Pharmacokinetics + Ligand Binding Flashcards
Spare receptors/Receptor reserve
Shown by ‘parallel shifting’ on a graph of response vs agonist concentration, when increasing amounts of NON-competitive antagonist is added (e.g. BCM which irreversibly alkylates muscarinic receptor). That means maximum response can still be reached in the presence of a NON-competitive antagonist if there are enough spare receptors.
Remember that partial agonist acts as a competitive antagonist, NOT a non-competitive antagonist.
With INCREASING doses of antagonist, maximum response will eventually be no longer attainable.
By measuring the dose ratio at the point at which the maximum response is just attainable, it is possible to estimate the percentage of receptors that are ‘spare’.
Existence of spare receptors increases the sensitivity of the tissue to the agonist.
Why dose ratio works
Makes no assumptions about the relationship between response and receptor occupancy. It only supposes that equal responses are produced by equal agonist occupancies.
EC50
Concentration of agonist that causes 50% of the maximum response.
Potency
Concentration of an agonist causing a particular magnitude of response.
E.g. A is more potent than B, A needs half the concentration of B to cause 20% of maximum response.
Note that potency is usually compared by comparing EC50 levels.
Stereoisomers can vary greatly in potency.
Competitive antagonist
With a FIXED concentration of COMpetitive antagonist, log agonist concentration vs response curve (which is SIGMOID) shifts to the right. Antagonist is SURMOUNTABLE simply by adding lots of agonist to compete off the antagonist.
Overcoming blockade by NON-competitive antagonist requires spare receptors, not simply more agonist.
Note that agonist concentration (NO LOG) vs response curve is hyperbolic.
Structure activity series
Determines the relative potencies (usually relative EC50 values) of structurally related compounds.
This aids receptor classification.
Dissociation constant (Kd)
Units M (mol/L). Ka units is M^-1 (L/mol). Kd=Bmax/2 So Kd is the concentration of agonist that produces 50% of the maximum specific binding/occupancy (maximum specific binding means all spare receptors are occupied).
Radioligand binding and Bmax
Binding of a radioligand (total binding) consists of a saturable component that - hyperbolic line (specific binding to the receptors) and an effectively linear non-saturable component (non-specific binding to non-receptor material).
In the presence of a large excess of unlabelled ligand the specific binding of the radioligand is almost completely abolished, but non-specific binding is almost unaffected.
So to find specific binding, use total binding minus non-specific binding.
Maximum specific binding (a.k.a binding site density) is Bmax. So Kd = agonist concentration at Bmax/2.
Affinity of competitive antagonist ‘K2’
In the presence of a FIXED concentration of competitive antagonist, a higher concentration ‘[D]2’ of agonist will be needed to occupy the same fraction of receptors and hence produce the SAME response.
So if the question gives you ‘response’ e.g. 50mm contraction and use of antagonist, then think dose ratio.
Affinity of a competitive ligand in radioligand experiment
FIXED concentration of radioligand is added, and then VARYING concentrations of unlabelled drug is added to displace the radioligand.
IC50 = Concentration of unlabelled drug that displaces 50% of SPECIFICALLY bound radioligand. When [U] = IC50, α=0.5 (fraction of receptor bound by unlabelled ligand).
Graph of bound radioligand vs LOG concentration of unlabelled ligand is sigmoid.
Partial agonist
On its own, it’s a weak agonist, usually with a low affinity constant. But this is not always true.
Partial agonist ALWAYS have a lower EFFICACY than full agonist.
When presented together with a full agonist, partial agonist acts as a COMPETITIVE antagonist (which can be competed off by high agonist concentrations - surmountable inhibition).
Hill slope
Does NOT equate to number of binding site per receptor. But it does suggest that there’s 1 binding site per receptor, since if there’s more than one binding site, binding sites are cooperative and Hill slope will be more than 1. (from supervision)
Equal to the number of molecules that must bind a receptor in order for it to open.
3 types of ligand binding question that can be asked
(1) Dose ratio question - NO LABELLED stuff. Look for changing amounts of COMPETITIVE antagonist in one column and changing amounts of agonist needed to produce SAME RESPONSE.
(2) IC50 question - Adding variable amounts of unlabelled drug to compete off a fixed amount of labelled drug. Look for column showing CHANGING AMOUNTS OF UNLABELLED DRUG.
(3) Scatchard plot - increasing amount of labelled drug and a fixed LARGE amount of unlabelled drug to remove non-specific binding. Look for column showing CHANGING AMOUNTS OF LABELLED DRUG and a LARGE amount of unlabelled drug.
Note on Bmax
Bmax can be given in mol of binding sites/gram of protein (see mock exam example).
Can then convert into mol/cell (question will give amount of protein in a cell).
Can also convert to number of binding sites/cell. Need to multiply by Avogardro’s constant (6.022x10^23).
First order kinetics/Linear elimination
Rate of elimination is directly proportional to plasma concentration.
Renal elimination by filtration ALWAYS follows first order kinetics, since it’s not enzyme dependent and CAN’T be saturated. But secretion via transporters can get saturated.
In 1st order kinetics, CLEARANCE IS CONSTANT. Also, route of administration will NOT affect clearance (e.g. take grapefruit juice and then inject by I.V. or take pill. Clearance for both should be the same).
Single compartment model
Absorption and distribution ignored.
Volume of distribution constant throughout.
