enzyme and enzyme kinetics

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AUSTRIA, BABYLYN C. ChE - 5201 11 93498 BIOCHEMICAL ENGINEERING PROBLEMS IN ENZYME KINETICS PROBLEM #1: Relation between Reaction Velocity and Substrate Concentration: Michaelis- Menten Equation a) At what substrate concentration will an enzyme with of 30 s -1 and a of 0.005 show one-quarter of its maximum rate? b) Determine the fraction of that would occur at the following substrate concentrations:[] = 1 2 , 2 , and 10 . Answers: a) Since = [] +[] and = 0.25(30 −1 ), = 7.5 −1 , we can substitute into the Michaelis-Menten equation to give = [] + [] 7.5 −1 = 30 −1 [] 5 + [] [ ] =1 .7 =1 .7 10 −3 b) We can arrange the Michaelis-Menten equation into the form = [] + [] Substituting [] = 1 2 into this equation gives =0 .33 Substituting [] = 2 into this equation gives =0 .67 Substituting [] = 10 into this equation gives =0 .91

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enzyme and enzyme kinetics

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Page 1: enzyme and enzyme kinetics

AUSTRIA, BABYLYN C. ChE - 5201

11 – 93498 BIOCHEMICAL ENGINEERING

PROBLEMS IN ENZYME KINETICS

PROBLEM #1:

Relation between Reaction Velocity and Substrate Concentration: Michaelis-

Menten Equation

a) At what substrate concentration will an enzyme with 𝑘𝑐𝑎𝑡 of 30 s-1 and a 𝐾𝑚 of

0.005 show one-quarter of its maximum rate?

b) Determine the fraction of 𝑉𝑚𝑎𝑥 that would occur at the following substrate

concentrations:[𝑆] =1

2𝐾𝑚, 2𝐾𝑚, and 10𝐾𝑚.

Answers:

a) Since 𝑉𝑜 =𝑉𝑚𝑎𝑥[𝑆]

𝐾𝑚+[𝑆] and 𝑉𝑜 = 0.25(30 𝑠−1), = 7.5𝑠−1, we can substitute into the

Michaelis-Menten equation to give

𝑉𝑜 =𝑉𝑚𝑎𝑥[𝑆]

𝐾𝑚 + [𝑆]

7.5 𝑠−1 =30 𝑠−1[𝑆]

5𝑚𝑀 + [𝑆]

[𝑆] = 1.7 𝑚𝑀 = 1.7 𝑥 10−3 𝑀

b) We can arrange the Michaelis-Menten equation into the form

𝑉𝑜

𝑉𝑚𝑎𝑥=

[𝑆]

𝐾𝑚 + [𝑆]

Substituting [𝑆] =1

2𝐾𝑚 into this equation gives

𝑉𝑜

𝑉𝑚𝑎𝑥= 0.33

Substituting [𝑆] = 2𝐾𝑚 into this equation gives 𝑉𝑜

𝑉𝑚𝑎𝑥= 0.67

Substituting [𝑆] = 10𝐾𝑚 into this equation gives 𝑉𝑜

𝑉𝑚𝑎𝑥= 0.91

Page 2: enzyme and enzyme kinetics

PROBLEM #2:

Properties of an Enzyme of Prostaglandin Synthesis

Prostaglandins are a class of eicosanoids, fatty acid derivatives with a variety of

extremely potent actions on vertebrate tissues. Prostaglandins are responsible for

producing fever and inflammation and its associated pain. They are derived from the 20-

carbon fatty acid arachidonic acid in reaction catalyzed by the enzyme prostaglandin

endoperoxide synthase. This enzyme, a cyclooxygenase, uses oxygen to convert

arachidonic acid to PGG2, the immediate precursor of many different prostaglandins.

a) The kinetic data given below are for the reaction catalyzed by prostaglandin

endoperoxide synthase. Focusing here on the two columns, determine the

𝑉𝑚𝑎𝑥 and 𝐾𝑚 of the enzyme.

Arachidonic

Acid (mM)

Rate of Formation of PGG2

(mW/min)

Rate of Formation of PGG2

with 10 mg/mL ibuprofen

(mW/min)

0.5 23.5 16.67

1.0 32.2 25.25

1.5 36.9 30.49

2.5 41.8 37.04

3.5 44.0 38.91

b) Ibuprofen is an inhibitor of prostaglandin endoperoxide synthase. By inhibiting the

synthesis of prostaglandins, ibuprofen reduces inflammation and pain. Using the

data in the first and third columns of the table, determine the type of inhibition that

ibuprofen exerts on the prostaglandin endoperoxide synthase.

