Official Ex-Notehall Study Guide BIOC 384 Block II Learning Objectives

8/2/2019 Official Ex-Notehall Study Guide BIOC 384 Block II Learning Objectives

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BIOC 384: EXAM II LEARNING OBJECTIVES


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Lecture XIII

Myoglobin, Hemoglobin

Terminology: ligand. Fractional saturation prosthetic group, cooperativity, protomer,binding site, allosteric (allosteric site, allosteric effector, allosteric regulation).Term Definition

Ligand An ion or molecule (usually) small that is bound by another molecule (usually

large).

Fractional Saturation Fraction of total binding sites on protein occupied by ligand, signified by Y.

Calculated by formula: =[!"]

!" ![!]=

[!]

!!![!].

Prosthetic Group Metal ion or organic or metallo-organic compound other than an amino acid thats

tightly bound to a protein (binding is covalent or tight non-covalent), required orthe proteins function/activity.

+ Cooperativity Affinity of binding sites is increased (positive cooperativity) or decreased

(negative cooperativity) upon the binding of a ligand to a binding site. Basically,

it is binding of a ligand to 1 binding site causes the effects of the properties (in theform of binding affinities) of other binding sites (on other subunits) of the same

protein molecule.

Protomer Structural unit of an oligomeric protein, can consist of one or several subunits

that assemble to form an oligomer. For example, hemoglobin is a dimer of two protomers.

Binding Site Site where specific, typically noncavalent interactions between molecularsurfaces occur. The essence of protein function/action is BINDING (recognition of

and interaction with other molecules). It is dependent on three major factors:

- Shape complementarity: Requiring lots of van der Waals interactions- Chemical complementarity: Requiring hydrogen bonds and ionic

interactions

- Hydrophobic effect: Hydrophobic ligand minimizes exposure to water bybinding in hydrophobic site in protein.

Allosteric (Allosteric site,

allosteric effector, allostericregulation)

Binding of ligand to one site on protein molecule affects binding properties/

affinities of another site on same protein molecule.Homotropic implies that the interacting sites all bind to the same ligand.

Heterotropic effects imply interacting sites binding to the same ligand.

Allosteric Site: Site other than the proteins active site.

Allosteric Effector/Modulator: Ligands that spur allosteric effects. They canaffect the equilibrium between the T and R states.

Allosteric Regulation: Regulation of an enzyme or other protein by binding an

effector molecule at the proteins allosteric site. Allosteric activators enhance theproteins activity, while allosteric inhibitors decrease the proteins activity.

Briefly describe the tertiary structure of Mb and Hb subunits (the globin fold),

explain how the helices are designated, and the roles of the proximal and distal His

residues in heme and oxygen binding.Myoglobin and hemoglobin subunits are similar in polypeptide sequence and contain a heme, and their overallstructure of subunits are similar. More notably, they have a similar tertiary structure, yet a different

quaternary structure. Myoglobin is a monomeric protein consisting of a single polypeptide chain with 153

amino acid residues, while hemoglobin is heterotetrameric, consisting of two

(141 AA residues) and two chains (146 AA residues). The quaternary structureof hemoglobin confers then allosteric properties. Myoglobin can bind to oxygen as

well as release. Hemoglobins function in vertebrate erythrocytes is to transportO2 form lungs to tissues and transport CO2 from tissues to lungs. Myoglobin ismainly the storage element within muscle tissue. It binds to O2 delivered to tissue

by blood and stored until needed as terminal electron acceptor for energy


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metabolism, and is the main intracellular transport for oxygen. To be a potent carrier, it needs to be able tocarry as well as release its ligand. The proximal and distal histidine residues in the heme group are typically the

main point of interest in myoglobin and hemoglobin. They play a role in the cooperativity of hemoglobin andassociated metal complex.

Remember that hemoglobin, as a structure, is very compact, with almost no empty space inside. It is mostly

(~75%) -helical, the rest mostly turns and loops (at the surface). There are 8 helices, designated A-H from Nto C terminus. The helices at the surface are amphipathic and the helices are termined by proline. The polar R-

groups are mostly at the surface, except HIS (F8)/fifth ligand (proximal) and HIS (E7)-near sixth coordinationsite but distal. As a quaternary structure it is a tetramer with two and two subunits in adult (and different at

different stage of life). Noncovalent bonds stabilize this structure. The / interactions dominate.

Write a general protein-ligand binding/dissociation reaction in both the association

and dissociation directions. What is the mathematical relationship between the

association and dissociation equilibrium constants?We know that the general reaction for ligand binding is: [] + [] []. From there we can determine theequilibrium expressions in both the association and dissociation directions. From there, we also know that the

association constant is the reciprocal of the dissociation constant.

Direction Equilibrium Expression Binding

Strong Weak

Association Constant!=

[]!"

[]!"[]!"

Large Small

Dissociation Constant! =

[]!"[]!"

[]!"=

1

!

Small Large

Describe how and where in the structure of Mb and Hb O2 binds, including roles of

protein functional groups and heme, and the oxidation state of the heme Fe required

for O2 binding.Remember that myoglobin is the oxygen storage protein (a single subunit) in muscle.

It binds to oxygen in muscle cells and kept until needed. Hemoglobin is the oxygentransport protein in the blood. It circulates in red blood cells, and binds to oxygen inthe lungs and releases it to oxygen-requiring tissues. The affinity for oxygen is

modulated, involving tight binding in lungs, but easy release when needed in tissues.The modulation relies on the allosteric properties of hemoglobin.

Oxygen binds to the heme group of hemoglobin and myoglobin. Heme is known also asiron protoporphyrin IX, with the bound Fe2+ iron. Each iron has six coordination sites,one of which can be occupied by O2. Note that the iron acts as a prosthetic group to

pick up the oxygen. If iron was a lone actor, it would be immediately reactive and spurgreater damage to the organism, particularly humans.

Know how to interpret the O2 binding curves [Y (fractional saturation) vs. pO2], i.e.

NON cooperative ligand binding to a protein or cooperative ligand binding and

explain what is meant by cooperativity. On both curves, indicate the value of P50, the

pO2 at which fractional saturation of protein with O2 is 0.5. In what part of the

cooperative binding curve (what part of the [ligand] concentration range) is protein

predominantly in the low binding affinity conformation, and in what part of the

ligand concentration range is the predominant form the high binding affinity

conformation?We can utilize the association and dissociation equilibria to our advantage in

fractional saturation. Fractional saturation (signified as Y or ) is the fraction of totalbinding sites on protein occupied by ligand. We can determine the fraction saturation

by the following formula: =!"#$"#%!"#$!!""#$%&'!"!#$!"#$"#% !"#$! =

[!"]

!" ! !=

[!]

!!![!]. This shows that as

[ligand] is added, more proteins are bound to ligand. When the fraction saturation is

exactly 0.5, the concentration of ligand will be equal to the dissociation constant

( = !).

Myoglobin binding to oxygen follows the same protein ligand interactions. However, since O2 is a gad it is moreconvenient to utilize partial pressures of oxygen (in the gas phase above the solution), than concentrations of O2


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dissolved in the solution, in the following formula: ! + !. Thus, we can utilize the same methods to

determine the dissociation constant: ! =!" [!!]

[!"#!], and we can further utilize this new equation to determine the

fractional saturation: =[!"#!]

!" ![!"#!]=

[!!]

!!![!!]=

!"!

!!"!!"!,! = !". We know that myoglobin is an oxygen

storage protein in tissues, and 1 oxygen binding site per myoglobin allowing for a hyperbolic binding curve. Freemyoglobin implies myoglobin with empty binding sites, and myoglobin with oxygen imply occupied sites. Thus atp50 is the pO2 that gives 50% saturation.

Explain how Hb works physiologically (in vivo), i.e., how cooperativity in O 2 binding

to Hb facilitates loading of O2 in the lungs and unloading of O2 in the tissues.

Include the role of the R state (oxy onformation) and the T state (deoxy

conformation) of Hb.Remember that heme binds to oxygen. Free heme binds to carbon monoxide 20,000 times better than oxygen.Thus, the protein regulates access to the heme. The heme in myoglobin binds

to carbon monoxide only 200 times better. A distal histidine can offset thebonding angle. It adopts a slight angle because His E7 sterically blocks the

perpendicular arrangement.

Hemoglobin is allosteric. There are 4 possible oxygen sites per hemoglobin.Oxygen is bound cooperatively, thus spurring a sigmoid binding curve (in

comparison to a nonallosteric protein which has a hyperboidal bindingcurve). This is due to the fact that myoglobin has only one oxygen-bindingsite and thus no opportunity for interaction, while hemoglobin has possible

interactions and communication between binding sites. There iscommunication between different ligand binding sites on same multimeric

protein molecule, in the form of structural (conformational) changes. Thereare two interconvertible conformational states of Hb, known as the T state

and the R state. Oxygen binding is regulated by 2, 3 diphosphoglycerate(DPG). Oxygen binding is pH dependent protons and carbon dioxide

promote the relase of oxygen, while oxygen promotes the release of protonsand carbon dioxide.

