L* Michael Berger (Center for Brain Research, Medical University Vienna, Austria): Ligand/Receptor.

L*

http://cwx.prenhall.com/horton/medialib/media_portfolio/09.html

Michael Berger (Center for Brain Research, Medical University Vienna, Austria):

Ligand/Receptor Interaction

Wenn Du mit anderen ein Schiff bauen willst,

Antoine de Saint Exupery

Wenn Du mit anderen ein Schiff bauen willst,beginne nicht, mit Ihnen Holz zu sammeln,


Wenn Du mit anderen ein Schiff bauen willst,beginne nicht, mit Ihnen Holz zu sammeln,

sondern wecke in Ihnen die Sehnsuchtnach dem großen weiten Meer.


What is a receptor?

What is a receptor?

A physical target mediating the physiological effect of a drug.

What is a receptor?


What is a ligand?

What is a receptor?


What is a ligand?A substance that (strongly) binds to a tissue.

What is a receptor?



What is an agonist?

What is a receptor?



What is an agonist?

A substance that causes an effect, an active change in the target tissue.

What is a receptor?



What is an agonist?


What is an antagonist?

What is a receptor?



What is an agonist?



A substance that blocks the effect of an agonist

What is a receptor?



What is an agonist?




What is a transmitter?

What is a receptor?



What is an agonist?




What is a transmitter?

A natural agonist released by a cell and acting on a neighboring cell.

B + L BL

Association:

KA =[BL]

[B] . [L]

B + L BL

Association:

KA =[BL]

[B] . [L]

KA: association equilibrium constant

B + L BL

Association:

KA =[BL]

[B] . [L]

KA: association equilibrium constant

BL B + L

Dissociation:

KD =[BL]

[B] . [L]

KD: dissociation equilibrium constant

B + L BL

Association:

KA =[BL]

[B] . [L]

KA: association equilibrium constantdimension: (concentration)-1

BL B + L

Dissociation:

KD =[BL]

[B] . [L]


B + L BL

Association:

KA =[BL]

[B] . [L]

KA: association equilibrium constantdimension: (concentration)-1

BL B + L

Dissociation:

KD =[BL]

[B] . [L]

KD: dissociation equilibrium constantdimension: concentration

B + L BL BL B + L

Association: Dissociation:

B + L BL BL B + L


Strong binding: equilibrium is

on right side on left side

B + L BL BL B + L




KA =[BL]

[B] . [L]>> 1 KD =

[BL]

[B] . [L]<< 1

B + L BL BL B + L




KA =[BL]

[B] . [L]>> 1 KD =

[BL]

[B] . [L]<< 1

ln KA positiv ln KD negativ

Van't Hoff: ΔGo = - RT . ln KA = + RT . ln KD

ΔGo: change in free enthalpy (Gibbs energy)R: universal gas constant, 1.987 cal/(Mol . °K) or 8.314 J/(Mol

. °K)T: absolute temperature

B + L BL BL B + L




KA =[BL]

[B] . [L]>> 1 KD =

[BL]

[B] . [L]<< 1

ln KA positiv ln KD negativ

The Van‘t Hoff equation allows the calculation of the free enthalpy change of a reaction from the reaction‘s equilibrium constant:

ΔGo (20 °C)

KA KD kcal/Mol kJ/Mol

107 M-1 10-7 M -9.4 -39.2

108 M-1 10-8 M -10.7 -44.8

109 M-1 10-9 M -12.0 -50.3

Van't Hoff: ΔGo = - RT . ln KA = + RT . ln KD

ΔGo: change in free enthalpy (Gibbs energy)R: universal gas constant, 1.987 cal/(Mol . °K) or 8.314 J/(Mol

. °K)T: absolute temperature

Examples for the change in free enthalpy Go in various reactions

Glucose + 6 O2 6 CO2 + 6 H2O -686 H2 + ½ O2 H2O -46 ATP ADP + Pi -7.3

ΔGo (kcal/Mol)



ΔGo (kcal/Mol)

In these reactions, Go is reduced (exergonic processes)



ΔGo (kcal/Mol)

Bond dissociation energies

HO-H HO· + ·H 118CH3CH2-H CH3CH2· + ·H 101CH3-CH3 CH3· + ·CH3 90

In these reactions, Go is reduced (exergonic processes)

Gibbs & Helmholtz: ΔGo = ΔHo – T . ΔSo

The free enthalpy change ΔGo of a reaction is composed of 2 terms:



change in enthalpy



change in enthalpy

change in entropy, multiplied by absolute temperature



change in enthalpy

change in entropy, multiplied by absolute temperature

ln KD = ΔGo / RT = ΔHo / RT - ΔSo / R

• KD measured at various temperatures

• ln KD plotted against 1/T

! attention ℮ ∫ ∑ mathematics ∂ ∞ √ % attention !





