Coordinated Control of Activated Glycolysis and Respiration in Tumor Cells 219
where F includes fructose diphosphate and triose phosphates and Vs includes AMP formation and conversion to ADP. Assuming complete coupling of both oxidative and glycolytic phosphorylation:
dDldt c_ 2 V PFK - 4 V LDH - 6 V R + Vs
(1)
dPldt
- 2 VLDH - 6VR + Vs
(2)
where D = ADP and P = intracellular Pi concentration, and V LDH is in
glucose equivalents (1/2dL(dt). Since there are 2 possible variables for each
of the 4 velocities, there are 16 possible combinations and solutions. To
limit the choices, it will be assumed that V R = KRD(D) and Vs = KSD(D)
+ 50 in all cases, as suggested by independent evidence (for V R cf. [5]).
Vs contains an additive term, 50, since it extrapolates to a substantial rate
at zero ADP; the relationship between Vs and A TP is more consistent, but
ATP may be considered a linear function of ADP (1). The remaining combinations are then: case (,7), V PFK = KFD(D), V LDH = KLD(D); case (b),
VPFK=KFD(D), VLDH = KLP(P); case (c), VPFK=KFP(P), V LDH =
KLD(D); and case (d), T 'PFK = KFP(P), VLDH = KLP(P). If either ADP
or Pi are unevenly distributed within the cell, the same equation will apply
so long as the concentration in a given compartment is proportional to the
total. These can be substituted into ego 1 and 2 and the resultant solved for
ADP.
Case (a) requires only eq. 1:
D= 1 I r { (r Do + 50) e r t - 50}
(3)
where r = (2KFD + KSD - 4KLD - 6KRD), and Do = ADP at t = o.
Cases (b)-(d) require both eq. 1 and 2 and reduce to the same general
form:
d 2 D
dD
r- A --- + B D + C = 0
(4)
d t 2
d t
(A2>4B)
D=5 1 e- at +5 2 e-/3t-C/B
(5)
where c< = 1(2 (A + V~4~4B-) ; ~ = 1/2 (A - VA2 - 4B)
(A2<4B)
D=53e-li2Atsin(1/2j/4B~A2)t-CfB
(6)
51-53 are constants of integrations.
Note that (A2 < 4B) produces an oscillating system. In case (b), A =
(6KRD + 2KLP - 2KFD - KSD); B = 2KLP(KsD - 2KFD - 6KRD); C =
2KLP50· In (c), A --- (6KRD + 4KLD - KSD); B = 2K FP (2KLD - 6KRD -
KSD); C = -2K FP 5 0 . In (d), A = (6K RD + 2KLP - K SD ); B = 2(KFP -
KLP)(6KRD - K SD ); C = 2 (K LP - K FP ) 50· From Fig. 1, in (min)-1:
KRD, 2.10; K SD , 9.1; K LD , 5.0; K FD , 11.1; K LP , 0.67; K FP , 1.57.5 0 = 5.9
fLmoles (ml cells Imin, and Do = 0.72 fLmoles (ml cells. Solutions are as
follows:
where F includes fructose diphosphate and triose phosphates and Vs includes AMP formation and conversion to ADP. Assuming complete coupling of both oxidative and glycolytic phosphorylation:
dDldt c_ 2 V PFK - 4 V LDH - 6 V R + Vs
(1)
dPldt
- 2 VLDH - 6VR + Vs
(2)
where D = ADP and P = intracellular Pi concentration, and V LDH is in
glucose equivalents (1/2dL(dt). Since there are 2 possible variables for each
of the 4 velocities, there are 16 possible combinations and solutions. To
limit the choices, it will be assumed that V R = KRD(D) and Vs = KSD(D)
+ 50 in all cases, as suggested by independent evidence (for V R cf. [5]).
Vs contains an additive term, 50, since it extrapolates to a substantial rate
at zero ADP; the relationship between Vs and A TP is more consistent, but
ATP may be considered a linear function of ADP (1). The remaining combinations are then: case (,7), V PFK = KFD(D), V LDH = KLD(D); case (b),
VPFK=KFD(D), VLDH = KLP(P); case (c), VPFK=KFP(P), V LDH =
KLD(D); and case (d), T 'PFK = KFP(P), VLDH = KLP(P). If either ADP
or Pi are unevenly distributed within the cell, the same equation will apply
so long as the concentration in a given compartment is proportional to the
total. These can be substituted into ego 1 and 2 and the resultant solved for
ADP.
Case (a) requires only eq. 1:
D= 1 I r { (r Do + 50) e r t - 50}
(3)
where r = (2KFD + KSD - 4KLD - 6KRD), and Do = ADP at t = o.
Cases (b)-(d) require both eq. 1 and 2 and reduce to the same general
form:
d 2 D
dD
r- A --- + B D + C = 0
(4)
d t 2
d t
(A2>4B)
D=5 1 e- at +5 2 e-/3t-C/B
(5)
where c< = 1(2 (A + V~4~4B-) ; ~ = 1/2 (A - VA2 - 4B)
(A2<4B)
D=53e-li2Atsin(1/2j/4B~A2)t-CfB
(6)
51-53 are constants of integrations.
Note that (A2 < 4B) produces an oscillating system. In case (b), A =
(6KRD + 2KLP - 2KFD - KSD); B = 2KLP(KsD - 2KFD - 6KRD); C =
2KLP50· In (c), A --- (6KRD + 4KLD - KSD); B = 2K FP (2KLD - 6KRD -
KSD); C = -2K FP 5 0 . In (d), A = (6K RD + 2KLP - K SD ); B = 2(KFP -
KLP)(6KRD - K SD ); C = 2 (K LP - K FP ) 50· From Fig. 1, in (min)-1:
KRD, 2.10; K SD , 9.1; K LD , 5.0; K FD , 11.1; K LP , 0.67; K FP , 1.57.5 0 = 5.9
fLmoles (ml cells Imin, and Do = 0.72 fLmoles (ml cells. Solutions are as
follows: