Statistical Thermodynamics: Ligand Binding Based on Cantor & Schimmel,

Chapt. 15

LIGAND EQUILIBRIA Macroscopic and microscopic equilibrium constants

Titration of Glycine (dissociation of two proton ligands)

Macroscopic Equilibria:

GH2K1 GH + H+

GH K2 G- + H+

where:

K1 =[GH][H+ ]

[GH2 ] (pK1 = 2.35)

K2 =[G- ][H+ ]

[GH] (pK2 = 9.78)

[1]

Microscopic forms:

GH2 = +H3NCH2COOHGH = +H3NCH2COO- + H2NCH2COOH

G- = H2NCH2COO-

[2]

Microscopic equilibria:

Substituting Eq. 2 into Eq. 1:

K1 =[ +H3NCH2COO- ] + [H2NCH2COOH]{ }[H+ ]

[ +H3NCH2COOH]= k1 + k2

K2 =[H2NCH2COO- ][H+ ]

[ +H3NCH2COO- ] + [H2NCH2COOH]{ } =1

1k3+ 1

k4

[3]

Long chain diacid protonation (proton binding at two equivalent sites)

Macroscopic equilibria:

A(H+ )2K1 A(H+ ) + H+

A(H+ ) K2 A + H+

Macroscopic dissociation constants:

K1=[A(H+)][H+][A(H+)2 ]

K2=[A][H+][A(H+)]

[4]

Microscopic equilibria:

H+AH+ k AH+ + H+

H+AH+ k H+A + H+

AH+ k A + H+

H+A k A + H+

Relationships between macroscopic and microscopic species:

A = AA(H+ ) = AH+ + H+AA(H+ )2 = H+AH+

[5]

Substituting Eq 5 into 4:

K1 = 2kK2 = k / 2K1 = 4K2

Interesting! Even though the actual affinity of each site for H+ is identical, the apparent affinity of the second binding reaction is four times smaller than the first. This statistical effect looks like negative cooperativity, but it's not!

Ligand Binding at Multiple Identical Independent Sites Calculating the number of microscopic species (multiplicity of binding modes)

Macromolecule M with n binding sites for ligand L with a microscopic dissociation constant of k

M + L M1

M1 + L M2

. . . . . . . . .

Mn-1 + L Mn

Mi is the macroscopic state of macromolecule with i ligands bound. For example, if n=4 then M2 is:

M2 =L LL L

+ L LL L

+ L LL L

+ L LL L

+ L LL L

+ L LL L

Thus, there are six ways to put two ligands in four sites. In general:

W (n,i) = n!(n − i)!i!

[6]

Calculation of ν (moles of L bound per mole of M)

ν =i[Mi ]

i=0

n

∑

[Mi ]i=0

n

∑ [7]

Macroscopic dissociation constants:

K1 =[M0 ][L]

[M1]

Ki =[Mi-1][L]

[Mi ]

Kn =[Mn-1][L]

[Mn ]

[8]

Solving for [Mi]:

[Mi ] =[Mi-1][L]

Ki

=[M0 ][L]

i

K jj=1

i

∏ [9]

Distinction between k (microscopic) and K (macroscopic) dissociation constants:

k =

L LL L

⎡

⎣⎢

⎤

⎦⎥[L]

L LL L

⎡

⎣⎢

⎤

⎦⎥

=

L LL L

⎡

⎣⎢

⎤

⎦⎥[L]

L LL L

⎡

⎣⎢

⎤

⎦⎥

=

L LL L

⎡

⎣⎢

⎤

⎦⎥[L]

L LL L

⎡

⎣⎢

⎤

⎦⎥

=

whereas:

K1 =

L LL L

⎡

⎣⎢

⎤

⎦⎥[L]

L LL L

⎡

⎣⎢

⎤

⎦⎥ +

L LL L

⎡

⎣⎢

⎤

⎦⎥ +

L LL L

⎡

⎣⎢

⎤

⎦⎥ +

L LL L

⎡

⎣⎢

⎤

⎦⎥

Relationship between k and Ki determined by the multiplicity of state i:

Ki =W (n,i −1)W (n,i)

k = n! (i −1)!(n − i +1)!n! i!(n − i)!

k = i!(n − i)!(i −1)!(n − i +1)!

k = in − i +1

k [10]

Substituting Eq. 10 into Eq. 9:

[Mi ] = [M0 ](n − j +1)

j⎡

⎣⎢

⎤

⎦⎥

j=1

i

∏⎧⎨⎩

⎫⎬⎭[L] / k( )i [11]