This leads to the cooperativity of hemoglobin. Remember that the hemoglobin is involved in the transport of

oxygen and carbon dioxide. Thus in the lungs, it has to have a high affinity to load the oxygen before deliveringto the tissues. However, at the tissues, it needs to have a low affinity because it needs to release oxygen. The

cooperativity enhances oxygen delivery and release by hemoglobin. The following describes the relativesaturations at the lung, tissue, and the difference.

Saturation Hb Cooperative NoncooperativeLung 98% 60%

Tissue 32% 22%

=Lung-Tissue 66% 38%

Briefly describe the structural change that occurs when O2 binds to the heme of a

subunit of Hb, including: (1) what in the heme structure triggers the protein

structural change when O2 binds, (2) how that first protein structural change is

communicated to other subunits to change the quaternary structure and the O 2 of the

other subunits, and (3) effect of the quaternary structural change on size of the

central cavity.The structural change occurring involves the cooperativity of the

structure. When there is no O2 binding (such as in deoxyhemoglobin),

the heme iron lies slightly outside porphyrin plane, bound (coordinated)to an N of proximal HIS (F8) residue. When O2 binds to the iron, Fe2+

moves into plane of heme, pulling with it HIS F8. Oxygen bindinginitiates structural changes.

At the tertiary level, there is conformational

change in one subunit. It causes structural

changes in interface between 11 and 22 protomers. This triggers that quaternary

change in whole hemoglobin tetramer. At the quaternary level, 11 shifts relative to

22 and rotates approximately 15 degrees. Thus when it binds, it decreases in size of

the central cavity. Binding pulls the histidine towards the heme moving F helix. This


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a small change that is a major player in the proteins function. Essentially,this is the allostery involved in hemoglobin. If there was a mutation from the

histidine to glycine, the cooperativity would become absent, because there isno bond between the glycine and ion. There would be no communication, and

thus consequently no allostery. Deoxyhemoglobin (the T state) goes tooxyhemoglobin (the R state). The allostery (switching the T and R states) is

the main contributor to the sigmoidal shape of the binding curve. Theconversion from T to R is characteristic of allosteric proteins. When there is

no oxygen present, the equilibrium lies far toward the T state (where there isa weak oxygen binding). The 11 shifts relative to 22 and rotates by approximately 15. On the opposite end

(when there is oxygen binding), the equilibrium is shifted toward the R state. Oxygen binding is a molecularswitch that induces a change in oxygen binding affinity over the whole population of Hb molecules in solution.

High affinity curve is where the R state predominates. The low affinity curve is at the point of low oxygen, wherethe T state is predominant. Remember, the T state (deoxyhemoglobin) is where the is no oxygen binding, while

the R state (oxyhemoglobin) is when the oxygen is bound.

The cooperativity of the hemoglobin can be modeled into a concerted or sequential element. The following table

discusses the models:

Model Diagram Description

Concerted Everything T state or everything in R state. All subunits are

in the same conformation in equilibria.

Sequential Break symmetry of hemoglobin. As O2 binds, there is aconformational change.

Explain the effect of 2,3-bisphosphoglycerate on the affinity of mammalian Hb for

oxygen, and describe where on the Hb molecule 2, 3-BPG binds, how many molecules

of 2,3-BPG bind to one Hb tetramer, and predominantly by what type of noncovalent

interactions the 2,3-BPG is bound. Does 2,3-BPG bind to the R state or the T state of

Hb?The affinity for hemoglobin for oxygen can be controlled by external effectors,

which affect the equilibrium between the T and R states. Favoring the T stateassists hemoglobin in unloading its bound oxygen. This is catalyzed by 2, 3-Bis-

phosphoglycerate (BPG). BPG decreases the affinity of hemoglobin for oxygen,

assisting it to unload the oxygen molceules where they are needed. This leads toa phenomenon known as the Bohr effect. The Bohr effect is when carbon dioxideand hydrogen ions also decrease the affinity of hemoglobin for oxygen, so that a

larger fraction of its oxygen can be delivered in the tissues. Thus, one cansurmise that the protons, carbon dioxide, and 2, 3-

bisphosphoglycerate all decrease oxygen affinity.BPG is the negative allosteric regulator found inred blood cells, and highly anionic (having a -5charge). On the surface of hemoglobin, it is not

very useful because it will bind to the water and

possibly become protonated, losing its charge.

The mechanism is simply put based on BPG. BPG binding atthe central cavity promotes: (1) Structural changes that

promote the T form, stabilizing deoxyhemoglobin, and (2)decreased affinity for oxygen. More oxygen binding sites

must be occupied in order to induce the R-T transition.Remember, the T state is a low-affinity state with low oxygen

concentration. The R state is a high-affinity state withhigher oxygen concentration. The 3 charged groups from

each chain in central cavity help bind BPG by ionicinteractions. 2, 3-BPG binds in the central cavity ofhemoglobin, and prefer to bind to the T state in order to help

shift from R state to T state. BPG allows for more oxygen tobe released due to such decreased binding affinity. Though it

is found in both lungs and tissues, but has little effect in thelungs but a much profound effect in the tissues, where

oxygen needs to be released.

) ) ) ) ) ) )

) ) ) ) ) )

) ) ) ) ) )

) ) ) ) ) )

) )

ionic)interac8on)

with)Asp)94))

) ) ) ) )

) ) ) ) ) )

) ) ) ) ) ) ) )) ) ) ) ) )


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This is particularly seen in other animals, but also in the early fetus. During early development, the human fetus

expresses different and hemoglobin genes. These are similar but not identical to the hemoglobin genesexpressed in adults. Fetal hemoglobin has a higher affinity for oxygen because the fetus has to extract oxygen

from maternal erythrocytes. This ability is found in the tertiary and quaternary structures of the hemoglobin.

At the quaternary level, fetal hemoglobin does not have the subunits found in adult hemoglobin, but subunits.

At the tertiary level, the positively charged histidine residues found in the adult hemoglobin is essentiallyreplaced by the hydrophilic serine, meaning that the ionic interactions are not present. This allows stabilization

into the R state from the absence of ionic interactions.

But also remember that 2,3-BPG is not the only relevant chemical in question. Itis important to recognize that carbon dioxide (CO2) and H+ ions are also negative

allosteric effectors of hemoglobin. In active muscle cells, oxygen is rapidlyconsumed and carbon dioxide and protons are produced. As the pH is lowered

there is a decrease in binding affinity, and that when [CO2] is added, there is also a decrease in binding affinityfrom stabilization of the T state. Combination of the two will reveal that it will have the furthest affinitydecrease. The curves can be seen, from the effects, shifting to the right.

The structural basis of the Bohr effect is found in the ionic interaction between His 146 and Asp 94 and the

covalent interaction of the N-termini of Hb and carbon dioxide. The ionic interactions are what stabilize the Tstate. If His 146 becomes deprotonated, then the T state destabilizes. But remember, that the pKa can occur

alongside conformation change. At a lower pH, His 146 is protonated. The covelent interaction involves carbondioxide reacting with the positively charged N-termini to form a negatively charged carbamate group while

releasing a proton. This leads to ionic interactions at the / interface.

This can lead to an overall understanding of the allostery and Bohr effects. Remember that at inhalation where[O2] is the highest, hemoglobin (in the R state) will bind to oxygen

to the tissue capillaries and then be transported (by erythrocytes)to the tissue capillaries where it will be conform to the T state (andconsequently release oxygen). From metabolism, the proton and

carbon dioxide concentrations will increase and then be bound to

the T state to the lungs. Deprotonation of His 146 will occur and

carbon dioxide will then be release through exhalation, allowingreversion to the R state.

This can later lead to a clinical application in terms of elements

such as mutant hemoglobin and the role of altitude in oxygenbinding. At high altitude states, the concentration of 2, 3 BPG

increases and the oxygen-binding curve will shift right. For mutant hemoglobin, thismainly involves exploration of the structure-function relationships in a protein.

Mutant hemoglobins can encompass silent

mutations, but there are mutations that can spurserious diseases. For example, sickle hemoglobin(HbS) is when there is a change in normalresidue 6 from negatively charged glutamic acidto hydrophobic valine can spur seriousconsequences.

In sickle cell anemia, the change from glutamic acid to valine creates a

hydrophobic patch, which spurs aggregation of hemoglobin. This causes red

cells to distort and block capillaries. In diseases such as -thalassemia, the chains are affect and the

hemoglobin does not become cooperative. In sickle hemoglobin (HbS), the hydrophobic valine on subunit of 1tetramer sticks to patch on T state subunit of another tetramer. The two subunits spur two knows, causing 2

hydrophobic patches. This propagates aggregation and can cause long rigid fibers. Such aggregation can causered blood cell sickling. The long fibers distort shape of red blood cells, sickle-shaped cells clog the capillaries and

poke holes in cells and cause a low hemoglobin concentration.

The following table will outline the diseases in a simplified form:

Type of Disorder Affected part of

Protein

Description

Sickle Cell Anemia Hemoglobin Q6 is mutated to V6, creating a hydrophobicpatch to allow hemoglobin to aggregate,

distorting and block capillaries

-Thalassemia -Chains Hemoglobin does not become cooperative

-Thalassemia -Chains Forms insoluble aggregates resulting in few redcells.

But for sickle hemoglobin patients, all is not lost. This allows protection from such diseases such as malaria.

