5. Enzymology Flashcards
coenzyme
A coenzyme is a substrate for its enzyme, and acts as a transient carrier of specific chemical (functional) groups.
In this role, they are used up during the reaction, but are usually kept at a constant, steady state level in the cell. Coenzymes do not form a permanent part of the enzymes structure.
Most coenzymes are derived from vitamins.
cofactors
Cofactors are non-protein groups that bind to the enzyme which are required for activity. Cofactors are not used up during the reaction.
Most cofactors are prosthetic groups or metal ions.
What are two important organic molecules that serve as cofactors?
- Flavin (prosthetic group involved in oxidation-reduction reactions)
- Heme (metalloorganic prosthetic group involved in oxygen binding)
Even though heme is considered a cofactor, why is hemoglobin not considered an enzyme?
Hemoglobin is not an enzyme, as it does not catalyze any chemical reaction.
oxidoreductase
Enzymes that transfer electrons (hydride ions or H atoms).
transferases
Enzymes that group transfer reactions.
hydrolases
Enzymes that perform hydrolysis reactions (transfer of functional groups to water).
lyases
Enzymes that add groups to double bonds, or form double bonds by removal of groups.
isomerases
Enzymes that transfer groups within molecules to yield isomeric forms.
ligases
Enzymes that form C-C, C-S, C-O, and C-N bonds by condensation reactions coupled to ATP cleavage.
enzyme
Enzyme is a catalyst that speeds up chemical reactions so that they occur at a useful rate. As a catalyst, the enzyme is not consumed during the reaction.
active site
The distinguishing feature of enzymes as catalysts is that the reaction takes place within the confines of a “pocket” on the enzyme called the active site.
The molecule that binds within the active site is called the substrate.
The active site is formed from amino acids that _______________________ and _______________________.
- Specifically recognize the substrate
- Accelerate the chemical reaction.
These two features are distinct, i.e. specificity does not guarantee rate acceleration and vice versa.
Describe the enzymatic activity of triose phosphate isomerase.
Triose phosphate isomerase in glycolysis catalyzes the conversion of glyceraldehyde-3-phosphate into dihydroxyacetone phosphate via an (unstable) endiol intermediate.
Describe what V0 is and deduce the units of V0.
The initial velocity, or V0, is the amount of product produced per unit time at the start of the reaction for a certain concentration. In layman terms, it means the rate when you’ve just combined the enzyme and substrate, and the enzyme is catalyzing the reaction as fast as it can at that particular substrate concentration. During this time, the product concentration is increasingly linearly.
K has units of s-1 whereas [S] has units of M. Thus, V has units of concentration/time, e.g., M/s.
What occurs to an enzyme at steady state conditions?
Steady state is when the rate at which the substrate is taken up by the active site is equal to the rate at which product is released by the active site.
In layman terms, the flux in is equal to the flux out.
What occurs to the active site (and therefore V) when there are high concentrations of substrate?
At high enough substrate concentrations, all of the active sites are converting S into P. This indicates that the enzyme is saturated, and the rate of reaction will no longer increase when additional substrate is added.
The velocity increases from V0 to Vmax.
What occurs to the active site (and therefore V) when there are low concentrations of substrate?
At low substrate concentrations, only a fraction of active sites are converting S into P.
What does Km stand for?
Why is Km useful?
The substrate concentration that gives you a rate that is halfway to Vmax is called the Km (0.5 x Vmax).
Km is a useful measure of:
- how quickly the reaction rate increases with substrate concentration.
- an enzyme’s affinity for its substrate.
Acetylcholine binds to acetylcholinesterase (Km = 9 x 10-5), whereas CO2 binds to carbonic anhydrase (Km = 1.2 x 10-2).
Which is the more efficiency or “better” enzyme? Explain why.
Acetylcholinerase.
- Smaller Km needs less substrate to engage half of the active sites, and is therefore considered to be a better enzyme.
- Larger Km needs more substrate to engage one half of the active sites.
What does [S] = Km indicate?
When [S] = Km, exactly 50% of active sites are converting S into P.
This graph is of an enzyme that is not under steady state conditions. Explain why V0 for the enzymes are considered the tangents of the curve.
When the substrate concentration is not under steady state conditions, V0 is the maximum possible reaction velocity or the initial velocity at a particular substrate concentration S, before S starts depleting.
A tangent to each curve taken at time 0 defines the initial velocity, V0, of each reaction.
What is Kcat?
What is the equation for Kcat?
Kcat is the number of substrate molecule each enzyme site converts to product per unit time.
