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Transcript Document 7507642

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

Ligand/Receptor Interaction

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

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,

Antoine de Saint Exupery

Wenn Du mit anderen ein Schiff bauen willst, beginne nicht, mit Ihnen Holz zu sammeln, sondern wecke in Ihnen die Sehnsucht nach dem großen weiten Meer.

Antoine de Saint Exupery

What is a

receptor

What is a

receptor

A physical target mediating the physiological effect of a drug.

What is a

receptor

A physical target mediating the physiological effect of a drug.

What is a

ligand

What is a

receptor

What is a

ligand

A physical target mediating the physiological effect of a drug.

A substance that (strongly) binds to a tissue.

What is a

receptor

What is a

ligand

A physical target mediating the physiological effect of a drug.

A substance that (strongly) binds to a tissue.

What is an

agonist

What is a

receptor

What is a

ligand

A physical target mediating the physiological effect of a drug.

A substance that (strongly) binds to a tissue.

What is an

agonist

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

What is a

receptor

A physical target mediating the physiological effect of a drug.

What is a

ligand

A substance that (strongly) binds to a tissue.

What is an

agonist

What is an

antagonist

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

What is a

receptor

A physical target mediating the physiological effect of a drug.

What is a

ligand

A substance that (strongly) binds to a tissue.

What is an

agonist

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

What is an

antagonist

A substance that blocks the effect of an agonist

What is a

receptor

A physical target mediating the physiological effect of a drug.

What is a

ligand

A substance that (strongly) binds to a tissue.

What is an

agonist

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

What is an

antagonist

A substance that blocks the effect of an agonist What is a

transmitter

What is a

receptor

A physical target mediating the physiological effect of a drug.

What is a

ligand

A substance that (strongly) binds to a tissue.

What is an

agonist

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

What is an

antagonist

A substance that blocks the effect of an agonist What is a

transmitter

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

Association: B + L BL K A = [BL] [B]

[L]

Association: B + L BL K A = [BL] [B]

[L] K A : association equilibrium constant

Association: B + L BL

K A

= [BL] [B]

[L] K A : association equilibrium constant Dissociation: BL B + L

K D

= [B]

[L] [BL] K D : dissociation equilibrium constant

Association: B + L BL

K A

= [BL] [B]

[L] K A : association equilibrium constant dimension: (concentration) -1 Dissociation: BL B + L

K D

= [B]

[L] [BL] K D : dissociation equilibrium constant

Association: B + L BL

K A

= [BL] [B]

[L] K A : association equilibrium constant dimension: (concentration) -1 Dissociation: BL B + L

K D

= [B]

[L] [BL] K D : dissociation equilibrium constant dimension: concentration

Association: B + L BL Dissociation: BL B + L

Association: B + L BL Dissociation: BL B + L Strong binding: equilibrium is on right side on left side

Association: B + L

Dissociation:

B + L Strong binding: equilibrium is on right side

K A

[BL]

[B]

[L] >> 1

K D

= on left side [B]

[L]

[BL]

<< 1

Association: B + L BL Dissociation: BL B + L Strong binding: equilibrium is on right side

K A

[BL]

[B]

[L] ln K A positiv >> 1

K D

= on left side [B]

[L]

[BL]

ln K D negativ << 1

Association: B + L BL Dissociation: BL B + L Strong binding: equilibrium is on right side

K A

[BL]

[B]

[L] ln K A positiv >> 1

K D

= on left side [B]

[L]

[BL]

ln K D negativ << 1 Van't Hoff:

ΔG o

= - RT

ln K A = + RT

ln K D ΔG o : change in free enthalpy (

ibbs energy) R: universal gas constant, 1.987 cal/(Mol

°K) or 8.314 J/(Mol

°K) T: absolute temperature

The Van‘t Hoff equation allows the calculation of the free enthalpy change of a reaction from the reaction‘s equilibrium constant: 10 7 10 10 K 8 9 A M -1 M M -1 -1 10 10 K 10 -7 -8 -9 D M M M

ΔG o

(20 ° C) kcal/Mol kJ/Mol -9.4

-10.7

-12.0

-39.2

-44.8

-50.3

Van't Hoff:

ΔG o

= - RT

ln K A = + RT

ln K D ΔG o : change in free enthalpy (

ibbs 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 G o various reactions in Glucose + 6 O H 2 + ½ O 2 ATP 2 6 CO 2 + 6 H 2 O H 2 O ADP + P i -7.3

ΔG o

(kcal/Mol) -686 -46

Examples for the change in free enthalpy G o various reactions in Glucose + 6 O H 2 + ½ O 2 ATP 2 6 CO 2 + 6 H 2 O H 2 O ADP + P i -7.3

ΔG o

(kcal/Mol) -686 -46 In these reactions,

G o

is reduced ( exergonic processes)

Examples for the change in free enthalpy G o various reactions in Glucose + 6 O H 2 + ½ O 2 ATP 2 6 CO 2 + 6 H 2 O H 2 O ADP + P i -7.3