ΔHo < 0: exotherm(reaction mixture warms)

ΔSo > 0(order is decreased)

lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

●●●●

Most common case: The warmer (the lower 1/T), the weaker the affinity (the less negative lg KD).



55 °C 0 °C

lg KD = 0.434 . ΔHo/R . 1/T – 0.434 . ΔSo/R [0.434 = 1/ln10]






lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

●●●●




55 °C 0 °C


Intersection with ordinate gives information about ΔSo.






lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

●●●●




55 °C 0 °C


Intersection with ordinate gives information about ΔSo.

Slope allows access to ΔHo.





ΔSo < 0(order is increased)

lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

If order is increased, driving force is even more sensitive to high temperatures





●●●●

55 °C 0 °C

●●●●






lg KD

1/T-9 -

-6 -

-3 -

0 -

●●●●

‚‚ ‚‚0.001 0.002 0.003 0.004

lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

If order is increased, driving force is even more sensitive to high temperatures





●●●●

55 °C 0 °C

It may be difficult to obtain solid data that allow to decide, if ΔSo is > or < 0.






lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

lg KD

1/T-9 -

-6 -

-3 -

0 -

‚‚ ‚‚0.001 0.002 0.003 0.004

lg KD

1/T-9 -

-6 -

-3 -

0 -

●●●●

‚‚ ‚‚0.001 0.002 0.003 0.004

ΔHo > 0: endotherm(reaction mixture cools)

Endotherm binding is driven by decrease in order only; here, driving force increases with temperature.





●●●●

55 °C 0 °C

●●●●

Mechanisms contributing to ligand/receptor interaction:

1. Ionic interaction2. Hydrogen bonds3. Hydrophobic interaction4. Cation/ interaction5. Van der Waals interaction

attraction between 2 charges depends on

e1 . e2

D . r2

ionic interaction

r ... distanceD ... dielectric constant


vacuum ...1.0

hexane ...1.9

H2O ...78

e1 . e2

D . r2

ionic interaction



vacuum ...1.0

hexane ...1.9

H2O ...78

e1 . e2

D . r2

ionic interaction


In water, ionic interaction is hindered

by shells of water molecules surrounding

each ion.

hydrogen bondsB + L BL

Formation of a hydrogen bond is highly exergonic, yields 3-7 kcal/mol

hydrogen bonds

BH2O + LH2O BL + H2OH2O

B + L BL


However, enthalpy balance is poor:

hydrogen bonds


B + L BL



1. Break this bond.

hydrogen bonds


B + L BL



1. Break this bond.

2. Break this bond.

hydrogen bonds


B + L BL



1. Break this bond.

2. Break this bond.

3. Form this bond.

hydrogen bonds


B + L BL



1. Break this bond.

2. Break this bond.

3. Form this bond.

4. Form this bond.

hydrogen bonds


B + L BL



1. Break this bond.

2. Break this bond.

3. Form this bond.

4. Form this bond.

Hydrogen bond formation mainly driven by increase in entropy, since the water molecules “get more freedom“ (2 kcal per mol of water).

hydrophobic interaction

Molecules or parts of molecules („residues“) without charge, that cannot form a hydrogen bond, are called hydrophobic. They aggregate together to reduce the contact with water to a minimum.

HO

H

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1 2

3

4567

8

9 10

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12131415

16



HO

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1

23 4

5

6

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8910

11

12




• This example is nice, but wrong.

• Hydrogene bonds are never left open.

• In contact with an inert partner, water molecules are highly ordered.

• Reduction of contact area leads to reduced order.





