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Quiz 2 for Chem 452 - Fall 2012: Enzyme Kinetics and Oxygen Binding Proteins, Exercises of Physical Chemistry

Araullo University (AU)Physical Chemistry

A take-home quiz for chem 452 - fall 2012, focusing on enzyme kinetics and oxygen binding proteins. The quiz includes questions related to the michaelis-menten equation, enzyme inhibition, and the behavior of myoglobin and hemoglobin. Students are required to show all calculations to receive full credit.

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Chem 452 - Fall 2012 - Quiz 2

(Take Home, due Monday, 22. Oct)

You may discuss with others strategies for answering these questions, but what you hand in should

represent your own work. You must show all calculations to receive full credit. Units are very important.

1. According the Michaelis-Menten equation, what is the vo/Vmax ratio when [S] = 3 KM?

2. If KM = 3!mM, and vo = 35 μmol/(mL•s) when [S] = 3!mM, what is the velocity, vo, for the reaction

when [S] = 18!mM?

3. The following kinetic data were obtained for an enzyme in the absence of an inhibitor, and in the

presence of two different inhibitors, (A) and (B), each at a concentration of 10.0!mM. Assume the

total enzyme concentration, [E]T, is the same for each experiment.

[S] {mM}

without inhibitor

vo {µmol/(mL•s)}

with inhibitor A

vo {µmol/(mL•s)}

with inhibitor B

vo {µmol/(mL•s)}

0.0

1.0

3.6

3.2

2.6

2.0

6.3

5.3

4.5

4.0

10.0

7.8

7.1

8.0

14.3

10.1

10.2

12.0

16.7

11.3

11.9

__________________________________ Name

1

Starting with the Michaelis-Menten equation:

KM1+I

[ ]

KI

⎛

⎝

⎜⎞

⎠

⎟

Substitute [S]= 3 KM

!

vo

Vmax

=3!KM

1 KM+3!KM

=3

1+3

vo

Vmax

=3

4

Starting with the Michaelis-Menten equation:

vo=Vmax S

[ ]

KM+S

[ ]

We could substitute the values we have for KM, vo and [S] and

solve for Vmax or we could simply recognize that since both [S]

and KM equal 3"mM, vo = 35"μmol/(mL•s) must be the half-

maximum velocity, which makes Vmax = 70"μmol/(mL•s).

Since [S] = 18"mM is equal to 6 KM,

vo=Vmax S

[ ]

KM+S

[ ]

vo

Vmax

=6 KM

1 KM+6 KM

=6

1 + 6 =6

7 (see Problem 1)

vo=Vmax

6

7

⎛

⎝

⎜⎞

⎠

⎟=70

µ

mol/ mL•s

( )

6

7

⎛

⎝

⎜⎞

⎠

⎟=60

µ

mol/ mL•s

( )

Key

2/2

10/10

Partial preview of the text

Download Quiz 2 for Chem 452 - Fall 2012: Enzyme Kinetics and Oxygen Binding Proteins and more Exercises Physical Chemistry in PDF only on Docsity!

Chem 452 - Fall 2012 - Quiz 2

(Take Home, due Monday, 22. Oct)

You may discuss with others strategies for answering these questions, but what you hand in should

represent your own work. You must show all calculations to receive full credit. Units are very important.

1. According the Michaelis-Menten equation, what is the v o/ V max ratio when [S] = 3 K M?

2. If K M = 3 mM, and vo = 35 μmol/(mL•s) when [S] = 3 mM, what is the velocity, v o, for the reaction

when [S] = 18 mM?

3. The following kinetic data were obtained for an enzyme in the absence of an inhibitor, and in the

presence of two different inhibitors, (A) and (B), each at a concentration of 10.0 mM. Assume the

total enzyme concentration, [E]T, is the same for each experiment.

[S] {mM} without inhibitor vo {μmol/(mL•s)} with inhibitor A vo {μmol/(mL•s)} with inhibitor B vo {μmol/(mL•s)}

Name __________________________________

Starting with the Michaelis-Menten equation:

K M 1 +

[I ]

K I

Substitute [S]= 3 K M!

vo

V max

3 K M

1 K M + 3 K M

vo

V max

Starting with the Michaelis-Menten equation:

vo =

V max [ S ] K M + (^) [ S ]

We could substitute the values we have for K M, vo and [ S ] and

solve for V max or we could simply recognize that since both [S]

and K M equal 3 mM, v o = 35 μmol/(mL•s) must be the half-

maximum velocity, which makes V max = 70 μmol/(mL•s).

