- Email: [email protected]
Application of electrical analogs to metabolic control analysis: Linear pathways with multiple feedback
hr. J. Eiochem. Vol. 22, No. II, pp. 129S1301, Printed in Great Britain
0020-71 IX/90 S3.00 + 0.00 Pergamon Press plc
1990
APPLICATION OF ELECTRICAL ANALOGS TO METABOLIC CONTROL ANALYSIS: LINEAR PATHWAYS WITH MULTIPLE FEEDBACK h0K
K. SEN
Department of Chemical Engineering, California Institute of Technology, Pasadena, CA 91125, U.S.A. (Received
26 February
1990; received for publication
11 May 1990)
Abstract-l. Using electrical analogs, we have presented a systematic procedure for calculating the flux control coefficients of linear metabolic pathways with multiple feedback loops. 2. In this method, an electrical analog circuit is constructed first for the unregulated pathway. 3. This circuit is subsequently modified in a step-by-step fashion to take into account the effect of each feedback loop in the pathway. 4. An analog circuit consists of resistances which are connected in series (or parallel) with a voltage (or current) source. 5. The flux control coefficients of the enzymes are represented by voltages across (or currents through) the resistances and are determined by an application of Ohm’s law. 6. We have investigated the possible patterns in linear pathways with two feedback loops. 7. This is followed by an analysis of a linear pathway with an arbitrary pattern of feedback inhibition.
INTRODUCTION In an earlier paper @en, 1990a), we have shown that electrical analogs can be conveniently used for the calculation of flux control coefficients of metabolic pathways. In that paper we examined an unregulated linear pathway, and a linear pathway with a feedback loop. An electrical analog circuit for the control structure of a given pathway is designed in the following manner. The enzymes in the pathway are associated with resistances which are connected in series (or parallel) with a voltage (or current) source. The magnitudes of these resistances depend on the elasticity coefficients of the enzymes. The flux control coefficients of the various enzymes are characterized by voltages across (or currents through) the resistances and are determined by an application of Ohm’s law. In this paper we shall use electrical analogs to analyze the flux-control distribution in linear metabolic pathways with multiple feedback loops. First we will enumerate the possible configurations with two feedback loops. Both overlapping and non-overlapping patterns of feedback loops will be considered. This is followed by an examination of a linear pathway with an arbitrary pattern of feedback inhibition. We shall see that the electrical analogs for each of these pathways can be constructed by modifying the analog circuit of the unregulated pathway (i.e. in the absence of the feedback loops) in a step-by-step fashion. These modifications can be made from an inspection of the topology of the feedback loops in the pathway, and it is not necessary to invoke the summation or connectivity theorems explicitly. As in the earlier paper (Sen, 1990a), the present formulation lies within the premises of the Metabolic Control Theory of Kacser and Burns (1973) and Heinrich and Rapoport (1974).
For the pathways examined below, we shall present only the current-analog circuits (i.e. in which all the resistances are connected in parallel with a current source). The corresponding voltage-analog circuits for these pathways can be designed by connecting the resistances in series with a voltage source and choosing the magnitudes of the resistances as reciprocals of those in the current-analog circuits [see Sen (1990a) for details]. It should be mentioned here that control coefficients of metabolic pathways are customarily evaluated by writing the governing equations using summation and connectivity theorems in a matrix form and subsequently inverting the appropriate matrix [see, for example, Fell and Sauro (1985) and Westerhoff and Kell(1987)]. Recently, using concepts from graph theory, Sen (1990b, c) has presented an alternate approach for calculating these control coefficients. A diagrammatic method due to Hofmeyr (1990) is also available for control-pattern analysis of metabolic pathways. LINEAR PATHWAYSWITH MULTIPLE FEEDBACKLOOPS
As our first example, we consider a linear pathway in which two different enzymes, namely, E, and E, undergo feedback inhibition by the same metabolite, K,. S2+X,2+X,2+X,~X,~P
1
(a)
(4
wl
In order to represent the control structure of this pathway by a current-analog circuit, we first construct a current-analog circuit for the unregulated pathway, i.e., by ignoring the feedback loops. This is
1295
Asox K.
1296
shown in Fig. I which has been reproduced from our previous paper (Sen, 1990a). In this figure a resistance R, (i = 1, 2, 3, 4, 5) is associated with the enzyme E,, and the currents C, through C5 designate the flux control coefficients of the enzymes E, through E,, respectively. The resistances are connected in parallel with a current source of strength unity. The magnitudes of the resistances depend on the elasticity coefficients of the enzymes, as shown in the figure. The symbol c; denotes the elasticity coefficient of an enzyme E, with respect to a metabolite X,. Furthermore, c:. which is the elasticity coefficient of the enzyme E, due to inhibition by its product, is written as c:=r;, c: being the magnitude of the elasticity coefficient. The reader is referred to the previous paper (Sen, 1990a) for further details. Effects of the two feedback loops in pathway (A) can be incorporated into the circuit of Fig. I as follows. Consider each feedback loop separately. For definiteness, consider the pathway with only the outer feedback loop first. We shall refer to it as pathway (A’). Note that there is only one enzyme, E,, which is located outside this loop on its downstream side. Consequently, as shown in Part I, we need to modify the resistance in the path of the current C5 by adding a resistance, R”,, in parallel with R5 (Fig. 2). The magnitude of this resistance is selected in the following manner. We redraw the pathway (A’) by deleting the feedback loop and all the enzymes and metabolites which lie inside it, thereby treating the enzymes E, and E, as contiguous enzymes with X, as the adjoining metabolite.