Thus, it is attributed as evolutionary shift. One copy allows viability while providing protection from malaria.

) ) ) ) ) ) )

) ) ) ) ) )

) ) ) ) ) )

) ) ) ) ) )

) )

) )

) ) ))

) ) ) ) )

) ) ) ) ) )

) ) ) ) ) ) ) )

) ) ) ) ) )

) ) ) ) )) )

Lungs

Tissues

O2'is'released'

[H+])and)[CO2])higher

from)metabolism)

T)state)

Hb(O24)carries)O2)to)

8ssue)capillaries)

2,3_BPG)throughout)body)(in)

both)lungs)and)8ssues)__)lile)

effect)in)lungs,)but)enhances)

release)in)8ssues)(=>)T)state)CO2)binds)to)T)state)in)

8ssues)and)is)

transported)to)lungs)

HIS)146)is)protonated)

reversal'of'carbamate'formaFon

=>'loss'of'CO2'(exhaled)'

[CO2])lower)

[H+])lower)/)pH)higher)

deprotonaFon'of''His'146 R_state)

O2 (inhaled) '


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Lecture XIV

Introduction to Enzymes and Enzyme Kinetics

TerminologyTerm Definition

Rate Enhancement Factor by which catalyst increases rate of reaction. Calculated by:

= !!"#"$%&'(!!"#$%$&'()*

.

Cofactor Substance required to be present in addition to an enzyme for reaction to be

catalyzed.

Coenzyme Non-protein chemical compound that is bound to a protein and required to

proteins biological activity.

Prosthetic Group Metal ion or organic or metallo-organic compound other than an amino acid thatstightly bound to a protein (binding is covalent or tight non-covalent), required or

the proteins function/activity. + Catalyst Substance that is not consumed by the reaction itself, and speed or slow the

reactions progress. They operate under physiological conditions, work byforming complexes, but are not chemically altered by the reaction.

Activation Energy Energy that must be overcome in order for a chemical reaction to occur.

Transition State Configuration along reaction coordinate where the highest energy occurs, thus it

must either go to either product or reactant.

Transition State Analog Chemical compounds with a structure that resembles transition state of a

substrate in a chemical reaction.

Active Site Site of enzyme where a substrate binds to.

Enzyme-Substrate Complex Complex formed when substrate binds to enzyme.

Induced Fit Interaction between enzyme and substrate is weak, but weak interactions

produce conformational changes in the enzyme to achieve shapecomplementarity.

Initial Velocity Reaction velocity where the concentration of product is approximately zero.

Steady State State where

Vmax Maximum rate achieved by the system, at maximum (saturating) substrate

concentrations.

Km Concentration of substrate at which velocity is exactly one-half of the maximal

velocity.

kcat Reaction rate that is observed when the catalyst is present.

Turnover Number Number of substrate molecules converted into product by one enzyme moleculeper unit time, when enzyme is fully saturated with substrate.

Kes Rate constant for the dissociation of the enzyme-substrate complex

Enzyme Efficiency Refers to rate enhancement, in which by increasing the rates results in a more

efficiency chemical reaction.

Double Reciprocal Plot Graphical representation of the linear Lineweaver-Burk equation of enzyme

kinetics, which determines reciprocal velocity as a function of the reciprocal ofsubstrate concentration.

Reversible Inhibition(Competitive, pure

noncompetitive,uncompetitive)

Reversible inhibition binds via weak, non-covalent interactions, which can beeasily removed by dialysis or dilution. There are three types of reversible

inhibition:

1. Competitive: Involves competition for the active site, affecting KM.Binding and release of inhibitor from the active site forms anequilibrium with dissociation constant Ki.

2. Uncompetitive: Involves inhibitor binding only to the complex formedbetween enzyme and substrate. Km and Vmax are reduced by equal

amounts.3. Noncompetitive: Reduces activity of the enzyme by binding to a

completely different site on the enzyme. Changes the apparent Vmax andcreates a different intercept.

Irreversible Inhibition Bind very tightly to the enzyme by possibly even covalent bonds, and isessentially irreversible because it is a kinetically controlled process. Can

inactivate proteins by altering active site of the target. This can help understandwhich functional groups are required for enzyme activity.

Affinity Label Molecular similar to structure to a particular substrate for a particular enzyme.

Transition State Analog Molecules resembling the transition state, which are potential inhibitors of

enzymes.


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Suicide Inhibition

(Mechanism-BasedInhibitor)

Form of irreversible enzyme inhibition that occurs when an irreversible complex

(by covalent bonding) forms during normal catalysis. The substrates likemolecules that cannot complete the reaction. Forms a stable inhibitor-enzyme

complex.

Review

Term DefinitionEquilibrium Constant Ratio of concentrations when equilibrium is reached in a reversible reaction. It

occurs when the rate of the forward reaction equals the rate of the reversereaction.

Mass Action Ratio for aReaction

Ratio of product and reactant concentrations. Calculated by: =[!]![!]!

[!]![!]!.

Biochemical StandardConditions

Reaction conditions occurring at a pH of 7 and at 37C.

Standard Free EnergyChange

Gibbs free energy at equilibrium, calculated by: ! = ln!"

Actual Free Energy Change The change in Gibbs free energy that is dependent on two parameters: (1) Thestandard free energy change for the reaction and (2) the actual mass action ratio.

Calculated by: ! = !+ ln.

Describe the general properties of enzymes as catalysts that are especially importantfor their roles as biological catalysts.Enzymes are typically treated as biological catalyst because (1) they are not altered chemically in the reaction,(2) they operate under physiological conditions (at moderate temperatures, around a neutral pH, and a low

concentration in an aqueous environment, and (3) work by forming complexes with their substrates andconsequently binding a unique microenvironment for the reaction to proceed. The substrate is going to bind to

the enzyme. Any deviation away from physiological conditions can have an effect on structure.

Explain the effect of a catalyst on the rate of a reaction, and on the equilibrium

constant of a reaction. Express the velocity of a simple reaction in terms of the rate

constant and the concentration of the reactant.Catalysts can strongly affect the rate of a reaction. Some catalysts can strongly increase the rate of an reaction,

while others can strongly decrease the reaction rate. We can measure the rate of the reaction simply by

measuring the concentration of the reactant or product over time = [!"#$%#&%]!"#$

[!"#$%&']!"#$

. We can

also determine the rate enhancement as a quotient of the catalyzed and uncatalyzed rate constants

= !!"#"$%&'(!!"#$%$&'()*

.

Express the equilibrium constant of a reaction in terms of the equilibrium mass

action ratio.The equilibrium constant is simply when !"#$%&' = !"#$%#&% . Thus we also know that the equilibrium mass action

ratio is equal to the equilibrium constant ()!" = !" . Thus one can state that for a given chemical equation

+ + , with the mass action ratio =[!]![!]!

[!]![!]!, we can also state that !" =

[!]![!]!

[!]![!]!, when the products

and reactants are at equilibrium.

Express the equilibrium constant of a reaction in terms of the rate constants for theforward and reverse directions. (Note that equilibrium constants are symbolized

with upper case K and rate constants with lower case k.)

Well, if we have a reaction consisting of substrates and products going to equilibria, we can state that the

velocities are the products of the concentration and the rate constant. Thus, ! = ![]!" and ! = ![]!" . At

equilibrium, the reaction velocities are equal != ! ![]!" = ![]!" . Thus, we can then derivate that

!!

!!

=

[!]!"

[!]!"= !" . If the forward rate constant increases, so increase must the reverse rate constant go by the

* *

* * *

* * * * * * * * *

* * * * * * * * * ** * * **

** * * * * * * * * * ***** * * * * * * * * * **** * * * *

S P(S = Substrate) (P = Product)

kF

kR

& & & & & & & & & & &


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same factor. If the forward rate constant decreases, so decrease must the reverse reaction go by the samefactor.

If an enzyme increases the rate constant for the forward reaction by a factor of 108,

by what factor does it increase the rate constant for the back reaction? What is the

rate enhancement brought about by the catalyst for that reaction?Remember that the though the reaction pathway is changed, the thermodynamic properties (particularlyG)

should not change. Thus, increasing the rate constant for the forward reaction by a factor of 108 should allow anincrease of the rate constant for the back reaction by a factor of 10 8. The rate enhancement is 108 because the

enzyme increased the rate constant by 108.

Draw the free energy diagram of a hypothetical reaction and show how a catalyst may

increase the rate of the reaction, pointing out on the diagram G for the overall

reaction, G uncat, G cat.

We need to remember the thermodynamic variables involved in a reaction. We know that molecules have to

collide with enough energy in order for the reaction to occur. This amount of energy is known as the activationenergy, which is the energy required to cause the reaction to occure. The catalyst increases the rate of thereaction by decreasing the activation energy required in order to achieve the reaction. A lower energy barrier

implies a faster reaction. Remember that the enzymes change the pathway of a reaction, but not the change in

Gibbs free energy. The rate of the reaction is dependent on the activation energy.

