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QUESTION

# 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').