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?

1 Answer
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

#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.


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.


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^(')#.


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