Indicate (and name) the quantity on a free energy diagram (HINT: its a specific

kind ofG) that determines the magnitude of the rate constant for the reaction at a

given temperature. You dont have to memorize the equation relating this quantity to

k.The rate constant (k) is dependent on G, which denotes the Arrhenius activation energy or the free energy of

activation for the reaction. This consists of the difference in free energy between the transition state and thestarting state, which is the barrier over which the reaction must go in order to proceed. This can be calculated

by the following equation: =!!!

!

!!!

!" , where h is the Planck Constant, R is the gas constant, and kb is the

Boltzmann Constant. Though this equation should NOT be memorized, we can observe how we can increase thereaction rate. Increasing the temperature (T) can do this, but it leaves the possibility of the proteins

denaturation. However, understanding this equation can allow us to understand how to derive the rate

enhancement. = !!!

!!

!!!"#"$%&'(!

!"

!!!

!!

!!!"#$%$&'()*!

!"

=

!!"#$%$&'()*!

!!!"#"$%&'(!

!" =

!!"#"!"#$%

!!"#$%$&'()*.

* * * * *

= transition state

high*energy*barrier

=

slower*reac)on

lower*energy*barrier*

=*

faster*reac)on*

G=*Free*

energy*gap*

between*

reactants*

and*products*

at*

equilibrium*

Guncat

Gcat

Guncat

Gcat

= -

* * * * * * *

Enzymes*change*

the*pathway*of*a

reac)on*

*

Enzymes*bind*

)ghtly*to*the*

transi)on*state*


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10

What reaction parameter (kinetic parameter) do enzymes affect in order to increase

the rate?Remember that catalysts increase rate constants because the enzymes decrease G. They also decrease the

energy of the transition state, thus stabilizing the transition state, allowing for the reaction rate to increase.

Describe the properties of the active site.Active sites have five major properties:

1. Three-dimensional cleft with participating components from different parts of theprimary structure.

2. A small portion of the total available volume. The amino acid not at the active siteposition, but the key side chains and allow the flexibility of the active site.

3. The clefts/crevices/pockets which can exclude water but allow placement of polarside chains to create a microenvironment.

4. Enzyme-Substrate Complex are generally stabilized by weak interactions.5. Active site has specificity of binding, meaning that it has a defined arrangement of

atoms in an active site, via two models: (1) a rigid, inflexible active site, meaningthat the active site of the unbound enzyme is complementary in shape of thesubstrate (called the lock and key model), or (2) plastic, flexible active site,

meaning the enzyme changes shape upon substrate binding to achieve shapecomplementarity (known as the induced fit model).

Describe the different models existing for enzyme-substrate binding.Model Diagram Description

Lock and Key Active site of unbound enzyme is complementary inshape to the substrate: a rigid, inflexible active site

Induced Fit Enzyme changes shape upon substrate binding toachieve shape complementarity.

Write out a simple Michaelis-Menten kinetic mechanism for an enzyme-catalyzed

reaction.We can write out the Michaelis-Menten kinetic mechanism for an enzyme-catalyzed reaction as the following:

. Basically, we know that the enzyme and substrate have to combine to yield an enzyme-substrate complex, and from this complex a product is yielded (as a function of time).

Recognize the Michaelis-Menten equation, and sketch a graph of V0 vs. [S] for an

enzyme-catalyzed reaction that illustrates Vmax and Km.

The Michaelis-Menten equation is as follows: !=

!"#

[!]

!!![!], where [S] is the concentration

of the substrate, and KM is the Michaelis constant, the concentration of substrate where the

velocity is exactly half-maximal ! =!!"#

!. Based on the forward and reverse reactions, we

can determine the Michaelis constant as: !=

!!!!!!

!!

. If the Michaelis constant is must

larger than substrate concentration ! [] , then ! =!!"#[!]

!!

.

We need to remember that before the reaction occurs, the enzyme must bind to substrate toyield a noncovalent enzyme-substrate complex. Thus, the concentration of the enzyme-substrate must increase before product increases. Once the concentration of the enzyme-substrate stops increasing and reaches a steady-state, the concentration of product will

increase in a linear fashion until the concentration of substrate is depleted, where the rateof formation of product slows and equilibrium is reached.

** * * * * * * * * * * * * * *

*

* *

* * * *

* * *

* * * * *

, * * *

* * * *

* * * *

* *

* * * * * * *

E.Fischer,1890 * * *

*

* * * * * * * * * * * * * * *

*

* *

* * * *

* * *

* * * * *

* * * *

* * * *

* * * *

* *

* * * * * * *

* * * D.Koshland, 1958

* *

E + S ES E + P

k1

k-1

k2

* *


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11

Define Km in terms of the rate constants in the Michaelis-Menten kinetic mechanism;

give the operational definition of Km that holds no matter what the actual kinetic

mechanism is for a particular enzyme.

Based on the forward and reverse reactions, we can determine the Michaelis constant as: ! =!!!!!!

!!

. KM is the

Michaelis constant, the concentration of substrate where the velocity is exactly one-half of the maximal velocity.

! = !!"#!

. This is the operational definition that holds for ANY kinetic mechanism.

Explain the relationship of kcat to Vmax, and the relationship of Km to KES.The turnover number kcat is the number of substrate molecules converted into product by one enzyme moleculeper unit time, when enzyme is fully saturated with substrate. Each catalyzed reaction takes place in time equal

to 1/k2. A high kcat implies that the chemical rearrangement steps are rapid. We can calculate the turnover

number by the following equation: !"# = ! =!!"#

! !

. Most enzymes need high substrate concentrations to create

significant amounts of products. At low concentrations not much happens. Some enzymes acan work betterthan others under the stringent conditions of low substrate concentration found in cells. The most efficiencyenzymes turn over a substrate whenever one is encountered. The upper limit is how fast substrates can diffuse

into an enzyme.

State the units of Km, kcat, and Vmax.

Constant Definition UnitsKM Concentration of substrate at which velocity is

exactly one-half of the maximal velocity.Concentration, such as M.

kcat Reaction rate that is observed when the catalystis present.

Inverse time, such as s-1

Vmax Maximum rate achieved by the system, atmaximum (saturating) substrate concentrations.

Concentration over time, such as M/s.

Given a hyperbolic Vo/Vmax vs. [S] plot, explain the meaning of the ratio Vo/Vmax in

terms of occupied active site concentration and total active site concentration, and

find Km directly from the graph, including its units.The ratio VO/Vmax can be interpreted as the relative rate, where the reaction velocity VO is in terms of themaximal velocity Vmax. In terms of occupied site concentration and total site concentration, this is simply a

quotient of the occupied and total active site concentration. This can aid in determining the fraction of occupied

active sites, regardless of total active site concentrations. From any hyperbolic graph, we can also determine theMichaelis constant KM finding the concentration of substrate where the relative rate is 50%, or 0.5, because theKM is the concentration where VO is half of Vmax.

Given a plot of VO/Vmax vs. [S], find/calculate the value of Vmax and

Km from the plot.One can calculate the value of Vmax by knowing the Vmax is a horizontal asymptote becausethe total number of active sites limits the fractional saturation. This is where the change

in the initial velocity over the change in substance concentration is almost zero!"!

![!]= 0 . The KM can be

determined from the plot by finding the substrate concentration where the VO/Vmax is 0.5 or 50%. One is alsoable to manipulate the graph towards a linear function by a Lineweaver-Burk plot. The hyperbolic plot can be

manipulated in the following manner: ! = !"#[!]

!!![!]

!

!!

=

!!![!]

!!"#[!]=

!!

!!"#

!

[!]+

!

!!"#

. Inhibitors can alter KM and

Vmax, affecting function of the enzymes.

What two things are the parameter kcat/Km used to indicate?The parameter kcat/KM can indicate the (1) efficiency of an enzyme and (2) the specificity of enzyme. Why is thisindicative of efficiency? Most enzymes need a high [S] to create significant amounts of products. At low

concentrations not much happens. Some enzymes can work better than others under the stringent conditions oflow [S] found in cells. The most efficient enzymes turn over a substrate whenever one is encountered. The

upper limit is how fast substrates can diffuse to an enzyme. Why is it also indicative of specificity? If the ratio ishigh, then the enzymes side groups must favorably interact with the substrate.


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Explain competitive, uncompetitive, and pure noncompetitive inhibition in terms of a

diagram of linked reaction equilibria for formation of ES, EI, and (if it can form) EIS.Type of

Inhibition

Definition Reaction Diagram

Uncompetitive Binding to site in proximity of the substrate

Noncompetitive Binding at a regulatory site not near the active site.

Competitive Competition for active site with substrate

Explain how these 3 types of inhibition can be distinguished from each other

graphically: on a VO vs. [S] plot, and on a double reciprocal (Lineweaver-Burk) plot.Type of

Inhibition

Definition Interpretation VO vs. [S] Plot Lineweaver-Burk

PlotKM Vmax KM/VmaxCom-

petitiveCompetition foractive site with

substrate

Inc N/C Inc

Uncom-

petitive

Binding to site in

proximity of thesubstrate. A

reduction in equalamounts.

Inc Inc N/C

Noncom-petitive

Binding at aregulatory site not

near the active site.

