To study the dependence of the rate of an enzyme-catalyzed reaction on the subst

Question

To study the dependence of the rate of an enzyme-catalyzed reaction on the substrate concentration, a constant amount of enzyme was added to a series of reaction mixtures containing different concentrations of substrate. The following data were obtained for an enzyme in 10ml reaction mixtures.

[S](mol/L)   V0(umol/min)
   1 x 10-4       6
   2 x 10-4       11.5
   5 x 10-4       38
   8 x 10-4       52
   10 x 10-4       59
   15 x 10-4       72
   18 x 10-4       78
   25 x 10-4       88

a, By inspection, what is the Vmax for this concentration of enzyme? (Do not plot)

b. By inspection, what is the Km of this enzyme? (Do not plot)

There are no V0 values that are very close together to suggest a plateau at Vmax... I'm thinking Vmax would have to be 88, making Km the value at V0 of 44...but there isn't one listed

Explanation / Answer

The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction is Vmax and Km is the concentration of substrate which permits the enzyme to achieve half Vmax.

so,the Vmax is the rate as u suspected, that is 88.

Coming to the Km,

half V max is 44 and the concentration of substrate which permits this half Vmax is 8*10-4 mol/L(The Concentration listed in left hand side and corresponding to the nearest next value of the 44 V)

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