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: 𝑌 = [!"]

!" ![!]= [!]

!!![!].

Prosthetic Group Metal ion or organic or metallo-organic compound other than an amino acid that’s tightly bound to a protein (binding is covalent or tight non-covalent), required or the protein’s function/activity. 𝐻𝑜𝑙𝑜𝑝𝑟𝑜𝑡𝑒𝑖𝑛⟷ 𝐴𝑝𝑜𝑝𝑟𝑜𝑡𝑒𝑖𝑛 + 𝑃𝑟𝑜𝑠𝑡ℎ𝑒𝑡𝑖𝑐  𝐺𝑟𝑜𝑢𝑝

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 the form 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 αβ protomers.

Binding Site

Site where specific, typically noncavalent interactions between molecular surfaces 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 by

binding in hydrophobic site in protein. Allosteric (Allosteric site,

allosteric effector, allosteric regulation)

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. The substrate can also function as the regulatory molecule. Heterotropic effects imply interacting sites binding to the same ligand. The regulatory molecule is NOT the same as the substrate. Allosteric Site: Site other than the protein’s active site. Allosteric Effector/Modulator: Ligands that spur allosteric effects. They can affect the equilibrium between the T and R states. Allosteric Regulation: Regulation of an enzyme or other protein by binding an effector molecule at the protein’s allosteric site. Allosteric activators enhance the protein’s activity, while allosteric inhibitors decrease the protein’s 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 overall structure 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 α (141 AA residues) and two β chains (146 AA residues). The quaternary structure of hemoglobin confers then allosteric properties. Myoglobin can bind to oxygen as well as release. Hemoglobin’s function in vertebrate erythrocytes is to transport O2 form lungs to tissues and transport CO2 from tissues to lungs. Myoglobin is mainly the storage element within muscle tissue.

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It binds to O2 delivered to tissue by blood and stored until needed as terminal electron acceptor for energy metabolism, and is the main intracellular transport for oxygen. To be a potent carrier, it needs to be able to carry 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 and associated metal complex. Remember that hemoglobin, as a structure, is very compact, with almost no empty space inside. It is mostly (~75%) α-helical, the rest mostly β turns and loops (at the surface). There are 8 helices, designated A-H from N to 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 coordination site but distal. As a quaternary structure it is a tetramer with two α and two β subunits in adult (and different at different stage of life). Noncovalent bonds stabilize this structure. The α/β 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: [𝑃] + [𝐿] ↔ [𝑃𝐿]. From there we can determine the equilibrium 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 𝐾! =

[𝑃𝐿]!"[𝑃]!"[𝐿]!"

Large Small

Dissociation Constant 𝐾! =

[𝑃]!"[𝐿]!"[𝑃𝐿]!"

=1𝐾!

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 oxygen transport protein in the blood. It circulates in red blood cells, and binds to oxygen in the 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 as iron 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 spur greater 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 θ) is the fraction of total binding sites on protein occupied by ligand. We can determine the fraction saturation by the following formula: 𝑌 = !"#$"#%  !"#$!  !""#$%&'

!"!#$  !"#$"#%  !"#$!= [!"]

!" ! != [!]

!!![!]. 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 ( 𝐿 = 𝐾!).

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Myoglobin binding to oxygen follows the same protein ligand interactions. However, since O2 is a gad it is more convenient to utilize partial pressures of oxygen (in the gas phase above the solution), than concentrations of O2 dissolved in the solution, in the following formula: 𝑀𝑏𝑂! ⟺ 𝑀𝑏 + 𝑂!. Thus, we can utilize the same methods to determine the dissociation constant: 𝐾! =

!" [!!]

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

fractional saturation: 𝑌 = [!"#!]

!" ![!"#!]= [!!]

!!![!!]= !"!

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

storage protein in tissues, and 1 oxygen binding site per myoglobin allowing for a hyperbolic binding curve. Free myoglobin implies myoglobin with empty binding sites, and myoglobin with oxygen imply occupied sites. Thus at p50 is the pO2 that gives 50% saturation.

Explain how Hb works physiologically (in vivo), i.e., how cooperativity in O2 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 the bonding angle. It adopts a slight angle because His E7 sterically blocks the perpendicular arrangement. Hemoglobin is allosteric, exemplary of homotropic effects. 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 binding curve). This is due to the fact that myoglobin has only one oxygen-binding site and thus no opportunity for interaction, while hemoglobin has possible interactions and communication between binding sites. There is communication between different ligand binding sites on same multimeric protein molecule, in the form of structural (conformational) changes. There are 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 protons and 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 delivering to 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 relative saturations at the lung, tissue, and the difference.

Saturation Hb Cooperative Noncooperative Lung 98% 60%

Tissue 32% 22% Δ=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 O2 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 binding initiates structural changes.

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At the tertiary level, there is conformational change in one subunit. It causes structural changes in interface between α1β1 and α2β2 protomers. This triggers that quaternary change in whole hemoglobin tetramer. At the quaternary level, α1β1 shifts relative to α2β2 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 a small change that is a major player in the protein’s 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 is no bond

between the glycine and ion. There would be no communication, and thus consequently no allostery. Deoxyhemoglobin (the T state) goes to oxyhemoglobin (the R state). The allostery (switching the T and R states) is the main contributor to the sigmoidal shape of the binding curve. The conversion 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 is a weak oxygen binding). The α1β1 shifts relative to α2β2 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 molecular switch 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, where the 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 a conformational 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 state assists 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 to a phenomenon known as the Bohr effect. The Bohr effect is when carbon dioxide and 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 can

surmise that the protons, carbon dioxide, and 2, 3-bisphosphoglycerate all decrease oxygen affinity. BPG is the negative allosteric regulator found in red blood cells, and highly anionic (having a -5 charge). 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 at the 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 with higher oxygen concentration. The 3 charged groups from each β chain in central cavity help bind BPG by ionic interactions. 2, 3-BPG binds in the central cavity of hemoglobin, and prefer to bind to the T state in order to help shift from R state to T state. BPG allows for more oxygen to be released due to such decreased binding affinity. Though it is found in both lungs and

These)ionic)interac8ons)stabilize)the)T)state)

CO2)reacts)with)the)+)charged)N_term)to)form)_)charged)carbamate)groups)=>)ionic)interac8ons)at)the)α/β)interface)

ionic)interac8on)with)Asp)94))

At)lower)pH,)His)146)protonated

Structural)Basis)of)the)Bohr)Effect)

The)N_termini)of)Hb)chains)lie)at)an)interface)between)α)and)β)subunits)

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tissues, but has little effect in the lungs but a much profound effect in the tissues, where oxygen needs to be released. This is particularly seen in other animals, but also in the early fetus. During early development, the human fetus expresses different α and β hemoglobin genes. These are similar but not identical to the hemoglobin genes expressed 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 β subunits found in adult hemoglobin, but γ subunits. At the tertiary level, the positively charged histidine residues found in the adult hemoglobin is essentially replaced 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. It is important to recognize that carbon dioxide (CO2) and H+ ions are also negative allosteric effectors of hemoglobin. In active muscle cells, oxygen is rapidly consumed 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 affinity from stabilization of the T state. Combination of the two will reveal that it will have the furthest affinity decrease. 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 T state. 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 carbon dioxide 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 α/β 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 (and consequently 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 Β His 146 will occur and carbon dioxide will then be release through exhalation, allowing reversion 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 oxygen binding. At high altitude states, the concentration of 2, 3 BPG

increases and the oxygen-binding curve will shift right. For mutant hemoglobin, this mainly involves exploration of the structure-function relationships in a protein. Mutant hemoglobins can encompass silent mutations, but there are mutations that can spur serious diseases. For example, sickle hemoglobin (HbS) is when there is a change in normal residue 6 from negatively charged glutamic acid to hydrophobic valine can spur serious consequences.

