Enzyme Kinetics discussion

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 Enzyme kinetics exercises

1. Derive the Michaelis –Menten equation for the next model. Assume rapid equilibrium

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Enzyme kinetics exercises 1. Derive the Michaelis –Menten equation for the next model. Assume rapid equilibrium. 2. Obtain the Vmax, Km and inhibitor types for the next examples: Use the next table to calculate Km, Vmax and inhibitor type. Activity (mM/ s) [Substrate] (mM) Control Inbibitor A Inhibitor B Inhibitor C 0 0 0 0 0 0.5 9.09091 0.90082 7.68049 4.54545 1 16.66667 1.7854 12.48439 8.33333 2 28.57143 3.50754 18.1653 14.28571 5 50 8.32639 24.98751 25 10 66.66667 15.36098 28.56327 33.33333 20 80 26.59574 30.7645 40 50 90.90909 47.39336 32.25598 45.45455 100 95.2381 64.10256 32.78581 47.61905 Follow the next steps: a) Plot the direct data (Activity vs substrate)(100x100 squares). All data points and lines must fit in the assigned space b) Obtain the double reciprocals [Substrate] (1/mM) Control 1/Activity (s/ mM) Inbibitor A Inhibitor B Inhibitor C C) Plot the inverse or the activity vs the inverse of the substrate. All data points and lines must fit in the assigned space D) Calculate the Vmax and Km in each case and assign the inhibitor type Condition Vmax (mM/ s) Km (mM) Control Inhibitor A Inhibitor B Inhibitor C Inhibitor type Steps to Derive Michaelis- Menten Equation Case 1: Rapid equilibrium Write down the kinetic model kcat is rate limiting (very slow). Steps 1) Write the rate- dependence equation v =[ES] kcat .…………………………(Eq 1) 2) Obtain the mass balance equation of the enzyme concentration [E]T= [E] + [ES] …………………….(Eq 2) 3) Define reaction intermediates ([ES]) in terms of [E], [S] Under rapid equilibrium: 𝐾𝑠 = [ES]= [𝐸][𝑆] [𝐸𝑆] [𝐸][𝑆] 𝐾𝑠 = 𝑘−1 𝑘1 …………………………..(Eq 3) 4) Define Vmax When [ES]= [E]T Vmax =[E]T kcat .…………………….