[Lii]

So I've been working on this graph, and I now have a lineweaver-burk comparing an uninhibited solution to two inhibited solutions (with different concentrations of inhibitor). I'm supposed to now find the alpha values (through the equation Vo=(Vmax(S))/(αKm + α'(s))) but I cannot figure out what numbers to use. Previously if I had a problem using that equation, the numbers were given to me, but now I'm stumped :/ I tried plucking numbers from the graph, using an (S) value and the corresponding Vo, but the resulting numbers don't look right. Should I use the Vmax and Km of the uninhibited or the inhibited? And if it's the inhibited shouldn't it be related to the uninhibited somehow? I've been trying different numbers for quite awhile and I still don't get it.

[Babcock_Hall]

Have you tried to write the Lineweaver-Burk (double reciprocal) form of the equation you provided below?  Also, do I understand correctly that you were given more than one value each for velocity and for ?  Were you given particular values of also?

[Lii]

We had 2 different concentrations of inhibitor. For each inhibitor concentration, we have 6 values for (S) which correspond to 6 values for V. There is only one (I) value for these 6 values.

I have the 1/(S) plotted against the 1/V for my LB plot.

[Babcock_Hall]

If there are six values of S for one value of I, and if there are also six values of S for a second value of I, then you should be able to graph two straight lines, each with its own value of the slope and intercept.  If you have velocity versus substrate data for the uninhibited enzyme, then you actually have a third concentration of I, namely I = 0, and this provides a third straight line.  Can you reproduce here the form of the equation you are using?

Do you know the difference between competitive, uncompetitive, and noncompetitive inhibitions?  Your Michaelis-Menton equation could be taken to mean that the inhibitor is of one of these three types, but that inference is not certain (the values of alpha and alpha' will nail it down).  With respect to the question of how to find alpha and alpha', there are a couple of possibilities.  Are you expected to do this with the aid of graph paper, a calculator, or a statistics program such as ProStat?

[Lii]

By the graph it looks like a mixed inhibitor. I have all the straight lines in my LB plot, I just need to find α and α'.

I tried using Vo = Vmax(s) / (αKm + α'(S)) and solving for the alphas, but I couldn't figure out which (S) and Vo from the graph to use.

I'm using Excel for this.

[Babcock_Hall]

If one were doing this in actual research, one would use a program that could handle two independent variables.  However, I seem to recall a method of doing this calculation that involved replots of the slopes and intercepts.  This method produces two K_i values that are related to the two alpha values by a well-known formula.  See W. W. Cleland "Steady State Kinetics" in "The Enzymes" for some background. 

You should have three sets of slopes and three sets of intercepts, corresponding to the three values of that you have (including I = 0).  Now plot the slopes versus .  The negative of the x-intercept of this replot is K_is.  Next plot the intercepts versus .  The negative of the x-intercept of this replot is K_ii.  The slopes and intercepts could also be used to calculate the alphas without calculating the K_i values, I suppose.

[Lii]

I might be getting somewhere... Should alpha and alpha' be constant, or should they change with the changing concentration of inhibitor?

[Babcock_Hall]

First let me correct my previous message, which had an unintended formatting error.  I forgot that when using I's or B's in square brackets without spaces, the software interprets them as italics or bolding commands.  Corrected version of the second paragraph:

You should have three sets of slopes and three sets of intercepts, corresponding to the three values of that you have (including I = 0).  Now plot the slopes versus [ I ].  The negative of the x-intercept of this replot is K_is.  Next plot the intercepts versus [ I ].  The negative of the x-intercept of this replot is K_ii.  The slopes and intercepts could also be used to calculate the alphas without calculating the K_i values, I suppose.

Now with respect to your question, no*, the alpha values are constants.  If I did my algebra correctly,
alpha = (1 + [ I ]/K_is) and alpha' = (1 + [ I ]/K_ii).
EDT
*This statement is in error.  The alpha values are not constant.  See message below.

[Lii]

Hm, my thought was since it isn't competitive inhibition (there are different Vmax's) then the apparent Vmax should equal Vmax/α' (as it does in uncompetitive and mixed inhibition). But when I tried to calculate the α' from that equation I got 2 different numbers depending on the concentration...

[Babcock_Hall]

Have you tried the replot method?  I would be tempted to define each of the three intercepts as an apparent 1/V_max, which should change as [ I ] changes.  Of course the intercept with [ I ] = 0 could be thought of as the true 1/V_max.

Try inverting the equation you provided in the first post.  If I did my algebra correctly, the intercepts are (1/V_max)(1 + [ I ]/K_ii) which is equal to alpha'/V_max

Now that I think about it, yes, alpha and alpha' should change.  Besides the number 1, they consist of one constant and one variable, each.  I did not think of the concentration of the inhibitor before, only the inhibition constant; it is getting late here, and I did not think it through fully.

[Lii]

It's real data, and pretty bad data, I think :/

I replotted the graphs like you said, but now for finding alpha and alpha' where it's 1+(I/Ki), which of the three [ I ] am I supposed to use?

[Babcock_Hall]

With respect to the replots:  One plot should be of the slope of the original Lineweaver-Burk plot on the y-axis versus [ I ] on the x-axis, with n = 3, where n = the number of (x, y) pairs.  The other plot should be identical, except it should be of the intercept of the original L-B plot on the y-axis.  These two graphs are straight lines as long as the inhibition is linear, and they should have positive slopes.
EDT
I may have misunderstood your question.  Each nonzero value of [ I ] should produce a different value of alpha, and I really can't see why one alpha is better than another.  (Perhaps that is why I originally approached this problem in terms of the dissociation constants, which are what one would report in a paper.)  When [ I ] = 0, alpha or alpha' = 1, and you obtain the true values of 1/V_max and 1/K_m. 

[Lii]

Right, I plotted the I values against the slopes. I also plotted the I values against the intercepts.

But now in the equation of 1+(I/Ki), where Ki is the negative x-intercept of the line, is the I value the slope of the line?

EDIT: Oh... so the values of alpha are different depending on which I value I use?

[Babcock_Hall]

Yes the alpha values will be different for different values of [ I ].  Did you put the [ I ] along the x-axis or the y-axis?

[Lii]

I put the [ I ] along the x-axis.