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 α-thalassemia, the β chains are affect and the hemoglobin does not become cooperative. In sickle hemoglobin (HbS), the hydrophobic valine on subunit of 1 tetramer 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 cause red 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 hydrophobic patch to allow hemoglobin to aggregate,

distorting and block capillaries α-Thalassemia β-Chains Hemoglobin does not become cooperative β-Thalassemia α-Chains Forms insoluble aggregates resulting in few red

cells.

These)ionic)interac8ons)stabilize)the)T)state)

CO2)reacts)with)the)+)charged)N_term)to)form)_)charged)carbamate)groups)=>)ionic)interac8ons)at)the)α/β)interface)

ionic)interac8on)with)Asp)94))

At)lower)pH,)His)146)protonated

Structural)Basis)of)the)Bohr)Effect)

The)N_termini)of)Hb)chains)lie)at)an)interface)between)α)and)β)subunits)

Effec8ve)transport)due)to)combined)allosteric)effects)

Lungs

Tissues

O2'is'released'

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

T)state)

Hb(O2)4)carries)O2)to)8ssue)capillaries)

2,3_BPG)throughout)body)(in)both)lungs)and)8ssues))__)livle)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'β'His'146 R_state)

O2'(inhaled)'

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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.

Lecture XIV

Introduction to Enzymes and Enzyme Kinetics

Terminology Term Definition

Rate Enhancement Factor by which catalyst increases rate of reaction. Calculated by: 𝑅𝑎𝑡𝑒  𝑒𝑛ℎ𝑎𝑛𝑐𝑒𝑚𝑒𝑛𝑡 =   !!"#"$%&'(

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

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 protein’s biological activity.

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

the protein’s function/activity. 𝐻𝑜𝑙𝑜𝑝𝑟𝑜𝑡𝑒𝑖𝑛⟷ 𝐴𝑝𝑜𝑝𝑟𝑜𝑡𝑒𝑖𝑛 + 𝑃𝑟𝑜𝑠𝑡ℎ𝑒𝑡𝑖𝑐  𝐺𝑟𝑜𝑢𝑝 Catalyst Substance that is not consumed by the reaction itself, and speed or slow the

reaction’s progress. They operate under physiological conditions, work by forming 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 shape complementarity.

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 molecule

per 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 of

substrate concentration. Reversible Inhibition

(Competitive, pure noncompetitive, uncompetitive)

Reversible inhibition binds via weak, non-covalent interactions, which can be easily 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 an equilibrium with dissociation constant Ki.

2. Uncompetitive: Involves inhibitor binding only to the complex formed between 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 and creates a different intercept.

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

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

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

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

Suicide Inhibition (Mechanism-Based

Inhibitor)

Form of irreversible enzyme inhibition that occurs when an irreversible complex (by covalent bonding) forms during normal catalysis. The substrates like

molecules that cannot complete the reaction. Forms a stable inhibitor-enzyme complex.

Review Term Definition

Equilibrium 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 reverse

reaction. Mass Action Ratio for a

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

[!]![!]!.

Biochemical Standard Conditions

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

Standard Free Energy Change

Gibbs free energy at equilibrium, calculated by: ∆𝐺!° = −𝑅𝑇 ln𝐾′!"

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

Calculated by: ∆𝐺! = ∆𝐺!° + 𝑅𝑇 ln𝑄.

Describe the general properties of enzymes as catalysts that are especially important for 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 and consequently 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 𝑟𝑎𝑡𝑒  𝑜𝑓  𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = [!"#$%#&%]

!"#$  𝑜𝑟   [!"#$%&']

!"#$. We can

also determine the rate enhancement as a quotient of the catalyzed and uncatalyzed rate constants 𝑟𝑎𝑡𝑒  𝑒𝑛ℎ𝑎𝑛𝑐𝑒𝑚𝑒𝑛𝑡 = !!"#"$%&'(

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

Express the equilibrium constant of a reaction in terms of the equilibrium mass action ratio. The equilibrium constant is simply when 𝑣!"#$%&' = 𝑣!"#$%#&%. Thus we also know that the equilibrium mass action ratio is equal to the equilibrium constant (𝑄)!" = 𝐾′!" . Thus one can state that for a given chemical equation

𝑎𝐴 + 𝑏𝐵 ↔ 𝑐𝐶 + 𝑑𝐷, with the mass action ratio 𝑄 = [!]![!]!

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

[!]![!]!

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

and reactants are at equilibrium.

Express the equilibrium constant of a reaction in terms of the rate constants for the forward 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, 𝑣! = 𝑘![𝑆]!" and 𝑣! = 𝑘![𝑃]!". At equilibrium, the reaction velocities are equal 𝑣! = 𝑣!  → 𝑘![𝑆]!" = 𝑘![𝑃]!". Thus, we can then derivate that

Chemical*Kine)cs*•  For*the*reac)on*

• Rate constants are lower case k’s • Equilibrium constants are upper case K's.

Enzymes*do*NOT*alter*equilibrium*quan))es*K’eq*or*ΔG’.*Enzymes*DO*increase*rate*constants*(kF*and*kR)*and*thus*increase*rates*of*reac)ons.**

• ***Velocity*(rate)*of*forward*reac)on*=*vF*=*kF*[S]eq***• ***Velocity*(rate)*of*reverse*reac)on*=*vR*=*kR*[P]eq**• ***The*equilibrium*constant*?*

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

kR

€

kFkR

=[P]eq[S]eq

= Keq'

At&equilibrium,&vF&=&vR,&so&kF&[S]eq&=&kR&[P]eq&

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!!!!= [!]!"

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

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

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 (particularly ΔG) should not change. Thus, increasing the rate constant for the forward reaction by a factor of 108 should allow an increase of the rate constant for the back reaction by a factor of 108. 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 ΔG for the overall reaction, ΔG✜

uncat, ΔG✜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 activation energy, which is the energy required to cause the reaction to occure. The catalyst increases the rate of the reaction 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: it’s a specific kind of ΔG) that determines the magnitude of the rate constant for the reaction at a given temperature. You don’t have to memorize the equation relating this quantity to k. The rate constant (k) is dependent on ΔGŦ, 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 the starting state, which is the “barrier” over which the reaction must go in order to proceed. This can be calculated

by the following equation: 𝑘 = !!!

!𝑒!∆!!

!" , 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 the reaction rate. Increasing the temperature (T) can do this, but it leaves the possibility of the protein’s denaturation. However, understanding this equation can allow us to understand how to derive the rate

enhancement. 𝑅𝑎𝑡𝑒  𝑒𝑛ℎ𝑎𝑛𝑐𝑒𝑚𝑒𝑛𝑡 =  

!!!! !

!∆!!"#"$%&'(!

!"

!!!! !

!∆!!"#$%$&'()*!

!"

= 𝑒∆!!"#$%$&'()*

! !∆!!"#"$%&'(!

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

.

Enzymes*lower*the*ac)va)on*energy*

= transition state

high*energy*barrier*=*

slower*reac)on*

lower*energy*barrier*=*

faster*reac)on*

ΔG’=*Free*energy*gap*between*reactants*and*products*at*equilibrium*

€

ΔGuncatǂ

€

ΔGcatǂ €

ΔGuncatǂ

€

ΔGcatǂ = -

Essence*of*catalysis:*Stabiliza)on*of*the*transi)on*state

Enzymes*change*the*pathway*of*a*reac)on**Enzymes*bind*)ghtly*to*the*transi)on*state*

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 ΔG≠. 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 the primary structure.

2. A small portion of the total available volume. The amino acid not at the active site position, 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 polar side 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, meaning that the active site of the unbound enzyme is complementary in shape of the substrate (called the lock and key model), or (2) plastic, flexible active site, meaning the enzyme changes shape upon substrate binding to achieve shape complementarity (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 in shape to the substrate: a rigid, inflexible active site

Induced Fit

Enzyme changes shape upon substrate binding to achieve 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: 𝑉! = 𝑉!"#[!]

!!![!], 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 𝑉! =

!!"#

!. Based on the forward and reverse

reactions, we can determine the Michaelis constant as: 𝐾! = !!!!!!

