**Note: Please be sure all of the plots generated are formatted properly. Axis labels, units, figure legend, caption are required for full credit on all plots. Do not use chart titles but include figure numbers and captions.**

**[note that i already drives the differential l equation fro S ES p ALL U NEED TO DO IS FIND THE PLOT AND ANSWER PART A B c**

**i alredy wrote the code but there some eroor with and what i need help with and also with answering the other parts]**

- a)Solve the differential equations for [S], [ES], [P] using MATLAB.

Let k1 = 10,000 (M-s)-1, k-1 = 10 s-1, and k2 = 100 s-1. The initial enzyme concentration (Eo) is 0.5 mM and the initial substrate concentration is 2 mM.Note that you will solve for [S], [ES], and [P] using ODE45.Plot [S] vs. t and [P] vs. t

Hint: substitute [E]= [Eo] –[ES] in differential equation 1 and 2 solved in class to get to the equations that will be used in MATLAB

At time t=0, [P]=0, [S]= 2 mM, and [ES]=0

b) Use the code written above to calculate product at three different initial substrate concentrations.

[S]= 2, 1, 0.5 mM. Plot [P] vs. t graphs for all three concentrations on the same graph. Compare the [P] curve from each substrate concentration and comment on the trend.

c)Also plot the Michaelis-Menten relationship for rate of product formation as a function of substrate concentration using the k values above. Use initial substrate concentration=2*10^-3:0.001:1;

Plot three curves with three different k1 and k-1 conditions on the same graph.

- k1= 10000, k-1= 1000
- k1=10000, k-1=100
- k1=10000, k-1=10

Discuss why change in k-1 changes the rate of product formation.

## Solution:

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