Question

Properties of an Enzyme of Prostaglandin.

a) The kinetic data given below are for the reaction catalyzed by prostaglandin endoperoxide synthase. Focusing here on the first two columns, determine the Vmax and Km of the enzyme.

[Arachidonic acid (mM)]         Rate of Formation of PGG₂ (mM/min)          Rate of Formation of PGG₂ with 10

.5                                                                23.5                                             with 10mg/mL Ibuprofen (mM/min)

1.0                                                              32.2                                                               16.67

1.5                                                              36.9                                                                25.25

2.5                                                              41.8                                                               30.49

3.5                                                              44.0                                                               37.04

                                                                                                                                          38.91

(sorry the table is wonky)

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 prostaglandin endoperoxide synthase.

Please show every step and reasoning. I really want to learn this. Please and thank you.

Answer 1

1. substrate=Arachidonic acid,

Enzyme=Prostaglandin endoperoxide synthase

Product-pgg2

Inhibitor=ibuprofen

To calculate Vmax and km,   use Michaelis-Menten equation, r=Vmax [S]/km+[S]

Rearranging this equation we get,   1/r=1/Vmax + km/Vmax 1/[S]

By plotting 1/r vs 1/[S] (linear-burkweaver plot)

Slope=km/Vmax     ,intercept=1/Vmax

[S] mM	Rate of reaction=r=mM/min	1/[S] mM-1	1/r(min mM-1)
0.5	16.67	2	0.06
1.0	25.25	1	0.04
1.5	30.49	0.7	0.03
2.5	37.04	0.4	0.03
3.5	38.91	0.3	0.02

Plot of 1/r vs 1/S (see image)

Gives slope=km/Vmax=0.014

Intercept=1/Vmax=0.005

1/Vmax=0.005    gives       Vmax=1/0.005=200 mM/min

Km/Vmax=0.014       gives          km=Vmax*0.014=200*0.014=2.8 mM/min

2)Proceeding in similar way for inhibited reaction

Rate of reaction= Rate of formation of PGG2 with 10 mg/mL ibuprofen (mM/min)

[S] mM	Rate of reaction=r=mM/min	1/[S] mM-1	1/r(min mM-1)
0.5		2	0.04
1.0	32.2	1	0.03
1.5	36.9	0.7	0.03
2.5	41.8	0.4	0.02
3.5	44.0	0.3	0.02

Plot of 1/r vs 1/S (see image)

Gives slope=km/Vmax=0.006

Intercept=1/Vmax=0.005

1/Vmax=0.005    gives       Vmax=1/0.005=200 mM/min

Km/Vmax=0.03       gives          km=200*0.006=200*0.03=6.0 mM/min

This is Competitive inhibition

Comparing the values of Vmax of inhibited reaction and non-inhibited reaction,

Vmax or maximum velocity of the reaction is same but km decreased in inhibitor catalysed reaction from 2.8 mM/min to 1.2mM/min.

Types of inhibition,

1. competitive inhibition=the inhibitor binds to the enzyme only and not to the substrate-enzyme complex. So either inhibitor or substrate binds to the substrate at a time.Vmax is unaffected, but km decreases.

Step 1)E+SES, k1=rate of forward rxn,k2=rate of backward rxn

Or E+IEI ,k’ =rate of reaction

Step2) ESProduct rate constant=k3

Km=k2+k3/k1

2)non-competitive inhibition=Inhibitor has identical affinity for E and ES.

Vmax decreases,km =constant

3)mixed type=inhibitor binds to both E and ES with different affinities for them. Vmax decreases as ES catalysis is slowed and also interferes with substrate binding. Km increases