Is it possible for there to be an inhibition to have an increasing Vmax but decr

Question

Is it possible for there to be an inhibition to have an increasing Vmax but decreasing Km as shown here? The paper says that its a mixed type inhibition. Can someone please explain this chart to me? Thank you.

Below, I have copied and pasted everything th epaper said in relation to the chart:

Inhibition kinetics showed that sodium oleate was a mixed-type inhibitor of chrysin glucuronidation in RLM (Ki = 3.23 g/mL) and RIM (Ki = 18.8 g/mL) (Figure 3). Therefore, it may be a good excipient candidate for formulating a nanoemulsion with inhibitory potential toward UGT metabolism.

RLM RIM Chrysin 1.25 M 3 Chrysin 2.5 UM Chrysin 5 Chrysin 1.25 HM 3 Chrysin 2.5 M Chrysin 5 uM Chrysin 10 M -Chrysin 10 M 2 2 8 16 2 8 16 24 Sodium oleate, /ml Sodium oleate, ug/ml Figure 3. Inhibition kinetics of sodium oleate with chrysin glucuronidation by RLM (A) and RIM (B)

Explanation / Answer

First let us refresh the concept of mixed inhibition. Mixed inhibition is a type of enzyme inhibition where the inhibitor may bind to the enzyme molecule irrespective of whether or not the enzyme has already bound itself to the substrate. This is called mixed inhibition as the concept is a mixture of both competitive inhibition (where the inhibitor binds to the enzyme only if the substrate has not bound to the enzyme) and uncompetitive inhibition (where the inhibitor binds to the enzyme only if the substrate has bound to the enzyme). Non-competitive inhibition is sometimes considered as a special case of mixed inhibition.

In mixed inhibition there are 2 possibilities:

Thus, mixed inhibition can result in one of the following:

But in both cases inhibition decreases the apparent maximum enzyme reaction rate, that is Vappmax < Vmax.

Thus to answer your question, a mixed inhibition can ouccur with Km increasing or decreasing but Vmax should be seen to decrease.

Ki - inhibitor constant - indicates the potency of the inhibitor is. Ki represents the concentration required to produce half the maximum inhibition.

Graph explanation:

When we plot 1/V vs. inhibitor concentration, we get a group of intersecting lines.

In competitive inhibition, lines converge above x-axis, and the point of intersection gives -Ki

In non-competitive inhibition, lines converge on the x-axis, and the point of intersection gives -Ki

In mixed inhibition, lines converge in the negative x-axis, and the point of intersection gives -Ki

Please remember here we are plotting 1/V and not V directly. Hence the graph is showing an increase in the 1/V values, which means V is decreasing (inverse relation). This is what is shown in the graph.

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