ; NÞ not only interact with each other but
also with the electric potential field Cðx; y; zÞ generated
by the actin–troponin/tropomyosin system. If the dipole
molecule in the electric field is located so that its positive
charge ðqi Þ is at ðx; y; zÞ and its negative charge ðqi Þ at
1
The projection of the change dhi in the direction of the radius
vector r is dr ¼ mi dhi , i.e. dr=dhi ¼ mi . Let the projection of r in the
direction axis of h0 be r0 . Then m0 ¼ r0 =r, and thus dr0 ¼ m0 dr þ rdm0 .
On the other hand, the projection of dhi in the direction of h0 is
dr0 ¼ m0i dhi , and thus m0i dhi ¼ m0 mi dhi þ rdm0 , i.e. dm0 =dhi ¼
ðm0i m0 mi Þ=r. Hence, dðm0 =r2 Þ=dhi ¼ ð2m0 =r3 Þðdr=dhi Þ þ ð1=r2 Þðdm0 =
dhi Þ ¼ ðm0i 3m0 mi Þ=r3 .
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408
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ðx ‘i hi ; y mi hi ; z ni hi Þ, then the electrostatic energy
of the dipole molecule is
W p ½GðBÞ GðAÞ
ui ¼ qi cðx; y; zÞ qi cðx ‘i hi ; y mi hi ; z ni hi Þ
d
cðx; y; zÞ þ qi Oðh2i Þ
¼ qi hi
dhi
¼ ½GðBÞ GðAÞ ½UðBÞ UðAÞ.
and, in a mathematically similar way to Eqs. (20)–(21),
when hi becomes very small but qi increases so that
pi ¼ qi hi ¼ constant, we get
dc
qc
qc
qc
þ mi
þ ni
ui ¼ pi
¼ pi l i
.
(25)
dhi
qx
qy
qz
When the potential function Cðx; y; zÞ can be given by
Cðx; sÞ with two variables x and s as in Eq. (18), where
s ¼ sðy; zÞ is the radial distance from the surface of the
actin–troponin system, Eq. (25) may be written as
dc
qc
qc
þ si
ui ¼ pi
¼ pi l i
,
(26)
dhi
qx
qs
where si is the direction cosine of the dipole pi with the
radial distance direction from the actin–troponin system
perpendicular to the x-axis.
The electric energy of the dipole moment p0 with
respect to the electric potential field C can be written
similar to Eq. (25) or to Eq. (26) as
dc
qc
qc
þ s0
u0 ¼ p0
¼ p0 l 0
.
(27)
dh0
qx
qs
Altogether, with Eqs. (23)–(24) and (26)–(27), the
total electrostatic energy of the myosin system is
X dV 0 1 X dV j
dc
U¼
pi
þ
pi
þ p0
2
dh
dh
dh
i
i
0
i
i;j
X dc
þ
pi
,
ð28Þ
dhi
i
where the summation is taken over all i ði ¼ 1; 2; . . . ; NÞ
and over all pairs ði; jÞ of dipoles, but iaj. The factor 12 is
included in Eq. (28) because all pairs of dipoles ðpi ; pj Þ
are counted twice in the summation.
3.4. Total free energy and Earnshaw’s theorem
As it is possible for the changes in the electrostatic
energy U to be transformed into work W, we can
combine Eq. (28) with the Gibbs free energy G and
define the total free energy G of the myosin system as
follows:
G ¼ GðT; p; n1 ; . . . ; nm Þ þ Uðp0 ; V 0 ; m0 ; pi ; mi ; cÞ.
for any isothermal and isobaric process A ! B is
(29)
Hence, the total entropy of the myosin system can
be written as S ¼ qG=qT ¼ ðqG=qTÞ þ ðqU=qTÞ,
where the temperature dependence of U is in terms of
the dipole moments and in the dielectric constant r .
According to the second law of thermodynamics, the
maximum amount of work W taken out of the system
ð30Þ
In the completed hydrolysis reaction of ATP the
electrostatic energy term makes a cycle process, and
thus UðBÞ ¼ UðAÞ, and W p ½GðBÞ GðAÞ, as we
estimated before by Eq. (8).
As the actual reaction times are very short, we have
W ¼ 0 during the reactions, and hence from Eq. (30) for
any reaction step ðA ! BÞ we get
GðBÞ þ UðBÞpGðAÞ þ UðAÞ.
(31)
If state A is such that it gives the minimum value for
the total free energy G ¼ G þ U, then it is a stable
equilibrium. It is stable, i.e. it cannot change, because
being the minimum value, it should increase; this is,
however, impossible due to the inequality (31). Hence,
the minimum value of the total free energy G gives the
stable equilibrium state when T and p are kept constant:
G ¼ GðT; p; n1 ; . . . ; nm Þ
þ Uðp0 ; V 0 ; m0 ; pi ; mi ; cÞ ¼ min !
ð32Þ
Taking the total differential of G with constant T and
p, we get the following necessary condition for the
minimum of G:
dG ¼ dG þ dU ¼ 0.
(33)
There is an elegant classical theorem called Earnshaw’s theorem expressed originally as: ‘‘A charged
body placed in a field of electric force cannot be in stable
equilibrium’’. This means that as far as there is an
electric force or torque affecting the electric system with
distributed charges, such as in this case the dipole
moments, the system cannot be in stable equilibrium.
The only way to reach the stable equilibrium state is for
the dipole moments to move and rotate until the
electrostatic energy U reaches its minimum value, after
which the electric forces that we obtain from the
derivatives of U disappear. Mathematically the minimum of the electrostatic energy U means that dU ¼ 0 in
the stable equilibrium state. The proof of Earnshaw’s
theorem can be found in the book by Maxwell (1873,
p. 174).
Hence, we see from Eq. (33) that in the stable
equilibrium state also
dG ¼
m
X
qG
dnk ¼ 0,
qn
k
k¼1
(34)
which is the same as Eq. (5) that we have already used
for the equilibrium states.
ARTICLE IN PRESS
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4. Analysis of the electric energy and ATPs hydrolysis
cycle
4.1. Electric energy between ATP and restrained water
molecules
Let us now suppose that the dipole p0 is fixed within
the myosin structure as well as the dipole pi so that the
dipole pi can rotate around its own center. The direction
of the lowest energy of the dipole pi with respect to the
dipole p0 is reached when the direction of pi ðhi Þ is
parallel to the electric field ðgrad V 0 Þ, i.e.
dV 0
grad V 0
ðminÞ ¼ pi grad V 0 dhi
jgrad V 0 j
"
2 2 #1=2
qV 0
qV 0
qV 0 2
¼ pi
þ
þ
, ð35Þ
qx
qy
qz
ui0 ðminÞ ¼ pi
where grad V 0 ¼ ðqV 0 =qx; qV 0 =qy; qV 0 =qzÞ.
Using Eq. (21) we get analogously with Eq. (23)
qV 0
q
p0 d 1
¼
qx
qx
4p0 r dh0 r
p0 1
¼ ð3mx m0 mx0 Þ,
ð36Þ
4p0 r r3
where mx ¼ cosðrxÞ; m0 ¼ cosðrh0 Þ and mx0 ¼ cosðxh0 Þ.
