Login

EadieHofstee plotModerator: BioTeam
5 posts
• Page 1 of 1
EadieHofstee plotHi there
Is the EadieHofstee plot just another way of representing the LineweaverBurk plot regarding inhibitors? Also when MORE inhibitor is present why on the LWBPlot does the gradient get steeper? I copied down the graph slopes in lectures but never really understood what they stood for..if the yaxis is 1/v and the xaxis is 1/[s] then it says slope =Km...is this right or have I copied it down wrong? Can someone please give me a brief description so I can get the gist of it and then understand it when I read the textbook Thanks biology_06er
Re: EadieHofstee plot
Yes, the EadieHofsteeplot is another form of the LineweaverBurkplot that will result in straight line. To answer your questions, we're going to do some maths using the MichaelisMenten kinetics: Assume we have an enzyme E and a ligand L. Then the following equilibrium occurs: E + L <> EL with dissociation constant Kd = [E][L]/[EL] We can rearrange this equation using [E]total = [E]free + [EL] = Bmax (maximum ability to bound) Now we multiply with [L]: Bmax[L] = [E][L] + [L][EL] > Bmax[L] = [E][L] * [EL]/[EL] + [L][EL] > Bmax[L] = [EL]*Kd + [EL][L] > thus the concentration of enzyme bound to its ligand is: [EL] = Bmax[L] / ( Kd + [L] ) Now let's go to the situation where an inhibitor is present. The inhibitor can also bind to the enzyme: E + I <> EI. Therefore the dissociation constant Kd for de ligand, the normal substrate for the enzyme, gets higher to a new value Kd* equal to Kd*(1 +[I]/Ki). What consequences does this have on the LBplot? Let's first derive it using the general equation for a straight line: y = ax + b The LBplot is the plot where 1/RL is plotted against 1/L where for convenience say RL = B from bound. We have the equation B = Bmax[L] / (Kd + [L]) and thus: y = Bmax * x / (Kd + x) Thus: 1/y = (Kd + x) / Bmax * x 1/y = Kd/ (Bmax * x) + 1/Bmax 1/y = 1/x * Kd/Bmax + 1/Bmax 1/B = 1/[L] * (Kd/Bmax) + 1/Bmax Thus the LBplot has as the slope Kd/Bmax and the intersect with the 1/Baxis is 1/Bmax and the intersect with the 1/[L]axis will be at the value 1/B is 0 hypothetically of course): 1/[L] * (Kd/Bmax) = Bmax 1/[L] = 1/Kd. Thus why does the gradient gets steaper when there is a competitive inhibitor present: A competitive inhibitor causes the dissociation constant to rise, since less substrate can bind. Thus: the new Kd, Kd* is higher then Kd: Kd* > Kd. The gradient is given by Kd / Bmax and because Kd = Kd* > Kd*/Bmax > Kd/Bmax and thus is the gradient steaper. Note that nothing happens with Bmax, because when [L] >> [I] (5 * [I]) B = Bmax. When a NONcompetitive inhibitor is present, Bmax* < Bmax and thus both the gradient and the intersect are affected: The gradient becomes steaper and the intersect lays higher on the 1/Baxis. I hope this bunch of maths helpes you to give insight in the enzyme kinetics. Note that the same is valid for drugs binding to a receptor or enzyme. Note that the same reasoning is valid to derive the so called Scatchard or EadieHofsteeplot (B/L against B). Another method you didn't mention is the Hanesplot (L/B against L)
Re: EadieHofstee plotI have to plot the Jmaxand the Kt of GLUT4 in the absence and presence of insulin. The problem is that I am unsure how to plot this on a HOFSTEEEADIE chart. I am desperate to know how I plot the information, as my assignment is due in to my tutor on the 20th of this month of May 2010.
Re: EadieHofstee plotWhat is the Km value from a EAdieHofstee plot when the Vmax is 0.0049 and the Vmax/Km is 2.11e4? Thanks!
Re: EadieHofstee plot
To hard to divide? http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
5 posts
• Page 1 of 1
Who is onlineUsers browsing this forum: No registered users and 1 guest 
© BiologyOnline.org. All Rights Reserved. Register  Login  About Us  Contact Us  Link to Us  Disclaimer & Privacy  Powered by CASPION