BIO 360 Chapter : Module 3_ Lesson 6 Enzyme Kinetics

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15 Sep 2022

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Kinetics study of the rate of reactant compounds. Rate of enzymatic reaction is affected by : Understand catalytic mechanism , find effective inhibitors and understand regulation of activity. _ michaels constant step i rate of formation and breakdown of es ; governed by rate constants. K - i [ es] -114 > [ es] K, ( [f- +] - [ es])[ s] (cid:8869) step > rate of formation and breakdown "s equal > steady-state. Assumption k , ( [ e- +] - [ es ])[ 57=14 _ , [ es] -1142 [ es] - y - - - - - mita"s" enzyme -1substrate tv " Gaggi _ ,j a enzyme + product gress. Enzyme (e) combines with its substrate to form an enzyme substrate ( es) complex in a fast reversible step : Es complex then breaks down / na slower second step -10 yield the-free enzyme and the reaction product cp) - - - -1m - = - -

Related Questions

a) The Km of an enzyme of an enzyme-catalyzed reaction is 7.5 ïM. What substrate concentration will be required to obtain 65% of Vmax for this enzyme? (same enzyme was used in part a and b) (6 points)

b) Calculate the Ki for a competitive inhibitor whose concentration is 7 x10-6 M. The Km in the presence of inhibitor was found to be 1.2x10-5 M. (6 points)

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Explain how the following mechanisms regulate enzyme activity.

Part A

Covalent modification

a)The product of a series of reactions acts as an inhibitor for an earlier reaction.

b)Hormones control the synthesis of enzymes.

c)A regulator binds to the enzyme at a site other than the active site. This binding changes the shape of the enzyme and alters the catalytic ability of the enzyme.

d)An inhibitor binds reversibly to the enzymesubstrate complex, blocking the binding of the second substrate to the active site.

e)The activity of an enzyme is influenced by the addition or removal of a group that is covalently bonded to the enzyme.

f)An inhibitor forms covalent bonds to the active site, permanently blocking it.

Part B

Genetic control

a)The product of a series of reactions acts as an inhibitor for an earlier reaction.

b)Hormones control the synthesis of enzymes.

c)A regulator binds to the enzyme at a site other than the active site. This binding changes the shape of the enzyme and alters the catalytic ability of the enzyme.

d)An inhibitor binds reversibly to the enzymesubstrate complex, blocking the binding of the second substrate to the active site.

e)The activity of an enzyme is influenced by the addition or removal of a group that is covalently bonded to the enzyme.

f)An inhibitor forms covalent bonds to the active site, permanently blocking it

Part C

Allosteric regulation

a)The product of a series of reactions acts as an inhibitor for an earlier reaction.

b)Hormones control the synthesis of enzymes.

c)A regulator binds to the enzyme at a site other than the active site. This binding changes the shape of the enzyme and alters the catalytic ability of the enzyme.

d)An inhibitor binds reversibly to the enzymesubstrate complex, blocking the binding of the second substrate to the active site.

e)The activity of an enzyme is influenced by the addition or removal of a group that is covalently bonded to the enzyme.

f)An inhibitor forms covalent bonds to the active site, permanently blocking it.

Part D

Feedback inhibition

a)The product of a series of reactions acts as an inhibitor for an earlier reaction.

b)Hormones control the synthesis of enzymes.

c)A regulator binds to the enzyme at a site other than the active site. This binding changes the shape of the enzyme and alters the catalytic ability of the enzyme.

d)An inhibitor binds reversibly to the enzymesubstrate complex, blocking the binding of the second substrate to the active site.

e)The activity of an enzyme is influenced by the addition or removal of a group that is covalently bonded to the enzyme.

f)An inhibitor forms covalent bonds to the active site, permanently blocking it

a) The Km of an enzyme of an enzyme-catalyzed reaction is 7.5 ïM. What substrate concentration will be required to obtain 75% of Vmax for this enzyme? (same enzyme was used in part a and b) (6 points) b) Calculate the Ki for a competitive inhibitor whose concentration is 7.5 x10-6 M. The Km in the presence of inhibitor was found to be 1.25x10-5 M. (6 points)

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BIO 360 Chapter : Module 3_ Lesson 6 Enzyme Kinetics

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