sakai-Enzymes Flashcards
What are the properties of the active site of an enzyme regarding its size, structure, and specificity?
The active site of an enzyme takes up a relatively small part of the enzyme.
It is a cleft formed by amino acids that are apart from each other in the primary structure.
The specificity for the substrates is high and can include optical configuration.
[human enzymes use mainly L-amino acids and D-carbohydrates. It is not clear
whether the compounds were first or the enzymes]
What is the difference between a holoenzyme and an apoenzyme?
A holoenzyme contains the apoenzyme plus the necessary coenzyme.
Some apoenzymes need metal ions in order to become the holoenzyme.
What happens after the substrate binds to the active site? What is an enzymatic transition state complex? Do all reactions go via a transition state?
After the substrate binds to the active site of the free enzyme, an enzyme-substrate
complex (ES) is formed.
The transition state complex is formed from the ES-complex by additional bonds. All reactions (catalyzed or uncatalyzed) have to go via a transition state.
Discuss the action of enzymes related to the G of a reaction and related to the energy of activation! [What can change the G of a reaction?]
Enzymes do not change the G of the reaction. Enzymes reduce the energy of activation.
[The G of the reaction can be changed by concentrations of substrates
and products or coenzymes.]
Is the reaction catalyzed by lactate dehydrogenase (LDH) reversible? Which coenzyme (cosubstrate) is needed for LDH in order to form lactate from pyruvate?
Lactate dehydrogenase catalyzes a reversible reaction, the direction depends
on the concentrations of substrates, coenzymes and products.
LDH needs NADH when pyruvate is used as substrate and in this direction, LDH forms lactate and NAD+ as products.
Explain oxidation and reduction related to electrons and hydrogen. Is pyruvate oxidized or is it reduced to lactate? Why are coenzymes absolutely necessary for oxidoreductases?
Simplified: Oxidation means donation, loss of electrons or hydrogen,
reduction means uptake, gain of electrons or hydrogen.
The structure of pyruvate is changed to the structure of lactate by uptake (gain) of
hydrogen (from NADH + H+). Therefore, pyruvate is reduced to lactate.
In this reaction, NADH is oxidized to NAD+
In the oxidation-reaction mechanism, there are always two compounds needed, one compound that is oxidized and another compound that is reduced at the same time.
When the substrate is oxidized, then the coenzyme is reduced. When the substrate is reduced, then the coenzyme is oxidized.
What is the velocity? Why is Vmax reached at very high substrate concentrations?
The velocity represents the product formation over time. It is expressed in
concentration (M/min), [whereas the rate of a reaction is expressed in actual
amounts (g/min) of products formed over time].
In specific experiments, the velocity points are measured under conditions
that the enzyme concentration is kept at a constant level in all vials, but the
respective substrate concentrations are increased in different vials.
At very high substrate concentrations, there is no free enzyme left and all the
enzyme is bound in the ES complex. With that, no more ES can be
formed, even at a further increased substrate concentration.
As the ES complex is needed for the product formation, a maximal product formation over time (Vmax) is reached.
In reactions that follow the Michaelis-Menten kinetics, what is Km? Can it be
calculated using the Michaelis-Menten equation?
Km is the Michaelis constant and it is the substrate concentration at half Vmax.
It indicates the affinity of the enzyme for the respective substrate.
Km can be derived from a graph where the velocity is plotted against the substrate
concentrations. It can also be calculated via the Michaelis-Menten equation.
The Michaelis-Menten equation is
Initial velocity V= Vmax [S] divided by Km + [S]
[when the equation is mathematically changed to the initial velocity at half
Vmax, then Km is the respective substrate concentration]
Describe Michaelis-Menten kinetics and explain the steady state assumption.
The Michealis-Menten kinetics was described over 100 years ago originally for
one substrate and a reversible binding of the enzyme to the substrate.
The enzyme forms reversibly an ES-complex which can dissociate back to enzyme and substrate, or can also dissociate into product and free enzyme.
Enzyme and substrate are in fast equilibrium with their ES complex, and it is assumed that the ES concentration does not change (steady state kinetics)
Is Vmax or is Km affected by the enzyme concentration in the assay?
The enzyme concentration changes Vmax but it does not change Km which is a
constant for a particular substrate for the investigated enzyme.
