1 / 33

Welcome to Introduction to Bioinformatics Wednesday, 16 October Metabolic modeling

Welcome to Introduction to Bioinformatics Wednesday, 16 October Metabolic modeling. Table of Contents First exam: Rules of the game Sep 27, SQ4: Write subroutine to print score array PS1M.8: Probability of evolution. Trypanosoma brucei Causative agent of sleeping sickness. Life Cycle.

onofre
Download Presentation

Welcome to Introduction to Bioinformatics Wednesday, 16 October Metabolic modeling

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Welcome toIntroduction to BioinformaticsWednesday, 16 OctoberMetabolic modeling • Table of Contents • First exam: Rules of the game • Sep 27, SQ4: Write subroutine to print score array • PS1M.8: Probability of evolution

  2. Trypanosoma bruceiCausative agent of sleeping sickness Life Cycle Central Nervous SystemDeath

  3. Entry of glucose Phosphorylation Breakdown to triose phosphates Asi AMP-P-As Conversion to pyruvate Release of pyruvate AMP-P +Asi Trypanosoma bruceiDependence on glycolysis Treatment Arsenate(AsO4 = Asi) Competitivewith Pi

  4. Trypanosoma bruceiDependence on glycolysis Treatment Something more specific? inhibitor1 Choice #1Starve the cell

  5. Trypanosoma bruceiDependence on glycolysis Treatment Something more specific? Choice #1Starve the cell Choice #2Stuff the cell pyruvate inhibitor2

  6. [I] 0 1 5 10 25 [ATP] [glucose]added Trypanosoma bruceiTest ideas for inhibitors inhibitor1 We knowcharacteristicsof enzymes

  7. One day (1/20) 32P 32S 1 atom of 32S20 atoms of 32P per day [32P] k = d[32S] / dt rateconstant concentration rate of change Characteristics of enzymesAnalogy of radioactivity How often does 32S appear? 1/20 of an atom per day? [X atoms of 32P] = __atoms of 32S per day

  8. One day (1/20) 32P 32S [32P] k = d[32S] / dt rateconstant concentration rate of change Characteristics of enzymesAnalogy of radioactivity Which quantity is an intrinsic characteristic?

  9. One day One day (1/3000000) (1/20) [32P] [14C] k k = d[14N] / dt = d[32S] / dt rateconstant rateconstant concentration concentration rate of change rate of change Characteristics of enzymesAnalogy of radioactivity

  10. [32P] k = d[32S] / dt rateconstant concentration rate of change Characteristics of enzymesChemical reactions AMP-P-As AMP-P +Asi

  11. = d[Asi] / dt rate of change = d[AMP-P-Asi] / dt = – rate of change k rateconstant Characteristics of enzymesChemical reactions AMP-P-As AMP-P +Asi [AMP-P-As] = d[AMP-P ] / dt concentration rate of change =

  12. = d[AMP-P-Asi] / dt = – rate of change = d[S] / dt [S] = – A differential equation k k rateconstant Characteristics of enzymesChemical reactions AMP-P-As AMP-P +Asi [AMP-P-As] concentration

  13. = d[AMP-P-Asi] / dt = – rate of change [S] = S0 e -k(t-to) Its solution? = d[S] / dt Check: k rateconstant Characteristics of enzymesChemical reactions AMP-P-As AMP-P +Asi [AMP-P-As] concentration = d[S] / dt = –k[S] A differential equation = S0 (–k) e -k(t-to) = -k S0e -k(t-to) = -k [S]

  14. [S] = S0 e -k(t-to) = d[S] / dt = –k[S] Characteristics of enzymesChemical reactions

  15. [S] = S0 e -k(t-to) Δt Characteristics of enzymesChemical reactions [S] = S0

  16. [S] = S0 + Δt d[S] / dt = d[S] / dt = –k[S] Characteristics of enzymesChemical reactions S0 Δt

  17. Characteristics of enzymesChemical reactions (Program)

  18. [S] = S0 e -k(t-to) = d[S] / dt = –k[S] Characteristics of enzymesChemical reactions

  19. Slope0 = -k[S0] [S1] = S0 + Δt d[S] / dt Slope1 = -k[S1] = d[S] / dt = –k[S] Characteristics of enzymesChemical reactions [S] = S0 + Δt d[S] / dt S0 Use average of Slope0 and Slope1 Δt

