Solutions to Exercises in Chapter 8 8.1. Screen the literature or Internet for large metabolites that are not proteins, peptides, RNA, or DNA. Solution: Large metabolites can be found in several other classes of molecules. Prominent examples are lipids and carbohydrates. Phospholipids like cardiolipins are large. Also, if glycosylation occurs, phospholipids and sphingolipids may become relatively large. A molecule is often large if it is a polymer, that is, an assembly of many (small) monomers. Prominently, carbohydrate polymers may reach very high molecular weights; an example is cellulose, the most abundant organic polymer on Earth. Another large carbohydrate is glycogen. An example from a different class is lignin, a potentially very large polymer, consisting of specific alcohols, that makes woody materials hard. It is the Earth’s second most abundant organic compound. 8.2. Describe (8.8) and (8.9) in words. Solution: Equation (8.8) describes the change in the substrate–enzyme complex over time as a balance of three processes. This change is driven by material produced from the binding of substrate S to the enzyme E, with rate k1, and by the disintegration of the complex (ES), which either returns substrate (unbinding, with rate k−1) or leads to product formation (catalysis, with rate k2). The latter two processes return free enzyme.

Equation (8.9) describes the dynamics of product formation. It is exclusively driven by the disintegration of the complex (ES) (with rate k2) toward P. The mechanism does not account for any utilization of P, or for an additional influx of substrate, and all material will eventually leave S and accumulate in the product pool P. 8.3. Confirm with mathematical means that no material is lost or gained in the Michaelis–Menten process. Solution: To show that no material is lost or gained, we may show that the derivative of (i.e., change in) the total pool of materials is zero. Remember that we are allowed to add differential equations to each other. First, the amount of free substrate S, plus substrate complexed with the enzyme (ES), plus product P does not change with time:

d[S + (ES)+ P]

dt= −k1SE + k−1(ES)+ k1SE − (k−1 + k2 )(ES)+ k2(ES) = 0.

 

Furthermore, the total amount of enzyme does not change with time:

d[E + (ES)]

dt= −k1SE + k−1(ES)+ k2(ES)+ k1SE − (k−1 + k2 )(ES) = 0.

Therefore, no material is being lost or gained by the system. 8.4. Formulate a differential equation for E in the Michaelis–Menten process. Compare this equation with (8.8) and interpret the result. Solution: There are only two forms of the enzyme, namely free (E) and bound to substrate (ES). We can either study the processes that generate free enzyme and formulate the following equation:

dEdt

= −k1SE + k−1(ES)+ k2(ES).

Or, we could formalize that the total amount of enzyme ET does not change, which implies

d[E + (ES)]

dt= 0,

and therefore

dEdt

= − d(ES)dt

.

Of course, both approaches should return the same result. In both cases, the initial condition is ET –(ES)(t0). 8.5. Derive (8.10) from the ODE system (8.7)–(8.9) yourself or find the derivation in the literature or Internet. For the derivation, make use of the quasi-steady-state assumption and of the fact that the sum E + (ES) is constant. Solution: The algebraic form of the Michaelis–Menten rate law derives from (8.8) and (8.9), the quasi-steady-state assumption, and the formulation of P as the rate vp. The result is

0 = k1SE − (k−1 + k2 )(ES),

vp = k2(ES).

 

The assumption that E + (ES) is constant allows us to replace (ES) with the difference between the constant quantity ET (total enzyme) and E. Simple algebra then leads to the so-called Briggs–Haldane formulation of the rate as

vp =

k1k2ET Sk−1 + k2 + k1S

.

It is customary to divide by k1 and then to rearrange the parameters and define the Michaelis constant

K M =

k−1 + k2

k1

and the maximum velocity

Vmax = k2ET . With these new parameters, the Michaelis–Menten rate law has the familiar form

vp =

VmaxSK M + S

.

8.6. Simulate (8.7)–(8.10) and compare P with vp in (8.10) for different initial values and parameter values. Why are no initial values needed in (8.10)? Solution: A crucial point to notice is that the solutions of (8.7)–(8.9) are functions of time, whereas vP in (8.10) is a function of S. Having said that, S is a function of time, so that we can plot vP as a function of time as well. Thus, we solve the differential equations, additionally compute vP(S) and P , which equals k2(ES)(t), and plot S, vP, and P in the same graph.

