Free energies and transition state analogues Flashcards

1
Q

Free energy perturbation

A

A method used in computational chemistry for calculating free energy changes from molecular dynamics simulations

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2
Q

How can different drug candidates be ranked more accurately than with the results from empirical scoring functions?

A

Using improved methods that calculate relative binding free energies (DeltaDeltaGbind)
Free energies cannot be easily calculated from empirical scoring functions

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3
Q

How could we go about obtaining the relative binding constants of 2 drug molecules binding (non-competitively) at the same receptor?

A

We could evaluate the relative binding constants as the ratio of the equilibrium constants (Kdrug2/Kdrug1)
We could obtain this from experimentally measured values
However, often in drug discovery, by way of reducing the amount of synthesis, we want to predict the relative binding before the second compound has been synthesised

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4
Q

What is the ratio of the equilibrium constants equal to?

A

Kdrug2/Kdrug1 = Kreceptor/Ksolution

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5
Q

Calculating Kreceptor/Ksolution

A

Computationally tractable by means of an ‘alchemical’ perturbation of the first molecule into the second (even though this is experimentally impossible)
This is done twice - once in solution and once when bound to the receptor site
Because the equilibrium constants and their ratios are related to changes in free energy (DeltaG = -RTlnKeq), this computational procedure is known as a “free energy perturbation”

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6
Q

How can we ‘perturb’ one molecule into another?

A

By gradually changing the values of the molecular mechanics parameters that describe it - i.e. the equilibrium bond length (b0) and the force constant (Kb)
e.g. cyclohexane to cyclohexane
b0 changes from 1.54 to 1.34 A in 10 steps of 0.02 A
Kb changes from 100 to 150 kcal/mol/A^2 in 10 steps of 10 kcal/mol/A^2
Can then perform molecular dynamics at each step to equilibrate the whole system at the chosen temperature

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7
Q

Why is each step in the perturbation process small?

A

So that there is minimal perturbation of the system

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8
Q

Relating free energy perturbation to binding free energies of drug molecules

A

Each step in the perturbation is associated with a change in free energy
The total change in free energy (on going from molecule 1 to molecule 2) is estimated by adding up all the small energy changes in all the steps
For drug candidates, by mutating the first molecule into the second molecule twice (once in solution and once when bound to the receptor), we can calculate Kreceptor/Ksolution and hence Keq2/Keq1

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9
Q

How do enzymes catalyse chemical reactions?

A

By stabilising transition states and, thus, reducing activation barriers

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10
Q

Transition state analogue

A

A molecule that mimics the transition state of an enzyme-catalysed reaction but cannot react any further (inhibitor)
i.e. ‘freezes’ an enzyme-catalysed reaction

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11
Q

Transition state analogue approach to rational drug design

A

A valuable approach to drug design in the absence of protein structure

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12
Q

Why are transition state analogues such powerful enzyme inhibitors?

A

Exquisitely specific

Required in extremely small doses

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13
Q

What information do we need to design a transition state analogue?

A

The structure of the transition state

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14
Q

How can we obtain the structure of the transition state?

A

If we know the mechanism of the reaction

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15
Q

Mechanism of amide bond hydrolysis in the active site of renin

A

Draw

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16
Q

How can we gain insight into reaction mechanisms for designing transition state analogues?

A

Programs such as Spartan can calculate the energy of the structures on the potential energy diagram (i.e. transition states, intermediates etc)
From this data we can calculate the activation barrier
If the activation barriers are calculated for the different possible mechanisms of a given reaction, the one with the lowest energy barrier is the preferred mechanism

17
Q

Another method for gaining mechanistic insight

A

Kinetic isotope effects

= klight/kheavy

18
Q

What is a primary kinetic isotope effect?

A

When a bond to the isotopically labelled atom is being formed or broken in the rate-determining step of a reaction
e.g. E2 elimination
kH/kD = 6.7
Consistent with an E2 elimination in which the C-H/C-D bond is broken in the rate determining step, because more energy is required to break a C-D bond than a C-H bond