SEN
In the above representation, the elasticity coefficient (6;) of an enzyme E, with respect to a metabolite X, is denoted by a dotted line with an arrow pointing from the metabolite to the enzyme. Note that t: is the elasticity coefficient of the enzyme E, due to feedback inhibition. The inverse ratio of the elasticity coefficients of the enzymes E, and E,, with an appropriate sign is taken as the value of R’, The sign of the elasticity coefficients is decided by the following convention: if the elasticity coefficient under consideration points in the forward (i.e. downstream) direction of the pathway, then it is used without a change of sign; if on the other hand, the elasticity coefficient points in the upstream direction of the pathway, then it is used with its sign reversed. Recall also that the elasticity coefficients due to product inhibition and feedback inhibition have negative values. In the expression for R;, we have written (I= -51 where E: is the absolute value of the e;asticit;‘coefficient of enzyme E, due to feedback inhibition. The effect of the inner feedback loop in pathway (A) can be included in the current-analog circuit of Fig. 1 in a similar manner by considering the pathway with this feedback loop only. Since ES is the only enzyme which lies outside this feedback loop on the downstream side, we introduce another resistance Rt in the path of the current C5, as shown in Fig. 2. For assigning its value, we redraw the pathway on hand
(Al)
c, _
R,= 1
A 1
b----4 Fig. I. A current-analog circuit for the control structure of an unregulated linear pathway with five enzymes.
l-----vFig. 2. A current-analog
circuit for the control structure of pathway (A).
Use of electrical analogs in metabolic control analysis by deleting the feedback loop (b) and all its intervening enzymes and metabolites, as depicted in the following representation. Here the enzymes E, and E, are treated as contiguous enzymes with X4 as the adjoining metabolite, and 6: represents the elasticity coefficient of enzyme E, due to feedback inhibition.
s -
x1 -
x2 -
X4 -
P
(A2)
In this representation, consider the enzyme E5 and all the enzymes and metabolites upstream from ES; form the product of the ratios of the elasticity coefficients of each contiguous pair of these enzymes in regard to their adjoining metabolites, with an appropriate sign. The sign is fixed by the convention adopted in the preceding paragraph. The reciprocal of this product is taken as the value of Rk (see Fig. 2). We have also written 6: = --E3 with E: > 0. To designate these additional resis&ces, we have used alphabatical superscripts referring to the feedback loops which contribute to the respective resistances. The flux control coefficients of the enzymes E, through ES, which are represented by the currents C, through C5, respectively, in Fig. 2 are determined as follows. The equivalent resistance (R) of the circuit in Fig. 2 is given by
1297
We now examine a situation in which the same enzyme, say, E, , is inhibited by two different metabolites, X2 and X,. SJ!+X,4X2~X34,&.Ii,P
(B)
A current-analog circuit for this pathway is shown in Fig. 3, which is constructed from Fig. 1 with the following modifications. The effect of the outer feedback loop (a) is incorporated in the same manner as was done in the case of pathway (A), resulting in the resistance R;. To take into account the effect of the inner feedback loop (c), notice that there are three enzymes E,, E4 and E5, located downstream of this feedback loop. Accordingly, we add three resistances Rs, Ri and R;, respectively, in the paths of the currents C, C4 and C5. To select the magnitudes of these resistances, we redraw the pathway by deleting the feedback loop (c) and all its intervening enzymes and metabolites as shown in the following representation. (Of course, the outer feedback loop does not enter into the present consideration.)
,
s -
x2 +
x3
-
P
x4 -
1 -=1+L+-L+L+_L+L+L (1) 5 R; R;’ R R, Rz 3 4 Since the source current is unity, the voltage across the resistances has the value V = R, according to Ohm’s law. The currents (i.e. the flux control coefficients) are therefore equal to C, = Ti/T,
i = 1, 2, 3, 4
(Bl)
(2a)
and (2b) where (2c)
(24
(2f)
Cl_ (29 are the conductances
of the various resistors, and
r=r,+r,+r,+r,+r,+r;+r~ is the total conductance
of the circuit.
R,= 1 0
J
1
*v-----i Fig. 3. A current-analog circuit for the control structure of pathway (B).
1298
ASOK
K. SEN
To select the value of R;, we consider the enzyme E, and all the enzymes and metabolites which are located upstream from E5 in this representation. The reciprocal of the product of the ratios of the elasticity coefficients of each successive pair of these enzymes is taken as the value of R’,. The resistances RS (and Ri) are chosen in a similar fashion by considering the enzyme E3 (and E4) and all the enzymes and metabolites upstream from E, (and E4) in the representation (Bl). Applying Ohm’s law in the circuit of Fig. 3, the flux control coefficients of the enzymes in pathway (B) are found to be c, = r, /r,
(3a)
c2 = r2 /P,
(3b)
c, = (r, + rf)/P, c4
=
(r,
+
r:p,
c5
=
(r,
+
r;
(3c) WV
+
I-g/F
W
where
r2+ r3+ r; + r4 +r:+r,+r;+r;
F = rl +
(W
Cz_
with
(39 and the expressions for r, through rs and r; are given in equations (2c-h). Our next example is a pathway in which two enzymes are inhibited by two different metabolites, as shown below.