Vmax is affected by the enzyme concentration in the assay, as more enzyme leads to more product formation over time. Vmax is larger in an assay that contains more enzyme.
[That is why the enzyme concentration has to be kept constant and only the substrate concentration is increased in the Michelis-Menten kinetics in order to determine Km.]
Km is a constant and is not affected by the enzyme concentration in the assay.
Km is the substrate concentration at half Vmax (no matter how large the
respective Vmax is in these experiments) and represents the affinity of the
enzyme for the tested substrate.
[The enzyme has a high affinity for the substrate when half Vmax is reached at a low substrate concentration, whereas when the enzyme has a low affinity for the substrate, a higher substrate concentration is needed to reach half Vmax.]
As an example, the enzyme A has a smaller Km than enzyme B for the same substrate. Does enzyme A have a higher or smaller affinity for the substrate than enzyme B? Explain!
Enzyme A with the smaller Km for the same substrate, has a higher affinity for the substrate than enzyme B. Less substrate is needed to achieve half Vmax
[an example is hexokinase (enzyme A) which has a much smaller Km for glucose than glucokinase (enzyme B).
Hexokinase, which is found in most cells, has a much higher affinity for glucose than glucokinase which is found mainly in the liver and -cells of pancreas. Glucokinase is a special type of hexokinase that has a large Km and large Vmax for glucose.]
What is the significance of the Km for a specific substrate?
Km is a constant and once established for the enzyme, it can be used in different laboratories as indicator of the affinity of the enzyme for this substrate. Also, the affinity for similar substrates can be discussed and compared, including drugs.
Describe an allosteric enzyme. Do allosteric enzymes follow Michaelis-Menten kinetics and show a hyperbolic curve? If not, how does the curve look like? Is
Vmax reached?
Allosteric enzymes consist of several protein subunits and show cooperative substrate binding. Their kinetics do not show a hyperbolic curve, the curve is
sigmoidal.
As it is an enzymatic catalysis, a Vmax is reached where all enzyme is in the ES
complex. The substrate concentration at half Vmax, is named K 0.5 (instead of Km), and is an indicator for the affinity of the allosteric enzyme for the substrate.
What is the reason for the sigmoidal shape of the curve for allosteric enzymes?
The velocity suddenly increases at higher substrate concentrations due to the cooperative substrate binding.
[compare to cooperative oxygen binding in the protein hemoglobin]
Back to Michaelis-Menten kinetics. Does a competitive inhibitor directly interfere with the formation of the ES complex, or does it directly interfere with the product formation? Can a competitive inhibition be overcome at high substrate concentrations? What is an apparent Km?
A competitive inhibitor interferes directly with the formation of the ES complex and competes with the substrate for binding to the enzyme. The inhibited product formation is the result of less ES formation due to competition.
[ The experiments are performed with constant enzyme concentrations and
constant inhibitor concentrations, whereas only the substrate concentration is
increased.]
At very high substrate concentrations, less inhibitor molecules can form the EI complex as more S is competing for formation of the ES complex and even the already formed EI complex is changed to the ES complex in reversible inhibitions. The inhibitor molecules are by far outnumbered.
With that, the competitive inhibition can be overcome at high substrate concentrations.
Km is a constant and represents the affinity of the enzyme for the substrate. In
competitive inhibition, we find a larger, apparent Km as the inhibitor interferes
with the normal formation of the ES complex and the substrate concentration at
half Vmax is higher.
The inhibitor does not change the affinity for the substrate, but more substrate is
needed in this condition to result in half Vmax due to the competitive inhibitor
Describe the binding and action of a noncompetitive inhibitor. Do you find an apparentVmax or an apparent Km?
A noncompetitive inhibitor binds to the enzyme or to the ES complex at another site which is not the catalytic site for the respective substrate.
Product formation is not possible from the ESI complex, and the Vmax of the uninhibited reaction is not reached, even at high [S]. Thereached apparent Vmax
is smaller.
Km, the substrate concentration at half Vmax, is the same, as the inhibitor does
not interfere with the formation of the ES complex.
[The half Vmax of the uninhibited reaction and the half apparent Vmax of the inhibited reaction are both achieved at the same substrate concentration.]