In a similar way we can express the other derivatives
and as a result we get
qV 0 2
qV 0 2
qV 0 2
þ
þ
qx
qy
qz
2
p0
1
¼
½ð3mx m0 mx0 Þ2
4p0 r r6
þ ð3my m0 my0 Þ2 þ ð3mz m0 mz0 Þ2 2
p0
1
½1 þ 3m20 ,
¼
4p0 r r6
ð37Þ
where we have used the identities m2x þ m2y þ m2z ¼
1; m2x0 þ m2y0 þ m2z0 ¼ 1 and mx mx0 þ my my0 þ mz mz0 ¼ m0
(the cosine of the angle between lines r and h0 with
direction cosines mx , my , mz and mx0 , my0 , mz0 , respectively). Substituting Eq. (37) into Eq. (35) we obtain for
the minimum, and correspondingly for the maximum
value,
pi p0 1
ð1 þ 3m20 Þ1=2 ,
4p0 r r3
pp 1
ui0 ðmaxÞ ¼ þ i 0 3 ð1 þ 3m20 Þ1=2 .
4p0 r r
ui0 ðminÞ ¼ ð38Þ
Independent of the orientation of the dipole pi the
amount of electric energy stored with dipoles p0 and pi is
always between the minimum and maximum values of
Eq. (38), which also means that
ð1 þ 3m20 Þ1=2 pm0i 3m0 mi pð1 þ 3m20 Þ1=2 .
(39)
409
Let us assume that the dipole moments pi are
uniformly distributed along a circle r ¼ constant. The
mean value of the cosine term in Eq. (38) can then be
defined as
ð1 þ 3m20m Þ1=2
Z p=2
1
½1 þ 3cos2 a0 1=2 da0 ffi 1:542,
p=2 0
ð40aÞ
which is the approximative value of the elliptic integral.
The corresponding mean cosine is m0m ¼ 0:678 and the
angle a0m ¼ 0:827 ðradÞ ¼ 47:3 .
Let us assume that the dipole moments pi are
uniformly distributed around the ATP-binding site
ðr0 X1:5 nmÞ up to the actin-binding site (r1 p4:0 nm;
Geeves and Holmes, 1999). Then we can define the mean
distance rm for Eq. (38) as
Z r1
ðp=4Þðr21 r20 Þ
ðp=2Þr dr
,
r3m
r3
r0
which gives
rm ¼ ½r0 r1 ðr1 þ r0 Þ=21=3 .
(40b)
With the numerical values of r0 ¼ 1:5 nm and r1 ¼
4:0 nm above we get rm ¼ 2:54 nm.
Substituting the numerical values given in Section 3.1
into Eq. (38) we obtain
ui0 ðmaxÞ ¼ ui0 ðminÞ ¼
¼
pi p0 1
ð1 þ 3m20m Þ1=2
4p0 r r3m
8:85 1030 7:68 1028
1
4p 8:854 1012 78:5 ð2:54 109 Þ3
1:542 ¼ 7:32 1023 J,
ð41Þ
and if the number of water molecules is N ¼ 720, we get
E Nui0 ðmaxÞ ¼ 32 kJ=mol.
Hence, we get an important estimation for the energy
storage variation between dipole moments p0 and pi
when only constrained water molecules are included:
X dV 0
pi
p þ E,
(42)
Ep
dhi
i
where E ¼ 32 kJ=mol.
4.2. Electric energy of the myosin head with respect to the
actin
Near the surface of the actin–tropomyosin system the
electric potential C can be approximated by Eq. (18),
from which the electric field can be derived as
qc
¼ c0 keks sin kx
qx
(43)
qc
c
¼ c0 keks cos kx þ 1 ,
qs
R0
(44)
where the coordinate system is fixed with the actin.
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410
Let us consider a dipole pi which is located in the
myosin head at point xi on the x-axis and at distance si
from the surface of the actin. Suppose that the dipole pi
makes an angle gi with the x-axis and an angle di with
the radial direction of the actin, i.e. with the s-direction
(Fig. 8). Thus, from Eqs. (26) and (43)–(44) we get for
the energy of the dipole pi in the electric field of the actin
pi
dc
¼ pi c0 keksi ½sin kxi cos gi þ cos kxi cos di dhi
c
pi 1 cos gi .
ð45Þ
R0
Assuming that the dipoles pi are equal in size, pi ¼ p,
and that approximately half of the dipoles ðN=2Þ are
located in the half of the myosin head that is in contact
with the actin, we may write the total electric energy as
X
X
dc
¼ pc0 k
eksi ½sin kxi cos gi
dhi
i
c X
þ cos kxi cos di p 1
cos di
R0 i
pi
i
¼ E 0 ðsin kx0 cos gm þ cos kx0 cos dm Þ
E 1 cos dm1 ,
ð46Þ
where
N ksm X ksi
e
¼
e ;
2
i
E1 ¼
E0 ¼
N
pc0 keksm ,
2
N c1
p ,
2 R0
ð47Þ
and the mean angles gm , dm and dm1 are defined as
X
eksi sin kxi cos gi ,
ðN=2Þeksm sin kx0 cos gm ¼
i
(48a)
ðN=2Þeksm cos kx0 cos dm ¼
X
eksi cos kxi cos di ,
i
(48b)
ðN=2Þ cos dm1 ¼
X
cos di .
(48c)
i
The average location of the dipoles pi in the myosin
head in the x-axis is marked by x0 in Eq. (46).
As we do not know the angles gi and di of the dipoles
pi , we cannot determine the mean angles gm , dm and dm1
from Eq. (48). However, with certain assumptions we
can derive from Eq. (46) some important relations of the
interactions between the myosin head and the actin.
Let us assume that the myosin head attaches to the
actin tangentially as shown in Fig. 8 and that the dipoles
pi are uniformly distributed along the myosin head.
Then the mean distance sm from the cylindrical surface
of the actin–tropomyosin system (with radius R0 ) to the
myosin head defined by Eq. (47) can be calculated as
Z H1
pffiffiffiffiffiffiffiffiffiffi
2
1
2
ksm
e
¼
ekð R0 þy R0 Þ dy,
(49)
H1 0
where H 1 ¼ 4 nm (the height of half of the myosin
head), R0 ¼ 5:5 nm and k ¼ 2p=a ¼ 1:142 nm1 . By
numerical integration we obtain from this for the mean
distance sm ¼ 0:37 nm, and thus E 0 ¼ 3:0 1020 J ¼
18 kJ=mol and E 1 ¼ 2:3 1021 J ¼ 1:4 kJ=mol.
The electric energy of the dipole moment p0 can be
written as in Eq. (46):
dc
¼ E p0 ðsin kx0 cos g0 þ cos kx0 cos d0 Þ
dh0
E p1 cos d0 ,
ð50Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
where s0 ¼ R20 þ H 21 R0 ¼1:30 nm, E p0 ¼ p0 c0 keks0
p0
¼ 2:5 1021 J ¼ 1:5 kJ=mol and E p1 ¼ p0 c1 =R0 ¼
5:6 1022 J ¼ 0:3 kJ=mol, where we have used the value
Fig. 8. Myosin head in tangential contact with the actin system (Bagshaw, 1994). The angles and distances used in the theory.
ARTICLE IN PRESS
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of p0 ¼ 7:68 1028 Asm. From these numerical values
we see that the electric energy of Eq. (46) is much more
important than that of Eq. (50).