Figure 4 depicts a current-analog circuit for this pathway. In comparison to Fig. 1, this figure has three additional resistances. The resistance R; is, as before, the contribution of the outer feedback loop (a). The resistances Rt and Rt are the consequences of the inner feedback loop (d). To select the magnitudes of these resistances, we consider the pathway with only the inner feedback loop. Next we delete this feedback loop, together with the enzyme E, and the metabolite X2 which lie inside the loop, and represent the pathway as follows.
R,= 1
5_
(3g)
Fig. 4. A current-analog
circuit for the control structure of pathway (C).
representation, we look at the enzyme E4 (or E,) and all the enzymes upstream of E4 (or E,); Rj (or Ri) is the reciprocal of the product of the ratios of the elasticity coefficients of each consecutive pair of these enzymes with respect to their adjoining metabolites. Application of Ohm’s law now yields the following expressions for the flux control coefficients of the various enzymes.
ci = rip+, i=l, 2, 3, c, = (r, + r: j/f, c5= (r, + r; + r;)/P
(4a)
W) (&I
with
(Je) and
P = r, + r2+ r3+ r4+ r: i-r,+r;+r$
Here c: = -S: is the elasticity coefficient of the enzyme E2 due to feedback inhibition. In the above
WI
In each of the pathways considered above, the two feedback loops are overlapping, i.e. there are some enzymes or metabolites in common between the two
1299
Use of electrical analogs in metabolic control analysis
loops. The following pathway shows another configuration with two overlapping feedback loops.
with r+ = f, + l-2 + r, + r,
(4
+r:+r,+r;+r;.
1 SEI.X,~XX2E,X,4X4~P
T
(D)
@I
Observe that the feedback loop (b) in this pathway is the same as the inner feedback loop in pathway (A), and the feedback loop (d) is the same as the inner feedback loop in pathway (C). Therefore, for constructing a current-analog circuit of this pathway from Fig. 1, we simply add a resistance Rg in the path thecurrent CS, due to the Presence theloop (b), and include two resistances G and Rf in the paths of the currents C4 and C,, respectively, as a consequence of the loop (d). The resulting current-analog circuit is portrayed in Fig. 5. Using Ohm’s law in this figure, the flux control coefficients are derived as
of
of
C,=T,/T+,
i= I, 2, 3,
c, = (r, + r:)/r*, cs = (r, + r; + Tf)/T
R, C2
_
5
_
64 (5b)
l
(5~)
04
The expressions for the conductances F, through TJ and Fk are given in equations [2(c-i)], whereas the expressions for fi and rt are shown in equations
(4~4.
In the pathway shown below there are two nonoverlapping feedback loops, i.e. there are no enzymes or metabolites in common between the two loops. El S~X,“;,qX,“~~P
(E)
(b) (c) The feedback loops (b) and (c) in this pathway are the same as the inner feedback loops in pathways (A) and (B), respectively. In view of this, a current-analog circuit for this pathway can be designed from that of the unregulated pathway (viz. Fig. I) in the following manner. Presence of the feedback loop (b) contributes a resistance, Ri, in the path of the current CI, and the feedback loop (c) requires the addition of shunt resistances Rs, & and RS, respectively, in the paths of the currents C,, C, and C,, as shown in Fig. 6. Furthermore, since the two feedback loops,
= t$i$
R,- 1
C,_
R,= 1
wvvv
0 1 Fig. 5. A current-analog
circuit for the control structure of pathway (D).
Fig. 6. A current-analog
circuit for the control structure of pathway (E).
ASOK
1300
K. SEN
(b) and (c), are non-overlapping and there is an enzyme E, located downstream from the right feedback loop (c), we add another resistance, Ry, in the path of the current C5. The magnitude of this resistance is chosen as follows. We redraw the pathway (E) by deleting both feedback loops (b) and (c) and all the enzymes and metabolites which are located inside each of these feedback loops. <: 6; 4 4 El+-, r -*EJ+- 1 r -+E, 1I II S-X,-X,-P
(El)
In Fig. 6, application of Ohm’s law leads to the following results for the flux control coefficients. c, = r, /r **,
(6a)
c* = rz/r**,
(6b)
c3=(r, + r;yr**, c4=(r, +r:)/r **, c, =(r, + r; + r; +rpyr**.
