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# The Km of a Michaelis Menten enzyme for a substrate is 1*10^-4M. At substrate conc. of 0.2M vo=43uM/min for a certain enzyme concentration. However with a substrate conc. of 0.02, vo has the same value. Show that this observation is accurate?

Michaelis-Menten Expression (MME): V0= Vmax*[S] / (Km + [S]) ——————————————————— We know regardless of [S], Km= 1x10^-4 for this enzyme. Observation A ([S] = 0.2M), MME: 4.3x10^4 = 0.2*Vmax_a / (1x10^-4 + 0.2) Observation B ([S] = 0.02M), MME: 4.3x10^4 = 0.02*Vmax_b / (1x10^-4 + 0.02) ——————————————————— Therefore: 0.2*Vmax_a / (1x10^-4 + 0.2) = 0.02*Vmax_b / (1x10^-4 + 0.02) If you simplify, you get: Vmax_a/Vmax_b = 0.2001/0.201 = 1.00 (3.s.f) Therefore the Vmax is almost identical in both circumstances. ——————————————————— Both circumstances use the same Vmax and Km values to achieve the same V0 therefore the observation is feasible and accurate. We can derive from this that substrate concentrations at, and over, [S]=0.02M do not result in an increase in V0; the Vmax has been reached.

Ali Rajani
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