ΔG o

(kcal/Mol) -686 -46 In these reactions,

G o

is reduced ( exergonic processes) Bond dissociation energies HO-H CH 3 CH 2 -H CH 3 -CH 3 HO

H CH 3 CH 2

CH 3

+ CH 3

H 101 90 118

The free enthalpy change

ΔG o

of a reaction is composed of 2 terms: Gibbs & Helmholtz:

ΔG o

= ΔH o – T

ΔS o

The free enthalpy change

ΔG o

of a reaction is composed of 2 terms: Gibbs & Helmholtz:

ΔG o

ΔH o

– T

ΔS o change in enthalpy

The free enthalpy change

ΔG o

of a reaction is composed of 2 terms: Gibbs & Helmholtz:

ΔG o

= ΔH o – T

. ΔS o

change in enthalpy change in entropy, multiplied by absolute temperature

! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

The free enthalpy change

ΔG o

of a reaction is composed of 2 terms: Gibbs & Helmholtz:

ΔG o

= ΔH o – T

ΔS o change in enthalpy change in entropy, multiplied by absolute temperature ln K D = ΔG o / RT = ΔH o / RT ΔS o / R • K D measured at various temperatures • ln K D plotted against 1/T ! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

ΔH o < 0: exotherm (reaction mixture warms) lg K D 0 55 °C 0 °C ΔS o > 0 (order is decreased) -3 -6 -9 ‚ ‚ ‚ ●●● ● ‚ 0.001 0.002 0.003 0.004

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

ln K D = ΔG o / RT = ΔH o / RT ΔS o / R • K D measured at various temperatures • ln K D plotted against 1/T lg K D = 0.434

ΔH o /R

1/T – 0.434

ΔS o /R ! attention ℮ ∫ ∑

mathematics

[0.434 = 1/ln10] ∂ ∞ √ % attention !

! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

ΔH o < 0: exotherm (reaction mixture warms) lg K D 0 55 °C 0 °C ΔS o > 0 (order is decreased) -3 -6 -9 ‚ ‚ ‚ ●●●● ‚ 0.001 0.002 0.003 0.004

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

Intersection with ordinate gives information about ΔS o .

ln K D = ΔG o / RT = ΔH o / RT ΔS o / R • K D measured at various temperatures • ln K D plotted against 1/T lg K D = 0.434

ΔH o /R

1/T – 0.434

ΔS o /R ! attention ℮ ∫ ∑

mathematics

[0.434 = 1/ln10] ∂ ∞ √ % attention !

! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

ΔH o < 0: exotherm (reaction mixture warms) lg K D 0 55 °C 0 °C ΔS o > 0 (order is decreased) -3 -6 -9 ‚ ‚ ‚ ●●●● ‚ 0.001 0.002 0.003 0.004

1/T ln K D = ΔG o / RT = ΔH o / RT ΔS o / R Most common case: The warmer (the lower 1/T), the weaker the affinity (the less negative lg K D ).

Intersection with ordinate gives information about ΔS o .

Slope allows access to ΔH o .

• K D measured at various temperatures • ln K D plotted against 1/T lg K D = 0.434

ΔH o /R

1/T – 0.434

ΔS o /R ! attention ℮ ∫ ∑

mathematics

[0.434 = 1/ln10] ∂ ∞ √ % attention !

! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

ΔH o < 0: exotherm (reaction mixture warms) lg K D 0 55 °C 0 °C ΔS o > 0 (order is decreased) -3 -6 -9 lg K D ‚ ‚ ‚ ●●●● ‚ 0.001 0.002 0.003 0.004

0 1/T ΔS o < 0 (order is increased) -3 -6 -9 ‚ ‚ ‚ ●●● ● ‚ 0.001 0.002 0.003 0.004

1/T If order is increased, driving force is even more sensitive to high temperatures ln K D = ΔG o / RT = ΔH o / RT ΔS o / R • K D measured at various temperatures • ln K D plotted against 1/T lg K D = 0.434

ΔH o /R

1/T – 0.434

ΔS o /R [0.434 = 1/ln10] ! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

ΔH o < 0: exotherm (reaction mixture warms) lg K D 0 55 °C 0 °C ΔS o > 0 (order is decreased) -3 -6 -9 lg K D ‚ ‚ ‚ ●●●● ‚ 0.001 0.002 0.003 0.004

0 1/T ΔS o < 0 (order is increased) -3 -6 -9 ‚ ‚ ‚ ●●● ● ‚ 0.001 0.002 0.003 0.004

1/T If order is increased, driving force is even more sensitive to high temperatures ln K D = ΔG o / RT = ΔH o / RT ΔS o / R It may be difficult to obtain solid data that allow to decide, if ΔS o is > or < 0.

