Enzyme Kinetics Practice Problems PDF

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These are problems and solutions for enzyme kinetics using Michaelis Menton Kinetics in Biochemical Engineering. Great for solving/optimizing enzyme bioreactors...

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Assignment 1 Solutions Problem 3.1

a.) Equilibrium approach * r values are the same as k values all are rate constants

r1 (S)( E ) = r2 ( ES)1

(1)

r3 ( ES) 1 = r4 ( ES) 2

(2)

( E0 ) = ( E ) + ( ES)1 + (ES) 2

(3)

r r3

from (2)

( ES)1 = 4 ( ES) 2

(4)

r2r4 ( ES) 2 r1 r3 ( S)

(5)

from (1) + (4) ( E ) =

Substitute (4) and (5) into (3):

æ r r4 + r1 r4 (S ) + r1 r3 (S ) ö ÷÷ ( ES) 2 r1 r3 ( S) è ø r1 r3 (E 0 )(S ) \ ( ES ) 2 = r2 r4 + r1 r4 ( S) + r1 r3 ( S)

( E 0 ) = çç 2

rate =

r5 E 0 S Km 2 ( Km1 +S ) +S

where Km 1 =

r2 r , Km 2 = 4 r1 r3

b.) Quasi–steady–state approach

d ( ES ) 1 dt d ( ES ) 2

= r1 ( E)( S) - r2 ( ES) 1 + r4 ( ES) 2 - r3 ( ES) 1 » 0

(6)

= r3 ( ES) 1 - r4 ( ES) 2 - r5 ( ES) 2 » 0

(7)

dt from (7)

( ES)1 =

r4 + r5 ( ES) 2 r3

(8)

ær r +r r +r r ö from (6) + (8): (E ) = ç 2 4 2 5 3 5 ÷ (ES )2 ç ÷ r1 r3 (S ) è ø substitute (8) and (9) into (3):

1

(9)

æ r r4 + r2 r5 +r3 r5 +r1 r4 (S ) +r1 r5 (S ) +r1 r3 (S ) ö ES ÷÷ ( )2 r1 r3 (S ) è ø æ ö r1 r3 (E 0 )(S ) ( ES)2 = çç ÷÷ è r2 r4 + r2 r5 + r3 r5 + r1 r4 (S ) + r1 r5 (S ) + r1 r3 (S ) ø r5 E 0 S \ rate = ( Km2 + r5 r3 )( Km1 + S) + r5 r1 +S r r where Km1 = 2 and Km2 = 4 r1 r3

( E0 ) = çç 2

Problem 3.2 * r values are the same as k values all are rate constants

E +S d ( ES ) dt rate =

r1 r-1

( ES)

r2 r-2

E +P

= r1 ( E )(S) + r-2 ( E )( P ) - r-1 (ES ) - r2 (ES ) » 0 d ( P) dt

= r2 ( ES) - r- 2 ( E )( P )

( E0 ) = ( E ) + ( ES) , from (1): ( E ) =

(2)

\ ( E ) = ( E 0 ) - ( ES)

r-1 + r2 (ES ) r1 (S ) +r-2 (P )

Substitute (3) and (5) into (2):

r1 r2 E0 ( S) + r2 r-2 E0 ( P ) - r - 2 ( E 0 - ( ES)) ( P) r- 1 + r2 + r1 (S ) + r- 2 ( P )

Substituting (ES) in terms of E0:

rate =

r1 r2 E ( S) - r-1 r-2 E0 ( P) r-1 + r2 + r1 (S ) + r- 2 (P )

LetVS = r2 E0 and VP = r-1 E0

rate =

(3) (4)

æ r (S ) + r- 2 (P ) + r- 1 + r2 ö Combine (3) + (4): ( E 0 ) = ç 1 ES ç ÷÷ ( ) r1 (S )+ r-2 (P ) è ø E 0 ( r1 (S ) + r-2 (P ) ) \ ( ES) = r1 (S ) + r-2 (P ) + r-1 +r2

rate =

(1)

r1 Vs ( S) - r-2 VP ( P ) r-1 + r2 + r1 (S ) + r- 2 (P )

2

(5)

Divide top and bottom by (r-1 + r2) and let 1 r1 1 r and = = -2 1 K m r-1 + r 2 K P r- 1 + r 2

\ rate =

(V

s

K1m ) S - ( VP K P ) P S P 1+ 1 + K m KP

Problem 3.3

k-1 + k2 = 4.5´ 10-5 M k1

a)

K¢m =

b)

E 0 = 10-6 M S = 10 -3 M V=

Vm + S K1m + S

but Vm = r2 E 0

\

V = 9.58´ 10-4 Msec-1

Problem 3.4

Plot 1/V versus 1/S (Lineweaver – Burk plot) y – intercept =

Hydration:

1 1 -1 , x – intercept = = Vm k 2 E 0 Km

1 40 = + 4 ´103 V (Co 2 )

1 V = 2.50 ´ 10- 4 M sec- 1 = 4000M -1 sec ® m Vm r2 = 8.93 ´10 4sec - 1 -1 - 4000 = = -100M-1 ® K m = 0.01M Km 40 Dehydration:

1 163.15 = + 1.31´ 104 V ( HCO 3 )

3

1 Vm = 7.61´ 10-5 M sec-1 -1 4 1.31 10 M sec = ´ ® 4 1 Vm r2 = 2.72 ´10 sec -1 - 1.31´ 10 4 - 1 = M ® K m = 1.24 ´ 10- 2 M Km 163.15

Problem 3.5 a.)

Plot 1/V versus 1/S:

-1

1 1 = 9.525 + 0.60 V S

-0.60 = -0.063M-1 \ K m,app = 16.0g S / L K m,app 9.525 Since Km,app> Km the inhibitor is competitive. =

b.)

æ ( I) ö K m,app = K m ç1 + ÷ è KI ø ( I) = 1.37g I L KI = K m,app K m - 1 4...