Simple enzymatic reaction


 

Problem 1 (25 points):
A simple enzymatic reaction:

exhibits a Lineweaver-Burk line that goes through the following coordinates of two points: (0,0.2) and (1,0.8).
Let s0 be the initial concentration of S, c0 be the initial concentration of C and e0 be the initial concentration of the enzyme, E. We also know that k1 = 10k-1, s0 = 100, e0 = 10, c0 =0 and p0 =0.
Find the only correct statement (below) and explain or demonstrate why.
k1 ≈ 1.724
Vmax = 0.2
Km = 3
Km = 5
Develop the dynamical formulas for this reaction and simulate the full enzyme reaction (complex formation phase in one simulation done over a short time horizon, followed by the full product formation simulation done over a longer time horizon).
Finally develop the Michaelis – Menten equation for this reaction and plot this result against the full dynamical formulas on the same plot. Plot the graphics, as before, in the short-time horizon and long-time horizon). Explain the differences you see and why they exist.

Problem 2 (25 points):

You are investigating a new enzyme through Michaelis Menten kinetics, and you have found that it has a maximum velocity of 6.0 μM/s when the [E] is 1.5 x 10-9 M based on the following data:

Initial velocity (V0) [S]0
1.0 μM/s 20 mM
2.0 μM/s 50 mM
3.0 μM/s 100 mM
4.0 μM/s 200 mM
5.0 μM/s 500 mM

Determine Km of this enzymatic reaction (Hint: before going to do a lot of calculations, STOP! THINK! – if it doesn’t come to you, go ahead and do a lot of calculations. I suggest MS Excel for the calculations)
Plot the full Michaelis-Menten reaction using either Matlab or Simulink

 

Problem 3 (25 points): (Multiple Choice Questions requiring Explanations)
Which statements about enzyme catalyzed reactions are NOT true? Explain your answer and explain why the statements that are not true are false (Remember that part of your grade is technical writing skills).
Enzymes form complexes with their substrates
Enzymes lower the energy of activation for chemical reactions
Enzymes change the equilibrium constant (Keq or Kss) for chemical reactions
Many enzymes undergo conformational changes when binding with substrate
For enzymatic reactions to occur at the “active site” of enzymatic proteins, the 3D orientation of substrate molecules is an important feature of the reaction
Enzyme inhibition can always be overcome by adding more substrate
Problem 4 (25 points)
Consider the following set of data and answer the following questions:

a. Plot the data (the velocities are the initial velocities) on a Lineweaver-Burk plot (be sure to label axes)
b. Determine the Km of the reaction without an inhibitor
c. Determine the Vmax of the reaction without an inhibitor
d. The second set of velocities represents the rate of the reaction when
an inhibitor is added. Plot these data on the same graph as above
and determine the new Km and Vmax and the type of inhibitor
(competitive, non-competitive).
e. Can the effects of the inhibitor be over-ridden by adding more
substrate? Why?

 

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