As an example, use k1 = 4, k−1 = 0.5, k2 = 2, S(0) = 9, (ES)(0) = P(0) = 0, E = ET = 1. Solving the system and expressing vP as a function of time leads to the plot below. We can see that P and vP are very similar; the x-axis in both cases is time. The two plots in the bottom panel are simply zoomed in. They demonstrate the quality of the approximation.

 

Equation (8.10) does not require an initial value as it is not a differential equation, but an

explicit function of S. 8.7. Search biochemistry books, papers, or the Internet for at least three different modes of enzyme inhibition. Compare and report their features. Solution: Enzyme inhibition can occur in different forms. The inhibitor may compete with the substrate (competitive inhibition), bind to (compete with) the enzyme–substrate complex (ES) (uncompetitive inhibition), or bind to both (mixed inhibition). In the case of partial inhibition, the complex (ESI) can allow residual product formation. The rate laws have some similarity with the Michaelis–Menten rate law, but are more complicated. For instance, in the case of competitive inhibition, the KM in the denominator becomes KM(1 + I/KI), where I is the inhibitor concentration and KI is a specific inhibition constant that characterizes the process. A different class of modulations of enzyme activity is allosteric modulation, including allosteric inhibition. In this case, the inhibitor binds to a site of the enzyme that is not its active site, but changes the active site so that its affinity to the substrate is reduced. The describing rate law is often a Hill function. Four books discussing different types of inhibition are listed below.

0 5 10

0

5

10S vP Pdot

0 0.25 0.5

0

5

10S vP Pdot

4 6 8

0

1

2S vP Pdot

 

Savageau MA. Biochemical Systems Analysis: A Study of Function and Design in Molecular Biology. Addison-Wesley, 1976. Segel LA. Biological Kinetics. Cambridge University Press, 1991. Schulz AR. Enzyme Kinetics: From Diastase to Multi-enzyme Systems. Cambridge University Press, 1994. Gutfreund H. Kinetics for the Life Sciences: Receptors, Transmitters and Catalysts. Cambridge University Press, 1995.

8.8. Select one mode of enzyme inhibition and find the corresponding mathematical formulation of a rate law. Plot vP against S for several inhibitor concentrations and compare the plots with the corresponding plot from the uninhibited reaction. Solution: As an example, consider competitive inhibition. The standard rate law in this case is

vP =

VmaxSK M (1+ I / KI )+ S

.

We can immediately see that the rate law is the same as the Michaelis–Menten rate law for I = 0 and that increasing inhibitor concentrations “inflate” the KM value; KI is the inhibition constant. As an example, the figure below shows four plots of vP against S for I = 0.1, 1, 2, and 10. The other parameters are Vmax = 4, KM = 2, and KI = 1.

 

8.9. Mine databases and the literature to reconstruct a chain of reactions describing the production of homoserine from aspartate. Search for rate laws and kinetic parameter values characterizing the reaction steps. Solution: One short reaction chain is

L-aspartate ð 4-aspartyl-phosphate ð aspartate 4-semialdehyde ð homoserine

The enzymes involved are

aspartate kinase (EC:2.7.2.4) aspartate-semialdehyde dehydrogenase (EC:1.2.1.11) homoserine dehydrogenase (EC:1.1.1.3)

BRENDA presents several KM and KI values, which depend on the organism and the conditions of the assay. 8.10. Search different data sources and compare kinetic parameter values and modulators characterizing or affecting the pyruvate kinase reaction. Write a report. Solution: It is nearly impossible to investigate this question comprehensively, as there are dozens of different pyruvate kinases from many organisms, and each enzyme has different KM, KI, inhibitors, and optimal pH values and temperatures. A good source to start the investigation is BRENDA. Typically, the KM values for the main substrate PEP and the secondary substrate ADP are in the following ranges: Typical range of KM for PEP: [0.3 µM, 4.43 mM] Typical range of KM for ADP: [0.03 mM, 14.1 mM] A specific discussion of the kinetic features of pyruvate kinase can be found in an old article by Collins and Thomas (1974). In this paper, the authors show how FBP and phosphate regulate pyruvate kinase in Streptococcus lactis (now called Lactococcus lactis).

Collins LB & Thomas TD. Pyruvate kinase of Streptococcus lactis. J. Bacteriol. 120 (1974) 52–58 (www.ncbi.nlm.nih.gov/pmc/articles/PMC245729/).

