From the Michaelis-Menten model for an enzyme-catalyzed reaction, the amount of the substrate
present at time
satisfies the differential equation
where
m
and
M
are positive constants. Solving this, one obtains
where
0
is the (nonzero) amount at the beginning. To find
, for any given value of
, it is necessary to solve this nonlinear equation, and how this might be done is considered in this exercise. How to find the numerical solution of this differential equation is considered in Exercise 7.20.
(a) As a function of
, sketch the two functions
M
ln(/
0) and
0−
m
−. Do this for
= 0 and for
> 0. Use this to explain why there is only one solution of (2.30).
(b) Use part (a) to explain why the solution satisfies 0
≤
0.
(c) Suppose (2.30) is to be solved using Newton’s method. What does (2.10) reduce to in this case? Based on parts (a) and (b), what would be a good choice for the starting point in this case? Make sure to explain why.
(d) Based on parts (a) and (b), what would be a good choice for a starting interval when using the bisection method to solve (2.30)? Make sure to explain why.
(e) Suppose (2.30) is to be solved using the secant method. What does (2.24) reduce to in this case? Based on parts (a) and (b), what would be a good choice for the two starting points in this case? Make sure to explain why.
(f) It is found that for sucrose,
m
= 0.76 mM/min and
M
= 16.7 mM [Johnson and Goody, 2011]. Also, assume that
0
= 100 mM. Use one of the above methods from (c)–(e) to find the value of
at
= 1 min, at
= 10 min, and at
= 100 min. Your answers should be correct to at least four significant digits. Make sure to state which method was used, why you made this choice, and what error condition you used to stop the calculation.
(g) Using the parameter values given in part (f), and one of the above numerical methods from (c)–(e), plot
for 0 ≤
≤ 1000.
(h) Explain how it is possible to produce the plot in part (g), from (2.30), without having to use a numerical solver to find
.
Exercise 7.20
This problem concerns the Michaelis-Menten equation
where
m
and
M
are positive constants. The initial condition is
(0) =
0. The biochemical applications of this equation, and how to solve it, were considered in Section 2.1.1 and in Exercise 2.17. In this exercise you are to solve this problem using the trapezoidal and RK4 methods.
(a) Solving this problem, one finds that the exact solution satisfies
Verify that if
satisfies the above equation, then it is a solution of the IVP.
In the following questions, assume that
m
= 0.76 mM/min,
M
= 16.7 mM, and
0
= 100 mM.
(b) Plot
for 0 ≤
≤ 500. An easy way to do this is to rewrite the equation in part (a) as
Picking values
j
, with 0
j
≤
0, the above equation can be used to find the corresponding
j
. With this, the requested plot can be obtained by plotting the values for the (
j
,
j
)’s.
(c) On the same axes, plot the exact and the two numerical solutions for 0 ≤
≤ 500 in the case of when
= 20.
(d) Redo (c) for
= 5,
= 10, and
= 40 (there should be one graph for each M).
(e) Plot the max error
∞
, defined in (7.44), as a function of
for each method, using
= 10, 20, 40, 80. The two curves should be in the same log-log plot.
(f) Compare the two methods based on your results from parts (c)–(e). This includes ease of use, speed of calculation, accuracy of results, and apparent stability.