Since [S] = 18 mM is equal to 6 K M ,

vo = V max^ [ S^ ]

K M + [ S ]

vo V max

6 K M

1 K M + 6 K M

(see Problem 1 ) v (^) o = V max^6 7

⎝⎜^

⎠⎟^

= (^) ( 7 0 μmol/ (^) (m L•s)) 6 7

⎝⎜^

⎠⎟^

= 60 μmol/ (m L•s)

Key 2/ 2/ 10/

a. Determine V max and K M for the uninhibited

without inihibitorwithout inihibitor with inibitor Awith inibitor A with inhibitor Bwith inhibitor B

[S] vo [S] vo [S] vo

0.0 mM 0.0 μmol/(mL•s) 0.0 mM 0.0 μmol/(mL•s) 0.0 mM 0.0 μmol/(mL•s)

1.0 mM 3.6 μmol/(mL•s) 1.0 mM 3.2 μmol/(mL•s) 1.0 mM 2.6 μmol/(mL•s)

2.0 mM 6.3 μmol/(mL•s) 2.0 mM 5.3 μmol/(mL•s) 2.0 mM 4.5 μmol/(mL•s)

4.0 mM 10.0 μmol/(mL•s) 4.0 mM 7.8 μmol/(mL•s) 4.0 mM 7.1 μmol/(mL•s)

8.0 mM 14.3 μmol/(mL•s) 8.0 mM 10.1 μmol/(mL•s) 8.0 mM 10.2 μmol/(mL•s)

12.0 mM 16.7 μmol/(mL•s) 12.0 mM 11.3 μmol/(mL•s) 12.0 mM 11.9 μmol/(mL•s)

1/[S] 1/vo 1/[S] 1/vo 1/[S] 1/vo

1.000 1/mM 0.280 (mL•s)/μmole 1.000 1/mM 0.309 (mL•s)/μmole 1.000 1/mM 0.392 (mL•s)/μmole

0.500 1/mM 0.160 (mL•s)/μmole 0.500 1/mM 0.189 (mL•s)/μmole 0.500 1/mM 0.224 (mL•s)/μmole

0.250 1/mM 0.100 (mL•s)/μmole 0.250 1/mM 0.129 (mL•s)/μmole 0.250 1/mM 0.140 (mL•s)/μmole

0.125 1/mM 0.070 (mL•s)/μmole 0.125 1/mM 0.099 (mL•s)/μmole 0.125 1/mM 0.098 (mL•s)/μmole

0.083 1/mM 0.060 (mL•s)/μmole 0.083 1/mM 0.089 (mL•s)/μmole 0.083 1/mM 0.084 (mL•s)/μmole

Determined valuesDetermined values

K M = 6.0 mM ( K M)app = 3.5 mM K M = 6.0 mM

V max = 25.0 μmol/(mL•s) ( V max)app = 14.6 μmol/(mL•s) ( V max)app = 17.9 μmol/(mL•s)

K I = K I = 14.0 mM K I = 25.0 mM

(Inhibitor decreases both K M and Vmax, by

1 + [I^ ]

K I

⎠⎟^ )

Unncompetitivve inhibition

Noncompetitivve inhibition

(Inhibitor increases K M by

1 + [I^ ]

K I

⎠⎟^ and leaves Vmax,

unchanged)

b. What is the turnover number for Hexokinase under these conditions?

turnover number = k cat =

V max

[ E ]total

μmol

mL•s

nmol

mL

186 x1 0 −^6

mol

mL•s

3. 0 x1 0 −^9

mol

mL

= 62 , 000 / s

c. What is the catalytic efficiency for Hexokinase under these conditions?

catalytic efficiency =

k cat

K M

62 , 000 /s

3. 0 x1 0 −^4 M

= 2. 1 x 108 / (^) (M • s)

d. Does Hexokinase display “catalytic perfection” under these conditions?

The catalytic efficiency is greater than 10^8 /(M•s), which puts it in the range that qualifies it to be

considered “catalytic perfection”.

e. What determines the ultimate speed limit of an enzyme-catalyzed reaction? That is, what is it that

imposes a physical limit on catalytic perfection?

