Question

In: Chemistry

The data below were collected for an enzyme-catalyzed reaction. [S] (mM) 0.25 0.50 1.00 2.00 V0...

The data below were collected for an enzyme-catalyzed reaction.

[S] (mM)

0.25

0.50

1.00

2.00

V0 (mM*s-1)

2.00

3.33

5.00

6.67

Determine Vmax and Km for this reaction.

Expert Solution

In the question above they have given different substrate concentration and value of V₀ at each substrate concentration so we can plot Lineweaver–Burk plot which is called as double reciprocal plot to get values of Km and Vmax for above reaction,

According to Michealis-Menten equation ,

V= Vmax[S]/Km+[S]

Taking the reciprocals gives ,

1/V = Km/Vmax[S]+1/Vmax

The Lineweaver–Burk plot was widely used to determine important terms in enzyme kinetics, such as Km and Vmax, before the wide availability of powerful computers and non-linear regression software. The y-intercept of such a graph is equivalent to the inverse of Vmax; the x-intercept of the graph represents ?1/Km

plotting 1/v against 1/[S] give a straight line:

y intercept = 1 / Vmax

gradient = Km / Vmax

x intercept = -1/ Km

Now y intercept is 0.1 from the graph,

Therefore 0.1 = 1/Vmax; Vmax= 10 mMS^-1

And x Intercept is = - 1 from the graph,

Therefore -1 = -1/Km; Km = 1 mM

queen_honey_blossom answered 1 year ago

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