complexation & protein binding. learning topics complexation mechanisms methods for...

Complexation & Protein Binding

Learning topics

• Complexation mechanisms• Methods for determination of stoichimiometric ratio and

stability constant • Protein binding• Analysis of protein binding• Scatchard method

ComplexationDefinition: usually non-covalent interactions between two or

more compounds that are capable of independent existence.

Mechanisms: classic donor-acceptor (Lewis acid-base), but any intermolecular forces may be considered:

van der Waals< induced dipolar <dipolar<hydrogen binding

Physico-chemical properties that may be altered:• Chemical (pH), instrumental (UV/IR spectra, conductance) • Formulation (solubility, partitioning, stability, drug delivery by

ion-exchange resins e.g. Delsym, Tussionex, Kayexalate, Renagel, Cholestyramine)

• Pharmacokinetic – permeability (bioavailability), protein binding (distribution)

Cation exchanger resins in drug delivery

Types of complex formation

• Metal ion complexes (metal ion coordination with ligands): monodendate, multidendate (chelate)

• Organic complexes (due to electrostatic, charge transfer, hydrogen binding, van der Waals forces or other hydrophobic interaction e.g. п- п stacking): molecular complexes, donor-acceptor complexes, charge-transfer complexes

• Inclusion / Occlusion complexes (a certain type of molecular complexes with unique properties)

Metal ion Complexes

Monodentate (Inorganic): • Inner-sphere complex: [Cu(NH3)4]2+ .2Cl- , Cu2+ (metal ion,

unfilled orbitals), 4 (coordination number), Cl- (counter ion), NH3 (ligand, electron pairs), orbital configuration dsp2 (square planar)

• Outer-sphere complex: [Fe(CN)6]3-, orbital configuration d2sp3 (octahedral)

• Biologic Examples: iron complexes (hemoglobin, cytochrome c, myoglobin), platinum complexes (cisplatin, carboplatin), copper complexes (hemocyanin, superoxide dismutase, cytochrome oxidase), cobalt (vitamin B12), zinc complexes (insulin, carbonic anhydrase, carboxypeptidase)

Bidendate: EDA in aminophiline, carboxylate

Multidentate (Chelate): EDTA (hexadentate)

Outer sphere vs. inner sphere

Organic complexes

• Charge transfer (donor-acceptor) complexes (complete or partial charge transfer, develop new band in spectra e.g. trinitrobenzene and benzene п bonding)

• Molecular complex (No charge transfer, contribution of weak van der Waals, dipole-dipole, hydrogen binding)

Inclusion/Occlusion Complexes

the molecular size matters1. Channel lattice type: starch for iodine2. Layer type: Bentonite, graphite3. Clathrate (cage-like lattice): hydroquinone cage for

warfarin4. Molecular sieves: zeolites (aluminosilicate) for

organic solvent, H2O

Mono and Macromolecular Inclusion/Occlusion

ComplexesCyclodextrins: cyclic oligosaccarides 6, 7 or 8

glucopyranose (α, β, γ) naturally by Bacillus macerans amylase from starch sized from 5-8A; applied for:

drug solubilization, flavor stabilization, long-acting deodorants, changing from liquid to solid state, prevention of incompatibilities, improvement of organoleptic properties, enantiomer separation, and bioavailability enhancement.

QuestionDue to low aqueous solubility and nephrotoxicity of β-CD, what kinds of synthetic CDs were developed ?

Complexation parametersstability constant,

stoichimiometric ratio

CD

CDK

CDCnD

nn

a

n

Analysis of stability constantsphase-solubility profile

• By Higuchi and Connors:

)1

.(][

))(

(

)(

)(

)(

0

000

000

0

000

0

0

0

n

a

n

a

na

nn

a

n

n

n

nn

a

n

SK

SnKCSS

n

SSCS

n

SS

K

CD

CDK

n

SSCCDCC

n

SSCD

SD

CDnSS

CD

CDK

CDCnD

Example

• Please calculate intrinsic solubility of PABA and stability constant of the β-CD complexes.

• Assumption 1:1 complex

Concentrations (mM)

β-CD PABA0.50 1.741.00 2.231.50 2.722.00 3.212.50 3.703.00 4.194.00 5.175.00 6.16

13

0

0

CD-

0

000

30.41)98.01(*10*25.1

98.0

)1(

25.1

25.198.0

C vs.