………(Eq 4) Do the math 5) Divide Eq 1 by Eq 2 𝑣 [E]T [𝐸𝑆] 𝑘𝑐𝑎𝑡 = [E] + [ES] 6) Substitute [ES] from Eq 3 [𝐸][𝑆] 𝐾𝑐𝑎𝑡 𝑣 = 𝐾𝑠 [𝐸][𝑆] [E]T 𝐸 + 𝐾𝑠 7) Move [E]T to the numerator and substitute Vmax from Eq 4 [𝐸][𝑆] [𝐸][𝑆] [𝐸]𝑇 𝐾𝑐𝑎𝑡 𝑉𝑚𝑎𝑥 𝐾𝑠 𝐾𝑠 𝑣= = [𝐸][𝑆] [𝐸][𝑆] 𝐸 + 𝐾𝑠 𝐸 + 𝐾𝑠 8) Multiply Numerator and Denominator by E [𝐸][𝑆] [𝑆] 𝑉𝑚𝑎𝑥 ( 𝐾𝑠 ) 𝐸 𝑉𝑚𝑎𝑥 𝐾𝑠 𝑣= = [𝐸][𝑆] [𝑆] 1 + 𝐾𝑠 (𝐸 + 𝐾𝑠 ) 𝐸 9) Multiply Numerator and Denominator by Ks 𝑣= [𝑆] 𝑉𝑚𝑎𝑥 (𝐾𝑠 ) 𝐾𝑠 (1 + [𝑆] ) 𝐾𝑠 𝐾𝑠 = 𝑉𝑚𝑎𝑥 [𝑆] 𝐾𝑠 + [𝑆] 10) Obtain the double reciprocal 1 𝐾𝑠 + [𝑆] 𝐾𝑠 𝑆 𝐾𝑠 1 1 = = + = + 𝑣 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 11) When 1/v =0 0= 𝐾𝑠 1 1 + 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 1 𝐾𝑠 1 = 𝑉𝑚𝑎𝑥 𝑉𝑚𝑎𝑥 [𝑆] 1 𝑉𝑚𝑎𝑥 1 = = [𝑆] 𝑉𝑚𝑎𝑥 𝐾𝑠 𝐾𝑠 Case 2: Steady state Write down the kinetic model Steps 1) Write the rate- dependence equation v=[ES] kcat .…………………………(Eq 1) 2) Obtain the mass balance equation of the enzyme concentration [E]T= [E] + [ES] …………………….(Eq 2) 3) Define reaction intermediates ([ES]) in terms of [E], [S] Under steady state: 𝑑[𝐸𝑆] = 0 = 𝑘1[𝐸][𝑆] − (𝑘– 1[𝐸𝑆] + 𝑘𝑐𝑎𝑡[𝐸𝑆]) 𝑑𝑡 𝑘1[𝐸][𝑆] = [𝐸𝑆](𝑘– 1 + 𝑘𝑐𝑎𝑡) [𝐸𝑆] = [𝐸][𝑆] 𝐾𝑚 [𝐸𝑆] = 𝑘1[𝐸][𝑆] (𝑘– 1 + 𝑘𝑐𝑎𝑡) 𝐾𝑚 = (𝑘– 1 + 𝑘𝑐𝑎𝑡) 𝑘1 .…………………….………(Eq 3) 4) Define Vmax When [ES]= [E]T Vmax= [E]T kcat .…………………….………(Eq 4) Do the math 5) Divide Eq 1 by Eq 2 𝑣 [E]T [𝐸𝑆] 𝑘𝑐𝑎𝑡 = [E] + [ES] 6) Substitute [ES] from Eq 3 [𝐸][𝑆] 𝐾𝑐𝑎𝑡 𝑣 = 𝐾𝑚 [𝐸][𝑆] [E]T 𝐸 + 𝐾𝑚 7) Move [E]T to the numerator and substitute Vmax from Eq 4 𝑣= [𝐸][𝑆] [𝐸][𝑆] 𝐾𝑚 = 𝑉𝑚𝑎𝑥 𝐾𝑚 [𝐸][𝑆] [𝐸][𝑆] 𝐸 + 𝐾𝑚 𝐸 + 𝐾𝑚 [𝐸]𝑇 𝐾𝑐𝑎𝑡 8) Multiply Numerator and Denominator by E [𝐸][𝑆] [𝑆] 𝑉𝑚𝑎𝑥 ( 𝐾𝑚 ) 𝐸 𝑉𝑚𝑎𝑥 𝐾𝑚 𝑣= = [𝐸][𝑆] [𝑆] 1 + (𝐸 + 𝐾𝑚 ) 𝐸 𝐾𝑚 9) Multiply Numerator and Denominator by Ks 𝑣= [𝑆] 𝑉𝑚𝑎𝑥 (𝐾𝑚) 𝐾𝑚 [𝐸][𝑆] (1 + 𝐾𝑚 ) 𝐾𝑚 = 𝑉𝑚𝑎𝑥 [𝑆] 𝐾𝑚 + [𝑆] 12) Obtain the double reciprocal 1 𝐾𝑚 + [𝑆] 𝐾𝑚 𝑆 𝐾𝑚 1 1 = = + = + 𝑣 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 Case 3: Competitive Inhibition (Rapid equilibrium) Write down the kinetic model Steps 1) Write the rate- dependence equation v=[ES] kcat .…………………………(Eq 1) 2) Obtain the mass balance equation of the enzyme concentration [E]T= [E] + [ES] + [EI] …………………….(Eq 2) 3) Define reaction intermediates ([ES]) in terms of [E], [S] Under rapid equilibrium: [𝐸][𝑆] 𝐾𝑠 = [𝐸𝑆] [ES]= 𝐾𝑖 = [EI]= [𝐸][𝑆] 𝐾𝑠 …………………………..(Eq 3) [𝐸][𝐼] [𝐸𝐼] [𝐸][𝐼] 𝐾𝑖 …………………………..(Eq 4) 4) Define Vmax When [ES]= [E]T Vmax=[E]T kcat .