Enzyme kinetics
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 victor
 King Cobra
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Enzyme kinetics
I was given this question by my professor:
1. [S] 3 μM  v = 10.4 μmol/min
2. [S] 5 μM  v = 14.5 μmol/min
3. [S] 10 μM  v = 22.5 μmol/min
4. [S] 30 μM  v = 33.8 μmol/min
5. [S] 90 μM  v = 40.5 μmol/min
Question:
a. what is the value of Vmax and Km?
b. make a double recripocal plot (LineweaverBurk) 1/v vs 1/[S].
My answer:
I use the MichaelisMenten equation to solve this problem, but I found difficulties in using it because there are two unknow factors which are Vmax and Km, thus it can't be solved by this equation. Then I wrote that Vmax as the highest rate among those 5 which is 40.5 μmol/min
. After that I solve the Km by using the equation where:
v = Vmax*[S] / Km + [S]
I put one of the test above in this equation (take example for test no.3) so:
22.5 = 40.5 * 10.10^6 M / Km + 10.10^6 M
then,
Km = 40.5*10^5  22.5*10^5 / 22.5
Km = 8 μM (the unit of Km is in concentration unit)
Am I right in solving this equation? I really need help on this.....
1. [S] 3 μM  v = 10.4 μmol/min
2. [S] 5 μM  v = 14.5 μmol/min
3. [S] 10 μM  v = 22.5 μmol/min
4. [S] 30 μM  v = 33.8 μmol/min
5. [S] 90 μM  v = 40.5 μmol/min
Question:
a. what is the value of Vmax and Km?
b. make a double recripocal plot (LineweaverBurk) 1/v vs 1/[S].
My answer:
I use the MichaelisMenten equation to solve this problem, but I found difficulties in using it because there are two unknow factors which are Vmax and Km, thus it can't be solved by this equation. Then I wrote that Vmax as the highest rate among those 5 which is 40.5 μmol/min
. After that I solve the Km by using the equation where:
v = Vmax*[S] / Km + [S]
I put one of the test above in this equation (take example for test no.3) so:
22.5 = 40.5 * 10.10^6 M / Km + 10.10^6 M
then,
Km = 40.5*10^5  22.5*10^5 / 22.5
Km = 8 μM (the unit of Km is in concentration unit)
Am I right in solving this equation? I really need help on this.....
Q: Why are chemists great for solving problems?
A: They have all the solutions.
A: They have all the solutions.
you cant say that Vmax is 40.5 because you dont know what happens when [S] gets larger.
You have to plot it (Lburke plot), u cant solve it with the menten equation because Vmax is approached asymptotically and u dont get a difinitive value.
if u plot it u get a straight line, if it obey michaelismenten kinetics (obviously it does, because otherwise u would have a mean teacher ).
the slope and the two intercepts will give u the answer. (just to not answer your question completly)
I may be wrong about why u cant solve it, but im pretty sure this is the way to solve it. Good luck
You have to plot it (Lburke plot), u cant solve it with the menten equation because Vmax is approached asymptotically and u dont get a difinitive value.
if u plot it u get a straight line, if it obey michaelismenten kinetics (obviously it does, because otherwise u would have a mean teacher ).
the slope and the two intercepts will give u the answer. (just to not answer your question completly)
I may be wrong about why u cant solve it, but im pretty sure this is the way to solve it. Good luck
 MrMistery
 Inland Taipan
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yeah, i don't get what the problem is. draw the graph and get the menten constant and Vmax from there. or better yet use EnzymeMaster or a similar program and have the computer do it for you
"As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife"  J. Craig Venter
 Dr.Stein
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As I told you yesterday in my lab, Mrs. Soekarti ever gave me this question for Basic Biochemistry Class of my Postgraduate Program. None of eleven students including me could make the correct answer thus she told us how to make it Unfortunately, I have no idea where I keep my exercise book for this. I will try to search for it, when I find it I will tell you as soon as possible
 victor
 King Cobra
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 Joined: Sat Apr 30, 2005 12:01 pm
 Location: Yogyakarta, Indonesia..