 

8.11. Use KEGG or BioCyc to determine edges representing the production of glucose 6-phosphate in a stoichiometric map. Solution: In KEGG, nine reactions are shown that use glucose 6-phosphate:

alpha-D-Glucose 1-phosphate 1,6-phosphomutase ATP:alpha-D-glucose 6-phosphotransferase alpha-D-Glucose 6-phosphate phosphohydrolase Polyphosphate:D-glucose 6-phosphotransferase UDPglucose:D-glucose-6-phosphate 1-alpha-D-glucosyltransferase Protein-N(pai)-phosphohistidine:sugar N(pai)-phosphotransferase alpha-D-Glucose 6-phosphate ketol-isomerase Sucrose 6-phosphate fructohydrolase ADP:D-glucose 6-phosphotransferase

8.12. Extract from KEGG and BioCyc all pathways using glucose 6-phosphate as a substrate. Are these pathways the same in E. coli, baker’s yeast, and humans? Solution: For example, using KEGG, we obtain for E. coli the following pathway maps:

 

KEGG also shows that E. coli may use glucose 6-phosphate (G6P) in starch, galactose, and inositol metabolism and for the synthesis of butirosin and neomycin (see the map for streptomycin biosynthesis). A few other pathways directly or indirectly use G6P. Finally, G6P is involved with E. coli’s two-component sensing system and with its PTS system, which is an alternative to glycolysis. Pathways like glycolysis are well conserved, but humans, for instance, have no two-component sensing system and no PTS system. Furthermore, some genes or enzymes have not been found in one or the other organism; they are white rather than shaded green in the KEGG map. Take for instance galactose metabolism. According to the KEGG maps, humans have the enzyme glucose 6-phosphatase [EC: 3.1.3.9], but this enzyme has not been identified in E. coli. E. coli, by contrast, has the enzyme galactonate dehydratase [EC: 4.2.1.6], which allows it to return galactonate to glycolysis. Similar analyses can be done in BioCyc and for yeast. 8.13. Mine databases and the literature to reconstruct a metabolic pathway system describing the production of lysine. Search for kinetic parameter values for as many steps as possible. Note: this is an open-ended problem with many possible solutions. Solution: A first step may be to launch KEGG and specify lysine, as well as an organism, such as E. coli. The search quickly shows a pathway panel for lysine biosynthesis, which reveals a complete path from aspartate to lysine. By contrast, if we specify Saccharomyces cerevisiae, the same path is not possible. Similarly, Homo sapiens does not have a biosynthesis pathway for lysine, which should not surprise us, because lysine is an essential amino acid that mammals cannot make and have to acquire through their diet (Tomé and Bos 2007). Metabolic engineers use Corynebacterium glutamicum to produce several hundred thousand tons of lysine per year, and indeed, selecting this organism, KEGG shows a path from aspartate (see below). Aspartate can be made from oxalocetate, which is a component in the TCA cycle.

 

The first enzyme in the conversion of aspartate into lysine is aspartate kinase [EC:

2.7.2.4]. BRENDA contains a lot of information about this enzyme, including its crystal structure. For E. coli, the KM for aspartate is between 0.51 and 2.1 mM, depending on conditions, but no value is listed for C. glutamicum. Further investigation does yield specific activity levels (Schrumpf et al. 1992).

Blombach B & Seibold GM. Carbohydrate metabolism in Corynebacterium glutamicum and applications for the metabolic engineering of L-lysine production strains. Appl. Microbiol. Biotechnol. 86 (2010) 1313–1322. Tomé D & Bos C. Lysine requirement through the human life cycle. J. Nutr. 137 (2007) 1642S–1645S. Schrumpf B, Eggeling L & Sahm H. Isolation and prominent characteristics of an L-lysine hyperproducing strain of Corynebacterium glutamicum. Appl. Microbiol. Biotechnol. 37 (1992) 566–571.

 

8.14. Given the known influxes (blue) and effluxes (green) in the figure below (where units are mmol L−1 min−1), compute the internal fluxes (red) of the system. Is the solution unique? Discuss issues of uniqueness generically and provide examples and counterexamples. Is the system in a steady state? Is the flux information sufficient to determine the concentrations of the metabolites A, …, H?

Solution: We can approach this problem either by tinkering or by using linear algebra. In the former case, we know, for instance, that F5 and F4 must be 8 units for D and then C to be in balance. In the latter case, we construct linear equations that balance fluxes at all nodes. Such equations are as follows:

Balance at

ABCDEFGH

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

=

20− F1 − F2

5+ F1 − F3 − F4

F4 − F5

F5 −8F2 + F3 − F6

F6 − F7 − F8

F7 − F9 − 7F8 + F9 −10

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

.

