Biol/Chem 5310

Lecture: 17

October 24, 2002

Summary of Michaelis-Menten Kinetics

If a plot of v vs. [S] is hyperbolic, revealing saturation ( or more easily recognized, if 1/v vs. 1/[S] is a straight line) the enzyme is said to obey Michaelis-Menten kinetics with respect to that substrate.

For a hyperbolic curve, mathematically:

i.e. the rate is equal to something "a" times [S] divided by something "b" plus [S]

• a is equal to Vmax
• b is equal to Km, the concentration of substrate when v = 1/2 Vmax
• at High [S], where [S]>>Km, then v = Vmax, a constant with no dependence on [S]
• at low [S], where [S]<<Km, then v =(Vmax/Km)[S]
• in this case, the rate is proportional to [S] with apparent first order rate constant of Vmax/Km

Remember, Km is the substrate concentration at which the rate is half-maximal. In general it is a function of several individual rate constants. In the simple scheme we have considered, shown below,

In the event that k-1 >> k2 , equilibrium in binding will occur, and Km will be equal to Ks, where

Km is not in general equal to the dissociation constant of the substrate, although it correlates with it:

High Km --> loose binding

Low Km --> tight binding

The catalytic constant kcat is defined as

Vmax divided by the total enzyme concentration ([E]T). It has units of (sec)-1; , so it is a first order rate constant. It is sometimes called the turnover number, because it reflects how many times the enzyme can turnover per second. In the simple scheme above, k cat is equal to k 2 , because Vmax = k 2 [E]T

In general, kcat will be a function of the rate constants in the reaction scheme.

A special case exists when [S] << Km

In this case [E] is approximately equal to [E]T

because [ES] << 0

so,

so k cat / Km is an apparent second order rate constant for E and S

This is sometimes called the affinity constant, its value reflects the enzyme's catalytic efficiency.

High efficiency will be associated with k2 >> k-1

i.e. product formation over release of substrate (back reaction)

The limit of efficiency will be limited by the rate of enzyme encountering the substrate. This is the "diffusion-controlled limit" which is ~108 - 109 M-1 s-1

Enzymes with k cat / Km > 108 M-1 s-1, such as acetylcholinesterase, are sometimes called "perfect enzymes"

Reversible Inhibitors: 3 general categories

1) Competitive: binds at the active site, competes with substrate. Only one can occupy the binding site at a time. Competitive inhibitors must look like the substrate, but are unable to be converted to products. They are usually the same size or smaller than the substrate. Example, malonate is a competitive inhibitor of succinate dehydrogenase, which converts succinate to fumarate.

{see structures}

Here [E]T = [E] + [ES] + [EI]

where

EI cannot bind substrate.

Define a :

a is equal to or greater than 1

It can be seen that the effect of a competitive inhibitor is to raise the apparent Km, but does not affect the Vmax.

Vmax is unaffected by a competitive inhibitor.

The apparent Km is increased by a competitive inhibitor.

See the Animation of Fig.12-7

2) Mixed Inhibition (Noncompetitive is a special case)

A mixed inhibitor binds to both E and ES, not at the substrate binding site:

Note that there are 2 Ki's. If they have the same value, i.e. the inhibitor binds equaly well to E and ES, then a mixed inhibitor is usually called a noncompetitive inhibitor.

and

In the presence of a mixed inhibitor, the following Lineweaver-Burk plot is obtained:

Both the apparent Km and the apparent Vmax are changed by the inhibitor:

or:

If , then the inhibitor is called noncompetitive, and a different looking Lineweaver-Burk plot is obtained:

In this case, the presence of the inhibitor does not affect the x-intercept.

But the Vmax is reduced by the inhibitor:

See the Animation of Fig. 12-9

3) Uncompetitive inhibition occurs when the inhibitor binds only to the ES complex:

Both Vmax and Km are reduced by the same factor, a'

In a Lineweaver-Burk plot, this leads to parallel lines:

See the Animation of Fig. 12-8

In general, Ki's can be determined, if a  values can be calculated from the plots, and if the concentration of Inhibitor is known.

For example, if a parallel line is obtained, as shown above, and Km=6mM, and Km(app)=4mM:

then, a'=1.5

If [I]=10 mM, then solve 1.5 = (1+10/x)

so x=Ki' = 20 mM (Notice that the units of Ki's come out as the units of I that are put in.)

Regulation of Enzyme Activity

1) Control of enzyme levels

• gene expression, on or off

2) Control of enzyme activity: fine tuning, up and down, example aspartate transcarbamoylase

carbamoyl phosphate   +   aspartate      ------->         N-carbamoyl aspartate   +         phosphate

N-carbamoyl aspartate is committed to go (in 6 steps) on to cytidine triphosphate (CTP), a DNA precursor.

A --> B --> C --> D --> E --> F --> G

• Each arrow represents an enzyme-catalyzed reaction
• The final product shown, G, inhibits the first reaction A --> B
• This is called Feedback Inhibition
• In a committed pathway it is more efficient to regulate at the first step.
• Under other conditions, it may be useful to activate the first enzyme,rather than inhibit it.

How does this occur? Allostery

Consider Aspartate transcarbamoylase, ATCase:

The crystal structure was first solved by William Lipscomb

ATCase has 12 subunits (see Chime link):

• 6 C-catalytic in 2 trimers
• 6 R-regulatory in 3 dimers
• ATCase shows positive cooperativity with respect to the substrate aspartate
• CTP and ATP are allosteric effectors
• CTP is an inhibitor
• ATP is an activator
Because of base pairing in DNA, the levels of purines (A and G) and pyrimidines (C and T) must be equal.
In the absence of regulatory subunits, the catalytic subunits are non-cooperative with respect to aspartate, and not affected by ATP and CTP
• ATP stabilizes the R-state form
• CTP stabilizes the T-state form
In the R-state, 2 domains of the catalytic subunits are bought closer together, to facilitate catalysis

Try the Chapter 12 quiz.