1 inhibit3.mcd. Substrate Inhibition. Instructor: Nam Sun Wang

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1 Subtrate Inhibition Intructor: Nam Sun Wang inhibitmcd Mechanim Enzyme combine with a ubtrate molecule for form a complex, which lead to product The active enzyme complex ES can further combine with a ubtrate molecule S to form an inactive form ESS, which doe not lead to product Thi i decribed chematically a follow k k S + E ES P + E k i The nd part i equivalent to the following notation + k S k k 4 ES + S ESS ESS (inactive) k 4 Derivation of Reaction Rate Expreion with Equilibrium Aumption Given dp/dt=rate=v v k ES Conervation of enzyme pecie E 0 E ES ESS Equilibrium Aumption: k S E k i ES k S ES k 4 ESS We have 4 eqn, and we can chooe to olve for any of the four unknown -- E, ES, ESS, and v ) Find the analytical expreion (via Math SmartMath ) E 0 k 4 k i k 4 k S k 4 k i k S k Find( E, ES, ESS, v) k 4 k S E 0 k 4 k S k 4 k i k S k S k k E 0 k 4 k S k 4 k i k S k k k 4 k S E 0 k 4 k S k 4 k i k S k Thu, the lat row i the analytical expreion for v k v k 4 k S E 0 k S k 4 k 4 k i S k k Thu, the above form i tranformed into the Michaeli-Menten form by defining: v m k E 0 k i v( ) v m k k 4 k

2 inhibitmcd Derivation of Reaction Rate Expreion with the Quai-Steady State Aumption Given dp/dt=rate=v v k ES Conervation of enzyme pecie E 0 E ES ESS Quai-Steady State Aumption (applicable to the intermediate pecie ES and ES ): d ES/dt = 0 k S E k i ES k ES k S ES k 4 ESS 0 d ESS/dt = 0 k S ES k 4 ESS 0 (Our work i done, jut let Mathcad do the ret) Find the analytical expreion (via Math SmartMath ) k E i k 0 k 4 k 4 k S k 4 k i k 4 k k S k Find( E, ES, ESS, v) k 4 k S E 0 k 4 k S k 4 k i k 4 k k S k S k k E 0 k 4 k S k 4 k i k 4 k k S k k k 4 k S E 0 k 4 k S k 4 k i k 4 k k S k Thu, the lat row i the analytical expreion for v (although it doe not look very compact) k v k 4 k S E 0 k 4 k S k 4 k i k 4 k k S k Thu, the above form i tranformed into the Michaeli-Menten form by defining: v m k E 0 k i k v v m k k 4 k The reaction rate expreion how ubtrate inhibition

3 inhibitmcd Calculation of the Maximum Point The rate goe through a maximum when dv/d=0 Let' find where thi maximum lie Here, we mark "" in the above equation and chooe Symbolic Differentiate on Variable 0 v m v m Chooe Symbolic Simplify yield: K 0 v m m K m K i Since there i only one equation, mark "" in the above equation and chooe Symbolic Solve for Variable yield: Since the ubtrate concentration mut be non-negative, we chooe the econd olution max Finally, the maximum rate i v max =v( max ) v max v m max max max Copy " " into the clipboard, mark " max " in the lat equation, and chooe Symbolic Subtitute for Variable yield: K v m max v m K m Further chooing Symbolic Simplify yield: v m v max

4 4 inhibitmcd Alternately, we can find v max by olving the following two equation imultaneouly v Given m max max v max max max Find max, v max K v i m K m K i v max i the econd element in the above olution vector -- copy and pate it K v i max v m K m K i Chooe Symbolic Simplify yield the ame anwer a before (a it hould be) v m v max

5 5 inhibitmcd Half-Maximum Calculation There are two point at which v v max v m v m Since there i only one equation, mark "" and chooe Symbolic Solve for Variable yield: 4 K i 4 K i 5 K m 8 5 K m 8 Alternately, we can call Math SmartMath Thi method work when there are multiple equation (Of coure, it work for jut one equation, a well) Given v m v m Find( ) 4 5 K m 8 K i 4 K Copy each term out and chooe Symboli Simplify yield the following two ubtrate concentration: 4 8 K m 4 8 K m

6 Numerical Plot of Subtrate Inhibition Rate Expreion Model parameter v m Subtrate inhibition rate expreion v m v( ) Maximum value v m max v max max = v max = 0 Half-maximum value 6 inhibitmcd 4 8 K m = K m = 479 Plot for ranging over 0, v max v( ) 0 0 v max Note that purely numerical calculation i much more abbreviated -- page of ymbolic derivation i reduced to jut the next few line 0 initial gue Maximum point: Given d d v( ) 0 max Find( ) max = Mathcad v7 fail to find a olution with the "Given-Find" block, but "root" i ok d max root, d v( ) max = 0999 v max v max v max = 0 Half-maximum point: Given v( ) v max half Find( ) half = 009

7 7 inhibitmcd Repeat with another initial gue to find the econd olution Given v( ) v max half Find( ) half = 479

8 5 inhibitmcd 5 i K m 8