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Complexation & Protein Binding Complexation • A complex is a species formed by the reversible or irreversible association of two or more interacting molecules or ions. • In the context of this course, it will be used to characterize the covalent or noncovalent interactions between two or more compounds that are capable of independent existence. • Complexes have been usually referred to as coordination compounds. (inorganic-inorganic) Co+3 + 6NH3=Co(NH3)6+3 (organic-inorganic) Ca-tetracycline (organic-organic) Inclusion Type (organic-organic) Inclusion Type (organic-inorganic) Complexation • • • • • Complexes, according to the classic definition, result from a donor-acceptor mechanism or Lewis acid-base reaction between two or more different chemical constituents. A Lewis acid is a molecule or ion that accepts an electron pair to form a covalent bond. The acceptor, or constituent that accepts a share in the pair of electrons, is frequently a metallic ion, although it can be a neutral ion. A Lewis base is a molecule or ion that donate a pair of unshared electrons by which the base coordinates with the acid. Any nonmetallic atom or ion, whether free or contained in a neutral molecule or in an ionic compound, that can donate an electron pair may serve as the donor. Metal ions are electron acceptors so that are capable of binding such ligands. Examples include Co3+ , Ni2+, Ag+, Fe2+ and Cr3+. Examples of electron donors include :NH3, H2O:, CN:- and Cl:- can act as ligands. Donor Acceptor Complex: Different than staring materials!! • Once complexation occurs, the physical and chemical properties of the complexing species are altered (solubility, stability, partitioning, energy absorption, and emission and conductance) • Complex formation usually alters the physical and chemical properties of the drug. For examples: (1) chelates of tetracycline with calcium are less water soluble and are poorly absorbed Milk; antacid, iron-supplement..etc Complex: Different than staring materials!! • Once complexation occurs, the physical and chemical properties of the complexing species are altered (solubility, stability, partitioning, energy absorption, and emission and conductance) • Complex formation usually alters the physical and chemical properties of the drug. For examples: (2) theophylline complexed with ethylenediamine to form aminophylline , which is more water soluble (for parenteral and rectal administration) Theophylline Aminophylline (Water-soluble) Complex: Different than staring materials!! • Once complexation occurs, the physical and chemical properties of the complexing species are altered (solubility, stability, partitioning, energy absorption, and emission and conductance) • Complex formation usually alters the physical and chemical properties of the drug. For examples: (3) cyclodextrins are used to form complexes with many drugs to increase their water solubility. Hydrophobic drug Hydrophilic exterior Hydrophobic interior Classification of Complexes Complexes may be divided broadly into two classes depending on whether the acceptor component is a metallic ion or an organic molecule; these are classified according to one possible arrangement in the following table. A third class, the inclusion / occlusion compounds, involving the entrapment of one compound in the molecular framework of another is also included in the table. 1. Metal Ion Complexes: a. Inorganic Type b. Chelates 2. Organic molecular Complexes 3. Inclusion/occlusion Complexes a. channel lattice type b. layer type c. clathrates Classification of Complexes • Intermolecular forces involved in the formation of complexes are the • van der Waals forces of dispersion • dipolar and induced dipolar types • Hydrogen bonding provide a significant force in some molecular complexes • coordinate covalence is important in metal complexes • Charge transfer and hydrophobic interaction are also introduced during the discussion. Metal Complexes: Inorganic Complexes • Example: – – – – [Co(NH3)6]3+ .3ClThe ammonia molecules in the hexamineocobalt III chloride is called the ligand. The ligands are coordinated to the cobalt ion. The coordination number of the cobalt ion is the number of ammonia molecules coordinated to the ion (6). Each ligand donates a pair of electrons to form a coordinate covalent link between itself and the central ion. Co3+ + 6(:NH3) = [Co(NH3)6]3+ Metal Complexes: Chelates – Chelates are complexes that typically involve a ring-like structure formed by the interaction