Enzyme kinetics

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    • #5893
      victor
      Participant

      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 (Lineweaver-Burk) 1/v vs 1/[S].

      My answer:
      I use the Michaelis-Menten 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? πŸ˜• I really need help on this…..

    • #55969
      oppox
      Participant

      you cant say that Vmax is 40.5 because you dont know what happens when [S] gets larger.

      You have to plot it (L-burke 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 michaelis-menten kinetics (obviously it does, because otherwise u would have a mean teacher πŸ™‚ ).

      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 πŸ™‚

    • #55984
      MrMistery
      Participant

      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 πŸ˜€

    • #56013
      oppox
      Participant

      enzymemaster? no no get your ruler out and draw oldschool πŸ™‚

    • #56052
      Dr.Stein
      Participant

      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 πŸ˜† 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 πŸ˜‰

    • #56076
      oppox
      Participant

      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).

    • #56079
      victor
      Participant

      I got the Lineweaver-Burk 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 Michaelis-Menten 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… πŸ™

    • #56148
      victor
      Participant

      ahahahaha….I got the answer…:lol: πŸ˜† πŸ˜† 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^-5

      So 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 = 44.942

      AVG Vmax = 44.4658

      I’ve also made the double-reciprocal (Lineweaver-Burk) 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)…. :mrgreen:

    • #56196
      oppox
      Participant

      then u know that in the future u can trust the L-B plot and it is more simple to use then the average thing u did πŸ˜‰

    • #56210
      victor
      Participant

      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 L-B plot….more simple, no need counting and more accurate in predicting the enzyme kinetics…:lol:

    • #87727
      madmarycash
      Participant

      I don’t understand how you can get the Km and the Vmax by using the L-B plot… πŸ™
      I tried plotting the values and the line came out weird (not straight).

    • #87744
      MrMistery
      Participant

      are you using a line plot instead of a scatter plot?

    • #87753
      blcr11
      Participant

      You are plotting the reciprocals of v and s, right? The data given here looks pretty linear when you plot 1/v on the y-axis vs 1/s on the x-axis. (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 y-intercept 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.

    • #87757
      blcr11
      Participant

      Sorry, that should be X-intercept = -1/Km, not y-intercept. 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 expression-though 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 double-reciprocal (Lineweaver-Burk) plots.

      http://www.graphpad.com/help/Prism5/pri … enzyme.htm

    • #88695
      setugoyal
      Participant

      the ques is pretty simple in fact….
      1/v=Km/(Vm[S]) + 1/Vm

      so 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….:)

    • #95141
      fengjili
      Participant

      i think you can use Origin to make [V]-[S] graph
      then use β€œnonlinear fitting”and choose β€œGrowth/Sigmoidal–Hil”
      fix n=1,fit then you will get the Vmax and K

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