INF-5610, Matematiske modeller i medisin Eksamen Forelesere: Det - PowerPoint PPT Presentation

INF-5610, Matematiske modeller i medisin Eksamen Forelesere: Det blir oppgitt seks temaer, senest to uker før eksamen Glenn Terje Lines (glennli@ifi.uio.no) Det skal forberedes 20 min foredrag for hvert tema Joakim Sundnes (sundnes@ifi.uio.no) På eksamen trekkes et av disse temaene og holde foredraget Temaer: Det vil også bli stilt spørsmål fra andre temaer Hjertecellenes egenskaper Strømflyt i hjertet og i kroppen, EKG Modeller for disse fenomener Numeriske metoder for disse modellene To obligatorisk oppgaver – p. 1 – p. 2 Forelesningsplan, del I Forelesningsplan, del II Anatomi, om celler og hjertet. Campell, KS Bidomene modell. KS Kap. 11. Presentasjon av oblig. Grunnleggende biophys proseeser. KS Kap. 2 Presentasjon av en anvendelse Ionekanaler. KS Kap. 3 Modeller for mekanikk. Eksiterbarhet og progagering av signal. KS Kap. 4 Numeriske metoder: fra modell til software, splitting. Nevroner og cell-koplinger KS Kap. 7&8 Numeriske metoder: prinsipp bak ODE og PDE løsere. Calcium dynamikk KS Kap. 5 – p. 3 – p. 4

Levels of modeling The Cell Membrane Body Consist of a bilipid layer Organ Embedded proteins for transport control Tissue Selectively permeable Cell Maintains concentration gradients Organelles Has a transmembrane potential Proteins – p. 5 – p. 6 The Cell Membrane The Cell Membrane – p. 7 – p. 8

Two types of transmembrane flow Transmembrane flow Passive: Diffusion along the concentration gradient Through the membrane (H 2 O , O 2 , CO 2 ) Through specialized channels (Na + , K + , Cl − ) Carrier mediated transport Active: Energy driven flow against the gradients ATP driven pumps (Na + − K + , Ca 2+ ) Exchangers driven by concentration gradients (Na + − Ca 2+ ) – p. 9 – p. 10 Active Transport Mathematical models of chemical reactions – p. 11 – p. 12

A two way reaction The Law of Mass Action The reverse reaction may also take place: Chemical A and B react to produce chemical C: k + k A + B − → C − → A + B C ← − k − The rate constant k determines the rate of the reaction. It can be The production rate is then: interpreted as the probability that a collision between the reactants produces the end results. d [ C ] = k + [ A ][ B ] − k − [ C ] If we model the probability of a collision with the product [A] [B] dt we get the law of mass action: At equilibrium when d [ C ] /dt = 0 we have: d [ C ] = k [ A ][ B ] k − [ C ] = k + [ A ][ B ] (1) dt – p. 13 – p. 14 Enzyme Kinetics k If A + B − → C is the only reaction involving A and C then Characteristics of enzymes: d [ A ] /dt = − d [ C ] /dt Made of proteins so that Acts as catalysts for biochemical reactions [ A ] + [ C ] = A 0 (2) Speeds up reactions by a factor > 10 7 Substituting (2) into (1) yields: Highly specific [ B ] Often part of a complex regulation system [ C ] = A 0 K eq + [ B ] where K eq = k − /k + . Notice that [ B ] = K eq = ⇒ [ C ] = A 0 / 2 and [ B ] → ∞ = ⇒ [ C ] → A 0 – p. 15 – p. 16

Reaction model of enzymatic reaction Mathematical model of enzymatic reaction Applying the law of mass action to each compound yields: k 1 − → k 2 S + E C − → P + E ← − d [ S ] k − 1 = k − 1 [ C ] − k 1 [ S ][ E ] + J S with dt d [ E ] S: Substrate = ( k − 1 + k 2 )[ C ] − k 1 [ S ][ E ] dt E: Enzyme d [ C ] = k 1 [ S ][ E ] − ( k 2 + k − 1 )[ C ] C: Complex dt d [ P ] P: Product = k 2 [ C ] − J P dt Here we also supply the substrate at rate J S and the product is removed at rate J P . – p. 17 – p. 18 Equilibrium In equilibrium we have d [ E ] = 0 Note that In equilibrium dt that is d [ S ] /dt = d [ E ] /dt = d [ C ] /dt = d [ P ] /dt = 0 ( k − 1 + k 2 )[ C ] = k 1 [ S ][ E ] it follows that that J S = J P . Since the amount of enzyme is constant we have Production rate: [ E ] = E 0 − [ C ] J = J P = k 2 [ C ] This yields E 0 [ S ] [ C ] = K m + [ S ] with K m = k − 1 + k 2 and E 0 is the total enzyme concentration. k 1 Production rate: d [ P ] [ S ] dt = k 2 [ C ] = V max K m +[ S ] , where V max = k 2 E 0 . – p. 19 – p. 20