between a partial ring of atom and a metal. – In chelates, ligands are usually organic molecules, known as chelating agents, chelators, chelants or sequestering agents. – When a ligand provides one group for attachment to the central ion, then its called monodentate. – Molecules with two or three groups are called bidentate and tridentate respectively (multidentate or polydentate). – If a metal ion binds to two or more sites on a multidentate ligand, a cyclic complex is formed; this cyclic complex is known as a chelate. Oxaliplatin, Bi-dentate EDTA( hexadentate) +M Gadolinium chelate octadentate Metal Complexes: Chelates • Some of the bonds in a chelate may be ionic or of the primary covalent type, while others are coordinate covalent links. • The formation of chelate complexes is controlled by stringent steric requirements on both the metal ion and the ligand. Only cis-coordinated ligands – ligands adjacent on a molecule – will be readily replaced by reaction with a chelating agent. Metal Complexes: Natural Chelates!! • Many biologically important molecules (e.g. hemoglobin, insulin, cyanocobalamine, chlorophyll) are chelates. Porphyrin ring square planar • Other biological chelates include albumin, the most common plasma protein which acts as a carrier of various metal ions (Cu2+ and Ni2+) and small molecules in the blood. Metal Complexes: Vitamin B12, Cyanocobalamin Metal Complexes: EDTA • EDTA is a synthetic chelating agent used to sequester ions (iron and copper) that catalyzes oxidative degradation reactions in drug preparation. • EDTA is also widely used to sequester and remove calcium ions from hard water. • treating mercury and lead poisoning; remove excess iron from the body. EDTA( hexadentate) +M EDTA: Ethylenediaminetetraacetic acid 2.3. Metal Complexes: • The chelating properties of procainamide (Sodium channel blocker, Class IA antiarrhythmic) has been used as an assay for its content in pharmaceutical preparations. Complex formation with Cu2+ results in a colored compound that can be measured by visible spectrophotometry. Thus calorimetric methods to assay procainamide in injectable solutions is based on the formation of a 1:1 complex of procainamide with cupric ion at pH 4 to 4.5. 2.3. Metal Complexes: b. Chelates: • Tetracycline antibiotics are capable of acting as chelating agents and binding a variety of polyvalent metal ions (Fe2+, Mg2+, Al3+, Bi3+ ). • The complexation results in changes in both the drugs’ and the metal ions’ physical and chemical properties. • The complexation between tetracycline antibiotics and metal ions either in food (cabbage) or in pharmaceutical preparations (iron containing supplements) has been found to reduce both the solubility and bioavailability of the antibiotics. 2.3. Metal Complexes: b. Chelates: • Tetracyclines are contraindicated in pediatric patients since they are prone to tetracycline complexation of calcium in teeth and bones resulting in teeth discoloration and bone growth problems. http://www.medicinescomplete.com/mc/rem/2012/images/c33-fig-33-4.png 2.4. Organic Molecular Complexes: • An organic molecular complex is made of constituents held together by weak forces of: 1) Hydrogen bonds energy of attraction is less than 5kcal/mol. the distance between these components is usually greater than 5 A. Weaker the coordination or covalent or chelates.. 2.4. Organic Molecular Complexes: • An organic molecular complex is made of constituents held together by weak forces of: 1) van der Wall’s forces Induced Dipole Dipole For example, the polar nitro groups of trinitrobenzene induce a dipole in the readily polarizable benzene molecule and the electrostatic interaction that results leads to complex formation. 2.4. Organic Molecular Complexes: • An organic molecular complex is made of constituents held together by weak forces of: 1) van der Wall’s forces Dipole Dipole Benzocaine Caffeine 2.4. Organic Molecular Complexes: • The incompatibility of certain polymers used in suspensions, emulsions, ointments and suppositories and certain drugs may be due to the formation of organic molecular complexes. The incompatibility may be manifested as a precipitate, flocculate, delayed biological absorption, loss of preservative action, or other undesirable physical, chemical, and pharmacologic effect. 