!!. If the Michaelis constant is must

larger than substrate concentration 𝐾! ≫ [𝑆] , then 𝑉! =

!!"#[!]

!!.

We need to remember that before the reaction occurs, the enzyme must bind to substrate to yield 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 rate of formation of product slows and equilibrium is reached.

*5*–*Specificity*of*binding:*defined*arrangement*of*atoms*in*an*ac)ve*site,*2*models**

Ac)ve*sites*

Ac)ve*of*the*unbound*enzyme*is*complementary*in*shape*to*the*substrate:*rigid,*inflexible*ac)ve*site*

Enzyme*changes*shape*upon*substrate*binding*to*achieve*shape*complementarity*

LOCK*and*KEY*model* INDUCED*FIT*model*

E.*Fischer,*1890* D.*Koshland,*1958*

*5*–*Specificity*of*binding:*defined*arrangement*of*atoms*in*an*ac)ve*site,*2*models**

Ac)ve*sites*

Ac)ve*of*the*unbound*enzyme*is*complementary*in*shape*to*the*substrate:*rigid,*inflexible*ac)ve*site*

Enzyme*changes*shape*upon*substrate*binding*to*achieve*shape*complementarity*

LOCK*and*KEY*model* INDUCED*FIT*model*

E.*Fischer,*1890* D.*Koshland,*1958*

MichaelislMenten*Equa)on*

E + S ES E + P k1

k-1

k2

€

V0 =Vmax[S]

KM + [S]

€

KM =k−1 + k2k1

Michaelis*constant*

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: 𝐾! = !!!!!!

!!. KM is the

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

!!"#

!. 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 molecule per 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: 𝑘!"# = 𝑘! =

!!"#

! !. Most enzymes need high substrate concentrations to create

significant amounts of products. At low concentrations not much happens. Some enzymes acan work better than others under the stringent conditions of low substrate concentration found in cells. The most efficiency enzymes 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 Units

KM 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 catalyst is present. Inverse time, such as s-1 Vmax Maximum rate achieved by the system, at maximum

(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 the maximal 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 the Michaelis constant KM finding the concentration of substrate where the relative rate is 50%, or 0.5, because the KM 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 because the 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 also able to manipulate the graph towards a linear function by a Lineweaver-Burk plot. The hyperbolic plot can be manipulated in the following manner: 𝑉! = 𝑉!"#

[!]

!!![!]→ !

!!= !!![!]

!!"#[!]= !!

!!"#

!

[!]+ !

!!"#. 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 this indicative 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 of low [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 is high, then the enzymes’ side groups must favorably interact with the substrate.

12

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 Plot KM Vmax KM/Vmax

Com-petitive

Competition for active site with

substrate

Inc N/C Inc

Uncom-petitive

Binding to site in proximity of the

substrate. A reduction in equal

amounts.

Inc Inc N/C

Noncom-petitive

Binding at a regulatory 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 on Km 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 while leaving 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 KM/Vmax is unchanged.

Relative to the values in the absence of an inhibitor, a noncompetitive inhibitor will leave KM unchanged 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 state that 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 to counter disease.

Lecture XV

Enzyme Catalysis

Uncompe))ve*inhibitors*

KM*/*Vmax*unchanged

KM*and*Vmax*are*reduced*by*equal*amounts

High*[S]*does*not*overcome*effect*of*I

Inhibitor binds only to the ES complex

NonlCompe))ve*inhibitors*

•  The*enzyme*can*bind*substrate*in*the*ac)ve*site,*and*inhibitor*in*a*regulatory*site.*

•  The*binding*and*release*of*the*inhibitor*from*the*regulatory*site*forms*an*equilibrium*with*dissocia)on*constant*Ki.*

•  The*inhibitor*can*bind*to*either*E*or*the*ES*complex.*When*I*is*bound,*the*enzyme*is*in*a*state*with*a*slower*cataly)c*rate.*But*substrate*affinity*is*unaffected*and*KM*is*unchanged.*

•  So*the*enzyme*cannot*achieve*Vmax,*but*rather*a*lower*apparent*maximal*velocity*

binding*and*release*of*I*from*the*ac)ve*site*forms*an*equilibrium*with*dissocia)on*constant*Ki*

Compe))ve*inhibitors*

13

Terminology Term 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 193

point 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 the Bronsted-Lowry definitions) to the substrate during the reaction.

Covalent A group on the enzyme becomes covalently modified during reaction, e.g. by forming a covalent 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) or loosely bound (binding reversibly and dissociating from the enzyme). They typically stabilize a negative charge on the reaction intermediate and allow proper orientation of substrate via coordination, to allow for binding at specific geometry. The metal ion also facilitates formation of nucleophile, and even polarizes 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 a proton, 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

14

a proton in catalysis then has to accept a proton in catalytic mechanism for catalyst to be regenerated in original conjugate 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) by donating 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 nucleophilic functional 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 catalytic mechanism. 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 enzyme

gaining 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 typically

deprotonated. Covalent Catalyst Rate acceleration by transient

formation 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 the ionization state of the substrate and possibly spur variation of the protein’s structure. The optimum pH in protein/substrate interactions is the pH at which the substrate maximally binds to substrate. The substrate attaches 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 enzyme’s activity (as a function of pH) – be able to interpret 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 the protein’s 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.

Op1mum(pH(and(protein/substrate(interac1ons((

VCOOV:(picks(up(H+( VNH3+:(loses(H+(

Op1mum(pH(around(7(A(substrate(ajaches(itself(to(the(enzyme(via(two(ionic(bonds.((

pH(lower(than(7(V(acidic pH(higher(than(7(V(basic

Ionic(bonds(between(the(substrate(and(the(enzyme(cannot(be(formed.((If(those(bonds(were(necessary(to(ajach(the(substrate(and(ac1vate(it(in(some(way,(then(at(lower(pH(or(higher(pH,(the(enzyme(won't(work.(

VNH3+:(no(change VCOOV:(no(change(

Op1mum(pH(and(protein/substrate(interac1ons((

VCOOV:(picks(up(H+( VNH3+:(loses(H+(

Op1mum(pH(around(7(A(substrate(ajaches(itself(to(the(enzyme(via(two(ionic(bonds.((

pH(lower(than(7(V(acidic pH(higher(than(7(V(basic

Ionic(bonds(between(the(substrate(and(the(enzyme(cannot(be(formed.((If(those(bonds(were(necessary(to(ajach(the(substrate(and(ac1vate(it(in(some(way,(then(at(lower(pH(or(higher(pH,(the(enzyme(won't(work.(

VNH3+:(no(change VCOOV:(no(change(

15

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 the process 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 cause changes 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: Hydrolysis Enhance Rate

✓ ✓

Carbonic Anhydrases ✓ Restriction Endonuclease:

Specificity, Hydrolysis ✓

Nucleoside Monophosphate Kinases: 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 the proteases. 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 carboxyl side of aromatic, large hydrophobic residues (right), and the amino-terminal side of the peptide bond to be cleaved (left). These residues are not easily accessible because the hydrophobic amino acids are in the core, and thus 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 experiments. Chemical labeling

16

experiments are used to figure out which residues are responsible for important steps in a reaction mechanism. One can use an irreversible inhibitor (group-specific agents) to label the particular functional group (even to its position). For example, in chymotrypsin, DIPF is utilized to find serine residues. It forms a covalent adult on Serine 195, which renders the enzyme inactive. Only Ser 195, out of the 28 serines in chymotrypsin, is labeled, suggesting that it is both especially reactive and that this reactivity is necessary for catalysis. Kinetic experiments can be utilized to figure out the steps involved in the reaction mechanism and the length of time in each step. The initial phase, the burst phase is typically associated with the release of product, thus it’s high rate. The second phase, the steady-state phase, is involved in waiting for the release of the intermediate. Subsequent reactions must wait for an active site to become available through release of an intermediate, giving

rise to the steady-state phase. This because there is time required in order for the product to arise from the intermediate. One can also use an affinity label in order to determine characterization of the mechanism. The affinity label should be complementary to the active site and chemically reactive toward specific side chain functional groups. One can take something that looks like substrate and can add something highly reactive. In chymotrypsin, deactivation of HIS 57 in this manner shows that it plays a role in catalysis. X-ray crystallography can allow inferences on mechanism. Determination of the crystal

structure of an enzyme provides a detailed description of the three-dimensional arrangement of the molecule and in particular of the active site. Crystal structure analysis reveals a catalytic triad, a group of 3 side chains, which are responsible for the peculiar reactivity of the serine. For chymotrypsin, it can be observed that there is interaction through disulfide bonds. The final way to determine the mechanism of a reaction is site-directed mutagenesis. Mutations along certain amino acid resides can help determine the amino acid’s role in the protein’s action. However, the same rules apply. If kcat is unchanged, then the mutation must occur at the specificity pocket. If KM is unchanged, the mutation must occur at the catalytic triad. Changes in the specificity in the specificity pocket will change the KM

along with the Vmax.