…………………….………(Eq 5) Do the math 5) Divide Eq 1 by Eq 2 𝑣 [E]T [𝐸𝑆] 𝑘𝑐𝑎𝑡 = [E]+ [ES]+[𝐸𝐼] 6) Substitute [ES] and [EI] from Eq 3 and Eq 4 [𝐸][𝑆] 𝑣 𝐾𝑠 𝐾𝑐𝑎𝑡 = [𝐸][𝑆] [𝐸][𝐼] [E]T 𝐸 + 𝐾𝑠 + 𝐾𝑖 7) Move [E]T to the numerator and substitute Vmax from Eq 4 [𝐸][𝑆] [𝐸][𝑆] 𝑉𝑚𝑎𝑥 𝐾𝑠 𝐾𝑠 𝑣= = [𝐸][𝑆] [𝐸][𝐼] [𝐸][𝑆] [𝐸][𝐼] 𝐸 + 𝐾𝑠 + 𝐾𝑖 𝐸 + 𝐾𝑠 + 𝐾𝑖 [𝐸]𝑇 𝐾𝑐𝑎𝑡 8) Multiply Numerator and Denominator by E [𝐸][𝑆] [𝑆] 𝑉𝑚𝑎𝑥 ( 𝐾𝑠 ) 𝐸 𝑉𝑚𝑎𝑥 𝐾𝑠 𝑣= = [𝑆] [𝐼] [𝐸][𝑆] [𝐸][𝐼] 1 + + (𝐸 + 𝐾𝑠 + 𝐾𝑖 ) 𝐸 𝐾𝑠 𝐾𝑖 9) Multiply Numerator and Denominator by Ks [𝑆] 𝑉𝑚𝑎𝑥 ( ) 𝐾𝑠 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝐾𝑠 𝑣= = = [𝑆] [𝐼] [𝐼] [𝐼] 𝐾𝑠 + [𝑆] + 𝐾𝑠 (1 + 𝐾𝑠 + 𝐾𝑖 ) 𝐾𝑠 𝐾𝑠 (1 + 𝐾𝑖 𝐾𝑖 ) + 𝑆 13) Obtain the double reciprocal [𝐼] 1 𝐾𝑠 1 1 = + (1 + ) 𝑣 𝑉𝑚𝑎𝑥 𝐾𝑖 [𝑆] 𝑉𝑚𝑎𝑥 Case 3: Uncompetitive Inhibition (Rapid equilibrium) Write down the kinetic model Steps 1) Write the rate- dependence equation v=[ES] kcat .…………………………(Eq 1) 2) Obtain the mass balance equation of the enzyme concentration [E]T= [E] + [ES] + [ESI] …………………….(Eq 2) 3) Define reaction intermediates ([ES]) in terms of [E], [S] Under rapid equilibrium: [𝐸][𝑆] 𝐾𝑠 = [𝐸𝑆] [ES]= 𝐾𝑖 = [𝐸][𝑆] …………………………..(Eq 3) 𝐾𝑠 [𝐸𝑆][𝐼] [𝐸𝑆𝐼] Substitute ES from Eq 3 𝐾𝑖 = [EI]= [𝐸][𝑆] [𝐼] 𝐾𝑠 [𝐸𝑆𝐼] = [𝐸][𝑆][𝐼] 𝐾𝑠𝐾𝑖 [𝐸][𝑆][𝐼] 𝐾𝑠 𝐾𝑖 …………………………..(Eq 4) 4) Define Vmax When [ES]= [E]T v=[E]T kcat .…………………….………(Eq 5) Do the math 5) Divide Eq 1 by Eq 2 𝑣 [E]T [𝐸𝑆] 𝑘𝑐𝑎𝑡 = [E]+ [ES]+[𝐸𝑆𝐼] 6) Substitute [ES] and [EI] from Eq 3 and Eq 4 𝑣 = [E]T [𝐸][𝑆] 𝐾𝑠 𝐾𝑐𝑎𝑡 [𝐸][𝑆] [𝐸][𝑆][𝐼] 𝐸 + 𝐾𝑠 + 𝐾𝑠 𝐾𝑖 7) Move [E]T to the numerator and substitute Vmax from Eq 4 [𝐸][𝑆] [𝐸][𝑆] [𝐸]𝑇 𝐾𝑐𝑎𝑡 𝑉𝑚𝑎𝑥 𝐾𝑠 𝐾𝑠 𝑣= = [𝐸][𝑆] [𝐸] [𝑆][𝐼] [𝐸][𝑆] [𝐸][𝑆][𝐼] 𝐸 + 𝐾𝑠 + 𝐾𝑠 𝐾𝑖 𝐸 + 𝐾𝑠 + 𝐾𝑠 𝐾𝑖 8) Multiply Numerator and Denominator by E [𝐸][𝑆] [𝑆] 𝑉𝑚𝑎𝑥 ( 𝐾𝑠 ) 𝐸 𝑉𝑚𝑎𝑥 𝐾𝑠 𝑣= = [𝑆] [𝑆][𝐼] [𝐸][𝑆] [𝐸][𝑆][𝐼] 1 + 𝐾𝑠 + 𝐾𝑠 𝐾𝑖 (𝐸 + 𝐾𝑠 + 𝐾𝑠 𝐾𝑖 ) 𝐸 9) Multiply Numerator and Denominator by Ks [𝑆] 𝑉𝑚𝑎𝑥 ( ) 𝐾𝑠 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝐾𝑠 𝑣= = = [𝑆] [𝑆][𝐼] [𝐼] [𝐼] 𝐾𝑠 + [𝑆] + 𝑆 𝐾𝑖 𝐾𝑠 + 𝑆 (1 + ) (1 + 𝐾𝑠 + ) 𝐾𝑠 𝐾𝑖 [𝐾𝑠] 𝐾𝑖 10) Obtain the double reciprocal [𝐼] 1 𝐾𝑠 1 1 = + (1 + ) 𝑣 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 𝐾𝑖 Case 4: Mixed Inhibition (Rapid equilibrium) Write down the kinetic model Steps 1) Write the rate- dependence equation v=[ES] kcat .