• K D measured at various temperatures • ln K D plotted against 1/T lg K D = 0.434

ΔH o /R

1/T – 0.434

ΔS o /R [0.434 = 1/ln10] ! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

ΔS o > 0 (order is decreased) ΔS o < 0 ! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

ΔH o < 0: exotherm (reaction mixture warms) lg K D lg K D ΔH o > 0: endotherm (reaction mixture cools) 0 0 55 °C 0 °C (order is increased) -3 -6 -9 lg K D ‚ ‚ ‚ ●●●● ‚ 0.001 0.002 0.003 0.004

0 -3 -6 -9 ‚ ‚ ‚ ●●● ● ‚ 0.001 0.002 0.003 0.004

1/T 1/T -3 -6 -9 ‚ ‚ ‚ ●●● ● ‚ 0.001 0.002 0.003 0.004

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

ln K D = ΔG o / RT = ΔH o / RT ΔS o / R • K D measured at various temperatures • ln K D plotted against 1/T lg K D = 0.434

ΔH o /R

1/T – 0.434

ΔS o /R [0.434 = 1/ln10] ! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

Mechanisms contributing to ligand/receptor interaction:

1. Ionic interaction 2. Hydrogen bonds 3. Hydrophobic interaction 4. Cation/ p interaction 5. Van der Waals interaction

ionic interaction

attraction between 2 charges depends on e 1

e 2 D

r 2 r ...

D ...

distance dielectric constant

ionic interaction

attraction between 2 charges depends on e 1

e 2 D

r 2 r ...

D ...

distance dielectric constant vacuum ...

hexane ...

H 2 O ...

1.0

1.9

ionic interaction

attraction between 2 charges depends on e 1

e 2 D

r 2 r ...

D ...

distance dielectric constant vacuum ...

hexane ...

H 2 O ...

1.0

1.9

78 In water, ionic interaction is hindered by shells of water molecules surrounding each ion.

hydrogen bonds

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

hydrogen bonds

B + L B  L Formation of a hydrogen bond is highly exergonic, yields 3-7 kcal/mol However, enthalpy balance is poor: B  H 2 O + L  H 2 O B  L + H 2 O  H 2 O

hydrogen bonds

B + L B  L Formation of a hydrogen bond is highly exergonic, yields 3-7 kcal/mol However, enthalpy balance is poor: B  H 2 O + L  H 2 O B  L + H 2 O  H 2 O 1. Break this bond.

hydrogen bonds

B + L B  L Formation of a hydrogen bond is highly exergonic, yields 3-7 kcal/mol However, enthalpy balance is poor: B  H 2 O + L  H 2 O B  L + H 2 O  H 2 O 2. Break this bond.

1. Break this bond.

hydrogen bonds

B + L B  L Formation of a hydrogen bond is highly exergonic, yields 3-7 kcal/mol However, enthalpy balance is poor: B  H 2 O + L  H 2 O B  L + H 2 O  H 2 O 2. Break this bond.

1. Break this bond.

3. Form this bond.

hydrogen bonds

B + L B  L Formation of a hydrogen bond is highly exergonic, yields 3-7 kcal/mol However, enthalpy balance is poor: B  H 2 O + L  H 2 O B  L + H 2 O  H 2 O 2. Break this bond.

1. Break this bond.

4. Form this bond.

3. Form this bond.

hydrogen bonds

B + L B  L Formation of a hydrogen bond is highly exergonic, yields 3-7 kcal/mol However, enthalpy balance is poor: B  H 2 O + L  H 2 O B  L + H 2 O  H 2 O 2. Break this bond.

1. Break this bond.

4. Form this bond.

3. 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.

H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H

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.

H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O 8 1 H O H 7 6 H H O 5 2 H 3 4 H O H H O H H O H H O H H O H H O 16 9 H H O H 15 14 H O 10 H 11 13 12 H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H

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.

H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H O H H O H H H O H 1 H O H H O 2 3 H H O H 12 H O H 11 10 H O H H O H H O H H O H H O H H O H 4 9 H O H O 5 6 7 8 H H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H H O H

hydrophobic interaction

• 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.

hydrophobic interaction

• 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

G o

by hydrophobic interaction is always due to the entropy term T

ΔS Gibbs & Helmholtz:

ΔG o

= ΔH o – T

. ΔS o

hydrophobic interaction

• 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

G o

by hydrophobic interaction is always due to the entropy term T

ΔS • Empirical rule: Δ

G o

water (in Ǻ 2 ).

= -0.03 x area hidden from

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

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

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

Van der Waals interaction

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

ΔG o

0.5 1.0 kcal/Mol (lower than hydrogen bond) A „good“ ligand undergoes 5-10 van der Waals contacts with his receptor.

hydrogen bond Van der Waals interaction

Van der Waals interaction

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

 1998 Leif Saul

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

formation of a hydrogen bond ...

loss of hydrogen bond with H 2 O...

kcal/mol - 5.0 + 5.0

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

formation of a hydrogen bond ...

loss of hydrogen bond with H 2 O...

preliminary balance: kcal/mol - 5.0 + 5.0

± 0

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

formation of a hydrogen bond ...

loss of hydrogen bond with H 2 O ...

2 H 2 O set free … Hydrophobic interaction … 8 van der Waals contacts … kcal/mol - 5.0 + 5.0

- 4.0

- 2.0

- 4.7

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

formation of a hydrogen bond ...

loss of hydrogen bond with H 2 O ...