(6~) (64 (6e)
Here
r**=r,+rz+r3+r;+r,+r: +r,+r;+rc,+rp,
(6f1
with
and the expressions for the remaining conductances in equations (6a)-(f) have been given earlier. Finally we examine a linear pathway containing both overlapping and non-overlapping patterns of feedback loops. As in pathway (E), the feedback loops (b) and (c) in this pathway are non-overlapping, whereas the feedback loops in any other pair overlap each other. (d) 1I gJ!+x,?+x,?+x,:x,JLP ,,
(d
,(aJ
(41,
(F)
Using the above pathway as an example, we will outline the general procedure of designing a current-analog circuit for a linear metabolic pathway with an arbitrary pattern of feedback inhibition. In this procedure, we first construct a current-analog circuit for the unregulated pathway, i.e. by ignoring all the feedback loops in the original pathway. Next we consider the original pathway with each feedback loop separately. In any of these pathways, if there are, say, II enzymes located downstream from the feedback loop, then in the circuit of the unregulated pathway, we add a resistance in the path of the current corresponding to each of these “downstream” enzymes. The magnitudes of the additional resistances are selected by representing the pathways in the manner discussed in the preceding paragraphs; see, for example, the representations (Al), (A2), (Bl)
and (Cl). Finally we see if there is any pair of non-overlapping feedback loops in the original pathway. If there is, then we consider the pathway with one such pair of feedback loops at a time and locate the enzymes which lie downstream of the right feedback loop of the pair. Accordingly, in the current-analog circuit we include a resistance in the path of each of the currents characterizing the “downstream” enzymes. The values of these resistances are assigned following the procedure described in connection with pathway (E). If there is a group of three or more feedback loops which are non-overlapping, then the foregoing technique can be extended in a systematic way. For the sake of brevity, these ramifications will not be discussed here. To design a current-analog circuit for pathway (F), we start with the current-analog circuit of the corresponding unregulated pathway; this is shown in Fig. 1. We then consider pathway (F) with the feedback loops (a), (b), (c) and (d), taken one at a time. Observe that there is only one enzyme downstream of the feedback loop (a). The effect of this feedback loop can be incorporated in Fig. 1 by an addition of the resistance R; in the path of the current C,. The enzyme E, is also the only enzyme downstream of the feedback loop (b); therefore a resistance, Rt, should be connected in parallel with R,. Next note that there are three enzymes, E,, E, and E,, which are located downstream of the feedback loop (c). Accordingly, the resistances R;, R; and RS should be added in the paths of the currents Cj, C, and C5, respectively. Furthermore, in view of the fact that the enzymes E, and E, lie downstream of the feedback loop (d), we should include the resistances Rj and Rt , respectively, in the paths of C, and C5. Finally, since the feedback loops (b) and (c) are non-overlapping, and the enzyme ES is located downstream of the right feedback loop (b), an additional resistance, Rp, must be connected in the path of C,. The complete analog circuit for this pathway is depicted in Fig. 7. Applying Ohm’s law, we find from Fig. 7 the following expressions for the flux control coefficients.
c, =r, p, c, =r2/P, c, =(r, + rgp, c, =(r, + r: +r,d)/P, c, =(r, + r; + r; +r; + rp +r;)/P,
(74 (7b) (7c)
6’4 (W
where
ra”=r,+r2+r3+r;+r,+r;+r: +r,+r;+r:+r:+rp+r$
m
DISCUSSION
In the previous section we have used electrical analog circuits for calculating the flux control coefficients of linear metabolic pathways with multiple feedback loops. The relative proportions of flux control of the enzymes in a given pathway can also be found directly from the corresponding analog circuit. To illustrate this, suppose that we want to
1301
Use of electrical analogs in metabolic control analysis
Similarly by combining the resistances Rr , Rt , Rs and RF, we obtain the equivalent resistance (Ry) in the
path of the current C,. We have
i=L+‘+‘+’ R;9
R,
R;
R;
RF’
Since the voltage is the same across all the resistors in the circuit, it follows that the ratio of the currents C, and C, is equal to the reciprocal of the equivalent resistances R;q and Ry. In other words, the ratio of the flux control coefficients of the enzymes E, and ES has the value C, RF9 r,+l-; -=-= r,+r!+r;+rF’ C, R?
(10)
which amounts to
The above ratio can also be obtained by dividing the expressions for C, and C5 as given in equations (6~) and (6e), respectively. The structure of the various pathways examined in this paper can be simplified by grouping certain enzymes together. The effective elasticity coefficients of the group of enzymes can be determined from the corresponding analog circuit by combining the appropriate resistances. The procedure for doing this was described in the previous paper (Sen, 1990a) and will not be repeated here. Acknowledgement-I would like to thank Professor J. E. Bailey for stimulating discussions. REFERENCES
.
1
-v-------cI Fig. 7. A current-analog
.
circuit for the control structure of pathway (F).
determine the ratio of the flux control coefficients of the enzymes E3 and E, in pathway (E). This ratio can be deduced from the circuit of Fig. 6 as follows. The equivalent resistance (R;9) in the path of the current C, in this circuit is given bv
(8)
Fell D. A. and Sauro H. M. (1985) Metabolic control and its analysis. Additional relationships between elasticities and control coefficients. Eur. J. Biochem. 148, 555-561. Heinrich R. and Rapoport T. (1974) A linear steadystate treatment of enzymatic chains. General properties. Control and effector strength. Eur. J. Biochem. 42, 89-102. Hofmeyr J. H. S. (1990) Control-pattern analysis of metabolic pathways. Flux and concentration control in linear pathways. Eur. J. Biochem. 186, 343-354. Kacser H. and Burns J. A. (1973) The control of flux. Symp. Sot. exp. Biol. 27, 65-104. Sen A. K. (1990a) Application of electrical analogs to control analysis of simple metabolic pathways. Biochem. J. (accepted for publication). Sen A. K. (199Ob) Metabolic control analysis: an application of signal flow graphs. Biochem. J. 269, 141-147. Sen A. K. (199Oc)Topological analysis of metabolic control. Math. Biosci. (accepted for publication). Westerhoff H. V. and Kell D. B. (1987) Matrix method for determining steps most rate-limiting to metabolic fluxes in biotechnological processes. Biitech. Bioengng 30, 101-107.