In any steady state, these balances must be zero at all nodes. Linear algebra tells us that there is no unique solution to this problem for two reasons: (1) there are 9 variables and 8 equations, which gives us 1 degree of freedom; and (2) the set of equations A, …, E and the set F, G, H are linearly dependent, because adding these equations in each case yields F6 = 17. This equal outcome increases the degrees of freedom. The consequences of one degree of freedom can be demonstrated with F, G, and H. We know that F6 = 17. If we were to set F9 = 0, we would have F7 = 7 and F8 = 10. However, we could also select something else for F9, such as F9 = 2, which would yield F7 = 9 and F8 = 8.

 

Indeed, any non-negative value smaller or equal to 17 could be chosen for F9. Similarly, if we keep F9 variable, there are some choices for F7 or F8. The second degree of freedom is due to a redundancy among A, B, and E. For instance, we could chose F1 = 15, which would yield F2 = 5 and F3 = 12. But F1 could be chosen differently between 0 and 20. The two “redundancy loops” are independent of each other. The system is assumed to be in a steady state. However, it is possible for the system to start outside this steady state and to approach a stable steady state after some time. The dynamics (e.g., speed) of such a trajectory cannot be determined from the given information. There is not even enough information to determine the intracellular concentrations. 8.15. Design models for the pathway in Figure 8.5. Represent the production of A as 1·D−0.5. For all simulations, start at the steady state (which you have to compute first) and double the efflux from D at t = 10. In the first set of simulations, use the MMRL for the reactions that use A, B, C, or D as substrate. Choose different combinations of Vmax and KM values as you like. Repeat the simulations with power-law rate laws. Summarize your findings in a report. Solution: With arbitrarily selected parameter values, a PLAS file may look as follows: A' = D^-0.5 - Va A / (Ka + A) B' = Va A / (Ka + A) - Vb B / (Kb + B) C' = Vb B / (Kb + B) - Vc C / (Kc + C) D' = Vc C / (Kc + C) – p * Vd D / (Kd + D) A = 1 B = 1 C = 1 D = 1 Va = 8 Ka = 2 Vb = 6 Kb = 4 Vc = 5 Kc = 1 Vd = 10 Kd = 4 p = 1 t0 = 0 tf = 10 hr = .1 The first task should be the computation of the steady state, which is most easily done by simulation (see the figure below). It has the values ASS = 0.3866414, BSS = 1.101685, CSS = 0.3497294, DSS = 0.5953253.

 

We use this steady state as initial values for further simulations. To study the doubled

efflux from D, we set p = 2 at t = 10. The doubled efflux should reduce the concentration of D, and, because D feeds back on the initial reaction, A, B, and finally C should be affected as well. It is hard to predict what the new steady state is. For instance, one could argue that the (invisible) precursor of the pathway is unchanged, but the efflux is increased, so that the total mass should decrease. Indeed, without inhibition, D decreases and the other three variables are unaffected (results not shown). However, with inhibition, the situation is more complex. As before, the pool D is decreased, and this reduces the efflux. Furthermore, the inhibition of the input by D is now weaker (because there is less D) so that actually more material is flowing into the system. All in all, the total mass (A + B + C + D) at the beginning is 2.43 and at the end it is increased to 2.92. Although the system is simple, these types of predictions about the dynamics and the specific distribution of mass among the four pools are difficult, because they depend on all the parameters in the system. The simulation result is as follows:

0 5 10

0

1

2A B C D

0 10 20

0

1

2A B C D

 

For a power-law system with arbitrary parameter values, the PLAS file might be as follows:

A' = D^-0.5 - ba A^ha B' = ba A^ha - bb B^hb C' = bb B^hb - bc C^hc D' = bc C^hc - p*bd D^hd A = 1 B = 1 C = 1 D = 1 ba = 2 ha = .5 bb = 1.5 hb = .7 bc = 2 hc = .2 bd = 3 hd = .8 p = 1 t0 = 0 tf = 20 hr = .1 The dynamics toward the steady state is generally similar, but the specific pool sizes are of course different (see the figure below). They could be made exactly equivalent, if we computed the parameters per approximation of the Michaelis–Menten system at the steady-state operating point (see Chapter 4).

