The Relationship Between Dose & Response

Question 1

What 3 things does the relationship between dose & response depend on?

Answer 1

1. Drug catabolism (how the drug is broken down by metabolic processes)
2. The relationship between drug binding & tissue response
3. The concentration-dependence of the drug binding to its site of action (the relationship between drug concentration and receptor occupancy)

Question 2

What is the law of mass action?

Answer 2

The rate of a reaction is proportional to the concentration of its reactants.

    k+1  A+R ——> AR Rate of formation of AR is directly proportional to [A][R] Rate = k+1 [A][R]

    k-1 AR ——> A+R  Rate of breakdown of AR is directly proportional to [AR] Rate = k-1 [AR]

Question 3

Which relationship does the law of mass action apply to?

Answer 3

The relationship between drug concentration & receptor occupancy

Question 4

What is the Hill-Langmuir equation?

Answer 4

pAR = [A] / ([A] + KA)

Question 5

What is KA, what are its units and how do you calculate it?

Answer 5

KA = dissociation equilibrium constant

Units of concentration (e.g. M)

KA = K-1/K+1
(KA = dissociation rate constant / association rate constant)

Question 6

What is K+1, what are its units?

Answer 6

K+1 is the association rate constant (rate constant for the formation of AR)

Units - M-1 s-1

Question 7

What is K-1, what is its units?

Answer 7

K-1 is the dissociation rate constant (the rate constant for the breakdown of AR)

Units - s-1

Question 8

What is everything in the law of mass equation?

Answer 8

k+1
A + R <———-> AR
k-1

Rate of formation of AR = k+1 [A][R]
Rate of breakdown of AR = k-1 [AR]

[A] = conc of unbound (free) agonist
[R] = conc of (free) receptor with no ligand bound
k+1 = association rate constant (units = M^-1, s^-1)
k-1 = dissociation rate constant (units = s^-1)

Question 9

What is the the equation for the law of mass action at equilibrium?

Answer 9

k+1 [A][R] = k-1 [AR]

Question 10

Why do we find the PROPORTION of activated receptors ?

Answer 10

We find the proportion instead of the concentration of activated receptor, as regardless of the concentration, if the proportion is small, then the concentration is unlikely to produce any response

Question 11

How do we find the proportion of active receptors?

Answer 11

[R]T = [R] + [AR]
[Total Receptors] = [free receptors] + [agonist-bound receptors]

pR = [R]/[R]T
Proportion of free receptors = [free receptors] / [total receptors]

pAR = [AR] / [R]T
Proportion of active receptors = [agonist-bound receptors] / [total receptors]

Question 12

How do we re-write the equation of the law of mass action at equilibrium using the proportion of receptors instead of the concentrations?

Answer 12

k+1 [A][R] = k-1 [AR]

—> k+1 [A] pR = k-1 pAR

Question 13

Why is there no clear distinction between the free concentration of agonist [A] and the total concentration of agonist [A]T?

Answer 13

Because this equation is applied to cases where [A] doesn’t significantly change as drug receptor complexes are formed
The drug is considered to be in EXCESS so that the concentration doesn’t change that much if a small amount of it combines with the receptors
So [A] is known

Question 14

How does the logarithmic vs. Linear scale of graphs of receptor occupancy over [A] differ?

Answer 14

Logarithmic - produces a sigmoid-shaped curve
Linear - slope decreases as [A] increases and the curve forms part of a rectangular hyperbola

Question 15

What is everything in the equations:

[R]T = [R] + [AR]
pR = [R] / [R]T
pAR = [AR] / [R]T

Answer 15

[R]T = total conc of receptors
pR = proportion of free (unbound) receptors
pAR = proportion of receptors with agonist bound

Question 16

Write out how the Hill-Langmuir equation is derived

Question 17

What is the equilibrium ASSOCIATION constant?

Answer 17

The inverse of the equilibrium dissociation constant:
K’A = k+1/ k-1

Question 18

Why is [A] often assumed to be approximately equal to [A]T?

Answer 18

We assume that the drug is in excess so that the combination of some molecules with receptors does not significantly change the concentration, in such a way that

[A]+[AR] = [A]T ~ [A]

Question 19

When [A] = KA, what does pAR equal?

Answer 19

When [A] = KA…

pAR = 0.5

Question 20

What is the Hill plot?

Answer 20

Where the Hill-Langmuir equation is rearranged to give:
log(pAR / 1-pAR) = log[A] - logKA

X axis = log[A]
Y axis = log(pAR / 1-pAR)
Y-intercept = -logKA

Question 21

What is the gradient of the line formed in the Hill plot, in theory, and what is it called?

Answer 21

1 (unity)
Called the Hill coefficient / slope

Question 22

In what type of response is the Hill coefficient almost always above unity (m>1)?

Answer 22

In responses mediated by ligand-gated ion channels