)1

.(][

MslopeS

slopeK

mMS

xy

S

SK

SKCSS

a

PABA

a

a

6.195)1)106.4105.5(101(104.6

106.4105.5

eine]PABA][Caff[

caffeine]-PABA[

1105.5107.3

101102.8

complexin PABA

complexin caffeinen

)105.5 ,10(2.8 C

)105.5 ,10(1 B

)104.6 (0,A

2-2-2-2-

2-2-

2-2-

2-2-

2-2-

2-2-

-2

K

Analysis of complex formation Solubility

Excess PABA

Dissolved PABA

Caffeine

Method of continuous variation

and Job’s method

• A- maximum deviation from linear changes of an additive property (e.g. dielectric constant) at a specific mole ratio (n)

• B- plot of absorbance difference (as the additive property) against mole fraction of ligand is used to calculate n• Job’s method: if the magnitude of absorbance is proportional only to the concentration of the complex, then:

nlog[A]log[M]logK]log[MA

[M][A]

][MAK

MAnAM

n

nn

n

Analysis of stability constantsSpectroscopy

Due to charge transfer (e.g. additional UV absorption for iodine in electron donor solvents)

0

0

0

00

Donor0

Acceptor0

00

D

1.

ε.K

1

ε

1

A

A

D1

DAεA

complex), ofty absorptiviε(molar

.[complex]A

),(CD

),(CA

[complex])1(A

[complex]K

:Equation HildebrandBenesi

K

K

D

Analysis of stability constantspotentiometric titration

][logK][][1.5n if

] [logK][][0.5n if

)1 ( ][2log][

1][][MA 1n f

]][[

][

][][][

][2][

ion] metal[

ligand] bound[

][

][

2H)COOCHCu(NHCOOCH2NHCu

22

1

22

22

21

2

2

222

1

222232 _

ApMAMA

ApMMA

nifApA

Mi

AM

MAKK

MAMAM

MAMAn

AMA

MAKMAAMA

AM

MAKMAAM

-

ExamplePlease determine average number of ligand (n) and stability exponent if addition of 0.001 mole Cu2+ to 0.03 M glycine HCl (pKa=9.7) in 100mL sample causes shifts of 1.7 mL (pH=4) and 5 mL (pH=8) as titrated with 0.3 M NaOH.

82.1030.752.3log

30.7)100

1000*10*7.1*3.003.0log(47.9p[A]

52.3)100

1000*10*5*3.003.0log(87.9p[A]

NaOH])[]log([]HAlog[][

][

]][OH[

1.5)nat p[A]logK 0.5,nat p[A](logK

4)(pH 5.0001.0

3.0*10*7.1n

8)(pH 5.1001.0

3.0*10*5n

ion metal of moles

releasedproton of oles

ion] metal[

ligand] bound[

3

3

3

2121

3

3

initialHApHpKapHpKaAp

HA

AKa

KK

mn

Analysis of stability constants Distribution (Question)

• Question : I2+I-→I3-

H20 replaced by 0.125 M KI → Co (I2) = 0.1890, Cw (I2) = 0.0283.

K (stability constant) ?Kiodine (o/w) = 625

1972MK

0.096M0.0280.125free w,[KI]

0.028M0.00030.0283complexed w,]2[I

M4103.034625

0.189free w,]2[I

free[KI]free]2[I

[complex]K

I2

I2

H2O

CS2

Protein BindingDrugs almost bind to albumin (acidic) and α1-acid glycoprotein (basic)

To act as drug reservoir and sustain disposition / therapeutic or toxic effects

affinity) site binding(K, ].[DK1

].[DK.νr

sites) binding ofnumber ,(ν ].[DK1

].[DKν.

][P

[PD]r

occupancy) fractional (r, ].[DK1

].[DK

][P

[PD]r

:absorption isothermalLangmuir

].[D].K[P]).[DK[PD].(1

[PD])]].([P.[DK[PD]

[PD]][P][P

]].[D[P

[PD]K

PDDP

fi

fii

fa

fa

t

fa

fa

t

fatfa

tfa

tf

ffa

Protein bindingAffinity and capacity

Langmuir isothermPB capacity & affinity

Double reciprocal vs. Scatchard

].[Pν.K][DK][D

][D

:ionconcentrat and natureprotein unkown if

r.Kν.K][D

r : Scatchard

].[Dν.K].[Dr.Kr

ν

1

][D

1.