A higher Kcat indicates a more efficient enzyme.
Kcat = Vmax/[E0]
What is the Michaelis-Menten equation?
Enzymes that display a hyperbolic curve can often be described by an equation relating substrate concentration [S], initial velocity Km, and maximum velocity Vmax, known as the Michaelis-Menten equation.
What is tavg?
tavg is the average amount of time for the substrate to go through the active site.
T/F: Km should be thought of as the affinity of the substrate for the enzyme’s active site in the steady state.
True. Km should be thought of as the affinity of the substrate for the enzyme’s active site in the steady state, not as the affinity of the substrate for the active site at equilibrium.
What is the significance of kcat/Km?
Deduce the units of kcat/Km.
kcat/Km is an apparent 2nd order rate constant at lowest concentrations of S, and therefore sets the lower limit for any 2nd order elementary rate constant in the reaction mechanism.
kcat/Km has units of M-1s-1.
Draw a enzyme kinetics function indication Km and Vmax.
What is the inverse of the Michaelis-Menton function?
Be able to deduce the equation.
The inverse of the Michaelis Menten functionis called the double-reciprocal or Lineweaver-Burk plot.
Draw a Lineweaver-Burk plot, and indicate the following using the Lineweaver-Burk equation:
- x-axis
- y-axis
- slope
- y-intercept
What is considered to be “catalytic perfection?”
Catalytic perfection is displayed in enzymes that turnover substrate as fast as the enzyme and substrate can diffuse together.
That is, kcat/Km is close to the diffusion-controlled limit, which means as soon as the enzyme and substrate come together, the reaction occurs. Therefore, the rate-limiting step is the diffusion for these enzymes, and not the chemistry.
T/F: Catalyst do NOT alter the equilibrium, i.e., they affect ΔG0.
False. Catalysts do NOT alter the equilibrium of the reaction, i.e., they do NOT affect ΔG‘0.
How do catalysts speed up chemical reactions?
Be able to draw a kinetic graph that is uncatalyzed and enzyme-catalyzed.
Catalysts speed up chemical reactions by stabilizing the transition state (TS).
The activation energy ΔG‡ is between the two end states substrate and product is lowered by the catalyst, which increases the rate constant.
How do enzymes specifically recognize their substrates?
- The enzyme active site is formed from amino acid side chains that form many weak, non-covalent interactions with the substrate (hydrogen bonding, salt bridge, hydrophobic, etc.).
- Further, the shape of the active site accommodates the substrate molecule.
Thus, a potential substrate molecule must have the correct shape to fit into the active site, and importantly, all the H bonding donors/acceptors and charged groups must have precisely positioned partners within the active site.
Suppose that an enzyme binds to two substrates. During an enzyme-catalyzed reaction, why does the enzyme re-position the two substrates?
In the enzyme catalyzed reaction, binding of A and B to the enzyme’s active site positions the molecules precisely so that the functional groups of A and B can react. Therefore, the active site greatly increases the probability of a reaction.
In fact, by bringing two molecules together in an enzyme’s active site, we can expect rate enhancements on the order of 105 to 108.
What is the “effective concentration” of an enzyme?
The rate enhancement of an enzyme is considered to be the “effective concentration” of an enzyme.
The “effective concentration” is the hypothetical concentration the reactant would have to be, free in solution, to experiences the same collisional frequency.
What are the two models describing the interaction between a substrate and an enzyme?
- Lock and key theory
- Induced fit theory
What is the lock and key hypothesis?
In 1894, Emil Fisher hypothesized that enzymes are structurally complementary t their substrates, and fit together like a lock and key.
This theory was disproved because it could not explain how enzymes accelerate chemical reactions. In fact, if the enzyme was to bind strongly to the substrate, the enzyme-substrate would be lower in energy, and the activation barrier would be much higher than in an uncatalyzed reaction.
How does a catalyst decrease the activation barrier?
That is, should an enzyme bind strongly to a substrate or its transition state?
The main function of a catalyst is to decrease the activation barrier. Therefore, enzymes should bind strongly to the transition state, thus offsetting the higher activation energy with the binding energy.
Enzymes recognize substrates through multiple weak bonds and interactions. How are these bonds strengthened upon binding of the enzyme and substrate?
Weak interactions between the enzyme and substrate are optimized and strengthened in the transition state (TS), thus compensating the large activation barrier with the binding energy.
Draw enzyme kinetic graphs of the following reactions:
- No enzyme
- Lock and key theory
- Induced fit theory
What are two experiments that provide evidence for the induced fit theory?