N/C Inc Dec

What is the effect of a competitive inhibitor on Km and on Vmax (compared to the

values in absence of inhibitor)? What is the effect of an uncompetitive inhibitor onKm and on Vmax (compared to the values in absence of inhibitor)? What is the effect

of a noncompetitive inhibitor on Km and on Vmax (compared to the values in absence

of inhibitor)? Relative to the values in the absence of an inhibitor,

a competitive inhibitor will decrease KM whileleaving VMax unchanged.

Relative to the values in the absence of an inhibitor,an uncompetitive inhibitor will both decrease KM and

Vmax, but equally so that the ratio K M/Vmax isunchanged.

Relative to the values in the absence of an inhibitor, a noncompetitive inhibitor will leave KMunchanged while decreasing Vmax.

What is a transition state analog?A transition state analog is a potent inhibitor of enzymes, as they are molecules resembling the transition statethat are more tightly bound than the substrates. They do not undergo chemical reactions and can act as

inhibitors by blocking active sites. They are useful in understanding chemical mechanisms as well as drugs tocounter disease.

Lecture XV

Enzyme Catalysis

* *

* * *

* * * * * * *

* * * ** * *

* *

* * * * * * * ** * * * * * *

* * * * * * * ** * * * * * *

* * *

* * * * * * * * * ** * * * * * * * * *

* * * * * * *

* * * * * * *

*

* * * * * * ** * * * * *

b * * * * * * *

ac * * * * * *

d * * *

* *


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TerminologyTerm Definition

Proteolysis Directed degradation (digestion) of proteins by cellular enzymes known as

proteases.

Serine Protease Enzymes that cleave peptide bonds in proteins, which serine serves as the

nucleophilic amino acid at the active site.

General Acid Based on the Bronsted-Lowry definition, a proton donor.

General Base Based on the Bronsted-Lowry definition, a proton acceptor.

Catalytic Triad Three amino acid residues found inside the active site of certain protease enzymes,

namely serine, aspartate, and histidine.

Tetrahedral

Intermediate

Reaction intermediate in which the bond arrangement around an initially double-

bonded carbon atom has transformed from trigonal to tetrahedral.

Acyl-Enzyme

Intermediate

Intermediate formed from acylation of the substrate, formed by the attack of the

active site serine residue on a peptide bond in a protein substrate.

Nucleophile Species that donates an electron-pair to an electrophile to form a covalent bond.

Oxyanion Hole Region in space where the backbone amide hydrogens of serine 195 and glycine 193point into the active site. Remember, the binding site specificity pocket and

catalytic triad play a role in the chymotrypsin chemical machinery.

Discuss (briefly explain):

General concepts in catalysis related to protein-ligand binding

Different enzymes utilize different mechanisms to reduce activation energy and increase the rate of reaction.Some involve specific groups and chemical mechanisms that are dependent on the specific reaction. There are

three general points involved in catalysis:

1. Electrostatic Effects: Increase in strength of ionic interactions due to lowering dielectric constant,altering pK values, stabilizing the charged intermediate, etc.

2. Desolvation: Exclusion of water from the active site (due to hydrophobic groups)3. Induced Fit: Change in the conformation of the enzyme or substrate to optimize interactions.4. Enzymes bind transition state very tightly, tighter than substrate: Free energy of transition state

(peak of free energy barrier on reaction diagram) is lowered because its distortion (electrostatic or

structure) is paid for by tighter binding of transition state than of substrate.

Most enzymes use a combination of several mechanistic strategies to increase the reaction rate.

General catalytic mechanisms used by enzymes to increase the rates of chemical reactions.

Catalytic

Mechanism

Description

General Acid-Base

A group on the enzyme acts as an acid or base, donating or removing (according to theBronsted-Lowry definitions) to the substrate during the reaction.

Covalent A group on the enzyme becomes covalently modified during reaction, e.g. by forming acovalent bond to the substrate during the reaction.

Metal Ion The enzyme to facilitate a chemical rearrangement or binding step uses a metal ion.

Catalysis by

Approximation

The enzyme holds two substrates near in space and in precisely the correct spatial

orientation to optimize their reaction. Optimized by (1) proximity and (2) orientation.

What is the role of ion in metal-ion catalysis?In metal-ion catalysis, the metal ion is utilized to facilitate allow a chemical rearrangement or binding step. The

metal ion can be tightly bound (as a metalloenzyme or a prosthetic group) orloosely bound (binding reversibly and dissociating from the enzyme). They

typically stabilize a negative charge on the reaction intermediate and allowproper orientation of substrate via coordination, to allow for binding at specific

geometry. The metal ion also facilitates formation of nucleophile, and evenpolarizes the scissile bond. This allows it to become stable and retain shape in the

presence of water because the enzyme can unfold (placing the inside out).

Which type of functional group can act as a general-base,

general acid, nucleophile?Enzyme functional groups can act as Bronsted acids or bases to allow proton

transfer and facilitate bond cleavage. Remember that a Bronsted acid donates aproton, while the Bronsted base accepts a proton. They are oriented to the active

site to permit favorable interaction with bound substrate. The group that donates


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a proton in catalysis then has to accept a proton in catalytic mechanism for catalyst to be regenerated in originalconjugate acid form. Acids or bases promote organic reactions. Thus, one can conclude from this is that pH

plays a major role in the rate of a reaction.

Nucleophiles (or nucleus-loving) will form a chemical bond with an electro phile (the electron-loving) bydonating an electron pair. When a covalent bond is create, the rate acceleration occurs by transient formation of

the covalent catalyst-substrate bond. It requires a highly reactive group, consisting of the nucleophilicfunctional group on the enzyme to react with the substrate to form the covalent bond. It is typically stepwise,

and more reactive in the second step due to lower activation energy relative to the non-covalent catalyticmechanism. Thus, the enzyme alters the pathway of the reaction progression.

Given an enzyme mechanism, be able to identify which residue most likely acts as a

general acid, general base or covalent catalyst.Behavior Definition Indication

General Acid Proton Donor Electrophile receives a proton, with the enzymegaining a proton. (Enzyme attacks a proton.)

Functional groups are typically protonated.

General Base Proton Acceptor Nucleophile attacks a proton, removing a proton

from the enzyme. Functional groups are typicallydeprotonated.

Covalent Catalyst Rate acceleration by transientformation of a covalent catalyst-

substrate bond.

Bonding of catalyst with substrate (catalyst-substrate) bond.

How can pH affect enzyme activity? Why? What is the optimum pH of an enzyme?Remember that the substrate needs to bind to the enzyme in order for the catalytic effects to occur. The

ionization states of the amino acid residues are involved in the catalytic activity. Changes in pH can affect theionization state of the substrate and possibly spur variation of the proteins structure. The optimum pH in

protein/substrate interactions is the pH at which the substrate maximally binds to substrate. The substrateattaches itself to the enzyme via ionic bonding. Any change (such as increasing or decreasing the pH) that

disrupts ionic binding will disrupt ionic interactions.

We can observe this in the following table:

pH

relationship

Diagram Charge of Carboxyl Charge of Amino

pH = optimal -1 +1

pH < optimal 0 +1

pH > optimal -1 0

Given a graph showing an enzymes activity (as a function of pH) be able tointerpret the data, i.e. what is the optimum pH, what residues could be involved in

catalysis (see lysozyme pH profile).The graph of optimum pH graph is similar to a titration curve, because as the pH increases (similar to increasing

base) the activity of the enzyme can be observed. Remember that the optimum pH is the pH at which theproteins enzymatic activity is maximal (at 100%). It is possible to roughly estimate the pKas of certain

functional groups at the points where there is 50% activity. The ionization states of the residues explain the pH-activity profile of an enzyme because protonation or deprotonation can affect the ionic forces involved in the

enzyme-substrate interaction.

( ( ( (

(

(

VCOOV:(picks(up(H+((

( (

( ( ( (( ( ( ( ( ( ( ( (

( ((

pH(lower(than(7(V(acidic ( ( ( ( (

( ( ( ( ( ( ( ( ( ( ((

( ( ( ( ( ( ( ( ( ( ( ( ( (

( ( ( ( ( ( ( ( ( ( ( (

VNH3+:(no(

change ( ( (

( ( ( (

(

(

( ( ( (VNH3+:(

loses(H+(

( ( ( (

( ( ( ( ( ( ( ( (

( ((

( ( ( ( ( pH(higher(than(7(V(basic

( ( ( ( ( ( ( ( ( ( ((

( ( ( ( ( ( ( ( ( ( ( ( ( (

( ( ( ( ( ( ( ( ( ( ( (

( ( VCOOV:(no(change


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Given an amino acid pKa and a pH, predict whether it is likely to act as acid or base

catalyst.pKa relationship to

optimal pH

Protonation

State

Outcome of -COOH Outcome of NH2

pKa > optimal pH De-protonated Charged (-1) Uncharged (0)

pKa < optimal pH Protonated Uncharged (0) Charged (+1)

One is able to predict the action of the amino acid based on the amino acid residue.The pH will predict whether the amino acid residue will act as a proton donor or a

proton acceptor. Remember that the amino acid functional groups can act in theprocess as a proton donor or a proton acceptor.

How can mutations (micro-environment) affect enzyme

activity?Mutations, from a micro-environmental perspective, can affect enzyme activity

by altering the ionizable state of the substrate and the enzyme. One is able to predict the changes because of

what is needed to bind and what is needed for the reaction to proceed. Even conservative mutations can causechanges in enzyme activity by altering rate or affecting the intermediate.