ν.K

1

r

1 reciprocal Double

].[DK1

].[DKν.r

tabaf

b

aaf

fafa

fa

fa

fa

Protein bindingcalculations

Fraction of drug bound, β

Example

• Please calculate affinity and capacity indices of protein binding of a new anti-HIV drug

0.00 0.50 1.00 1.50 2.00 2.500

0.05

0.1

0.15

f(x) = − 0.060022266 x + 0.159872829R² = 0.996707969872418

r

r /

[Df]

(L

/nm

ole

)

65.20.06

0.159

06.0K

r.Kν.K][D

r

a

aaf

Protein BindingMethod of Analysis

• Dialysis (cut-off 12-14 KDa): Equilibrium vs. Dynamic mode• Ultrafiltration (cut-off 10-30 KDa)• Ultracentrifugation (20000-50000 rpm, Ficoll or cesium

chloride gradient may be required)• Gel filtration chromatography (Sephadex columns)• Capillary Electrophoresis• Spectroscopy methods (qualitative)

Equilibrium Dialysis

Dynamic Dialysis

I

II

t

t

t

t

S

S100100%PB

).dt,100

%PB-k.(1

][D

]d[D:II

k.dt,][D

]d[D :I

constantty permeabiliapparent k

Homework

• Martin’s Physical Pharmacy:• Problem 11.7• Problem 11.13• Problem 11.15

Chemical kinetics and drug stability

Ali Tamaddon, Ph.DDepartment of Pharmaceutics

Shiraz Faculty of Pharmacy

Topics

• Reaction rate and order• Zero, first and second order reactions• Half-life and shelf-life calculations• Complex reactions (reversible, parallel, series) • Michaelis-Menten nonlinear kinetic• Parameters influencing reaction rate (temp, pH, ionic

strength, solvent, dielectric constant)• Collision theory and Arrhenius equation• Stability testing methods (shelf-life estimation)

Kineticimplications in pharmaceutics

• Degradation reactions• Storage conditions, shelf-life, expiration date• Stability testing• Pharmacokinetics• …

Reaction rate

• k: reaction rate constant in elementary reactions

ba BAkrate

dt

Bd

bdt

Ad

arate

productbBaA

][][

][.

1][.

1

Reaction order• Reaction order: zero, first, second, etc.• depending on molecularity (number of molecules, ions,

atoms reacting simultaneously) in elementary reactions,

• but not in complex reactions

Zero order reactions

0

00.90

0

01/20

00

00

00

0

0

0

1.0 then t9.0C

5.0 then t5.0C

.

)0.(

.

sec)./(

k

CCif

k

CCif

t

x

t

CCk

tkCC

tkCC

dtkdC

lmolesk

kdt

dCrate

First order reaction

110.90

111/20

0

10

303.2

.

0

.0

10

10

1

11

1

105.09.0ln then t9.0C

693.05.0ln then t5.0C

)(log

302.2log

302.2303.2

.loglog

10.

.

.lnln

)0.(lnln

.

)(sec

.

1

1

kkCif

kkCif

xa

a

tC

C

tk

tkCC

CC

eCC

tkCC

tkCC

dtkC

dC

k

Ckdt

dCrate

tk

tk

First order reaction

Second order reaction1st type

200.90

201/20

20

20

22

112

22

.9

1 then t9.0C

.

1 then t5.0C

.11

)0.()1

(1

.

)sec..(

.

kCCif

kCCif

tkCC

tkCC

dtkC

dC

molelk

Ckdt

dCrate

Example• Please determine rate-constant

and half-life of a second-order reaction if the initial concentration of 0.05 M changed by 0.0088 M over 35 min.

min9.163.

1t

min./122.0

35).0088.005.0(*05.0

0088.0

)..(

.).(

,

..

.11

201/2

2

2

2

00

20

0

20

kC

molelk

txaa

xk

tkxaa

x

xCCCa

tkCC

CC

tkCC

Second order reaction2nd type

tkABA

B

A

B

tkABBA

AB

tkBA

AB

AB

tkxB

B

xA

A

AB

dtkxBxA

dx

xBxAkdt

dx

CCkrate BA

.).][]([][

][ln

][

][ln

.).][]([]][[

]][[ln

.]][[

]][[ln.

)][]([

1

)0.()]][

][ln()

][

][.[ln(

)][]([

1

.)]).([]([

)]).([].([

..

