From the data in the table, determine the following. Also determine if hydroxyme
ID: 59319 • Letter: F
Question
From the data in the table, determine the following. Also determine if hydroxymethlyaspartate is a competitive or noncompetitive inhibitor of the reaction.
Km for the uninhibited enzyme = uM
apparent Km for the inhibited enzyme = uM
the Vmax for the uninhibited enzyme = arbitrary units
the Vmax for the inhibited enzyme = arbitrary units
based on the above information the inhibitor is competitivenoncompetitive .
From the data in the table, determine the following. Also determine if hydroxymethlyaspartate is a competitive or noncompetitive inhibitor of the reaction.
Km for the uninhibited enzyme = uM
apparent Km for the inhibited enzyme = uM
the Vmax for the uninhibited enzyme = arbitrary units
the Vmax for the inhibited enzyme = arbitrary units
based on the above information the inhibitor is competitivenoncompetitive .
[S](uM) V, no inhibitor
(arbitrary units) V, with inhibitor
(arbitrary units) 9 2.11E-2 7.59E-3 45 7.74E-2 3.36E-2 136 0.140 7.86E-2 227 0.167 0.107 454 0.195 0.147
Explanation / Answer
** It would be better to make a graph and answer the question. Anyways, I will calculate it theoritically.
1/Vo= 1/vmax +Km/Vmax * 1/S (without inhibitor)
Now, Vmax for inhibited enzyme is 0.147 ; Km will come greater than 136 micromole graphically
and Vmax for un-inhibited enzyme is 0.195. Km will come to be greater than 45 and less than 136 micromole graphically.
For unihibited enzyme,
1/.02 = 1/0.195 +Km / 0.195 * (1/9)
44.87 = Km/ 1.75
Km= 78.74 micromole
For inhibited enzyme, 1/0.007 = 1/0.147 + Km / 0.147 *(1/9)
142.85- 6.80 = Km/ 1.32 = 179.5 micromole
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