Name enzymes that use metal ion catalysis, general acid-base catalysis, covalent

catalysis, catalysis by approximation or combination of those.Enzyme Catalytic Mechanism Utilized

Acid-Base Covalent Metal Ion Catalysis by Approximation

Serine Proteases: HydrolysisEnhance Rate

Carbonic Anhydrases

Restriction Endonuclease:Specificity, Hydrolysis

Nucleoside MonophosphateKinases: Group Transfer

Enolase: Two Step Reaction

Lysozyme

Explain why peptide bonds are kinetically stable in the absence of a catalyst, given

that equilibrium lies far in the direction of hydrolysis.Peptide bonds are kinetically stable in the absence of a catalyst, given that equilibrium lies far in the direction of

hydrolysis, mainly because they have the partial double bond (the resonance structure) that allows for stability.Though peptide hydrolysis is thermodynamically favored, it is kinetically slow. This is we have enzymes called

proteases to undergo proteolysis. Proteolysis is the directed degradation (digestion) of proteins by theproteases. There are four classes of proteases:

Type of Protease Catalytic Mechanism Utilized

Acid Base Covalent Metal Ion Catalysis by Approximation

Serine

Cysteine

Aspartic Acid

Metallo-proteases

One example of a protease is chymotrypsin, which is a digestive serine protease involved in the breakdown of

proteins and peptides so that their amino acids can be used. They particularly cleave proteins on the carboxylside of aromatic, large hydrophobic residues (right), and the amino-terminal side of the peptide bond to becleaved (left). These residues are not easily accessible because the hydrophobic amino acids are in the core, andthus difficult to access.

What type of methods can be used to characterize the mechanism

of an enzyme?Characterizating the mechanism of an enzyme involves (1) chemical labeling, (2)

kinetics experiments, (3) affinity labeling, (4) x-ray crystallography, and (5) site-directed mutagenesis


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Describe substrate binding, including the role and chemical nature of the specificity pocket in

chymotrypsin, and which peptide bond in the substrate (relative to the specificity group) will be cleaved.Substrate binding involves formation of a tetrahedral intermediate. The Ser-Oengages in a nucleophilic attack on carbonyl C of substrate, with stabilization by

aspartate. The O of the Ser-OH activated by H-bond to HIS in catalytic triad. Thishelps maintain perfect orientation of His and Ser in hydrogen bonded network

and facilitates H+ transfer by electrostatic stabilization of HisH+ after it hasaccepted the proton. His accepts the H from Ser-OH to become HisH+. This yields

a tetrahedral intermediate, which is a structure presumed similar to that of transition state for its formation andbreakdown.

The hydrophobic specificity pocket of chymotrypsin is the area of the active site responsible forchymotrypsins substrate specificity. Remember that chymotrypsin is specific to peptide bonds containing an

aromatic. The glycine (the smallest amino acid) residues line the pocket to allow bulky hydrophobic side chainsto fit in the binding site, whilst allowing Ser 189 to allow the reaction to occur via nucleophilic attack. The active

site is where the chemistry occurs, mainly to stabilize the transition state. The hydrophobic pocket is where thebinding occurs and is involved in the specificity.

Explain the role of each member of the catalytic triad in the reaction.

Amino Acid-

Position

Role in Reaction

SER 195 Involved in the direct nucleophilic attack of the carbonyl of the substrate, activated the

H-bond to HIS in catalytic triad.

HIS 57 General base catalyst, allowing facilitation of the removal of the S195 hydroxyl proton.

This allows formation of the alkoxide ion, the nucleophile.ASP 102 Orientation of the HIS 57 and stabilizes the protonated form.

Explain the role of the oxyanion hole in the mechanism.The oxyanion hole is an area in the active site of serine proteases that binds the transition

state particularly tightly. The active site binds oxyanion more tightly than the substrate.An additional hydrogen bond forms between tetrahedral oxyanion and enzyme groups

around it to form an oxyanion hole portion of the active site. The h-bond couldnt form tothe carbonyl oxygen. The H-bond, at this time, can form to carbonyl oxygen because of the

structural change (particularly lengthening of the C-O bond). The H-bonds to negativelycharged oxygen (or oxyanion) are stronger than to neutral O.

Describe which type(s) of general catalytic mechanisms are used by chymotrypsin.Chymotrypsin utilizes covalent catalytic as well as acid-base catalytic mechanisms.

Compare (very briefly, just the bottom line) the overall 3-dimensional structures

of chymotrypsin, trypsin, and elastase, and compare the substrate binding

specificities of those 3 enzymes, explaining the relationship of the specificity

site/pocket structure to the differences in substrate specificity.Mammalian serine proteases have nearly identical 3-dimensional folds. It has a 40%sequence identity, explaining the homologous nature of the the serine proteases. The

tertiary structure explains the same mechanism to hydrolyze peptide bonds. Withinthis family there are also proteolytic enzymes in the blood-clotting cascade.

However, delving into the active sites will allow for certain differences among these

enzymes. SER 189 of chymotrypsin is lined with glycines and only has the serine.Thus one can conclude that it has particular affinity for aromatic groups such asphenylalanine, tyrosine, and tryptophan. The ASP 189 of trypsin has a negativecarboxylic acid function group, thus allows the inference of favorable binding to

positively charged amino acids. Elastase has Valines at positions 190 and 216,

allowing small amino acids to enter. Proline cannot permit binding due to the fusedring structure. If there was a mutation in the catalytic triad, the reaction rate would

be significantly diminished but binding could still occur, meaning the KM would remainconstant while the kcat would decrease. The change of amino acid in the specificity

pocket can change the specificity of the molecule.

There are other unrelated classes of proteases, whose sequences and structures are unrelated to those ofchymotrypsin, nevertheless have the same spatial arrangement as HIS-ASP-SER. The same catalytic method

appears to have arisen independently at least three times in nature, implying convergent evolution. The enzymearose from a previous ancestor.


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How do 3 other classes of proteases (besides the serine proteases) generate

nucleophiles potent enough to attack a peptide carbonyl group?There are other classes of proteases that generate nucleophiles potent enough to attack a carbonyl group.

Class Nucleophile Potency

CysteineProteases

Cysteine Structural homolog of chymo

Aspartyl

Proteases

Water

molecule

ASP is involved in general base catalysis as well as orientation and

polarization of the substrate carbonym. Optimum pH is ~2, pKa of one ASPshifted from 4 to 1.2

Metallo-

proteases

Water

molecule

Metal and general bae catalysts to activate a water molecule.

Nucleoside

MonophosphateKinase

Phosphate

TransferTrue substrate is the magnesium ion complex of the NTP. It is bound to the

and phosphoryl groups and to four water molecules, including a conservedaspartate residue. The protein is needed to avoid side reactions. The

environment needs to isolate from water.

What is the P-loop, what is its amino-acid sequence, what is its role in catalysis?The P-loop is a motif widely used in nature, and involved in reactions such as

phosphoryl transfer, ATP synthesis, myosin, signal transduction proteins, etc. It isconserved and interacts with the phosphryl groups, and has a consensus sequence of

G-X-X-X-X-G-K. Its role in catalysis involves protein-substrate interactions. It is anATP or GTP binding motif found in many nucleotide-binding proteins, as well as

interactions with phosphate groups of the nucleotide and with a magnesium ion,coordinating and -phosphates. The magnesium ion is bound to and phosphorylgroups and four water molecules, including a conserved aspartate residue. The lysine

residue is especially important in nucleotide binding.

Such interactions allow for catalysis by approximation (induced fit). Once thenucleotide interacts with the P-loop, there are local structural rearrangements and the

closing down of the lid domain. This allows isolation of the active site from the aqueousenvironment, preventing transfers of phosphate to water. The lid closes and brings

the two substrates together and assures that the hydrolysis only occurs when the acceptor is proximal.

Knowing the mechanism of an enzyme i.e. active site residues and residues

responsible for specificity, be able to interpret/predict the effect of mutation on

enzyme activity, enzyme specificity.Active site residues rely on what substrate residues are being bound. The residues responsible for specificity inthe active site are those that make up the pocket of the enzyme. Mutation will change the enzymes activity byhow much the mutation affects the characteristics of the active site. If the mutation is not in the active site,

there is no change in enzyme activity. If the mutation changes polarity/charge of active site residues, theactivity will consequently diminish or cease completely in activity. A mutation can change enzyme specificity by

how much the mutation affects the characteristics of the pocket. If it changes the polarity/charge, the specificitywill be lost and unable to bond to correct substrate residues.

Affected Site of

Mutation

Effect on Enzyme Activity (kcat) Effect on Enzyme Specificty (KM)

Catalytic Triad Decreased Unchanged

Hydrophobic Pocket Unchanged Decreased

However, remember, anything that affects binding will affect the overall activity of the enzyme.

Why do we study enzyme mechanisms?Enzyme mechanisms are important because such knowledge allows us to produce medications that help counter

undesirable effects. Drugs are used to treat maladies ranging from headache to HIV infection are almost alwaysinhibitors of an enzyme. Knowledge of the enzyme mechanism helps in the designing and production of drugs.