0 10 20

0

1

2A B C D

 

Now starting at the steady state, setting p = 2, and looking at the dynamics and total sum reveals the following plot and a sum of first 2.29 and then 3.72:

As in the previous system, D drops first, followed by increases in the other pools. 8.16. Design a power-law model for the pathway in Figure 8.6. Ignore intermediates that are suggested by dotted lines and set all rate constants equal to 1. For all substrate degradation processes, use kinetic orders of 0.5. For inhibition effects, start with kinetic orders of −0.5. Subsequently, change the inhibition effects one by one, or in combination, to other negative values. Always start your simulations at the steady state. Summarize your findings in a report. Solution: The following is a possible PLAS implementation of the system: A' = p*C^gAC D^gAD - ba A^ha B' = ba A^ha - b1 B^hb1 C^-0.5 - b2 B^hb2 D^-0.5 C' = b1 B^hb1 C^gCC - bc C^hc D' = b2 B^hb2 D^gDD - bd D^hd A = 1 B = 1 C = 1 D = 1 ba = 1 ha = .5 b1 = 1 hb1 = .5 b2 = 1 hb2 = .5 bc = 1 hc = .5

0 10 20

0

1

2A B C D

 

bd = 1 hd = .5 gAC = -0.5 gAD = -0.5 gCC = -0.5 gDD = -0.5 t0 = 0 tf = 20 hr = .1 A plot starting at initial values of 1 is shown below. It seems that C is missing, but because of symmetry its trend line is the same as that of D and covered up by it.

From now on, we start at the steady state, which is

A = 2.522 B = 0.398 C = 0.631 D = 0.631

For instance, p = 2 at t = 2 raises the input persistently and leads to a general, yet different increase in the system variables (as seen in the figure below); again the time course of C is covered by that of D.

0 10 20

0

1.5

3A B C D

0 10 20

0

3

6A B C D

 

For the situations gAC = −1 or gAD = −1, the steady state is different, so we need to compute it. Then, starting with this state and setting p = 2 at t = 2 yields the result shown in the figure below. The two situations yield the same plot, because C and D have the same value and the same effect on the system. The plots for gCC = −1 (gDD = −0.5) and for gDD = −1 (gCC = −0.5) are similar, but the trend line of C becomes visible (not shown).

Of course, many combinations are possible. The two figures below show results for systems that exhibit damped oscillations. On the left is a system with stronger inhibition of the reaction between A and B (gAC = −2, gAD = −2, gCC = −0.5, gDD = −0.5). Interestingly, if the inhibition within the system is also stronger (gAC = −2, gAD = −2, gCC = −2, gDD = −2), the oscillations are better buffered.

If the reaction from A to B is inhibited even more strongly (gAC = −8, gAD = −2, gCC =

−0.5, gDD = −0.5), the system becomes unstable and may enter a stable limit cycle; see the figures below for time courses (left) and a so-called phase-plane plot (right), where B is plotted against A. It shows how the trajectory zooms inwards, but ends at the limit cycle.

0 10 20

0

3

6A B C D

0 10 20

0

3

6A B C D

0 10 20

0

3

6A B C D

 

Finally, if we additionally increase the internal feedback (gAC = −8, gAD = −2, gCC = −2,

gDD = −2), the limit cycle disappears and the system reaches a stable steady state again (as shown in the figure below). Thus, the roles of the “outer” and “inner” feedback are very different. The outer feedback tends to destabilize and the inner to stabilize the system.

8.17. How could one model the effect of altered gene expression on a metabolic pathway? Discuss different options. Solution: Genes are usually not included in metabolic pathway models (but see Chapter 11 for such a combination). Therefore, altered gene expression is usually modeled in an indirect fashion. If nothing further is known, one could assume as a default that a change of X% in the expression of a gene that codes for an enzyme would correspond to the same (or a similar) change in the enzyme activity itself. This activity is either explicitly modeled or “hidden” in a parameter. For instance, in Michaelis–Menten and Hill models, Vmax is the product of the catalytic rate constant kcat (= k2) and the total enzyme amount (which is equated with activity). In power-law models, the enzyme activity is often included in the corresponding rate constant or it is explicitly