2000

0

2000

0

20

0

00

20

0

0

0

00

200

002

2

Pseudo-order reactionsApparent rate constants

• Pseudo-zero order (example):• Aspirin hydrolysis in aqueous suspension (6.5%) follows

pseudo zero-order reaction at pH=6 and room temperature. What is the shelf-life? (Saspirin=0.33%), K1=4.5*10-6 sec-1

• Pseudo-first order:

daysk

C

lgCkkrate

Ckdt

dCrate

hydrolysisAA

app

solid

51.0

t

sec./105.1.

.

0

00.9

510

1

solution saturated

].[

then[A][B]

]][[

1

2

AKrate

if

BAkrate

productBA

app

Determination of reaction order

• Substitution method• Graphical method ((a-x) vs. t, ln(a-x) vs. t, 1/(a-x) vs. t) • Half-life method:

1)log(

loglog

loglog)1(log.

1

1

2

)2(2/1)1(2/1

2/1

)1(2/1

a

a

ttn

kantak

tn

Complex ReactionsReversible

• Example: epimerization of tetracycline, 32% iso-7-chlorotetracycline + 68% iso-7-chloro-4-epi-tetracycline in equilibrium, please calculate kf and kr if slope = 0.01 min-1

tkk

AA

AA

A

AA

A

B

kr

kfK

BkrAkfdt

Ad

BA

rf

eq

eq

eq

eq

eq

eq

303.2

)(log

].[].[][

0

0

1

1

min016.0

min007.0

01.0303.2

1.2

01.0303.2

1.232

68

kf

kr

krkr

krkfkr

kfK

Complex reactionparallel vs. series

• Parallel (e.g. general acid-base catalyzed reactions)

• Series or consecutive (a complex calculations!)

)1.(.

]).[(].[].[][

0

21

2121

2

1

ktii eA

K

kB

kkK

AkkAkAkdt

Ad

BA

BA

).()(

11.(

).()(

.

21

21

1

1221

02

12

101

0

21

tktk

tktk

tk

ekekkk

AB

eekk

kAB

eAA

BBA

Complex reactionMichaelis-Menten (Enzyme

catalyzed)

][

]][[][

][

])[]]([[

][][]])[[]([

][][][

]])[[]([][

1

32

321

32

1

31

2

Sk

SEES

ES

ESES

k

kkK

ESkESkSESEk

ESkESkdt

ESd

SESEkdt

ESd

PSESE

M

T

TM

T

T

kk

k

][

1.

1

V

1

:Burk-Lineweaver

)][

][.(

][

]][[.].[

max

33

SV

K

V

SK

SVV

Sk

SEkESkV

m

M

m

M

M

T

Line-weaver Burk equation

V (μg/l.min) 0.0350 0.0415 0.0450 0.0490 0.0505

C (M) 0.0025 0.0050 0.0100 0.0167 0.0333

Question: Please calculate KM

KineticFactors: A-Temperature

• Based on collision theory:

)11

.(2.303R

E-log

2.303RT

E-logAlogk

Equation) (.

.

..

on)Distributi (.

..

12

a

1

2

a

TTk

k

ArrheniuseAk

ePZk

NePZrate

BoltzmaneNN

NZPrate

RT

Ea

RT

Ea

tRT

Ea

RT

Ea

ti

i

Example

• Please determine the activation energy for the below data:

0.002 0.00220.00240.00260.0028 0.0030.544

0.545

0.546

0.547

0.548

0.549

0.55

0.551

0.552

f(x) = − 8.95655653247871 x + 0.571367059892067

1/T

Logk+

2

Absolute T, K k, year-1

450 0.0356

400 0.0354

361 0.0352

Accelerated Stability Testing

Arrhenius method• Shelf-life calculation based on logK in the elevated

temperature and extrapolation to normal condition

• Example: Please calculate shelf-life for the last slide corresponding example

Log (k+2)=0.541, then k=0.035t0.9=0.105/k=0.105/0.035=3 years

Accelerated Stability Testing

Amirjahed method

Q10 and shelf-life estimation

• Example: Reconstituted ampicillin suspension is stable for 14 days when stored in the refrigerator, please calculate shelf-life at room temperature

T

T

T

TTT

TTR

Ea

TR

Ea

TR

Ea

T

T

Q

TtTTt

Qk

kQ

e

eA

eAQ

k

kQ

)()(

.

.

190190

1010)(

)1

)10(

1(

1.

)10(

1.

