Enzyme kinetics
 This topic has 15 replies, 8 voices, and was last updated 11 years, 2 months ago by fengjili.

AuthorPosts


October 4, 2006 at 1:52 pm #5893victorParticipant
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/minQuestion:
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? π I really need help on this…..

October 4, 2006 at 3:25 pm #55969oppoxParticipant
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 π ).
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 π

October 4, 2006 at 6:27 pm #55984MrMisteryParticipant
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 π

October 5, 2006 at 9:56 am #56013oppoxParticipant
enzymemaster? no no get your ruler out and draw oldschool π

October 6, 2006 at 2:26 am #56052Dr.SteinParticipant
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 π³ thus she told us how to make it π 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 π

October 6, 2006 at 9:58 am #56076oppoxParticipant
did you solve it? just make two columns, one with 1/[S] and one with 1/V and then just plot those values (reverse it so u get a positive line).

October 6, 2006 at 12:51 pm #56079victorParticipant
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… π

October 8, 2006 at 3:31 am #56148victorParticipant
ahahahaha….I got the answer…:lol: π π I think this answer should be worth of my 3 hours struggling with this equation…..:lol:
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 = 0.723×10^5
Km32 = 1.230×10^5
Km43 = 1.006×10^5
Km54 = 0.990×10^5So the average Km should be : 0.98725×10^5
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 = 44.942AVG 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)….

October 9, 2006 at 6:55 am #56196oppoxParticipant
then u know that in the future u can trust the LB plot and it is more simple to use then the average thing u did π

October 9, 2006 at 9:50 am #56210victorParticipant
Yup…my professor also said that I’m too dilligent to to those average thing, but afterall, she appreciated my work….:lol: then next time, I’ll use LB plot….more simple, no need counting and more accurate in predicting the enzyme kinetics…:lol:

December 11, 2008 at 4:31 pm #87727madmarycashParticipant
I don’t understand how you can get the Km and the Vmax by using the LB plot… π
I tried plotting the values and the line came out weird (not straight). 
December 12, 2008 at 4:17 am #87744MrMisteryParticipant
are you using a line plot instead of a scatter plot?

December 12, 2008 at 2:59 pm #87753blcr11Participant
You are plotting the reciprocals of v and s, right? The data given here looks pretty linear when you plot 1/v on the yaxis vs 1/s on the xaxis. (Plotting the raw numbers of v vs s will give you a sigmoidal plot; not at all linear.) If you have a calculator that will do linear regression, enter your values of (1/s,1/v) as your (x,y) pairs. The slope of the regression will be Vm while the yintercept is 1/Km. You can estimate them graphically if you prefer, but regression is a bit more accurate, plus you can estimate the errors in the parameters from the residual sums of squares – with a little math, anyway. Sometimes, the highest substrate concetrations (those at or near saturation levels) will give reciprocals that fall off the line. When that happens, you probably shouldn’t use those data points to estimate the kinetic parameters. They won’t be reliable.

December 12, 2008 at 6:11 pm #87757blcr11Participant
Sorry, that should be Xintercept = 1/Km, not yintercept. And the slope is Km/Vmax not Vmax. Memory didn’t serve me very well this time and I was too lazy to derive it from the rate expressionthough that’s a useful excercise if you’ve never done it before.
This link has a small, but straightforward graph of the rate law and the doublereciprocal (LineweaverBurk) plots.
http://www.graphpad.com/help/Prism5/pri … enzyme.htm

January 26, 2009 at 6:42 pm #88695setugoyalParticipant
the ques is pretty simple in fact….
1/v=Km/(Vm[S]) + 1/Vmso by plotting the 1/[V] by 1/[S] graph, we may find the intercept on y axis….which is 1/Vm in this case….and then do the reciprocal to find out Vmax…..and then put this value in the above equation…..at a particular [S] and v….and find the value of Km….:)

November 20, 2009 at 9:44 am #95141fengjiliParticipant
i think you can use Origin to make [V][S] graph
then use βnonlinear fittingβand choose βGrowth/Sigmoidal–Hilβ
fix n=1,fit then you will get the Vmax and K


AuthorPosts
 You must be logged in to reply to this topic.