When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 450C lost 50% of its activity in 12 minutes, but when incubated at 450C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 minutes. Suggest why thermal denaturation of hexokinase was retarded in the presence of one substrates

When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 450C lost 50% of its activity in 12 minutes, but when incubated at 450C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 minutes. Suggest why thermal denaturation of hexokinase was retarded in the presence of one substrates

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter23: Fatty Acid Catabolism

Section: Chapter Questions

Problem 21P: Using the ActiveModel for enoyl-CoA dehydratase, give an example of a case in which conserved...

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When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 45 °C lost 50% of its activity in 12 min, but when incubated at 45 °C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 min. Suggest why thermal denaturation of hexokinase was retarded in the presence of one of its substrates.
When enzyme solutions are heated, there is a progessive loss of catalytic activty over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 45 degrees Celsius lost 50% of its activity in 12 minutes, but when incubated at 45 degrees Celsius in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 minutes. Suggest why thermal denaturation of hexokinase was retarded in the presence of one substrates.
Substrates and reactive groups in an enzyme’s active site must be precisely aligned in order for a productive reaction to occur. Why, then, is some conformational flexibility also a requirement for catalysis?
In a particular enzyme, an alanine residue is located in a cleft where the substrate binds. A mutation that changes this residue to glycine has little effect on activity; however, another mutation, which changes the alanine to a glutamate residue, leads to a complete loss of activity. Provide a brief explanation for these observations. As a follow-up on the previous question, you have another enzyme that you suspect binds its substrate in a similar site as the enzyme you are studying above. Assuming you have access to techniques whereby you could synthesize mutant proteins, outline experimental steps you would use to help you determine whether the two proteins recognize substrate by similar structural features.
In what way is the observed mode of action of hexokinase consistent with the induced-fit theory of enzyme action?
In enzyme kinetics, for the reversible with two central complexes mechanism, please provide complete proof that the rate equation is the equation below. The variables denoted with f indicate forward direction while the variables denoted with b indicate backward direction.
For a lot of enzymes that work on fatty acids, the rate determining step is the release of the product from the active site. This means that the activation energy for product release is much higher than the free energy of catalysis. What enthalpic or entropic contributions would make the activation energy for product release so high and explain?
You begin to study enzyme Z, which catalyzes a simple reversible reaction that interconverts compound S and compound P. You observe that the ∆G´° for the S to P conversion to be –6 kJ/mol, and that compound S has ∆G´° for binding to enzyme Z of –15 kJ/mol, while compound P has a ∆G´° for binding to enzyme Z of –13 kJ/mol. Please explain the effect of enzyme Z on conversion of S to P. (Your answer should include a graph qualitatively showing energy versus reaction progress; however, you still need to explain youranswer in words!) not sure how to make the correct graph.
When studying the mechanism of the enzymatic reaction, functional groups were found that ensure the connection of the enzyme molecule with the substrate and take a direct part in the act of catalysis. What are these areas of the enzyme formed by these groups called? What functional structures form them and why?
Why might the compound shown below act as a transition state analog of phosphoglucose isomerase? A drawing of the normal transition state for this enzyme is needed.
When free water molecules are excluded from an enzyme’s catalytic pocket, is a catalytic −OH group on an amino acid side chain a stronger or weaker nucleophile? Explain.
Using lysozyme as an example, what can an enzyme’s structure reveal about its catalytic mechanism?