0020-71 IX/90 S3.00 + 0.00 Pergamon Press plc
1990
APPLICATION OF ELECTRICAL ANALOGS TO METABOLIC CONTROL ANALYSIS: LINEAR PATHWAYS WITH MULTIPLE FEEDBACK h0K
K. SEN
Department of Chemical Engineering, California Institute of Technology, Pasadena, CA 91125, U.S.A. (Received
26 February
1990; received for publication
11 May 1990)
Abstract-l. Using electrical analogs, we have presented a systematic procedure for calculating the flux control coefficients of linear metabolic pathways with multiple feedback loops. 2. In this method, an electrical analog circuit is constructed first for the unregulated pathway. 3. This circuit is subsequently modified in a step-by-step fashion to take into account the effect of each feedback loop in the pathway. 4. An analog circuit consists of resistances which are connected in series (or parallel) with a voltage (or current) source. 5. The flux control coefficients of the enzymes are represented by voltages across (or currents through) the resistances and are determined by an application of Ohm’s law. 6. We have investigated the possible patterns in linear pathways with two feedback loops. 7. This is followed by an analysis of a linear pathway with an arbitrary pattern of feedback inhibition.
INTRODUCTION In an earlier paper @en, 1990a), we have shown that electrical analogs can be conveniently used for the calculation of flux control coefficients of metabolic pathways. In that paper we examined an unregulated linear pathway, and a linear pathway with a feedback loop. An electrical analog circuit for the control structure of a given pathway is designed in the following manner. The enzymes in the pathway are associated with resistances which are connected in series (or parallel) with a voltage (or current) source. The magnitudes of these resistances depend on the elasticity coefficients of the enzymes. The flux control coefficients of the various enzymes are characterized by voltages across (or currents through) the resistances and are determined by an application of Ohm’s law. In this paper we shall use electrical analogs to analyze the flux-control distribution in linear metabolic pathways with multiple feedback loops. First we will enumerate the possible configurations with two feedback loops. Both overlapping and non-overlapping patterns of feedback loops will be considered. This is followed by an examination of a linear pathway with an arbitrary pattern of feedback inhibition. We shall see that the electrical analogs for each of these pathways can be constructed by modifying the analog circuit of the unregulated pathway (i.e. in the absence of the feedback loops) in a step-by-step fashion. These modifications can be made from an inspection of the topology of the feedback loops in the pathway, and it is not necessary to invoke the summation or connectivity theorems explicitly. As in the earlier paper (Sen, 1990a), the present formulation lies within the premises of the Metabolic Control Theory of Kacser and Burns (1973) and Heinrich and Rapoport (1974).
For the pathways examined below, we shall present only the current-analog circuits (i.e. in which all the resistances are connected in parallel with a current source). The corresponding voltage-analog circuits for these pathways can be designed by connecting the resistances in series with a voltage source and choosing the magnitudes of the resistances as reciprocals of those in the current-analog circuits [see Sen (1990a) for details]. It should be mentioned here that control coefficients of metabolic pathways are customarily evaluated by writing the governing equations using summation and connectivity theorems in a matrix form and subsequently inverting the appropriate matrix [see, for example, Fell and Sauro (1985) and Westerhoff and Kell(1987)]. Recently, using concepts from graph theory, Sen (1990b, c) has presented an alternate approach for calculating these control coefficients. A diagrammatic method due to Hofmeyr (1990) is also available for control-pattern analysis of metabolic pathways. LINEAR PATHWAYSWITH MULTIPLE FEEDBACKLOOPS
As our first example, we consider a linear pathway in which two different enzymes, namely, E, and E, undergo feedback inhibition by the same metabolite, K,. S2+X,2+X,2+X,~X,~P
1
(a)
(4
wl
In order to represent the control structure of this pathway by a current-analog circuit, we first construct a current-analog circuit for the unregulated pathway, i.e., by ignoring the feedback loops. This is
1295
Asox K.
1296
shown in Fig. I which has been reproduced from our previous paper (Sen, 1990a). In this figure a resistance R, (i = 1, 2, 3, 4, 5) is associated with the enzyme E,, and the currents C, through C5 designate the flux control coefficients of the enzymes E, through E,, respectively. The resistances are connected in parallel with a current source of strength unity. The magnitudes of the resistances depend on the elasticity coefficients of the enzymes, as shown in the figure. The symbol c; denotes the elasticity coefficient of an enzyme E, with respect to a metabolite X,. Furthermore, c:. which is the elasticity coefficient of the enzyme E, due to inhibition by its product, is written as c:=r;, c: being the magnitude of the elasticity coefficient. The reader is referred to the previous paper (Sen, 1990a) for further details. Effects of the two feedback loops in pathway (A) can be incorporated into the circuit of Fig. I as follows. Consider each feedback loop separately. For definiteness, consider the pathway with only the outer feedback loop first. We shall refer to it as pathway (A’). Note that there is only one enzyme, E,, which is located outside this loop on its downstream side. Consequently, as shown in Part I, we need to modify the resistance in the path of the current C5 by adding a resistance, R”,, in parallel with R5 (Fig. 2). The magnitude of this resistance is selected in the following manner. We redraw the pathway (A’) by deleting the feedback loop and all the enzymes and metabolites which lie inside it, thereby treating the enzymes E, and E, as contiguous enzymes with X, as the adjoining metabolite.