Describe the main features of the chemical mechanism of hydrolysis of peptide bonds by chymotrypsin, including the following:

What is the “job” of the catalyst (the protease), i.e., what group needs to be made more susceptible to nucleophilic attack? The protease chymotrypsin has structural elements at the active site. It has a Serine 195, Histidine 57, and an Aspartic Acid 102. These three amino acids are in proximity of the active site and thus known as the catalytic triad. His 57 is a general base catalyst that facilitates the removal of the hydroxyl proton on Serine 195, allowing formation of an alkoxide ion, which is a strong nucleophile. The Aspartic Acid 102 allows orientation of the His 57 and stabilizes the protonated form due to its negative charge. Remember that the bond between the carboxyl side of aromatic, large hydrophobic residue and the amino-terminal side of the peptide residue is cleaved. The main features of the hydrolysis of peptide bonds via chymotrypsin involve covalent catalysis. Thus, it occurs in two stages: acylation and de-acylation. Acylation requires addition of the acyl group to the enzyme to yield an acyl-enzyme intermediate. Deacylation with water will yield the enzyme and a carboxylic acid product. The free enzyme is re-generated. The group that needs to be more susceptible to nucleophilic attack is the Serine 195 group. It is made more nucleophilic than usual with the assistance of nearby His residue as the general base. The nucleophilic attack is facilitated by the Histidine, acting as the general base. Acylation will allow the 2nd product to bind to serine and the 1st product is released. Acylation allows formation of the 1st Tetrahedral Intermediate. Serine attacks the carbonyl of the substrate in a nucleophilic fashion and allows stabilization by the Asp. The O of Ser-OH is activated by H-bond of the His on catalytic triad. This helps maintain perfect orientation of His and ser in hydrogen bonded network, and facilitates proton transfer by electrostatic stabilization of HisH+ after it has accepted the proton. complex also yields a tetrahedral intermediate with an oxyanion hole, which is covalently bonded to carbonyl C. HisN, acting as a general base, then facilitates the nucleophilic attack. At deacylation, the HisH+ acts as a general acid and donates proton back to the serine.

Chymotrypsin(Mechanism(

HisH+((general(acid)(donates(proton(back(to(Ser(O*

nucleophilic(ajack(facilitated(by(HisN:(ac1ng(as(general(base(*

Acylation: 2nd product bonded to SER and 1st product is released

Deacylation: 2nd product is released

nucleophilic(ajack(facilitated(by(HisN:(ac1ng(as(general(base(*

17

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-O engages 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. This helps maintain perfect orientation of His and Ser in hydrogen bonded network and facilitates H+ transfer by electrostatic stabilization of HisH+ after it has accepted 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 and breakdown. The hydrophobic “specificity pocket” of chymotrypsin is the area of the active site responsible for chymotrypsin’s 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 chains to 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 the binding 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 couldn’t form to the 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 negatively charged 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. Within this 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 as phenylalanine, tyrosine, and tryptophan. The ASP 189 of trypsin has a negative carboxylic 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 fused ring 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 remain constant 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 of chymotrypsin, 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 enzyme arose from a previous ancestor.

18

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 Cysteine

Proteases Cysteine Structural homolog of chymotrypsin

Aspartyl Proteases

Water molecule

ASP is involved in general base catalysis as well as orientation and polarization of the substrate carbonyl. Optimum pH is approximately 2, pKa

of one ASP shifted from 4 to 1.2 Metallo-

proteases Water

molecule Metal and general bae catalysts to activate a water molecule.

Nucleoside Monophosphate

Kinase

Phosphate Transfer

True substrate is the magnesium ion complex of the NTP. It is bound to the β and γ phosphoryl groups and to four water molecules, including a conserved

aspartate 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 is conserved 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 an ATP 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 β and γ-phosphates. The magnesium ion is bound to β and γ phosphoryl groups 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 the nucleotide 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 aqueous environment, 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 in the active site are those that make up the pocket of the enzyme. Mutation will change the enzyme’s activity by how 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, the activity 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 specificity will be lost and unable to bond to correct substrate residues.

Affected Site of Mutation

Effect on Enzyme Activity (kcat)

Effect on Enzyme Specificty (KM)

Effect on Rate of the Reaction (Vmax)

Catalytic Triad Decreased Unchanged Decreased Hydrophobic Pocket Unchanged or Decreased Decreased Decreased

However, anything that affects binding will affect the overall activity of the enzyme. Anything that can affect the binding may or may not change the kcat, because there is change to the specificity for that specific amino acid. A conservative mutation can induce the smallest change, while a nonconservative mutation can induce the largest change.

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 always inhibitors of an enzyme. Knowledge of the enzyme mechanism helps in the designing and production of drugs.

19

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 two identical 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 induced fit). HIV uses its symmetrical nature as a mechanism. Aided by general base catalysts, water attacks the carbonyl carbon, generating a tetrahedral intermediate. This yields a transition state. 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 pKa of 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 typically interfere with the protein. They form non-covalent complexes with enzymes but they bind so tightly that they can be considered irreversible inhibitos. Transition state analogs 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 a hydroxyl positioned next to a branch containing a benzyl group. Another medication, Crixivan has a conformation that approximates the 2-fold symmetry of the enzyme. Crixivan thus inhibits HIV protease without affecting normal cellular Asp proteases, which don’t have the 2-fold symmetry that HIV protease has.

What type of inhibitor is penicillin? How does penicillin work? Penicillin is a β-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 the peptidoglycan 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 consequently cell lysis from those weak linkages. The build up of the peptidoglycan precursors can cause activation of the cell wall hydrolases and autolysins, digesting the bacteria’s remaining peptidoglycan. Thus, this causes a rapid bacterial 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 in the 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 about biological catalysts, as scientists research biochemical pathways, and attempt to apply these catalysts to replace previous, 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-subunit complex.

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.

20

Feedback Inhibition A decrease in activity from the molecule binding to the enzyme. Cooperativity Phenomenon displayed by enzymes or receptors that have multiple binding sites

where affinity is increased or decreased. 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 enzyme’s activity. It

typically is an activator. Heterotropic

Effector/Regulator Regulatory molecule that is not also the enzyme’s 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 and binding 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. It will consequently shift to the T state.

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 real

substrate. Cascade Sequential vents that is consequent upon binding of ligand by a membrane

receptor, 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 enzyme’s 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 for nucleic 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. When CTP 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 show cooperativity.

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 shift from a less active state to a more active state. The conformational change is linked with ligand binding, yielding homotropic 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. Small molecule regulators can bind to the enzyme and change the threshold so as to adjust the activity to the required level. Allosteric regulation permits the rapid cycling of enzyme between more active and less active conformations (just association/dissociation of small molecules). They do not display the hyperbolic curve, and consequently does not follow Michaelis-Menten kinetics.