Cooperativity Mathematical model of cooperativ reaction Applying the law of mass action to each compound yields: k 1 − → k 2 S + E C 1 − → E + P ← − ds k − 1 = − k 1 se + k − 1 c 1 − k 3 sc 1 + k − 3 c 2 dt dc 1 = k 1 se − ( k − 1 + k 2 ) c 1 − k 3 sc 1 + ( k 4 + k − 3 ) c 2 k 3 dt − → k 4 S + C 1 C 2 − → C 1 + P ← − dc 2 = k 3 sc 1 − ( k 4 + k − 3 ) c 2 k − 3 with dt S: Substrate E: Enzyme C1: Complex with one S C1: Complex with two S P: Product – p. 21 – p. 22 Equilibrium Case 1: No cooperation Set dc 1 dt = dc 2 The binding sites operate independently, with the same rates k + dt = 0 , and use e 0 = e + c 1 + c 2 , and k − . k 1 and k 3 are associated with events that can happen in two ways, thus: K 2 e 0 s k 1 = 2 k 3 = k + c 1 = K 1 K 2 + K 2 s + s 2 k − 3 = 2 k − 1 = k − e 0 s 2 c 2 = K 1 K 2 + K 2 s + s 2 Which gives this reaction speed: V = 2 k 2 e 0 s K = k − + k 2 where K 1 = k − 1 + k 2 , K 2 = k 4 + k − 3 Reaction speed: K + s, k 1 k 3 k + V = k 2 c 1 + k 4 c 2 = ( k 2 K 2 + k 4 s ) e 0 s Note that this is the same as the reaction speed for twice the K 1 K 2 + K 2 s + s 2 amount of an enzyme with a single binding site. – p. 23 – p. 24

Case 2: Strong cooperation The Hill equation In general with n binding sites, the reaction rate in the limit will The first binding is unlikely, but the next is highly likely, i.e. k 1 is small, and k 3 is large. We go to the limit: be: s n V = V max k 1 → 0 , k 3 → ∞ , k 1 k 3 = const K n m + s n This model is often used when the intermediate steps are so unknown, but cooperativity suspected. The parameters K 2 → 0 , K 1 → ∞ , K 1 K 2 = const V max , K m and n are usually determined experimentally. In this case the rection speed becomes: k 4 e 0 s 2 s 2 V = m + s 2 = V max K 2 K 2 m + s 2 with K 2 m = K 1 K 2 , and V max = k 4 e 0 – p. 25 – p. 26 Carrier-Mediated Transport Uniport Some substances can not pass the membrane on their own, but Substrate S combines with a carrier protein C to form a complex are helped by a carrier protein. P . The protein has two conformal states. Model: k + k k − Types of transport: − → − → − → S i + C i P i P e S e + C e ← − ← − ← − Uniport: Transport of single substance k − k k + Symport: Transport of several substances k − → in same direction C i C e ← − k Antiport: Transport of several substances in opposite directions With symport and antiport the carrier molecule as several binding sites. – p. 27 – p. 28

Model for Carrier Mediated Transport, Uniport Size of flux in equilibrium Applying the law of mass action: The flow in equilibrium can be setting the derivatives to zero and solve for J . d [ S i ] = k − [ P i ] − k + [ S i ][ C i ] − J dt d [ S e ] This yields a system of six eq. and seven unknowns. = k − [ P e ] − k + [ S e ][ C e ] + J dt d [ P i ] = k [ P e ] − k [ P i ] + k + [ S i ][ C i ] − k − [ P i ] dt The amount of protein is conserved so we have: d [ P e ] = k [ P i ] − k [ P e ] + k + [ S e ][ C e ] − k − [ P e ] dt [ C i ] + [ C e ] + [ P i ] + [ P e ] = C 0 d [ C i ] = k [ C e ] − k [ C i ] + k − [ P i ] − k + [ S i ][ C i ] Solving for J in equilibrium then gives: dt d [ C e ] = k [ C i ] − k [ C e ] + k − [ P e ] − k + [ S e ][ C e ] J = 1 [ S e ] − [ S i ] dt 2 kKC 0 ([ S i ] + K + K d )([ S e ] + K + K d ) − K 2 d Here J is the influx of the glucose molecules (S). with K = k − /k + and K d = k/k + . – p. 29 – p. 30 Size of flux in equilibrium Model for symport Two different substances S and T are transported in the same J = 1 [ S e ] − [ S i ] 2 kKC 0 direction. The carrier C has m binding sites for S and n for T : ([ S i ] + K + K d )([ S e ] + K + K d ) − K 2 d k + k p k − Factors affecting the flux: − → − → − → mS i + nT i + C i P i P e mS e + nT e + C e ← − ← − ← − The amount of Carrier molecules C 0 k − k − p k + The rate constants k − → Substrate gradient C i C e ← − k – p. 31 – p. 32