  20. = d[S] / dt = –k[S] Characteristics of enzymesChemical reactions [S] = S0 + Δt d[S] / dt Slope0 = -k[S0] [S1] = S0 + Δt d[S] / dt Slope1 = -k[S1] S0 Use average of Slope0 and Slope1 Runge-Kutta method Δt

  21. Characteristics of enzymesEnzymatic reactions inhibitor1

  22. = d[G6P] / dt = – rate of change k rateconstant Characteristics of enzymesEnzymatic reactions Glucose-6-phosphate Fructose-6-phosphate [G6P] concentration k  0

  23. E-complex Characteristics of enzymesEnzymatic reactions G6P +E G6P·E F6P·E F6P +E

  24. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P +E1 E1-complex F6P +E1 k1r k1cr d d[G6P] / dt = d[E1] / dt = d[E1-complex] / dt = d[F6P] / dt = -[G6P] [E1] k1f + [E1-complex] k1r -[G6P] [E1] k1f + [E1-complex] k1r+ [E1-complex] k1cf - [F6P][E1] k1cr +[G6P] [E1] k1f - [E1-complex] k1r- [E1-complex] k1cf + [F6P][E1] k1cr +[E1-complex] k1cf - [F6P][E1] k1cr

  25. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P +E1 E1-complex F6P +E1 k1r k1cr k1if I+E1 E1-I-complex k1ir - [G6P] [E1] k1f + [E1-complex] k1r+[E1-complex] k1cf - [F6P][E1] k1cr - [I] [E1] k1if + [E1-I-complex] k1ir d[E1] / dt = d[I] / dt = d [E1-I-complex] / dt = d

  26. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P +E1 E1-complex F6P +E1 k1r k1cr d d[G6P] / dt = d[E1] / dt = d[E1-complex] / dt = d[F6P] / dt = -[G6P] [E1] k1f + [E1-complex] k1r -[G6P] [E1] k1f + [E1-complex] k1r+ [E1-complex] k1cf - [F6P][E1] k1cr +[G6P] [E1] k1f - [E1-complex] k1r- [E1-complex] k1cf + [F6P][E1] k1cr +[E1-complex] k1cf - [F6P][E1] k1cr

  27. 0 (steady state assumption) Characteristics of enzymesEnzymatic reactions k1f k1c G6P +E1 E1-complex F6P +E1 k1r d[E1-complex] / dt = d[F6P] / dt = +[G6P] [E1] k1f - [E1-complex] k1r- [E1-complex] k1c + [F6P][E1] k1cr +[E1-complex] k1c - [F6P][E1] k1cr

  28. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P +E1 E1-complex F6P +E1 k1r [G6P] [E1] k1f = [E1-complex] k1c - [E1-complex] k1r d[E1-complex] / dt = d[F6P] / dt = [G6P] ([Etotal]-[E1-complex]) k1f = [E1-complex] (k1c - k1r) +[G6P] [E1] k1f - [E1-complex] k1r- [E1-complex] k1c + [F6P][E1] k1cr +[E1-complex] k1c - [F6P][E1] k1cr

  29. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P +E1 E1-complex F6P +E1 k1r [G6P] [E1] k1f = [E1-complex] k1c - [E1-complex] k1r d[F6P] / dt = [G6P] ([Etotal]-[E1-complex]) k1f = [E1-complex] (k1c - k1r) [G6P] [Etotal] k1f [G6P] [Etotal][G6P] + (k1c -k1r) [G6P] + (k1c -k1r)/k1f [E1-complex] = = +[E1-complex] k1c - [F6P][E1] k1cr

  30. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P +E1 E1-complex F6P +E1 k1r d[F6P] / dt = +[E1-complex] k1c [G6P] [Etotal] k1c[G6P] + (k1c -k1r)/k1f = [G6P] [Etotal] k1c[G6P] + Km =

  31. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P+E1 E1-complex F6P +E1 k1r d[F6P] / dt = +[E1-complex] k1c [G6P] [Etotal] k1c[G6P] + (k1c -k1r)/k1f = [G6P] [Etotal] k1c[G6P] + Km = [G6P][Etotal] k1c[G6P] + Km Max d[F6P] / dt = (Vmax) = [Etotal] k1c

  32. Characteristics of enzymesEnzymatic reactions k1f k1cf G6P+E1 E1-complex F6P +E1 k1r [G6P] VMax[G6P] + Km d[F6P] / dt = [S] VMax[S] + Km v (velocity) = d[product] / dt =

More Related