Question

Consider the data collected for an enzyme-catalyzed reaction. [S] (MM) 2 (MM :5-1) 0.20 0.86 0.33 1.20 2.00 2.42 4.00 2.69 Determine Vmax and Km for this reaction. mMs-1 Km = mM

Answer 1

Answer:

The Lineweaver Burk equation is

1/V₀=(Km/V_max)(1/[S]) + 1/Vmax

y=mx+b

Here the plot 1/V₀ vs 1/[S] gives a straight line with a slope, m= Km/V_max and y-intercept, b=1/V_max

The calculations are shown below

[S](mM)	v0 (mM/s)	1/[S](mM-1)	1/v0(s/mM)
0.2	0.86	5	1.162790698
0.33	1.2	3.03030303	0.833333333
2	2.42	0.5	0.41322314
4	2.69	0.25	0.371747212

The plot is shown below

y = a + b*x No Weighting 1.38885E-6 Equation Weight Residual Sum of Squares Pearson's r Adj. R-Square 1 1.0 - Intercept Slope Value 0.32985 0.16647 Standard Error 6.27234E-4 2.13589E-4 1/10 3 1/[S]

The plot is a straight line with R²=1.

Slope=0.16647=Km/V_max and y-intercept=0.32985=1/V_max

V_max=1/0.32985=3.03 mM s^-1.

Km/V_max=0.16647

Km=0.166 47 x 3.03=0.505 mM.

Therefore V_max=3.03 mM s^-1.

Km=0.505 mM.

Please let me know if you have any doubt. Thanks