0 100 200

0

10

20A B C D

0 8 16

0

2

4

A

B

0 10 20

0

3

6A B C D

 

modeled with an independent variable. Thus, changes in these parameters or the independent variable may roughly correspond to changes in gene expression. If the gene codes for one among several isozymes, then the turnover rate of the overall reaction has to be adjusted in a proportional manner. Genes could also code for modulators, such as inhibitors. For instance, in the case of competitive inhibition, the inhibitor modulates the KM of a reaction. In power-law models, the inhibitor may be represented as an independent variable, whose value would be changed. If the change in gene expression alters the input to a system, then the change is rather straightforward to implement. 8.18. Compare the ascorbate pathway in M. pneumoniae with that in E. coli. Report your findings. Solution: If you have a large computer screen, or two screens, the easiest strategy is to open KEGG (or BioCyC) twice and to display the map of ascorbate metabolism for E. coli and for Mycoplasma pneumonia; note that the pathway databases have maps for different strains of these organisms; here we discuss E. coli K-12 MG1655 and M. pneumoniae M129. With both pathway displayed side-by-side, it is easy to compare (eco00053 versus mpn00053 in KEGG) which enzymes are shaded (known to exist) and which are not. We can see that generally more enzymes exist (or are known) in E. coli. In particular, E. coli possesses a degradation path beyond ascorbate 6-phosphate toward xylulose 5-phosphate, and thus the pentose phosphate pathway. It is unclear how M. pneumoniae would remove ascorbate 6-phosphate. Keep in mind that it is possible that no-one has looked for the missing enzyme, L-ascorbate 6-phosphate lactonase [EC: 3.1.1.-], or for another degradation path. 8.19. Find five pathways in E. coli that do not exist in M. pneumoniae. Solution: We already discussed the pathway from ascorbate to the pentose phosphate pathway in Exercise 8.19. Ideas for other differing pathways are maybe most easily found from the colored overall metabolic pathway maps eco01100 and mpn01100 in KEGG or similar maps in BioCyc. A comparison shows immediately where certain pathways are known or not known to exist. Examples are the terpenoid pathway, starch metabolism, glycerolipid metabolism, and aminosugar metabolism. Also, there are many differences in the interconversions of pentoses and glucuronate, which are displayed in specific pathway maps (eco00040 versus mpn00040 in KEGG).

Keep in mind, though, that some enzymes might exist, but that we have not confirmed them yet in one or the other organism.

 

8.20. Explore how much citric acid is being produced annually and which microorganism is primarily used for its large-scale industrial production. Search for models that have addressed citric acid production and review their key features. Solution: According to Wikipedia, the world-wide annual production in 2007 was approximately 1,600,000 tons. The main producer is the black fungus Aspergillus niger. Comprehensive early models were proposed by Torres (1994). These models described the pathway of citric acid production and excretion in A. niger. Torres’ group later published several papers suggesting the steady-state optimization of citric acid production in the same organism. Because of its importance, many other authors have modeled this fermentation process, using different techniques, such as simple kinetics or metabolic flux analysis. The following is a selection of articles:

Alvarez-Vasquez F, González-Alcón C & Torres NV. Metabolism of citric acid production by Aspergillus niger: model definition, steady-state analysis and constrained optimization of citric acid production rate. Biotechnol. Bioeng. 70 (2000) 82–108. Arzumanov TE, Sidorov IA, Shishkanova NV & Finogenova TV. Mathematical modeling of citric acid production by repeated batch culture. Enzyme Microb. Technol. 26 (2000) 826–833. Hu J & Wu P. Mathematical model for citric acid fermentation. Chin. J. Biotechnol. 9 (1993) 179–188. García J & Torres NV. Mathematical modelling and assessment of the pH homeostasis mechanisms in Aspergillus niger while in citric acid producing conditions. J. Theor. Biol. 282 (2011) 23–35. Sharifzadeh Baei MS, Mahmoudi M & Yunesi H. A kinetic model for citric acid production from apple pomac by Aspergillus niger. Afr. J. Biotechnol. 7 (2008) 3487–3489. Torres NV. Modeling approach to control of carbohydrate metabolism during citric acid accumulation by Aspergillus niger. I. Model definition and stability of the stable state. Biotechnol. Bioeng. 44 (1994) 104–111. Torres NV. Modeling approach to control of carbohydrate metabolism during citric acid accumulation by Aspergillus niger. II. Sensitivity analysis. Biotechnol. Bioeng. 44 (1994) 112–118. Torres NV, Voit EO & González-Alcón C. Optimization of nonlinear biotechnological processes with linear programming. Application to citric acid production by Aspergillus niger. Biotechnol. Bioeng. 49 (1996) 247–258.

 

Torres NV & Voit EO. Pathway Analysis and Optimization in Metabolic Engineering. Cambridge University Press, 2002. Tran-Dinh S, Hoerter JA, Mateo P, et al. A simple mathematical model and practical approach for evaluating citric acid cycle fluxes in perfused rat hearts by 13C-NMR and 1H-NMR spectroscopy. Eur. J. Biochem. 245 (1997) 497–504.

