# What does a mixed inhibitor (as opposed to a competitive or non/uncompetitive inhibitor) do to the slope and y-intercept of a Lineweaver-Burk, or Double-Reciprocal plot?

Apr 10, 2015

A mixed inhibitor changes both the slope and the y-intercept of a double-reciprocal plot.

In this case, you have to look at the equation that describes the Lineweaver-Burk plot for all the cases mentioned and figure out which one corresponds to the given criteria.

When no inhibitor is present, the Lineweaver-Burk equation looks like this

$\frac{1}{V} _ 0 = \underbrace{{K}_{m} / {V}_{\text{max")_(color(blue)("slope")) * 1/([S]) + underbrace(1/V_"max")_(color(green)("y-intercept}}}$

Now, when an uncompetitive inhibitor is present, the Lineweaver-Burk equation becomes

$\frac{1}{V} _ 0 = {K}_{m} / {V}_{\text{max" * 1/([S]) + (1 + ([I])/K_I^('))/V_"max}}$

As you can see, the y-intercept changes, i.e. it goes up by a factor of $1 + \frac{\left[I\right]}{K} _ {I}^{'}$, but the slope remains unchanged. When a competitive inhibitor is present, the Linewaver-Burk equation becomes

$\frac{1}{V} _ 0 = \frac{\left(1 + \frac{\left[I\right]}{K} _ I\right) \cdot {K}_{m}}{V} _ \text{max" * 1/([S]) + 1/V_"max}$

This time, the slope of the line changes by a factor of $1 + \frac{\left[I\right]}{K} _ I$, but the y-intercept remains unchanged. Finally, when a mixed inhibitor* is present, the equation becomes

$\frac{1}{V} _ 0 = \frac{\left(1 + \frac{\left[I\right]}{K} _ I\right) \cdot {K}_{m}}{V} _ \text{max" * 1/([S]) + (1 + ([I])/K_I^('))/V_"max}$

Now both the slope and the y-intercept change, the former by a factor of $1 + \frac{\left[I\right]}{K} _ I$, and the latter by a factor of $1 + \frac{\left[I\right]}{K} _ {I}^{'}$. SIDE NOTE The terms $\alpha$ and ${\alpha}^{'}$ are actually a different notation for $1 + \frac{\left[I\right]}{K} _ I$ and for $1 + \frac{\left[I\right]}{K} _ {I}^{'}$, respectively.