SEN
In the above representation, the elasticity coefficient (6;) of an enzyme E, with respect to a metabolite X, is denoted by a dotted line with an arrow pointing from the metabolite to the enzyme. Note that t: is the elasticity coefficient of the enzyme E, due to feedback inhibition. The inverse ratio of the elasticity coefficients of the enzymes E, and E,, with an appropriate sign is taken as the value of R’, The sign of the elasticity coefficients is decided by the following convention: if the elasticity coefficient under consideration points in the forward (i.e. downstream) direction of the pathway, then it is used without a change of sign; if on the other hand, the elasticity coefficient points in the upstream direction of the pathway, then it is used with its sign reversed. Recall also that the elasticity coefficients due to product inhibition and feedback inhibition have negative values. In the expression for R;, we have written (I= -51 where E: is the absolute value of the e;asticit;‘coefficient of enzyme E, due to feedback inhibition. The effect of the inner feedback loop in pathway (A) can be included in the current-analog circuit of Fig. 1 in a similar manner by considering the pathway with this feedback loop only. Since ES is the only enzyme which lies outside this feedback loop on the downstream side, we introduce another resistance Rt in the path of the current C5, as shown in Fig. 2. For assigning its value, we redraw the pathway on hand
(Al)
c, _
R,= 1
A 1
b----4 Fig. I. A current-analog circuit for the control structure of an unregulated linear pathway with five enzymes.
l-----vFig. 2. A current-analog
circuit for the control structure of pathway (A).
Use of electrical analogs in metabolic control analysis by deleting the feedback loop (b) and all its intervening enzymes and metabolites, as depicted in the following representation. Here the enzymes E, and E, are treated as contiguous enzymes with X4 as the adjoining metabolite, and 6: represents the elasticity coefficient of enzyme E, due to feedback inhibition.
s -
x1 -
x2 -
X4 -
P
(A2)
In this representation, consider the enzyme E5 and all the enzymes and metabolites upstream from ES; form the product of the ratios of the elasticity coefficients of each contiguous pair of these enzymes in regard to their adjoining metabolites, with an appropriate sign. The sign is fixed by the convention adopted in the preceding paragraph. The reciprocal of this product is taken as the value of Rk (see Fig. 2). We have also written 6: = --E3 with E: > 0. To designate these additional resis&ces, we have used alphabatical superscripts referring to the feedback loops which contribute to the respective resistances. The flux control coefficients of the enzymes E, through ES, which are represented by the currents C, through C5, respectively, in Fig. 2 are determined as follows. The equivalent resistance (R) of the circuit in Fig. 2 is given by
1297
We now examine a situation in which the same enzyme, say, E, , is inhibited by two different metabolites, X2 and X,. SJ!+X,4X2~X34,&.Ii,P
(B)
A current-analog circuit for this pathway is shown in Fig. 3, which is constructed from Fig. 1 with the following modifications. The effect of the outer feedback loop (a) is incorporated in the same manner as was done in the case of pathway (A), resulting in the resistance R;. To take into account the effect of the inner feedback loop (c), notice that there are three enzymes E,, E4 and E5, located downstream of this feedback loop. Accordingly, we add three resistances Rs, Ri and R;, respectively, in the paths of the currents C, C4 and C5. To select the magnitudes of these resistances, we redraw the pathway by deleting the feedback loop (c) and all its intervening enzymes and metabolites as shown in the following representation. (Of course, the outer feedback loop does not enter into the present consideration.)
,
s -
x2 +
x3
-
P
x4 -
1 -=1+L+-L+L+_L+L+L (1) 5 R; R;’ R R, Rz 3 4 Since the source current is unity, the voltage across the resistances has the value V = R, according to Ohm’s law. The currents (i.e. the flux control coefficients) are therefore equal to C, = Ti/T,
i = 1, 2, 3, 4
(Bl)
(2a)
and (2b) where (2c)
(24
(2f)
Cl_ (29 are the conductances
of the various resistors, and
r=r,+r,+r,+r,+r,+r;+r~ is the total conductance
of the circuit.
R,= 1 0
J
1
*v-----i Fig. 3. A current-analog circuit for the control structure of pathway (B).
1298
ASOK
K. SEN
To select the value of R;, we consider the enzyme E, and all the enzymes and metabolites which are located upstream from E5 in this representation. The reciprocal of the product of the ratios of the elasticity coefficients of each successive pair of these enzymes is taken as the value of R’,. The resistances RS (and Ri) are chosen in a similar fashion by considering the enzyme E3 (and E4) and all the enzymes and metabolites upstream from E, (and E4) in the representation (Bl). Applying Ohm’s law in the circuit of Fig. 3, the flux control coefficients of the enzymes in pathway (B) are found to be c, = r, /r,
(3a)
c2 = r2 /P,
(3b)
c, = (r, + rf)/P, c4
=
(r,
+
r:p,
c5
=
(r,
+
r;
(3c) WV
+
I-g/F
W
where
r2+ r3+ r; + r4 +r:+r,+r;+r;
F = rl +
(W
Cz_
with
(39 and the expressions for r, through rs and r; are given in equations (2c-h). Our next example is a pathway in which two enzymes are inhibited by two different metabolites, as shown below.
