Modeling biochemical reactions Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4500, Spring 2017 M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 1 / 10
Overview In biochemistry, 2+ species, or “reactants” can react if they come toegether and collide. Alternatively, one species can degrade. More is needed, though: correct orientation, enough energy, etc. Examples CH 4 � 2 O 2 Ý Ñ CO 2 � 2 H 2 O (burning of methane) H � � OH ✁ Ý Ñ H 2 O unfolded protein Ý Ñ folded protein 2 SO 2 � O 2 Ý Ý á Ý 2 SO 3 â Ý O 3 Ý Ñ O 2 � O 2 O 3 Ý Ñ 3 O 2 M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 2 / 10
Mass-action kinetics Classification of reactions: A Ý Ñ P : “uni-molecular” A � B Ý Ñ P : “bi-molecular” A � B � C Ý Ñ P : “tri-molecular” Law of mass-action kinetics A reaction rate is proportional to the probability of collision of reactants involved. Assume this probability is proportional to the concentration of each reactant R , denoted r R s . ODE model d r P s k A Ý Ñ P : ✏ k r A s dt d r P s k A � B Ý Ñ P : ✏ k r A sr B s dt d r P s k 1 Ý á A � B P : ✏ k 1 r A sr B s ✁ k 2 r P s â Ý dt k 2 M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 3 / 10
Mass-action kinetics Enzymes are proteins that catalyze reactions (up to 10 12 -fold!) An example Consider the following chemical reaction k 1 k 3 Ý á E � S ES Ý Ñ E � P â Ý k 2 E ✏ enzyme, S ✏ substrate, ES ✏ enzyme-substrate complex, and P ✏ product. d r ES s ✩ ✏ k 1 r E sr S s ✁ ♣ k 2 � k 3 qr ES s ✬ dt ✬ ✬ ✬ ✫ d r P s ✏ k 3 r ES s dt ✬ ✬ ✬ ✬ ✪ E 0 ✏ r E s � r ES s , E 0 ✏ initial enzyme concentration Assumptions E 0 is constant. Enzyme-substrate complex reaches equilibrium much earlier than the product does, so d r ES s ✓ 0. dt M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 4 / 10
Mass-action kinetics Goal Write the differential equation d r P s ✏ k 3 r ES s in terms of r S s , not r ES s . dt Since d r ES s ✓ 0, we can simplify the ODE for r ES s : dt d r ES s ✏ k 1 r E sr S s ✁ ♣ k 2 � k 3 qr ES s ✏ 0 . dt Upon solving for r E s , we get r E s ✏ ♣ k 2 � k 3 qr ES s k 1 r S s . Plugging this into E 0 ✏ r E s � r ES s and solving for r ES s : E 0 r S s r ES s ✏ k 2 � k 3 . � r S s k 1 Alas, we can write d r P s k 3 E 0 r S s ✏ V max r S s ✏ k 3 r ES s ✏ k 2 � k 3 dt K m � r S s . � r S s k 1 M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 5 / 10
Michaelis–Menten equation Recall the following chemical reaction: k 1 k 3 E � S Ý á Ý Ñ E � P ES â Ý k 2 E ✏ enzyme, S ✏ substrate, ES ✏ enzyme-substrate complex, and P ✏ product. Definition The Michaelis–Menten equation is one of the best-known models of enzyme kinetics. d r P s ✏ V max r S s and K m ✏ k 2 � k 3 where V max ✏ k 3 E 0 , K m � r S s , dt k 1 ❧♦♦♦♦♠♦♦♦♦♥ f ♣r S sq Remarks The “reaction rate”, f ♣r S sq , is a strictly increasing function of r S s . r S sÑ✽ f ♣r S sq ✏ V max , lim (biologically, the maximum reaction rate) f ♣ K m q ✏ 1 2 V max . The reaction rate f ♣r S sq is proportional to E 0 . M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 6 / 10
Michaelis–Menten equation Recall the following chemical reaction: k 2 k 3 Ý á E � S ES Ý Ñ E � P â Ý k 1 E ✏ enzyme, S ✏ substrate, ES ✏ enzyme-substrate complex, and P ✏ product. Further assumptions Substrate concentration is conserved: S 0 ✏ r S s � r ES s � r P s . E 0 ✦ S 0 , so r ES s ✦ r S s and r P s . d Together, this means S 0 ✓ r S s � r P s . Taking dt of both sides yields d r S s ✏ ✁ d r P s ✏ ✁ V max r S s k m � r S s . dt dt Usually, V max , K m , and S 0 are known quanities. This is now something we can easily solve, graph, analyze, etc. M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 7 / 10
Multi-molecule binding Consider a reaction where n molecules of a substrate S react with an enzyme E : k 1 k 3 Ý á E � nS ES n Ý Ñ E � P â Ý k 2 The enzyme-substrate complex here is ES n . By mass-action kinetics, d r ES n s ✩ ✏ k 1 r E sr S s n ✁ ♣ k 2 � k 3 qr ES n s ✬ dt ✬ ✬ ✬ ✫ d r P s ✏ k 3 r ES n s ✬ dt ✬ ✬ ✬ ✪ E 0 ✏ r E s � r ES n s , E 0 ✏ initial enzyme concentration As before, assume r ES n s reaches equilibrium much quicker than r P s and r S s : d r ES n s r E s ✏ ♣ k 2 � k 3 qr ES n s ✏ 0 ù ñ k 1 r S s n . dt Plugging this into E 0 ✏ r E s � r ES n s and solving for r ES n s yields E 0 r S s n ✏ V max r S s n d r P s r ES n s ✏ ù ñ k 2 � k 3 K m � r S s n . dt � r S s n k 1 M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 8 / 10
Multi-molecule binding Hill equation Given the chemical reaction k 2 k 3 E � nS Ý á Ý Ñ E � P ES n â Ý k 1 we derived the following ODE involving r P s and r S s : ✏ V max r S s n d r P s and K m ✏ k 2 � k 3 where V max ✏ k 3 E 0 , dt K m � r S s n , k 1 ❧♦♦♦♦♦♠♦♦♦♦♦♥ f ♣r S sq This is called the Hill equation with Hill coefficient n . Remarks The “reaction rate”, f ♣r S sq , is a strictly increasing function of r S s . r S sÑ✽ f ♣r S sq ✏ V max , lim (biologically, the maximum reaction rate) f ♣ K 1 ④ n m q ✏ 1 2 V max . The reaction rate f ♣r S sq is proportional to E 0 . n ✏ 1 is just the Michaelis–Menden equation. M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 9 / 10
Hill equations t n The following shows several “Hill functions” y ✏ 1 � t n , for various values of n . M. Macauley (Clemson) Modeling biochemical reactions Math 4500, Spring 2017 10 / 10
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