• Reduction of Go by hydrophobic interaction is always due to the entropy term T . ΔS







• Reduction of Go by hydrophobic interaction is always due to the entropy term T . ΔS

• Empirical rule: Δ Go = -0.03 x area hidden from water (in Ǻ2).


cation/ interaction

A molluscan acetylcholine (AcCh) binding protein, with high sequence homology to the AcCh binding site of the nicotinic receptor, has been crystallized. The binding pocket is surrounded by tyr and trp residues (Bejc et al. 2001, Nature 411: 269)

Van der Waals interaction

http://www.columbia.edu/cu/biology/courses/c2005/lectures/lec02_06.html

Two atoms „touching“ each other with their electron shells redistribute their charges, resulting in attraction.


hydrogen bond


range 3-4 Ǻ, turns into repulsion at shorter distances

contribution to ΔGo 0.5-1.0 kcal/Mol (lower than hydrogen bond)

A „good“ ligand undergoes 5-10 van der Waals contacts with his receptor.

1998 Leif Saul


The ensemble of van der Waals interactions is responsible for the key/lock nature of ligand/receptor interaction.

Example for the interaction of a hypothetical ligand with its receptor:

formation of a hydrogen bond ... - 5.0 loss of hydrogen bond with H2O... + 5.0

kcal/mol


formation of a hydrogen bond ... - 5.0 loss of hydrogen bond with H2O... + 5.0

kcal/mol

preliminary balance: ± 0


formation of a hydrogen bond ... - 5.0 loss of hydrogen bond with H2O ... + 5.02 H2O set free … - 4.0Hydrophobic interaction … - 2.08 van der Waals contacts … - 4.7

kcal/mol



kcal/mol

balance: -10.7

ΔGo (20 °C)

KA KD kcal/Mol kJ/Mol

107 M-1 10-7 M -9.4 -39.2

108 M-1 10-8 M -10.7 -44.8

109 M-1 10-9 M -12.0 -50.3



kcal/mol

balance: -10.7

How many receptors do we expect in a responsive tissue?

Which analytical tools will be necessary to detect them?


• Theoretical assumption: the tissue consists of cubes 10 µm x 10 µm x 10 µm

• Then, 1 mg tissue would consist of 100 x 100 x 100 = 106 cells

• Theoretical assumption: the tissue consists of cubes 10 µm x 10 µm x 10 µm

• Then, 1 mg tissue would consist of 100 x 100 x 100 = 106 cells

• If each cell bears 1 binding site, this would result in 106 binding sites / mg tissue

• 1 fMol = 6 x 1023-15 = 6 x 108 molecules

• 106 molecules = 1/600 fMol


Josef Loschmidt (1821-1895)

• The most common binding sites occur at densities of 10 to several 100 fMol/mg tissue.

• This is much more than 1/600 fMol/mg tissue.• Thus, receptor-bearing cells have not only 1, but several thousands of

binding sites.


Freeze-fracture analysis of AMPA receptors labelled with immuno gold antibodies (5 nm)

at the postsynaptic site on cerebellar

Purkinje cells (climbing fiber

input). Tanaka et al (2005) J

Neurosci 25:799

Labelling: Replacement of one or more protons by tritium (3H; molecule practically unchanged)

Marie & Pierre Curie


Marie & Pierre Curie


Radioactivity measured in

• Curie (Ci, mCi, µCi)

(the radioactivity of 1 g radium)

Henry Becquerel




• Becquerel (Bq, decays / s)

• dpm (decays / min)

1 Bq = 60 dpm




• Becquerel (Bq, decays / s)

• dpm (decays / min)

1 µCi = 2 220 000 dpm1 nCi = 2 220 dpm1 pCi = 2.22 dpm

Henry Becquerel

t ½ 1 10 100 1000 fMol

14C 5 730 y

3H 12.4 y

35S 87 d

131I 8 d

The shorter the half-life, the hotter the radioligand.


Comparison of 3H with other nuclides (1 radioactive atom / molecule)


How many dpm can be expected from 1 fMol 3H?

A rule of thumb is a principle with broad application that is not intended to be strictly accurate or reliable for every situation. (Wikipedia)


How many dpm can be expected from 1 fMol 3H?


How many dpm can be expected from 1 fMol 3H?• 1 fMol = 10-15 x 6 . 1023 = 6 . 108 molecules



• t½ = 12.3 y = 4 500 d = 108 000 h = 6.48 . 106 min



• t½ = 12.3 y = 4 500 d = 108 000 h = 6.48 . 106 min

General idea: Since we know that half of the radioactive nuclei will decay in 6.48 million minutes, we might obtain the number of nuclei decaying in 1 minute simply by dividing half of the number of nuclei by 6.48 millions.