4.3. The mutual interaction energy between strongly
constrained water molecules
Let us look at a dipole pj and the dipoles pi , which are
very close to it. To find the minimum energy we assume
that the nearest dipoles pi of pj are not randomly
orientated but parallelly orientated so that they form a
chain. Let us fix the coordinate system to the dipole pj so
that the dipole pj is at the origin and the axis hj points
out in the direction of the dipole pj . Hence, the dipoles pi
on the positive side of hj make angles ðhi hj Þ ¼ 0, ðrhi Þ ¼
0 and ðrhj Þ ¼ 0, and thus mij ¼ cosðhi hj Þ ¼ 1, mi ¼
cosðrhi Þ ¼ 1 and mj ¼ cosðrhj Þ ¼ 1. Correspondingly,
for the dipoles pi located on the same line of hj but on
the negative side of the dipole pj we have ðhi hj Þ ¼ 0,
ðrhi Þ ¼ p and ðrhj Þ ¼ p, and thus mij ¼ 1, mi ¼ 1 and
mj ¼ 1. So in both cases ðmij 3mi mj Þ ¼ 1 3 ¼ 2,
which is its minimum value as we see from Eq. (39):
mij 3mi mj X ð1 þ 3m2j Þ1=2 X 2 ð1pmj p1Þ.
If the dipoles make chains at equal distances d from
each other, and if the dipole moments are all equally
large ðpi ¼ pj ¼ pÞ, then
X dV j
1 X pi pj
pi
¼
ð1 3Þ
4p0 r i ðidÞ3
dhi
i
ð2Þ 2 1
1
1
1
¼
p
2
þ þ þ ,
ð51Þ
4p0 r d 3 13 23 33
where the factor 2 comes from the summation of
positive and negative sides of the hj -axis.
The sum
is the value of the Riemann zeta
Pof the series
3
function 1
1=k
ffi
1:202,
and hence
k¼1
X
i
pi
dV j
4:808 p2
.
4p0 r d 3
dhi
(52)
The remarkable thing in the summation of Eq. (51) is
that the series converges very rapidly; only a few terms
need to be taken into the summation. Physically it
means that the length of the chain is not important. We
can have several separate chains instead of one longer
one, and if the total number of dipoles in both cases is
the same, the energy will be approximately the same.
Let us assume that the strongly restrained water
molecules ðN ¼ 720Þ can nearly all make such chains
with each other in the myosin head. Then the lowest
value of the total electrostatic energy in the chains of the
dipoles is
1 X dV j
1 4:808 p2
pi
N
Ec.
2 i;j
2 4p0 r d 3
dhi
(53)
411
Substituting the distance d ¼ 0:57 nm (Section 3.1) into
Eq. (53) we obtain
E c ¼ 8:2 1021 J ¼ 51 kJ=mol.
If the dipoles pi and pj are randomly orientated, then
there will always be equal positive and negative energy
pairs and the total sum of the energy is zero. Hence, the
electrostatic energy variation between the random
state’s zero energy and the minimum energy is
1 X dV j
0X
p
XE c .
(54)
2 i;j i dhi
After the completed power stroke cycle and in the
absence of ATP, the strongly restrained water molecules
will reorientate back to their initial directions so that the
electric energy of Eq. (54) together with Eq. (46) gets the
minimum value. In an ideal case the configuration of the
myosin head can simultaneously change so that the total
electric energy U remains constant (see Fig. 9).
4.4. On the electric energy of weakly restrained water
molecules
Under the influence of the electric field E the energy
density u ðJ=m3 Þ in the electrostatic field can be written
as
u ¼ 12E D ¼ 12E ð0 E þ Pf þ PÞ
¼ 120 r E 2 þ 12E P,
ð55Þ
where D is the electric displacement field, Pf ¼ 0 ðr 1ÞE is the linearly dependent polarization field (field
induced dipole moment/unit volume) and P is the
polarization field generated by the dipole moments p0
and pi that are strongly fixed within the myosin system.
The weakly restrained water molecules rotate freely
around their fixed positions as normal free water.
Hence, the electric energy for a weakly restrained water
molecule can be estimated as
1 1
E 2
38
2
0 r E ffi 1:04 10 J
uf ¼
,
(56)
nw 2
V=m
where nw is the number of water molecules per unit
volume, i.e. nw ¼ 3:34 1028 1=m3 .
Another way to estimate the electric energy for a
weakly restrained water molecule is to use the Langevin–Debye equation (Reitz et al., 1993)
p2
E 2
uf ¼ a0 0 þ
,
E 2 ffi 0:64 1038 J
V=m
3kB T
(57)
where a0 is the electronic polarizability of a water cluster
for which Ghanty and Ghosh (2003) have calculated a
value a0 ¼ 1:12 1029 m3 . For the dipole moment of a
water molecule we have used the value p ¼ 8:85 1030 Asm (Section 3.1).
ARTICLE IN PRESS
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Fig. 9. Gibbs free energy and electric energy diagrams for actomyosin ATP hydrolysis. (a) One power stroke, (b) two power strokes
ðT ¼ ATP; D ¼ ADP; P ¼ Pi ; M ¼ myosin; A ¼ actinÞ.
The mean radial electric field (at r ¼ rm ¼ 2:54 nm)
induced by the dipole moment p0 can be derived from
Eq. (21):
qV 0
pm 1
p
1
¼ 0 0 3 p 0 3 ¼ 1:1 107 V=m.
qr
2p0 r rm 2p0 r rm
(58)
The mean electric field (at sm ¼ 0:37 nm) induced by
the potential field of the actin can be estimated from
Eqs. (43) to (44)
jgrad cjpc0 keksm þ c1 =R0 ¼ 1:0 107 V=m.
(59)
Combining Eqs. (58) and (59) we see that Ep2:1 107 V=m and hence using Eq. (56) we get an upper limit
estimation for the electric energy of weakly restrained
water molecules (N f ¼ 1410; Section 3.1):
E f ¼ N f uf p1410 1:04 1038 ð2:1 107 Þ2
¼ 6:5 1021 J ¼ 4 kJ=mol.
ð60Þ
That is a small amount compared with the energy
variations of strongly restrained water molecules, as we
see from Eqs. (42) and (46) and (54). Therefore, the
weakly restrained water molecules can be left out of the
terms of Eq. (28).
4.5. Electric energy changes of ATPs binding and
hydrolysis process
At each reaction step of ATP the electric dipole
moment and its direction may change: ðp0 ; h0 Þ !
ðp00 ; h00 Þ, or we may also write it as ðp0 ; V 0 Þ ! ðp00 ; V 00 Þ.
As a consequence of this, the electric energy stored in the
myosin system changes:
dc
0
0 dc
U U ¼ p0 0 p0
dh
dh0
" 0
#
X dV 0 X dV 0
0
þ
pi
pi
.
ð61Þ
dhi
dhi
i
i
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As the reaction time itself is very short, the directions of
the dipole moments pi do not have time to turn or to
move during the reaction, and therefore the other two
terms of Eq. (28), which do not contain p0 or V 0 , will
not change, and are not included in Eq. (61). The second
bracketed term in Eq. (61) is much more important
than the first bracketed term, as we can see by
comparing Eq. (42) with Eq. (50).
On the other hand, Eq. (31) is valid for any reaction
step, i.e.
Uðp00 ; V 00 ; m00 ; pi ; mi ; cÞ Uðp0 ; V 0 ; m0 ; pi ; mi ; cÞ
p ½GðT; p; n01 ; . . . ; n0m Þ GðT; p; n1 ; . . . ; nm Þ,
ð62Þ
or briefly U 0 Up ðG 0 GÞ ¼ DG. In the ideal
case, in the reversible process, there is an equality in Eq.