21

Homotropic effects involve the substrate binding to allosteric enzymes. This cooperative substrate binding and activation involves the substrate binding to one active site, altering substrate binding affinity and/or catalytic activity at other active sites on the same enzyme molecule. Extreme homoallostery can occur at concentrations of substrate below a critical. 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 binding affinity 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 favor the 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. ATCase’s quaternary structure consists of catalytic trimers in addition to the regulatory dimers. These regulatory dimers contain zinc domains, which bind to 4 cysteine. The structure, as a whole, can be dissociated, isolated and reconstitute. The regulatory domains regulate the activity of the enzyme. The full protein has six regulatory and six catalytic subunits. ATCase catalyzes the first step in a biosynthetic pathway to produce cytidine triphosphate (CTP). Cytosine triphosphate can inhibit the activity of ATCase by binding to the regulatory subunit of the protein, shifting the protein from the R state to the T state and consequently inhibiting its activity. This typically can be observed in physiological states. When CTP levels are high, there is no need to produce more pyrimidines, and the inhibition of ATCase slows down the pathway. As CTP levels fall, the inhibition is removed, and more pyrimidines can be synthesized. ATP is a purine nucleotide and not a product of the ATCase pathway. ATP is the cellular energy source. When ATP levels are high, the cell is metabolically very active and preparing to divide. Therefore, it must duplicate its DNA, and both ATP and CTP are needed for DNA synthesis. High ATP 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 low concentrations, it doesn’t have binding either, and doesn’t go into the T state. Thus, binding of the ligand must occur for the enzyme’s 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 on the 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 state increasing 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 of other ligands (regulatory signaling molecules) to sites different from the primary ligand and cause conformational changes that alter primary ligand binding. There are two types of regulation by heterotropic 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) heteroallostery occurs when the

22

binding causes the shift towards the T state. In summary, this can be described in the following table:

Type of Allostery Diagram Example Description Homotropic

Oxygen binding to

hemoglobin.

Binding of first substrate enhances the probability of second substrate binding.

Heterotropic

Binding of CTP to ATCase.

The effector molecule binds to site on regulatory subunit. This sends a message to the catalytic

subunit. Then, the substrate binds more or 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 another molecule to modify the activity of the enzymes. A specific enzyme catalyzes the modification reaction. The modifying group removed by a different catalytic enzyme. The enzyme can cycle between active and inactive (or more and less active) states. Phosphorylation and dephosphorylation are probably the most common means of regulating

enzymes, membrane channels, and virtually every metabolic process in eukaryotic cells. The enzyme is regulated by covalent modifications. Protein kinases are one of the largest protein families known. They are typically 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 common phosphoryl 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 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. A particular kinase always phosphorylates a residue in a specific sequence or a “consensus” sequence. Dephosphorylation involves the phosphate group being removed by hydrolysis of the phosphate ester. Protein phosphatases catalyze dephosphorylation.

Name (“generic” names) the types of enzymes that catalyze phosphorylation and dephosphorylation 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 a phosphate

group

Kinases Typically target serine, threonine, or tyrosine residues Target hydroxyl groups

Dephos-phoryl-

ation

Removal of a phosphate

group.

Phospha-tases

N/A

Generaliza'on*

Homotropic*Allosterism* Heterotropic*Allosterism*

Generaliza'on*

Homotropic*Allosterism* Heterotropic*Allosterism*

23

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 different enzyme 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 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 of energetics, 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 as long 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 can phosphorylate 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 target proteins 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 between starting 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 typically phosphorylate 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, so doesn’t 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 signaling molecule in both prokaryotic nad eukaryotic cells. cAMP is considered a second messenger, a type of signaling molecule whose production is under the control of other “messengers” such as hormones coming to the cell from the extracellular environment. Most effects of cAMP in cells are 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 and

two regulatory subunits (C2R2). The regulatory subunits are inhibitory. This form cannot phosphorylate the targets. The cAMP binding to the regulatory subunits makes the regulatory subunits dissociate from the catalytic subunits. This makes it a great example of integration of allosteric regulation and regulation by reverse covalent 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 the protein. 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 move closer 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 in three dimensions and are later activated by enzyme-catalyzed cleave (hydrolysis) of one or more specific peptide bonds. Zymogen

24

activation is an irreversible proteolytic-activating process. The strategy involved with this involves the prevention of indiscriminate proteolysis, and delivery only when necessary. The zymogens are stored in granules, relatively inert, and upon hormonal activations, the granules are dumped into the small intestines, spurring activation. The different digestive proteases have different substrate specificities, enabling the breakdown of a wide variety of peptides. A single enzyme known as trypsin activates all the zymogens. Trypsin is activated by enteropeptidase, which is secreted by cells lining 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 bond “after” 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 as its 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 are activated. This is not limited to the digestive system. Elastase activity is blocked by α1 antitrypsin, which is also known as antielastase since it is a better inhibitor of elastase than trypsin. Excess elastase activity destroys the alveolar walls in the lungs by digesting fibers and other connective tissue proteins. The Met 358 of the inhibitor is essential for binding to elastase. Smoking oxidizes this function amino acid residue and allows for no elastase binding, spurring emphysema.

Briefly describe the structural change that occurs upon the activation of chymotrypsinogen, including what changes occur in the active site. Chymotrypsin is not the initial product, but a product from an inactive enzyme (called chymotrypsinogen). With trypsin, chymotrypsinogen is then cleaved to yield π-chymotrypsin (an active form). When two dipeptides are removed, then α-chymotrypsin is yielded into an A, B, and C chain interlinked by an interchain disulfide bond. The effect of creating a break at isoleucine 16 is to yield an N-terminal at the B chain. The interaction at ASP 194 is ionic. The N-terminal of isoleucine 16 is turned inward, allowing binding.

Discuss the protective mechanism that keeps prematurely activated pancreatic digestive enzymes inside the acinar cells from autodigesting the pancreas, and describe/name an example.

The major protective mechanism that prevents auto-digestion of the acinar cells of the pancreas is the zymogens, the inactivated forms of the pancreatic enzymes. This allows it to be only activated when needed, particularly when food is present and required necessary breakdown. Examples are listed.

Explain how a cascade of catalysts (e.g., in PKA activation) results in amplification of a signal. The cascade of catalysts involves a series of activations and catalysis that cause amplification of the signal. One single activated protein kinase molecule can phosphorylate hundreds of target proteins in a very short time, producing a highly amplified effect. If target proteins themselves are enzymes activated by phosphorylation, each activated enzyme then can carry out many, many catalytic cycles on its substrate. The ultimate result is that a major multiplicative consequence between starting signal and final outcome several steps away. Essentially, it is a chain reaction in which phosphorylation of one reactant causes catalysis of a series of phosphorylations.

25

Define isozymes. Isozymes are enzymes that differ that have the same function, similar sequence, but different kinetics. They are expressed in different tissues to optimize for the particular metabolic needs. The distribution of different isozymes ofa given enzyme reflects at least four factors: (1) the different metabolic patterns in different organs, (2) different locations and metabolic roles for isozyme in the same cell, (3) different stage of development in embryonic and fetal tissues and in adult tissues, and (4) different responses of isozymes to allosteric modulators.

Explain the purpose of isozymes in metabolism. Isozymes are important in metabolism, particularly in lactate dehydrogenase. Mammals have two version of this enzyme, the H isozyme (found in Heart) and the M isozyme (found in skeletal muscle) which are extremely homologous (approximately 75% identity). The H4 tetramer (all H subunits) functions optimally under aerobic conditions, and has higher substrate affinity and can also be allosterically inhibited by pyruvate. The M4 tetramer (all four subunits have the M isozyme) functions optimally under anaerobic conditions, and has lower affinity for substrate and is not allosterically regulated. The mixed tetramers exhibit intermediate properties between the H4 and M4 tetramers, yielding a differential expression of the two isozymes that allow fine control over the LDH activity. At different developmental stages of the heart, the two isozymes of LDH are expressed to different degrees. In the early form the M form predominantes, while in the adult the H form predominates.

