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Report (any type)/Brief report
22/09/2021
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″https://studydaddy.com/”v_0## vs. ##1″https://studydaddy.com/”[S]##, you have a slope of ##K_M”https://studydaddy.com/”v_(max)## and a y-intercept of ##1″https://studydaddy.com/”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’)##.
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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