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Engineering of Biological Processes Lecture 6: Modeling metabolism Mark Riley, Associate Professor Department of Ag and Biosystems Engineering The University.

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Presentation on theme: "Engineering of Biological Processes Lecture 6: Modeling metabolism Mark Riley, Associate Professor Department of Ag and Biosystems Engineering The University."— Presentation transcript:

1 Engineering of Biological Processes Lecture 6: Modeling metabolism Mark Riley, Associate Professor Department of Ag and Biosystems Engineering The University of Arizona, Tucson, AZ 2007

2 Objectives: Lecture 6 Model metabolic reactions to shift carbon and resources down certain paths Evaluate branch rigidity

3 These two curves have the same v max, but their K m values differ by a factor of 2. Low K m High K m r 1 = vmax 1 S K m1 + S Low K m will be the path with the higher flux (all other factors being equal). Low K m also means a strong interaction between substrate and enzyme. Michaelis Menten kinetics

4 Example: Enhancement of ethanol production Want to decrease the cost Cheaper substrates –Greater number of substrates Not just glucose Higher rates of production  Y p/s Yield of product per substrate consumed  Y p/x Yield of product per cell

5 Species used Saccharomyces cerevisiae –Produces a moderate amount of ethanol –Narrow substrate specificity (glucose) Zymomonas mobilis –Produces a large amount of ethanol –Narrow substrate specificity (glucose) Escherichia coli –Broad substrate specificity –Low ethanol production –Much is known about its genetics

6 Goal Combine the advantages of ZM + EC

7 Ethanol production 1 st attempt: amplify PDC activity Resulted in accumulation of acetaldehyde. No significant increase in EtOH. Increase in byproducts from acetaldehyde 2nd attempt: amplify PDC activity & ADH (alcohol dehydrogenase) Gave a significant increase in EtOH

8 Ethanol K m = 0.4 mM Acetate K m = 2.0 mM Lactate K m = 7.2 mM K m = 0.4 mM This approach worked because of the large differences in K m ’s

9 Some definitions F tot = vmax 1 S K m1 + S + vmax 2 S K m2 + S Total flux Selectivity F1F1 F2F2 vmax 2 S K m2 + S vmax 1 S K m1 + S =

10 Selectivity So, to enhance r 1, we want a small value of K m1

11 Model conversion of pyruvate

12

13 Model production of ethanol

14 Ethanol K m = 0.4 mM

15 Ethanol K m = 1 mM

16 Ethanol K m = 10 mM

17 Glucose Glucose 6-Phosphate Fructose 6-Phosphate Fructose 1,6-Bisphosphate Glyceraldehyde 3-Phosphate Pyruvate Acetate Acetyl CoA Citrate  -Ketoglutarate Succinate Fumarate Oxaloacetate Phosphogluconate Glyceraldehyde 3-Phosphate Acetaldehyde 2-Keto-3-deoxy-6- phosphogluconate Glyceraldehyde 3-Phosphate + Pyruvate Lactate Ethanol Malate Isocitrate CO 2 +NADH FADH 2 CO 2 +NADH NADH GTP GDP+P i Phosphoenolpyruvate

18 Glucose Glucose 6-Phosphate Fructose 6-Phosphate Fructose 1,6-Bisphosphate Glyceraldehyde 3-Phosphate Pyruvate Phosphogluconate Phosphoenolpyruvate

19 GlucoseGlucose 6-Phosphate Fructose 6-Phosphate Fructose 1,6-Bisphosphate Pyruvate Simplified metabolism - upstream end of glycolysis ATP ADP v1 ATP ADP v2 v3 ATP ADP v4 v5 ATPADP v6 ADPATP v7 2 ADPATP + AMP v8 Additional reactions

20 How do you model this? What information is needed? –equations for each v –initial concentrations of each metabolite

21 Mass balances

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24 Metabolite profiles

25 Rates of reaction

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27 Reaction branch nodes P1P1 P2P2 I S J1 J2J3 J1 = J2 + J3 Flux of carbon Product yields are often a function of the split ratio in branch points (i.e., 20% / 80% left / right).

28 Types of reaction branch nodes (rigidity) Flexible nodes –Flux partitioning can be easily changed Weakly rigid nodes –Flux partitioning is dominated by one branch of the pathway Deregulation of supporting pathway has little effect on flux Deregulation of dominant pathway has large effect on flux Strongly rigid nodes –Flux partitioning is tightly controlled Highly sensitive to regulation

29 Types of reaction branch nodes P1P1 P2P2 I S - - Regulation Negative feedback

30 Flexible nodes The split ratio will depend on the cellular demands for the 2 products Can have substantial changes in the flux partitioning

31 Rigid nodes Partitioning is strongly regulated by end product activation and inhibition Deregulation of such a node can be very difficult to perform

32 P1P1 P2P2 I S - - Flexible node P1P1 P2P2 I S Strongly rigid node P1P1 P2P2 I S Weakly rigid node Regulation Negative feedback Regulation Positive feedback

33 Branch point effect Citrate Glyoxylate  -Ketoglutarate Isocitrate Lyase (IL) K m =604  M V max =389 mM/min Isocitrate Dehydrogenase (IDH) K m =8  M V max =126 mM/min Glyoxylate shunt (cells grown on acetate) For growth on acetate, Isocitrate = 160  M

34 Flux is very sensitive to [isocitrate] first order in IL, zero order in IDH 160  M When [S] = 50 uM, r IL = 110 uM/min r IDH = 20 uM/min When [S] = 160 uM, r IL = 120 uM/min r IDH = 60 uM/min

35 Branch point effect Citrate Glyoxylate  -Ketoglutarate Isocitrate Lyase (IL) K m =604  M V max =389 mM/min Dehydrogenase (IDH) K m =8  M V max =625 mM/min Glyoxylate shunt (cells grown on glucose) For growth on glucose, Isocitrate = 1  M V max had been =126 mM/min

36 Flux is not sensitive to [isocitrate] first order (but very low) in IL, first order in IDH 1  M Note that [S] is much lower than before.

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38 Which path controls the branch ratio? Citrate Glyoxylate  -Ketoglutarate Isocitrate Lyase (IL) K m =604  M V max =389 mM/min Dehydrogenase (IDH) K m =8  M V max =625 mM/min Glyoxylate shunt (cells grown on glucose) Under growth by glucose, Isocitrate = 1  M

39 Which path controls the branch ratio? The one that adapts to the available substrate controls the branch. This depends on the values of v max, K m, and [S] for each reaction.


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