Lecture XVII

Carbohydrates and Glycobiology

Describe the three classes of carbohydrates with respect to structure and function. Carbohydrates are another class of biomolecules that are involved in nutritional, structural, and informational aspects of biological processes. They are the most abundant molecules on earth. They are typically held to the empirical formula (CH2O)n where n is a value greater than three (n ≥ 3). The name itself refers to the 2:1 ratio of H and O (such as water), and consequently received the name carbohydrates. Carbohydrates function according to nutritional, structural, and informational demands. From there, there are three types of structural carbohydrates: (1) simple sugars, (2) complex carbohydrates (polysaccharide), and (3) glycoconjugates. Simple sugars consist of monosaccharides and disaccharides that function in energy conversion. They primarily are involved in energy conversion and storage, and also serve as building blocks for polysaccharides and glycoconjugates. There are many types of monosaccharides with chiral centers. The differences are held in the number of carbons and the chiral centers, particularly at the C-1 carbon and the C-5 carbon. Aldehyde containing monosaccharides are called aldoses, with the aldehyde at the C-1 carbon. However, there are ketone-containing monosaccharides known as ketoses, with the carbonyl group along the C-2 carbon. Sugars exhibit chiral centers which result in different enantiomers. Glyceraldehyde is the smallest monosaccharide. Most carbohydrates in nature are in the D-configuration, even if L-isomers also exist, usually conjugated to proteins or lipids. A monosaccharide will have an epimer. If the carbon backbone is large enough, the monosaccharide of five, six, or seven carbons are more stable in aqueous environments as cyclic structures than as open chains. They can be oxidized by a solution of copper, known as Fehling’s solution. By cyclization, the carbonyl C becomes a new chiral center known as the anomeric carbon. From this, there are two isomers, α (down) and β (up). The α is cis to the reference carbon (C-5) and β is trans to the reference carbon (C-5). Anomers can have different properties. Complex carbohydrates serve structural roles or function as storage forms of glucose. A common form is the disaccharide. Disaccharides are formed by a condensation reaction between two monosaccharides. They have a reducing and non-reducing end. The non-reducing end (per its name) is not does not react with the copper solution in a reduction reaction, while the reducing end does. There are many common disaccharides, and these sugars can be modified. Glycoconjugates consist of carbohydrate units linked to proteins or lipids.

Recognize and identify structural features of simple sugars: aldose, ketose, linear, ring, chiral centers, epimers, reducing end, anomers.

Structural Feature

Description Illustration

Carbohydrate,Func7ons,

Nutri7onal,

• Energy,storage,–,glycogen,,starch,

• Fuels,–,glucose,• Metabolic,intermediates,,

Structural,

• Cell,walls,–,bacteria,,plants,

• Exoskeleton,–,arthropods,

• Connec7ve,7ssues,,car7lage,,bone,

• Components,of,nucleo7des,

Informa7onal,

• Molecular,recogni7on,

• CellHcell,communica7on,

26

Aldose Monosaccharide with an aldehyde. The carbonyl is at the end of the carbon chain.

Ketose Monosaccharide with a ketone. The carbonyl is at

the second carbon.

Linear Description of the carbon backbone with not

cyclization.

Ring Cyclic form of the carbohydrate with the carbonyl

attacks the C-5 carbon.

Chiral Centers Optically active centers along the carbon backbone,

which can alter function by altering the functional groups on the centers.

Epimers Two sugars that different in configuration around a

single carbon.

Reducing End End that is reduced in a reduction reaction, and

consequently reactive with copper. This is opposite to a non-reducing end. Ketose sugars must be first converted to an aldose to have a reducing end, and

sugars must conver to open chain from a stable yclic form to be reducing.

Anomers Special type of epimer. They are stereoisomers of a cyclic saccharide that differs only in its configuration

at hemiacetal or hemiketal carbon, known as the anomeric carbon.

Identify linkages between dissacharides (e.g. 1-4, 1-4, etc.) Remember that disaccharides are formed by a condensation reaction between two monosaccharides. They will bind in an α-1,4 or a β-1,4 fashion. Number 1 (designated C-1) refers to the hemiacetal, or the reducing end of the monosaccharide. From this there are large diversity possible, by different monosaccharide units, the conformation of the anomeric carbon, and the linkage through different –OH groups. If the α-

hydroxyl (which is down) on the anomeric carbon binds with the C-4 carbon, it becomes a α-1, 4 linkage. If the β-hydroxyl (which is up) on the anomeric carbon binds with the C-4 carbon, it becomes a β-1, 4 linkages. This increases the diversity of the carbohydrates and its associated polymers. Such linkages will allow combining residues to provide polysaccharides. They can yield homopolysaccharides (same saccharides) and heteropolysaccharides (either having two monomer types, unbranched and multiple monomer types). It can either be branched or unbranched. Branching is important to allow an increase in nonreducing ends and glucose availability. Carbohydrates are the

Simple,sugars,

•  AldehydeHcontaining,monosaccharides,are,called,aldoses,,

,•  KetoneHcontaining,monosaccharides,

are,called,ketoses,,,•  The,carbonyl.atom.(C=O),is,either,

at:,–  the,end,of,the,carbon,chain,(aldose),–  the,second,carbon,(ketose),,

1, 1,

2, 2,

Simple,sugars,

•  AldehydeHcontaining,monosaccharides,are,called,aldoses,,

,•  KetoneHcontaining,monosaccharides,

are,called,ketoses,,,•  The,carbonyl.atom.(C=O),is,either,

at:,–  the,end,of,the,carbon,chain,(aldose),–  the,second,carbon,(ketose),,

1, 1,

2, 2,

Cycliza7on,of,Monosaccharides,

Monosaccharides,of,five,,six,or,seven,carbons,are,generally,more,stable,in,aqueous,solu7on,as,cyclic,structures,than,as,open,chains,,

can,be,oxidized,by,Cu2+,(reduce,Cu2+),,

no,oxida7on,

Cycliza7on,of,Monosaccharides,

Monosaccharides,of,five,,six,or,seven,carbons,are,generally,more,stable,in,aqueous,solu7on,as,cyclic,structures,than,as,open,chains,,

can,be,oxidized,by,Cu2+,(reduce,Cu2+),,

no,oxida7on,

Sugars,have,chiral,centers,which,result,in,different,enan7omers,,

•  Glyceraldehyde,is,the,smallest,monosaccharide,,

enan7omers,

,Most,carbohydrates,in,nature,are,in,the,DHconfigura7on,,though,LHisomers,also,exist,,usually,conjugated,to,proteins,or,lipids,,

Epimers,•  Two,sugars,that,differ,in,configura7on,around,a,single,carbon,

,Disaccharides,are,formed,by,a,condensa7on,reac7on,between,two,monosaccharides,,

•  Large,diversity,possible:,–  Different,monosaccharide,units,–  Anomeric,carbon,can,be,in,α,or,β,conforma7on,–  Linkage,through,different,–OH,groups,

(α1,,,,,,,,4),glycosidic,bond,,

does,not,react,with,copper,in,a,reduc7on,reac7on,

reac7ve,with,copper,

Monosaccharide,1,/,glycosidic,linkage,/,Monosaccharide,2,,Glc , , ,,,,,,,,,,(α,1,,,,,,,4),,,,,,,,,,,,,,,,,,,,,,,,,,,,,Glc,

Anomers,

•  By,cycliza7on,,carbonyl,C,becomes,a,new,chiral,center,–,

anomeric.carbon.•  2,different,isomers,can,be,formed,–,α,and,β.anomers.•  Anomers,can,have,different,proper7es,•  hbp://www.biochem.arizona.edu/classes/bioc462/462a/jmol/monosacc/sugar11.htm,

64%.36%.

αH”down”=,cis,to,reference,carbon,(CH5),,,,,,,,,,,βH”up”=,trans,to,reference,carbon,(CH5),,

Common,Disaccharides,

Polysaccharides,(Glycans),

Branching.