2 H 2 O set free … Hydrophobic interaction … 8 van der Waals contacts … balance: kcal/mol - 5.0 + 5.0

- 4.0

- 2.0

- 4.7

-10.7

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

formation of a hydrogen bond ...

loss of hydrogen bond with H 2 O ...

2 H 2 O set free … Hydrophobic interaction … 8 van der Waals contacts … balance: 10 7 10 8 10 9 K A M-1 M-1 M-1 K 10 -7 10 -8 10 -9 D M M M

ΔG o

(20 ° C) kcal/Mol kJ/Mol -9.4

-39.2

-10.7

-12.0

-44.8

-50.3

kcal/mol - 5.0 + 5.0

- 4.0

- 2.0

- 4.7

-10.7

How many receptors do we expect in a responsive tissue?

Which analytical tools will be necessary to detect them?

How many receptors do we expect in a responsive tissue?

• • 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 = 10 6 cells

How many receptors do we expect in a responsive tissue?

Josef Loschmidt (1821-1895) • • • • • 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 = 10 6 cells If each cell bears 1 binding site, this would result in 10 6 binding sites / mg tissue 1 fMol = 6 x 10 23-15 = 6 x 10 8 molecules 10 6 molecules = 1/600 fMol

How many receptors do we expect in a responsive tissue?

• • • 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

Which analytical tools will be necessary to detect them?

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

Which analytical tools will be necessary to detect them?

Radioactivity measured in • Curie (Ci, mCi, µCi) (the radioactivity of 1 g radium) Marie & Pierre Curie

Which analytical tools will be necessary to detect them?

Radioactivity measured in • Curie (Ci, mCi, µCi) • Becquerel (Bq, decays / s) • dpm (decays / min) 1 Bq = 60 dpm Henry Becquerel

Which analytical tools will be necessary to detect them?

Radioactivity measured in • Curie (Ci, mCi, µCi) • Becquerel (Bq, decays / s) • dpm (decays / min) 1 µCi = 2 220 000 dpm 1 nCi = 2 220 dpm 1 pCi = 2.22

dpm Henry Becquerel

14 C

Which analytical tools will be necessary to detect them?

Comparison of 3 H with other nuclides (1 radioactive atom / molecule) t ½

1 10 100 1000

5 730 y

fMol

3 H 35 S 131 I 12.4 y 87 d 8 d The shorter the half-life, the hotter the radioligand.

Which analytical tools will be necessary to detect them?

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

Which analytical tools will be necessary to detect them?

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

rule of thumb

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

Which analytical tools will be necessary to detect them?

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

• 1 fMol = 10 -15 x 6

10 23 = 6

10 8 molecules

Which analytical tools will be necessary to detect them?

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

• 1 fMol = 10 -15 x 6

10 23 = 6

10 8 molecules • t ½ = 12.3 y = 4 500 d = 108 000 h = 6.48

10 6 min

Which analytical tools will be necessary to detect them?

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

• 1 fMol = 10 -15 x 6

10 23 = 6

10 8 molecules • t ½ = 12.3 y = 4 500 d = 108 000 h = 6.48

10 6 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.

Which analytical tools will be necessary to detect them?

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

• 1 fMol = 10 -15 x 6

10 23 = 6

10 8 molecules • t ½ = 12.3 y = 4 500 d = 108 000 h = 6.48 • dpm = 6

10 8

0.5

/ 6.48

10 6 = 46

Which analytical tools will be necessary to detect them?

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

• 1 fMol = 10 -15 x 6

10 23 = 6

10 8 molecules • t ½ = 12.3 y = 4 500 d = 108 000 h = 6.48 • dpm = 6

10 8

0.5

/ 6.48

10 6 = 46

100 10 6 min 80 60 40 t 1/2 = 12.3 years 20 linear decay 0 0 5 10 15 20 years 0.5

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

Which analytical tools will be necessary to detect them?

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

• 1 fMol = 10 -15 x 6

10 23 = 6

10 8 molecules • t ½ = 12.3 y = 4 500 d = 108 000 h = 6.48 • dpm = 6

10 8

ln2 / 6.48

10 6 = 64

100 10 6 min 80 t 1/2 = 12.3 years 60 40 exponential decay 20 linear decay 0 0 5 10 15 20 years 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

Why the natural logarithm of 2?

! attention ℮ ∫ ∑

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∂ ∞ √ % attention !

Which analytical tools will be necessary to detect them?

A … number of radioactive nuclei k … decay constant dA/dt = -k

A ∫(1/A)dA = -k

∫dt ln(A/A o ) = -k

Δt ! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

! attention ℮ ∫ ∑

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∂ ∞ √ % attention !

Which analytical tools will be necessary to detect them?

A … number of radioactive nuclei k … decay constant dA/dt = -k

A ∫(1/A)dA = -k

∫dt ln(A/A o ) = -k

Δt A = A o . e -k.