2.5. Inclusion Compounds: • This class of complexes differ from the previously discussed classes in that they are mainly the result of the architecture of the molecules rather than their chemical affinity. • In this class of complexes, one of the constituents of the complex is trapped in the open lattice or cage like structure of the other to yield a stable arrangement. • Some times they are referred to as occlusion compounds. A) Channel Lattice Type starch and iodine B) Layer Type Bentonite Inorganic silicates 2.5. Inclusion Compounds: • This class of complexes differ from the previously discussed classes in that they are mainly the result of the architecture of the molecules rather than their chemical affinity. • In this class of complexes, one of the constituents of the complex is trapped in the open lattice or cage like structure of the other to yield a stable arrangement. • Some times they are referred to as occlusion compounds. C) Clathrates Co-Crystallization D) Monomolecular inclusion Compounds Hydrophobic drug Hydrophilic exterior Hydrophobic interior 2.5.4. Monomolecular inclusion Compounds: • In this class of inclusion compounds, a single guest molecule is entrapped in the cavity of one host molecule. • A representative example of such compounds is cyclodextrins. • Cyclodextrins are cyclic oligosaccharides containing a minimum of six D (+) glucopyranose units attached by an -1,4 linkage. • Cyclodextrins are produced from starch by the action of bacterial amylase. • The naturally occurring -CD, -CD and -CD contain 6, 7 and 8 units of glucose respectively. 2.5.4. Monomolecular inclusion Compounds: • The interior of the CD cavity is usually hydrophobic because of the CH2 groups, while the exterior of the cavity is hydrophilic because of the presence of the hydroxyl groups. • Complexation with CD does not ordinarily involve the formation of covalent bonds. Molecules of appropriate size and stereochemistry can be included in the cyclodextrin cavity by hydrophobic interaction. • The aqueous solubility of many lipophilic drugs is improved by complexation with CD and CD derivatives. 2.5.4. Monomolecular inclusion Compounds: • The bioavailability of many of these drugs has been improved as well. • CD has been used to improve the organoleptic characteristics (bitter taste) of oral liquid formulations. • Hydrophobic CD derivatives has been used as sustained release drug carriers. Ethylated -CD has been used to reduce the release rate of the water soluble drug diltiazem. 2.6. Methods of Analysis: • Different applications of complexes in pharmaceutical sciences require a quantitative knowledge of the complexation process and product. – The Stoichiometric Ratio of ligand to metal or donor to acceptor. – The Stability Constant of the formed complex. 2.6.1. Distribution Method: • Distribution of solute between two phases can be used to calculate the stability constant of complexes. • This depends on the fact that the distribution coefficient applies only to the species common to both phases. • Example: The complexation of iodine (I2) with potassium iodide (I-)can be represented by the following equilibrium: I2 + I- = [I -3 ]w K= [I ]w *[I 2 ]w I3- Water [I-]w + [I2]w Ko/w [I2]o Oil CS2 K [I-3]w 2.6.1. Distribution Method: • Determine Ko/w • Add KI to the aqueous layer with known amount Total [I-] • Determine [TOTAL iodine(free and complexed)] in each phase by titration with thiosulphate solution using starch as indicator • Determine [TOTAL iodine in oil] =[I2]o • Determine [TOTAL iodine in water] =[I2]w + [I-3]w Water [I-]w + • • • • Ko/w Total [I-] [I2]o [I2]w + [I-3]w [I2]w Ko/w [I2]o Oil CS2 K [I-3]w 2.6.1. Distribution Method: - Ko/w - Total [I-] - [I2]o - Total I2 in water [I2]w+[I-3]w [I 2 ]o K o/w = [I 2 ]w [I -3 ]w K= [I ]w *[I 2 ]w 1) From Ko/w and [I2]o, calculate [I2]w 2) From ([I2]w + [I-3]w) and [I2]w, calculate [I-3]w 3) complexed [I-3]w = complexed [I-]w (I2 in excess) 4) From Total [I-] and complexed [I-]w, calculate Free [I-]w =Total [I-] - complexed [I-]w Water [I-]w + [I2]w Ko/w [I2]o Oil CS2 K [I-3]w Example 10-2: Martin’s 6th ed. - Ko/w=652 - Total [I-]=0.125 - [I2]o=0.1896 - Total I2 in water [I2]w+[I-3]w=0.02832 1) From Ko/w and [I2]o, calculate [I2]w K o/w = [I 2 ]o [I 2 ]w 625 = 0.1896 [I 2 ]w 3) complexed [I-3]w = complexed [I-]w (I2 in excess) 4) From Total [I-] and complexed [I-]w, calculate Free [I-]w =Total [I-] - complexed [I-]w Complexed[I - ]w = [I -3 ]w = 0.02802 [I 2 ]w, free = 0.000303M 2) From ([I2]w + calculate [I-3]w [I- 3]w) and [I2]w, [I - ]w = Total[I - ]w - Complexed[I - ]w [I - ]w = 0.125- 0.02802 = 0.09698M Finally: Plug into the equation!! [I 2 ]w,complexed = [I 2 ]w,total - [I 2 ]w, free [I -3 ]w = 0.02832 - 0.000303 3 w [I ] = 0.02802 [I -3 ]w K= [I ]w *[I 2 ]w 0.02802 0.09698* 0.000303 