Figure 4 depicts a current-analog circuit for this pathway. In comparison to Fig. 1, this figure has three additional resistances. The resistance R; is, as before, the contribution of the outer feedback loop (a). The resistances Rt and Rt are the consequences of the inner feedback loop (d). To select the magnitudes of these resistances, we consider the pathway with only the inner feedback loop. Next we delete this feedback loop, together with the enzyme E, and the metabolite X2 which lie inside the loop, and represent the pathway as follows.
R,= 1
5_
(3g)
Fig. 4. A current-analog
circuit for the control structure of pathway (C).
representation, we look at the enzyme E4 (or E,) and all the enzymes upstream of E4 (or E,); Rj (or Ri) is the reciprocal of the product of the ratios of the elasticity coefficients of each consecutive pair of these enzymes with respect to their adjoining metabolites. Application of Ohm’s law now yields the following expressions for the flux control coefficients of the various enzymes.
ci = rip+, i=l, 2, 3, c, = (r, + r: j/f, c5= (r, + r; + r;)/P
(4a)
W) (&I
with
(Je) and
P = r, + r2+ r3+ r4+ r: i-r,+r;+r$
Here c: = -S: is the elasticity coefficient of the enzyme E2 due to feedback inhibition. In the above
WI
In each of the pathways considered above, the two feedback loops are overlapping, i.e. there are some enzymes or metabolites in common between the two
1299
Use of electrical analogs in metabolic control analysis
loops. The following pathway shows another configuration with two overlapping feedback loops.
with r+ = f, + l-2 + r, + r,
(4
+r:+r,+r;+r;.
1 SEI.X,~XX2E,X,4X4~P
T
(D)
@I
Observe that the feedback loop (b) in this pathway is the same as the inner feedback loop in pathway (A), and the feedback loop (d) is the same as the inner feedback loop in pathway (C). Therefore, for constructing a current-analog circuit of this pathway from Fig. 1, we simply add a resistance Rg in the path thecurrent CS, due to the Presence theloop (b), and include two resistances G and Rf in the paths of the currents C4 and C,, respectively, as a consequence of the loop (d). The resulting current-analog circuit is portrayed in Fig. 5. Using Ohm’s law in this figure, the flux control coefficients are derived as
of
of
C,=T,/T+,
i= I, 2, 3,
c, = (r, + r:)/r*, cs = (r, + r; + Tf)/T
R, C2
_
5
_
64 (5b)
l
(5~)
04
The expressions for the conductances F, through TJ and Fk are given in equations [2(c-i)], whereas the expressions for fi and rt are shown in equations
(4~4.
In the pathway shown below there are two nonoverlapping feedback loops, i.e. there are no enzymes or metabolites in common between the two loops. El S~X,“;,qX,“~~P
(E)
(b) (c) The feedback loops (b) and (c) in this pathway are the same as the inner feedback loops in pathways (A) and (B), respectively. In view of this, a current-analog circuit for this pathway can be designed from that of the unregulated pathway (viz. Fig. I) in the following manner. Presence of the feedback loop (b) contributes a resistance, Ri, in the path of the current CI, and the feedback loop (c) requires the addition of shunt resistances Rs, & and RS, respectively, in the paths of the currents C,, C, and C,, as shown in Fig. 6. Furthermore, since the two feedback loops,
= t$i$
R,- 1
C,_
R,= 1
wvvv
0 1 Fig. 5. A current-analog
circuit for the control structure of pathway (D).
Fig. 6. A current-analog
circuit for the control structure of pathway (E).
ASOK
1300
K. SEN
(b) and (c), are non-overlapping and there is an enzyme E, located downstream from the right feedback loop (c), we add another resistance, Ry, in the path of the current C5. The magnitude of this resistance is chosen as follows. We redraw the pathway (E) by deleting both feedback loops (b) and (c) and all the enzymes and metabolites which are located inside each of these feedback loops. <: 6; 4 4 El+-, r -*EJ+- 1 r -+E, 1I II S-X,-X,-P
(El)
In Fig. 6, application of Ohm’s law leads to the following results for the flux control coefficients. c, = r, /r **,
(6a)
c* = rz/r**,
(6b)
c3=(r, + r;yr**, c4=(r, +r:)/r **, c, =(r, + r; + r; +rpyr**.