Δt ! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

! attention ℮ ∫ ∑

mathematics

∂ ∞ √ % attention !

Which analytical tools will be necessary to detect them?

A … number of radioactive nuclei k … decay constant k is related to t ½ : dA/dt = -k

A ∫(1/A)dA = -k

∫dt ln( A/A o ) = -k

Δt A = A o . e -k.

Δt ln( ½ ) = -k

t ½ k = ln2 / t ½ ! attention ℮ ∫ ∑

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Which analytical tools will be necessary to detect them?

A … number of radioactive nuclei k … decay constant k is related to t ½ : for 1 min ( Δt = 1): dA/dt = -k

A ∫(1/A)dA = -k

∫dt ln(A/A o ) = -k

Δt A = A o . e -k.

Δt ln(½) = -k

t ½ k = ln2 / t ½ ΔA = k

1 = ln2 / t ½

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Which analytical tools will be necessary to detect them?

Therefore, it can be expected that 6

10 8 tritium nuclei (1 fMol) will emit 6

10 8

ln2 / 6.48

10 6 = 64 electrons / min.

A molecule labeled with one single 3 H has a specific radioactivity (short:

specific activity

) of 64 dpm / fMol.

Which analytical tools will be necessary to detect them?

Therefore, it can be expected that 6

10 8 tritium nuclei (1 fMol) will emit 6

10 8

ln2 / 6.48

10 6 = 64 electrons / min.

A molecule labeled with one single 3 H has a specific radioactivity (short:

specific activity

) of 64 dpm / fMol.

Remember: 1 µCi = 2 220 000 dpm 1 nCi = 2 220 dpm 1 pCi = 2.22

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

Which analytical tools will be necessary to detect them?

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

Which analytical tools will be necessary to detect them?

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

14 C

Which analytical tools will be necessary to detect them?

Comparison of 3 H with other nuclides (1 radioactive atom / molecule) t ½

1 10 100 1000

5 730 y 0.14

1.4

14 140

fMol

3 H 35 S 131 I 12.4 y 87 d 8 d 64 3.3*10 3 36*10 3 640 33*10 3 360*10 3 6.4*10 3 330*10 3 3.6*10 6 64*10 3 3.3*10 6 36*10 6 dpm / mg tissue

14 C

Which analytical tools will be necessary to detect them?

Comparison of 3 H with other nuclides (1 radioactive atom / molecule) t ½

1 10 100 1000

5 730 y 0.14

1.4

14 140

fMol

3 H 35 S 131 I 12.4 y 87 d 8 d 64 3.3*10 3 36*10 3 640 33*10 3 360*10 3 6.4*10 3 330*10 3 3.6*10 6 64*10 3 3.3*10 6 36*10 6 dpm / mg tissue most common experimental condition

Properties of 3 H

• can replace 1 H present in every organic molecule • does not change the properties of the labeled molecule (no isotope effect) • • t ½ b 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 3 H

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*

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*

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*

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*

polyethylene glycol

n = 6 000 – 8 000

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*

Analytical techniques

centrifugation

• filtration • slice autoradiography B + L* BL*

Analytical techniques

filtration

• slice autoradiography B + L* BL*

Analytical techniques

slice autoradiography

B + L* BL* L* coating glass L* L* L* L* L* L* L* L* L* L* L*

Saturation & non-specific binding

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 T o S a v e t h is t e m p la t e ,C h o o s e F ile :T e m p la t e :T e m p la t e S a v e .

BL (fMol) 4000 3000 2000 favorable conditions 1000 0 0 500 no analysis possible 1000 L (nM) 1500 2000

Saturation & non-specific binding T o S a v e t h is t e m p la t e ,C h o o s e F ile :T e m p la t e :T e m p la t e S a v e .

BL (fMol) 4000 At low nM concentrations, most of the radioligand L is bound to saturable high affinity sites.

3000 2000 favorable conditions 1000 0 0 500 no analysis possible 1000 L (nM) 1500 2000

Saturation & non-specific binding T o S a v e t h is t e m p la t e ,C h o o s e F ile :T e m p la t e :T e m p la t e S a v e .

BL (fMol) 4000 At low nM concentrations, most of the radioligand L is bound to saturable high affinity sites.

3000 2000 favorable conditions 1000 0 0 500 no analysis possible 1000 L (nM) 1500 2000 At high concentrations, the linearly rising non-specific binding will dominate, and specific binding can no longer be detected.

BL 150 100 50 0 0

Saturation & non-specific binding

BL B + L K

= [B]

[L] [BL] K D : dissociation equilibrium constant 10 L 20 30 With increasing [L] more binding sites are occupied (BL) and free sites (B) are lost. The sum BL + B =

B M

remains constant.

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 10 L 20 30

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 10 L 20 30 K

= (

B M

- [BL])

[L] [BL] solve for [BL]:

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 10 L 20 30 K

= (

B M

- [BL])

[L] [BL] solve for [BL]: K D

[BL] =

B M .

[L] – [BL]

[L]

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 10 L 20 30 K

= (

B M

- [BL])

[L] [BL] solve for [BL]: K D

[BL] =

B M .

