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YEAR 1 - Physiology and Pharmacology (Anton) > 4. Principles Of Drug (& Hormone) Action (HT) > Flashcards

4. Principles Of Drug (& Hormone) Action (HT) Flashcards

1
Q

Define drug selectivity and specificity.

A
  • Selectivity - The degree to which the drug acts on a given target compared to other targets.
  • Specificity - Relates to the number of different mechanisms involved. Examples of specific drugs include atropine (a muscarinic receptor antagonist). A non-selective drug has its effect by multiple mechanisms.
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2
Q

Can a drug be completely specific?

A

No, so increasing the dose of drugs may result in binding to off-target molecules.

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3
Q

What equations represent agonist-receptor and antagonist-receptor interactions?

A
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4
Q

What forces affect the affinity of a drug for a receptor?

A
  • Electrostatic forces
  • Hydrogen bonds
  • Van der Waals forces
  • Hydrophobic bonds
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5
Q

The probability that a drug occupies the binding site depends on…

A
  • Affinity
  • Drug concentration
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6
Q

State the law of mass action.

A

The rate of a reaction is proportional to the concentrations of the reacting substances.

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7
Q

In drug interaction calculations, what symbols may be used for:

  • Unoccupied receptors
  • Agonists
  • Occupied receptors
  • Antagonists
A
  • Unoccupied receptors - R
  • Agonists - A
  • Occupied receptors - AR
  • Antagonists - B
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8
Q

What are these symbols and what is the relationship between them: Ntot , NAR , NR

A

Ntot = NAR + NR

  • Ntot = Total number of receptors
  • NAR = Number of occupied receptors
  • NR = Number of unoccupied receptors
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9
Q

What are these symbols and what is the relationship between them: [R]tot , [AR] , [R]

A

[R]tot = [AR] + [R]

  • [R]tot = Concentration of total number of receptors
  • [AR] = Concentration of occupied receptors
  • [R] = Concentration of unoccupied receptors

However, this is assuming that we know the concentrations of the receptor, which is only the case if the receptor is soluble and we know its molecular weight. Many receptors are embedded in the membrane and are therefore not soluble.

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10
Q

What are these symbols and what is the relationship between them: pAR , pR

A

1 = pAR + pR

  • pAR = Proportion of receptors that are occupied
  • pR = Proportion of receptors that are unoccupied
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11
Q

What are pAR and pR equal to?

A
  • pAR = NAR / Ntot
  • pR = NR / Ntot
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12
Q

What does the p in pAR stand for?

A
  • It stands for proportion
  • However, it is usally referred to as occupancy
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13
Q

In pharmacology, is it preferable to think in terms of occupanies and absolute receptor numbers (e.g. NAR = pAR x Ntot) or receptor concentrations (e.g. [R] and [AR])?

A

It is preferable to think in terms of occupanies and absolute receptor numbers because receptor concentrations may be difficult (or impossible) to define.

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14
Q

Describe how the equation Ntot = NAR + NR can be used to derive other equations.

A
  • Ntot = NAR + NR
  • Dividing each term by the volume (V) gives concentrations:
  • [R]tot = [AR] + [R]
  • However, it is difficult to determine concentrations since part of the drug may not be soluble or we might not know the molecular weight. Therefore, we can also divide the first equation by the total number of receptors (Ntot):
  • 1 = pAR + pR
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15
Q

For a unimolecular transition (involving only one reactant and product), what are the units for the forwards rate constant?

A

s-1

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16
Q

For a bimolecular transition (involving one reactant and two products), what are the units for the forwards rate constant?

A

M-1s-1

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17
Q

What is the symbol for the forwards and backwards rate constant?

A
  • Forwards: k+1 or kon
  • Backwards: k-1 or koff
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18
Q

Write an equation for the rate of the forwards reaction for these reactions.

A
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19
Q

Write the expressions for the rate of the forward and backward reactions.

A
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20
Q

What is the equilibrium dissociation constant in drug binding, what are the units and what is the symbol for it?

A
  • KA
  • It is equal to: KA = k-1 / k+1
  • Units: M (or mol/L)

It is the concentration of the drug which results in 50% of the receptors being occupied.

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21
Q

State the Hill-Langmuir equation.

A
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22
Q

Derive the Hill-Langmuir equation.

A

Assuming equilibirum, when the rate of the forward reaction is the same as the rate of the backward reaction:

23
Q

What are the units of the KA (equilibrium dissociation constant)?

A

M (or mol/L)

24
Q

Interpret what the KA (equilibrium dissociation constant) in drug binding tells you practically and why.

A
  • It is the concentration of the drug which results in 50% of the receptors being occupied
  • This can be seen when [A] is equal to [KA] in the Hill-Langmuir equation, so that pAR = 0.5
  • It is a measure of agonist affinity for the given receptor
25
Q

The higher the affinity of the drug for the receptor, the … the value of KA.

A

Lower

26
Q

How can KA be determined from a graph of pAR against drug concentration?

A

It is the drug concentration that results in half of the maximal pAR.

27
Q

Draw the arithmetic and semi-log graphs of receptor occupancy against drug concentration.

A
28
Q

When plotting a graph of receptor occupancy against drug concentration, what is the advantage of a semi-log plot?

A
  • Easy to see the midpoint (for calculating KA)
  • Wide range of concentrations can be seen
29
Q

Describe how the relationship between the amount of drug bound and drug concentration may be demonstrated graphically.

A
30
Q

What is a Scatchard plot and what can be found from it?

A

It is a graphical representation of: B/[A] = Bmax/KA - B/KA

31
Q

What is non-specific drug binding?