When an enzyme is catalytically perfect, the reaction rate has become dependent on the rate at which the

substrate is able to diffuse into the active site. This rate places an upper limit of 10^8 /(M•s) to 10^9 /(M•s) on

the catalytic efficiency of an enzyme catalyzed reaction. Beyond this, there is nothing that evolution can do

to further increase the rate at which the enzyme catalyzes the reaction.

f. In a sentence, describe Hexokinase based on its Enzyme Commission (EC) number. For example,

the EC number for the enzyme Chymotrypsin is 3.4.21.1, which tells us that Chymotrypsin

(3.4.21. 1 ) is a hydrolase ( 3 .4.21.1) and serine type endopeptidase (3.4. 21 .1) that cleaves peptide bonds (3. 4 .21.1).

The E.C. number for Hexokinase is 2.7.1.1, which tells us that Hexokinase (2.7.1. 1 ) is a transferase

( 2 .7.1.1) that transfers a phosphate group (2. 7 .1.1) to an alcohol group as the acceptor (2.7. 1 .1) to

produce a phosphate ester.

5. Both myoglobin and hemoglobin function as oxygen binding proteins,

a. Each contains an Fe2+^ ion, which desires to interacts with six ligands. Describe the six ligand

interactions that an Fe2+^ ion in oxymyoglobin.

The six ligand from an octahedral around the Fe2+. Four of the ligands are the nitrogens provided by the

porphyrin ring and all lie in the same ring. The fifth ligand is a nitrogen from a histidine side chain (the

proximal histidine) and the six ligand is used bind the oxygen molecule.

b. The distal histidine , while not one of the ligands for the Fe2+^ ion, nonetheless plays some

important roles with respect to oxygen binding by hemoglobin. Describe two of these.

The distal histidine’s side chain imidazole sits near the site where the O 2 binds. (1) It hydrogen bonds to the

bound O 2 and helps prevent it being release as a destructive super oxide radical (O 2 - ), which would also

leave the iron as Fe3+, and kill the myoglobin as well. (2) The distal histidine also forces the sixth ligand that

binds to the heme group to bind at an angle. This is not an issue for the intended oxygen ligand, which

prefers to bind at an angle, but it, in particular, lowers the affinity for the toxic carbon monoxide (CO), which

prefers to bind straight on, at right angles to the heme group.

c. Using the axes provided below, illustrate how the binding of oxygen to myoglobin differs from

that for hemoglobin. Draw your curves showing myoglobin with a P 50 of 5 torr and showing

hemoglobin with a P 50 of 25 torr (Be sure to label your curves.)

5/

Mb

Hb

d. Explain how the behaviors illustrated above optimize myoglobin and hemoglobin for their

different physiological roles.

The oxygen-binding role for Hb is to circulate in the blood and pick up O 2 in the lungs and deliver it to the

tissues, such as muscles, where it passes the O 2 over to a Mb molecule, which then holds on to it until

needed. The binding curve for Hb shows that it can become nearly fully saturated with O 2 when in the lung,

where the pO 2 for oxygen is around 100 torr. As the O 2 -bound Hb moves out the the tissues, the pO 2 levels

fall. In the process, the Hb’s affinity for O 2 falls off more rapidly than that for Mb. This helps to optimize Hb’s

ability to then transfer its O 2 cargo to an awaiting Mb molecule.

e. If the pO 2 in the lungs is 100 torr, and the pO 2 in active muscles is 25 torr, assuming a Hill

coefficient of n = 2.8 for hemoglobin, what percentage of the O 2 picked up by the hemoglobin in

the lungs will be released to the myoglobin in the muscles?

The fraction bound by both Mb and Hb are described by the following equations:

For Mb: Y = pO^2 P 50 + pO 2

and for Hb: Y = (^ pO^2 )

n ( P 50 )n + ( pO 2 )n^ , where n is the Hill coefficient. In the lung, Mb: Y = pO 2 P 50 + pO 2

100 torr 5 torr + 100 torr

Hb: Y = ( p O 2 ) n (^ P 50 )n^ +^ ( p O 2 )n^

( 1 00 torr)^2.^8

( 2 5 torr)^2.^8 + ( 1 00 torr)^2.^8

In the muscles, Mb: Y = pO 2 P 50 + pO 2

25 torr 5 torr + 25 torr

Hb: Y = ( pO 2 ) n (^ P 50 )n^ +^ ( p O 2 )n^

Quiz 2 for Chem 452 - Fall 2012: Enzyme Kinetics and Oxygen Binding Proteins, Exercises of Physical Chemistry

Related documents

Partial preview of the text

Download Quiz 2 for Chem 452 - Fall 2012: Enzyme Kinetics and Oxygen Binding Proteins and more Exercises Physical Chemistry in PDF only on Docsity!