(62). If DGo0, we then see that U 0 2U can be positive,
meaning that the electric energy can be charged by the
chemical reaction energy. On the other hand, if DG40,
then necessarily U 0 Up DGo0, which means that
the electric field energy is converted to ‘‘help’’ the
reaction to proceed in a ‘‘non-spontaneous’’ ðDG40Þ
direction. (An example of this is the removal of ADP
from the acto-myosin system ðA M D ! A M þ DÞ,
Fig. 3).
Let us then consider the hydrolysis cycle of ATP
shown in Fig. 3. The first step of the cycle is ATPs
binding to the actomyosin system ðT þ A M !
A M TÞ, for which the Gibbs free energy change
was estimated as DG ¼ 33:4 103 J=mol as shown in
Fig. 3. (With aATP ¼ 4 103 the corresponding value is
DG ¼ 36:0 103 J=mol; Kodama, 1985.)
In ATPs binding step the electrical change of the
system is
ðp0 ¼ 0; V 0 ¼ 0Þ ! ðp00 ¼ 230D; V 00 Þ,
413
molecules with the same dipole moment p0 is given by
Eq. (42) (DU II p64 kJ=mol, Fig. 9). As the ATP
hydrolysis process is a cyclic process, at the end of it
the electric energy has to have the same value again as at
the beginning of the process. Therefore, after the power
stroke process, or possibly after several power stroke
processes, a change is required for the dipole moment
p0 , which can then return the electric energy of the
myosin system to its initial value. This is done by the
ATP hydrolysis reaction, which reduces the dipole
moment ðp0 ! p00 Þ by the dissociation of the phosphate
group from the actomyosin system ðA M D P !
A M D þ PÞ and for which we have the remarkable
Gibbs free energy change DG ¼ 22:8 103 J=mol as
shown in Fig. 3. (With ap ¼ 2 103 the corresponding
value is DG ¼ 21:1 103 J=mol; Kodama, 1985.)
Substituting Eq. (61), and using only its second
bracketed term, which is much more significant, into
Eq. (62), we get for the hydrolysis reaction
X dV 0 X dV
0
0
pi
pi
dhi
dhi
i
i
p DG ¼ 22:8 103 J=mol.
ð64Þ
On the other hand, the left-hand side of Eq. (64) can be
estimated with the aid of Eqs. (41) and (42) as follows:
X dV 0 X dV 0
0
pi
pi
dh
dhi
i
i
i
ffiN
p 1
1=2
½ðp00 Þð1 þ 3m02
0m Þ
4p0 r r3m
ðp0 Þð1 þ 3m20m Þ1=2 "
#
p00 ð1 þ 3m00m 2 Þ1=2
¼E 1
,
p0 ð1 þ 3m20m Þ1=2
ð65Þ
which on the basis of Eq. (61) means that
0
U U ¼
p00
dc X dV 00
þ
pi
dh0
dhi
i
pðE p0 þ E p1 Þ þ E ¼ 33:8 kJ=mol,
ð63Þ
where we have also used Eqs. (42) and (50) with the
numerical values of E p0 ¼ 1:5 kJ=mol; E p1 ¼ 0:3 kJ=mol
and E ¼ 32 kJ=mol. As the maximum value of ðU 0 UÞ
in Eq. (63) is so close to ðDGÞ presented above,
this means that theoretically the whole Gibbs free
energy change of ATPs binding to the actomyosin
system can be stored in the form of electrostatic energy,
presuming that the dipole moments pi are properly
orientated before ATPs binding (DGðbindingÞ ¼ DU I
in Fig. 9).
The electric energy charged in the myosin system can
then be converted into work by reducing the electric
energy U. From Eq. (30) we see that for each electrical
working step W p DU. The lowest value for the
electric energy between ATP and constrained water
where p00 is the dipole moment of ADP and m00m is the
mean direction cosine of the dipole p00 . This can differ
from m0m as it is not only the dipole moment (p0 ) that
changes when the phosphate group is removed from
ATP, but also the direction ðh0 Þ of the dipole moment
that is changed. The mean cosine of the orientation
angle of the dipole moment p0 ¼ 230 D before the
reaction is according to Eq. (40a) m0m ¼ 0:678 ð47:3 Þ.
After the hydrolysis reaction the dipole moment is p00 ¼
90 D and the angle of the dipole orientation is changed
about 15 (Fig. 5). Hence, m00m ¼ ½cosð47:3 þ 15 Þþ
cosð47:3 þ 15 Þ=2 ¼ 0:655. Substituting these numerical values (with E ¼ 32 kJ=mol, Eq. (42)) into Eq. (65),
we obtain
X dV 0 X dV 0
kJ
0
,
(66)
pi
pi
¼ 19:7
mol
dh
dh
i
i
i
i
which is close to the value of the Gibbs free energy
change ðDGÞ of the hydrolysis reaction, Eq. (64).
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Hence, the electric dipole moment is decreased
appropriately so that the hydrolysis reaction energy
ðDGÞ can be converted into electric form (DU III in
Fig. 9). This makes it possible to return the electric
energy to its initial value, which is necessary in order to
remove the electric dipole p00 from the system. After the
hydrolysis reaction, a certain amount of work can
be done with the aid of the smaller dipole p00 (DU IV in
Fig. 9). After that the ADP molecule is removed and the
ATP hydrolysis cycle is completed.
In Fig. 9 we have illustrated the whole hydrolysis
cycle with the corresponding changes of the electric
energy in the myosin system. In Fig. 9 we have also
illustrated the process when the hydrolysis cycle consists
of two power strokes. If the change of DU for one power
stroke cycle is much less than the energy U charged in
the system, then it is possible that the myosin head can
perform several power strokes before the electric energy
U is discharged. However, after the hydrolysis reaction
only one power stroke is possible. Fig. 9 represents the
ideal process, where the efficiency of the hydrolysis cycle
corresponds to the theoretical maximum, i.e. W ¼
DG, where DG is the Gibbs free energy change of the
completed ATP hydrolysis reaction.
5. Combining electric theory with the mechanics of the
acto-myosin system
5.1. General formulae of electric forces and torques
between ATP and restrained water molecules
The decrease in the electrically stored interaction
energy ðui0 Þ between the dipole p0 and the myosin
structure fixed dipole pi , Eq. (23), is the primary energy
source for the mechanical rotation of the myosin head.
Applying the principal of virtual work, we derive
formulae for the electric forces in the general case
from Eq. (23) and then in the minimum energy state
from Eq. (38). The mathematical method is the same
that Maxwell (1873) used for magnets.
Let us choose an arbitrary direction of hk and let us
move the dipole pi by the distance dhk in this direction.
Applying the principle of virtual work, we get from
Eq. (23) the force F k acting on the dipole pi parallel to
the line hk :
dui0
pi p0
d3
1
¼
Fk ¼ dhk 4p0 r dh0 dhi dhk r
pi p0 d 1
¼ ðm 3m0 mi Þ
4p0 r dhk r3 0i
pp 1
¼ 3mk i 0 4 ðm0i 5m0 mi Þ
4p0 r r
pp m
pp m
ð67Þ
þ 3m0k i 0 4i þ 3mik i 0 40 ,
4p0 r r
4p0 r r
where mk ¼ cosðrhk Þ; m0k ¼ cosðh0 hk Þ and mik ¼ cosðhi hk Þ.2
On the other hand, the force F k can be compounded
of three forces F r , F 0 and F i that are in the directions of
r, h0 and hi , respectively. Using this component
expression the force F k can be written as
F k ¼ F r mk þ F 0 m0k þ F i mik .