…………………………(Eq 1) 2) Obtain the mass balance equation of the enzyme concentration [E]T= [E] + [ES] + [EI] + [ESI] …………………….(Eq 2) 3) Define reaction intermediates ([ES]) in terms of [E], [S] Under rapid equilibrium: [𝐸][𝑆] 𝐾𝑠 = [𝐸𝑆] [ES]= 𝐾𝑖 = [EI]= [𝐸][𝑆] [𝐸][𝐼] [𝐸𝐼] [𝐸][𝐼] …………………………..(Eq 4) 𝐾𝑖 𝑎𝐾𝑖 = [EI]= …………………………..(Eq 3) 𝐾𝑠 [𝐸][𝑆][𝐼] 𝑎𝐾𝑠 𝐾𝑖 [𝐸][𝑆][𝐼] 𝑎𝐾𝑠𝐾𝑖 …………………………..(Eq 5) 4) Define Vmax When [ES]= [E]T v=[E]T kcat .…………………….………(Eq 5) Do the math 5) Divide Eq 1 by Eq 2 𝑣 [E]T [𝐸𝑆] 𝑘𝑐𝑎𝑡 = [E]+ [ES]+[𝐸𝐼]+[𝐸𝑆𝐼] 6) Substitute [ES], [EI] and ESI from Eq 3 and Eq 4 [𝐸][𝑆] 𝑣 𝐾𝑠 𝐾𝑐𝑎𝑡 = [𝐸][𝑆] [𝐸][𝐼] [𝐸][𝑆][𝐼] [E]T 𝐸 + + + 𝐾𝑠 𝐾𝑖 𝑎𝐾𝑠 𝐾𝑖 7) Move [E]T to the numerator and substitute Vmax from Eq 4 [𝐸][𝑆] [𝐸][𝑆] [𝐸]𝑇 𝐾𝑐𝑎𝑡 𝑉𝑚𝑎𝑥 𝐾𝑠 𝐾𝑠 𝑣= = [𝐸][𝑆] [𝐸][𝐼] [𝐸] [𝑆][𝐼] [𝐸][𝑆] [𝐸][𝐼] [𝐸][𝑆][𝐼] 𝐸 + + + 𝐸 + + + 𝐾𝑠 𝐾𝑖 𝑎𝐾𝑠 𝐾𝑖 𝐾𝑠 𝐾𝑖 𝑎𝐾𝑠 𝐾𝑖 8) Multiply Numerator and Denominator by E [𝐸][𝑆] [𝑆] 𝑉𝑚𝑎𝑥 ( 𝐾𝑠 ) 𝐸 𝑉𝑚𝑎𝑥 𝐾𝑠 𝑣= = [𝐸][𝑆][𝐼] [𝑆] [𝐼] [𝑆][𝐼] [𝐸][𝑆] [𝐸][𝐼] 1 + 𝐾𝑠 + 𝐾𝑖 + 𝐾𝑠 𝐾𝑖 (𝐸 + + + )𝐸 𝐾𝑠 𝐾𝑖 𝑎𝐾𝑠 𝐾𝑖 9) Multiply Numerator and Denominator by Ks [𝑆] 𝑉𝑚𝑎𝑥 (𝐾𝑠 ) 𝐾𝑠 𝑉𝑚𝑎𝑥 [𝑆] 𝑉𝑚𝑎𝑥 [𝑆] 𝑣= = = [𝑆] [𝐼] [𝐼] [𝐼] [𝐼] [𝐼] [𝑆][𝐼] 𝐾𝑠 + [𝑆] + 𝐾𝑠 +𝑆 (1 + 𝐾𝑠 + 𝐾𝑖 + ) 𝐾𝑠 𝐾𝑖 𝑎𝐾𝑖 𝑆 (1 + 𝐾𝑖 ) + 𝐾𝑠 (1 + 𝑎𝐾𝑖 ) 𝑎[𝐾𝑠] 𝐾𝑖 10) Obtain the double reciprocal [𝐼] 1 [𝐼] 1 𝐾𝑠 1 = + (1 + ) (1 + ) 𝑣 𝑉𝑚𝑎𝑥 𝐾𝑖 [𝑆] 𝑉𝑚𝑎𝑥 𝑎𝐾𝑖
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Running head: ENZYME KINETICS

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Enzyme Kinetics
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Enzyme Kinetics

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Enzyme kinetics exercises
1. Derive the Michaelis –Menten equation for the next model. Assume rapid equilibrium.

Forward Rate Constant For The First Reaction Is K+1 Or Simply K1
Backward Rate Constant For The First Reaction Is Is K-1
Forward Rate Constant For The Second Reaction Is K+2 Or Simply K2
Backward Rate Constant For The Second Reaction Is Is K-2
We Assume That The Second Step...


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