Activity Number 2.docx

GROUP 6 Daquiog, Justine Decena, Clarissa Mamba, Rhea Singson, Mariella

PROBLEM 2.1 In order to measure the enzyme activity and the initial rate of reaction, 5 mL of cellobiose (100mumol/mL) and 44 mL of buffer solution were placed in a stirred vessel. The reaction was initiated by adding 1 mL of enzyme (beta-glucosidase) solution which contained 0.1 mg of protein per mL. At 1, 5, 10, 15, and 30 minutes 0.1 mL of sample was removed from the reaction mixture and its glucose content was measured. The results were as follows: TIME

GLUCOSE CONCENTRATION

(min)

(µmol/mL)

1

0.05

5

0.23

10

0.38

15

0.52

30

1.03

REQUIRED: a. What is the activity of the β-glucosidase in units/mL of enzyme solution and in umts/mg protein? A unit is defined as the enzyme activity which can produce Imumol of product per minute. b. What is the initial rate of reaction?

SOLUTION: a. Total Volume = 44 + 5 + 1 = 50 mL

Concentration vs Time 1.2

y = 0.033x + 0.0391 R² = 0.9973

1

C

0.8 0.6 0.4 0.2 0 0

5

10

15

20

25

t Based from the graph above; y = 0.033x + 0.0391 m = 0.033mumol/ml.min 0.033*50=1.65 mumol/min Activity of the β-glucosidase, A1 =

1.65mumol / min .1mg / ml * .1ml

A1 =165 units/mg protein

A2 =

1.65mumol / min 1mlenzyme

A2 = 1.65 units/ml of enzyme

b. m = Initial rate of reaction m = 0.033mumol/mL.min

30

35

PROBLEM 2.15 Eadie (1942) measured the initial reaction rate of hydrolysis of acetylcholine (substrate) by dog serum (source of enzyme) in the absence and presence of prostigmine (inhibitor), 1.5 x 10-7 mol/ L and obtained the following data: Substrate Concentration, Cs (mol/L) 0.0032 0.0049 0.0062 0.008 0.0095

Initial Reaction Rate Absence of Prostigmine 0.111 0.148 0.143 0.166 0.2

Rate (mol/L-min) Presence of Prostigmine 0.059 0.071 0.091 0.111 0.125

Cs/r (Uninhibited)

Cs/r (Inhibited)

0.028828829 0.033108108 0.043356643 0.048192771 0.0475

0.05423729 0.06901408 0.06813187 0.07207207 0.076

REQUIRED: a. Is prostigmine competitive or noncompetitive inhibitor? b. Evaluate the Michaelis-Menten kinetic parameters in the presence of inhibitor by employing the Langmuir Plot. SOLUTION: a. Prostigmine as competitive or noncompetitive inhibitor

Langmuir Plot of Hydrolysis of Acetylcholine 0.09 0.08

y = 2.9883x + 0.0489

0.07 0.06 y = 3.3133x + 0.0191

0.05

Cs/r

0.04

Without Inhibitor

0.03

With Inhibitor

0.02 -KMI -0.02

-KM -0.015

-0.01

0.01 0 -0.005 -0.01 0

0.005

0.01

0.015

Cs

Based from the graph, the values of Km with and without the presence of inhibitor is not equal. Therefore, prostigmine is a competitive inhibitor.

b. Langmuir Equation (With inhibitor):

K Cs 1  Cs  m r rmax rmax y  2.9883x  0.0489

1 rmax

 2.9883

rmax  0.3346

mol L  min

K mi  0.0489 rmax K mi  0.01636

mol L

PROBLEM 2.19 The initial rate of reaction for the enzymatic cleavage of deoxyguanosine triphosphate was measured as a function of initial substrate concentration as follows (Kornberg et al., J. BioI.Chern., 233, 159, 1958): Substrate Concentration (µmol/L)

Initial Reaction Rate (µmol/Lmin)

6.7 3.5 1.7

0.3 0.25 0.16

1/Cs

1/r

0.149254 3.333333 0.285714 4 0.588235 6.25

REQUIRED: a. Calculate the Michaelis-Menten constants of the above reaction. b. When the inhibitor was added, the initial reaction rate was decreased as follows: Substrate Concentration (µmol/L)

Inhibitor (µmol/L)

6.7 3.5 1.7

146 146 146

Initial Reaction Rate (µmol/Lmin) 0.11 0.08 0.06

1/r

9.090909 12.5 16.66667

Is this competitive inhibition or noncompetitive inhibition? Justify your answer by showing the effect of the inhibitor graphically. [Contributed by Professor Gary F. Bennett, The University of Toledo, Toledo, OH] SOLUTION: a. Michaelis-Menten constants of the below reaction Substrate Concentration (µmol/L)

