Welcome to Introduction to Bioinformatics Wednesday, 16 October

Welcome to Introduction to Bioinformatics Wednesday, 16 October

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 Central Nervous System Death Trypanosoma brucei Dependence on glycolysis Entry of glucose Phosphorylation

Breakdown to triose phosphates Treatment Arsenate (AsO4 = Asi) Competitive with Pi Asi AMP-P-As Conversion to pyruvate Release of pyruvate AMP-P +Asi Trypanosoma brucei Dependence on glycolysis inhibitor1 Treatment

Something more specific? Choice #1 Starve the cell Trypanosoma brucei Dependence on glycolysis Treatment Something more specific? Choice #1 Starve the cell Choice #2 Stuff the cell pyruvate inhibitor2 Trypanosoma brucei Test ideas for inhibitors [I]

[ATP] 0 1 5 10 25 [glucose]added We know characteristics of enzymes inhibitor1 Characteristics of enzymes Analogy of radioactivity One day (1/20) P S

32 32 How often does 32S appear? [X atoms of 32P] [32P] concentration k rate constant = 1/20 of an atom per day? 1 atom of 32S per day 20 atoms of 32P __atoms of 32S per day = d[32S] / dt

rate of change Characteristics of enzymes Analogy of radioactivity One day (1/20) P 32 S 32 Which quantity is an intrinsic characteristic? [32P] concentration k rate constant

= d[32S] / dt rate of change Characteristics of enzymes Analogy of radioactivity One day (1/20) [32P] k = d[32S] / dt concentration rate constant rate of change One day (1/3000000) [14C] concentration

k = d[14N] / dt rate constant rate of change Characteristics of enzymes Chemical reactions AMP-P-As [32P] concentration AMP-P +Asi k = d[32S] / dt rate constant rate of change

Characteristics of enzymes Chemical reactions AMP-P-As [AMP-P-As] concentration AMP-P +Asi k = d[AMP-P ] / dt rate constant rate of change = d[Asi] / dt rate of change = = d[AMP-P-Asi] / dt rate of change Characteristics of enzymes Chemical reactions

AMP-P-As [AMP-P-As] concentration A differential equation AMP-P +Asi k = = d[AMP-P-Asi] / dt rate constant rate of change [S] k = = d[S] / dt Characteristics of enzymes Chemical reactions AMP-P-As AMP-P +Asi

[AMP-P-As] concentration A differential equation Its solution? Check: k = = d[AMP-P-Asi] / dt rate constant rate of change = d[S] / dt = k[S] [S] = S0 e -k(t-to) = d[S] / dt = S0 (k) e -k(t-to) = -k S0 e -k(t-to) = -k [S] Characteristics of enzymes Chemical reactions

1 [S] 0.8 [S] = S0 e -k(t-to) 0.6 0.4 = d[S] / dt = k[S] 0.2 0 0 1 2 3

t (sec) 4 5 Characteristics of enzymes Chemical reactions 1 [S] = [S] 0.8 [S] = S0 e -k(t-to) 0.6 S0 0.4

0.2 tt 0 0 1 2 3 t (sec) 4 5 Characteristics of enzymes Chemical reactions 1

[S] = S0 + tt d[S] / dt [S] 0.8 0.6 S0 0.4 0.2 = d[S] / dt = k[S] tt 0 0 1 2 3

t (sec) 4 5 Characteristics of enzymes Chemical reactions (Program) Characteristics of enzymes [S]S Chemical reactions 1 0.9 0.8 0.7 0.6 0.5 0.4

0.3 0.2 0.1 0 [S] = S0 e -k(t-to) = d[S] / dt = k[S] 0 1 2 3 t t (sec) 4 5 Characteristics of enzymes

Chemical reactions = d[S] / dt = k[S] [S] = S0 + tt d[S] / dt 1 [S] 0.8 Slope0 = -k[S0] [S1] = S0 + tt d[S] / dt Slope1 = -k[S1] Use average of Slope0 and Slope1 0.6 S0 0.4 0.2 tt

0 0 1 2 3 t (sec) 4 5 Characteristics of enzymes Chemical reactions = d[S] / dt = k[S] [S] = S0 + tt d[S] / dt 1 [S]

0.8 Slope0 = -k[S0] [S1] = S0 + tt d[S] / dt Slope1 = -k[S1] Use average of Slope0 and Slope1 Runge-Kutta method 0.6 S0 0.4 0.2 tt 0 0 1 2

3 t (sec) 4 5 Characteristics of enzymes Enzymatic reactions inhibitor1 Characteristics of enzymes Enzymatic reactions Glucose-6-phosphate [G6P] concentration Fructose-6-phosphate k = = d[G6P] / dt

rate constant k0 rate of change Characteristics of enzymes Enzymatic reactions G6P + E G6PE F6PE E-complex F6P + E Characteristics of enzymes Enzymatic reactions