8.21. Compare kinetic parameters of the fermentation pathway in yeast that were presented in the work of Galazzo and Bailey and of Curto et al. with those that are listed in BRENDA and ExPASy. Summarize your findings in a spreadsheet or a report. Solution: The papers by Galazzo and Bailey and by Curto et al. contain a large number of parameters. As an example, consider the initial reaction of the pathway, catalyzed by hexokinase, which converts glucose into glucose 6-phosphate (see also Chapter 11). Actually, this reaction is catalyzed in yeast by three enzymes (hexokinase I and II, as well as glucokinase), which in Galazzo and bailey’s paper are not distinguished. The two substrates are glucose and ATP, with KM values of 0.11 and 0.1 mM, respectively. Specifying yeast and hexokinase in BRENDA returns enzyme EC: 2.7.1.1. The database contains plenty of information. For ATP, the KM is indeed given as 0.1 mM. For glucose, we find an array of values between 0.04 mM (for glucokinase) and 0.47 mM (hexokinase). Actually, for the combination of hexokinase I and II, the KM is listed as 0.12 mM, which is very close to Galazzo and Bailey’s value. As a second example, Galazzo and Bailey present the KM for UDPG in the glycogen synthase (EC: 2.4.1.11, 2.4.1.21) with a value of 1 mM. In BRENDA, it is listed as 0.4 mM or 2.4 mM, with a possible maximum of 7.2 mM. Here we are not so sure any more, due to the wide range. Finally, for phosphofructokinase (PFK; EC: 2.7.1.11), Galazzo and Bailey use a KM of 1 mM for the first substrate, fructose 6-phosphate (F6P), and of 0.06 mM for the second substrate, ATP. BRENDA does not offer values for yeast but only two references. For other organisms, the range of KM values is huge (F6P: between 0.011 and 7 mM (with a particular E. coli mutant PFK listed as 254 mM); ATP: between 0.005 and 0.7 mM). Thus, in some cases, the values are quite similar, but this is not always the case, and we sometimes find no information at all or quite a wide range.

 

8.22. Consider the model in the figure below, which we have already studied in Chapter 3 (Exercise 3.28). It describes the pentose pathway, which exchanges carbohydrates with different numbers of carbon atoms, as shown in the table below. For instance, two units of X1 are needed to produce one unit of X2 and X3, and each reaction V5,3 uses one unit of X5 to generate two units of X3. Note that X3 appears in two locations, but represents the same pool. Assume that all processes follow mass action kinetics and that the influx (V7,1) and the effluxes (V3,0 and CO2) are 20 mmol L−1 min−1 and that all other fluxes have a value of 10 mmol L−1 min−1. Confirm that these settings correspond to a consistent steady-state flux distribution. Assume the following steady-state concentrations: X1 = 10, X2 = 6, X3 = 12, X4 = 8, X5 = 8, X6 = 4, X7 = 20. Compute the rate constants of all processes.

Number of carbon atoms in each metabolite in the above pathway

Compound pool No. of carbon atoms X1 5 X2 7 X3 3 X4 4 X5 6 X6 6 X7 5

F6P (X5)

GAP (X3)

Xyl (X7)

Ri5P; Xy5P; Ru5P (X1)

S7P (X2) GAP (X3)

E4P (X4)

G6P (X6)

CO2

V5,6V5,6

V4,5V4,5

V1,2V1,2

V6,1V6,1

V7,1V7,1

V2,5V2,5

V5,3V5,3

V3,0

V3,0

 

Solution: We formulate equations by including in each term the variable(s) involved with a reaction, a rate constant and, where needed, a stoichiometric constant. The latter is needed in the reaction between X5 and X3, because one is a hexose and the other a triose. The equations are as follows:

X1 = k71X7 + k61X6 − k12 X1 − k45 X1X4 ,X2 = X1 − k25 X2 X3,X3 = X1 + k45 X1X4 − k25 X2 X3 − k30 X3,X4 = k25 X2 X3 − k45 X1X4 ,X5 = k25 X2 X3 + k45 X1X4 − 2k53 X5 − k56 X5,X6 = k56 X5 − k61X6.

