3. Quantification of drug-target interaction Flashcards

1
Q

What does the potency of a drug depend on?

A

The strength of the interaction with its target

Must consider

  • Kinetics (reversible/irreversible binding)
  • Binding site (orthosteric/allosteric)
  • Functional effect (activation/inhibition, competetive vs non/uncompetitive inhibition)
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2
Q

Define the molecular mechanisms of action and its paramaters

A

MMOA of a drug is the interaction between a drug and a target that leads to a specific outcome

Must consider:

  • Kinetics (reversible/irreversible binding)
  • Binding site (orthosteric/allosteric)
  • Functional effect (activation/inhibition, competetive vs non/uncompetitive inhibition)
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3
Q

What matters when componds have slow kiinetics?

A

Residence time

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4
Q

Define the 4 types of conformational mechanisms

A

Uncompetitive inhibition

The inhibitor binds the enzyme/receptor only when the natural ligand is bound. Potency is therefore proportional to the physiological concentration of the ligand.\

Noncompetitive inhibition

The inhibitor reduces the activity of the enzyme/receptor and binds to an allosteric site, whether it has the ligand bound or not

Agonism

Binding of compound activates the receptor or enzyme, can be full or partial

Allosteric modulation

Binding of compound will modiy the activity of a receptor or an enzyme, thereby shifting its substrate selectivity profile

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5
Q

Define the equilibium state in terms of ligand/receptor interaction

How is the EQUILIBRIUM dissociation constant calculated?

What is receptor occupancy and how is it calculated?

A

Equilibrium is when the rate of complex association is equal to the rate of dissociation

The equilibrium dissociation constant (Kd) is the ratio of kinetic constants k(off) and k(on)

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6
Q

Why can we assume the receptor occupancy is a function of just the total ligand concentration and Kd?

How can one calculate Y when this is the case?

What happens when Y is 50%?

What is cooperativity and how does it affect the function for Y?

A

Because receptor concentration is much lower than either the ligand concentration or Kd

Y = [Ltotal]/[Ltotal]+Kd

0.5 = X/X+Y

Y = X, Kd = [Ltotal]

Cooperativity occurs when ligands binding at one site influences binding at another site (eg hemoglobin)

Same equation as above, but all 3 terms are raised to the hill coefficient n, where positive cooperativity occurs when n > 1 and negative cooperativity occurs when n < 1

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7
Q

What is KI?

How do reversible and irreversible inhibition differ? What are their effects on the equation for Y?

A

Koff/Kon but for an inhibitor

Reversible: produce a binding isotherm with apparent equilibrium constant Kapp, which depends on both KI, KL (equilibrium constant of ligand), and [I]

Y = [L]/[L] + Kapp

Kapp = KL(1 + ([I]/KI))

Irreversible: Reduce proportion of active receptors without changing ligand KL

Y = [L]/([L] + KL) * [Runmodified]/[Rtotal]

Reversible inhibitors keep the same shape, but shift if to the right (KD -> Kapp)

Irreversible inhibitors keep the same KD, but Lower the maximal response

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8
Q

Define a two-state model of receptor activation

How do you find the Kapp for a two state model when the ligand is an agonist? An inverse agonist? A partial agonist? A neutral agonist?

A

Two-state receptors have two states, one that binds agonist (RA) and one that does not bind (RI)

An agonist will typically bind the active state with a constant Kd, whifting the equilibrium towards the active form

Kint = [RA]/[RI]

Kapp = Kd (1+1/Kint) for agonist

An invers agonist binds the inactive state, shifting the equilibrium away from the active form

Kapp = Kd (1+Kint) for inverse agonist

A partial agonist binds both forms, with different affinities (end result depends on the affinities for each state)

Kapp = (Kint + 1)/(Kint/KA + 1/KI) for partial agonist

Where KA = [RA][A]/[RA:A], KI = [RI][A]/[RI:A], and neither can be inferred from receptor activation curves

A neutral agonist binds equally well to active and inactive conformations, but will antagonize the action of another agonist (competitive inhibition)

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9
Q

How can enzyme kinetics be described (i.e. with what equation)? What does this equation relate? What are its 3 assumptions?

A

Michaelis-Menten equation, relates initial rate of product formation (v0) to the substrate concentration

(The initial rate of an enzymatic reaction depends on how much substrate is present)

v0 = Vmax[S]/KM+[S]

Where KM is the [S] where V0 is 0.5 * Vmax

and Vmax = kcat[E][S]

Assumptions:

Free diffusion

Enzyme concentration much smaller than substrate concentration or KM

Irreversibility [S] >> [P], dG << 0

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10
Q

How does a competitive inhibitor affect enzyme kinetics?
What does it depend on?