Substituting Eq. 11 into Eq. 7:

ν =

i[M0 ](n − j +1)

j⎡

⎣⎢

⎤

⎦⎥

j=1

i

∏⎧⎨⎩

⎫⎬⎭[L] / k( )i

i=1

n

∑

[M0 ]+ [M0 ](n − j +1)

j⎡

⎣⎢

⎤

⎦⎥

j=1

i

∏⎧⎨⎩

⎫⎬⎭[L] / k( )i

i=1

n

∑=

i (n − j +1)j

⎡

⎣⎢

⎤

⎦⎥

j=1

i

∏⎧⎨⎩

⎫⎬⎭[L] / k( )i

i=1

n

∑

1+ (n − j +1)j

⎡

⎣⎢

⎤

⎦⎥

j=1

i

∏⎧⎨⎩

⎫⎬⎭[L] / k( )i

i=1

n

∑

Notice that:

(n − j +1)j=1

i

∏ =n!

(n − i)!

so:

(n − j +1)

jj=1

i

∏ =n!

(n − i)!i!=W (n,i) [13]

Substituting Eq. 13 into Eq. 12:

ν =iW (n,i) [L] / k( )i

i=1

n

∑

1+ W (n,i) [L] / k( )ii=1

n

∑ [14]

The denominator of Eq. 14 is the binomial expansion of (1+[L]/k)n:

1+ W (n,i) [L] / k( )ii=1

n

∑ = 1+ [L] / k( )n [15]

Mathematical trick! Differentiate Eq. 15 with respect to [L]/k:

iW (n,i) [L] / k( )i−1i=1

n

∑ = n 1+ [L] / k( )n−1 [16]

Multiply both sides by [L]/k:

iW (n,i) [L] / k( )ii=1

n

∑ = n [L] / k( ) 1+ [L] / k( )n−1 [17]

Substituting Eq. 17 for the numerator of Eq. 14 and Eq. 15 for the denominator of Eq. 14:

ν =n [L] / k( ) 1+ [L] / k( )n−1

1+ [L] / k( )n=n [L] / k( )1+ [L] / k( )

[18]

Linearize to get the Scatchard Equation:

ν 1+ [L] / k( )

[L]= n / k

ν[L]

= n / k −ν / k [19]

Scatchard plot for a macromolecule with n independent and identical binding sites:

ν[L]

=n / k

1+ [L] / k( )= n / k −ν / k

0

2

4

6

8

10

12

0 1 2 3 4 5

!/[

L]

(M-1

)

!

Slope= -1/k

Intercept= n/k

Intercept= n

Scatchard plot for a macromolecule with n independent and identical binding sites:

Scatchard plot for two classes of independent binding sites with different affinities

ν[L]

=n / k

1+ [L] / k( )= n / k −ν / k

ν[L]

=ni / ki

1+ [L] / ki( )i∑

0

2

4

6

8

10

12

0 1 2 3 4 5

!/[

L]

(M-1

)

!

Slope= -1/k

Intercept= n/k

Intercept= n0

10

20

30

40

50

60

70

0 0.5 1 1.5 2 2.5 3 3.5 4

!/[L] (M

-1)

!

Intercept= n1/k

1+n

2/k

2

Intercept= n1+n

2

The Hill Constant

For ligand binding that is infinitely cooperative over part of saturation range, i.e., an all-or-none reaction:

M0 + nL Kn Mn

K n =[M0 ][L]n

Mn

[20]

where K is the apparent dissociation constant for each of the interacting sites. For this situation:

ν =n[Mn ]M0 +Mn

=n[L]n /Kn

1+ [L]n /Kn

ν[L]

=n[L]n−1 /Kn

1+ [L]n /Kn

[21]

The fractional saturation (0→1) is:

y = [L]n /Kn

1+ [L]n /Kn [22]

In reality, infinitely cooperative binding is not observed. However, in the saturation range of 25% to 75% semiempirical equations in the form of Eq.s 20 – 22 can be used:

ν =n[L]αH /KαH

1+ [L]αH /KαH

ν[L]

=n[L]αH−1 /KαH

1+ [L]αH /KαH

y = [L]αH /KαH

1+ [L]αH /KαH

[23]

where αH (Hill constant) is between 1 (no cooperativity between sites) and n (perfect cooperativity between sites)

Steepness at the midpoint sensitive to αH :

d ln[y / (1− y )]{ }

d(ln[L])=αH

at half-saturation ([L]1/2):

d[y / (1− y )]

d[L]⎛⎝⎜

⎞⎠⎟ y=1/2

=αH,1/2

[L]1/2