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I got the LineweaverBurk graph, but I get confused with the intercept line (y line) which said is showing the rate of Vmax of the reaction. When I plot the graph, I found that the highest v (not Vmax) which is 40.5 is not placed on the intercepting y line.....so, I think that 40.5 is not the Vmax yet.
But my professor said that I can find the both Vmax and Km by using the MichaelisMenten equation. She said that, by comparing two reaction on the question and substituting here and there, we can obtain Vmax or Km....but, until now, I still can't solve it....ahh, I have a really bad math...
But my professor said that I can find the both Vmax and Km by using the MichaelisMenten equation. She said that, by comparing two reaction on the question and substituting here and there, we can obtain Vmax or Km....but, until now, I still can't solve it....ahh, I have a really bad math...
Q: Why are chemists great for solving problems?
A: They have all the solutions.
A: They have all the solutions.
 victor
 King Cobra
 Posts: 2668
 Joined: Sat Apr 30, 2005 12:01 pm
 Location: Yogyakarta, Indonesia..
 Contact:
ahahahaha....I got the answer... I think this answer should be worth of my 3 hours struggling with this equation.....
So, the Vmax is 44.4658
I do it by using the elimination method by comparing 2 reactions which is 2;1, 3;2, 4;3, and 5;4. then I get each Km for those reaction I compared.
Km21 = 0.723x10^5
Km32 = 1.230x10^5
Km43 = 1.006x10^5
Km54 = 0.990x10^5
So the average Km should be : 0.98725x10^5
Then, by using this average Km, I insert it into each reaction equation to obtain each Vmax and then the average Vmax can be obtained.
Vmax1 = 44.624
Vmax2 = 43.130
Vmax3 = 44.713
Vmax4 = 44.923
Vmax5 = 44.942
AVG Vmax = 44.4658
I've also made the doublereciprocal (LineweaverBurk) graph and I got that the y intercept line (which means the Vmax) is about 1/44 (1/v). so, say it...am I correct about this question? (Ihope so)....
So, the Vmax is 44.4658
I do it by using the elimination method by comparing 2 reactions which is 2;1, 3;2, 4;3, and 5;4. then I get each Km for those reaction I compared.
Km21 = 0.723x10^5
Km32 = 1.230x10^5
Km43 = 1.006x10^5
Km54 = 0.990x10^5
So the average Km should be : 0.98725x10^5
Then, by using this average Km, I insert it into each reaction equation to obtain each Vmax and then the average Vmax can be obtained.
Vmax1 = 44.624
Vmax2 = 43.130
Vmax3 = 44.713
Vmax4 = 44.923
Vmax5 = 44.942
AVG Vmax = 44.4658
I've also made the doublereciprocal (LineweaverBurk) graph and I got that the y intercept line (which means the Vmax) is about 1/44 (1/v). so, say it...am I correct about this question? (Ihope so)....
Q: Why are chemists great for solving problems?
A: They have all the solutions.
A: They have all the solutions.
 victor
 King Cobra
 Posts: 2668
 Joined: Sat Apr 30, 2005 12:01 pm
 Location: Yogyakarta, Indonesia..
 Contact:
Yup...my professor also said that I'm too dilligent to to those average thing, but afterall, she appreciated my work.... then next time, I'll use LB plot....more simple, no need counting and more accurate in predicting the enzyme kinetics...
Q: Why are chemists great for solving problems?
A: They have all the solutions.
A: They have all the solutions.

 Garter
 Posts: 1
 Joined: Thu Dec 11, 2008 4:24 pm
Re: Enzyme kinetics
I don't understand how you can get the Km and the Vmax by using the LB plot...
I tried plotting the values and the line came out weird (not straight).
I tried plotting the values and the line came out weird (not straight).
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