Question

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?

Answer 1

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

`1/V_0 = underbrace(K_m/V_"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

`1/V_0 = K_m/V_"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 + ([I])/K_I^(')`, but the slope remains unchanged.

![https://biochemanics.wordpress.com/2013/04/07/reversible-inhibition/]()

When a competitive inhibitor is present, the Linewaver-Burk equation becomes

`1/V_0 = ((1 + ([I])/K_I) * K_m)/V_"max" * 1/([S]) + 1/V_"max"`

This time, the slope of the line changes by a factor of `1 + ([I])/K_I`, but the y-intercept remains unchanged.

![http://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Competitive_Inhibitor]()

Finally, when a mixed inhibitor* is present, the equation becomes

`1/V_0 = ((1 + ([I])/K_I) * K_m)/V_"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 + ([I])/K_I`, and the latter by a factor of `1 + ([I])/K_I^(')`.

![https://daniellaharewood.wordpress.com/2013/04/14/227/]()

SIDE NOTE The terms `alpha` and `alpha^(')` are actually a different notation for `1 + ([I])/K_I` and for `1 + ([I])/K_I^(')`, respectively.