At first glance, the simplest approach to determining how quickly a drug combines with its receptors might seem to be to measure the rate at which it acts on an isolated tissue, but two immediate problems arise. The first is that the exact relationship between the effect on a tissue and the proportion of receptors occupied by the drug is often not known and cannot be assumed to be simple, as we have already seen. A maximal tissue response only rarely corresponds to half-maximal receptor occupation. We can take as an example the action of the neuromuscular blocking agent tubocurarine on the contractions that result from stimulation of the motor nerve supply to skeletal muscle in vitro. The rat phrenic nerve–diaphragm preparation is often used in such exper-iments. Because neuromuscular transmission normally has a large safety margin, the contractile response to nerve stimulation begins to fall only when tubocurarine has occupied on average more than 80% of the binding sites on the nicotinic acetylcholine receptors located on the superficial
logit[ ]p loge p
= p
−
⎛
⎝⎜ ⎞
⎠⎟ 1
p= e a bx + − +
1
1 ( )
p
p ea bx 1 − = +
logit[ ]p log p
p a bx
= e
−
⎛
⎝⎜ ⎞
⎠⎟ = + 1
p z
K z
b
= b
+
muscle fibers. So, when the twitch of the whole muscle has fallen to half its initial amplitude, receptor occupancy by tubocurarine in the surface fibers is much greater than 50%.
The second complication is that the rate at which a ligand acts on an isolated tissue is often determined by the diffusion of ligand molecules through the tissue rather than by their combination with the receptors. Again taking as our example the action of tubocurarine on the isolated diaphragm, the slow development of the block reflects not the rate of binding to the receptors but rather the failure of neuromuscular transmission in an increasing number of individual muscle fibers as tubocurarine slowly diffuses between the closely packed fibers into the interior of the preparation.
Moreover, as an individual ligand molecule passes deeper into the tissue, it may bind and unbind several times (and for different periods) to a variety of sites (including receptors). This repeated binding and dissociation can greatly slow diffusion into and out of the tissue.
For these reasons, kinetic measurements are now usually done with isolated cells (e.g., a single neuron or a muscle fiber) or even a patch of cell membrane held on the tip of a suitable microelec-trode. Another approach is to work with a cell membrane preparation and examine directly the rate at which a suitable radioligand combines with, or dissociates from, the receptors that the membrane carries. Our next task is to consider what binding kinetics might be expected under such conditions.
1.3.2 INCREASESIN RECEPTOR OCCUPANCY
In the following discussion, we continue with the simple model for the combination of a ligand with its binding sites that was introduced in Section 1.2.1 (Eq. (1.1)). Assuming as before that the law of mass action applies, the rate at which receptor occupancy (pAR) changes with time should be given by the expression:
(1.18)
In words, this states that the rate of change of occupancy is simply the difference between the rate at which ligand–receptor complexes are formed and the rate at which they break down.*
At first sight, Eq. (1.18) looks difficult to solve because there are no less than four variables:
pAR, t, [A], and pR. However, we know that pR = (1 – pAR). Also, we will assume, as before, that [A] remains constant; that is, so much A is present in relation to the number of binding sites that the combination of some of it with the sites will not appreciably reduce the overall concentration.
Hence, only pAR and t remain as variables, and the equation becomes easier to handle.
Substituting for pR, we have:
(1.19)
Rearranging terms,
(1.20)
This still looks rather complicated, so we will drop the subscript from pAR and make the following substitutions for the constants in the equation:
* If the reader is new to calculus or not at ease with it, a slim volume (Calculus Made Easy) by Silvanus P. Thompson is strongly recommended.
