Question

Explain why each of the following data sets from a Lineweaver Burk plot are not individually ideal for determining KM for an enzyme catalysed reaction that follows Michaelis-Menten kinetics.

Set A	1/[S] (mM⁻¹)	1/v0 (𝛍M⁻¹.s)
	0.5	2.4
	1.0	2.6
	1.5	2.9
	2.0	3.1

Set B	1/[S] (mM⁻¹)	1/v0(𝛍M⁻¹.s)
	8	5.9
	10	6.8
	12	7.8
	14	8.7

Short answer

These are not ideal values for Lineweaver Burk plots.

Step-by-step solution

Michaelis Menten constant

The rate at which the enzyme's substrate reacts with the enzyme and creates the products is used to calculate the reaction rate. TheMichaelis-Menten constantis KM. The KMvalue is not a direct measure of the enzyme's affinity for the substrate, but rather the concentration of S, where the maximum velocity is half. The highest rate at which the enzyme catalyses the reaction is known as Vmax.

KM and Vmax

The substrate concentration should be both less and bigger than the KM value to achieve an optimum Lineweaver-Burk plot. The specified sets A and B are represented in a graph with substrate concentration [S] on the X-axis and beginning velocity v on the Y-axis. To connect sets A and B, the line is extrapolated. The graph is used to determine the KM values. The concentration of substrate in set A is larger

Graphical representation for the Lineweaver-Burk plot

Since the set A and set B has substrate concentration which greater or lesser than KM. So, the given sets arenot ideal values for Lineweaver-Burk plots.