Chem 452 - Fall 2012 - Quiz 2

(Take Home, due Monday, 22. Oct)

You may discuss with others strategies for answering these questions, but what you hand in should

represent your own work. You must show all calculations to receive full credit. Units are very important.

1. According the Michaelis-Menten equation, what is the v o/ V max ratio when [S] = 3 K M?

2. If K M = 3 mM, and vo = 35 μmol/(mL•s) when [S] = 3 mM, what is the velocity, v o, for the reaction

when [S] = 18 mM?

3. The following kinetic data were obtained for an enzyme in the absence of an inhibitor, and in the

presence of two different inhibitors, (A) and (B), each at a concentration of 10.0 mM. Assume the

total enzyme concentration, [E]T, is the same for each experiment.

Name __________________________________

Starting with the Michaelis-Menten equation:

K M 1 +

K I

Substitute [S]= 3 K M!

vo

V max

3 K M

1 K M + 3 K M

vo

V max

Starting with the Michaelis-Menten equation:

vo =

We could substitute the values we have for K M, vo and [ S ] and

solve for V max or we could simply recognize that since both [S]

and K M equal 3 mM, v o = 35 μmol/(mL•s) must be the half-

maximum velocity, which makes V max = 70 μmol/(mL•s).

Since [S] = 18 mM is equal to 6 K M ,

vo = V max^ [ S^ ]

K M + [ S ]

6 K M

1 K M + 6 K M

⎝⎜^

⎠⎟^

⎝⎜^

⎠⎟^

= 60 μmol/ (m L•s)

a. Determine V max and K M for the uninhibited

without inihibitorwithout inihibitor with inibitor Awith inibitor A with inhibitor Bwith inhibitor B

[S] vo [S] vo [S] vo

0.0 mM 0.0 μmol/(mL•s) 0.0 mM 0.0 μmol/(mL•s) 0.0 mM 0.0 μmol/(mL•s)

1.0 mM 3.6 μmol/(mL•s) 1.0 mM 3.2 μmol/(mL•s) 1.0 mM 2.6 μmol/(mL•s)

2.0 mM 6.3 μmol/(mL•s) 2.0 mM 5.3 μmol/(mL•s) 2.0 mM 4.5 μmol/(mL•s)

4.0 mM 10.0 μmol/(mL•s) 4.0 mM 7.8 μmol/(mL•s) 4.0 mM 7.1 μmol/(mL•s)

8.0 mM 14.3 μmol/(mL•s) 8.0 mM 10.1 μmol/(mL•s) 8.0 mM 10.2 μmol/(mL•s)

12.0 mM 16.7 μmol/(mL•s) 12.0 mM 11.3 μmol/(mL•s) 12.0 mM 11.9 μmol/(mL•s)

1/[S] 1/vo 1/[S] 1/vo 1/[S] 1/vo

1.000 1/mM 0.280 (mL•s)/μmole 1.000 1/mM 0.309 (mL•s)/μmole 1.000 1/mM 0.392 (mL•s)/μmole

0.500 1/mM 0.160 (mL•s)/μmole 0.500 1/mM 0.189 (mL•s)/μmole 0.500 1/mM 0.224 (mL•s)/μmole

0.250 1/mM 0.100 (mL•s)/μmole 0.250 1/mM 0.129 (mL•s)/μmole 0.250 1/mM 0.140 (mL•s)/μmole

0.125 1/mM 0.070 (mL•s)/μmole 0.125 1/mM 0.099 (mL•s)/μmole 0.125 1/mM 0.098 (mL•s)/μmole

0.083 1/mM 0.060 (mL•s)/μmole 0.083 1/mM 0.089 (mL•s)/μmole 0.083 1/mM 0.084 (mL•s)/μmole

Determined valuesDetermined values

K M = 6.0 mM ( K M)app = 3.5 mM K M = 6.0 mM

V max = 25.0 μmol/(mL•s) ( V max)app = 14.6 μmol/(mL•s) ( V max)app = 17.9 μmol/(mL•s)

K I = K I = 14.0 mM K I = 25.0 mM

(Inhibitor decreases both K M and Vmax, by

1 + [I^ ]

K I

⎠⎟^ )

Unncompetitivve inhibition

Noncompetitivve inhibition

(Inhibitor increases K M by

1 + [I^ ]

K I

⎠⎟^ and leaves Vmax,

unchanged)

b. What is the turnover number for Hexokinase under these conditions?

turnover number = k cat =

V max

μmol

mL•s

nmol

mL

186 x1 0 −^6

mol