• t½ = 12.3 y = 4 500 d = 108 000 h = 6.48 . 106 min

• dpm = 6 . 108 . 0.5 / 6.48 . 106 = 46

General idea: Since we know that half of the radioactive nuclei will decay in 6.48 million minutes, we might obtain the number of nuclei decaying in 1 minute simply by dividing half of the number of nuclei by 6.48 millions.



• t½ = 12.3 y = 4 500 d = 108 000 h = 6.48 . 106 min

• dpm = 6 . 108 . 0.5 / 6.48 . 106 = 46

0.5 would be correct, if the decay rate would be the same for the whole decay period.

0 5 10 15 200

20

40

60

80

100

linear decay

t1/2

= 12.3 years

%

orig

inal

act

ivity

years



• t½ = 12.3 y = 4 500 d = 108 000 h = 6.48 . 106 min

• dpm = 6 . 108 . ln2 / 6.48 . 106 = 64

0.5 would be correct, if the decay rate would be the same for the whole decay period. However, radioactive decay follows an exponential law; therefore, 0.5 must be replaced by ln2 = 0.69.

0 5 10 15 200

20

40

60

80

100

exponential decay

linear decay

t1/2

= 12.3 years

%

orig

inal

act

ivity

years

Why the natural logarithm of 2?


A … number of radioactive nucleik … decay constant

dA/dt = -k . A∫(1/A)dA = -k . ∫dtln(A/Ao) = -k . Δt








A = Ao . e-k.Δt






A = Ao . e-k.Δt

k is related to t½: ln(½) = -k . t½

k = ln2 / t½






A = Ao . e-k.Δt

k is related to t½: ln(½) = -k . t½

k = ln2 / t½

for 1 min (Δt = 1): -ΔA = k . A . 1 = ln2 / t½ . A


Therefore, it can be expected that 6 . 108 tritium nuclei (1 fMol) will emit

6 . 108 . ln2 / 6.48 . 106 = 64 electrons / min.

A molecule labeled with one single 3H has a specific radioactivity (short: specific activity) of 64 dpm / fMol.

1 µCi = 2 220 000 dpm1 nCi = 2 220 dpm1 pCi = 2.22 dpm


A molecule labeled with one single 3H has a specific radioactivity (short: specific activity) of 64 dpm / fMol.

Remember:

64 dpm / fMol = 28.8 pCi / fMol = 28.8 Ci / mMol

Therefore, it can be expected that 6 . 108 tritium nuclei (1 fMol) will emit

6 . 108 . ln2 / 6.48 . 106 = 64 electrons / min.

64 dpm / fMol = 28.8 pCi / fMol = 28.8 Ci / mMol


t ½ 1 10 100 1000 fMol

14C 5 730 y 0.14 1.4 14 140

3H 12.4 y 64 640 6.4*103 64*103

35S 87 d 3.3*103 33*103 330*103 3.3*106

131I 8 d 36*103 360*103 3.6*106 36*106

dpm / mg tissue



t ½ 1 10 100 1000 fMol

14C 5 730 y 0.14 1.4 14 140

3H 12.4 y 64 640 6.4*103 64*103

35S 87 d 3.3*103 33*103 330*103 3.3*106

131I 8 d 36*103 360*103 3.6*106 36*106

most common experimental condition

dpm / mg tissue



Properties of 3H

• can replace 1H present in every organic molecule

• does not change the properties of the labeled molecule (no isotope effect)

• t½ 12.4 y

• decay (emits electrons)

• radiation reaches in air 6 mm, in liquid and tissue 6 µm

• relatively safe to work with (no shielding required)

• the only risk is incorporation of > 1 mCi

• only reliable method of counting:

Properties of 3H

• can replace 1H present in every organic molecule

• does not change the properties of the labeled molecule (no isotope effect)

• t½ 12.4 y

• decay (emits electrons)

• radiation reaches in air 6 mm, in liquid and tissue 6 µm

• relatively safe to work with (no shielding required)

• the only risk is incorporation of > 1 mCi

• only reliable method of counting: liquid scintillation

Analytical techniques

equilibrium dialysis

non-equilibrium techniques for receptors in solution

• gel filtration• charcoal adsorption• precipitation• adsorption to glass fiber filters

non.-equilibrium techniques for particulate receptors

• centrifugation• filtration• slice autoradiography

B + L* BL*







BL*

B

L*

L*

B + L* BL*







B + L* BL*







B + L* BL*

polyethylene glycoln = 6 000 – 8 000







B + L* BL*

glass

coating

L*

L* L*L*

L*

L*

L*

L*

L*

L*

L*L*







B + L* BL*

Saturability: a radioligand can only be displaced if the target density is low.

