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# How does one calculate ##K_M## and ##v_max## from the 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')##.