How do I make a Lineweaver-Burk (Double Reciprocal) Plot?

The equation for the Lineweaver-Burk plot is gotten by doing as the alternative name suggests... taking the reciprocal.

GENERAL REACTION

#\mathbf(E + S stackrel(k_1)(rightleftharpoons) ES stackrel(k_2)(->) E + P)#
#color(white)(aaaaaa)^(\mathbf(k_(-1)))#

LINEWEAVER-BURK PLOTS WITHOUT INHIBITOR

#\mathbf(v_0 = (v_max[S])/(K_M + [S]))#

where #v_max = k_2[E]_"total"#, #k_2# is the observed rate constant for the conversion of the enzyme-substrate complex to the free enzyme and the product, and #[E]_"total"# is the total concentration of the enzyme (free, complexed, whatever).

So naturally, you reciprocate as follows:

#1/(v_0) = (K_M + [S])/(v_max[S])#

#1/(v_0) = (K_M)/(v_max[S]) + cancel([S])/(v_maxcancel([S]))#

#color(blue)(1/(v_0) = (K_M)/(v_max)1/([S]) + 1/(v_max))#

Once you plot #1"/"v_0# vs. #1"/"[S]#, you have a slope of #K_M"/"v_(max)# and a y-intercept of #1"/"v_(max)#. You can solve it from there.

Of course, this is assuming that there is no inhibitor. If there is an inhibitor, then you can have either of the following reactions:

ENZYME INHIBITION

#\mathbf(E + I rightleftharpoons EI)#

#K_I = ([E][I])/([EI])#

where #K_I# is the dissociation constant for the #EI# complex into the free enzyme and the inhibitor.

#\mathbf(ES + I rightleftharpoons ESI)#

#K_I' = ([ES][I])/([ESI])#

where #K_I'# is the dissociation constant for the #ESI# complex into the #ES# complex and the inhibitor.

The resultant Lineweaver-Burk equations are still basically identical, other than the fact that now we'd use #K_M^"app"# and #v_max^"app"#, which have different definitions in each case and are used in place of #K_M# and #v_max#, respectively.

LINEWEAVER-BURK EQUATIONS FOR INHIBITION

Competitive Inhibition (binds only to free enzyme):

#color(blue)(1/(v_0) = (alphaK_M)/(v_max)1/([S]) + 1/(v_max))#

where #alpha = 1 + ([I])/(K_I)#.

Note that here, #K_M^"app" = alphaK_M#, and #v_max^"app" = v_max#.

Uncompetitive Inhibition (binds only to #ES# complex):

#color(blue)(1/(v_0) = (K_M)/(v_max)1/([S]) + alpha/(v_max))#

where #alpha = 1 + ([I])/(K_I)#.

Note that here, #K_M^"app" = (K_M)/(alpha)#, and #v_max^"app" = (v_max)/(alpha)#.

Pure Non-Competitive Inhibition (binds onto enzyme and #ES# complex with equal affinity):

#color(blue)(1/(v_0) = (alphaK_M)/(v_max)1/([S]) + (alpha)/(v_max))#

where #alpha = 1 + ([I])/(K_I)#.

Note that here, #K_M^"app" = K_M#, and #v_max^"app" = (v_max)/(alpha)#.

Mixed Non-Competitive Inhibition (binds onto enzyme and #ES# complex with different affinities):

#color(blue)(1/(v_0) = (alphaK_M)/(v_max)1/([S]) + (alpha')/(v_max))#

where #alpha = 1 + ([I])/(K_I)# and #alpha' = 1 + ([I])/(K_I')#.

Note that here, #K_M^"app" = (alphaK_M)/(alpha')#, and #v_max^"app" = (v_max)/(alpha')#.