,formed,from,~3H20,sugar,residues,,

Informa7onal,Molecules,

27

most abundant on Earth, and the most abundant for is cellulose, a polysaccharide. Naming involves the nonreducing monosaccharide name – numbers in linkage – reducing monosaccharide name.

Describe starch and glycogen structure and function. Starch and glycogen are storage forms of glucose, functioning as short term energy reserves and hydrolyzed by α-amylase that cleaves α-1 4 glycosidic bonds.. Starch is a combination of amylose and amylopectin and found in plants. It has a pyramid-esque structure, with progressive branching pattern widening as it gets away from the protein. It has about 35 non-reducing ends per 1,000 glucose units. Glycogen is another type of glucose homopolymer, which serves as a short-term energy reserve for animals. It has a circular branching pattern of glucose molecules surrounding the covalently bonded protein glycogenin. It has about 100 non-reducing ends per 1,000 glucose units. More glucose can be retrieved form glycogen due to the increased branching and higher number of reducing ends. There are also other types of polysaccharides. Many polysaccharides contribute to structural integrity of the organism. For example, glycosaminoglycans are found in connective tissue and are involved in many functions.

Three types of glycoconjugates: glycoprotein, proteoglycan, peptidoglycan. Describe structures, branching, attachment to proteins or peptides (N-linked and O-linked). Carbohydrates can be attached to other biomolecules, yielding compounds known as glycoconjugates. 50% of all human proteins are glycoproteins, and approximately 1% of the human genome encodes enzymes required for the synthesis and degradation of carbohydrates and glycoconjugates. The building blocks of glycan groups on most glycoconjugates are a combination of at least eleven different monosaccharides. They can be attached to other biomolecules as branched and unbranched structures. The general structure and function of the glycan groups is not known, because there is no sugar code identified. This is unfortunately due to many factors. For one, the monosaccharide addition to proteins and lipids by glycosyltransferase enzymes is not a template-directed process. Glycan structures have been found to be similar, but not identical, between molecules of the same protein, owing to stochastic events affected by the cellular environment. It plays strongly in cell signaling and immunity, with one cell presenting information through a glycoconjugate to another, either by intrinsic regulation (where the two similar cells recognize each other) or by extrinsic regulation (with the bacteria cells presenting the glycoconjugate to immune cell in the body, inducing the immune response to attack). It is finally technically challenging to decipher complex glycan structures because of limitations in glycan analytical methods and instrument sensitivity. There are three types of glycoconjugates: (1) glycoprotein, (2) proteoglycan, and (3) peptidoglycan. These types can be described in the following table:

Glyco-conju-gate

Description Function Structure Branching Attachment to Proteins or Peptides (N-linked

and O-linked) Glyco-

proteins Protein with

sugar attached (with protein

as major component)

Carbohydrate part can influence protein folding, protein stability, facilitate

protein targeting (localization) within a cell, and mediate recognition by

other proteins

N-linked or O-linked depending on the amino acid. N-linked utilized

ASN while O-linked utilizes SER or THR.

Glycosylation is specific and requires activity of

enzymes called glycosyltransferases.

Proteo-glycans

Sugar with protein

attached (sugar is

major component)

Major component of all extracellular matrices, and mediate ligand interaction with specific receptors of

the cell surface. They also bind to extracellular

proteins/signaling molecules to alter their

activities.

Branching is from compounds binding to the

core protein that is anchored to the membrane.

One or more sulfated glycosaminoglycan is joined covalently to a

membrane protein or a secreted protein. They bind to glycoproteins through O or N-linked

glycosidic bonds.

Peptido-glycans

Peptidoglycan sugar with

peptide attached (sugar is

major component)

Polymer consisting of sugars and amino acids that forms a mesh-like

layer outside the bacteria, forming the cell wall.

They utilize O-glycosidic bonds.

Glycoproteins,•  Sugar,link,to,protein:,OHlinked,or,NHlinked,

SER.THR.

ASN.

Protein glycosylation is highly specific and requires the activity of enzymes called glycosyltransferases,

Glycoproteins,•  Sugar,link,to,protein:,OHlinked,or,NHlinked,

SER.THR.

ASN.

Protein glycosylation is highly specific and requires the activity of enzymes called glycosyltransferases,

Proteoglycans,

•  One,or,more,sulfated,glycosaminoglycan,are,joined,covalently,to,a,membrane,protein,or,a,secreted,protein,

•  Major,component,of,all,extra,cellular,matrices,

•  Mediate,ligand,interac7on,with,specific,receptors,of,the,cell,surface,

•  Bind,to,extracellular,proteins/signaling,molecules,to,alter,their,ac7vi7es,

Proteoglycans,

•  One,or,more,sulfated,glycosaminoglycan,are,joined,covalently,to,a,membrane,protein,or,a,secreted,protein,

•  Major,component,of,all,extra,cellular,matrices,

•  Mediate,ligand,interac7on,with,specific,receptors,of,the,cell,surface,

•  Bind,to,extracellular,proteins/signaling,molecules,to,alter,their,ac7vi7es,

Pep7doglycans,form,bacterial,cell,walls,,

hexosamine,polysaccharide:,repea7ng,,disaccharide,MurNAcH(βH1,4)HGlcNAc,,units,that,can,be,cleaved,by,the,enzyme,lysozyme,,

hexosamine,polysaccharide,chains,are,,covalently,linked,through,oligopep7des,,(which,contain,both,D,and,L,amino,acids),,

Pep7doglycans,form,bacterial,cell,walls,,

hexosamine,polysaccharide:,repea7ng,,disaccharide,MurNAcH(βH1,4)HGlcNAc,,units,that,can,be,cleaved,by,the,enzyme,lysozyme,,

hexosamine,polysaccharide,chains,are,,covalently,linked,through,oligopep7des,,(which,contain,both,D,and,L,amino,acids),,

28

Linear sugar structure, but branching is from the

linkage through oligopeptides.

More details on N-Linked and O-Linked: N-linked utilizes the amide of ASN to bind to the saccharide. It has a preferred sequence of Asn-X-Ser/Thr. O-linked oligosaccharides utilize the hydroxyl of Ser or Thr to bind to saccharide, and do not have a preferred sequence.

Glycoproteins are information rich molecules. Many proteins secreted by eukaryotic cells are glycoproteins, in the form of antibodies, milk proteins, certain hormones, interferons, and many cell surface receptors. One major form of glycoproteins that perform this function is called lectins. They are proteins that bind to carbohydrates with high affinity and specificity. They often appear on the surface of cells, and mediate cell-cell recognition, signaling, and adhesion. They bind to carbohydrate with high specificity and moderate to high affinity. They target newly synthesized proteins (intracellular) and are important in the development of human disease. It is also found on the surface of viruses and involved in the attachment to the host cell.

Describe glycosyltransferases and how genetic differences in this enzyme result in immunological incompatibility.

Glycosylation involves making a sugar link to a protein, either O-linked or N-linked. The protein glycosylation is a highly specific process and requires the activity of enzymes called glycosyltransferases. Glycosyltransferases remove UDP (uracil diphosphate) from the glycan group and connects to the protein residue. After the first sugar is connected to the protein, glycan groups can be added to create a branch or linear structure. Genetic differences in the expression and activity of glycosyltransferases can account for immunological incompatibility between individuals. Determination of the blood type has to do with what glycosyltransferase enzymes are present. Immunological incompatibility can lead to donor-receptor complications in transplants and blood transfusion due to

tissue rejection. The human ABO blood groups are determined by genetic variants of the enzyme α1-3-N-acetylgalactosaminyltransferase. It attaches either a GaINAc or GAI sugar residue to the H antigen (glycan) present on glycoproteins and glycolipids on red blood cells. If the glycosyltransferase enzyme is absent or nonfunctional, then the H antigen glycan group is not modified, which presents Type O blood. Type A blood has anti-B antibodies and Type B has anti-A antibodies. Glycosyltransferase enzyme A and B can also co-exist, yielding Type AB blood.

