Competitive inhibtuon, Vmax no inhibition = 2.94 (mol/L)/min; V max . 3mM inhibi

Question

Competitive inhibtuon, Vmax no inhibition = 2.94 (mol/L)/min; V max . 3mM inhibitor = 2 38 (mol/l,)/min; V max . 5mM Inhibitor = 2.0 (mol/L)/min B) mixed inhibition. Vmax no inhibition = 2 (mol/LJ/min; VM 3mM inhibitor = 2.38(mol/L)/min; Vm 5mM Inhibitor = 2.94(mol/L)/min pufcHtt-compctitive inhibition Vmax no inhibition c 2.0(mol/L)/min; V max 3mM inhibitor = 2.38(mol/L)/min; V max = 5mM inhibitor = 2.94 (mol/L)/min D) competitive Jnhihumn, Vmax no inhibition = 2.0 (mol/L)/min; 3mM lnhlbitm = 2.38 (mol/LJ/min: V max 5mM inhibitor = 2.94 (m o l/ l) /m in L; ) pure non-compeliliw i nhibition Vmax no inhibition = 2.94 (mol/L)/mm; 3mM inhibitor =2.38 (mol/L)/min; V max . 5mM inhibitor = 2.0 (m o l/L )/m in

Explanation / Answer

The above shown results are (in graph) for,

(i) Non-competitive inhibitor : Non-competitive inhibitors have same Km but different Vmax which is the case here.

(ii) Vmax for each are thus,

E) pure non-competitive inhibition. Vmax no inhibition = 2.94 mol/L/min; Vmax 3mM inhibitor = 2.38 mol/L/min; Vmax 5mM inhibitor = 2.0 mol/L/min

Reactions with no-inhibitors will have a higher Vmax in case of non-competitive inhibition scenario.

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