G6P + E1 k1f k1r E1-complex k1cf k1cr F6P + E1 dd[G6P] / dt = -[G6P] [E1] k1f + [E1-complex] k1r d[E1] / dt = -[G6P] [E1] k1f + [E1-complex] k1r + [E1-complex] k1cf - [F6P] [E1] k1cr d[E1-complex] / dt = +[G6P] [E1] k1f - [E1-complex] k1r - [E1-complex] k1cf + [F6P] [E1] k1cr d[F6P] / dt =

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

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

F6P + E1 dd[G6P] / dt = -[G6P] [E1] k1f + [E1-complex] k1r d[E1] / dt = -[G6P] [E1] k1f + [E1-complex] k1r + [E1-complex] k1cf - [F6P] [E1] k1cr d[E1-complex] / dt = +[G6P] [E1] k1f - [E1-complex] k1r - [E1-complex] k1cf + [F6P] [E1] k1cr d[F6P] / dt = +[E1-complex] k1cf - [F6P] [E1] k1cr Characteristics of enzymes Enzymatic reactions G6P + E1 k1f k1r

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

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

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

G6P + E1 E1-complex d[F6P] / dt = +[E1-complex] k1c = = [G6P] [Etotal] k1c [G6P] + (k1c - k1r)/k1f [G6P] [Etotal] k1c [G6P] + Km k1cf F6P + E1 Characteristics of enzymes Enzymatic reactions G6P + E k1f

k1r 1 E1-complex k1cf d[F6P] / dt = +[E1-complex] k1c = = Max d[F6P] / dt = (Vmax) [G6P] [Etotal] k1c [G6P] + (k1c - k1r)/k1f [G6P] [Etotal] k1c [G6P] + Km [G6P] [E ] k c [G6P] + K total

1 m = [Etotal] k1c F6P + E1 Characteristics of enzymes Enzymatic reactions G6P + E 1 k1f k1r d[F6P] / dt = v (velocity) = E1-complex

k1cf [G6P] VMax [G6P] + Km d[product] / dt = [S] VMax [S] + Km F6P + E1

Recently Viewed Presentations

  • World War I, aka "The Great War" Long and Short-term Causes

    World War I, aka "The Great War" Long and Short-term Causes

    First Moroccan Crisis (1905) Wilhelm II sends fleet down to Morocco makes a speech to advocate for Moroccan independence. As a response, the Algerciras Conference was held to discuss French claims of Morocco. Everybody supported France, except for Austria-Hungary. In...
  • Presentation - California

    Presentation - California

    Closing Documents from Departments. 1. Executed legal documents—Site Lease, Facility Lease, Continuing Disclosure Agreement. 2. Closing Certificate of Department (Exhibit 5) 3. Opinion of Counsel to Department (Exhibit 6) 4. Department Certificate to Tax Certificate (Exhibit 7) 5. Tenant Certificate...
  • The Veldt - LT Scotland

    The Veldt - LT Scotland

    Explain what this relationship makes us think/feel/realise about relationships, families, technology etc. Relationship (Highlight these key quotes and select at least 2-3 and analyse fully) 'They were awfully young, Wendy and Peter, for death thoughts.' p4 'Or Peter's set it...
  • Modernism in the 20th Century - Blogs@Baruch

    Modernism in the 20th Century - [email protected]

    Modernism in the 20th Century. Literary Themes and Styles. Doubt, conflict, and strife. Protagonists are often troubled with who they are and their place in the world. Fragmentation of the interior and exterior.
  • Biology of Insects - Mr. Robertson's Classroom

    Biology of Insects - Mr. Robertson's Classroom

    Biology of Insects Their body parts and life cycles "insecticum" Latin word meaning "cut into" Insects are "cut into" three distinct parts Head Thorax Abdomen The Head Functions Securing food Sensory perception Mouth parts Antennae Eyes The Thorax Used for...
  • CT Urography and applications in uroephithelial tumors

    CT Urography and applications in uroephithelial tumors

    Arial Wingdings Times New Roman Beam 1_Beam Photo Editor Photo CT Urography and applications in uroephithelial tumors IVP (intravenous pyelography) CT Urography (CTU) CTU at Sheba Protocol CTU Normal CTU - Axial images MPR MIP 3D volume rendering CTU -...
  • Certification of Ownership

    Certification of Ownership

    RLA's & Easements. RLA & Easement - requirements for Certification. At the present time, RLA & Easement requests are not required for Certification Requests. If SPC feels an agency is not actively pursuing RLA's & Easement's for their projects, they...
  • Maturity - Units 3 & 4 Psychology

    Maturity - Units 3 & 4 Psychology

    Results. Following exposure to the model in the movie, each child was placed individually in a room that had many toys and a BoBo doll. The child's behaviour was then observed through a one-way mirror to see whether they imitated...