(68)
Comparing Eqs. (67) and (68) we see that
3pi p0 1
ðm 5m0 mi Þ,
4p0 r r4 0i
3p p 1
3p p 1
F 0 ¼ i 0 4 mi ; F i ¼ i 0 4 m0 .
4p0 r r
4p0 r r
Fr ¼
ð69Þ
When the dipole pi is orientated so that the minimum
value of Eq. (38) is reached, the axis of the dipole pi is in
the same plane as the dipole p0 . Suppose that we now
have the minimum state and let us then move the dipole
pi in this plane in the direction of hk by a step dhk . As
before, we get the force by the principal of virtual work:
dui0 ðminÞ
pp d 1
2 1=2
¼ i 0
ð1
þ
3m
Þ
Fk ¼ 0
dhk
4p0 r dhk r3
"
#
p i p0 3
1 þ 4m20
m0
¼ m m .
4p0 r r4 ð1 þ 3m20 Þ1=2 k ð1 þ 3m20 Þ1=2 0k
ð70Þ
On the other hand, the force can be written as
F k ¼ F r mk þ F 0 m0k
(71)
and hence, comparing Eqs. (70) and (71) we get
Fr ¼ pi p0 3 1 þ 4m20
,
4p0 r r4 ð1 þ 3m20 Þ1=2
F0 ¼ þ
pi p0 3
m0
.
4
4p0 r r ð1 þ 3m20 Þ1=2
ð72Þ
The force F r in Eq. (72) is always negative, which means
that the myosin system tends to a state where the ATP
molecule attracts all the restrained water molecules
around it.
Due to the mathematical abstraction of the dipole
moment into one point (derivation of Eq. (21)) the
forces derived above are not able to give the electric
torques in a general case. Therefore, we also need to
derive the electric torques by the principle of virtual
work from the electric energy. Let us assume that the
2
As
d=dhk ½ð1=r3 Þðm0i 3m0 mi Þ ¼ ð3=r4 Þðdr=dhk Þðm0i 3m0 mi Þ þ
ð1=r3 Þ½dm0i =dhk 3ðdm0 =dhk Þmi 3m0 ðdmi =dhk Þ we get Eq. (67) by
substituting
here
dr=dhk ¼ mk ,
dm0i =dhk ¼ 0,
dm0 =dhk ¼
ðm0k m0 mk Þ=r and dmi =dhk ¼ ðmik mi mk Þ=r (as in the derivation of
Eq. (23)).
ARTICLE IN PRESS
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415
Fig. 10. (a) Electric forces and torques between ATP and restrained water molecules at the minimum state of electric energy. (b) Orientation of the
dipoles pi around p0 at the minimum state of electric energy. (c) Torques affecting the myosin head.
dipole p0 makes an angle a ¼ ðh0 rÞ with the position
vector r of the dipole pi , i.e. m0 ¼ cos a, and that the
dipole pi makes angle b ¼ ðrhi Þ with the position vector
r, i.e. mi ¼ cos b. To simplify the analysis, let us assume
that the dipole moments p0 and pi are in the same plane,
which means that ðh0 hi Þ ¼ ðh0 rÞ þ ðrhi Þ ¼ a þ b, i.e.
m0i ¼ cosða þ bÞ.
The myosin head can be separated by two sites, the
actin-binding site and the catalytic site, which receives
ATP, and these two sites can rotate around a narrow
junction O. Let the dipole p0 be placed at the junction O
and let the orientation of the dipole p0 be parallel with
the x-axis as shown in Fig. 10. Mathematically the
dipoles pi around p0 are located in two sectors:
0oaop=2 (sector I, the catalytic site) and 04a4
p=2 (sector II, the actin-binding site).
The rotational movement of the dipole pi can be
thought of as being composed of two rotations with the
angle changes of a and b. The change in the electric
energy ui0 of Eq. (23) can be written as
"
#
qui0
qui0
da þ
db
dui0 ¼ qa b
qb a
¼ M ai da þ M bi db,
ð73Þ
where the electric torques are defined as
qui0
M ai qa b
¼
pi p0 1
½sinða þ bÞ 3 sin a cos b,
4p0 r r3
qui0
qb a
pp 1
¼ i 0 3 ½sinða þ bÞ 3 cos a sin b.
4p0 r r
ð74Þ
M bi ð75Þ
According to these definitions, the electric torques M ai
and M bi are positive if the electric forces tend to increase
the angles a and b, respectively. The angles a and b are
defined to be positive in the clockwise direction.
The decrease in the electric energy ðdui0 Þ is converted
into mechanical work with the aid of the torques M ai
and M bi :
dW ¼ dui0 ¼ M ai da þ M bi db.
(76)
The mechanical work can be transferred to the
surroundings by the actin or it can be stored in other
parts of the myosin system. The discharge process of the
electric energy ui0 , Eq. (23), depends on the torques M ai
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416
and M bi which in turn depend on how the angles a and b
change during the rotation. In the following we will
study the electric torques and the rotational motion of
the myosin head with different possible angle changes of
a and b.
5.2. Analysis of electric torques with different angle
changes of dipoles
Rotation with steepest gradient: Let us then assume
that the angles a and b change in such a manner that the
decrease in the electric energy ui0 is maximal, i.e.
qui0
qui0
;
ðda; dbÞ is parallel with ,
(77)
qa
qb
which, using the definitions of Eqs. (74)–(75), means
that
da
db
¼
.
M ai M bi
(78)
Using Eqs. (73) and (78) the change of electric energy
can now be written as
dui0 ¼ ðM ai da þ M bi dbÞ ¼ ðM 2ai þ M 2bi Þ
¼ ðM 2ai þ M 2bi Þ
db
.
M bi
da
M ai
ð79Þ
As dui0 o0 we see from Eq. (79) that if da40, then
necessarily M ai 40, and vice versa, if dao0, then
necessarily M ai o0. Similarly, if db40, then necessarily
M bi 40, or if dbo0, then necessarily M bi o0. Hence, in
both cases, every dipole pi independent of its angles
ða; bÞ transfers positive rotational work according to Eq.
(76).
As a simple illustration of the process of Eq. (78) let
us consider a dipole pi for which at some instant of time
a ¼ b. Due to the symmetry of ui0 ; ui0 ða; bÞ ¼ ui0 ðb; aÞ,
we see from Eqs. (74) and (75) that M ai ¼ M bi when
a ¼ b. Hence, the rotation of the dipole pi around p0
proceeds if ‘‘we were tightening the wheel nut of a car’’;
the rotation angle of a corresponds to the rotation of b,
both in the same direction and by an equal amount,
da ¼ db.
General rotation theory and total torque: In a general
case let us suppose that the dipole pi has turned the
myosin structure by an angle dj around the joint O.
Then the work done by the electric forces will be
ðdui0 =djÞdj, and the total moment of the electric
forces turning the myosin structure will be
dui0
qui0 da qui0 db
¼
dj
qa dj
qb dj
da
db
þ M bi
.
¼ M ai
dj
dj
Mi ð80Þ
Due to the elastic structure of the myosin head, the
term db=dj in Eq. (80) can be greater than one, i.e.
the water molecule can turn more than the myosin
structure turns and in this way increase the total torque
effectively.