Initial Reaction Rate (µmol/Lmin)

6.7 3.5 1.7

0.3 0.25 0.16

1/Cs

0.149254 3.333333 0.285714 4 0.588235 6.25

Lineweaver-Burk Plot of Enzymatic Cleave of Deoxyguanosine Triphosphate 7 6

y = 6.7758x + 2.2168 1/r

5 4 3 2 1 0

0.1

0.2

0.3

0.4 1/Cs

Lineweaver-Burk Equation:

1 Km 1 1   r rmax Cs rmax y  6.7758 x  2.2168

1 rmax

 2.2168

1/r

0.5

0.6

0.7

rmax  0.451

mol L  min

Km  6.7758 rmax K m  3.0566

mol L

b. Plotting the graph of the reaction with and without inhibitor using Lineweaver-Burk Equation

Lineweaver-Burk Plot of Enzymatic Cleavage of Deoxyguanosine Triphosphate 18 16 14

y = 16.68x + 7.0637 R² = 0.9754

12 10

1/r

8 6 4

y = 6.7758x + 2.2168 R² = 0.9922

2 0 -0.6

-0.4

-0.2

-2 0

0.2

0.4

0.6

0.8

1/Cs Without inhibitor

WIth inhibitor

LINEWEAVER-BURK EQUATION With Inhibitor Without Inhibitor rmax Km rmax Km 0.141569 2.361369 0.4511007 3.056568 Based from the graph, the values of Km (with and without inhibitor) are almost the same. Moreover, the y-intercepts, rmax of the Lineweaver-Burk for both inhibited and uninhibited are not equal. Therefore, it is a noncompetitive inhibition.

PROBLEM 2.20 The effect of an inhibitor on an enzyme reaction was studied by measuring the initial rates at three different initial inhibitor concentrations. The obtained Michaelis-Menten kinetic parameters are as follows: Inhibitor µmol/L 0 2 4 6

rmax µmol/L min 0.70 0.20 0.11 0.08

KM µmol/L 5 5 5 5

REQUIRED: a. Write the kinetic model for this enzyme reaction. b. Derive the rate equation. State your assumptions for any simplification of the equation. c. Estimate the value of inhibition kinetic parameter.

SOLUTION: a. Kinetic model for this enzyme reaction Since the given has constant KM shown in the table given table, then the enzyme reaction is non-competitive inhibition reaction. The kinetic model would be: K1

E +S ⇔ ES K 2

K3

E+I ⇔ EI K4 K5

EI + S ⇔ EIS K6 K7

ES + I ⇔ ESI K 8

K9

ES → E + P

b. Derive the rate equation. State your assumptions for any simplification of the equation. Assumptions:  

The dissociation constant for the first equilibrium reaction is the same as that of the third equilibrium reaction. The dissociation constant for the second equilibrium reaction is the same as that of the fourth equilibrium equation.

The two equilibrium reactions, k2 k  K S  6  K IS k1 k5 k4 k  K I  8  K SI k3 k7

If the slower reaction, the product formation step, determines the rate of the reaction according to Michaelis-Menten assumption, the rate can be expressed as:

rp  k9 ES 

(1)

E0   E   ES   EI   ESI 

(2)

k9 ES  E   ES   EI   ESI 

(3)

The enzyme balance gives

Divide (1) by (2),

rp

E0 



Applied Law of Mass Action

KS 

K 2 E S  E S    ES   ES  K1 KS

(4)

KI 

K 4 E I  E I    EI   EI  K3 KI

(5)

KS 

K 8 ES I  ES I    ESI   ESI  K7 KI

(6)

Substitute (4), (5), (6) into (3),

rp

k9



E S  K

E 0  E   E S   E I   ES I  KS

rp

KI

k9



KI

E S  K

E 0  E   E S   E I   E S I  KS

KI

KS KI

Eliminate [E],

S  rp



EO  k9

KS

S   I   S I  1 KS

KI

KS KI

Substitute rpmax  E 0  k 9

S  rp rpmax



KS S   I   S I  1 KS KI KS KI

Multiply numerator and denominator by K S p rp r pmax



S  K S I  S I  K S  S    KI KI

Rearranging

rp r pmax rp r pm ax

 KS  

S  K S I  KI

 S  

S 

S I  KI

 I    S   1  I   K S  1   K I  K I   