enzyme kinetics

Question 1

Catalyst

Answer 1

a compound that increases the rate of a chemical reaction (e.g. making and breaking covalent bonds) without being changed in the process. Thus, enzymes accelerate specific reactions needed in a biological system. They do so by forming a specific three-dimensional structure with an active site.

Question 2

Enzyme characteristics

Answer 2

They are (primarily!) proteins with specific structures and active sites formed when the protein folds into its three dimensional shape. The formation of a highly ordered active site makes the enzyme very specific for its substrates. Some use specific cofactors or coenzymes. Coenzymes provide chemical groups for the reaction and are “used up.” Cofactors are not. Enzymes are classified and named by what they do.

Question 3

cofactor

Answer 3

a non-protein chemical compound that is required for the protein’s biological activity. These proteins are commonly enzymes, and cofactors can be considered “helper molecules” that assist in biochemical transformations.

Question 4

coenzyme

Answer 4

a complex organic or metalloorganic cofactor, most of which are derived from vitamins and from required organic nutrients in small amounts. Metabolism involves a vast array of chemical reactions, but most fall under a few basic types of reactions that involve the transfer of functional groups. These group-transfer intermediates are the loosely bound organic coenzymes. Each class of group-transfer reaction is carried out by a particular cofactor, which is the substrate for a set of enzymes that produce it, and a set of enzymes that consume it. An example of this are the dehydrogenases that use nicotinamide adenine dinucleotide (NAD+) as a cofactor.

Question 5

prosthetic group

Answer 5

A coenzyme or cofactor that is tightly bound to the enzyme

Question 6

holoenzyme

Answer 6

An inactive enzyme without the cofactor is called an apoenzyme, while the complete enzyme with cofactor is called a holoenzyme.

Question 7

reaction equilibria

Answer 7

DG’o =-RTlnKeq When the energy of product is lower than substrate,
S will spontaneously convert into P. However, deltaG’° says nothing about the
rate at which this occurs.The rate of the reaction is dependent upon the activation energy, deltaG‡

Question 8

Transition State (TS) Theory

Answer 8

there is a single path (sequence of chemical structures) for S to convert to P, and the TS is the highest energy structure along this path. The TS is equally likely to either proceed to form P or decay back to form S. The TS’s existence is very short lived – on the order of a molecular bond vibration. Thus, the rate of a reaction depends on the rate at which the ground state structures can form the TS. It has nothing to do with deltaG°’, rather, it is related to deltaG‡.

Question 9

How enzymes work

Answer 9

The enzyme has decreased DG‡ (stabilized the transition state). Thus, the energetic “barrier” between the substrate and products is lower. This accelerates the RATE of the reaction. But, the equilibrium of the reaction HAS NOT CHANGED.

Question 10

rate constant (k)

Answer 10

k = (κT/h) e-DG‡/RT where κ= Boltzmann constant h = Planck’s constant
T = temperature. Therefore, since Velocity (V) = k [S], as the rate constant increase, the velocity of the reaction increases

Question 11

How do enzymes lower DG‡?

Answer 11

1. Binding energy between enzyme (E) and substrate (S) to form the ES complex. Multiple, weak noncovalent interactions that provide specificity as well as catalysis. Enzymes are optimized to bind the transition state. 2. Increase “local concentration” of substrates. 3. Rearrangement of covalent bounds in specific active site chemistry. (eg acid-base, covalent, and metal ion catalysis)

Question 12

how enzymes stabilize the transition state structure

Answer 12

Enzyme active sites make specific non-covalent interactions with the substrate (hydrogen bonding, salt bridges, hydrophobic, etc.). These specifically interact with the transition state.

Question 13

How enzymes increase “effective concentration”

Answer 13

The ES complex is essentially a single unit rather than a sum of many interactions as might occur in bulk solution. Both are entropically unfavorable, but a pre-formed active site is less so. Further, the active site orders multiple substrates, thus increasing their “effective concentration.” The active site can bring together multiple substrates, therefore greatly increases reaction probability

Question 14

acid-base catalysis

Answer 14

Proton donors and acceptors, i.e. acids and base may donate and accept protons in order to stabilize developing charges in the transition state.This typically has the effect of activating nucleophile and electrophile groups, or stabilizing leaving groups.

Question 15

Covalent catalysis

Answer 15

Covalent catalysis involves the substrate forming a transient covalent bond with residues in the active site or with a cofactor. This adds an additional covalent intermediate to the reaction, and helps to reduce the energy of later transition states of the reaction. The covalent bond must, at a later stage in the reaction, be broken to regenerate the enzyme.

Question 16

Metal ion catalysis

Answer 16

A metal such as Cu2+ or Zn2+ can also stabilize the TS. The metal must be able to be bound to the charged intermediate and hence the TS. The tetrahedral oxyanion intermediate of the reaction of an electrophilic carbonyl C can interact with a metal if there is an O on an adjacent atom which can help coordinate the metal ion. This charge stabilization of the developing negative in the TS and the full negative in the intermediate is often called electrostatic catalysis. This method is likely to be found in many enzymes since nearly 1/3 of all enzymes require metal ions.

Question 17

Measuring enzymatic parameters

Answer 17

step 1: Measure initial velocity, Vo, at one [S] step 2: Repeat your measurement at many [S] step 3: Create Vo vs. [S] plot

Question 18

Km

Answer 18

the substrate concentration at which the reaction rate is half of Vmax. The Km is not the same as a binding constant (Kd). The Km can sometimes be a reflection of the binding affinity of E towards S. Indeed, typically Km values are approximately equal to intracellular S concentrations. But this is not always the case

Question 19

Vmax

Answer 19

the maximum rate achieved by the system, at maximum (saturating) substrate concentrations.