d
dAR A R AR
( ) p [ ]
t =k+1 p −k p−1
d
dAR A AR AR
( )
[ ]( )
p
t =k+1 1−p −k p−1
d
dAR A A AR
( )
[ ] ( [ ])
p
t =k+1 − k−1+k+1 p
a = k+1[A]
b = k–1 + k+1[A]
Hence,
This can be rearranged to a standard form that is easily integrated to determine how the occupancy changes with time:
Integrating,
We can now consider how quickly occupancy rises after the ligand is first applied, at time zero (t1 = 0). Receptor occupancy is initially 0, so that p1 is 0. Thereafter, occupancy increases steadily and will be denoted pAR(t) at time t:
Hence,
Replacing a and b by the original terms, we have:
(1.21)
Recalling that k–1/k+1 = KA, we can write:
t1 = 0 p1 = 0 t2 = t p2 = pAR(t)
d d p
t = −a bp
dp d
a bp t
t t p
p
− =
∫
∫
12
1 2
loge a bp ( )
a bp− b t t
−
⎛
⎝⎜ ⎞
⎠⎟ = − −
2 1
2 1
log ( )
( )
( ) ( )
e
bt
bt
a bp t
a bt
a bp t
a e
p t a
b e
⎧ −
⎨⎩
⎫⎬
⎭= −
− =
= −
−
− AR
AR
AR 1
p t k
k k e k k t
AR
A A
( ) [ ]A [ ]
( [ ])
= ++
{
−}
− +
− −++ 1
1 1
1 1 1
p t
K e k k t
AR
A
A A
( ) [ ]A [ ]
( [ ])
= +
{
1− − −1++1}
When t is very great, the ligand and its binding sites come into equilibrium. The term in large brackets then becomes unity (because e–∞ = 0) so that
We can then write:
(1.22) This is the expression we need. It has been plotted in Figure 1.3 for three concentrations of A.
Note how the rate of approach to equilibrium increases as [A] becomes greater. This is because the time course is determined by (k–1 + k+1[A]). This quantity is sometimes replaced by a single constant, so that Eq. (1.22) can be rewritten as either:
pAR(t) = pAR(∞)(1 – e–λt) (1.23)
or
pAR(t) = pAR(∞)(1 – e–t/τ) (1.24)
where
λ = k–1 + k+1[A] = 1/τ
where τ (tau) is the time constant and has the unit of time; λ (lambda) is the rate constant, which is sometimes written as kon (as in Chapter 5) and has the unit of time–1.
FIGURE 1.3 The predicted time course of the rise in receptor occupancy following the application of a ligand at the three concentrations shown. The curves have been drawn according to Eq. (1.22), using a value of 2 × 106 M–1sec–1 for k+1 and of 1 sec–1 for k–1.
pAR KA A ( ) [ ]A
∞ = [ ] +
pAR( )t =pAR( ){∞ 1−e−(k−1+k+1[A])t}
1.3.3 FALLSIN RECEPTOR OCCUPANCY
Earlier, we had assumed for simplicity that the occupancy was zero when the ligand was first applied. It is straightforward to extend the derivation to predict how the occupancy will change with time even if it is not initially zero. We alter the limits of integration to
Here, pAR(0) is the occupancy at time zero, and the other terms are as previously defined.
Exactly the same steps as before then lead to the following expression to replace Eq. (1.22):
(1.25) We can use this to examine what would happen if the ligand is rapidly removed. This is equivalent to setting [A] abruptly to zero, at time zero, and p(∞) also becomes zero because eventually all the ligand receptor complexes will dissociate. Eq. (1.25) then reduces to:
(1.26) This expression has been plotted in Figure 1.4.
The time constant, τ, for the decline in occupancy is simply the reciprocal of k–1. A related term is the half-time (t1/2). This is the time needed for the quantity (pAR(t) in this example) to reach halfway between the initial and the final value and is given by:
For the example illustrated in Figure 1.4, t1/2 = 0.693 sec. Note that τ and t1/2 have the unit of time, as compared with time–1 for k–1.
FIGURE 1.4 The predicted time course of the decline in binding-site occupancy. The lines have been plotted using Eq. (1.26), taking k–1 to be 1 sec–1 and pAR(0) to be 0.8. A linear scale for pAR(t) has been used on the left, and a logarithmic one on the right.
t1 = 0 p1 = pAR(0) t2 = t p2 = pAR(t)
pAR( )t =pAR( ) {∞ + pAR( )0 −pAR( )}∞ e−(k−1+k+1[ ])A t
pAR( )t = pAR( )0e−k t−1
t1 2 k1 0 693
/
= .
−
It has been assumed in this introductory account that so many binding sites are present that the average number occupied will rise or fall smoothly with time after a change in ligand concentration; events at single sites have not been considered. When a ligand is abruptly removed, the period for which an individual binding site remains occupied will, of course, vary from site to site, just as do the lifetimes of individual atoms in a sample of an element subject to radioactive decay. It can be shown that the median lifetime of the occupancy of individual sites is given by 0.693/k–1. The mean lifetime is 1/k–1. The introduction of the single-channel recording method has made it possible to obtain direct evidence about the duration of receptor occupancy (see Chapter 6).