Appendix VI
129
APPENDIX VI
Derivation of relaxation times for a three step series mechanism
The reaction between S (polymer) and D (dye) can be described by a three step series mechanism as
(VI.1)
where S and D denote respectively the free sites of the nucleotide and the free dye, and D,S, DS
Iand DS
IIdifferent bound forms.
Such a system can theoretically display three different relaxation effects, each one being characterised by its relaxation time λ.
The (VI.1) process is composed of three thermodynamically independent reactions
D + S ' D,S (VI.2)
D,S ' DS
I(VI.3)
DS
I' DS
II(VI.4)
The three reciprocal relaxation times 1/τ
k= λ
kare related to the equilibrium concentrations and to the rate constants of the three steps trough the determinantal equation (Maggini et al, 1997)
(VI.5)
where g
ij= g
jiand the g
ijcoefficients are related to the equilibrium concentrations by the equations (Castellan, 1963)
S]
[D, 1 [D]
1 [S]
g
11= 1 + + (VI.6)
r1g11- λk r1g12 r1g13 r2g21 r2g22 - λk r2g23 r3g31 r3g32 r3g33 - λk
= 0
S + D k
1D,S k
2DS
IDS
IIk
-2k
3k
-3k
-1Relaxation times equations
130
S]
[D,
g
12= − 1 (VI.7)
0
g
13= (VI.8)
] [DS
1 S]
[D, g 1
I
22
= + (VI.9)
] [DS g 1
I
23
= − (VI.10)
] [DS
1 ]
[DS g 1
II I
33
= + (VI.11)
where [S], [D], [D,S], [DS
I] e [DS
II] are the equilibrium concentrations of the S, D, D,S, DS
Ie DS
IIspecies respectively.
Eq. (VI.5) is a cubic equation with respect to λ which leads to quite complicate expression for λ
1, λ
2and λ
3.
Nevertheless, the problem can be greatly simplified if the three kinetic effects take place in very different time ranges, that is λ
1>> λ
2>> λ
3. If this is the case, like it happens indeed for the systems here investigated, three distinct expressions can be used for each relaxation time deriven from the general expression (Castellan, 1963)
1 k
k k k
D D r
−
=
λ (VI.12)
where D
k(D stays for Determinant) are given by
D
0= 1 (VI.13)
D
1= g
11(VI.14)
(VI.15)
D2 =
g11 g12 g21 g22
Appendix VI
131 (VI.16)
whereas r
kare the exchange rate of the reactions (VI.2), (VI.3) and (VI.4).
r
1= k
1[D][S] = k
-1[D,S] (VI.17)
r
2= k
2[D,S] = k
-2[DS
I] (VI.18)
r
3= k
3[DS
I] = k
-3[DS
II] (VI.19)
We can, then, obtain the final expressions for the reciprocal relaxation times
1 1
11 1 0
1 1
r g k ([D] [S]) k
D
r D = = + +
−=
=
11
1
λ /τ (VI.20)
2 1
2 1 11
2 2 12 22 2 11
21 12 22 2 11 1
2 2
k
[S]) ([D]
K 1
[S]) ([D]
k K g
r g g g r
g g g r g D
r D +
−+ +
= +
−
− =
=
=
=
22
1
λ /τ (VI.21)
(where K
1= k
1/k
-1)
3 2
1 3 2 1 21 12 22 11
21 13 23 11 32 33 3 2 3 3 3
3
k
]) S [ ] D )([
K 1 ( K 1
]) S [ ] D ([
k ) K 1 ( K g
g g g
) g g g g g ( g D r r D
1 +
−+ +
+
+
= +
⎭ ⎬
⎫
⎩ ⎨
⎧
−
− −
=
= τ
=
λ (VI.22)
(where K
2= k
2/k
-2).
The above relationships have been derived for a mechanism of ordinary reactions.
It has been demonstrated (Jovin and Striker, 1977) that when one of the reactant is a linear polymer the expression for g
11should be modified in
S]
[D, 1 [D]
1 [S]
(r) '
g11 =−f + +
(VI.23)
D3 =
g11 g12 g13 g21 g22 g23 g31 g32 g33
Relaxation times equations