Answers:

a) Calculate the reciprocal values for the data, as in parentheses below, and prepare

a double-reciprocal plot to determine the kinetic parameters.

[S] (mM) (1/[S]

(mM-1))

𝑽𝒐 (mM/min)

(𝟏/𝑽𝒐 (min/mW))

𝑽𝒐 with 10 mg/mL

ibuprofen(mM/min)

(𝟏/𝑽𝒐 (min/mW))

0.5 (2.0) 23.5 (0.043) 16.67 (0.06)

1.0 (1.0) 32.2 (0.0321) 25.25 (0.0396)

1.5 (0.67) 36.9 (0.027) 30.49 (0.0328)

2.5 (0.4) 41.8 (0.024) 37.04 (0.027)

3.5 (0.27) 44.0 (0.023) 38.91 (0.0257)

Page 3: enzyme and enzyme kinetics

From the graph,

𝑉𝑚𝑎𝑥 = 51.55 mM/min

𝐾𝑚 = 0.598 mM

Solving for 𝑉𝑚𝑎𝑥 and 𝐾𝑚 using linear regression:

1

𝑉=

𝐾𝑚

𝑉𝑚𝑎𝑥(

1

𝑆) +

1

𝑉𝑚𝑎𝑥

𝑦 = 𝑚𝑥 + 𝑏

where =1

𝑣 , 𝑥 =

1

𝑆 , and 𝑚 =

𝐾𝑚

𝑉𝑚𝑎𝑥

Using the linear regression, the following values are obtained:

A = b = 0.019371

B = m = 0.011604

Substituting the values of b and m to solve for 𝑉𝑚𝑎𝑥 and 𝐾𝑚:

𝐴 =1

𝑉𝑚𝑎𝑥

0.019371 = 1

𝑉𝑚𝑎𝑥

𝑉𝑚𝑎𝑥 = 51.6245 mM/min

Page 4: enzyme and enzyme kinetics

and, 𝑚 =𝐾𝑚

𝑉𝑚𝑎𝑥

0.011604 =𝐾𝑚

51.6245

𝐾𝑚 = 0.59907 𝑚𝑀

b) Ibuprofen acts as a competitive inhibitor. The double reciprocal plot (with inhibitor)

shows that, in the presence of ibuprofen, the 𝑉𝑚𝑎𝑥 of the reaction is unchanged

(the intercept on the the 1/𝑉𝑜 axis is the same) and 𝐾𝑚 is increased (1/𝐾𝑚 is closer

to the origin).

PROBLEM #3:

Determination of 𝑲𝒎

An enzyme is discovered that catalyzes the chemical reaction

SAD HAPPY

A team of motivated researchers set out to study the enzyme, which they call

happyase. They find that the 𝑘𝑐𝑎𝑡 for happyase is 600𝑠−1. They carry out several

experiments.

When [𝐸𝑡] = 20 𝑛𝑀 and [𝑆𝐴𝐷] = 40 µ𝑀, the reaction velocity, 𝑉𝑂, is 9.6 µMs-1.

Calculate 𝐾𝑀 for the substrate SAD.

Answer:

We know 𝑘𝑐𝑎𝑡, [𝐸𝑡], [𝑆], and [𝑉𝑂]. We want to solve for 𝐾𝑀. Substituting the

known values allows us to solve for 𝐾𝑀.

𝑉𝑂 =𝑘𝑐𝑎𝑡[𝐸𝑡][𝑆]

𝐾𝑚 + [𝑆]

9.6 µ𝑀𝑠−1 =(600𝑠−1)(0.020µ𝑀)(40µ𝑀)

𝐾𝑚 + 40µ𝑀

9.6 µ𝑀𝑠−1 =480µ𝑀2𝑠−1

𝐾𝑚 + 40µ𝑀

Solving for 𝐾𝑚 gives,

𝐾𝑚 = 10µ𝑀

Reference:

CourseSmart International E-Book for Principles of Biochemistry

by David L. Nelson, Michael M. Cox