BME 100 Computer project 2_nigel

# BME 100 Computer - BME 100 Computer Project 2 BME 100 Computer project 2 Nigel Chou 1 since units of dM i dt nM therefore units of ko s nM dm s 1 s

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BME 100 Computer Project 2 BME 100 Computer project 2 Nigel Chou 1) since units of s nM = dt dM i , therefore units of s 1 , s nM = = m o d k j K , j P and j M are nM (given) The terms in j K and j P cancel out, therefore units of s nM = k Since units of s nM = dt dP i , therefore units of s 1 = = p p d k These results are summarized in table 1. Table 1: Dimensions of parameters Parameter Units K j nM P j nM M i nM k nM s -1 k 0 nM s -1 d m s -1 k p s -1 d p s -1 m m d d dt t d 1 = = τ dt dM K dt dm M m K i j i i i j 1 = = dt dM K dt dm M m K i j i i i j 1 = = , 0 1 1 ) ( L LR C k K k = - and LR R R N N N - = 0 ) ( , We can find the rate of change of LR n with : ( 29 LR LR LR R LR R LR R LR R LR R R LR R LR R R LR LR R L R LR R LR n n K N N N N N K N N N N K k N N k N N K k k N k N C k N d dt dt dN N d dn - - = - - = - = - = - = = - - - - - ) 1 ( ) ( ) ( ) ( ) ( ) ( 1 ) ( ) ( 1 ) ( ) ( 1 ) ( 1 0 0 0 0 0 1 0 1 0 1 1 1 0 1 0 0 1

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BME 100 Computer Project 2 Hence the dimensionless form of equation 1 is: LR LR LR LR n n K d dn - - = ) 1 ( τ (1) Using the additional relations: 0 * * ) ( ' P M M N K K = , 0 ) ( ' P M M N K K = , 0 2 0 1 2 ) ( ' ) ( R P N k N k k - = , ' ) ( ) ( * 2 0 1 * 2 2 k N k N k P T E - = and 1 * = + P P n n We can find the rate of change of * P n with : * * * * 2 * * 2 * * * * 2 2 0 * * 0 * * 2 0 0 0 2 * * 0 * * 2 0 0 2 1 * 0 * * 2 0 1 0 0 2 0 1 0 1 * * * 2 2 0 * 0 * ' ' 1 ' ) 1 ( ' ' ' ' ' ) ( ' ) ( ' ) ( ' ) ( ) ( ' ' ) ( ' ' ) ( ) ( ) ( ' 1 *' ) ( ' ) ( ' ) ( ) ( ) ( ' ) ( ) ( 1 1 * ) ( ) ( ) ( 1 ) ( 1 1 1 2 1 p M p p M p LR p M p p M p LR p P M p P p P M p P R LR P M p P P M p P T E R P M p P p P M p P T E R p P P M P T E P M P T E P p p p n K n k n K n n k n K n k n K n n k N N K N N k N N K N N N N k N K N N k N K N N N N k k N K N N k N k N K N N N N k N k N k N K N N k N K N N k N d dt dt dN N d dn + - - + - = + - + = + - + = + - + = + - + = + - + = = - - - - Hence the dimensionless form of equation 2 is: * * * * 2 * * 2 * ' ' 1 ' ) 1 ( ' p M p p M p LR p n K n k n K n n k d dn + - - + - = (2) 2) At t = 0, no ligand has bound to receptor, hence there is no LR complex. Thus: 0 ) ( 0 ) ( 0 0 0 = = = = R R LR t LR N N N n and 1 ) ( ) ( ) ( 0 0 0 0 = = = = R R R LR t R N N N N n At t = 0, none of protein P has been phosphorylated since there is no E 1 (the activated kinase domain of the receptor complex) to phosphorylate it. Thus: 0 ) ( 0 ) ( 0 0 * 0 * = = = = R P p t p N N N n and 1 ) ( ) ( ) ( 0 0 0 0 = = = = R P P p t p N N N N n 3) At steady state, 0 = dt dn LR and 0 * = dt dn P 2
BME 100 Computer Project 2 Since τ is proportional to t, 0 ) 1 ( = - - = = LR LR LR LR LR n n K dt dn d dn τ Thus 0 = - - LR LR LR LR n n K K LR LR LR K K n + = 1 Similarly for n P* , 0 ' ' 1 ' ) 1 ( ' * * 2 * * 2 * * = + - - + - = = p M p p M p LR P P n K n k n K n n k dt dn d dn Substituting for LR n in this equation, we get 0 ' ' 1 ' ) 1 ( 1 ' * * * * 2 * * 2 = + - - + - + P M P P M P LR LR n K n k n K n K K k 0 ' 0 . 1 1 ' ) 1

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## This lab report was uploaded on 04/09/2008 for the course BME 100 taught by Professor Yuan during the Fall '07 term at Duke.

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BME 100 Computer - BME 100 Computer Project 2 BME 100 Computer project 2 Nigel Chou 1 since units of dM i dt nM therefore units of ko s nM dm s 1 s

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