The sum of internal forces of the myosin system is
zero, and also the sum of the internal moments of forces
M i is zero (otherwise the dipole moment p0 would
rotate):
X
X
Mi þ
M i ¼ 0,
(81)
i2I
i2II
where the dipole system is divided into two rotatable
sectors I and II as shown in Fig. 10. However, as the
rotational angles are different for those two parts
ðdjNO adjOA Þ, the internal moments of forces can do
positive internal work, which can be written as
!
!
X
X
dW ¼
M i djNO þ
M i djOA
i2I
i2II
¼ MðdjNO djOA Þ ¼ Mdjo ,
ð82Þ
where the change of the total torsion angle of the myosin
head is defined as djo djNO djOA (see Fig. 2). The
work can also be written as
X dui0 X dui0 dW ¼
djNO þ
djOA
djNO
djOA
i2I
i2II
!
X
q X
¼
ðdui0 Þ ¼ ui0 djo .
ð83Þ
qjo
i
i
Hence, the total electric torque can be expressed either
by Eq. (82) or by Eq. (83) as
!
X
X
q X
M¼
Mi ¼
ðM i Þ ¼ ui0
qjo
i
i2I
i2II
!
q X dV 0
¼ pi
.
ð84Þ
qjo
dhi
i
In a general case it is complicated to estimate the electric
torque M as it depends on how the dipoles pi are
orientated and how they turn around during the
rotation process of the myosin head. However, we can
derive a formula for the torque M in an important
special case, at the minimum state of electric energy,
which is useful for understanding the scissors-shaped
motion of the myosin head.
Total torque at the minimum state of electric energy:
Let us assume that the dipole pi has rotated with respect
to the angle b into such an orientation that its energy ui0
has decreased to the minimum value (Eq. (38))
ui0 ðminÞ ¼ pi p0 1
ð1 þ 3cos2 aÞ1=2 .
4p0 r r3
(85)
ARTICLE IN PRESS
M.J. Lampinen, T. Noponen / Journal of Theoretical Biology 236 (2005) 397–421
We can now easily see that at the minimum state of
electric energy, the electric field ðgrad V 0 Þ is parallel
with the dipole pi , which means that the electric torque
M bi vanishes: sin a cos b 2 cos a sin b ¼ 0, i.e. tan a ¼
2 tan b. Hence, if M bi ¼ 0, then
½cosða þ bÞ 3 cos a cos b2
¼ ½cosða þ bÞ 3 cos a cos b2
þ ½sin a cos b 2 cos a sin b2
¼ 1 þ 3cos2 a
from which we get m0i 3m0 mi ¼ cosða þ bÞ 3 cos a
cos b ¼ ½1 þ 3cos2 a1=2 , and thus from Eq. (23) we
obtain Eq. (85). When the dipole moment p0 is in the
direction of the x-axis and y is perpendicular to x, then
m0 ¼ cos a; mx ¼ m0 ; mx0 ¼ 1,
qffiffiffiffiffiffiffiffiffiffiffiffiffi
my ¼ 1 m20 ; my0 ¼ 0
grad V 0 ¼ ðqV 0 =qx; qV 0 =qyÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ þ constant ð3m20 1; 3 1 m20 m0 Þ ð86Þ
according to which the gradients of V 0 have been drawn
in Fig. 10.
At the minimum state of energy the torque M mi can
be derived from Eq. (85)
dui0 ðminÞ
pp 13
sin 2a
¼ i 0 3
,
da
4p0 r r 2 ½1 þ 3cos2 a1=2
which is in fact generated by the force of F 0 (Eq. (72))
which we see as follows:
3pi p0 1 cos a sin a
r.
4p0 r r4 ð1 þ 3cos2 aÞ1=2
Eq. (87) is equivalent to M ai of Eq. (74) when the torque
M bi of Eq. (75) is zero. Eq. (87) shows that although the
dipole orientation with respect to the angle b has
reached the minimum value of electric energy, there still
exists a torque M mi as the electric energy tends to the
lowest energy, also with respect to the angle a.
When the dipoles pi are distributed symmetrically in
the parts of I and II ðN I ¼ N II ¼ N=2Þ, we get from Eq.
(87) an estimation for the electric torque at the
minimum state of electric energy
Mffi
N pi p0 1 3
sinð2am Þ
2 4p0 r r3m 2 ½1 þ cos2 am 1=2
As the electric energy decreases, the closer the dipole
pi comes to the actin surface (see Eq. (45)), a radial
inwards-directed electric force is exerted, which bends
the myosin head closer to the actin surface. Applying the
principle of virtual work, the mean radial force F s can
be derived from Eqs. (46) and (47) as
"
#
q X dc
N
Fs ¼ p
¼ pc0 k2 eksm
qsm i i dhi
2
½sin kx0 cos gm þ cos kx0 cos dm ,
(87)
M mi ¼ ðF 0 sin aÞr ¼ we get M ¼ 16 pN nm. Substituting the mean angle of
sector II, am ¼ 47:3 , into Eq. (88) we obtain
þ16 pN nm (as expected according to Eq. (81)), which
means that the electric forces tend to rotate the sector II in
a clockwise direction whereas the sector I tends to rotate in
a counter-clockwise direction. Hence, the two rotable parts
I and II tend to rotate in opposite directions reducing the
angle jo as if we had a scissors-shaped myosin head shown
in Fig. 2. The change of the angle jo in the power stroke
process illustrated in Fig. 2 is Djo ¼ 64 110 ¼
46 ¼ 0:80 rad, and thus, according to Eq. (82), the
work done by the torque M during the power stroke is
W ¼ MDjo ¼ ð16 pN nmÞ ð0:80 radÞ ¼ 13 pN nm.
(The work done by the constant force F (¼ 3:4 pN,
Section 2.1) on the actin during the power stroke is
3:4 pN 5:5 nm ¼ 19 pN nm).
5.3. Electric forces between the myosin head and the
actin– tropomyosin system
and we get from Eq. (36)
M mi ¼ 417
(88)
from which, with the numerical values of sector I
(rm ¼2:54 109 m; am ¼0:827 ðradÞ¼47:3 ; N=2¼360),
ð89Þ
where sm is the mean distance of the myosin head and
the actin–tropomyosin system. When F s o0, then
the force is opposite to the s-direction and attracts the
myosin head towards the actin. The force F x needed
to cause the myosin head to slide over the actin
in the direction of the x-axis can be derived from
Eq. (46) as
"
#
q X dc
N
Fx ¼ p
¼ pc0 k2 eksm
qx0 i i dhi
2
½cos kx0 cos gm sin kx0 cos dm .
ð90Þ
The myosin head has to reach such a distance sm and
place x0 during the power stroke that the force F x is
equal to or greater than the external force, otherwise the
myosin head slides over the actin. But this is not the
crucial criterion: more important is the electric
(contact) torque, which prevents the myosin head from
sliding over the actin by rotation. Let M A be the torque,
which is reckoned to be positive as the electric
forces tend to increase the rotation angle jA . Then we
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418
get from Eq. (46)
defined as
!
X dc
q
dc
MA ¼ pi
þ p0
qjA
dhi
dh0
i
!
!
X dc
q
q X dc dx0
ffi pi
pi
¼
qjA
dhi
qx0
dhi djA
i
i
N
pc0 k2 eksm l OA cos jA
2
½cos kx0 cos gm sin kx0 cos dm ,
¼ þ
ð91Þ
where the position of x0 depends on the rotation angle
jA . When the contact point A between the myosin head
and the actin is fixed, as is assumed when no sliding
takes place, and the myosin head turns virtually by djA ,
then dx0 ¼ l OA cos jA djA , where l OA is the distance
between A and O.