Describe the basic features of peptidoglycans in bacterial cell walls. What enzyme does penicillin target? What are two mechanisms of antibiotic resistance to β-lactam antibiotics? Peptidoglycans are the main constituent of bacterial cell walls. The cell walls consist of hexosamine polysaccharide, which are repeating disaccharide MurNAC-(β-1,4)-GlcNAc units that can be cleaved by the enzyme lysozyme. Hexosamine polysaccharide chains are covalently linked through oligopeptides (which contain both D and L amino acids). Formation of the bacteria cell wall involves a terminal glycine residue of the pentaglycine bridge and a terminal D-Ala-D-Ala unit. The pentaglycine chain attacks the peptide bond between the two D-alanine residues, yielding a Gly-D-Ala Crosslink and a D-Ala. Another reaction that can yield this is also a transpeptidation reaction, in which there is a nucleophilic attack of the hydroxyl group of serine on the

substrate, making a covalent acyl-enzyme intermediate. From this information, we can understand the mechanism of certain antibiotics, such as penicillin. Penicillin is a β-lactam antibiotic that targets peptidoglycan synthesis through targeting transpeptidase (acting as an inhibitor). Peptides are linked by transpeptidase. We can note from the structure that penicillin has a similar conformation to R-D-Ala-D-ala peptide. The normal catalytic mechanism makes covalent intermediate with penicillin, but enzyme-penicillin derivative cannot continue. This inhibitor is covalently attached on the enzyme. Thus, one can infer that penicillin is both a transition-state analog and a suicide substrate.

,Gene7c,differences,in,the,expression,and,ac7vity,of,glycosyltransferases,account,for,immunological,incompa7bility,between,individuals,,

,•  The,human,ABO,blood,groups,are,

determined,by,gene7c,variants,of,the,enzyme,α1H3HNHacetylgalactosaminyltransferase,

•  abaches,either,a,GalNAc,or,Gal,sugar,residue,to,the,H,an7gen,(glycan),present,on,glycoproteins,and,glycolipids,on,red,blood,cells,,

•  If,the,glycosyltransferase,enzyme,is,absent,or,nonfunc7onal,,then,the,H,an7gen,glycan,group,is,not,modified:,O,type,blood,,,

Forma7on,of,the,bacteria,cell,wall,

Pentaglycine,chain,abacks,the,pep7de,bond,between,two,DHalanine,residues,

Transpep7da7on,reac7on:,

CrossHlink,formed,

nucleophilic,aback,of,enzyme,Ser–OH,on,substrate,,making,a,covalent,acylHenzyme,intermediate,

29

Ultimately, one can say that penicillin (the first antibiotic discovered) interferes with the synthesis of the bacterial wall.

Unfortunately, with the increase in distribution and consumption of antibiotics, there came a rise of antibiotic resistance. There are two mechanisms of antibiotic resistance to β-lactam antibiotics: (1) expression of enzymes that inactivate the antibiotic (such as β-lactamase, and (2) expression of variant transpeptidase enzymes that do not bind the antibiotic. Inactivation by enzyme expression involves the bacteria secreting an enzyme (β-lactamase) in conjunction with transpeptidase in order to inactivate the antibiotic. The enzyme β-lactamase will break open the ring of the penicillin derivative,

spurring inactivation. Ultimately, the transpeptidase is intact and the bacterium in question survives. Expression of variant transpeptidase enzymes that do not bind the antibiotic involves production of a different conformation of the protein that is performs the same function, but the antibiotic cannot bind and consequently inhibit. Thus, the bacterial cells will live. This is found in methicillin resistant bacteria, and MRSA is a growing health problem, increasing from 2-63% of total Staph infections

between 1974 and 2004. One can simplify this with a parable of the exes. Suppose an ex-boyfriend wants to keep a stalking ex-girlfriend away from getting into the house (or vice versa, it doesn’t matter). The ex-boyfriend can keep her out by either (1) adding another key lock to the door or (2) changing the lock on the current door lock. This is the same for bacteria and various forms of penicillin. The bacteria’s ultimate goal is survival, and it will utilize strategies to allow itself and its progeny to survive.

Describe two general biochemical methods in glycobiology.

There are two major biochemical methods in glycobiology: (1) mass spectrometry and (2) lectin array. Mass spectrometry is an analytical technique that measures the mass-to-charge ratio of the charged particles. From this data, the composition and structure of the molecules can be inferred. For example, we can allow glycan identification by mass spectrometry by first purification by liquid chromatography and the mass spectrometry. The glycans will allow inferring on what are the glycoconjugates of the membrane. Liquid chromatography will allow proper sorting based on size. Larger molecules are released later in the columns. The samples are treated with different enzymes associated with cleavage of monosaccharides. It ultimately allows the generation of a map of large molecules that were not cleaved and shorter molecules that were cleaved. Lectin arrays can allow us to analyze high-throughput screenings of glycans, using either naturally occurring lectins or artificialmonoclonal antibodies. They are immobilized on a certain chip and incubated with a fluorescent glycoprotein sample. Such assays can identify glycan-binding interactions associated with cellular phenotypes. It gives information on which types of glycans based on interactions that are formed. Antibody arrays have the same procedure but utilize antibodies specific to proteins, sugars, or bot. They can screen large number of molecules at one time.

Give an example of how glycobiology can be used to help in understanding disease. They can be used to detect glycome differences between biopsies isolated from normal and cancerous liver tissues. One can isolate glycoproteins from a tumor biopsy, attach the fluorescent label, and analyze with the lectin array. At the same time, one can also isolate glycoproteins from the normal tissue, attach fluorescent label, and analyze in the same manner. One can make a comparison of the glycan structure between the two samples. One glycoprotein has the same glycan structure in both samples, while the other two glycoproteins are

different. Such comparisons will allow us to further understand changes in normal and diseased tissue.

Describe how glycan mimetics can be used as drugs. Glycan mimetics can be utilized to a pharmaceutical advantage to disrupt glycan-binding interactions between pathogenic bacteria and host cells. The strategy is dependent knowledge of the mechanism. If

Glycan,mime7cs,can,be,used,as,drugs,to,disrupt,glycan,binding,interac7ons,between,pathogenic,

bacteria,and,host,cells,

If,glycan,groups,on,the,bacterial,cell,surface,mediate,abachment,to,a,host,cell,lec7n,protein,,then,a,lec7n,,array,is,used,to,iden7fy,compounds,(Drug,2),that,disrupt,this,interac7on.,

bacterium,

lec7n,

drug,1, drug,2,

If,the,mechanism,of,bacterial,binding,to,host,cells,is,mediated,by,a,bacterial,lec7n,protein,,than,a,glycan,,array,is,used,to,iden7fy,compounds,(Drug,B),that,disrupt,this,interac7on.,

bacterium,

lec7n,

drug,B,drug,A,

Biochemical,methods,in,glycobiology,Glycan'identification by'mass's pectrometry

Glycan'identification by'liquid'ch romatography Glycoconjug ate'anal ysis'using'a ntibody'array

Glycoconjug ate'anal ysis'using'le ctin'array

Releaseglycans

Labelglyco:

conjugates

Cellularglycoconjugates

high,throughput,arrayHbased,screening,assays,,to,iden7fy,glycan,binding,interac7ons,associated,,with,cellular,phenotypes.,,

iden7fica7on,of,glycan,group,structures,on,purified,glycoproteins,using,liquid,chromatography,and,mass,spectrometry,

30

glycan groups on the bacterial cell surface mediate attachement to a host cell lectin protein, then a lectin array is used to identify compounds that disrupt this interaction, which is shown on the left illustration. On the other hand (as shown as the right illustration), if a bacterial lectin protein mediates the mechanism of bacterial binding to host cells, than a glycan array is used to identify compounds that disrupt this interaction. In summary, the glycan mimetics can disrupt glycan binding between bacteria and host cells by aiming at two targets: either with the drug binding to the host molecule’s lectin where bacteria would bind or the drug binding to bacteria’s lectin where it would bind to the host lectin.