A

The inhibitor prohibits the binding of the natural substrate (only E, E:S or E:I can exist)

There will be an increase in the apparent Km

This change is proportional to how much inhibitor [I] is used, and how strong the inhibitor is [KI]

v0 = Vmax[S]/[S]+KM(1+[I]/KI)

or

v0 = Vmax[S]/[S]+KMapp

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11
Q

How does a noncompetitive inhibitor effect enzyme kinetics?

What does it depend on?

A

The noncompetitive inhibitor will bind the allosteric site and prevent funcitoning. E, EI, ES, and ESI are all possible complexes, but only ES transforms the substrate into a product

Leads to a decrease in apparent Vmax

Depends on inhibitor concentration [I] and inhbition constant [Ki]

v0 = Vmax/1+[I]/KI x [S]/[S] + Km = Vmaxapp[S]/[S] +Km

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12
Q

How does an uncompetitive inhibitor effect enzyme kinetics?

What does it depend on?

A

Uncompetitive inhibitors bind the E:S complex and prevent function of the enzyme.

Addition of an uncompetitive inhibitor changes the shape of the reaction curve, as inhibitory effect is proportional to substrate concentration

At high [S} the net effect is a reduction in apparent Vmax

Change is proportional to [I] and KI

v0 = Vmax[S]/S + Km

Slightly different from uncompetition

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13
Q

Between Competitive and non/uncompetitive mechanisms, which are more potent?

Which do we see more of in drug development, and why?

A

Competitive and noncompetitive mechanisms achieve greater inhibition at high substrate concentrations ([S]>Km) compared to competitive mechanisms at the same [I]/KI ratio

However, it is very hard to develop non-competitive inhibitors, and they are often discovered by chance. Humans are better at coming up with competitive inhibitors and thus most drugs that inhibit enzyme function tend to be competitive

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14
Q

Define the residence time and how it is used

Why is it important in determining the efficacy of an inhibitor?

A

Residence time is 1/koff

Used to calculate half life of the molecule

koff (and thus residence time) were found to be a better predictor of inhibitor potency than KI

This is because KI implies equilibrium, which is hardly the case in the human body

Picture: different drug’s efficacy plotted against, different variables, Clear correlation in residence time

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15
Q

Is the relationship between receptor/enzyme occupancy and biological response linear?

Name an example of when it is not

Describe ED50, TD50 and Therapeutic index

A

Not always.

eg, the effector may be activated downstream of a receptor through a signalling cascade. Amplification through seccond-messengers will lead to a non-linear relationship

ED50 = the dose at which 50% of a therapeutic effect is obtained, pertains to maximal efficiency

TD50 = the dose at which 50% of a toxic effect is obtained

Therapeutic index is TD/ED

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16
Q

What is used to quantify a synergistic effect between two compounds? How is this done?

How do you calculate the extent of synergy?

What mechanisms are involved in synergy?

A

Via an isobologram (see photo)

First graph: Drugs A and B treat the same disease. Graph effect/conc for each drug individually: A has EC50 of 250, and B has EC50 of 20

Second graph, Dose of B plotted over dose of A. Dashed line from EC50 of each as intercepts, represents no effect. This line is called the isobole.

Test at varying concs of dose A and B in combination and plot which dose combination gives the same effect as the isobole. This gives the curve

This is typically done by fixing one dose for A and varying doses of B

Any curve below the dashed line represents positive synergy

The Combination index {CI) can determine the extent of synergy.

CI = D1/Dx1 + D2/Dx2

Where Dx1 and Dx2 are the doses required from drug 1 or 2 to achieve X effect alone

D1 and D2 are the doses of 1 and 2 in combination to achieve x effect

Syngergy can work through many different mechanisms, such as using different mechanisms to achieve the same effect more efficiently. The most common mechanism is to target two enzymes that converge on the same pathway

17
Q

What are the two methods learned in class to measure interactions between molecules?

BIG question

A

Surface plasmon resonance (SPR)

Done in real time

Light is reflected off a gold surface, and refracted light is measured. When a molecule is bound (a “ligand”), the refracted light changes frequency

Running a flow channel over the “ligand” and passing an “analyte” over will cause the two to bind, leading to a change in measured refraction.

This change is directly proportional to bound mass, and Resonance Units can be measured to calculate kon and koff

Stoichiometry can be found with relevant data (Rmax is maximum response from an analyte, Ri is ammount of ligand immobilized in RUs)

Florescene polarization (FP)

aka Fluorescence anisotropy

A small molecule is emits polarized light (fluorphore) when irradiated only when it does not move (i.e. if bound in a large complex)

Choice of flurophore depends on expected vaiation in molecular weight of the interacting molecules, the fluorescence lifetime changes the change in polarization during a change in molecular weight (you want a smaller lifetime for a small change in weight)

You want a small flurophore (large change in weight when binding) in low concentrations (measurements are quite sensitive), and you want to test with very little concentration of analyte (100 fold below while still being above Kd) so you can accurately calculate Kd

18
Q
A