K = 954 K= 2.6.2. Solubility Method: Solutions of the complexing agent in various concentrations Excess solid (Drug) In stoppered containers Bottles are agitated in a constant temperature bath until equilibrium is attained Aliquot portions of the supernatant liquid are removed and analyzed Point B 2.6.2. Solubility Method: Saturation of the Drug-Ligand (solubility limit) Molar Concentration of the Drug Point C All excess solid Drug converted to Drug-Ligand complex Point A Solubility of the Drug Molar Concentration of the Ligand Solubility profile of a drug in the presence of a complexing agent 2.6.2. Solubility Method: • Point A in the previous graph represents the intrinsic solubility of the drug in water. • As we add the ligand, the drug complexes with it and more solid drug is withdrawn into solution to maintain the free drug concentration constant, resulting in increased total drug concentration. • Consequently the concentration increases to reach point B. • At point B the system is saturated with respect to both the drug and the Point B complex. Molar Concentration of the Drug Saturation of the Drug-Ligand (solubility limit) Point C All excess solid Drug converted to Drug-Ligand complex Point A Solubility of the Drug Molar Concentration of the Ligand 2.6.3. Solubility Method: • In the plateau BC, the complex continues to form and precipitate, however, the total concentration does not change because of the presence of excess solid. • At point C, all the excess solid has been exhausted and turned into the complex. • The decline in the total concentration is due to the formation of higher complexes with lower solubility. Point B Molar Concentration of the Drug Saturation of the Drug-Ligand (solubility limit) Point C All excess solid Drug converted to Drug-Ligand complex Point A Solubility of the Drug Molar Concentration of the Ligand Calculation of the ratio of the constituents (B-C plateau ) : 1. The concentration of the drug entering the complex during this plateau is the solid drug that dissolves during this period = total amount of solid added initially __ the amount dissolved at point B 2. calculate the concentration of the ligand entering the complex throughout the plateau (B-C). m[ Drug ] n[ Ligand ] [ Drug m Ligand n ] m n m Drug0 y2 y2 Ration Ration Drug n x2 x1 y1 B C X1 X2 A Drug0 y2 x2 x1 Ligand Calculation of the stability constant (A-B segment) : – The concentration of the complex = Total Conc of Drug at B-Conc of Drug at A – The concentration of the free drug is constant through AB= Conc at A. – The concentration of the free ligand= originally added ligand concentration to the system- the concentration of the complex. m[ Drug ] n[ Ligand ] [ Drug m Ligand n ] [ Drug m Ligand n ] ( assume : n m 1) [ Drug ]m * [ Ligand ]n [ Drug Ligand ] K [ Drug ] * [ Ligand ] [ Drug Ligand ] y L y1 B C yL2 K A Drug y1 [ Drug ] Cons tan t y1 [ Ligand ] xL [ Drug Ligand ] xLL Ligend Example 10-3 (Martin’s 6th ed.): PABA (drug) and caffeine (ligand) Given [ PABA]0 0.073M Initial PABA concentration added to the system in excess y1 0.0458M Solubility of PABA in water y2 0.055M Concentration of PABA (free+complexed) in water x1 0.017 M x2 0.035M y2 Q1: Ratio? y1 C X1 X2 A PABA m n m Drug0 y2 0.073 0.055 0.18M Ration Drug m[ Drug ] n[ Ligand ] [ Drug m Ligand n ] B n x2 x1 0.035 0.017 0.18M Drug0 y2 0.18 Ration 1 x2 x1 0.18 Ligand Caffeine Example 10-3 (Martin’s 6th ed.): PABA (drug) and caffeine (ligand) Given y1 0.0458M xL 0.01M Solubility of PABA in water Total Caffeine concentration added to the system up to (xL, yL) y L 0.0531M Concentration of PABA (free+complexed) in water Q2: K? B PABA [PABA]+[Caff ]- - - - > [PABA - Caff ] yyL2 [PABA - Caff ] y1 K= A [PABA]*[Caff ] [PABA - Caff ] = yL - y1 = 0.0531- 0.0458 = 0.0073M [PABA] = Cons tan t = y1 = 0.0458M [Caff ] = x L -[PABA - Caff ] = 0.01- 0.0073 = 0.0027M 0.0073 K= = 59 0.0458* 0.0027 xLL C Home work • The problems to be solved in relation to the solubility method from the physical pharmacy book – 4th edition are: 11-4, 11-5, 11-6 – all from page 279. If you have the fifth edition, solve the following problems in relation to solubility: 11-4, 11-5, 11-6 – all from page 715. • Example 11-3 page 265 in the fourth edition from the physical pharmacy book. If you have the fifth edition, see example 11-3 page 283. • If you have the sixth edition of the physical pharmacy book by Martin, see Example 10-3 page 212 and solve problem 10-3 in relation to the solubility method. Protein Binding: interaction of drugs with plasma proteins (albumin) Antibody-antigen recognition (immunity) enzyme-substrate interaction drug binding to receptor Protein Binding: The binding of drugs to proteins in the body can affect their actions by: – Affecting the drug distribution throughout the body. – Affecting the activity of the drug by reducing amount of free drug available to bind the receptor site. – Retard the excretion of the drug and increase its half life. Protein Binding: • Since proteins are molecules composed of different types of amino acids, the interactions between proteins and small molecules can occur through one or more than one of the followings: – Hydrogen bonding – Electrostatic interactions – van der waals interaction – Hydrophobic interactions Schematic representation of hydrophobic interaction Protein Binding Pf + Df PD • The two most important parameters of protein binding are: – Affinity of binding , expressed using the association constant and is a measure of the strength of interaction between the protein and drug molecule. [PD] K= [P] f [D] f • K: Association constant • [PD]: The concentration of formed protein-drug complex = concentration of bound drug • [P] f: The concentration of free protein(unbound) • [D]f: The concentration of free drug (unbound) – Ratio of bound drug to total proteins = [ D ]bound [ PD ] r [ P ]total [ P ]total Protein Binding: Equilibrium Dialysis Dialysis tubing or sac: selective diffusion of small molecules through semipermeable membranes equilibrium dialysis Protein Binding: Equilibrium Dialysis • • • The protein (e.g. serum albumin or other protein under investigation) at a specific concentration and drug in various concentrations are placed in a tied cellulose semipermeable tubing (dialysis bag or sac). The sac is placed in a beaker with proper media to simulate the physiological one. Drug only start to diffuse and ultimately it will reach equilibrium (concentration of free drug in sac=concentration of free drug in dialysate=[D]f ). If binding occurs, the drug concentration in the sac [D]total containing the protein is greater at equilibrium than the concentration of the drug in the vessel outside the sac [D]f . Samples are collected and analyzed to obtain the concentration of free and bound drug. Change this K= Analyze T0 [P]t [PD] [P] f [D] f Obtain [D]f in the beaker [D]f (in eq. with sac!!) [DP]= [D]b=[D]t-[D]f [P]t (stay constant) [P]f= [P]total -[DP] Teq Experimental Setup for equilibrium dialysis for analysis of protein-ligand interaction Protein Binding: Equilibrium Dialysis K [ PD ] ........consider : [ P ] f [ P ]t [ PD ] [ P] f [ D] f K [ PD ] ([ P ]t [ PD ])[ D ] f [ PD ] K [ D ] f ([ P ]t [ PD ]) [ PD ] K [ D ] f [ PD ] K [ D ] f [ P ]t K [ D] f [ PD ] r [ P ]t 1 K [ D ] f K [ D ] free [ D ]bound r [ P ]total 1 K [ D ] free This equation assumes the K[D f ] presence of one binding site, r=v in case of the presence of v 1+ K[D f ] independent binding sites, V: # of drugs bind to a single protein the equation becomes Maximum binding capacity Langmuir isotherm and the double reciprocal plot: Langmuir isotherm v r=v K[D] f 1+ K[D] f r 1 1+ K[D] f = r vK[D] f Double reciprocal plot y = bx + a 1/r [D]f (mole/L) 1 1 1 = + r vK[D] f v V: # of drugs bind to a single protein Maximum binding capacity slope = 1/vK y-intercept = 1/v 1/[D]f (L/mole) Example The number of binding sites and the association constant for the binding of sulfamethoxypyridazine to albumin at pH 8 can be obtained from the following data: [D] bound [P]total 0.23 0.46 0.66 0.78 [D] f x10-4 0.10 0.29 0.56 1.00 r= where [Db] is the concentration of drug bound, also referred to as [PD], and [Pt] is the total protein concentration. What values are obtained for the number of binding sites, v, and for the association constant, K? Example D 0.00001 0.000029 0.000056 0.0001 r 0.23 0.46 0.66 0.78 1/D 100000 34482.7586 17857.1429 10000 1/r 4.34782609 2.17391304 1.51515152 1.28205128 Slope Y-intercept when X=0.0 X-intercept when Y=0.0 R square 3.415e-005 ± 6.598e-007 0.9437 ± 0.03554 -27635 0.9993 5 4 y-Intercept=1/v v=1.06 (~ 1) 1/r 3 2 1 Slope=1/vk K=2.93X104 L/mole 0 0 50000 100000 1/[D]f (L/mole) 150000 Home work The following data were obtained in the in vitro binding study of naproxen, with human serum albumin at 37 C. a) Plot the data according to the double-reciprocal model of the proteindrug binding b) After linear egression analysis (linear fitting using excel or any other software), calculate the binding constant K and the number of binding sites (v) of naproxen to albumin Answers: v=2.0; K=8.33X108 L/mole