Question 20

Michaelis–Menten kinetics

Answer 20

V=[Vmax [S]]/[[S] + Km ]

Question 21

kcat

Answer 21

the rate limiting step of any enzyme catalyzed reaction. kcat is the “turnover number” for an enzyme. It measures the number of substrate molecules turned over (into product) per enzyme molecule, per second. If several rate constants are partially rate limiting (generally, within an order of magnitude of the slowest rate constant), then kcat does not correspond to any single rate constant, but is a composite rate constant. The inverse is the amount of time for each subsrate molecule to be processed

Question 22

kcat/Km

Answer 22

is a useful way to compare enzymes. the larger this number, the more efficient this enzyme

Question 23

Reversible inhibition

Answer 23

types include: competitive: Bind only to E (enzyme without substrate). Uncompetitive: Bind only to ES complex. Mixed: Bind to both the E and ES complex. noncompetitive inhibitors are a type of mixed.

Question 24

competitive inhibition

Answer 24

the substrate and inhibitor cannot bind to the enzyme at the same time, as shown in the figure on the left. This usually results from the inhibitor having an affinity for the active site of an enzyme where the substrate also binds; the substrate and inhibitor compete for access to the enzyme’s active site. This type of inhibition can be overcome by sufficiently high concentrations of substrate (Vmax remains constant), i.e., by out-competing the inhibitor. However, the apparent Km will increase as it takes a higher concentration of the substrate to reach the Km point, or half the Vmax. Competitive inhibitors are often similar in structure to the real substrate. kcat is unaffected while the apparent Km increases as [I] increases. Therefore Km changes but Vmax does not change

Question 25

uncompetitive inhibition

Answer 25

the inhibitor binds only to the substrate-enzyme complex, it should not be confused with non-competitive inhibitors. This type of inhibition causes Vmax to decrease (maximum velocity decreases as a result of removing activated complex) and Km to decrease (due to better binding efficiency as a result of Le Chatelier’s principle and the effective elimination of the ES complex thus decreasing the Km which indicates a higher binding affinity). All measurable parameters are affected by an uncompetitive inhibitor (kcat, Vmax & Km).

Question 26

Methotrexate

Answer 26

is a common cancer drug. Looks a lot like the true substrate

Question 27

mixed inhibition

Answer 27

the inhibitor can bind to the enzyme at the same time as the enzyme’s substrate. However, the binding of the inhibitor affects the binding of the substrate, and vice versa. This type of inhibition can be reduced, but not overcome by increasing concentrations of substrate. Although it is possible for mixed-type inhibitors to bind in the active site, this type of inhibition generally results from an allosteric effect where the inhibitor binds to a different site on an enzyme. Inhibitor binding to this allosteric site changes the conformation (i.e., tertiary structure or three-dimensional shape) of the enzyme so that the affinity of the substrate for the active site is reduced. All measurable parameters are usually affected by a mixed inhibitor (kcat, Vmax & Km).

Question 28

noncompetitive inhibition

Answer 28

a form of mixed inhibition where the binding of the inhibitor to the enzyme reduces its activity but does not affect the binding of substrate. As a result, the extent of inhibition depends only on the concentration of the inhibitor. Vmax will decrease due to the inability for the reaction to proceed as efficiently, but Km will remain the same as the actual binding of the substrate, by definition, will still function properly. kcat and Vmax are affected,
but the Km remains unchanged.

Question 29

Irreversible inhibitors

Answer 29

Irreversible inhibitors usually covalently modify an enzyme, and inhibition can therefore not be reversed.

Question 30

Lineweaver–Burk plot

Answer 30

The Lineweaver–Burk plot was widely used to determine important terms in enzyme kinetics, such as Km and Vmax, before the wide availability of powerful computers and non-linear regression software. The y-intercept of such a graph is equivalent to the inverse of Vmax; the x-intercept of the graph represents −1/Km. It also gives a quick, visual impression of the different forms of enzyme inhibition.

Question 31

Allosteric regulation of enzyme

Answer 31

the regulation of a protein by binding an effector molecule at a site other than the protein’s active site. The site the effector binds to is termed the allosteric site. Allosteric sites allow effectors to bind to the protein, often resulting in a conformational change. Effectors that enhance the protein’s activity are referred to as allosteric activators, whereas those that decrease the protein’s activity are called allosteric inhibitors. Allosteric regulations are a natural example of control loops, such as feedback from downstream products or feedforward from upstream substrates. Long-range allostery is especially important in cell signaling. Allosteric regulation is also particularly important in the cell’s ability to adjust enzyme activity.

Question 32

Covalent modification of the enzyme

Answer 32

active and inactive form of the enzyme are interconverted by covalent modification of their structures which are catalysed by other enzymes. This type of regulation consists on the addition and elimination of some molecules which can be attached to the enzyme. The most important groups that work as modificators are phosphoryl, methyl, uridyl, adenylyl and adenosine diphosphate ribosyl. These groups are joined to or eliminated from the protein by other enzymes. The most remarkable covalent modification is phosphorylation.

Question 33

Binding of another regulatory protein

Answer 33

the binding of another protein regulates the activity of the enzyme

Question 34

Proteolytic cleavage of the enzyme

Answer 34

Some enzymes need to go through a maturation process to be activated. A precursor (inactive state, better known as zymogen) is first synthesized, and then, by cutting some specific peptide bonds (enzymatic catalysis by hydrolytic selective split), its 3D conformation is highly modified into a catalytic functional status, obtaining the active enzyme.