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To which protease class does HIV protease belong? Describe the quaternary structure

and symmetry of the HIV protease and where in the quaternary structure the active

site residues are located.HIV protease is an ASP protease, a homodimer (containing twoidentical subunits, each contributing an aspartate to active site),and particularly cut peptide bonds between phenylalanine and

proline. The active site residues are near the core, and the

protein closes down after substrate binds (exhibiting an inducedfit). HIV uses its symmetrical nature as a mechanism. Aided bygeneral base catalysts, water attacks the carbonyl carbon,

generating a tetrahedral intermediate. This yields a transitionstate. The tetrahedral intermediate collapses, and the amino acid leaving group is protonated as it is expelled.The active site has a pocket to bind aromatic groups next to the bond to be cleaved. One asparatate function

group is protonated while the other is deprotonated. It is pH-dependent, showing a pKa as 4.5 while having pKaof each dimer as approximately 3.4 and 5.8, respectively.

Explain why transition state analogs are potent inhibitors.To counter HIV protease, medications called HIV proteases inhibitors typicallyinterfere with the protein. They form non-covalent complexes with enzymes but they

bind so tightly that they can be considered irreversible inhibitos. Transition stateanalogs are potent inhibitos mainly because they mimic the tetrahedral intermediate,

but allow it to be stable enough for irreversibility. The remainder of the structure was designed to fit into and

bind to various crevices along the surface of the enzyme. HIV protease inhibitos share a core structure of ahydroxyl positioned next to a branch containing a benzyl group. Another medication, Crixivan has aconformation that approximates the 2-fold symmetry of the enzyme. Crixivan thus inhibits HIV protease

without affecting normal cellular Asp proteases, which dont have the 2-fold symmetry that HIV protease has.

What type of inhibitor is penicillin? How does penicillin work?Penicillin is a-lactam antibiotic that inhibits the formation of peptidoglycan cross-links in the bacterial cell

wall, but no effect on cell wall degradation. It binds to the enzyme (DD-transpeptidase) that links thepeptidoglycan molecules in bacteria. Thus, there is no cross-link replacement, as the bacteria still has enzymes

that hydrolyze the peptidoglycan cross-links, weakening the cell wall of the bacteria. It will thus consequentlycell lysis from those weak linkages. The build up of the peptidoglycan precursors can cause activation of the cell

wall hydrolases and autolysins, digesting the bacterias remaining peptidoglycan. Thus, this causes a rapidbacterial death.

Have an appreciation for the commercial use of enzyme in the chemical and

pharmaceutical industries.Biological catalysts have been utilized, even though society did not know what caused such catalysis, such as inthe fermentation of fruit juices and the baking of bread. The applications of biocatalysts to commercial processes

will continue as long as there are consumer needs and diseases. However, there is still much to know aboutbiological catalysts, as scientists research biochemical pathways, and attempt to apply these catalysts to replaceprevious, possibly inefficient catalysts.

Lecture XVI

Regulation of Enzymes

Terminology (some are review):Term Definition

Quaternary Structure Arrangment of multiple folded or coiled proteins molecules in a multi-subunitcomplex.

Multimeric Protein Group of two or more associated polypeptide chains.

Ligand Substance that forms a complex with another molecule to yield a biological effect.

Binding site Region of protein specific to a ligand to form a chemical bond.

Feedback Inhibition A decrease in activity from the molecule binding to the enzyme.

Cooperativity Phenomenon displayed by enzymes or receptors that have multiple binding siteswhere affinity is increased or decreased.


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Cooperative Binding Behavior occurring from the affinity for its ligand changes with the amount of

ligand already bound.

Allosteric Regulation of en enzyme or other protein by binding an effector molecule at the

allosteric site.

Homotropic

Effector/Regulator

Substrate for its target enzyme and regulatory molecule of enzymes activity. It

typically is an activator.

Heterotropic

Effector/Regulator

Regulatory molecule that is not also the enzymes substrate. It can either be an

activator or inhibitor of the enzyme.

Allosteric Activator(Positive Heterotropic

Effector/Regulator)

Binding of one ligand enhances attraction between substrate molecules andbinding sites. Will shift towards the R state.

Allosteric Inhibitor(negative Heterotropic

Effector/Regulator)

Binding of one ligand decreases the affinity for substrate to other active sites.

Cyclic Adenosine

Monophosphate (cAMP)

Second messenger important in biological processes, which is derived from

adenosine triphosphate and utilized in signal transduction.

Consensus Sequence Most common nucleotide or amino acid at a particular position during alignment of

sequences.

Pseudosubstrate Protein with similar structure of an enzyme, which consequently blocks the

enzyme activity by making the enzyme bind to the pseudosubstrate than the realsubstrate.

Cascade Sequential vents that is consequent upon binding of ligand by a membranereceptor, which uses intermediates to amplify the cellular response.

Reciprocal Regulation Method of regulation by which one process is inhibited while an opposing process is

activated by reversible phosphorylation.Zymogen Inactive enzyme precursor that is activatated by proteolytic means.

Define feedback inhibition and describe how ATCase uses it as a regulatory

mechanism.Biological processes are carefully regulated. Groups of enzymes work together in sequential pathways to work

at the proper time and place. There are several methods of regulating the activity: (1) allosteric control, (2)multiple forms of enzymes, (3) reversible covalent modification, (4) proteolytic activation.

Feedback inhibition is a decrease of the enzymes activity from the product binding to the enzyme. ATCase is the

enzyme that catalyses the first step in a biosynthetic pathway that produces pyrimidine nucleotides needed fornucleic acids, energy storage, and enzyme cofactors. ATCase is essentially inhibited by the end product of the

pathway. When CTP levels are low and more isneeded, the activity of ATCase increases to make more. WhenCTP is abundant, however, the pathway is shut down. ATCase displays sigmoidal kinetics, and thus substrate

binding to one active site converts enzyme to R state and increasing their activity. The active sites showcooperativity.

Briefly explain the allosteric regulation of ATCase, including its quaternary

structure, its role in metabolism, and how its activity is regulated by allosteric

inhibition and activation. Include the physiological rationale for the inhibition and

activation.Allosteric control involves the activity being controlled by modulating the levels of small signaling molecules.Allosteric enzymes have multiple subunits, which exert influence on one another in the complex. The binding of

substrate at one site affects the affinity for substrates at other sites, eventually causing a conformational shiftfrom a less active state to a more active state. The conformational change is linked with ligand binding, yieldinghomotropic or heterotropic effects.

Allosteric enzymes do not follow Michaelis-Menten kinetics. The activity increases so steeply above a

threshold so that a small change in substrate concentration causes the large change in activity. Smallmolecule regulators can bind to the enzyme and change the threshold so as to adjust the activity to the requiredlevel. Allosteric regulation permits the rapid cycling of enzyme between

more active and less active conformations (just association/dissociation ofsmall molecules). They do not display the hyperbolic curve, andconsequently does not follow Michaelis-Menten kinetics.

Homotropic effects involve the substrate binding to allosteric enzymes. This

cooperative substrate binding and activation involves the substrate bindingto one active site, altering substrate binding affinity and/or catalytic activity

at other active sites on the same enzyme molecule. Extreme homoallosterycan occur at concentrations of substrate below a critical.


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Heterotropic effects involve the binding of other ligands (regulatory signaling molecules) to different sites from

the primary ligand (regulatory site). This can cause conformational changes that alter primary ligand bindingaffinity or catalytic activity. Positive regulation typically involves ligands known as activators, which allow the

enzyme to favor the R state. Negative regulation typically involves inhibitors, which allow the enzyme to favorthe T state. Heterotropic effectors bind to a different site from the active site.

Outline the structural effects of binding of CTP and PALA to ATCase.ATCases quaternary structure consists of catalytic trimers in addition to the regulatory dimers. Theseregulatory dimers contain zinc domains, which bind to 4 cysteine. The

structure, as a whole, can be dissociated, isolated and reconstitute. Theregulatory domains regulate the activity of the enzyme. The full protein has

six regulatory and six catalytic subunits. ATCase catalyzes the first step in abiosynthetic pathway to produce cytidine triphosphate (CTP). Cytosine

triphosphate can inhibit the activity of ATCase by binding to the regulatorysubunit of the protein, shifting the protein from the R state to the T state and

consequently inhibiting its activity. This typically can be observed inphysiological states. When CTP levels are high, there is no need to produce

more pyrimidines, and the inhibition of ATCase slows down the pathway. AsCTP levels fall, the inhibition is removed, and more pyrimidines can besynthesized. ATP is a purine nucleotide and not a product of the ATCase

pathway. ATP is the cellular energy source. When ATP levels are high, thecell is metabolically very active and preparing to divide. Therefore, it must

duplicate its DNA, and both ATP and CTP are needed for DNA synthesis. HighATP levels can override the inhibitory effects of CTP.

N-(Phosphoneacetyl)-L-Aspartate (PALA) can be utilized to identify active sites via a competitive inhibitor,

because it will bind to the same active site. PALA binding causes a conformational change from the T(tense/absence of substrate)) state to the R (relaxed/presence of substrate) state. PALA is a homotropic

allosteric enzyme. At high concentrations, the enzyme does not work, as the enzyme is in the T state. At lowconcentrations, it doesnt have binding either, and doesnt go into the T state. Thus, binding of the ligand must

occur for the enzymes shift from the T state to the R state.