Experiment 1: Some TS analogs bind the enzyme with 102 to 106 fold higher affinity than the substrate.
Experiment 2: Monoclonal antibodies were produced to recognize TS analogs. These mAbs werethen able to catalyze the cleavage of esters with 102 to 103 fold rate enhancements!
What are the three types of enzymatic catalysis?
- General acid/base catalysis (vs. specific)
- Metal ion catalysis
- Covalent catalysis
Describe a general base catalysis.
In specific base catalysis, a water hydroxide anion serves as the base.
Describe a general acid catalysis.
In specific acid catalysis, a water hydronium ion serves as the acid.
What kind of catalysis is taking place in this picture?
Describe what is occuring.
General base catalysis
- Uncatalyzed hydrolysis of an ester involves an unstable charge distribution in the TS.
- This reaction can be catalyzed by the acetate ion, by stabilizing the unfavorable charge distribution that develops on the attacking water molecule in the TS. This is called general-base catalysis. In specific base catalysis, a water hydroxide anion serves as the base.
What kind of catalysis is taking place in this picture?
Describe what is occuring.
General acid catalysis
- Uncatalyzed hydrolysis of an acetal (a carbon with two single bonded oxygen atoms) involves an unstable charge distribution in the TS.
- This reaction can be catalyzed by the acetic acid, by stabilizing the unfavorable charge distribution that develops on the oxygen atom in the TS.
Describe how the following amino acid side chains can function in general acid/base catalysis:
- Glu/Asp
- Lys/Arg
- Cys
- His
- Ser
- Tyr
How do metal ions facilitate catalysis?
Metal ion cofactors can be positioned to stabilize charged TS and/or intermediates.
What are the two requirements for covalent catalysis?
- The catalyst must be a better nucleophile than water
- The covalent intermediate must be less stable to hydrolysis than the substrate
Nucleophiles are chemical species that donates an electron pair to form a chemical bond in relation to a reaction. In order to be a better nucleophile, the chemical species must have more electrons.
How does covalent catalysis affect the enzyme kinetics graph?
In covalent catalysis, the reaction has been “broken up” into pieces, each with a lower ∆G≠ than the uncatalyzed reaction.
Again, the ∆G0 has not been affected.
Describe why chymotrypsin, an enzyme found in the digestion system, is considered a serine protease.
Chymotrypsin is considered a serine protease, because serine acts as the catalyst at its active site.
The hydroxyl group of the serine amino acid serves as the nucleophile (covalent catalysis), and specifically cleaves peptide bonds that are adjacent to hydrophobic amino acids.
Chymotrypsin displays which of the following enzymatic catalysis methods?
- General acid/base catalysis (vs. specific)
- Metal ion catalysis
- Covalent catalysis
What are the seven steps of chymotrypsin enzymatic catalysis?
- E meets S
- Formation of the ES complex
- Short-lived intermediate and backbone cleavage
- Acyl-enzyme intermediate
- Fate of the intermediate
- Short-lived intermediate and second product
- Enzyme-product 2 complex
What is the first step of chymotrypsin?
What is the second step of chymotrypsin?
What is the third step of chymotrypsin?
What is the fourth step of chymotrypsin?
What is the fifth step of chymotrypsin?
What is the sixth step of chymotrypsin?
What is the seventh step of chymotrypsin?
What are allosteric enzymes?
Certain enzymes function as regulatory checkpoints along various pathways.
Their function is considerably more complex because they contain multiple active sites.
Allosteric enzymes fine-tune a pathway.
What is the difference between homoallosteric and heteroallosteric enzymes?
Homoallosteric enzymes are enzymes act as both a regulator and a substrate of a pathway.
Heteroallosteric enzymes are enzymes that regulate a pathway, but are not a substrate of the pathway.
What is the typical structure of allosteric enzymes?
How are they activated?
Allosteric enzymes are typically made up of both catalytic domains (C) and regulatory domains (R).
Positive effector molecules activate catalysis at the catalytic domain by binding at the regulatory domain.
An allosteric enzyme may also be activated by its own substrate-homotropic cooperativity.
Describe the kinetic curve of such an enzyme.
A sigmoidal velocity curve is the result.
Note the nonlinear velocity curve at low S values.
A heterotropic positive or negative modulator can significantly influence an allosteric enzyme.
Describe the kinetic curve of such an enzyme. Include positive and negative modulation.
Note how much the Km can change by simply adding a positive or negative modulator.
Why is allosteric modulation so important?
Small changes in effector or substrate concentrations can result in large changes in enzyme activity. That is, conversion of S to P can be regulated like a switch.
