NON-LINEAR EFFECTS OF LOWINTENSITY PHYSICAL AGENTS ON
ENZYMATIC REACTIONS IN
BIOLOGICAL OBJECTS
V.A. Monich, S.L. Malinowskaya,
K.A. Sokolova
Enzymes catalyze biochemical reactions. The reactions
convert substances into products. An enzyme molecule (Enz)
attaches to a substance molecule (Sbs) and modify the
molecular bonds. As a result, a product (Pro) molecule
formation becomes much more easier:
Sbs + Enz -> (SbsEnz) -> Pro + Enz
Math models of chemical reactions is based on the law of
mass. According to this law, rates of reactions of two or more
reactants are proportional to products of their concentrations if
a temperature of the medium remains constant
We use the following assumptions: 1)the concentrations of the
substrates, the enzymes and their products are dependent only
on time and are not dependent on the spatial coordinates, 2)the
amounts of all these substances are large.
A part of the complexes (SbsEnz) may destruct, so the
synthesis process with the rate r11,
Sbs + Enz -> (SbsEnz) has the retroaction,
(SbsEnz) -> Sbs + Enz with the rate, r12 and the one-sided
process,
(SbsEnz) -> Pro + Enz has the rate r2.
dCEnz /dt = - r11CEnzCSbr + r12CEnz Sbr + r2 CEnz Sbr
dCEnz Sbr /dt = r11CEnzCSbr - r12CEnz Sbr - r2 CEnz Sbr
dCSbr /dt = - r11CEnzCSbr + r12CEnz Sbr
dCPro /dt = r2 CEnzSbr
This system can be transformed:
(1) + (2) dCEnz /dt + dCEnz Sbr /dt = 0
(1)
(2)
(3)
(4)
(5)
CEnz + CEnz Sbr = (CEnz) 0 , the initial concentration of the enzyme (6)
(1) - (3) - (4) = 0,
CEnz /dt = dCSbr /dt + dCPro /dt , after integration we receive,
(7)
CEnz = CSbr + CPro + K (the constant of integration)
(8)
In a quasistationary approximation all components of the reactions reach
their stationary state instantly. Particularly, the Enz Sbr complex reaches its
stationary state, that is, the
dCEnz Sbr /dt ≈ 0
(9)
The (2) and the (7) yield to
r11CEnzCSbr - r12CEnz Sbr - r2 CEnz Sbr ≈ 0 → CEnz Sbr =
= r11 /( r12 + r2 ) CEnzCSbr
CEnz Sbr = rMi CEnzCSbr ,
(10)
where rMi = r11 /( r12 + r2 )
The eq. (6) and (10) yield, CEnz = (CEnz) 0 - CEnz Sbr → CEnz Sbr =
rMi ((CEnz) 0 - CEnz Sbr ) CSbr →
→ CEnz Sbr ≈ rMi ((CEnz) 0 CEnz Sbr /(1 + rMi CSbr)
(11)
Eq (5) yields, dCEnz /dt ≈ - dCEnz Sbr /dt , this result and the eq. (7) give,
dCSbr /dt = dCEnz /dt - dCPro /dt → dCSbr /dt = - dCEnz Sbr /dt - dCPro /dt
→
dCEnz Sbr /dt ≈ 0 (from the eq. (9)) → dCSbr /dt ≈ - dCPro /dt = r2 CEnzSbr
(see eq. 4) and the eq. (11) yields dCSbr /dt = - r2 rMi (CEnz) 0 CSbr /(1 + rMi
CSbr) = - Ceff rMi CSbr /(1 + rMi CSbr)
→ dCSbr /dt = - Ceff CSbr /(rM + CSbr),
(12)
where Ceff = r2 (CEnz) 0 and rM = 1 /rMi
The speed of change of the substrate as a function of the substrate
concentration (see the equation 12) is represented in the Fig.1. This
function really shows how an activity of the biochemical reaction is
dependent on the substrate concentration. The speed saturates at the
level Ceff = r2 (CEnz) 0 that depends on the origin concentration of the
enzyme and the rate (r2) of the Enzyme-Substrate complex
disintegration. The measure of the substrate concentration value is
rM.
Fig. 1
The speed of change of the substrate as a function of the substrate
concentration
So, the speed of the biochemical reaction can be modulated by
changing the initial amount of the enzyme concentration, (CEnz) 0 , by
modulation of r11, r12, the action rates and the retroaction rates of the
reaction Sbr + Enz → SbrEnz and also by r2, the rate of the final
reaction, (SbrEnz) → Pro + Enz .
It is useful to decrease the number of variables in the basic equations
and introduce dimensionless variables. Consider again the equations (2)
and (3)
dCEnz Sbr /dt = r11CEnzCSbr - r12CEnzSbr - r2 CEnzSbr
dCSbr /dt = - r11CEnzCSbr + r12CEnzSbr .
taking into account that CEnz = (CEnz) 0 - CEnzSbr (see eq. 6), we get
the following system of equations:
dCEnz Sbr /dt = r11(CEnz) 0 CSbr - (r11 C Sbr + r11 + r2 )CEnzSbr (13)
dCSbr /dt = - r11(CEnz) 0 CSbr + (r11 CSbr + r12)CEnzSbr
(14)
After division both sides of the equations by the product,
(CEnz) 0 (CSbr) 0 r11, and introducing the new variables v, u, T and the
parameter, p,
(v = CEnzSbr /(CEnz)0 , u = CSbr / (CSbr)0 , T = t r1 (CEnz)0,
p = (CEnz)0 /(CSbr)0 ), we get :
p dv/dT = u – [u + (r12 + r2)/ /(r11CSbr)0]x
(15)
du/dTt = - u + [u + r12 / r11(CSbr)0)]x
(16)
Fig. 2
Development of the set of concentrations of catalyzed biochemical
reaction components with time
Eeffect of red light on a level of activity of SOD in
the heart tissues
Con: the control group data, L1: the data for the
subgroups effected by He-Ne light, L2: the data for
the subgroups effected by luminescent light
THANK YOU!