Sketch plots of VO vs. [S] for an allosteric enzyme that illustrate positive homotropic

regulation and positive and negative heterotropic regulation, with ATCase as an

example. Specifically, sketch (all on the same axes) for ATCase: VO vs. [aspartate]

curves with no heterotropic regulators present, with an allosteric inhibitor present,

and with an allosteric activator present.Homoallostery occurs through cooperative substrate binding and activation. The substrate binding to one active

site alters the substrate binding affinity and/or catalytic affinity at other active sites onthe same enzyme molecule. Homotropic allosterism occurs when the binding of the first

enhances the probability of consecutive substrate binding.

Allosteric enzymes do not follow the hyperbolic curve, but follow the sigmoidal curve.From this, one can conclude that allosteric enzymes do not follow the Michaelis-Menten

kinetics either. Substrate binding to one active site converts enzyme to R stateincreasing their activity, and the enzymes collectively exist in a mixture of the R-state

and T-state. The sigmoid represents the mixture of two Michaelis-Menten enzymes.Positive homotropic regulation occurs when substrate binding to one active site alters

the substrate binding affinity at other active sites on the same enzyme.

Meanwhile, heteroallostery occurs when the binding ofother ligands (regulatory signaling molecules) to sites

different from the primary ligand and causeconformational changes that alter primary ligand

binding. There are two types of regulation byheterotropic effectors, positive and negative. Positive

(or activating) allostery occurs when the binding yields a shift (or favoring)towards the R state of the enzyme. Negative (or inhibiting) heteroallosteryoccurs when the binding causes the shift towards the T state.

In summary, this can be described in the following table:

Type of Allostery Diagram Description


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Homotropic Binding of first substrate enhances the probability of second

substrate binding.

Heterotropic The effector molecule binds to site on regulatory subunit. This sends

a message to the catalytic subunit. Then, the substrate binds moreor less readily depending on whether the effector is positive or

negative.

Describe in general terms how cells carry out reversible covalent modification of

enzymes, and how the modification would be removed.Covalent modifications involve the covalent attachment of anothermolecule to modify the activity of the enzymes. A specific enzyme

catalyzes the modification reaction. The modifying group removed by adifferent catalytic enzyme. The enzyme can cycle between active and

inactive (or more and less active) states. Phosphorylation anddephosphorylation are probably the most common means of regulating

enzymes, membrane channels, and virtually every metabolic process in eukaryotic cells. The enzyme isregulated by covalent modifications. Protein kinases are one of the largest protein families known. They aretypically involved in the addition of the phosphoryl group. In phosphorylation, the phosphoryl transfer involves

ATP or another nucleoside triphosphate to a protein. Adenosine triphosphate (ATP) is the most commonphosphoryl group donor, and it is extremely variable (the Gibbs Free Energy is extremely negative), so this is

essentially irreversible. Protein kinases catalyze phosphorylation. Kinases are very specific not only for localsequence but also for three-dimensional structure around it, and phosphorylate only a single target protein or a

small number of closely related target proteins. Some protein kinases are multifunctional and canphosphorylate many different target proteins. A particular kinase always phosphorylates a residue in a specific

sequence or a consensus sequence. Dephosphorylation involves the phosphate group being removed byhydrolysis of the phosphate ester. Protein phosphatases catalyze dephosphorylation.

Name (generic names) the types of enzymes that catalyze phosphorylation anddephosphorylation of proteins, specify what types of amino acid functional groups are

generally the targets of phosphorylation, and show the structure of such an enzyme

functional group before and after phosphorylation.Reaction Description Generic

Enzyme

Target Amino Acid Functional Groups Structure of Enzyme (before

and after)

Phos-phoryla-

tion

Addition of aphosphate

group

Kinases Typically target serine, threonine, or tyrosineresidues Target hydroxyl groups

Dephos-

phoryl-ation

Removal of a

phosphategroup.

Phospha

-tases

N/A

Explain whether the dephosphorylation reaction is actually the chemical reverse of

the phosphorylation reaction, and if not, what type of reaction the dephosphorylation

represents.Dephosphorylation is not the chemical reverse of the phosphorylation reaction, because it requires a differentenzyme and an additional reactant for the reaction to occur. The dephosphorylation reaction represents more of

a hydrolysis reaction because it requires water and protein phosphatase for the reaction to occur. This reaction

*

* * * *

*

* * * *


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is also extremely unfavorable, as the Gibbs free energy is negative, and thus a reverse process could not occur(and ATP cannot by synthesized in this manner).

List the reasons why phosphorylation is such an effective control mechanism.So it leads to the question of why phosphorylation is important. Phosphorylation is important because ofenergetics, efficiency, and amplification. The substantial amount of energy in the phosphate bond can strongly

affect conformation equilibria (energetics). Phosphorylation can be accomplished in seconds and can last for aslong as needed. By regulating the phosphorylation and dephosphorylation steps, the activity of the target can be

adjusted to synchronize with a physiological process (efficiency and timeliness). A single kinase canphosphorylate and activate hundreds of target molecules resulting in a large effect from a small stimulus. One

single activated protein kinase can phosphorylate hundreds of target proteins in a very short time. If targetproteins themselves are enzymes activated by phosphorylation, each activated enzyme then can carry out

many, many catalytic cycles on its substrate. This yields a result of a major multiplicative effect betweenstarting signal and final outcome several steps away.

What is the consensus sequence for PKA? Which residues are involved?Protein kinases are very specific not only for local sequence but also for three-dimensional structure around it,and phosphorylate only a single target protein or a small number of closely related target proteins. Some

protein kinases are multifunctional and can phosphorylate many different target proteins. Kinases typicallyphosphorylate a residue in a specific sequence of a consensus sequence. In PKA, there are two consensus

sequences: (1) Arg-Arg-X-Ser-Z or (2) Arg-Arg-X-Thr-Z, where X is a small amino acid residue and Z is a large

hydrophobic residue. Protein kinase A binds other substrate protein sequences with a much lower affinity, sodoesnt phosphorylate them very often.

Explain the regulation of protein kinase A (PKA) activity by cAMP, including

quaternary structural changes in PKA triggered by cAMP binding. How does the

term pseudosubstrate relate to the role of the regulatory subunits in PKA?Allosteric effects of a small signaling molecule regulate protein kinases themselves often. Cyclic AMP activates

protein kinase A. Cyclic AMP is an important intracellular signalingmolecule in both prokaryotic nad eukaryotic cells. cAMP is considered a

second messenger, a type of signaling molecule whose production isunder the control of other messengers such as hormones coming to the

cell from the extracellular environment. Most effects of cAMP in cellsare mediated by this one effect activation of protein kinase A. The

binding of cAMP alters quaternary structure of protein kinase A. The

PKA inactive form (without cAMP bound) has two catalytic subunits andtwo regulatory subunits (C2R2). The regulatory subunits are inhibitory. This form

cannot phosphorylate the targets. The cAMP binding to the regulatory subunitsmakes the regulatory subunits dissociate from the catalytic subunits. This makes it agreat example of integration of allosteric regulation and regulation by reversecovalent modification.

The pseudosubstrate sequence of PLA functions as an inhibitor. The ATP-Mg2+

complex and part of the inhibitor is bound in deep cleft between 2 lobes of theprotein. ATP is bound more to one lobe, while the inhibitor binds more to the other

lobe. An induced fit occurs through substrate peptide binding, in that the lobes movecloser together. The alanine is in the pseudosubstraate sequence, while the serine is

in the actual substrate sequence.

Describe the general mechanism by which zymogens are activated active enzymes.Zymogens (or proenzymes) are biologically synthesized,

catalytically inactive precursor polypeptide chains. They fold inthree dimensions and are later activated by enzyme-catalyzed

cleave (hydrolysis) of one or more specific peptide bonds. Zymogenactivation is an irreversible proteolytic-activating process. Thestrategy involved with this involves the prevention of

indiscriminate proteolysis, and delivery only when necessary. Thezymogens are stored in granules, relatively inert, and upon

hormonal activations, the granules are dumped into the smallintestines, spurring activation. The different digestive proteases

have different substrate specificities, enabling the breakdown of a wide variety of peptides. A single enzyme


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known as trypsin activates all the zymogens. Trypsin is activated by enteropeptidase, which is secreted by cellslining the digestive tract. In turn, trypsin activates the other zymogens.

With these enzymes, there are also inhibitors of proteolytic enzymes. PTI binds very tightly to trypsin and it is

extremely difficult to dissociate the complex. Pancreatic trypsin inhibitor is a substrate, but the peptide bondafter the lysine is cleaved only VERY slowly. The combination of very tight binding and very slow catalytic

turnover makes PTI a very effective inhibitor. It is inhibited due to the change in shape of the complex as well asits tight binding.

Give examples of enzymes and proteins that are derived from zymogens and the

biological processes they mediate.The gastric and pancreatic zymogens are mainly involved in digestion.Once again they are synthesized and stored in granules, and (once

activated) are secreted into the small intestine, where they areactivated.

This is not limited to the digestive syste

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