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Modeling Chemical Systems BIOE 4200. Chemical Systems Every physiologic systems depends on multiple chemical reactions Examples include hormone-receptor.

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Presentation on theme: "Modeling Chemical Systems BIOE 4200. Chemical Systems Every physiologic systems depends on multiple chemical reactions Examples include hormone-receptor."— Presentation transcript:

1 Modeling Chemical Systems BIOE 4200

2 Chemical Systems Every physiologic systems depends on multiple chemical reactions Examples include hormone-receptor interactions, enzyme-substrate binding, and blood gas dynamics The diffusion of reactants or products within tissues affects this process Chemical reaction models may be needed depending on the level of detail

3 Chemical Reactions Differential equations can be derived directly from reaction stoichiometry Forward and reverse reaction rates (k f and k d ) may depend on environmental factors such as temperature or pressure Enzymes or catalysts can also affect k f and k d

4 Chemical Reactions Can make assumptions to simplify equations Abundant species: suppose [A] >> [B] and [A 2 B], then [A] will be relatively constant throughout reaction Stoichiometry: when [B] increases, [A 2 B] must decrease and vice versa, so [B] + [A 2 B] = constant Let [A] = C A and [B] + [A 2 B] = C t

5 Hemoglobin and Oxygen Hemoglobin molecules carry O 2 in blood Each molecule has 4 heme groups Each heme group binds to one O 2 molecule Binding of O 2 molecule changes affinity for next O 2 molecule Changes in binding affinity are represented by different rate constants at each step

6 Hemoglobin Equations

7 Use state variables – x 0 = [Hb 4 ] – x 1 = [Hb 4 O 2 ] – x 2 = [Hb 4 O 4 ] – x 3 = [Hb 4 O 6 ] – x 4 = [Hb 4 O 8 ] Gases in solution are represented by their partial pressure – PO 2 = [O 2 ] Can also substitute for sum of Hb 4 molecules – sum(x i ) = constant

8 Diffusion Particles will flow between fluid regions that have different particle concentration Concentration difference between two points can serve as a “driving force” for particle flow Flow is proportional to concentration difference: Q = k(C 2 – C 1 ) Analogous to flow of fluid in pipe due to pressure difference: Q = k(P 2 – P 1 ) The proportionality constant k is determined by the resistance to particle flow – cross section area, distance between C 1 and C 2, how many other particles in fluid, etc.

9 Diffusion The flow of particles can alter the concentration gradient Q = k(C 2 – C 1 ) represents the flow of particles from C 2 to C 1 dC 1 /dt & dC 2 /dt are proportional to Q – Q in units of particles / time – C 1 & C 2 in units of particles / volume – dC 1 /dt & dC 2 /dt in units of particles / volume / time – Proportionality constant has units of 1 / volume Can rewrite equation as dC 1 /dt = -dC 2 /dt = k(C 2 – C 1 ) – Proportionality constant has units of 1 / time

10 Scaling and Gain Some processes directly transform input to output (y = ku) No state equation is needed Useful to model electrical amplification Example: op-amp in inverted configuration has gain V out /V in = -R 2 /R 1 – Scaling factor k has no units Useful for transforming input units to output units Example: digital speedometer measures MPH, output is V – Scaling factor k has units V/MPH

11 Integration Some processes are needed to transform a rate (input) to an absolute amount (output) – Convert from mg/sec to mg – Convert from velocity to displacement Original equation is Can represent this using state equations:

12 Differentiation Some processes are needed to transform an absolute amount (input) to a rate (output) – Convert from mg to mg/sec – Convert from displacement to velocity Original equation is Cannot represent this using state equations! Will be able to represent this using transfer functions (later in course) Not a big deal – don’t usually need to convert variables this way


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