[L] – [BL]

[L] [BL] . ([L] + K D ) =

B M

. [L]

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 10 L 20 K

= (

B M

- [BL])

[L] [BL] solve for [BL]: 30 K D

[BL] =

B M .

[L] – [BL]

[L] [BL] . ([L] + K D ) =

B M

. [L]

Langmuir

isotherm [L] [BL] =

B M .

[L] + K D

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 10 L 20 Irving Langmuir 1881-1957 Nobel price 1932 K

= (

B M

- [BL])

[L] [BL] solve for [BL]: 30 K D

[BL] =

B M .

[L] – [BL]

[L] [BL] . ([L] + K D ) =

B M

. [L]

Langmuir

isotherm [L] [BL] =

B M .

[L] + K D

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 10 L 20 Irving Langmuir 1881-1957 Nobel price 1932 K

= (

B M

- [BL])

[L] [BL] solve for [BL]: 30 K D

[BL] =

B M .

[L] – [BL]

[L] [BL] . ([L] + K D ) =

B M

. [L]

Langmuir

isotherm [L] [BL] =

B M .

[L] + K D wrong!

BL 150 100 50 0 0

Saturation & non-specific binding

= [B]

[L] [BL] replace [B] by

B M

– [BL]: K

= (

B M

- [BL])

(L o – [BL]) [BL] 10 L 20 30 correct

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Saturation & non-specific binding

K D = [B]

[L] [BL] replace [B] by

B M

– [BL]: K D = (

B M

- [BL])

(L o – [BL]) [BL] solve for [BL]: K D

[BL] = B M

L o – B M

[BL] – [BL]

L o + [BL] 2 10 L 20 30 ! attention ℮ ∫ ∑

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BL 150 100 50 0 0 ! attention ℮ ∫ ∑

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Saturation & non-specific binding

10 L 20 K

= [B]

[L] [BL] replace [B] by

B M

– [BL]: 30 K

= (

B M

- [BL])

(L o – [BL]) [BL] solve for [BL]: K D

[BL] = B M

L o – B M

[BL] – [BL]

L o + [BL] 2 [BL] 2 – [BL]

( B M + L o + K D ) + B M

L o = 0 ! attention ℮ ∫ ∑

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BL 150

Saturation & non-specific binding

Sweet memories… [B]

[L] K

= [BL] replace [B] by

B M

– [BL]: 100 50 0 0 10 L 20 30 K

= (

B M

- [BL])

(L o – [BL]) [BL] solve for [BL]: K D

[BL] = B M

L o – B M

[BL] – [BL]

L o + [BL] 2 [BL] 2 – [BL]

( B M + L o + K D ) + B M

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BL 150

Saturation & non-specific binding

Sweet memories… [B]

[L] K

= [BL] replace [B] by

B M

– [BL]: 100 50 0 0 K

= (

B M

- [BL])

(L o – [BL]) [BL] 10 L solve for [BL]: K D

[BL] = B M

L o – B M

[BL] – [BL]

L o + [BL] 2 20 30 [BL] 2 – [BL]

( B M + L o + K D ) + B M

L o = 0 [BL] = ½

{ B M + L o + K D - [( B M + L o + K D ) 2 – 4

B M

L o ] ½ } ! attention ℮ ∫ ∑

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BL 150

Saturation & non-specific binding

Sweet memories… [B]

[L] K

= [BL] replace [B] by

B M

– [BL]: 100 50 0 0 K

= (

B M

- [BL])

(L o – [BL]) [BL] solve for [BL]: K D

[BL] = B M

L o – B M

[BL] – [BL]

L o + [BL] 2 10 20 30 [BL] 2 – [BL]

( B M + L o + K D ) + B M

L o = 0 L [BL] = ½

{ B M + L o + K D - [( B M + L o + K D ) 2 – 4

B M

L o ] ½ } In this case, quantities L o and K D are not entered as concentrations, but as moles in the respective volume chosen, in the same units as B M .

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BL 150

Saturation & non-specific binding

Sweet memories… [B]

[L] K

= [BL] replace [B] by

B M

– [BL]: 100 50 0 0 K

= (

B M

- [BL])

(L o – [BL]) [BL] solve for [BL]: 3 times more ligand than receptors at K D concentration (8% loss) 10 20 L 30 K D

[BL] 2 [BL] = B M – [BL]

(

B L M o – + L B M o

+ K [BL] – [BL]

D ) + B M

L o L o = 0 + [BL] [BL] = ½

{ B M + L o + K D - [( B M + L o + K D ) 2 – 4

B M

L o ] ½ } 2 In this case, quantities L o and K D are not entered as concentrations, but as moles in the respective volume chosen, in the same units as B M .

