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Chapter 12: Q8CP (page 373)

Why can’t enzyme kinetics prove that a particular enzyme mechanism is correct?

Short Answer

Expert verified

This is just one rate-determining enzymatic step, this reaction appears to be designed as a single catalytic step with an apparent unimolecular rate constant kcat.

Step by step solution

01

Introduction

In enzyme-catalyzed processes, the rate of catalysis does not change in response to the increasing substrate. The beginning rate is calculated over a wide range of substrate concentrations. The enzyme beginning rate, on the other hand, rises with increasing substrate concentrationsuntil it reaches Vmax, the maximum rate.

02

The enzyme mechanism is incorrect

The M-M or Michaelis-Menten equationestablishes a first reaction between the enzyme and the substrate in order to form the enzyme-substrate complex. The rate of enzymatic reaction increases when the substrate concentration increases up to aVmax as a result, the rate of reaction is no longer dependent on the complex. The reaction takes the form of an order 0 unimolecular reaction.

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Most popular questions from this chapter

Compare and contrast the actions of an allosteric effector, a competitive enzyme inhibitor, and a non-competitive inhibitor.

Explain the structural basis for cooperative substrate binding and allosteric control in ATCase.

How does pure non-competitive inhibition differ from other forms of inhibition?

Ethanol in the body is oxidized to acetaldehyde (CH₃CHO) by liveralcohol dehydrogenase (LADH). Other alcohols are also oxidized byLADH. For example, methanol CH₃OH), which is mildly intoxicating, isoxidized by LADH to the quite toxic product formaldehyde (CH₂O). Thetoxic effects of ingesting methanol (a component of many commercialsolvents) can be reduced by administering ethanol. The ethanol acts as acompetitive inhibitor of the methanol by displacing it from LADH. Thisprovides sufficient time for the methanol to be harmlessly excreted bythe kidneys. If an individual has ingested 100 mL of methanol (a lethaldose), how much 100 proof whiskey (50% ethanol by volume) must beimbibe to reduce the activity of his LADH toward methanol to 5% of itsoriginal value? The adult human body contains -40 L of aqueous fluidsthroughout which ingested alcohols are rapidly and uniformly mixed.The densities of ethanol and methanol are both 0.79 g .cm-3. Assumethe KM values of LADH for ethanol and methanol to be 1.0 X 10⁻³Mand 1.0 X 10⁻²M, respectively, and that KI = KM for ethanol.

Calculate the half-life, in years, for the reaction2X→Y when the starting concentration of X is6μM and the rate constant is 3.6×10‐3M‐1s‐1.
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