Other examples for saturability: Langmuir isotherme (mono-molecular layer on a surface), enzyme reaction rate (Michaelis-Menten).

http://www.steve.gb.com/science/membranes.html

Saturation & non-specific binding

0 500 1000 1500 20000

1000

2000

3000

4000

favorableconditions

no analysispossible

To Save this template,C hooseFile:Template:Template Save.

L (nM)

BL (fMol)


0 500 1000 1500 20000

1000

2000

3000

4000

favorableconditions

no analysispossible


L (nM)

BL (fMol)

At low nM concentrations,most of the radioligand Lis bound to saturablehigh affinity sites.


0 500 1000 1500 20000

1000

2000

3000

4000

favorableconditions

no analysispossible


L (nM)

BL (fMol)

At low nM concentrations,most of the radioligand Lis bound to saturablehigh affinity sites.

At high concentrations,the linearly rising

non-specific bindingwill dominate, and

specific bindingcan no longerbe detected.


BL B + L KD =[BL]

[B] . [L]



With increasing [L] more binding sites are occupied (BL) and free sites (B) are lost. The sum

BL + B = BM

remains constant.0 10 20 300

50

100

150

BL

L


KD =[BL]

[B] . [L]

replace [B] by BM – [BL]:

0 10 20 300

50

100

150

BL

L


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . [L]

solve for [BL]:

0 10 20 300

50

100

150

BL

L


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . [L]

solve for [BL]:

0 10 20 300

50

100

150

BL

L

KD . [BL] = BM . [L] – [BL] . [L]


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . [L]

solve for [BL]:

0 10 20 300

50

100

150

BL

L

KD . [BL] = BM . [L] – [BL] . [L]

[BL] . ([L] + KD) = BM . [L]


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . [L]

solve for [BL]:

[BL] = BM .[L]

[L] + KD Langmuir isotherm

0 10 20 300

50

100

150

BL

L

KD . [BL] = BM . [L] – [BL] . [L]

[BL] . ([L] + KD) = BM . [L]


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . [L]

solve for [BL]:

[BL] = BM .[L]


0 10 20 300

50

100

150

BL

L

KD . [BL] = BM . [L] – [BL] . [L]

[BL] . ([L] + KD) = BM . [L]

Irving Langmuir1881-1957

Nobel price 1932


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . [L] wrong!

0 10 20 300

50

100

150

BL

L

solve for [BL]:

[BL] = BM .[L]


KD . [BL] = BM . [L] – [BL] . [L]

[BL] . ([L] + KD) = BM . [L]

Irving Langmuir1881-1957

Nobel price 1932


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL]) correct

0 10 20 300

50

100

150

BL

L


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL])

solve for [BL]:

KD . [BL] = BM . Lo – BM . [BL] – [BL] . Lo + [BL]2

0 10 20 300

50

100

150

BL

L




KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL])

solve for [BL]:


0 10 20 300

50

100

150

BL

L



[BL]2 – [BL] . (BM + Lo + KD) + BM . Lo = 0


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL])

solve for [BL]:

0 10 20 300

50

100

150

BL

L



Sweet memories…


[BL]2 – [BL] . (BM + Lo + KD) + BM . Lo = 0


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL])

solve for [BL]:

[BL] = ½ . {BM + Lo + KD - [(BM + Lo + KD)2 – 4 . BM . Lo]½}



0 10 20 300

50

100

150

BL

L

Sweet memories…


[BL]2 – [BL] . (BM + Lo + KD) + BM . Lo = 0


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL])

solve for [BL]:



In this case, quantities Lo and KD are not entered as concentrations, but as moles in the respective volume chosen, in the same units as BM.

