Given below are steady-state enzyme kinetic data (initial rates at various subst

Question

Given below are steady-state enzyme kinetic data (initial rates at various substrate concentrations) for a particular enzyme.

[S], mM           vo, mM/min

            1.5                      4.2

            3.0                      6.2

            5.0                      7.7

            8.0                      8.9

            12                       10.3

a. Use a Lineweaver-Burk (double reciprocal) plot to determine the values of Km and V from the above data – show your work, including a copy of the plot.

b. In the above experiment, the enzyme concentration used was 1.3 mM (1.3 x 10-6 M). Give the values of kcat (in s-1) and kcat/Km (in M-1s-1) for the enzyme in the above reaction.

Explanation / Answer

V= Vmax[S] /(KM+S)

1/V= (KM+S)/VmaxS

1/V=( 1/S)KM/Vmax +1/Vmax

so a plot of 1/V Vs 1/S ( Lineweaver-Burk plot) gives intercept of 1/Vmax and slope of KM/Vmax

The intercept is 1/Vmax= 0.080

Vmax= 1/0.080=12.5 mM/min

KM/Vmax( slope)= 0.237

KM= 0.237*12.5=2.962 /mM

KCat= turn over number = Vmax/Et where Et= Enzyme concentration = 12.5/1.3 = 9.61/min= 9.61/60sec =0.160/sec

KCa/KM= 0.160/2.962/sec.mM= 0.0540/sec.mM= 0.0540/10-3 sec.M=54/M.sec

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