ENZYMES - KINTEICS

KINETICS

What is Michaelis-Menten kinetics and what is its primal use?

Since enzyme-catalyzed reactions are saturable (the velocity of the reaction is limited by the amount of available enzymes), their rate of catalysis does not show a linear response to increasing substrate – because the substrate will be turned into product during the cause of the reaction. The problem is solved by measuring the initial rate of the reaction in relation to different substrate concentrations As [S] gets higher, the enzyme becomes saturated with substrate and the rate reaches Vmax, the enzyme's maximum rate. Thus : Vmax for an enzyme can be estimated through this procedure. Furthermore, kinetics provide information of what kind of enzyme we're dealing with and what it actually does in the cell.

A full account of the derivation of the Michaelis & Menten - equation and its different terms can be found here.

Hexokinase I (in muscle) and hexokinase IV (in liver) both catalyze phosphorylation of Glucose to G6P. Their respective Km(glucose) are 0,5 mM and 10 mM for isoenzymes respectively. Both glucose levels are kept relatively constant at approximately 5 mM.

Which of those two enzymes could be regulated by substrate level control? Motivate!

First of all : what is Km, what is substrate level control?

What is Km? = ½ of Vmax

What is V0? = initial velocity of an enzyme catalyzed reaction.

Ok, so let´s look at the Km of these two isoenzymes; 0,5 mM and 10mM respectively. It is quite clear that Hexokinase I will be “constantly saturated”, given that the glucose level is 5 mM.

Hexokinase IV on the other hand has a Km of 10 mM, that is “Room for more”. If you add more than 5 mM - Hexokinase IV will “work harder”. But Hexokinase I is saturated already at 1 mM (2 * Km = Vmax), and thus will not be effected even if you add vast amounts of substrate.

Consider two isoenzymes that catalyze the same reaction : A → B.

Enzyme I has a kcat of 25/s and a Km of 1 mM.

Enzyme II has a kcat of 1/s and a Km of 0,04 mM.

Which one of the enzymes EI and EII will catalyze the reaction with the highest rate if :

a) [A] = 10 mM?

b) [A]= 0,005 mM?

Well, bitches – I honestly don't know if I'm right or wrong here, but here's what I think :

In the first case both enzymes are saturated because the substrate concentration is well over the Km for both of them. Thus, we must use the specificity constant; kcat/Km.

25/ 1 and 1/0,04 equals 25 in BOTH cases – thus the enzymes are equally efficient at the substrate concentration of 10 mM.

In the second case neither enzyme is saturated! However, E2 is a little closer to saturation than E1, because the latter has a higher Km. Thus : E2 works faster in this case, as it is closer to the substrate-concentration at Vmax and Km than E1! I'll try to double check this, but I do think I might just be right.

What different kinds of inhibition exists?

Competitive, un-competitive, non-competitive and mixed. Note that un-competitive and non-competitive inhibition are two different things!!!

What is a competitive inhibitor?

An inhibitor that is reversibly bound to the active site; it competes with the substrate!

Hence it is often structurally similar to the substrate.

What is the kinetic effect of a competitive inhibitor?

Since competitive inhibitors occupy the active site and hence blocks it, more substrate is required for saturation. This is evident since Km is INCREASED whereas Vmax STAYS THE SAME! The enzyme is still able to catalyze the reaction at the same velocity as before the presence of an inhibitor – BUT it takes more substrate to get there.

What defines a un-competitive inhibitor?
It binds to a DIFFERENT site than the active site, and ONLY to the ES-complex. When the un-competitive inhibitor binds to the active site, the ES-complex will be unable to change the substrate into a product!

How can un-competitive inhibition be detected through kinetics?

It lowers Vmax AND Km! This may seem strange, since a lowered Km suggest a higher affinity between enzyme and substrate. The observation is readily explained if Le Chateliers principle is considered. Since the inhibitor binds to the ES-complex, the equilibrium between substrate and enzyme is disturbed, and will be pushed to work against the change, toward the formation of more ES! Hence, kinetics will show lower Vm and lower Km – a clear indication of uncompetitive inhibition!

What is a competitive inhibitor?

An inhibitor that is reversibly bound to the active site; it competes with the substrate!

Hence it is often structurally similar to the substrate. Many important pharmaceutical agents are competitive inhibitors.

What is the kinetic effect of a competitive inhibitor?

Since competitive inhibitors occupy the active site and hence blocks it, more substrate is required for saturation. This is evident since Km is effected, whereas Vmax is NOT! The enzyme is still able to catalyze the reaction at the same velocity as before the presence of an inhibitor – BUT it takes more substrate to get there.

What defines a non-competitive inhibitor?
Non-competitive inhibitors bind at an allosteric site on the enzyme but the active site is unblocked. Thus, the K_m value is unchanged while Vmax is lowered. However, pure non-competitive inhibitors (substrate has the same affinity for E and E-I) are more or less unknown. With a mixed non-competitive inhibitor, the affinity of the E-I complex for the substrate is not the same as the unbound enzyme. In this case, not only is V_max lowered, but K_m is also raised. The Lineweaver-Burk plot will show changes in the x-intercept and increasing slope.

What is mixed inhibition?

It binds to a different site than the substrate (it is thus NON-competitive), but may bind to BOTH ES AND FREE E!

What is the kinetic sign of mixed inhibition?

Raised Km, lowered Vmax!