10

)10(10

days 875.016

14)5()25(

164

9090

10

20

1010

T

T

T

Q

CtCt

QQ

KineticFactors: B-Medium Polarity

• Solvent polarity: polar (high solubility parameter) solvents generally accelerate reactions if the product is more polar than the reactants and vice versa.

PolarityDielectric constant

• Dielectric constant (ε): charge storation i.e. ε= C/C0

• Water (78.5)• Methanol (31.5)• Ethanol (24.3)• Acetone (19.1)• PEG400 (12.5)• Ether (4.3)• Ethyl acetate (3.0)

KineticFactors: C-Dielectric

Constant• Opposite sign ions vs. similar sign ions:• By increasing dielectric constant:

– K decreases for opposite signs– K increases for similar signs

• For reaction of ion and neutral molecules:– K increases with decreasing dielectric constant

KineticFactors: D-ionic strength

• Primary salt effect (non-catalytic salts):

• If zA and zB have the same sign, then zAzB is positive and the rate constant increases with ionic strength.

• 2. If zA and zB have different signs, zAzB is negative and the rate constant decreases with ionic strength.

• 3. If one of the reactants is uncharged, zAzB is zero and the rate constant is

independent of the ionic strength.

BA

ii

BA

BA

BA

BAZZ

ZZ

ZZZZ

ZZkk

Z

BABAkk

kk

BA

BAK

PBABA

BA

BA

BABA

02.1loglog

zc2

1μ

Eq.) Huckel(Debye 51.0log

)loglog(logloglog

.

.]][[

][

0

j

1

2ii

2

0

0

)(

)(

KineticFactors: E-pH/Buffer

• Specific acid-base catalysis:

][][0 OHkHkkk OHHobs

KineticFactors: E-pH/Buffer

• General acid-base catalysis: rate constant furthermore depends on the acidity constant and the concentration of buffer reagents (kA[acid], kB[Base]).

Questions

• 15-4• 15-19• 15-22

Q15.4

• Zero order rate constant for degradation of the colorant in a multisulfa preparation. Please determine Ea, K25C, and the predicated life for changing absorbance from initial value of 0.470 to 0.225.°C k 1/T ln K

40 0.00011 0.003194888 -9.1150350 0.00028 0.003095975 -8.1807260 0.00082 0.003003003 -7.1062170 0.00196 0.002915452 -6.23481

ln k= 24.2 - 10428.2 (1/T) -Ea/R=10428.2Ea = 1.9872 * 10428.2 = 20754 cal /moleT=298 °K, then k= 2.05231E-05At 25°C, 0.225=0.470-2.05 * 10-5tt= 11937.8h

Q15.19

• The following data are for the decomposition of 0.056 M glucose at 140 C, please calculate K0 and KH

kobs (hr-1) [H3O+]0.00366 0.0108

0.0058 0.01970.00818 0.02950.01076 0.03940.01217 0.0492

kobs=k0+kH*[H3O+]k0=0.001(hr-

1)kH=0.227 (hr-1)

0 0.01 0.02 0.03 0.04 0.05 0.060

0.002

0.004

0.006

0.008

0.01

0.012

0.014

f(x) = 0.2276388274645 x + 0.0013485740478R² = 0.992288609004898

[H3O+]

k

Q15.22• The hydrolysis of the prostaglandin in aqueous solution

was studied by varying the buffer. The dependence of kobs and pH is given as follows, do the buffer system influence hydrolysis? Please compute KH+ and KOH

-.

Buffer pHKobs * 107 (sec-1) log Kobs

HCl 1.15 3360-

3.47366

Formate 2.99 35-

5.45593

Formate 3.21 22.1-

5.65561

Formate 3.22 21.6-

5.66555

Phosphate 6.51 18.6-

5.73049

Phosphate 6.57 21.4-

5.66959

Phosphate 7.21 84.5-

5.07314

Carbonate 9.22 8350-

3.07831if specific acid-base catalysis:kobs=k0+kH*[H3O+]+kOH*[OH-]at low pH, then log Kobs = log kH

- pHintercept= -2.25 kH= 5.62 * 10-3

at high pH, then kobs=KOH.Kw/[H3O+]

log Kobs= log kOH.kw +pHintercept=-12.11

if kw=12.63 * 10

-12 then KOH=6.1 0

0 1 2 3 4 5 6 7 8 9 10

-7-6-5-4-3-2-10

pHL

og

k