From the following data on an enzymatic reaction based on competitve inhibition,
ID: 853818 • Letter: F
Question
From the following data on an enzymatic reaction based on competitve inhibition, determine (a) Km for the enzyme and (b) K1 for the inhibitor-enzyme comlpex.
substrarate concentration produce per hour, ug
mM no inhibitor 6mM inhibitor
2 139 88
3 179 121
4 213 149
10 313 257
15 379 313
Explanation / Answer
Km is expressed in units of concentration, usually in Molar units. Km is the concentration of substrate that leads to half-maximal velocity. Km = 1/2 Vmax
Velocity = V = Vmax*[S] / ([S]+Km)
Vmax is the limiting velocity as substrate concentrations get very large. Vmax (and V) are expressed in units of product formed per time. If you know the molar concentration of enzyme, you can divide the observed velocity by the concentration of enzyme sites in the assay, and express Vmax as units of moles of product formed per second per mole of enzyme sites. This is the turnover number, the number of molecules of substrate converted to product by one enzyme site per second. In defining enzyme concentration, distinguish the concentration of enzyme molecules and concentration of enzyme sites (if the enzyme is a dimer with two active sites, the molar concentration of sites is twice the molar concentration of enzyme).
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