An reaction follows michaelis menten enzyme kinetics. Km=.001 M, and the initial concentration of substrate=0.840993 M. After 2 seconds, 5% of the substrate has been converted to product.

 

How much CO2 will be converted after 10, 30, and 60 seconds?

 

I know the answer is 25% after 10 seconds, 75% after 30 seconds, and 100% after 50 seconds, but I don't know how to get these numbers using the m-m equations: v=d[P]/dt=Vmax*/ +Km

Just looking at the answers, it seems like the reaction rate just stays the same, and 5% is converted for each 2 seconds the reaction goes. However, I feel like this a huge simplification...what was the point of including Km and the initial substrate concentration if this was the case? Is there a way to do the problem that utilizes the m-m equation?

Thanks so much!

This is probably too late but like a working example??

 

 

 

[math]

 

\begin{array}{lcl}

 

R_0 = \frac{k_2_0[E]_0}{_0 + K_m} & \; & ( \; 1.1 \; ) \: Michaelis - Menten \; Rate \; Law \\

 

\\ \\ \\

 

Where \; _0 >> K_m \; then \; R_0 = k_2[E]_0 = R_{max} & \; & ( \; 1.2 \; ) \\

 

\\ \\ \\

 

\frac{1}{R_0} = \frac{1}{R_{max}} + \frac{K_m}{R_{max}} \frac{1}{_0} & \; & ( \; 1.3 \; ) \: Lineweaver - Burk \; Equation

 

\end{array}

 

[/math]

If you work it out your v and Vmax should be relatively similar.

 

Just work the % conversion rate they gave you to M/second and calculate your Vmax, you can then work from there just using the times that they gave you. i.e. If xM/second then how much in 10 seconds? And then just convert that back to percentage.

Hi I am new in here, could you please help in how I can find the Vo when I know only the time and the absorbance for an enzyme?

A good place to start is to define what you mean by V_0. Then you can see what information is needed.