Task on the Michaelis-Menten equation?
KM Hexokinase: 1mM, KM Glucokinase: 11mM, Substrate concentration: 1mM How much higher is the rate of hexokinase?
(A) 2x (B) 4xM (C) 6x (D) 8x (E) 10x
If I convert the Michaelis-Menten equation to vmax, I get a value of 2 for hexokinase and 12 for glucokinase. This also gives me the correct answer, which is C, i.e., 6 times faster. The only faster enzyme here would actually be glucokinase, not hexokinase. On the other hand, this cannot be true, since the enzyme with a smaller Michaelis-Menten constant must be faster.
(2 votes)
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V = (v)max. * [S])/(KM + [S]
For hexokinase:
V = (v)max. * 1 mM)/(1 mM + 1 mM) = vmax. * 1 mM)/2 mM = 1/2*vmax.
For the glucokinase:
V = (v)max. * 1 mM)/(11 mM + 1 mM) = vmax. * 1 mM)/12 mM = 1/12 * vmax.
Under the assumption that vmax. is identical for both enzymes, the enzyme reaction in the hexokinase would then be 6-fold faster, since 1/2=6*1/12 is.
The answer is C, i.e. 6x faster for hexokinase compared to glucokinase, The fact that the glucokinase has a higher KM means that it works more slowly at a constant substrate concentration and the hexokinase works faster due to its lower KM The vmax values support this observation
Okay, just how is the solution/acquisition route?