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An inhibitor is added to an enzyme-catalyzed reaction at a concentration of 26.7 μM. The Vmax remains constant at 50.0 μM/s, but the KM increases by 150%. Calculate the Ki for this inhibitor.

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mickymike92

Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

What is the Ki for the inhibitor?

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

  • Kma = Km/(1 + [I]/Ki)

Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

where:

  • Kma is the apparent Km due to inhibitor
  • Km is the Km of the enzyme-catalyzed reaction
  • [I] is the concentration of the inhibitor

Solving for Ki:

where

[I] = 26.7 μM

Km = 1.0

Kma = (150% × 1 ) + 1 = 2.5

Ki = 26.7 μM/{(2.5/1) - 1)

Ki = 53.4 μM

Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

Learn more about enzyme inhibition at: https://brainly.com/question/13618533

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