All pools are in a steady state; X7 is an independent variable and therefore constant. Substituting the steady-state values for all X’s, we obtain a system of linear equations for the k’s. The following is a solution:

k7,1 = 0.5, k1,2 = 1, k2,5 = 1/7.2 ≈ 0.1389, k3,0 = 1/1.2 ≈ 0.8333, k4,5 = 0.125, k5,3 = 0.625, k5,6 = 1.25, k6,1 = 2.5.

A PLAS file with the system is shown below. It also contains the right-hand sides, which

are useful for checking whether the fluxes are in balance. X1' = k71 X7 + k61 X6 - k12 X1 - k45 X1 X4 X2' = k12 X1 - k25 X2 X3 X3' = k12 X1 + k45 X1 X4 - k25 X2 X3 - k30 X3 X4' = k25 X2 X3 - k45 X1 X4 X5' = k25 X2 X3 + k45 X1 X4 - 2 k53 X5 - k56 X5 X6' = k56 X5 - k61 X6 X1 = 10 X2 = 6 X3 = 12 X4 = 8 X5 = 8 X6 = 4 X7 = 20 k12 = 1 k25 = 1/7.2 k30 = 1/1.2 k45 = 0.125

 

k53 = 0.625 k56 = 1.25 k61 = 2.5 k71 = .5 v1 = k71 X7 + k61 X6 - k12 X1 - k45 X1 X4 v2 = k12 X1 - k25 X2 X3 v3 = k12 X1 + k45 X1 X4 - k25 X2 X3 - k30 X3 v4 = k25 X2 X3 - k45 X1 X4 v5 = k25 X2 X3 + k45 X1 X4 - 2 k53 X5 - k56 X5 v6 = k56 X5 - k61 X6 t0 = 0 tf = 10 hr = 0.1

The system is indeed in a steady state, but this steady-state solution is not unique with

respect to the k’s. For instance, we could multiply all k’s with 10 and also receive a valid solution, because at steady state, all equations equal zero. Thus, any positive multiplier is permissible. Multiplication with 10 would speed up the dynamics 10-fold, but the steady state would be the same. Furthermore, the system of linear equations in the k’s is underdetermined: it contains six equations in eight unknowns. Therefore, there are other solutions that are not scaled from the solution above. 8.23. Using the model in Exercise 8.22 with the same flux distribution, is it possible that the system can have different steady-state concentrations? If not, explain. If the answer is yes, construct one or more examples. Write a brief report on your findings. Solution: Here, we are looking for steady states in terms of X’s. The answer depends on the independent input (X7). If we assume that X7 is fixed, then the system with a given flux distribution has as many variables as equations (although one has to be careful with such a statement, because the system is not linear). It has a unique and stable nontrivial steady state, where all eigenvalues have negative real parts. Simulations confirm these statements. However, if we are allowed to change X7, other steady states are possible, even with the same rate constants. To construct an example, note that X2 and X4 always appear in a product. Leave these two variables unchanged and multiply all other variables (including X7) by 10 (or any other positive value). The system is again in a steady state, because all equations are multiplied with this value.

 

8.24. Using the results from Exercise 8.22, perform a simulation that starts at the steady state and models a bolus perturbation to X1. Make a prediction of the model responses before you simulate and summarize your findings in a brief report. Solution: Many simulations are possible. If we double X1 at time t = 2, all variables respond with an immediate increase, except for X4, which decreases, because its reaction with X1 toward X3 and X5 uses up material. The system then returns to its old steady state, as shown in the following figure:

If we permanently change the input X7, the system assumes a new steady state. If we

double all variables, including X7, but leave X2 and X4 unchanged, the solution immediately settles in a new steady state (see Exercises 8.23 and 8.25). However, if we implement the same alteration again, but without changing the input X7, the system returns to the original steady state (not shown). 8.25. Using the results from Exercises 8.22–8.24, perform two simulations that start at the steady state and model either a persistent or a bolus perturbation to X7. Make a prediction of the model responses before you simulate and write a brief report on your findings. Solution: For a persistent increase in X7, the system assumes a new steady state (as shown in the figure below for doubling X7 at time t = 2). This is expected, as the system is stable and the input, but not the efflux, of the system is altered.

0 5 10

0

10

20X1 X2 X3 X4 X5 X6

 

For a temporary bolus (X7 = 40 between t = 2 and t = 5), the system tries to reach the

steady state shown in the figure above, but then returns to the original steady state (see the figure below). This should have been expected as well, as the system is returned to its original input state at time t = 5.

0 10 20

0

15

30X1 X2 X3 X4 X5 X6

0 10 20

0

15

30X1 X2 X3 X4 X5 X6