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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 π‘˜π‘π‘Žπ‘‘ of 30 s-1 and a πΎπ‘š of

0.005 show one-quarter of its maximum rate?

b) Determine the fraction of π‘‰π‘šπ‘Žπ‘₯ that would occur at the following substrate

concentrations:[𝑆] =1

2πΎπ‘š, 2πΎπ‘š, and 10πΎπ‘š.

Answers:

a) Since π‘‰π‘œ =π‘‰π‘šπ‘Žπ‘₯[𝑆]

πΎπ‘š+[𝑆] and π‘‰π‘œ = 0.25(30 π‘ βˆ’1), = 7.5π‘ βˆ’1, we can substitute into the

Michaelis-Menten equation to give

π‘‰π‘œ =π‘‰π‘šπ‘Žπ‘₯[𝑆]

πΎπ‘š + [𝑆]

7.5 π‘ βˆ’1 =30 π‘ βˆ’1[𝑆]

5π‘šπ‘€ + [𝑆]

[𝑆] = 1.7 π‘šπ‘€ = 1.7 π‘₯ 10βˆ’3 𝑀

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

π‘‰π‘œ

π‘‰π‘šπ‘Žπ‘₯=

[𝑆]

πΎπ‘š + [𝑆]

Substituting [𝑆] =1

2πΎπ‘š into this equation gives

π‘‰π‘œ

π‘‰π‘šπ‘Žπ‘₯= 0.33

Substituting [𝑆] = 2πΎπ‘š into this equation gives π‘‰π‘œ

π‘‰π‘šπ‘Žπ‘₯= 0.67

Substituting [𝑆] = 10πΎπ‘š into this equation gives π‘‰π‘œ

π‘‰π‘šπ‘Žπ‘₯= 0.91

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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

π‘‰π‘šπ‘Žπ‘₯ and πΎπ‘š 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)

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From the graph,

π‘‰π‘šπ‘Žπ‘₯ = 51.55 mM/min

πΎπ‘š = 0.598 mM

Solving for π‘‰π‘šπ‘Žπ‘₯ and πΎπ‘š using linear regression:

1

𝑉=

πΎπ‘š

π‘‰π‘šπ‘Žπ‘₯(

1

𝑆) +

1

π‘‰π‘šπ‘Žπ‘₯

𝑦 = π‘šπ‘₯ + 𝑏

where =1

𝑣 , π‘₯ =

1

𝑆 , and π‘š =

πΎπ‘š

π‘‰π‘šπ‘Žπ‘₯

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 π‘‰π‘šπ‘Žπ‘₯ and πΎπ‘š:

𝐴 =1

π‘‰π‘šπ‘Žπ‘₯

0.019371 = 1

π‘‰π‘šπ‘Žπ‘₯

π‘‰π‘šπ‘Žπ‘₯ = 51.6245 mM/min

Page 4: enzyme and enzyme kinetics

and, π‘š =πΎπ‘š

π‘‰π‘šπ‘Žπ‘₯

0.011604 =πΎπ‘š

51.6245

πΎπ‘š = 0.59907 π‘šπ‘€

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

shows that, in the presence of ibuprofen, the π‘‰π‘šπ‘Žπ‘₯ of the reaction is unchanged

(the intercept on the the 1/π‘‰π‘œ axis is the same) and πΎπ‘š is increased (1/πΎπ‘š is closer

to the origin).

PROBLEM #3:

Determination of π‘²π’Ž

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 π‘˜π‘π‘Žπ‘‘ for happyase is 600π‘ βˆ’1. They carry out several

experiments.

When [𝐸𝑑] = 20 𝑛𝑀 and [𝑆𝐴𝐷] = 40 ¡𝑀, the reaction velocity, 𝑉𝑂, is 9.6 Β΅Ms-1.

Calculate 𝐾𝑀 for the substrate SAD.

Answer:

We know π‘˜π‘π‘Žπ‘‘, [𝐸𝑑], [𝑆], and [𝑉𝑂]. We want to solve for 𝐾𝑀. Substituting the

known values allows us to solve for 𝐾𝑀.

𝑉𝑂 =π‘˜π‘π‘Žπ‘‘[𝐸𝑑][𝑆]

πΎπ‘š + [𝑆]

9.6 Β΅π‘€π‘ βˆ’1 =(600π‘ βˆ’1)(0.020¡𝑀)(40¡𝑀)

πΎπ‘š + 40¡𝑀

9.6 Β΅π‘€π‘ βˆ’1 =480¡𝑀2π‘ βˆ’1

πΎπ‘š + 40¡𝑀

Solving for πΎπ‘š gives,

πΎπ‘š = 10¡𝑀

Reference:

CourseSmart International E-Book for Principles of Biochemistry

by David L. Nelson, Michael M. Cox