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BL 150

Saturation & non-specific binding

Sweet memories… [B]

[L] K

= [BL] replace [B] by

B M

– [BL]: 100 50 0 0 K

= (

B M

- [BL])

(L o – [BL]) [BL] solve for [BL]: 3 times more receptor than ligand at K D concentration (57% loss) 10 20 30 L K D

[BL] 2 [BL] = B M – [BL]

(

B L M o – + L B M o

+ K [BL] – [BL]

D ) + B M

L o L o = 0 + [BL] [BL] = ½

{ B M + L o + K D - [( B M + L o + K D ) 2 – 4

B M

L o ] ½ } 2 In this case, quantities L o and K D are not entered as concentrations, but as moles in the respective volume chosen, in the same units as B M .

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Saturation & non-specific binding

BL 150 100 50 0 0 A realistic saturation function is a composite of 2 simultaneous processes: 1: non-specific binding It is sufficient to measure 2 points; extrapolation of L  0 results in the blank of the measuring method (*).



10 L 20 

Saturation & non-specific binding

BL 150 100 50 0 0

10 A realistic saturation function is a composite of 2 simultaneous processes: 1: 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.

20 30 L

Saturation & non-specific binding

BL 150 100 50 0 0

10 A realistic saturation function is a composite of 2 simultaneous processes: 1: 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 ( †).

20 30 L

Saturation & non-specific binding

Mathematical combination of both processes: BL 150 100 50 0 0 † 

10 L 20 30



Saturation & non-specific binding

BL 150 100 50 0 0 Mathematical combination of both processes:

1: non-specific binding [BL] =

B U .

[L] [L] + K U † 

10 L 20 

Saturation & non-specific binding

BL 150 100 50 0 0 Mathematical combination of both processes: † 

10 L 20 

1: non-specific binding [BL] =

B U .

[L] [L] + K U 2: specific binding [L] [BL] =

B S .

[L] + K S

Saturation & non-specific binding

BL 150 100 50 0 0 Mathematical combination of both processes: † 

10 L 20 30



1: non-specific binding [BL] =

B U .

[L] [L] + K U 2: specific binding [L] [BL] =

B S .

[L] + K S K U (~mM) >> K s (nM)

Saturation & non-specific binding

Mathematical combination of both processes: 1: non-specific binding [L] BL 150

[BL] =

B U .

[L] + K U 2: specific binding 100 † [L] 

[BL] =

B S .

50  [L] + K S 0 0

10 20 30 K U (~mM) >> K s (nM) L At reasonable ligand concentrations, [L] + K U a linear function of [L]: ~ K U and non-specific binding is [L]

B U

[BL] =

B S .

[L] [L] + K S K U

Saturation & non-specific binding

BL 100 The most important value, the specific binding, is not directly accessible. It must be calculated by substracting the non-specific binding from total binding.

50 0 -9 NB -8 -7 -6 log[I] -5

Saturation & non-specific binding

BL 100 The most important value, the specific binding, is not directly accessible. It must be calculated by substracting the non-specific binding from total binding.

50 0 -9 NB -8 -7 -6 log[I] -5 The non-specific binding NB is measured as bound ligand that is impossible to displace, even by high concentrations of potent displacers.

Saturation & non-specific binding

Strategies to keep non-specific binding low:

Saturation & non-specific binding

Strategies to keep non-specific binding low: • choose a

biological source

with a high density of high-affinity binding sites

Saturation & non-specific binding

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 K D

( a few hundred to a few thousand dpm will be sufficient as result)

Saturation & non-specific binding

thin layer chromatography

Saturation & non-specific binding

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 K D ( 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

Saturation & non-specific binding

rinsing procedure

filters, respectively of pellets and

Radioligands for excitatory amino acid (EAA) receptors

Classification of glutamate receptors ionotropic receptors metabotropic receptors

Radioligands for excitatory amino acid (EAA) receptors

Classification of glutamate receptors ionotropic receptors metabotropic receptors NMDA receptors non-NMDA receptors Group I Group II Group III

Radioligands for excitatory amino acid (EAA) receptors

Classification of glutamate receptors ionotropic receptors metabotropic receptors NMDA receptors non-NMDA receptors Group I Group II Group III AMPA receptors kainate receptors

Radioligands for excitatory amino acid (EAA) receptors

Classification of glutamate receptors ionotropic receptors NMDA receptors non-NMDA receptors AMPA receptors kainate receptors Schmid et al (2009) PNAS 106:10320

Radioligands for excitatory amino acid (EAA) receptors

N H HOOC COOH N-methyl-D aspartic acid (NMDA) H 2 N COOH L-glutamic acid (S)-1-aminopropane-1,3 dicarboxylic acid COOH H 2 N COOH D-Aminophosphonovaleric acid PO 3 H 2 H 2 N COOH PO 3 H 2 CGP 39653 (E)-2-Amino-4-propyl-5 phosphono-3-pentenoic acid