Substituting the numerical values ðN ¼ 720; p ¼ 8:85
1030 Asm; C0 ¼0:0125 V ; k ¼1:142 109 m1 ; sm ¼0:37
109 m; cos jA ¼1; l OA ¼4 109 mÞ into Eqs. (89)–(91)
we get the following values for the amplitude terms:
N
F^ s ¼ F^ x ¼ pc0 k2 eksm ¼ 34 pN,
2
N
^ A ¼ pc0 k2 eksm l OA ¼ 136 pN nm.
M
2
The greatest distance from the contact point A to the
junction of the tail T is 14.5 nm. Hence, assuming that the
maximum supporting contact torque is the amplitude
^ A ¼ 136 pN nm, we get for the maximum load
M
F max ¼ 136 pN nm=14:5 nm ¼ 9:4 pN, which is quite close
to the estimation F max ¼ 8 1 pN given by Morel (1991).
5.4. Estimation of elastic stiffness torques with electric
theory
The constrained water molecules are fixed by strong
hydrogen-bond-type molecular forces with the proteins.
The successive dipoles tend to orientate parallel to each
other in order to minimize their mutual electric energy.
In the rest state, when other electric forces are absent or
small, the myosin head tends to deform into a
geometrical shape that gives the minimum value for
the electric energy of Eq. (54). This is analogous with the
surface tension of water that deforms the droplets into a
spherical shape in order to minimize the surface energy.
The electric forces arising from the potentials of V 0
and C, together with the external load force, deform the
myosin head and change the orientation angles of the
dipoles. As a consequence of this, the term ðmij 3mi mj Þ,
the minimum of which is ð2Þ, becomes greater and the
electric energy of Eq. (54) increases, which requires
work. The torsional deformations are opposed by the
elastic stiffness torques M SO and M SN affecting the
rotational joints O and N, respectively, which can be
M SO
M SN
"
#
q 1 X dV j
¼
p
,
qjO 2 i;j i dhi
"
#
q 1 X dV j
¼
p
.
qjN 2 i;j i dhi
ð92Þ
Suppose that the stiffness torques M SO and M SN
increase linearly with the corresponding rotational
angles and that the rotation of Djmax corresponds to
the zero of the energy of Eq. (54). Then the internal
deformation work is ðM smax =2ÞDjmax ¼ 8:2 1020 J ¼
82 pN nm and, assuming that Djsmax ffi p=2, we get an
estimation: jM SO j þ jM SN jpM s max ffi 104 pNnm. As
there are two rotational joints O and N in the myosin
head system, the stiffness torques M SO and M SN depend
on the distribution of restrained water molecules
around O and N. The protein structure itself can also
have significant stiffness forces, as demonstrated by
Howard (2001).
5.5. On the Brownian stochastic motion of the myosin
head
During the power stroke the contact point A is fixed
within the actin. As the myosin head rotates and the
dipoles pi turn during the power stroke, the mean
position x0 with the angles gm and dm of Eq. (91) change.
Let us suppose that the myosin head reaches the
position ðx00 ; g0m ; d0m Þ where the moment M A vanishes,
i.e. ½cos kx00 cos g0m sin kx00 cos d0m ¼ 0. Then also according to Eq. (90), F x ¼ 0 and the myosin head can
slide over the actin freely around the position
ðx00 ; g0m ; d0m Þ. From Eqs. (46) and (91) we see that when
M A ¼ 0 the electric energy has the minimum value.
Hence, the myosin head can stay in the detached
position ðx00 ; g0m ; d0m Þ, vibrating freely around it in
Brownian motion until it receives enough thermal
fluctuation energy to overcome the energy barrier
around it. The solution to the problem of escape over
a potential barrier is the classical Arrhenius formula
(Gardiner, 1984), according to which the escape time is
proportional to the height of the energy barrier in
exponential form. To estimate the height of the energy
barrier from Eq. (46) we see first that
ðsin kx00 cos g0m þ cos kx00 cos d0m Þ2
¼ ðsin kx00 cos g0m þ cos kx00 cos d0m Þ2
þ ðcos kx00 cos g0m sin kx00 cos d0m Þ2
¼ cos2 g0m þ cos2 d0m ,
and thus ðsin kx00 cos g0m þ cos kx00 cos d0m Þ ¼ ðcos2 g0m þ
cos2 d0m Þ1=2 . On the other hand, as j cos p
g0mffiffijp1
and
ffi
j cos d0m jp1, the maximum variation is 2. Hence,
when the myosin head is in the minimum energy state,
ARTICLE IN PRESS
M.J. Lampinen, T. Noponen / Journal of Theoretical Biology 236 (2005) 397–421
the height of
barrier
around it according to
pffiffithe
ffi energy p
ffiffiffi
Eq. (46) is 2 2E 0 ¼ 2 2 3 1020 J ¼ 8:5 1020 J
¼ 21kB T, where kB ¼ 1:381 1023 J=K (Boltzmann’s
constant), T ¼ 298:15 K and E 0 is calculated from Eq.
(47) with the mean distance sm ¼ 0:37 nm. The result of
21kB T is quite close to the values that have been used in
stochastic models for the energy barrier, e.g. in the work
of Buonocore and Ricciardi (2003).
Since
always
jp
sin
ffiffiffi kx0 cos gm þ cos kx0 cos dm j
pj sin kx0 j þ j cos kx0 jp 2, by combining Eqs. (46)
and (50) we obtain a general estimation for the variation
of the electric energy
pffiffiffi
dc X dc
2ðE 0 þ E p0 Þpp0
þ
pi
dh0
dhi
i
pffiffiffi
p 2ðE 0 þ E p0 Þ,
ð93Þ
where
pffiffiffi
2ðE 0 þ E p0 Þ ¼ 4:6 1020 J ¼ 28 kJ=mol ¼ 11kB T.
transformed into mechanical work
"
!
X dc
q
dc
djA
pi
þ p0
F dxp dU ¼ qjA
dhi
dh0
i
!
q X dV 0
þ
pi
djo
qjo
dhi
i
!
q 1 X dV j
djo
þ
p
qjo 2 i;j i dhi
!
#
q
1 X dV j
þ
p
djN
qjN 2 i;j i dhi
¼ M A djA þ M djO þ M SO djO
þ M SN djN ,
The work done by the myosin on the actin is
dW ¼ F dx, where dx is the movement of the actin
under the force F. We can now express the work in terms
of mechanics and in terms of thermodynamics by
electric theory.
Multiplying the kinematic (2) by the force F and then
using Eq. (3) we obtain purely on the basis of mechanics
F dx ¼ M N djTN þ ðM O M N Þ djNO
þ ðM A M O Þ djOA
¼ M N djN þ M O djo þ M A djA ,
ð94Þ
where djN ¼ djTN djNO ¼ dðjTN jNO Þ, djO ¼
djNO djOA ¼ dðjNO jOA Þ and djA ¼ djOA . The
angles of the joints N and O are defined as jN ¼
p þ ðjTN jNO Þ and jO ¼ p þ ðjNO jOA Þ as shown
in Fig. 2.
According to Eq. (30) the work can also be written in
terms of thermodynamics as
F dxp dG ¼ ðdG þ dUÞ.
(95)
As the instants of the reaction steps are very short, the
myosin head does not have enough time to move
ðdx ¼ 0Þ, and hence during the actual instants of the
reaction steps
dG ¼ dG þ dUp0,
(96)
which is equivalent to Eq. (31) (Eq. (96) is also valid for
the stochastic process when the myosin is detached from
the actin, because then no work is done by the myosin
system on the actin.)
