Download 2. Complexation and Protein Binding

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