Eadie-Hofstee plot

[biology_06er]

Hi there

Is the Eadie-Hofstee plot just another way of representing the Lineweaver-Burk 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 y-axis is 1/v and the x-axis 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

[sdekivit]

Hi there

Is the Eadie-Hofstee plot just another way of representing the Lineweaver-Burk 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 y-axis is 1/v and the x-axis 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

Yes, the Eadie-Hofstee-plot is another form of the Lineweaver-Burk-plot that will result in straight line.

To answer your questions, we're going to do some maths using the Michaelis-Menten 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 LB-plot? Let's first derive it using the general equation for a straight line:

y = ax + b

The LB-plot 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 LB-plot has as the slope Kd/Bmax and the intersect with the 1/B-axis 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 NON-competitive 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/B-axis.

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 Eadie-Hofstee-plot (B/L against B). Another method you didn't mention is the Hanes-plot (L/B against L)

[lightbeing]

I 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 HOFSTEE-EADIE 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.

[adaffara]

What is the Km value from a EAdie-Hofstee plot when the Vmax is 0.0049 and the Vmax/Km is 2.11e-4? Thanks!

[JackBean]

What is the Km value from a EAdie-Hofstee plot when the Vmax is 0.0049 and the Vmax/Km is 2.11e-4? Thanks!

To hard to divide?