Dividing the process in this way for reaction instants
and for deformation processes, the energy for the power
stroke is taken from the electric energy U and
ð97Þ
where we have used Eqs. (84) and (91)–(92). Hence,
F dxp dU ¼ M A djA þ ðM SO þ MÞ djO
þ M SN djN .
5.6. Work done by the electric torques of the myosin
system
419
ð98Þ
Comparing Eq. (98) with Eq. (94) we see that the
torque M O of Eq. (94) consists of two parts,
M O ¼ M SO þ M. The torque M SO is the stiffness
torque given by Eq. (92) and the torque M is generated
by the interaction of the dipoles p0 and pi and defined by
Eq. (84). When the myosin head is in the detached
position, M A ¼ M SN ¼ 0 and also M O ¼ M SO þ
M ¼ 0, which means that the ATP-driven torque M
and the stiffness torque M SO keep each other in the
equilibrium.
In Section 2.2 we estimated for an example of the
power stroke that M A ¼ 36 pN nm, M O ¼ 25 pN nm
and M N ¼ 14 pN nmð¼ M SN Þ. If we use the numerical
value of Eq. (88) for M ¼ 16 pN nm, we have M SO ¼
41 pN nm and for the total stiffness torque we get
jM SN j þ jM SO j ¼ 55 pN nm, which is in accordance
with the numerical range ðp104 pN nmÞ given in Section
5.4.
The torque M is negative, which means that it tends to
decrease the angle jo , i.e. djNO o0 and djOA 40 as
djO ¼ djNO djOA . On the other hand, the torque
M A is positive, which means that it tends to increase the
angle jA , i.e. djA ¼ djOA 40. Hence, both the torques
M and M A try to turn the body OA in the clockwise
direction (djOA 40, Fig. 2), and thus increase the
force F.
Each torque is important in different stages of the
process giving a contribution to the work. The torques
are not constant during the power stroke, but just to
give a simple illustration let us calculate the work
done by the constant torques above and with the
angle changes shown in Fig. 2 for the power
stroke: M A DjA ¼ 36 0:50 ¼ 18 pN nm; M SO Djo ¼
41 ð0:80Þ ¼ 33 pN nm; MDjo ¼ ð16Þ ð0:80Þ ¼
13 pN nm; M SN DjN ¼ 14 1:2 ¼ 17 pN nm. Thus,
altogether the work done by the torques during the
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M.J. Lampinen, T. Noponen / Journal of Theoretical Biology 236 (2005) 397–421
power stroke is þ15 pN nm. However, for the
completed cycle only the torques M and M A give a
positive network.
To see this, let us study a completed cycle process of
the myosin head: attachment to the actin, a power
stroke, detachment from the actin and return back to
the original detachment position. The work done in
the ideal (reversible) process, when we have equality in
Eq. (98), is
I
I
I
W¼
M A djA þ M SO djO þ M djO
I
þ
M SN djN ,
ð99Þ
where the integrations are taken around closed cycles
along the corresponding rotation angles. The stiffness
torques M SO and M SN are derivatives of continuously
differentiable functions and, therefore, as we see from
Eq. (92), we have
"
#
I
I
q 1 X dV i
p
djO ¼ 0,
M SO djO ¼ qjO 2 i;j i dhj
"
#
I
I
q 1 X dV i
M SN djN ¼ p
djN ¼ 0
qjN 2 i;j i dhj
as the electric energy term returns to its initial value
after the completed cycle.
Thus, the network is determined only by the torques
M A and M:
I
I
(100)
W ¼ M A djA þ M djO .
These integrals are not zero because of the discontinuities of the electric energy functions of Eq. (61) at the
reaction steps. Combining Eq. (61) with Eq. (91) we
obtain
!
I
I
X dc
q
dc
M A djA ¼ pi
þ p0
qjA
dhi
dh0
i
X
dc
¼
D p0
,
ð101Þ
dh0
where the summation is taken over all reaction steps.
Similarly, combining Eq. (61) with Eq. (84) we obtain
!
I
I
q X dV 0
M djo ¼ pi
djo
qjo
dhi
i
"
#
X X dV 0
¼
D
pi
.
ð102Þ
dhi
i
On the other hand, for each reaction step we have
Eq. (62), and, thus, by summing up the steps we get for
the ideal (reversible) cycle process
I
I
W ¼ M A djA þ M djo ¼ DG,
(103)
where DG is the Gibbs free energy change for the
completed hydrolysis reaction of ATP. From Eqs. (42),
(50), (101) and (102) we see that the main part of
the work for the whole cycle is done by the torque M of
Eq. (84).
6. Conclusions
The mathematical analysis of the electric dipole
theory is based on Eq. (29), which can be expressed as
X dV 0 1 X dV j
G ¼ GðT; p; n1 ; . . . ; nm Þ þ
pi
þ
p
dhi 2 i;j i dhi
i
|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}
U TW
X
dc
dc
þ p0
þ
pi
.
dh0
dh
i
i
|fflffl{zfflffl} |fflfflfflfflffl
ffl{zfflfflfflfflfflffl}
U AT
U W W
ð104Þ
U AW
The term U TW contains the electrostatic energy
owing to the mutual interaction of electric dipoles of
ATP (T) and water (W) molecules fixed within the
myosin system. The numerical calculations in Section
4.5 show that the Gibbs free energy G changes of ATPs
binding ðDG ffi 33 kJ=molÞ and hydrolysis reaction
ðDG ffi 23 kJ=molÞ can be converted to the electric
energy form nearly by the term U TW (DU TW (binding
of ATP) p32 kJ=mol, DU TW (hydrolysis of ATP)
ffi 20 kJ=mol). The decrease in the electric energy
U TW towards the minimum generates an internal
electric torque M (jMjX16 pN nm; Section 5.2), which
performs the main part of the work for the whole cycle
(Eq. (103)). After the hydrolysis reaction of ATP the
remaining dipole moment of ADP is appropriately
smaller, enabling the removal of ADP from the myosin
head so that the cycling process can be repeated.
The numerical calculations presented in Section 4.5
also show that the number of water molecules restrained
strongly in the myosin system ðN ffi 720Þ seems to be
quite optimal with respect to the dipole moment of
ATP—if the number N were much below this, the
chemical energy changes (ATPs binding and hydrolysis
reaction) could not be stored into the form of electric
energy with a good efficiency.
The electrostatic energy term U W W describes the
elastic energy of the myosin system stored between
restrained water molecules and its derivatives give the
internal elastic stiffness torques for the structure. The
electric energy of the ATP molecule in the potential field
of the actin (A) is given by the term U AT and,
correspondingly, the electric energy of restrained water
molecules by the term U AW . The electric interaction
ARTICLE IN PRESS
M.J. Lampinen, T. Noponen / Journal of Theoretical Biology 236 (2005) 397–421
between the strongly restrained water molecules and the
actin produces a contact force and a contact torque
which prevent the sliding and make it possible to
transfer the work to the actin. After the power stroke,
the myosin head reaches a position where the electric
contact force and torque with the actin are zero and it is
possible to return the myosin head to the initial position
by stochastic Brownian motion. From the electric
energy term U AW we have estimated the maximum
force per myosin head (F max ¼ 9:4 pN; Section 5.3) and
also the (maximum) energy barrier (21kB T; Section 5.5)
for the stochastic Brownian motion.
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