Ca 2+ Na + Mg 2+ Mg 2+

outside membrane

-70 mV Mg 2+ Mg 2+

Ca 2+ Mg 2+ Na + Mg 2+ glu Mg 2+ gly

outside membrane

glu Mg 2+ AMPA receptor -70 mV

Ca 2+ Mg 2+ Na + Mg 2+ glu Mg 2+ gly

outside membrane

glu Mg 2+ -30 mV

Ca 2+ Mg 2+ Na + Mg 2+ glu Mg 2+ gly

outside membrane

glu Mg 2+ -10 mV

Ca 2+ Mg 2+ Na + Mg 2+ glu Mg 2+ gly

outside membrane

glu Mg 2+ 0 mV

Ca 2+ Mg 2+ Na + Mg 2+ glu Mg 2+ gly

outside membrane

glu Mg 2+ 0 mV

Radioligands for excitatory amino acid (EAA) receptors

Cl H 2 N COOH OH O N H O Cl Cl HOO C N H COOH Glycine L-701.324 ( a phenyl quinolinone) MDL-105.519 (an indole carboxylic acid) NH MK-801 *

Radioligands for excitatory amino acid (EAA) receptors

Cl H 2 N COOH OH O N H O Cl Cl HOO C N H COOH Glycine NH MK-801 * L-701.324 ( a phenyl quinolinone) * N N O MDL-105.519 (an indole carboxylic acid) [ 3 H]GSK-931.145

radioligand for the glycine transporter GlyT-1 (Herdon et al 2010 Neuropharmacol 59:558)

Radioligands for excitatory amino acid (EAA) receptors

O H N H OH H COOH H 2 N * * O N COOH OH kainic acid ( a pyrrolidine) AMPA ( a -

mino-3-hydroxy 5-

ethylisoxazol-4-

ropionic

cid) H 2 N COOH H COOH LY-354.740 (a bicyclo[3.1.0]hexan)

Radioligands for excitatory amino acid (EAA) receptors

HOOC H O H N H OH H COOH O H N H OH H COOH kainic acid ( a pyrrolidine) H 2 N * * O N COOH OH AMPA ( a -

mino-3-hydroxy 5-

ethylisoxazol-4-

ropionic

cid) Grant et al (2010) Neurotox Terat 32:132 H 2 N COOH H COOH LY-354.740 (a bicyclo[3.1.0]hexan)

Radioligands for excitatory amino acid (EAA) receptors

O H N H OH H COOH H 2 N H 2 N COOH O N OH O N OH Muscimol Ibotensäure H 2 N * * O N COOH OH H 2 N COOH H COOH kainic acid ( a pyrrolidine) AMPA ( 5-

a m

cid) a -

mino-3-hydroxy ethylisoxazol-4 LY-354.740 (a bicyclo[3.1.0]hexan)

ropionic

Radioligands for excitatory amino acid (EAA) receptors

O H N H OH H COOH H 2 N * * O N COOH OH kainic acid ( a pyrrolidine) AMPA ( a -

mino-3-hydroxy 5-

ethylisoxazol-4-

ropionic

cid) H 2 N COOH O S O H 2 N COOH O H COOH H COOH LY-404.039 LY-379.268

H 2 N COOH H COOH LY-354.740 (a bicyclo[3.1.0]hexan)

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*

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* ... applied to weak ligands (K D nM) > 20 ● 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 incubation vials.

to the rinsed

The most important binding techniques

... are all non-equilibrium techniques for particulate receptor preparations: B + L* BL* • Centrifugation •

Filtration over glass fiber filters

• Slice autoradiography ... Can only be applied to high affinity ligands (K D < 20 nM) ● you need a vacuum filter box

The most important binding techniques

... are all non-equilibrium techniques for particulate receptor preparations: B + L* BL* • Centrifugation •

Filtration over glass fiber filters

• Slice autoradiography ... Can only be applied to high affinity ligands (K D < 20 nM) ● you need a vacuum filter box or better a

harvester

● for radioligands with amino group, the glass fiber filter must be soaked polyethylenimine ● in 0.3%

The most important binding techniques

B + L* BL* ... are all non-equilibrium techniques for particulate receptor preparations: scintillator

L* L* L* L*

• Centrifugation •

Filtration over glass fiber filters

• Slice autoradiography ... Can only be applied to high affinity ligands (K D < 20 nM) ● you need a vacuum filter box or better a

harvester

● for radioligands with amino group, the glass fiber filter must be soaked polyethylenimine in 0.3% ● for best results, filter should be shaken in scintillation cocktail for 30 min.

The most important binding techniques

... are all non-equilibrium techniques for particulate receptor preparations: B + L* BL* L* coating glass L* L* L* L* L* L* L* L* L* L* L* • • Centrifugation • Filtration over glass fiber filters

Slice autoradiography

The most important binding techniques

... are all non-equilibrium techniques for particulate receptor preparations: B + L* BL* • • Centrifugation • Filtration over glass fiber filters

Slice autoradiography

The most important binding techniques

... are all non-equilibrium techniques for particulate receptor preparations: B + L* BL* • • Centrifugation • Filtration over glass fiber filters

Slice autoradiography

... 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 phosphoscreen slices to film or ● evaluation by co exposure of stripes containing known amounts of radioactivity.

Document 7507642

Transcript Document 7507642

Ligand/Receptor Interaction

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