A

The binding of a ligand to something other than its designated receptor such as various other receptors, or different types of transporters in the cell membrane.

32
Q

Describe how total binding, non-specific binding and specific binding can be investigated experimentally.

Include graphs.

A

Total binding:

  • Use radioactively-labelled drugs to bind to homogenous receptors.
  • Wash away any excess drug molecules.
  • Then measure the remaining radioactivity to estimate binding.

Non-specific binding:

  • Non-specific refers to the binding of a ligand to something other than its designated receptor such as various other receptors, or different types of transporters in the cell membrane
  • It can be determined by using a saturating concentration of non-labelled drugs to bind to homogenous receptors, then adding the labelled drug.
  • After washing out the unbound drugs, the bound radioactivity is measured, which is a measure of non-specific binding.

Specific binding:

  • This is equal to the total binding minus the non-specific binding.
33
Q

Give the equation for the response to a drug at a given concentration.

A

Note how this is the Hill-Langmuir equation multiplied by the efficacy.

34
Q

What is the EC50 and what does it depend on?

A
  • It is the drug concentration that produces the half-maximal response
  • It depends on both the affinity and efficacy
35
Q

In drug binding and response studies, what is the difference between the KA and EC50?

A
  • KA is the drug concentration required to produce half-maximal receptor occupancy
  • EC50 is the drug concentration required to produce half-maximal response
36
Q

Is EC50 or KA usually lower? Why?

A
  • EC50 is usually lower
  • This is due to “spare receptors”
    • Number of receptors present is larger than the number required to produce the full response
    • Duration of the cell/tissue/organ response is longer than the duration of the drug-receptor interaction (so once part of the drug has unbound, there may still be a full response)
37
Q

Do full agonists all have the same KA?

A

No, they may have different KA values.

38
Q

What are the 3 types of antagonist you need to know about?

A
  • Competitive reversivle antagonist
  • Competitive irreversible antagonist
  • Non-competitive antagonist
39
Q

How does a competitive reversible antagonist work and how does it affect the dose-response curve of the agonist?

A
  • Binds to the receptor but triggers no response (has affinity, but no efficacy)
  • Prevents the binding of the agonist
  • Shifts the dose-response curve to the right, so that EC50 is increased, but the maximal response is unchanged
40
Q

What is rA?

A
  • The dose ratio
  • It is the ratio of the EC50 of the dose-response curves for an agonist with and without a competitive reversible antagonist present
  • It is the ratio by which [A] has to increase to overcome the competition by B
41
Q

Write an equation for the receptor occupancy for a receptor with both an agonist and competitive reversible antagonist present.

A
42
Q

What indicates the strength of a competitive reversible antagonist?

A
  • The KB - the equilibrium constant for the antagonist binding to the receptor.
  • Experimentally, the KB quantifies the affinity of an antagonist for a receptor, with a low KB value indicating a strong antagonist.
43
Q

Give an equation for the rA (not in terms of EC50s).

A

rA = ([B]/KB)+1

44
Q

A low KB value indicates a … antagonist.

A

Strong

45
Q

What two factors does rA depend on and what does this show?

A

As per the equation rA = ([B]/KB) + 1 :

  • [B] - The concentration of the antagonist
  • KB

This shows that KB quantifies the strength of the antagonist. Note that rA does not depend on [A] or KA.

46
Q

State the Schild equation.

A

log(rA - 1) = log[B] - logKB

47
Q

How can the strength of a competitive reversible antagonist by determined experimentally?

A
  • The KB value is a measure of antagonist strength (low KB = high strength)

Two ways of calculating KB:

  • Plotting two semi-logarithmic dose-response curves; one without the antagonist and one with a known concentration of antagonist. By comparing the two graphs’ EC50 values, the rA can be calculated, after which the KB can be calculated using rA = [B]/KB + 1.
  • In order to obtain higher accuracy in the value for KB, a Schild plot can be plotted, which is a double logarithmic graph of log(rA - 1) against log[B], obtained by rearranging the previous equation. It is a graphical representation of log(rA - 1) = log[B] - log(KB). A regression line is drawn and the y-intercept is equal to -log(KB), from which the potency of the antagonist can be deduced.
48
Q

Draw the two graphs involved in drawing a Schild plot.

A

The graph on the left shows the data gathered to plot the Schild plot on the right.

49
Q

How does a competitive irreversible antagonist work and how does it affect the dose-response curve of the agonist?

A
  • Forms covalent bonds with the receptor binding site, so that it is bound irreversibly and the agonist cannot bind
  • Graph shifts downwards and never reaches maximal response
  • EC50 is not affected
50
Q

Does increasing the agonist concentration overcome the effects of a competitive irreversible antagonist?

A

No, because the antagonist does not unbind after it has bound.

51
Q

How does a non-competitive antagonist work and how does it affect the dose-response curve of the agonist?

A
  • Bind to a site separate from the agonist binding site
  • The curve shifts downwards, so that the maximal response is not reached
  • The EC50 may or may not be increased
52
Q

Explain constitutive receptor activity.

A
  • A receptor which is capable of producing a biological response in the absence of a bound ligand is said to display “constitutive activity”.
  • This occurs because a receptor may exist in many potential conformations, which can include an active and inactive form.
  • The constitutive activity of a receptor may be blocked by an inverse agonist, which shifts the receptor to the inactive form.
53
Q

On what type of receptors may an inverse agonist act and what is the effect on the dose-response curve?

A
  • It acts on receptors with constitutive activity (where the drug displays baseline activity without any drug bound)

YEAR 1 - Physiology and Pharmacology (Anton) (13 decks)

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