(6~) (64 (6e)
Here
r**=r,+rz+r3+r;+r,+r: +r,+r;+rc,+rp,
(6f1
with
and the expressions for the remaining conductances in equations (6a)-(f) have been given earlier. Finally we examine a linear pathway containing both overlapping and non-overlapping patterns of feedback loops. As in pathway (E), the feedback loops (b) and (c) in this pathway are non-overlapping, whereas the feedback loops in any other pair overlap each other. (d) 1I gJ!+x,?+x,?+x,:x,JLP ,,
(d
,(aJ
(41,
(F)
Using the above pathway as an example, we will outline the general procedure of designing a current-analog circuit for a linear metabolic pathway with an arbitrary pattern of feedback inhibition. In this procedure, we first construct a current-analog circuit for the unregulated pathway, i.e. by ignoring all the feedback loops in the original pathway. Next we consider the original pathway with each feedback loop separately. In any of these pathways, if there are, say, II enzymes located downstream from the feedback loop, then in the circuit of the unregulated pathway, we add a resistance in the path of the current corresponding to each of these “downstream” enzymes. The magnitudes of the additional resistances are selected by representing the pathways in the manner discussed in the preceding paragraphs; see, for example, the representations (Al), (A2), (Bl)
and (Cl). Finally we see if there is any pair of non-overlapping feedback loops in the original pathway. If there is, then we consider the pathway with one such pair of feedback loops at a time and locate the enzymes which lie downstream of the right feedback loop of the pair. Accordingly, in the current-analog circuit we include a resistance in the path of each of the currents characterizing the “downstream” enzymes. The values of these resistances are assigned following the procedure described in connection with pathway (E). If there is a group of three or more feedback loops which are non-overlapping, then the foregoing technique can be extended in a systematic way. For the sake of brevity, these ramifications will not be discussed here. To design a current-analog circuit for pathway (F), we start with the current-analog circuit of the corresponding unregulated pathway; this is shown in Fig. 1. We then consider pathway (F) with the feedback loops (a), (b), (c) and (d), taken one at a time. Observe that there is only one enzyme downstream of the feedback loop (a). The effect of this feedback loop can be incorporated in Fig. 1 by an addition of the resistance R; in the path of the current C,. The enzyme E, is also the only enzyme downstream of the feedback loop (b); therefore a resistance, Rt, should be connected in parallel with R,. Next note that there are three enzymes, E,, E, and E,, which are located downstream of the feedback loop (c). Accordingly, the resistances R;, R; and RS should be added in the paths of the currents Cj, C, and C5, respectively. Furthermore, in view of the fact that the enzymes E, and E, lie downstream of the feedback loop (d), we should include the resistances Rj and Rt , respectively, in the paths of C, and C5. Finally, since the feedback loops (b) and (c) are non-overlapping, and the enzyme ES is located downstream of the right feedback loop (b), an additional resistance, Rp, must be connected in the path of C,. The complete analog circuit for this pathway is depicted in Fig. 7. Applying Ohm’s law, we find from Fig. 7 the following expressions for the flux control coefficients.
c, =r, p, c, =r2/P, c, =(r, + rgp, c, =(r, + r: +r,d)/P, c, =(r, + r; + r; +r; + rp +r;)/P,
(74 (7b) (7c)
6’4 (W
where
ra”=r,+r2+r3+r;+r,+r;+r: +r,+r;+r:+r:+rp+r$
m
DISCUSSION
In the previous section we have used electrical analog circuits for calculating the flux control coefficients of linear metabolic pathways with multiple feedback loops. The relative proportions of flux control of the enzymes in a given pathway can also be found directly from the corresponding analog circuit. To illustrate this, suppose that we want to
1301
Use of electrical analogs in metabolic control analysis
Similarly by combining the resistances Rr , Rt , Rs and RF, we obtain the equivalent resistance (Ry) in the
path of the current C,. We have
i=L+‘+‘+’ R;9
R,
R;
R;
RF’
Since the voltage is the same across all the resistors in the circuit, it follows that the ratio of the currents C, and C, is equal to the reciprocal of the equivalent resistances R;q and Ry. In other words, the ratio of the flux control coefficients of the enzymes E, and ES has the value C, RF9 r,+l-; -=-= r,+r!+r;+rF’ C, R?
(10)
which amounts to
The above ratio can also be obtained by dividing the expressions for C, and C5 as given in equations (6~) and (6e), respectively. The structure of the various pathways examined in this paper can be simplified by grouping certain enzymes together. The effective elasticity coefficients of the group of enzymes can be determined from the corresponding analog circuit by combining the appropriate resistances. The procedure for doing this was described in the previous paper (Sen, 1990a) and will not be repeated here. Acknowledgement-I would like to thank Professor J. E. Bailey for stimulating discussions. REFERENCES
.
1
-v-------cI Fig. 7. A current-analog
.
circuit for the control structure of pathway (F).
determine the ratio of the flux control coefficients of the enzymes E3 and E, in pathway (E). This ratio can be deduced from the circuit of Fig. 6 as follows. The equivalent resistance (R;9) in the path of the current C, in this circuit is given bv
(8)
Fell D. A. and Sauro H. M. (1985) Metabolic control and its analysis. Additional relationships between elasticities and control coefficients. Eur. J. Biochem. 148, 555-561. Heinrich R. and Rapoport T. (1974) A linear steadystate treatment of enzymatic chains. General properties. Control and effector strength. Eur. J. Biochem. 42, 89-102. Hofmeyr J. H. S. (1990) Control-pattern analysis of metabolic pathways. Flux and concentration control in linear pathways. Eur. J. Biochem. 186, 343-354. Kacser H. and Burns J. A. (1973) The control of flux. Symp. Sot. exp. Biol. 27, 65-104. Sen A. K. (1990a) Application of electrical analogs to control analysis of simple metabolic pathways. Biochem. J. (accepted for publication). Sen A. K. (199Ob) Metabolic control analysis: an application of signal flow graphs. Biochem. J. 269, 141-147. Sen A. K. (199Oc)Topological analysis of metabolic control. Math. Biosci. (accepted for publication). Westerhoff H. V. and Kell D. B. (1987) Matrix method for determining steps most rate-limiting to metabolic fluxes in biotechnological processes. Biitech. Bioengng 30, 101-107.