By quantitative balance, the total amount of Enzyme is [E] 0= [E] + [EI] + [ES] + [ESI].
By using a=1+[I]/KI and a′=1+[I]/K′I, it is followed by [E]0=[E]a+[ES]a′
This equation can be written like this, [E]0=(Km[ES])/([S]0)a + [ES]a′=[ES]( aKm/[S}0+a'), because of Km=[E][S]/[ES] and [S]≈[S]0.
V=kb [ES] =kb [E] 0/ (aKm/[s] 0+a'). Kb [E] 0 is Vmax. This is why V=Vmax/(a^'+aKm/[S]0).
This equation can be rearranged like this, 1/V= a'/Vmax+(aKm/Vmax)1/[S]0, which is the Lineweaver-Burk plots.
Inhibition mode Competitive Uncompetitive Noncompetitive
The Lineweaver-Burk plot.
(Fig 9) The Lineweaver-Burk plots The enzyme inhibition can be explained by Gibbs energy. At the standard state, which is [E]= …show more content…
[S]= [P]=[I]=1M with temperature is 25 Celsius degree and pH is 7, the Gibbs free energy changes look like these.
∆Gs°′= -RTln (1/Ks)
∆GES°′ŧ=-RT (lnkBT/h-lnkcat)
∆GE°′ŧ=- RT (lnkBT/h-lnkcat/Ks)
In these case, ŧ is noted as transition state, R is a gas constant, kB is Boltzmann constant, h is plank’s constant, T is absolute temperature, and Ks is the equilibrium constant of E+S↔ES.
This is can be plotted at the graph. (Fig 10) Standard Gibbs energy profile
The standard state Gibbs energy can be changed following the variation of concentration of inhibitors.
[E]I denotes the concentration of enzyme when competitive inhibitor exists. [E]I= [E]-[EI]
KI= [E]i[I]/[EI]
[E]I= [E]- ([E]i[I])/KI
[E]I= [E]/a (a=1+ [I]/KI)
The Gibbs energy change of substrates is followed by
∆Gs′= ∆Gs°′-RTln([E]i[I]/[EI] )
∆Gs′= ∆Gs°′-RTln([E][I]/[EI] )/a
∆Gs′= ∆Gs°′+RTlna. (This is because at the standard state [E] = [S] = [ES] =1 M)
[ES]I is the concentration of the enzyme-substrate complex when uncompetitive inhibitors exist. At that time, [ES] I = [ES]-[ESI]
Following with this,
KI′=[ES]i[I]/[ESI]
[ES]I= [ES]- [ES]i[I]/Ki
[ES]I =[ES]/a′ (a′=1+[I]/K′I)
∆Gs′= ∆Gs°′-RTln {([E][I]/[ES]i)}
∆Gs′= ∆Gs°′- RTln {([E][I]/[ES] )a'}
∆Gs′= ∆Gs°′- RTlna′(this is because at the standard state, [ES]= [E]= [S])
When noncompetitive inhibitor exists, the [ES] I and [E] I are same value mentioned above. Also, the value of a is equal to a′.
Following with this,
∆Gs′= ∆Gs°′-RT{([E]i[I]/[ES]i)}
∆Gs′= ∆Gs°′-RT{(([E][I]/[ES] …show more content…
)/a)a}
∆Gs′= ∆Gs°′ (this is because it is the standard state.)
The changes of ∆GES′ŧ and ∆GE°′ŧ are related to whether formation of ES is favorable or not.
Inhibition mode Competitive Uncompetitive Noncompetitive
Gibbs energy changes
(kcal/mol)
The dark blue shows the changes of Gibbs energy changes as no binding with I.
The light blue shows Gibbs energy changes when I bind with E or ES.
(Fig 11) The Gibbs energy changes when inhibitors exist The Application of Enzyme Inhibition
Enzyme inhibition is very practical in the field of pharmacology in order to dealing with diseases.
For instances, inhibition of calystegines can be one method of dealing with the diabetes mellitus type 2. The difference between type 1 and type 2 is type 2 diabetes develops slowly. It is following the pre-diabetes status such as obesity, high blood pressure, and increased blood glucose level after the meal because of less insulin sensitivity. For these patients, glucosidase inhibitors like acarbose and miglitol, which inhibit maltase and sucrose, are applied as medicine. Also, calystegines A3 and B2 from potatoes can play a role in inhibitors for maltase and sucrase. (Fig 12) the structures of substrates and inhibitors
At the Caco-2 cell, the inhibition of sucrase and maltase by calystegines A3 and B2 give us this result. Maltase Sucrase Vmax[pkat/mg] Km[mM] R2 KI[µM] Vmax[pkat/mg] Km[mM] R2 KI[µM]
No inhibitor 827±12 7.3±0.5 0.9938 412±23 11.1±2.5 0.9570
Acarbose 5µM 725±12 15.6±0.8 0.9980 4.4 166±11 18.2±3.9 0.9678 7.8 calystegines A3 476µM 939±23 11.1±1.1 0.9913 d 298±7 12.5±1.1 0.9939 227±47
Calystegines B2 476µM 1016±22 25.4±1.5 0.9980 582±144 236±8 25.4±2.3 0.9948 55±12
(Fig 13) Maltase and sucrase inhibition by calystegines A3 and
B2
This table shows that sucrase is inhibited by calystegines A3 and B2 much more severely than maltase. It means we can control the level of sucarse easily by using inhibitors. Conclusion Enzymes play essential role in our body. By understanding enzyme kinetics and inhibition, we can treat many diseases and get lots of advantages, even though it is still obscure and unknown. Therefore, researching about enzyme kinetics has infinite possibility in our life.