0 10 20 300

50

100

150

BL

L

Sweet memories…



[BL]2 – [BL] . (BM + Lo + KD) + BM . Lo = 0


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL])

solve for [BL]:




0 10 20 300

50

100

150

BL

L

3 times more ligand than receptors at KD concentration (8% loss)

Sweet memories…



[BL]2 – [BL] . (BM + Lo + KD) + BM . Lo = 0


KD =[BL]

[B] . [L]


KD =[BL]

(BM - [BL]) . (Lo – [BL])

solve for [BL]:




0 10 20 300

50

100

150

BL

L

3 times more receptor than ligand at KD concentration (57% loss)

Sweet memories…



[BL]2 – [BL] . (BM + Lo + KD) + BM . Lo = 0


A realistic saturation function is a composite of 2 simultaneous processes:

0 10 20 300

50

100

150

*

1

BL

L

1: non-specific binding

It is sufficient to measure 2 points; extrapolation of L 0 results in the blank of the measuring method (*).




It is sufficient to measure 2 points; extrapolation of L 0 results in the

blank of the measuring method (*).

0 10 20 300

50

100

1502

*

1

BL

L

2: specific binding

... Is sitting on the non-specific binding, obtained as difference between total and non-specific binding.




It is sufficient to measure 2 points; extrapolation of L 0 results in the

blank of the measuring method (*).2: specific binding

... Is sitting on the non-specific binding, obtained as difference between total and non-specific binding (†).

0 10 20 300

50

100

150

†

2

*

1

BL

L


0 10 20 300

50

100

150

†

2

*

1

BL

L

Mathematical combination of both processes:


0 10 20 300

50

100

150

†

2

*

1

BL

L



[BL] = BU .[L]

[L] + KU

2: specific binding


0 10 20 300

50

100

150

†

2

*

1

BL

L



[BL] = BU .[L]

[L] + KU

[BL] = BS .[L]

[L] + KS

2: specific binding


0 10 20 300

50

100

150

†

2

*

1

BL

L



[BL] = BU .[L]

[L] + KU

[BL] = BS .[L]

[L] + KS

KU (~mM) >> Ks (nM)

2: specific binding


0 10 20 300

50

100

150

†

2

*

1

BL

L



[BL] = BU .[L]

[L] + KU

[BL] = BS .[L]

[L] + KS

KU (~mM) >> Ks (nM)

At reasonable ligand concentrations, [L] + KU ~ KU and non-specific binding is a linear function of [L]:

[BL] = BS .[L]

[L] + KS

BU

KU

. [L]+

-9 -8 -7 -6 -50

50

100

NB

BL

log[I]


The most important value, the specific binding, is not directly accessible. It must be calculated by substracting the non-specific binding from total binding.

-9 -8 -7 -6 -50

50

100

NB

BL

log[I]


The most important value, the specific binding, is not directly accessible. It must be calculated by substracting the non-specific binding from total binding.

The non-specific binding NB is measured as bound ligand that is impossible to displace, even by high concentrations of potent displacers.


Strategies to keep non-specific binding low:



• choose a biological source with a high density of high-affinity binding sites




• select a radioligand concentration around the expected KD ( a few hundred to a few thousand dpm will be sufficient as result)





• use a clean radioligand; if necessary, any radioligand can be purified easily by thin layer chromatography






• If you filter your samples and if you use a radioligand with an amino group, pre-treat the glass fiber filters with polyethylene imine






• If you filter your samples and if you use a radioligand with an amino group, pre-treat the glass fiber filters with polyethylene imine

• optimise the rinsing procedure of pellets and filters, respectively

Classification of glutamate receptors

ionotropic receptors metabotropic receptors

Radioligands forexcitatory amino acid (EAA) receptors




NMDAreceptors

non-NMDAreceptors

Group I Group II Group III




NMDAreceptors

non-NMDAreceptors

AMPAreceptors

kainatereceptors

Group I Group II Group III



ionotropic receptors

NMDAreceptors

non-NMDAreceptors

AMPAreceptors

kainatereceptors

Schmid et al (2009) PNAS 106:10320


COOH

COOH

NH2

L-glutamic acid(S)-1-aminopropane-1,3-dicarboxylic acid

COOH

PO3H2

NH2

COOH

PO3H2

NH2D-Aminophosphonovaleric acid

CGP 39653(E)-2-Amino-4-propyl-5-phosphono-3-pentenoic acid

COOHNH

HOOC

N-methyl-D-aspartic acid (NMDA)

Mg2+

Mg2+

-70 mV

Na+

Ca2+

Mg2+

Mg2+

outside

membrane

inside

Mg2+

Mg2+

-70 mV

Na+

Ca2+

Mg2+

Mg2+

outside

membrane

inside

glu gly glu

AMPA-receptor

Mg2+

-30 mV

Na+

Ca2+

Mg2+

Mg2+

outside

membrane

inside

glu gly gluMg2+

-10 mV

Na+

Ca2+

Mg2+

Mg2+

outside

membrane

inside

glu gly glu

Mg2+

Mg2+

0 mV

Na+

Ca2+

Mg2+

Mg2+

outside

membrane

inside

glu gly glu

Mg2+

Mg2+


COOHNH2

Glycine

NH

O

OH

O

Cl

L-701.324 ( a phenyl quinolinone)

NCOOH

CHOO

H

Cl

Cl

MDL-105.519 (an indole carboxylic acid)

MK-801

NH

*


COOHNH2

Glycine

NH

O

OH

O

Cl

L-701.324 ( a phenyl quinolinone)

NCOOH

CHOO

H

Cl

Cl

MDL-105.519 (an indole carboxylic acid)

MK-801

NH

*

N

O

N*

[3H]GSK-931.145radioligand for the glycine transporter GlyT-1

(Herdon et al 2010 Neuropharmacol 59:558)


OH

O

NH

COOHH

Hkainic acid ( a pyrrolidine)

COOHNH2

ON

OH

**

AMPA (-Amino-3-hydroxy-5-methylisoxazol-4-propionic acid)

H COOH

COOH

NH2

LY-354.740 (a bicyclo[3.1.0]hexan)


OH

O

NH

COOHH


COOHNH2

ON

OH

**


H COOH

COOH

NH2


Grant et al (2010) Neurotox Terat 32:132

HOOC

H

OH

O

NH

COOHH

H


OH

O

NH

COOHH


COOHNH2

ON

OH

**


H COOH

COOH

NH2


NO

COOH

OH

NH2

NO OH

NH2

Muscimol Ibotensäure


OH

O

NH

COOHH


COOHNH2

ON

OH

**


H COOH

COOH

NH2

LY-354.740 (a bicyclo[3.1.0]hexan)O

H COOH

COOH

NH2S

H COOH

COOH

NH2

O

O

LY-404.039 LY-379.268

The most important binding techniques

... are all non-equilibrium techniques for particulate receptor preparations:

• Centrifugation• Filtration over glass fiber filters• Slice autoradiography

B + L* BL*




B + L* BL*

... applied to weak ligands (KD > 20 nM) ● you need a high speed refrigerated certrifuge ● plastic vials must support 40 000 x g ● after centrifugation, pellet and inner wall needs rinsing ● scintillation cocktail added directly to the rinsed incubation vials.




B + L* BL*

... Can only be applied to high affinity ligands (KD < 20 nM) ● you need a vacuum filter box




B + L* BL*

... Can only be applied to high affinity ligands (KD < 20 nM) ● you need a vacuum filter box or better a harvester ● for radioligands with amino group, the glass fiber filter must be soaked in 0.3% polyethylenimine ●




B + L* BL*

... Can only be applied to high affinity ligands (KD < 20 nM) ● you need a vacuum filter box or better a harvester ● for radioligands with amino group, the glass fiber filter must be soaked in 0.3% polyethylenimine ● for best results, filter should be shaken in scintillation cocktail for 30 min.

L* L*L* L*

scintillator




B + L* BL*

glass

coating

L*

L* L*L*

L*

L*

L*

L*

L*

L*

L*L*

... applied to frozen slices prepared in a cryostat / microtom (10-20 µm) ● tissue must be shock-frozen (-40 °C) in dry ice / isopentane ● slices taken up to coated glass slides ● for incubation, you can use..




B + L* BL*

... applied to frozen slices prepared in a cryostat / microtom (10-20 µm) ● tissue must be shock-frozen (-40 °C) in dry ice / isopentane ● slices taken up to coated glass slides ● for incubation, you can use a jar or...




B + L* BL*

... applied to frozen slices prepared in a cryostat / microtom (10-20 µm) ● tissue must be shock-frozen (-40 °C) in dry ice / isopentane ● slices taken up to coated glass slides ● for incubation, you can use a jar or simply a droplet on the slide ● expose dried slices to film or phosphoscreen ● evaluation by co-exposure of stripes containing known amounts of radioactivity.

L* Michael Berger (Center for Brain Research, Medical University Vienna, Austria): Ligand/Receptor.

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Transcript of L* Michael Berger (Center for Brain Research, Medical University Vienna, Austria): Ligand/Receptor.